QUESTION:
suppose there were a crazy
architect who wanted to build a conical monument or perhaps working
structure on the tip instead of at the wider base, and without any
outside support beams. could the building be supported by a massive
internal gyroscope?

ANSWER:
Yes.

QUESTION:
Am I right that planets tend
to be generally spherical? (Not perfect spheres, of course, but for the
most part sphere-like). Why is this? Your explanation of how/why fire
manifests as sphere in space seems related, but wouldn't the properties
of the matter which make up a planet have an effect on the shape of the
planet? To this end, is it possible to have a hotdog-shaped planet, or
will rotational and orbital forces (and others?) always collect a
planet's matter into a sphere?

ANSWER:
The main reason is that astronomical objects
are usually formed by smaller pieces coming together. Initially, very
small stuff (dust size) sticks together by colliding and since this is
a random process, is just as likely to stick a little piece on any part
of a larger piece with which it collides; therefore the initial "seeds"
tend to be roughly spherical. Then, when a piece gets big enough to
have significant gravity, it starts "sucking up" neighboring smaller
pieces to grow but, because the gravitational force is spherically
symmetric (i.e. it is equally strong in any direction the same
distance from the object), it tends to maintain a basically
spherical shape as it grows. If it were hot dog shaped, gravity would
pull harder on the ends than the sides and it would tend back toward
spherical shape. I read a good analogy somewhere: Suppose you want to
build a building 1000 stories high, essentially a nonspherical bump on
our essentially spherical earth. You would likely fail because gravity
would cause it to collapse under its own weight. The earth is actually
slightly oblate, that is fatter at the equator than the poles, because
it is rotating and the centrifugal force pulls it out at the equator.

QUESTION:
If density plays a factor in
the strength of an objects gravitational field, why is it not part of
Newton's law of gravitation?

ANSWER:
The universal law of gravitation is strictly
valid only for point masses, not extended mass distributions. What you
then do is to imagine two bodies each made up of an infinite number of
infinetesmal point masses, find the force between each pair, and add
them all up to get the net force (if you know calculus, this process is
called integration). That is where the density comes into play since
the (infinetesmal) mass at a point is proportional to the density
there. Interestingly, if the density of each object is spherically
symmetric (i.e. the density depends only on how far you are from
its center) and if the shapes are spheres, you can actually treat each
as a point mass with all the mass at the center and get the right
answer.

QUESTION:
Why do clouds float? Water
vapor is heavier than air. And what determines their altitude?

ANSWER:
Actually, clouds are not water vapor
but either water droplets or ice crystals. If it were simply water
vapor, it would mix with the air and diffuse about sort of uniformly.
In fact, the amount of water vapor in the air is what determines
humidity. Clouds stay aloft for the same reason that dust motes
floating around, also heavier than air: air drafts push them around.
For more information go to the
weather network .

QUESTION:
Is Einstein's theory of general relativity correct? It's
taught in high schools (well, special relativity) and universities but
there's a web site, http://www.relativitychallenge.com/index.htm, that
claims there are mathematical errors in it. Also, are new theories
(such as the string theory) based on relativity or do they assume that
it's wrong?

ANSWER:
What does correct mean? All observations regarding
gravity are in accord with the general theory. But, there are some
important predictions of the theory, black holes and gravity waves for
example, for which we have indirect evidence but not direct evidence.
Is it possible that someday some experiment may not be in accord with
the theory? Of course! This theory cannot be the final word because
nobody has been able to reconcile it with quantum mechanics, one of the
other most important theories of nature. Special relativity is another
issue and is generally considered certainly correct since it simply
describes the nature of space and time. Regarding your website, I
certainly will not critique it because I know that the world is full of
persons with their own personal theories of the universe desperately
seeking attention for their ideas. I will acknowledge that no theory is
immune from mathematical errors which, even if present, do not
necessarily negate the validity of the theory.

QUESTION:
If you're driving a car at 100km/h on a flat, straight
road, and the passenger is flying a remote controlled plane at 100km/h
beside the car; what would happen if the plane is steered so that it
flys inside the car window? Would it appear to hover, as it's speed,
relative to the ground below is the same as the car, or smash into the
windshield, as it's speed relative to the car has increased to 200km/h?

ANSWER:
Before coming inside, the airplane is at rest relative
to the car, so if it keeps flying it would hover when it entered. But,
is that possible? How does the airplane fly? By moving through the air.
Before entering the car, the airplane sees a 100 km/hr "wind" going
opposite his direction of flight (assuming that the air outside the car
is still); if the plane were at rest relative to the air, e.g.
if there were a real 100 km/hr tailwind, the plane would drop to the
ground! But that is exactly what the situation would be when the plane
entered the car—it would suddenly be at rest relative to the air and
drop to the floor. If it were a helicopter instead of an airplane,
which flys by moving its wings (the rotor) through the air instead of
moving the air over the wing, it would hover inside the car.

QUESTION:
What are all of the types of particles in physics? The
ones I know of are:

electrons
photons
phonons
Could you give me
the complete list with short descriptions?

ANSWER:
What is a particle? How about a grain of sand? How
about an atom or a nucleus? There are more than 150 particles known in
elementary particle physics, but physicists no longer think of them as
"elementary" because they can be thought of as being built from more
elementary building blocks. One important example of those building
blocks is the quarks. The whole issue is quite complicated, more than I
can answer in a forum like this one. I recommend you do some reading.
One good web site which does a good job at disentangling the whole mess
is Wikepedia.com .

QUESTION:
If an airplane's wing shape lets it fly. (curved allows
more air
to flow over it thus air pushes up on the flat part of the wing at
least
that's what they tell me...)
How does an airplane fly upside down? Wouldn't the air
pushing the flat part of the wing force it into the ground?

ANSWER:
The standard textbook explanation of how airplanes fly
is a considerable oversimplification. In addition to the Bernoulli
effect to which you refer, the "angle of attack" is also important. I
have included details in a previously answered
question.

QUESTION:
Is it theoretically possible to make a magnet-only powered
motor capable of spinning forever?

ANSWER:
I guess that if you could make something without
friction it could spin forever. You wouldn't even need the magnet. But
in the real world, perpetual motion machines are forbidden by the
second law of thermodynamics.

QUESTION:
What would happen if a car was traveling at the speed of
light, and then turned on it's headlights? I read this question in a
magazine but didn't understand the answer too well. Can you try to
explain it please?

ANSWER:
This is easy to answer: a car cannot travel at the
speed of light.

QUESTION:
A question I have to research...... A light source is 2m
below the surface of the water in a calm pool. Find the radius of the
circle through which the light travels from the water into the air.
Take the refractive index of water as 4/3 and air as 1.

ANSWER:
Since this sounds like homework, I won't work it out
but I will tell you the idea. Light coming straight up from the source
hits the surface perpendicularly (zero angle of incidence, relative to
the normal to the surface) but light not straight up strikes at some
angle of incidence other than zero. As you go farther and farther away
from the straight up point, the angle gets bigger and bigger.
Eventually the angle becomes greater than the critical angle and no
light can escape.

QUESTION:
I have a simple question concerning electorn orbit
changes. How fast do electrons stay in an unstable orbit before
dropping to their basic orbit?

ANSWER:
There is no single answer to this question because it
depends on what atom you look at and specifically on which orbits are
involved in the transition. Typical lifetimes are on the order of
nanoseconds (billionths of a second).

QUESTION:
What is gravity - I am told it is a force that pulls
objects together - but how and why?

ANSWER:
This is a question about which whole volumes have been
written. It is one of nature's four fundamental forces and it is, by
far, the weakest force in nature. The reason why gravity is so weak is
one of the great unsolved problems of physics. In simple, classical
terms, gravity is a force which is caused and felt by objects which
possess a property called gravitational mass which, as far as
we can tell, is any material thing. You might find the statement that
it is weak to be surprising since it is the force of which we are all
most aware. But consider this: the weight of a pin is the graviational
force which the whole earth exerts on it but that force is
easily balanced by a small magnet (which uses the electromagnetic
force, another of the four). If you want to probe more deeply you need
the theory of general relativity. Here the idea is that the presence of
mass actually causes the space around it to warp and this warp of space
results in masses wanting to move toward each other. An often-used
analogy is to imagine a bowling ball placed in the center of a
trampoline which causes there to be a sag in the center; now place a
marble on the trampoline and, of course, it rolls toward the bowling
ball.

QUESTION:
I had the following question on my physics test and could
not figure it out. A child holds a sled weighing 77.0N at rest on a
frictionless incline at 30.0 degrees. Find a) the magnitude of the
force the child must exert on the rope, and b) the magnitude of the
force of the incline exerts on the sled. Answer a) 38.5 N b) 66.7N

ANSWER:
There are three forces on the sled, the force due to
the child (F ), the weight of the sled (W ),
and the force from the incline (N ). Since the incline is
frictionless, N has only a component perpendicular to
the incline; F has only a component parallel to the
plane. Hence N must be equal to the magnitude of the component
of the weight perpendicular to the incline (77cos[30]=66.7) and F must
be equal to the magnitude of the component of the weight parallel to
the incline (77sin[30]=38.5).

QUESTION:
This goes all the way back to high school (as well as
college) physics. What is friction? It was always taught to me as
simply something that exists between two surfaces and the friction
between two surfaces (their relative coeficient of friction) must be
empirically determined. That is how I was taught about friction. It
seems to me with our deep understanding of physics down to sub-atomic
and maybe string level, SOMEONE has had to come up with a better theory
of friction than what I learned. It there an post-Newtonian theory of
friction and is there a way to calculate the coeficient of friction
between two surfaces?

ANSWER:
I have previously answered a question similar to
yours. Link here .

QUESTION:
I am currently reading "The Elegant Universe" by Brian
Greene. In his description of the "horizon problem" of cosmology,
Greene describes how the uniform background radiation is too uniform
according to the standard model of the big bank because light would not
have time to travel between two currently distant regions of space no
matter how close to the moment of the big bang we go. I don't
understand this because it either suggests that a.) the two regions of
space have traveled with a relative velocity of GREATER than the speed
of light or b.) that there must have been an initial displacement
between those two regions of space sufficiently large to account for
light being unable to travel between the two regions. Neither of these
seem consistent with other aspects of physics in the standard big bang
theory. What am I missing here? Am I just not understanding Greene's
explanation?

ANSWER:
from L. A. Magnani:
The horizon problem is indeed a problem for STANDARD
big bang cosmology (i.e., the version developed in the 1950's - 1980's.

The way out is to invoke a brief
period of "inflation" during the early Universe when spacetime expands
much more rapidly than the expansion rate we infer from galaxy
redshifts today.

This inflationary explanation
was proposed in the mid 1980's by Alan Guth and others and is supposed
to be produced by a phase transition in the vaccum of some kind or
other - the cause of the inflationary epoch is still a matter of
debate.

The confusion for this person, I
think, is arising from mixing up the sound speed (which is what
is necessary to establish thermal equilibrium between two regions) with
the speed of light. The inflation does not have to occur at
greater than the speed of light. It just has to occur at a
velocity greater than the sound speed of the medium to effectively
thermally decouple opposite sides.

From J.-P. Caillault:
The commonly accepted solution to the horizon problem is
Inflation, which was when the early universe must have expanded
exponentially (faster than the speed of light, but this doesn't
violate relativity since it's the universe itself which was
expanding, not anything moving within it). Most
cosmologists now accept Inflation as part of the "standard big
bang theory," but this NYU person is probably thinking of the big
bang paradigm that prevailed prior to the introduction of the
Inflationary idea (by Alan Guth in the late 1970s).

from L. A. Magnani:
I think JP is right. The thermal speed may not be relevant
because
the coupling is between photons scattering off plasma, rather than
what goes on in a gas if the particles are doing the energy exchange.

But the expansion of
spacetime can go on at faster than the speed
of light - something that is recognized also in standard big bang
cosmology.

QUESTION:
Please explain insimple terms what E=MC2 stands for.

ANSWER:
This means that mass (m ), which
measures the inertia of a quantity (and to which its weight is
proportional) is just a form of energy. The amount of energy (E )
is enormous because the factor c ^{2} is the square of
the speed of light and c =3x10^{8} m/s=186,000
miles/second is a huge number. To give an example, suppose that you
could completely change a pound of something into energy. The amount
you would have would be about 4x10^{16 } Watt-seconds which is
about 10 billion kilowatt-hours; this amounts, approximately, to the
total energy output of all nuclear reactors in the US in 4 days!

QUESTION:
Two twins, Bill and Ben are 22.0 years old and they leave
Earth for a distant planet 8 light years away. The twins depart at the
same time on Earth, and travel in different space ships. Bill travels
at 0.9c, while Ben travels at 0.5c. What is the difference between
their ages when Ben arrives on the new planet?

ANSWER:
This sounds suspiciously like a homework problem to
me! But I thought it was particularly interesting so I answered it
anyway! To see the solution, link
here .

QUESTION:
In an idealized case when no air resistance and
engine-fuel factors are considered would the same plane travel the
distances Vienna - Tokio and Tokio - Vienna for the same time? Why?

ANSWER:
I'm not sure what you are getting at here. For
starters, for no air resistance the plane could not fly! The most
important issue in time differences in long distance flights is
head/tail winds, but without air resistance, we would ignore these.
Let's assume, rather than no air resistance, equal resistance for any
direction (perfectly still air). Then, if there are identical airspeeds
in any direction there would be identical groundspeeds, so the answer
to your question would be that the times would be equal.

QUESTION:
Relativistic theory says mass increases to infinity as
speed of light is approached. Yet accelerators routinely accelerate
particles to near light speed ( 99 percent in some cases) without the
particles ever getting anywhere near infinite mass. Why? And if there
is some mass increase, what is the largest ever recorded and at what
"speed"?

ANSWER:
You should look at masses relative to the rest mass.
The mass of a particle traveling with a speed of .9999c is
about 76.6 times its rest mass; this is about how fast an electron with
kinetic energy 6 GeV at the CEBAF accelerator at Jefferson Lab
travels. If you were to make the energy 1000 times larger, the
speed would only increase by about .01%. Records of largest mass are
not kept. If you find the highest speeds recorded for a mass m _{0} ,
then the mass will be given by m =m _{0} [1-v ^{2} /c ^{2} ]^{-1/2} .

QUESTION:
If temperature is defined as the average kinetic energy of
molecules in a mass, then why is there not a universal molar specific
heat for all substances in all states?

ANSWER:
In fact, the molar specific heat for most solids at
temperatures near room temperature is nearly constant as you suggest.
For a gas, however, if it can change volume it can therefore do work
and so only part of the heat (energy) added increases the temperature
and part comes out of the system as work.

QUESTION:
Suppose you have a one inch perfectly square bar of length
L resting on a flat perfectly rigid surface. You have a roller of
weight W resting at some point on the bar. The roller is a cylinder so
its contact with the bar is a line perpendicular to the length
direction and parallel to the surface. What is the vertical force at
each point along the bar due to the weight? (The point is the end of a
line across the bottom of the bar that is perpendicular to the length
and the question is about the force on that line.)
A practical application for this is the determination of the weight of
a roller required to compress bonding tape to stick a bar to a sheet of
metal. A particular pressure is required to cause the adhesive to
adhere properly.

ANSWER:
For the application you have in mind, your model
(contact being a line) is too simple to yield a meaningful answer. The
force which the cylinder exerts on the bar is W and that force
is (ideally) distributed evenly across the line. However, pressure is
defined as force/area and since the area of a line is zero, the
pressure is infinite! That should give you no problem sticking the
tape! To compute the pressure you need to let the line have some width w ;
then, L being the width of the rod (and length of your line),
the area of contact will be wL and the (average) pressure would
be W /(wL ). This is still a simplification since the
force is likely to vary across the distance w , but that is
maybe more detail than you want (or that I could probably deduce!)

QUESTION:
I would like to know what cold is. I would like to know
what is happening on the atomic level that makes (we'll say) air, cold
and pass on that coldness.
I know that heat is a form of radiation and that the molecules get
excited and knock eachother around, but what do the molecules of cold
do and how do they do it? Is it atomic spin, do they crowd other atoms
and force them to slow down?

ANSWER:
Cold is not a noun, it is an adjective. What you want
to know is what is the difference between something cold and something
hot; and you want to understand on a microscopic (atomic) level. Let's
talk about the air in the room. As you probably know, all the molecules
in the air move around and their velocities have a distribution, some
fast, some slow, some moderate, etc . But you could, in
principle, measure all the speeds and take the average. Now, the higher
the average speed is, the hotter the gas is. That is what hot
means —speedy atoms, on average. If you want to get more technical, you
can talk about the average kinetic energy of the molecules in the room,
kinetic energy being mv ^{2} /2 where m is the
mass of the molecule and v is its speed. Now, temperature (that
number we usually use to measure hot or cold) is merely a measure of
the average kinetic energy of the molecules in the room.

QUESTION:
why does "fire" behave the way it does in space? why
does it not "go up" like fire on earth, but instead look like a sphere
or something? also, he says that he's heard it's more "dangerous"
[sic] [unpredictable?] in space and never goes out? is any of
that latter claim true?

ANSWER:
This is a good question and you can find it discussed
many places on the web. Two are
Wonderquest and
Scientific American . I will add my two cents' worth: How could a
flame go up if there is no direction identifiable as up? If all
directions are equivalent, then a physical phenomenon like a flame
would of necessity be spherically symmetric like the space. Regarding
the "never goes out" part, a flame is just a chemical reaction and if
either the fuel or the oxygen runs out, the flame will stop. Regarding
"dangerous", on earth the "updraft" of the flame allows oxygen to be
"sucked in" at the point of combustion making for more efficient
burning, so that would likely be a more dangerous situation. I believe
that something burns more quickly in gravity for this reason.

QUESTION:
While driving on an interstate highway the other day, I
was idly day-dreaming about driving my vehicle up onto the ramps of a
trailer that was being towed directly ahead of me. I was traveling at
70 MPH relative to the ground. The trailer ahead of me was moving at a
speed slightly slower than me. As I approached, I thought - my wheels
are turning at a rate that moves me along at 70MPH relative to the
highway pavement, but if I start to drive up onto the ramps of the
trailer ahead of me (which was traveling at - say- 68MPH relative to
the pavement), once my turning wheels hit the ramps, I would be
instantly accelerated to 138MPH - the sum of the trailer's speed
relative to the ground plus my speed relative to the ground - clearly
faster than I had been traveling an instant before I went onto the
trailer.
Now - what if I was traveling at the speed of light, and the vehicle in
front of me was traveling at just under the speed of light? Why
wouldn't my speed suddenly exceed the speed of light as soon as I drove
up onto the ramps?

ANSWER:
Where did you get the idea that you would be
accelerated to 138 mph? Your speed with respect to the road would
remain 70 or close to it. Your front wheels would have to make a little
adjustment because it would be like the road suddenly started moving
backward at 2 mph so your wheels would have to start spinning as if you
were going 72, but that would probably be no big deal (but your tires
might give a short squeal like when you brake or accelerate quickly).
Now, anything which you can understand using everyday examples like the
one you give is not applicable to objects which are traveling near the
speed of light; that is one of the important truths about the theory of
special relativity, that our intuition is not a good measure of what is
reasonable if we are in a regime (high speeds) where we have no
experience.

QUESTION:
Time moves slower for someone who is moving very fast than
for someone else at rest.' How can this be said if motion is
relative? Example: Two people are floating out in space, and person A
sees person B zip by him very fast. B of course sees A zip by him. If
it is correct to say that A is moving and B is at rest, and correct to
say B is moving and A is at rest, how can you decide who is moving
slower through time, and thus ageing slower than the other?

ANSWER:
To each observer the other's clock running slow. They
do not appear to be running slow, they are running
slow. This is very puzzling as you note in your question, but it is
also very true. The trouble is that you have the standard intuitive
feeling for time, that there is some ablsolute and correct clock with
respect to which all others can be compared. In fact, time is such that
there is no problem with A seeing B's clock being slow and vice
versa . There is no way that they can sit down in a room together
and look at the two clocks to find out who is right because they are
not in the same reference frame. To do that, one or the other would
have to accelerate and the theory of special relativity is not
applicable to accelerating frames. You can understand time dilation
without talking about acceleration by studying the "twin
paradox" which I have discussed in answer to a previous question.

QUESTION:
Can an object have positive acceleration and be slowing
down?

ANSWER:
Yes, of course. Acceleration, like velocity, is a
vector quantity and so it has a sign in a particular coordinate system.
(I will restrict myself to one dimensional examples for clarity.)
Suppose that an object has a velocity v _{1} =-2 m/s and
then, five seconds later, has a velocity v _{2} =-1 m/s;
this object is moving in the negative direction (direction of
decreasing coordinate) and is "slowing down". Now let's compute its
average acceleration, the change in velocity over the elapsed time: (v _{2} -v _{1} )/t=(-1-(-2))/5=+0.2
m/s^{2} , a positive acceleration! The best known example of
this is an object thrown straight up into the air: the object slows
down on the way up and speeds up on the way down, but the acceleration
is always the same.

QUESTION:
What is the rate of speed at the point where the two
blades of a scissor meet as they close?

ANSWER:
That is determined entirely by how fast you "snip",
how big the scissors are, and how far from the pivot point the mesh
point is. For example, suppose the scissors are initially open 30
degrees ( p /6 radians) and
close in 1/10 of a second; the average angular velocity of one blade
relative to the other would be w =p /6/(1/10)=5.24 radians/s. If the
blades extended from 1 to 10 cm from the axis, then the relative speed
of one blade with respect to the other would be calculated as v=Rw =5.24 to 52.4 cm/s where R is the
distance from the pivot.

QUESTION:
how can we prove that earth can be considered as an
inertial frame of reference to a good approximation?

ANSWER:
The definition of an inertial frame is one in which Newton's
first law is true. That is, if an object has zero net force on it, it
will move with constant velocity (i.e. move with constant speed
in a straight line or be at rest). The degree to which this is true
determines the degree to which it is an inertial frame. A rough measure
would be to ask what is the maximum acceleration of the surface of the
earth; to calculate this I will ignore accelerations due the motion of
the earth around the sun and the motion of the sun around the center of
the galaxy since these are demonstrably smaller than the acceleration
due to the earth's rotation. Consider an object with mass 1 kg at the
equator. It is sitting "at rest" on a scale which reads, of course, 9.8
N. But, does it really? The weight of the object is 9.8 N but the scale
reads the normal force between the mass and the scale, and these cannot
be exactly equal because the object is accelerating because it is
moving in a circle of radius R _{E} =6.4x10^{6} m
with a speed of v =2 p R _{E} /T= 2 p (6.4x10^{6} m)/(1 day)=465
m/s; so the acceleration is a=v ^{2} /R _{E} =0.034
m/s^{2} . So the scale would read a "weight" of 9.77 N for an
object whose weight was actually 9.8 N. If you can tolerate errors on
the order of 0.3%, you may consider the earth to be an inertial frame
of reference.

QUESTION:
What would happen if the two slit experiment were done
with electrically neutral particles like neutrinos or neutrons?

ANSWER:
The interference has nothing to do with electrical charge,
only with the wave-like properties of the interfering particles. In
order to be able to observe interference, the slit spacing should not
be large compared to the deBroglie wavelength of the particles. The
wavelength is l= h /p where h
is Planck's constant and p is the linear momentum of the
particle. Once you know the wavelength and the slit spacing d ,
the analysis proceeds just as if you were doing the Young's double-slit
problem for light, d sin q = nl .
n =0,1,2... for maxima.

QUESTION:
I am having a VERY difficult time with this question and I
don't know why. I have gotten 4 different answers! I am
just not sure which formula to use, or how to start anymore.
Please help. The question:
Part 1: Suppose you throw a 3 kg ball straight up at 40
m/s. Using energy conservation, calculate how high the ball would
go if there was no air resistance.
Part 2: Suppose that the ball actually reached a maximum height
of only 75 m. How much energy was lost due to friction with the
air on the way up?

ANSWER:
Part 1:
E _{1} =mv ^{2} /2=2400 J
E _{2} =mgy= 3x9.8xy =29.4y
but, E _{1} =E _{2} , so y =81.6
m.

Part 2:
E _{3} =mgy'= 3x9.8x75=2205 J
therefore, D E =E _{3} -E _{1} =-195
J.
195 joules of energy was lost.

QUESTION:
Since the earth is being continuously bombarded by cosmic
rays, why can't we develop some method of harnessing the energy from
these particles?

ANSWER:
Many of these are either trapped in the Van Allen radiation
belts by the magnetic field of the earth or lose much of their energy
interacting with the atmosphere. Although I am not an expert, I believe
that the total energy content would not be large and efforts would be
much more productively directed at harnessing energy reaching us from
the sun.

QUESTION:
Why does an object gain mass the faster it travels? Where
does this mass come from? Do scientists believe there is any way to get
around this and somehow travel at or past the speed of light?

ANSWER:
Mass is inertia, that is resistance to acceleration. When we
say that the mass increases, this simply means that it gets harder and
harder to accelerate a particular object as it speeds up. So, think of
mass as inertia, not "stuff" and then you won't have to worry about
where that "stuff" came from. Another way to think about the problem is
to ask how much energy it takes to accelerate something up to some
particular speed; it would take an infinite amount of energy to
accelerate anything all the way up to the speed of light. Special
relativity, on which all this is based, is very well verified by many
experiments, so no scientists believe that there is any way to
accelerate something beyond or even to the speed of light. There is
some discussion of tachyons which are particles traveling faster than
the speed of light, and they would behave in predictable ways. The
problem is, that they must have always been there since you can't get
there from here! Nobody has ever observed a tachyon and they (in my
opinion) likely do not exist.

QUESTION:

What will happen if the
centripetal, gravitational force of Earth exerted on the moon becomes
stronger ?

ANSWER:
Suppose that the moon is in a roughly circular orbit with the
current gravitational force. If the force were to suddenly increase,
the orbit would become elliptical and the moon would come much closer
to the earth. This is shown to the right where the the moon's current
circular orbit is shown and then if the force increased at the point
where the ellipse is tangent to the circle, the new orbit would be the
elliptical (egg-shaped) orbit. The earth is not drawn to scale here.

QUESTION:
In electrostatics we learn
that the answers we get when computing voltages are independent of the
definition of ground potential. What I mean by this is that the
equations apparently "don’t care" if the ground we define to be Zero
Volts is actually at Zero, or -pi volts, or + one gadzillion megavolts.
Only voltage differences matter. So, is there a mathematical property
embedded or encapsulated within the equations of electrostatics which
ensures the physics will be invariant upon changes in the definition of
the zero voltage? If so, what is this property called? I ask this
because I am familiar with the idea that the equations of physics (for
instance, Maxwell's equations) "don't care" what the velocity of the
laboratory is in which those equations are derived, and in a certain
sense this gives rise to special relativity. Thus my curiosity about
this "don't care" condition involving voltages, and what the more
fundamental principles involved might be.

ANSWER:
The two examples you allude to are different kinds of
"relative". Special relativity, which ensures that Maxwell's equations
are the same in any reference frame, has one frame moving relative
to another. Voltages are defined by considering one point in space relative
to another. Voltage has the property you refer to (only the difference
counts) because of the way it is defined. Voltage is related to energy
(and work) and energy also has this property that the total energy is
not relevant (because potential energy may be arbitrarily zeroed), only
the change in energy. Consider an electric charge q and it
takes an amount of work W_{ab} to move it from point a
to point b. Then the potential energy difference is U_{b} -U_{a} =-W_{ab}
(which is a definition); because this is the definition, you
see that what U itself is is not well defined (e.g. if
I add an arbitrary constant, say 5,000 J to both U_{b} and
U_{a} , the equation is still true). Finally, the electric
potential V is defined to be U/q , so V_{b} -V_{a}
is just a measure of the work necessary to move a one Coulomb charge
from one point in space to another but V_{b} and V_{a}
themselves have no meaning.

QUESTION:
I am a high school junior
and I got stuck with this question while doing a prep test. And this is
the one -A braking system for a roller coaster is designed to stop it
over a distance of 15 m when the coaster enters the stopping area with
a speed of 40 km/h. What acceleration must the braking system provide?

ANSWER:
First, convert km/hr to m/s: 40 (km/hr) x (1000 m / 1 km) x (1 hr/ 3600 s) = 11.11 m/s. Now,
write the equations of motion for constant acceleration:
x=x_{0} +v_{0} t+at^{2} / 2
v=v_{0} +at
Now, taking the origin to be
where the braking begins, x_{0} = 0 and v_{0} = 11.11
m/s. Now, let x =15 m and v =0 m/s:
15=11.11t+at^{2} / 2
0=11.11+at
Solving these two equations yields a =-4.11 m/s^{2}
and t =2.7 s.

QUESTION:
The faster you move through
space, the slower you move through time. I heard someone say on a
television program, that this astronaut, out of all the astronauts in
the world has been in space the longest. They said that because he has
spent so much time circling the earth at such high speeds, he has
traveled into the future a tiny little bit. My question is this: If we
are always moving trough space because of the earth's many movements,
isn't what they said false? Moving with the earth does count when
concidering your actual speed through space, doesn't it?

ANSWER:
What matters here is relative speed (relativity refers to how
things are in one system relative to another). So, he ages less rapidly
than you because he is moving relative to you.

QUESTION:
My 7th grade Honors Math
teacher want my class and I to find out what is the space shuttle
acceleration launch speed? My math teacher said that it woud be less
then 7 mph.

ANSWER:
Well, there is a little problem here: "acceleration launch
speed" doesn't mean anything in physics. The acceleration is one thing
and the speed is another. Speed (velocity) tells how fast the height is
changing and acceleration tells how fast the speed is changing. Maybe
your teacher means "what is the speed of the shuttle at the instant
when launch begins?" If that is so, then the answer is zero. The
average acceleration during launch is about 33 m/s/s which is about 74
mi/hr/s which means that, on average, the speed increases by 74 mi/hr
each second over the 8.5 minutes it takes to reach orbital speed. Of
course, the acceleration is smaller at the start of the launch. At the
end of the first second, for example, suppose that the whole thing has
lifted one foot up; the average acceleration would be about 1.36
mi/hr/s and the speed would be 1.36 mi/hr.

QUESTION:
Many years ago I worked for
eight months in a medical radiation laboratory that serviced medical
linear accelerators. The output from the "linac" was 4Mev and with this
being the case I used the equation E=hf to determine that the photons
being emitted must lie in the gamma radiation section of the
electromagnetic spectrum.. I was however informed that it was x-rays
the "linac" was emitting and was also told gama radiation and x-rays
are the same anyway. I have not dealt with these subjects for
many many years so can you tell me..

Was I right to use the
equation E=hf to determine the frequency of these photons and does it
work out the gamma radiation is the correct output?
Are gamma and x-rays the same
thing?
ANSWER:

You
would have been right if each proton (I presume it was a proton linac
with the protons having 4 MeV kinetic energy) had stopped and all its
kinetic energy were carried off by a single photon. In fact, the proton
beam is smashed into a piece of metal where it interacts many time with
atoms, giving each atom a little of its kinetic energy. Then each atom
deexcites producing an x-ray.
They
are not absolutely defined, so they can overlap a bit, i.e. a
low energy gamma ray and a high energy x-ray might have the same
energy. However, physicists often make the identification by where the
photon comes from: gamma rays come from nuclei and x-rays come from
atoms.
QUESTION:
Why do air bags in cars
reduce the chaces of injury in accidents?

ANSWER:
It is all Newton's second law. Your momentum is proportional
to your speed. To stop you, that is to take away your momentum, a force
must be exerted. The more quickly you change your momentum, the greater
the force which is required. Imagine jumping out of four-story
building. When you hit the ground your momentum disappears in an
instant and, for that to happen, you are subject to an enormous upward
force by the ground; it is this force which hurts or kills you.
However, if there is a big soft pillow that you land on, the effect is
that it takes longer to stop you and so you will experience a much
smaller force over the stopping time; therefore you are much less
likely to be injured. I think that you can make the translation of this
example to the airbag.

QUESTION:
Was looking at a program on
discovery channel that depicted a spacecraft propelled by use of a
solar sail, the interesting piece was that it was propelled by a
powerfull laser on a space station. The question is If a laser powfull
enough is fired in a direction and is powerfull enough to propel the
space craft why would'nt it (the laser that is) be pushed in the
opposite direction to the way in which the light is travelling.

ANSWER:
It would, but if you think of the space station as being
attached to the earth, the whole earth/space station system would
recoil but with negligible velocity because the net mass is so huge.

QUESTION:
Relating to free-falling
objects, if a object is thrown upwards at a velocity (v), when that
object reaches it maximun height, the velocity of the object = 0.
My question is, if you could instantaneously observe the object
while at max height (duration when V=0) then, wouldn't the acceleration
of that object be 0? I understand that g is a constant and never
changes, but wouldn't the acceleration of that object change?

ANSWER:
Acceleration has nothing to do with velocity. Rather it
measures the rate of change of velocity. The acceleration is
proportional to the force on an object and if the force is the object's
weight only, it will have always a constant acceleration which is
pointing down (acceleration is a vector). Acceleration essentially
tells you what the velocity will be a short time later, so if v=0 now,
v will be a small velocity downward a short time later. If the
acceleration were zero, the velocity would not change and the object
would stay at rest forever.

QUESTION:
If you skimmed a one
molecule thick layer of water off the surface of the earth's oceans how
much water would you have?

ANSWER:
The surface area of the earth is about 200 million square
miles and about 70% is water, so 140 million square miles; that is
about 3.6x10^{14} m^{2} . The diameter of a water
molecule is about 3 angstroms=3x10^{-10} m. So the volume is
11x10^{4} m^{3} which is about 30 million gallons. It
would fit in a cube of about 50 yards on a side.

QUESTION:
The textbooks of
physics state that 1 coulomb is a charge equal to 6.242x10^{18} electronic charges, and that the charge
of one electron is 1.602x10^–19 C. My question is: How did
the number 6.242x10^{18} come into
existence? What is its history? Did this number originate from a
measured quantity, that is, experimentally, or is it dirived
mathematically?

ANSWER:
What you are actually asking here is: "How is a Coulomb
defined and how can the charge, in Coulombs, of an electron be
measured?" (not to put words in your mouth, or anything!) It is
somewhat circuitous since the thing which is defined is the unit of
current, the Ampere (A), and the Coulomb (C) is defined in terms of the
Ampere. If you have two very long parallel wires each carrying equal
current I and separated by 1 m, the force per unit length (N/m,
newtons per meter) is 2 x 10^{-7} N/m when I= 1 A; that
is an operational definition of the Ampere. Now, a Coulomb is the
amount of charge which passes through a wire carring 1 A of current in
one second (s), so 1 A=1 C/s. That defines 1 C. Now, as you know,
electric charges exert forces on each other. It may be determined that
the force F (in N) felt by a particle with charge q _{1}
(in C) due to a charge q _{2}
(in C) which is a distance r (in m) away is F= 9x10^{9} (q _{1} q _{2} /r ^{2} );
this is called Coulomb's law. Now that you know the force law, you can
find the charge on an electron by measuring the force between two
electrons separated by a known distance. This charge turns out to be
1.6x10^{-19} C. If that is the number of coulombs per electron,
then the number of electrons per coulomb is simply the reciprocal,
1/1.6x10^{-19} =6.24x10^{18} .

QUESTION:
What would be the final
speed of an electron as it passes though a field generated by 1V
potential difference and expressed in km/s. We would assume the
same conditions as those in which an electron would gain an energy of 1
eV. Is it possible to determine this speed experimentally?

ANSWER:
Technically, one should use relativity to answer this
question but 1 V inparts, as we shall see, a velocity which is much
smaller than the speed of light, so I will use classical physics. You
are right, the kinetic energy of the electron will be 1 eV and since
1eV=1.6x10^{-19} J and the mass of an electron is 9.1x10^{-31
} kg, we can write that 1.6x10^{-19} =9.1x10^{-31} v^{2} /2.
Solving, v =5.93x10^{5} m/s=593 km/s (which is much
smaller than the speed of light, 3x10^{8} m/s). And, yes, of
course, it is possible to measure this experimentally.

QUESTION:
What is the difference
between Linear velocity and Angular velocity?

ANSWER:
Linear velocity measures the rate of change by virtue of
translation (moving in a line). For example, when we speak of a car
going with a velocity 60 miles/hour it is going that fast down the
road. Angular velocity measures the rate of change by virtue of
rotation. For example, the earth is rotating on its axis with an
angular velocity of 1 revolution/24 hours. The wheels of the car have
both linear and angular velicity.

QUESTION:
If I set up a laser that
sends a beam out to a mirror and then the beam is reflected back upon
itself, is it possible to adjust the the distance between the laser
source and the mirror so that one could see interference effects?

ANSWER:
Yes. You can set up a standing wave. But you could not "see"
it because the nodes would only be half the wavelength of the light
apart. This is a technique which is used to create a diffraction
grating out of light.

QUESTION:
Why is parking a car in a
measured space easier while reversing than when moving forwad?

ANSWER:
I don't know if this is really physics; more common sense. It
is because you steer with your front wheels. If you steer your front
end into a parking space, the rear end is left outside and there is no
way to get it in. If you steer so that your rear end goes in first (you
have to go in reverse to do this) then your front end is left outside
but now you can steer it in.

QUESTION:
Can a person get shocked
from the electrical charge that comes up from the ground during a
lightning strike or is it from the charge coming from the cloud?

ANSWER:
Usually the bottom of a cloud is negatively charged, so when
lightening occurs it will result in a large electric current of
electrons flowing to the ground. This is what will kill you (I find
"shock" too mild a word!) Also, you will get burned by the extremely
hot plasma which is the path through which the electrons flow. There is
a lot more detail at
http://science.howstuffworks.com/lightning.htm

QUESTION:
What causes light bulbs to
glow? Is it the gas inside? Do different kinds of bulbs
contain different gases? Like neon gas in neon light bulbs.

ANSWER:
Usually when we refer to "light bulbs" we are talking about
"incandescent" lights. Here electric current is passed through a very
thin filament (wire) and it becomes white hot; that is the source of
the light. The bulb is filled with an inert gas so that the wire will
not burn as it would in air. There is a disadvantage to this kind of
light, however —only a small fraction of the energy it uses is
converted to light, only about 10%; most of the rest of the energy
becomes heat. "Flourescent" lights are much more efficient. These
devices have a mercury vapor inside them which is caused by a very high
voltage across the ends of the tube to emit radiation; unfortunately,
most of this radiation is in the ultraviolet region which is not
visible. To remedy this, the tube is coated with a material called a
phosphor. When ultraviolet radiation strikes the phosphor, it is
absorbed and reemitted as visible light. (You may have seen a "black
light" which makes some clothing, posters, etc. glow; it is an
ultraviolet light and the the glowing things are phosphors.) If you
fill the tube with other gasses it will often glow with visible colors
without a phosphor, e.g. neon will glow orange.

QUESTION:
If I hold a bicycle wheel (with an axle) that the wheel is free to spin
about, and I hold the two ends of the axle in each hand, how would I
would I find the minimum rpm (rev/min) that would allow me to hold it
in one hand & it not fall to the ground? There has to be a minimum
value for this horizontial gyroscope's angular velocity.

ANSWER:
This is a very complicated question. In fact the top (I will
refer to your wheel as a top) begins to drop the instant you let go of
it regardless of how fast it is spinning and then "nutates" as it
precesses. However, there is a simple formula which tells you the
minimum angular velocity of the wheel (which is only valid for the
angle q
with the vertical
unequal to 90^{o} ): S _{min} =[4mgI_{s}
cos q
]^{1/2} /I where
I_{s} is the moment of inertia of the top about
its symmetry axis and I is the moment of inertia about an axis
passing through the pivot point and perpendicular to the symmetry axis.
For example, a top straight up spins in a vertical direction until the
the angular speed drops below S _{min} = [4mgI_{s} ]^{1/2} /I
and then falls. You should get a book on intermediate
mechanics (e.g . Marion and Thornton or Fowles and Cassiday) to
study this very beautiful problem.

QUESTION:
Why is the difference between the deviation produced by a prism onto
red light and blue light called angular dispersion, and that of yellow
light called mean deviation?

ANSWER:
I will venture a guess. Red and blue light are approximately
the longest and shortest wavelengths of the visible spectrum, and the
angle between them is therefore a measure of the total dispersion of
the system over the visible spectrum. Yellow light is in the middle of
the spectrum and so its deviation is about halfway between red and
blue, so this is the approximate mean (average).

QUESTION:
I read on your website that electrons flow on the surface of a
wire/conductor for AC currents, known as the "skin effect". I was
wondering why this happens for AC currents and not for DC currents?

ANSWER:
The thing which pushes the electrons out to the surface is
the magnetic field. There is always a magnetic field for any current,
but its effect on DC currents is small. However, if the current changes
with time, then so does the magnetic field. It is beyond the scope of
this site to work out the details, but new phenomena appear with time
varying magnetic fields which result in much less negligible effects of
the magnetic fields on the electrons; this does not become important
until the frequencies are radio frequencies (MHz or more).

QUESTION:
Why are objects as seen using mirrors closer than they appear to be?

ANSWER:
The blunt answer is that they generally are not. If you stand
in front of a mirror, your image is precisely as far behind the mirror
as you are in front of it. You are probably referring to the sideview
mirrors in cars which have a warning imprinted on them about objects
being closed than they appear. The reason is that the mirror is convex
rather than being flat like your bathroom mirror; convex means that the
mirror has a curvature such that it is a portion of the outside of a
sphere. (A concave mirror has a curvature such that it is a portion of
the inside of a sphere.) The reason for this is that you will get a
wider view of what is behind and beside you. I cannot give you a
tutorial on optics here, but you can read about it in any elemtary
physics text or many web sites, for example
here .

QUESTION:
Is there any instance (hypothetical or not) that only one of the four
fundamental forces is at work or acting?

ANSWER:
Since most particles in nature have mass, gravity is always
at work there (even if negligible). Furthermore, if you are in a region
of space which contains mass, even a massless particle (nowadays only
photons are thought to be massless). But there is a hypothetical
situation. Suppose that you have an electromagnetic wave propogating
through totally empty space. Then there will be electric and magnetic
fields so only the electromagnetic force exists. You could split hairs,
of course, and say that it is a field, not a force, which is in the
space through which the wave travels.

QUESTION:
Ok, i know what E=MC^{2} is, but do you have a DETAILED
description explaining it ? Do you have any examples of it that i can
teach to a senior (College) class ?

ANSWER:
It basically says that mass is a type of energy. You need to
know, of course, what energy is. In a nutshell, energy is what changes
about something if you do work on it, that is if you push on it over
some distance. For example, if you push hard on a baseball at rest over
a couple of feet (i.e. pitch) you do work and impart to the
baseball kinetic energy which it did not have before. If you take a
baseball on the floor and lift it up to a table top, the kinetic energy
has not changed but work has been done lifting it; here you have
imparted gravitational potential energy to the baseball.

With that
said, let us give an example of an experiment which could be done to
prove that mass and energy are interchangeable. Suppose that we take an
atomic nucleus, for example the nucleus of the most common isotope of
carbon which consists of six protons and six neutrons, and rip it all
apart into its 12 constituent pieces. Will this take work? Of course,
because otherwise this nucleus would not exist since there would be
nothing holding it together. Before we rip it apart we should measure
its mass; I will call that M _{C} . After we have
finished, we have done an amount of work W and have six
protons, each of mass M _{p } and six neutrons, each of
mass M _{n} . Does M _{C} =6M _{p} +6M _{n} ?
Someone who has studied chemistry is very likely to answer
affirmatively to this question but the answer is no and it is not a
hypothesis, it can easily be done. In fact the mass of the sum of the
parts is larger than the mass of the nucleus and E=Mc ^{2}
gives the result: W =(6M _{p} +6M _{n-} M _{C} )c ^{2} .
The mass gained is not some trivially small amount —it is on the order
of 1%. Nuclear energy, of course, is where the energy from nuclear
power plants and nuclear bombs comes from.

QUESTION:
If spring scales (bathroom scales) measure weight (force), and Dr.'s
scales measure mass, why do I "weigh" the same on both the spring scale
and the balance in the Dr.'s office?

ANSWER:
Both scales measure weight, they just do it in different
ways. The spring scale measures the force necessary to compress (or
stretch) a spring by a certain amount; knowing the properties of the
spring, the scale can be calibrated. The doctor's scale is essentially
a balance where your weight is compared with a known weight; the idea
of torque is also used where perhaps 1/10 your weight is needed to
balance your weight. Weight is a force, and mass is, conceptually, a
very different thing: mass measures the resistance (inertia) which
something has to acceleration when you push on it. Because of one of
the most fortuitous "accidents" of nature, it just so happens that
weight is exactly proportional to mass, so measuring weight turns out
to be equivalent to measuring mass. The "accident" is that inertial
mass is precisely the same as gravitational mass (which is a property
of matter which measures how strongly its gravitational attraction to
other bodies is). We now understand that this is not an accident; the
theory of general relativity fully explains this equivalence.

QUESTION:
I teach AP physics in a high school in michigan, and can't seem to
reconcile these two facts: The electric field due to an infinite
conducting sheet with surface charge density sigma is
E=sigma/Epsilon_0. If I introduce an oppositely charged infinite
conducting sheet facing the original, by superposition, I get that the
field between them should be double in strength, i.e. E=
2*sigma/epsilon_0. However, gauss's law, using a cylinder with one flat
face between the sheets and one face within one of the conducting
sheets still gives me E=sigma/epsilon_0. Where is the flaw in my logic?
When I look at the field lines, I see that the oppositely charged
infinite sheet doesn't introduce more, since every positive charges
field line on the positive sheet must end on a negative charge, either
at infinity or on the negative sheet, but that doesn't explain to me
why superposition doesn't seem to work here.?

ANSWER:
The problem you are having is rather straightforward. You are
correct in saying that with two sheets the field is twice as large
between the plates; however, the field outside the plates, also by your
superposition argument, is zero. Thus, when Gauss's law is applied
there is no flux leaving the surface outside, which gives twice the
field inside: e _{0} E _{1} *(2*A )=s
A with one plate
and e _{0} E _{2} *A =s
A with two, so E _{2} =2*E _{1}

QUESTION:
Since the orbital period of a satellite in near-earth orbit is much
less than 24 hours, why does the earth itself rotate only at that rate?
If the earth had formed from a collection of infalling particles,
wouldn't they have been rotating at the average orbital period based on
their distance from the centre of mass?

ANSWER:
The orbital velocity has nothing to do with the earth's
rotation. Suppose that when the earth formed it did so from a large
number of rocks all at rest. Each would fall toward the center of mass
and the resultant earth would have no rotation; the near-earth orbit
would still be the same, though, because it depends only on the mass
and radius of the earth. The real key to understanding how the earth
rotates is to understand that how it ends up depends on how it starts
and the operative concept is angular momentum. Angular momentum of the
earth is the same as it was before the earth was formed; as the
distribution of mass changes the angular momentum stays the same but
the angular velocity changes. If the present day earth were suddenly to
shrink to half its current radius, the length of a day would shorten by
a factor of four, 1 day = 6 hours.

QUESTION:
Is the amount of matter in the universe constant? A related question is
can new matter be created?

ANSWER:
Matter and energy are interchangeable, so matter can be
created by adding energy to a system. The best known example is called
pair production: a photon (quantum of light) may spontaneously create
an electron/positron pair (a positron is the antiparticle of the
electron). Another example is that the mass of the nucleus of an atom
is less than the mass of all its neutrons and protons, so when that
nucleus was made (probably in some star) a little bit of matter
disappeared from the universe. Obviously, the amount of matter in the
universe is not constant, but the amount of energy, we believe, is.

QUESTION:
While helping my daughter in grade 5 with a wind power project I was
wondering how to measure in a simple way the wind speed of the fan. We
thought to try the approach where you suspend a ping pong ball from a
thread. A table exists which relates angle of swing to wind speed.
However I was wondering about the physics of it.

If you
have a ping pong ball suspended from a 30 cm thread and the ping pong
ball weights 0.0027 KG then if a wind blows the string at an angle of
30 degrees from the vertical, then what would be the wind speed. What
formula would you use if you ignore the aerodynamic effects of the wind
going around the ball etc. I am helping Raeann with her wind power
project. So we have a room fan that we use to drive a wind turbine
(propeller hooked to a motor). It would be nice to measure the wind
speed of the fan. It is expensive to buy a real anemometer so people on
the net have published a table that relates ping pong ball angle to
wind speed. However I was interested if you could calculate this. So
the force downwards is mg for the ping pong ball. The tension onthe
thread would have a downward force and a sideward force component. The
sideways force would have to be matched by the pressure of the wind.
Wind pressure would include the density of air and the cross section
area of the ping pong ball I would imagine. Also there is the potential
energy of the ball lifting up so many meters would be matched by the
kinetic energy of the ball. So any thoughts. [Questioner also included
data which came from http://marsville.enoreo.on.ca/mission/challenges/anemometer.htm
.

Angle
kph
90 0.00
85 9.30
80 13.20
75 16.30
70 19.00
65 21.60
60 24.00
55 26.40
50 29.00
45 31.50
40 34.40
35 37.60
30 41.50
25 46.20
20 52.30 .]

ANSWER:
If you plot your data, angle as a function of wind speed, it
will not be particularly enlightening. Before plotting anything you
should think about the physics. This is the simple pendulum problem
except with a horizontal force which keeps the ball at a particular
angle. I will not do the details which you can get in any elementary
physics text; I will give the results. Let us call the (horizontal)
force of the wind on the ball F , the (vertical) weight of the
ball W , and the tension (along the string) in the string T,
and the angle the string makes with the horizontal q . Then, solving this
problem we find that T =W/ sin q and F=T cos q =W cos q /sin q . To understand the
physics, therefore, you should plot F as a function of cos q /sin q . I have done
this in the plot on the right. The black crosses are the data, the red
line is a fit. In essence, what you find by fitting the data is that
this is almost a perfect parabola, that is the force is proportional to
the wind speed squared.

If you
want to now calculate the force of the wind on the ball, it is
approximately F=W 0.001 v ^{2} where v
is the speed of the wind in km/hr. Once you know the force, you can
deduce the angle q
=arctan[W/F]=arctan[1000/v ^{2} ].
For example, if v= 24, q =
arctan[1.74]=60.1 degrees, in pretty good agreement with the data
above. It is interesting that the length of the string is irrelevant;
also, you do not need to know the weight of the ball as long as you
have the quoted data. Probably more useful to you would be the
inverse of this equation, v =[1000/tan q
]^{1/2} ; for
example, if the angle is 30 degrees, v= 41.6 km/hr.

For
common wind speeds on things about the size of ping pong balls, wind
resistance is roughly proportional to speed squared. This is not always
the case and it can also be proportional to the wind speed or to some
combination of linear and quadratic.

So now
you understand things. Perhaps the simplest thing to do is just take
the given data as the "calibration" of your instrument and then, having
measured the angle, interpolate.

QUESTION:
Someone in an internet forum claims that Einstein's Theory of General
Relativity shows that a geocentric model of the Universe is entirely
equivalent as a heliocentric one. Is he right?

ANSWER:
First, it is the solar system which we should talk about, not
the universe. Geocentric has a specific meaning, namely that the earth
sits still and the sun goes around us. But, as we all know, this is not
a possible explanation using the laws of classical physics. What this
person was probably referring to is that the principle of general
relativity states that the laws of physics are the same in all frames
of reference. That is, you may equally well understand the motion of
the solar system from the perspective of a coordinate system tied to
the earth as tied to the sun. This does not mean that geocentric and
heliocentric are equivalent but rather that the question of who is at
the center is meaningless.

QUESTION:
Have scientists been able to accelerate a particle past the speed of
light? Or at least up to the speed of light?

ANSWER:
A "massless particle" necessarily must travel with the speed
of light (like a photon, the particle associated with light itself).
But, if a particle has any mass, it may become arbitrarily close to and
below the speed of light, but never equal to or greater. The easiest
way to understand this is to understand that the mass of a particle
increases as its speed increases in such a way that the mass approaches
infinity when speed approaches that of light. It therefore would
require an infinite amount of energy to accelerate a particle to the
speed of light and, of course, there is not an infinite amount of
energy in the entire universe.

QUESTION:
I am a sophmore in high school and i have a question that i thought of
while in chemistry. Is it possible to trap light(laser beam would
probable work best) using mirrors that would continually bounce the
light off each other without letting the light escape? If this is
possible will the light still be there even after the original light
source was shut off?

ANSWER:
Yes, that is possible but you need to devise a mirror which
is perfectly reflective and that is not such a trivial thing.
Think about a one-dimensional trap, two plane mirrors one meter apart.
Suppose that they are 99% efficient at reflecting the light (much
better than your bathroom mirror). And, you have trapped a beam
of light in there (easiest to think of it as a very short pulse moving
back and forth. Each time it reflects it will loose 1% of its
intensity. Now, the speed of light is 3x10^{8} m/s, so
the time between when the pulse leaves one mirror until it hits the
other is 0.33x10^{-8} s, that is it has 300,000,000 collisions
per second. If it loses 1% each collision, there will not be much left
after a second.

QUESTION:
What causes say, wood or metal, to be nd and break? If I were to put a
board on bricks and hit it hard/fast enough it would break because it
causes shear (I believe) but what would cause the board to break, say I
was in space and I hit it extremely hard? It would definetely still
break but nothing is pushing on the outsides of the board so why
wouldn't the board just go forward rather than bend and break?

ANSWER:
Suppose you have a board of length 2L and you
exert a force F in the center. Then there will be a
torque FL about one end. You should think of this torque
which breaks the board. If the ends of the board are held fixed,
there will be four forces on the board, a force N up on each
end of the board, the applied force F , and the weight W
as shown in the figure. So, you can see, the torque about one end
is (F+W )L- 2NL= (F+W-2N )L . Now,
if you are in empty space, the forces N and W go away,
but there is still a torque about the end due to F . So
pushing on the object will do two things: accelerate it (because of the
unbalanced force) and deform it (due to the unbalanced torque).

QUESTION:
A friend and I were discussing ballistics at a 1000 yard target
shooting match and need some expertise on a question. If two bullets
leaving the same caliber rifle with the same ballistic coeffiecents are
fired with the only difference being the weight of the bullet (for
example, 300 gr versus 150 gr), which bullet will incur the most wind
deflection?

ANSWER:
I know little about ballistics. As best as I can tell (with a
cursory internet search), a ballistic coefficient tells what
the air drag on a bullet is for a particular velocity. So, imagine that
you have two bullets which have the same speed and therefore experience
the same force F due to air friction. Newton's second law tells
us that a=F /m , so the one with less mass has a greater
acceleration and so it will lose its velocity more quickly. For the
same reason (a=F /m ), the lighter bullet is likely to have
a larger muzzle velocity (the speed it exits from the rifle) if the
force propelling the two bullets is the same. The air friction force F
depends on the speed v , probably approximately like F=cv ^{2}
where c is a constant (probably related to your ballistic
coefficient). Therefore, the lighter bullet probably experiences a
bigger F than the heavier one. Looks to me like the heavier
bullet wins on both fronts.

QUESTION:
My husband and I may get divorced over this question! We both
have our positions, so maybe you can help us out. It's extremely
cold here in Calgary, about -20 celcius, and this came up on the way
home after starting up a very cold car. When starting the car,
after the temperature gage starts to indicate that the engine is
warming up, my husband cranks the heat full blast. It's my
position that if he were to keep it at a low setting, the air coming
out of the vent would be warmer, just not as much of it. If we
select the high setting, the temperature coming out of the vent will be
cooler. I think this is because it is forced air, and cold air is
being added to warm air from the engine that is not so warm yet,
thereby diluting it. Once the engine has warmed up enough, the
effect of the forced air created by cranking the heat is irrelevant
because the engine is very hot (meaning that the temperature of the air
coming out of the vent is the same at any setting, once the engine is
hot). He says that the setting of the fan does not change the
temperature coming out of the vents at the same engine temperature.
Which one of us is right????

ANSWER:
Well, it depends on what you want. If you want the air to be
as warm as possible coming in, you are likely right. On the other hand,
what you probably really want is to maximize the rate at which your car
is heating up and, in that case, your husband is likely right. I say
"likely" for the following reasons. Heat will be transferred from the
heating coils to the air passing over them at some rate and that rate
may or may not depend on the rate at which air is flowing over
them. One possible scenario is that the rate is about the same
regardless of whether the fan is high or low, i.e. maybe the
same amount of heat per second is achieved with either fast or slow air
flow; in this scenario, the air from the slower fan will be warmer than
from the faster fan but each will warm the car up in the same amount of
time because each carries the same number of calories or BTU or
whatever per second. It is my guess that your husband is right if you
want to heat the car up as soon as possible since I would guess that
fast air blowing across the heater coils would take the heat away
faster from the coils.

But, as
is often the case in science, there is nothing to take the place of a
measurement and that might be what you have to do to get a definitive
answer. Let us make up what an experiment might measure so you can see
how you could definitively do a measurement. The first thing you need
to know is that the energy contained in a gas is proportional to its absolute
temperature, i.e. E=a (T+ 273) where a is some
constant, E is the energy, and T is the celcius
temperature; the 273 is to convert T to absolute temperature
(-273 C is absolute zero). Suppose that the temperature of your air is
15 and your husband's is 5. Then, the same volume of air contains
energy in the ratio E _{wife} /E _{husband} =288/278=1.036
(you are winning so far!) But, the volumes of air are not the
same —your husband's method moves, let's say, twice as much air, so E _{wife} /E _{husband} =1.036/2=0.518,
your method now losing out by nearly a factor of two. This is not
definitive since it depends on the relative temperatures and air flow
volumes. I expect, as I said above, that the way to warm up the car the
fastest is your husband's.

QUESTION:
Hey, do different frequencies of light have different amounts of heat
energy attributed to them? In otherwords, is UV light hotter/cooler
than visible light?

ANSWER:
Any frequency of light may carry any amount of energy —that
is what the intensity of the light is. However, we know that light is
made up of many photons, each carrying the minimum energy that such a
frequency can carry. The energy E of a photon is
determined by by the frequency f by the relation E=hf where
h is Planck's constant (an extremely tiny number).
Therefore, one photon of UV carries more energy than one photon of
visible light because its frequency is higher. So, UV light of
the same intensity as light in the visible range has fewer photons.

QUESTION:
I was told to ask this question to a phyicist, so here goes. Where did
air come from?

ANSWER:
Well, how far back do we want to go? All heavy elements
(essentially heavier that hydrogen) were produced in stars and then,
when the star was "all burnt up", it exploded and sent all the heavy
elements flying into space and then they eventually come together again
to form planets, etc . (Scientists like to say that we are
all made of "star dust". Then, depending on the chemistry of the
planet, its temperature, and other factors, some of the planet will
become an atmosphere, i.e. gases will escape from the surface
somehow In some cases (like the moon) the gravity is not strong
enough to hold the atmosphere and it eventually "leaks" off into
space. In the case of the earth, there is virtually no hydrogen
or helium in the air because it has all leaked off. The detailed
composition of the atmosphere depends on chemistry and biology.
For example, it is thought that originally the earth had much more
carbon dioxide in its air but that evolution of green plants resulted
in there being much more oxygen now.

QUESTION:
Hi. I was just wondering if you could give me an explanation on
why cars cannont fly? I realize that the gravitional pull has an effect
on it, but I want to know more specifically all the reasons.

ANSWER:
Anything can fly. You simply need to exert an upward
force equal or bigger than the weight of the object. An airplane
has wings and the air is made to flow over the wings such that the air
pressure on the bottom is greater than the top so there is a force up
which, if big enough, can lift it off the ground. A car could fly
if you gave it some upward force; for example, lift it up with a
crane! Or fit it with wings and an engine to keep it moving
forward.

QUESTION:
I am trying to get an estimated maximum wind speed that it would take
to blow over a 500 lb security tower that stands 10 feet tall. Can you
help me find ways to determine this?

ANSWER:
You have not given enough information. The force which
the wind exerts depends on the geometry of the tower. Also, is
the tower anchored to the ground in any way? Look here where I will put a very
rough calculation. The answer, about 60 mi/hr, is about what
you would expect, and it is likely that any other calculation would not
be a much better predictor.

QUESTION:
being that energy is conserved, what becomes of a sound wave in a
vacuum (i.e. space)? where does the energy go that would
otherwise go to produce the sound?

ANSWER:
Imagine that you have, as an example, a vibrating reed.
It has, as you imply energy. As it vibrates, it loses energy in
several ways:

As you
note, the sound carries away energy.
As it
moves through the air, it experiences air friction which takes energy
away; this ends up as heat in the air and in the reed itself.
There
is "internal" friction because the reed is not perfectly elastic; for
example, if you have something like a piece of thin metal and
repeatedly bend and unbend it, it will get hot because of internal
friction.
Those
three will be the main modes of energy loss. Now, if you take
away the air, the first two modes of energy dissipation are no longer
available and so the reed will simply lose its energy more slowly, i.e.
the reed will vibrate much longer before it stops.

QUESTION:
Hi, my friend and I were discussing rolling objects down a ramp.
I said that if one object is a cylinder and the other is a sphere, with
both the same radius and mass, that they both would have the same speed
at the bottom of the ramp. But my friend said that no she thinks
the sphere would be going faster. We are very interested to find
out which is right?

ANSWER:
The rolling object with the smallest moment of inertia will
win the race (and hence be going faster at the bottom of the
ramp). A solid uniform cylinder has a moment of inertial I=mR ^{2} /2
and the solid uniform sphere has I= 2mR ^{2} /5.
So the sphere is the winner since 2/5<1/2; but it is a pretty close
race. You can try this experimentally but since it is so close,
the results will often not be definitive because of other factors
(rolling friction, nonuniformities, air friction, bumpy ramp, etc. ).
If you want to try it experimentally, try a race with a cylinder and a
hoop (hollow cylinder) which has a moment of inertial I=mR ^{2} .
Here the solid cylinder should be the clear winner since 1/2<1 by a
pretty good factor.

QUESTION:
how do waves "rob" energy from one another very rarely to form massive
"killer" waves that rise somewhere around 100 feet in the middle of the
ocean?

ANSWER:
I am not really sure here. Usually "killer waves" refer
to tsunamis (tidal waves) which are caused by earthquakes, volcanoes,
or other geological catastrophes. However, water waves, like any
others, are subject to the superposition principle which
states that if two or more waves come to the same place the net
disturbance will be the sum of all the individual disturbances.
Simplistically, think of two waves, each 40 ft high and one comes from
the southeast and one comes from the northeast. If they collide
someplace where they are in phase (both are up at the same time) then
an 80 ft high wave will appear there. However, if they collide
someplace where they are out of phase (one is up and one is down), that
point will be calm. It is also what happens with a lens which
focuses waves: it gets very bright where the light comes to a focus
because many waves are adding up. But it is not a case of
"robbing" energy; it is more a case of combining their forces.

QUESTION:
I just got my last test back in college physics and I got this question
wrong, and I want to find out what was the right way to do
it.
A pitcher accelerates a .14 kg. ball from rest to 42.5 m/s in .06
seconds.
a.) How much work does the pitcher do on the ball?
b.) What is the pitcher's power output during the pitch?
c.) Suppose the ball reaches 42.5 m/s in less than .06 seconds, Is the
power produced by the pitcher in this case more than, less than, or the
same as the power found in part b. Explain.

ANSWER:
a.) The work done is equal to the kinetic energy
change. Since the ball started at rest, W=mv ^{2} /2=126.4
J.
b.) The average power is the work divided by the time it takes to
deliver that energy. P=W /t= 2,107 W.
c.) If the same amount of energy is delivered in less time, the power
will be greater.

QUESTION:
If a puck slides across ice, and slows from 45m/s to 44 m/s in 25 m. ,
why does after another 25 m. does it slow to less than 43 m/s?

ANSWER:
The force on the puck is approximately constant and so its
acceleration is constant. Suppose that it takes a time t _{1}
to go that first 25 m. Then it will take it t _{1}
until the speed decreases to 43 m/s. But, it is, on average,
going more slowly during the second t _{1} and so it
will go less far than 25 m.

QUESTION:
Reflected light wave will have a phase change of 180 degrees at denser
medium, say when it travels from air to glass. The speed of light in
glass is smaller than that in air and we define glass as a denser
medium. For sound wave, its speed in air is smaller than that in glass.
Should we define air is a 'denser' medium for sound?

ANSWER:
For any kind of wave, reflection at a boundary will have a
phase change if the speed in the medium from which the wave is
reflected is smaller than the speed in the medium in which waves are
traveling.

QUESTION:
At the most fundamental level, exactly where does the energy from
fusion and fission come from? I know e=mc^2 describes how much energy,
but not the process itself. I know about the curve of nuclear binding
energy. E.g, when four hydrogen nuclei fuse, the resultant helium atom
has less mass and the excess is released as energy. But where exactly
does the energy come from? Is it correct to say the strong nuclear
force ultimately provides this? Or is simply an intrinsic process we
accept ("it just happens")? At the lowest level, is there a describable
mechanism by which matter stores energy, or by which the
mass->energy conversion releases energy?

ANSWER:
It does indeed come from E=mc ^{2} . And
yes, it comes from the strong interaction. The example you state
(4H going to 1He) is not a good one because it is incorrect because two
of the protons have to turn into neutrons + electrons which complicates
things (but happens ultimately). Better to fuse two deuterons
(nuclei of "heavy hydrogen" which consists of a bound neutron and
proton) into an alpha particle (a He nucleus). As you correctly
state, energy is released because the mass of an alpha particle is
smaller than the mass of two deuterons. It comes from the process
of their becoming bound together so, as you suggest, the strong force
is responsible. It is perhaps easier to understand to think of
the reverse process: in order to pull apart an alpha particle into two
deuterons, you must supply work, right? Where does the energy
that you put in go? It goes into mass.

QUESTION:
If two trains,one loaded with lead, the other empty, are travelling at
60 miles per hour on identical flat tracks and at the same time their
engines were put in neutral which one would travel further and why?

ANSWER:
It all depends on the friction which each train
experiences. Normally we think of friction between surfaces
sliding on each other as increasing as those surfaces are pressed to
gether harder. The wheels are not slipping and furthermore being
steel are not very deformable, so their contribution to the friction is
rather small and probably similar for the two trains. However,
there are bearings which have some friction and the friction will
surely get larger as you increase the force (coming from the train's
weight) on them. So, the heavier train will have more friction
and therefore go less far.

QUESTION:
I learned that the interior of canons when smooth provided less
accuracy. When we learned to machine a spiral on the interior it
increased the accuracy much like a football through is more
accurate if you put a spin on the ball when releasing it. Why
does setting a spin on a projectile increase the accuracy of that
projectiles aim?

ANSWER:
You are right —the rifling (which is what the machined
spirals are called) imparts spin to the projectile. Why should
that help accuracy? Well, if something is spinning it will
continue pointing in the same direction forever unless there is some
external torque on it (this is how a gyroscope works); this is called
conservation of momentum. It is well known that such a projectile
is more accurate. But the reason is not just that it is spinning
and not "tumbling". It has to do with the interaction with the
air; a tumbling projectile will tend to be deflected by the air it is
moving through more during its trajectory than one which is not
tumbling. If there were no air, any projectile, no matter how it
spun or tumbled, would be equally accurate since the center of mass
would move as if it were a simple point.

QUESTION:
why is Ke=1/2mv^2; especially if it takes onlytwice as much rocket fuel
to accelerate it to twice the speed?

ANSWER:
I do not know where you got the idea that twice the fuel
results in twice the speed. That is incorrect. But you are correct in
your implication that twice the fuel will not result in twice the
kinetic energy. The problem is that when you burn fuel, a certain
amount of energy is released; but, you also "throw out" the burnt fuel.
In order to conserve the momentum of the system, you cannot give all of
the released energy to the rocket —the spent fuel gets some of
it. So looking at the relation between energy released to the
energy gained by the rocket is not a useful thing to do. To
answer your question, though, the reason we define kinetic energy as mv ^{2} /2
is because if you do work W on a particle, its kinetic energy
increases by exactly W .

QUESTION:
In rotational motion of a rigid object why are torque, angular
momentum, angular velocity and angular acceleration as vector
quantities defined along the axis of rotation? Is it due to the
lack of a better way to express those values/quantities graphically or
is there real meaning? For example when calculating angular
momentum (L=rxp) the linear momentum can be given a value directed out
of the rotational plane into the third axis or parallel to the axis of
rotation. This would seem to mean that linear momentum would be
directed away from the plane of rotation and directed around the axis
of rotation. Like the linear momentum would make a spiral around
the axis of rotation.

ANSWER:
Well, first of all they are not all necessarily along the
axis of rotation, only angular velocity and angular momentum are.
Torque and angular acceleration must be in the same direction because
of Newton's second law. But torque about an axis is defined as
r x F where
r
is the "moment arm" and
F is the applied force.
Your definition of angular momentum is valid only for a point mass
which has linear momentum p in some well-defined direction. For a
rigid object, you must look at each infinitesmal part of the object (of
mass dm and velocity v having momentum d p = v dm and therefore angular
momentum d L = r xd p
) and then add up all
the infinitesmal angular momenta by integrating to get the total.
There is indeed real meaning since a complete description of rotational
physics results.

QUESTION:
Assume a V-slot is cut across the diameter of a turntable, and a ball
placed near the center. As the turntable spins, the ball moves away
from the center under the influence of an unbalanced centrifugal force.
Since "w" is constant (the ball stays in the V-slot), and "r" is
increasing (the ball is moving away from the center), then "v" must
also also be increasing (the rotational speed of the ball). Since "v"
is a squared term in the formula f = v * v / r, then the centrifugal
force "f" pulling the ball away from the center is INCREASING over
time. So the acceleration is not being caused by a constant force upon
a mass over time - as we normally think of acceleration - it is being
caused by an increasing force upon a mass over time. Can this be
(should this be) called "hyper-acceleration"? Also, since this ball
described above now has gained significant "radial momentum" (what is
the proper term for this?), the "exit" path of ball would NOT be
tangential - correct?

ANSWER:
You are analyzing the problem all wrong. For starters, there
is no such thing as centrifugal force (it is what is referred to in
mechanics as a fictitious force). I guess by w you mean the angular
velocity of the turntable. I am going to assume that the ball slides
frictionlessly in the slot; you could solve the problem if the ball
rolled without slipping, but it would be more complicated and obscure
the important features of the physics for you. The problem is identical
to a bead on a straight, smooth stick which is rotating. To use a
centrifugal force implies that you will work in the (accelerated) frame
at rest with the stick and if you do this, you must also introduce
something called the Coriolis force (another fictitious force). I take
it that your level is such that we should stick to the inertial frame
in which the stick rotates. Then there is no force along the stick so
the acceleration along the stick must be zero. This will be puzzling to
you because the bead starts not sliding out the stick but certainly
ends up sliding out. This is really a subtle problem which requires you
to work in polar coordinates. There is a force in the tangential
direction, however, which causes an acceleration in the tangential
direction. Again, this may be surprising to you because the
particle does not have an angular acceleration! The final result
is that the particle will, if it starts at some position r _{0} ,
then when it is a distance r from the axis of rotation it will
have a radial speed v _{r} = w [r ^{2} -r _{0} ^{2} ]^{1/2}
and a tangential speed of v _{t} =w r. So, you are
right, it will come off the end with a velocity which is not
tangential. If you want to see the details (you have to be familiar
with calculus and polar coordinates, send me another question and I
will give you the details.

ANSWER:
I just realized that I have previously answered this question
and the details are available here . That
problem looks a little different but upon reflection you will see that
it is identical to your problem.

QUESTION:
Could a person stand the pressure it would take to make a relatively
heavy gas (e.g. Argon) dense enough to enable bouyancy?

ANSWER:
Well, the density of our bodies is on the order of that of
water, so the density of the gas would have to be greater than water. I
don't think we could endure such a pressure.

QUESTION:
Ok, i know Bernoulli's equation 'says' if you reduce the cross
sectional area, you will increase the velocity of the fluid and
decrease the pressure. While i was watering my garden, i sometimes
squeeze the end of the hose to increase the velocity to make the water
go further. However, when i went to turn off the water from the tap,
the velocity of the water reduced even though i was decreasing the
cross sectional area by closing the valve. So, why does the velocity
increase when i squeeze the end of the hose, but reduce when closing
the valve, when both are reducing the area. Bernoulli's equation 'says'
the velocity should increase.

ANSWER:
When you are closing the valve, the water velocity through
the valve is increasing but the rate at which water is entering
the hose is decreasing. Hence the velocity through the hose
decreases.

QUESTION:
How does an inverted image of any object apperes in hot areas causeing
to feel like water pond in that area. or what happens to the ray of
light coming from that object to us after total internal refraction?

ANSWER:
I believe what you are asking is how a mirage is formed.
Basically what happens is that the layer of air near the ground becomes
much hotter than the air higher up so you have a thin layer of very hot
air which has a lower density and therefore a smaller index of
refraction. You may treat the layer of air as a slab of material from
which, since it has lower index of refraction, can have light reflected
from it so it acts like a mirror. It is really noticeable for a very
glancing angle of reflection and may be thought of as total internal
reflection.

The above
is one good way to think about what is going on qualitatively, but it
is not what really happens. There is not really a surface where index
of refraction changes from one value to another. Rather, the change
from hot to cool air is continuous over some small distance so you have
a continuously changing index of refraction. The light gets bent as is
moves through this region (that is, refraction and not reflection is
what happens) so that, if the angle of incidence is very glancing, the
light will get bent up and never actually hit the ground, so the effect
is the same as reflection.

QUESTION:
When you jump off a chair, you bend your knees on landing. i
understand that this reduces the shock to your body. But how so?

ANSWER:
When you hit the floor, the floor exerts an upward force on
you. Newton's second law says that the average force you
experience must be your mass times your average acceleration. The
average acceleration is defined to be your change in speed divided by
the time it takes your speed to change. What flexing your knees
does is to increase the time it takes to stop, thereby decreasing the
average force you experience. For example, if your mass is 90 kg
(about 200 lb) and you jump from a 1 m high chair, then you will hit
the floor with a speed of about 4.5 m/s. If you keep your knees
straight you might stop in about 1/10 second, so the force you feel
will be about 90x4.5/(1/10)=4050 Newtons which is about 910 pounds; but
if you flex you knees you might make the time be 1/4 second, decreasing
the average force you feel to 90x4.5/(1/4)=1620 Newtons=364 pounds.

QUESTION:
This is a basic question. f=ma. If I push on a concrete
wall, I am applying a force but there is no accleration of the
mass. What am I doing to the wall?

ANSWER:
For all intents and purposes, the mass of the wall is
infinite and so, however hard you press, the acceleration of the wall
will be zero. What you are doing to the wall is exerting a force on it
and the wall will experience this force. Of course, if the force
is big enough, the wall will be damaged by your force. For example, if
the wall is made of plaster and you hit it with a hammer you are likely
to knock a hole in the plaster with the force.

QUESTION:
How does a neutron change into a proton?

ANSWER:
The physical process is called beta decay. It is not
actually as simple as a neutron changing into a proton because that
would not conserve electric charge (a neutron has no charge, a proton
is positively charged). What actually happens is that the neutron
turns into a proton plus an electron plus an antineutrino. The
antineutrino is a nearly massless particle which interacts very weakly
with matter; therefore, although its presence was predicted
theoretically in the '30s, it was not experimentally observed until the
late '50s. For a classic description of beta decay, link here .
For a description which goes deeper and describes the decay in terms of
what is happening with the constituent quarks, link
here .

QUESTION:
how can i design and conduct an experiment to show that dark surfaces
are better absorbers of radiation than shiny or white or bright
surfaces?

ANSWER:
The easiest experiment is to take two identical objects and
paint one black and one white. Then put them in the sun for a little
while. Then feel them. The black will be warmer. If you want to get
more quantitative, get two bottles, one painted black and one painted
white. Fill the bottles with water, cork them, and put a thermometer
through a small hole drilled in each cork. Then place them both
in the sun and record the temperature of the water in each bottle as a
function of time (maybe every 5 minutes or so) and graph your results.

QUESTION:
I know that the speed of light can be measured in miles per second or
meters per second, etc., but is there any way of expressing the speed
of light according to one of the other physical constants? In other
words, since things like seconds and miles are arbitrary, how might
nature express the speed of light, if it could?

ANSWER:
All nature "knows" is that the speed of light is a
fundamental physical constant. As far as we know, it cannot be
expressed in terms of other fundamental constants. (Actually, that is
what "fundamental" means; if it could be expressed in terms of others
it would not be fundamental.) In order to quantify that, however,
one needs operational definitions of length and time. The history of
all this is interesting. Prior to 1960, the meter was defined to be the
length of a standard stick kept in a vault in Paris and the second was
defined defined as a fraction (1/86,400) of a "solar day"; with these
definitions, the speed of light could be measured. Then, in 1967, the
second was redefined by using what is called an atomic clock. With this
new definition of time, measurements of the speed of light could be
made again. Then, in 1983, the meter was redefined again, but now as
the distance which light travels in 1/299,792,458 seconds; you see that
what this definition does is to set the speed of light to be
299,792,458 meters/second in our system of units and tailor the length
of the meter to agree. It is all very nice because length is defined in
terms of time and a fundamental physical constant. You can read more
about these details here
if you are interested.

QUESTION:
Is time really a measurable property?

ANSWER:
Yes, of course. Anything for which there is an
operational definition may be measured. A second is defined
in terms of an atomic clock. What time is not, however, is a
universal quantity; that is, just because clocks run at one rate on the
earth does not mean that identical clocks which are not here run at the
same rate. For example, if a clock has a very large velocity
relative to the earth (large means not small compared to the speed of
light, 186,000 miles/second) it will run slower than our clocks.
This is the famous result of the theory of special relativity and is
very well verified experimentally.

QUESTION:
A horizontial U-tube of horizontial length (L) & vertical columns
(each column y and y in static equilibrum & open to atmosphere).
The U-tube is filled with water (or alcohol or whatever fluid) such
that the horizontial section is full & the columns of the tube are
filled to some height y. The U-tube & the fluid inside are now
given a purely horizontial acceleration to right which results in an
unbalanced force that causes the fluid in left column (end of the tube)
to rise vertically some amount delta y. Assume this is an ideal fluid
with no friction losses & ignore the U-tube's mass. My question is
this in F net = mass x acceleration...........2nd law, but what is the
fluid mass to be accelerated? Is it the entire mass or just the mass
contained in the horizontial length (L)? Since there's no vertical
acceleration, I think it's the mass in the horizontial section only and
not that contained in vertical columns. Correct?

ANSWER:
If you focus your attention in the fluid in the horizontal
section of the tube, it is being accelerated. What force is
accelerating it? The only force is that due to the pressure difference
across its two ends. That is why the fluid rises in one side and falls
in the other, to provide this pressure difference at their bottoms. The
mass of the fluid in the vertical parts of the tube is being
accelerated by the sides of the tube pushing on it.

QUESTION:
I understand that Rayleigh scattering is associated with particles
about the size of 1/10th the wavelength of the incident light and that
Mie scattering occurs from particles about the wavelength of light and
larger. I have two main, related questions:

Where
does the 1/10th wavelength value come from? Is this a real
constraint, or is it simply empirical reflecting the type of
experiments that people do to investigate Rayleigh scattering?
What
happens in the region between where Rayleigh scattering is dominant and
where Mie scattering is dominant? Is it a mixture of the two, or
is there another scattering mechanism occurring? Where is this
described mathematically (i.e., what can I use to model my data)?
ANSWER:
Calculation of scattering of electromagnetic radiation is one
of the classic topics of electricity and magnetism. It is very
complex and not easy to understand intuitively. As often happens
in theoretical physics, the results are most easily expressed
mathematically in certain limits, i.e. when one or more
parameters of the theory are very small or very large. Rayleigh
scattering is the expression of the scattering theory in what is called
the long-wavelength limit, meaning long compared to the size of the
scatterer (usually taken to be a sphere). How long is long?
It is not unusual to use one order of magnitude as a rule-of-thumb to
test bigness or smallness. There is certainly nothing magic about
1/10, it is just a place where you can expect a pure Rayleigh
scattering calculation to do pretty well. You should not think of
Rayleigh scattering and Mie scattering as two different things but as
the result of scattering theory in two different domains. I am
not an expert at these things, but as you now make the particles get
bigger, the theory becomes much more complicated and a smooth transiton
is made from the characteristics of Rayleigh scattering (where
intensity depends strongly on wavelength) to Mie scattering where it
does not; finally at very short wavelengths (large particle size) the
scattering just becomes geometrical optics. You can get some
detail on Rayleigh scattering at a Lawrence
Livermore site. You can do Mie scattering calculations here . Doing a Google search on Mie scattering or
Rutherford scattering will bring up a wealth of useful site.

QUESTION:
Regards rotation dynamics about a fixed axis, the torque & angular
velocity vectors are directed "along the axis of rotation" using the
right hand screw rule. But what is the direction of the angular
acceleration vector? I know from F = ma, that the acceleration of a
body is in the direction of the net force. Hence, I think it
(acceleration) should be in the direction of the net torque, which for
a disk is tanget to the surface assuming no slippage. Correct?

ANSWER:
Angular acceleration is not the same as usual acceleration
and so F =ma is not what you want to be
thinking about. Instead, Newton's second law takes its rotational
form t =Ia where I is the moment of
inertia. So, since this is a vector equation, the torque and
angular acceleration vectors must be in the same direction. Your
question is what is that direction? Now, you already know the
direction of the angular velocity vector and, if angular velocity
changes, there is an angular acceleration which is defined as a= [w _{2} -w _{1} ]/Dt where w _{2} is the angular
velocity at the end of the time interval D t
and w _{1} is the angular
velocity at the beginning of the time interval D t, that is the
angular acceleration points in the direction of the vector difference [w _{2} -w _{1} ] . But, this difference
is either parallel or antiparallel to the angular velocity direction
depending on whether the disk is speeding up (parallel to w _{1} ) or slowing down
(antiparallel to w _{1} ).

QUESTION:
Can it be proven that a massive particle/body, once accelerated to
close to light speed, would still adhere to the laws of physics as we
know them? Aside from light, I have not been able to find evidence of
any objects that travel remotely close to 180,000 miles per
second. I ask because the laws of physics that we know of seem to
break down as an object approaches a large massive body (black hole),
and I wondered if it has been proven that objects approaching light
speed would not behave similarly.

ANSWER:
The theory of special relativity is one of the best verified
(experimentally) of all physical laws. There are many things which are
not massless and move with velocities (relative to you) that are not
small compared to the speed of light. Any modern particle accelerator
has electrons, protons, or atomic nuclei which move with very large
velocities; a typical speed at an electron accelerator would be greater
than 99% the speed of light. Cosmic rays, which rain down on us all the
time have speeds comparable to the speed of light. The most distant
galaxies move with speeds more than half the speed of light. In
the vicinity of a black hole, you must invoke the general theory of
relativity which is the accepted theory of gravity.

QUESTION:
#1. you have two rectangular magnets (or four or more) each close
enough (and held in position) that they are exerting an equal force
against each other (same poles toward one another obviously). Now; what
is it like in the middle of the opposing fields? Is there a
"dead" center? Would it create properties that are creating either a
"nothingness" or a "greatness" of force??

ANSWER:
Almost always when you have two or more sources of any field,
there will be places where the field is zero. For example, if you
put two equal electric point charges some distance apart, the electric
field will be zero halfway between them. If you put two bar
magnets (which is what I presume you mean by "rectangular magnets") so
the bars are in a line with north poles facing each other as shown in
the figure above, then the field will be zero where the smiley face
is. But there is no particular profound significance to this; it
just means that if you put a north or south pole of another magnet
there that it will experience no force.

QUESTION:
A block of mass (M) is sitting on a turntable that's rotating with some
value of omega. A smooth cord runs from the block thru the center of
the turntable and connects to another block of mass (m)..M>m. The
upper block is at rest relative to the turntable(it doesn't move
sideways nor is it pulled to the turntable's center by the cord-smaller
block system). I want to understand the direction of the upper block's
frictional force, i.e. is it sideways due to the turntable's rotation
(where it has a torque) or is it along the radius (parallel to the
cord) and hence has no torque about the axis of rotation. I wanted to
apply the conservation of angular momentum, but force's direction is
not apparent. Can you help?

ANSWER:
First of all, if the angular velocity is constant, there is
no angular acceleration and therefore no net torque on M .
Thus there is no tangential (sideways as you say) component of the
frictional force. Since m is at rest, the tension in the
string is mg. But the net force on M must be
the centripetal force, F _{net} =Mv ^{2} /R=MR w ^{2} , the direction being
toward the center and R being the distance from the
center. Hence, F _{net} =mg+f , so f=MR w ^{2} -mg. So
the answer to your question depends on both R and w ; if MR w ^{2} -mg> 0, f
points radially in and if MR w ^{2} -mg< 0, f
points radially out; also, if MR w ^{2} -mg= 0, f
=0.

QUESTION:
1) how can positrons be artificially produced
2) how can anti-protons be artificially produced
3) how can both be stored for long periods of time

ANSWER:
Positrons are naturally occurring in nature and come from
radioactive nuclei which undergo b^{+} decay; such nuclei are
unstable because they have too many protons and a proton decays into a
positron, a neutron, and a neutrino. They are also found in
cosmic rays (which is where they were first observed). But, if
you need them in copious quantities, they are created in accelerators
by what is called pair production where a very fast charged particle
(often an electron) strikes a nucleus and spontaneously creates an
electron-positron pair. Antiprotons are made in a similar
fashion, by creating a proton-antiproton pair; this is usually done by
shooting high-energy protons at a metal target. Antimatter is
most often stored in what are called storage rings in which
magnetic fields cause the particles to move around donut-shaped tubes
which have been evacuated to very high vacuum. If antiparticles
come in contact with their particle counterparts they annihilate and
disappear in a tiny flash of energy.

QUESTION:
Is it possible that just like gravity is a function of mass, so too,
ANTI-gravity is a function of vacuum (or ABSENSE of mass)?

ANSWER:
I am not really sure what you are trying to ask here since
there is no such thing as antigravity that we know about. One
interpretation of you question would be would a massless object
experience no graviational force; it turns out that the best known
massless object, a photon (a bundle of electromagnetic energy, e.g.
light), does experience the gravitational force, so light is bent when
it passes a massive object (a star, for example). If there were
antigravity, it would more likely be due to the existance of a
different kind of mass (call it antimass) which, instead of being
attracted to usual mass would be repelled. This would be like the
electric force which can be either attractive or repulsive because
there are two kinds of electric charge (called positive and
negative).

QUESTION:
I have read that it would take an enormous amount of fuel for a rocket
to approach the speed of light because mass increases as the speed of
light is approached. My question is: why wouldn't the increased
in the mass of the fuel compensate for the increase in the mass of the
rocket?

ANSWER:
How would it compensate? The fuel is in the rest frame
of the rocket and, in that frame, has its rest mass. In a frame
in which the rocket is moving, the mass of the fuel is very large, but
this in not due to an increase in the amount of fuel; rather it
reflects that it took work (energy) to accelerate this fuel up to its
large speed. This work shows up as an apparent increase in mass.
One of the biggest problems in carrying a large amount of fuel with you
is that you need to accelerate much of it up to high speeds which is,
in some sense, wasted energy because the fuel is not your
payload.

FOLLOWUP
QUESTION:
What I had in mind was that it takes mass thrown out the back to
accelerate a rocket and as you accelerate you have more mass to throw
out the back. Another thing you said was "This work shows up as an
apparent increase in mass". I'm guessing that your word 'apparent' is
the key to why it would not compensate, am I right?

ANSWER:
The mass of fuel which the rocket sees is unchanged, so if he
throws out some of the mass the remainder of the rocket will acquire an
added momentum equal and opposite to that which thrown mass carried
away. The thrown out mass as seen from the "rest frame" is much bigger
so it carries much more momentum; however, the mass of the rocket is
bigger by the same fraction so the increase in speed will be the same
in both frames. When I say apparent it is because many physicists
prefer to think think of mass as the inertia of an object at rest (rest
mass) and, in order for conservation of momentum to be conserved
(thereby preserving Newton's laws), the momentum must be redefined as p=m _{0} v /[1-(v ^{2} /c ^{2} )]^{1/2}
rather than the usual p=m _{0} v . Here, p
is the momentum, m is the rest mass, v is the speed,
and c is the speed of light. So, you see, you may interpret
this redefinition of p to mean that the mass increases as m =m _{0} /[1-(v ^{2} /c ^{2} )]^{1/2}
and momentum is still mv . But in mechanics, the important
thing is momentum, not mass.

QUESTION:
At the top of a ramp there are a solid sphere, a hollow sphere, a
hollow cylinder, and a solid cylinder are all released at the top of a
ramp. They all have the same radius and mass. Which one reaches the end
of the ramp last? It was given that the hollow cylinder comes in
last, but that seems backward to me since the moment of inertia for the
hollow cylinder is the greatest.

ANSWER:
The kinetic energy at the bottom of the ramp must be shared
by the translational and rotational motions of the object. And
since the potential energy of all four objects are the same, the
kinetic energies at the bottom must be equal. To find out which
arrives last we need to figure out whose velocity at the bottom is
least. The kinetic energy is given by K= ^{1} /_{2} mv ^{2} +^{1} /_{2} I w ^{2} =^{1} /_{2} [m +(I /R ^{2} )]v ^{2} .
So the speed v at the bottom is smallest when the moment of
inertia I is the biggest (and vice versa ).

QUESTION:
I understand that the surface area of a car's tyre in contact with the
road surface is entirely dependant on the weight over the tyre and the
pressure within the tyre. A given weight with a given pressure will
provide a given SA. What I don't know is: If you add more weight
over the tyre does the tyre's surface area increase or does the tyre's
pressure increase? Or maybe a bit of both ? Does the tyre's elasticity
play a critical part?

ANSWER:
The key here is the deformation of the tire. For a
given tire, the harder you press it down the "flatter" it becomes so
the surface area in contact with the road increases. Similarly, a
tire with a high pressure will be less responsive to being flattened so
it makes sense that the area depends also on the pressure. If
adding more weight to the load actually changes the volume of the air
in the tire, then the pressure would increase accordingly.

QUESTION:
I have heard that there are several wavelengths of light that never
reach our planet. What if a material could be constructed with the
specifics to reflect that wavelength? What would we see? Everything, or
nothing at all?

ANSWER:
If you could make an object which reflected one wavelength
l and absorbed all
others, then if the light illuminating it did not include the
wavelength l , it would appear black.

QUESTION:
Can you help me understand what this term is or is not? It is used in
regard to calculation of moment of inertia, but texts seem to give it a
passing few words & never develop the concept. A study of
physical pendulums requires the calculation of moments of inertia (I),
so would (or could) this constant be used to compute the I? For
example, if the pendulum had a moveable disk mounted to a thin rod
& you wanted to know how far the disk should be moved (either
way) to achieve a certain period (T), can this be used in the problem
solving?

ANSWER:
As you probably know, the moment of inertia of a point mass m
a distance of s from an axis is ms ^{2} .
The radius of gyration is a number which expresses the moment of
inertia for any object about some axis in this form, that is I=mG ^{2}
where G is called the radius of gyration and m is the
mass of the object. For example, the moment of inertia of a
uniform disk of radius R and mass m about an axis
perpendicular to its surface and through its center is I=mR ^{2} /2=mG^{2} ,
so G =R /[2]^{1/2} . For the axis shifted by
an amount d , I=m (R ^{2} /2+d ^{2} )=mG^{2} ,
so G =[R ^{2} /2+d ^{2} ]^{1/2} .
It is not clear to me how introducing G would make problem
solving any easier.

QUESTION:
I am an armchair physicist with a Master's degree in Metallurgical
Engineering.

My
question pertains to Einstein's famous E=mc2 equation: Why
doesn't E = ONE-HALF mc2, which is the sum of the momenta of all of the
particles moving at the speed of light? Any elementary
physics student knows that kinetic energy (which, when 100% kinetic,
there is zero potential energy) = the sum of the momenta of the masses
moving at velocity v. In other words, the integral of mv of the
masses is 1/2 mv2.

Wouldn't
it follow that when all of a given mass's energy is completely
converted to energy, that it is ALL kinetic energy — whose sum
of all of the elementary constituent particles moving at the velocity
of light would equal 1/2 mc2. In light of string theory, I would
think this would be an obvious result.

Please
tell me why I am wrong — why am I off by a factor of 2? I have
asked physics professors at a number of universities, who give me
typical BS answers as "well, it's just a constant of the universe" (it
isn't), or "it's been proven experimentally" (it hasn't — even
man-made nuclear explosions have resulted in only a few percentage
points of the entire mass being converted to energy; I am not
sure that even the energy of matter-antimatter annihilations have been
measured to an accuracy to distinguish mc2 from 1/2 mc2).

If you
can explain the answer, I am intelligent enough that you can explain it
in technical terms.

If you
cannot explain the answer, it this a bit of original thinking that has
not been previously addressed?

ANSWER:
Well, gosh, you gotta get out of that armchair and learn
something about the theory of special relativity. Your
classically-based ideas of momentum and kinetic energy are wrong.
In my discussion, I will call the rest mass of a particle m
and the speed of the particle is v . Linear
momentum is not mv but rather p=mv [1-v ^{2} /c ^{2} ]^{-1/2} .
Note that as the speed approaches the speed of light, p becomes
infinite, not mc which is a whole lot less than infinity even
if c is a big number! Now, to calculate the kinetic
energy of the particle, we must calculate the work necessary to
increase the speed from zero to v . You do this by
integrating F dx ; if you now write F =dp /dt
(Newton's second law which is still valid as long as it is understood
that p is not mv ) and do some calculus and algebra, you
need to integrate mu [1-u ^{2} /c ^{2} ]^{-3/2} du
from u= 0 to u=v, and you get K=mc ^{2} {[1-v ^{2} /c ^{2} ]^{-1/2} -1}
for kinetic energy . We now make the identification that
the kinetic energy is the total energy minus the energy the particle
has by virtue of its mass (called the rest mass energy). So rest
mass energy of a particle is mc ^{2} and total energy of
a particle is mc ^{2} [1-v ^{2} /c ^{2} ]^{-1/2} .
To make this even more plausible, kinetic energy should reduce to its
classical value if v<<c ; this is easy to show using a
binomial expansion for the square root, [1-v ^{2} /c ^{2} ]^{-1/2} =1+v ^{2} /(2c ^{2} )+...,
so K=mv^{2} /2+...

QUESTION:
do AM and FM radio waves travel at the speed of light?

ANSWER:
Yes. Radio waves are electromagnetic radiation, just
like light but different wavelengths. All electromagnetic
radiation propogates with a speed of about 186,000 miles/second in
vacuum (the speed in air is very close, but slightly less).

QUESTION:
When a spring is compressed and suddenly reliesed it's going to
accelerate and will achieve maximum speed when it reaches its normal
lenght. From that point on it will decelerate. Is it possible to
calculate it's accelaration based on its physical caracteristics as
diameter, twist, wire diameter and material properties?

ANSWER:
I suppose it is possible if you are an engineer, but I
wouldn't know how to do it. What I do know, though, is the
physics. I would first do a measurement of the spring constant k ,
the length of the unstretched spring L, and the mass m .
Then I would compress it against a wall by an amount d and let
it go. If the floor is smooth (frictionless), the kinetic energy
of the spring would have to equal the potential energy the spring had
at the beginning, ^{1} /_{2} kd ^{2} .
However, the acceleration is not a single number because one end of the
spring is at rest. If you call the velocity of the free end of
the spring v _{0} (when the spring returns to
its unstretched length) and assume that the speed of each part of the
spring increases linearly from one end to the other, then v=v _{0} (x /L )
where x is the distance from the wall and L is the
unstretched length of the spring. Assuming that the mass of the
spring is uniformly distributed along its length, the kinetic energy of
a piece of length dx at position x will be ^{1} /_{2}
[(m /L )dx ][v _{0} (x /L )]^{2}
= ^{1} /_{2} [(mv _{0} ^{2} /L^{3} )x ^{2} dx ];
integrating, the total kinetic energy (which must equal the potential
energy at the beginning) is ^{1} /_{6} mv _{0} ^{2} =^{1} /_{2}
kd ^{2} so v _{0} =[3kd ^{2} /m ]^{1/2} .
This is not what you wanted, but it is what I can do!

The above
gives you the velocity of the end of the spring (as well as every other
point) when the spring has just reached its unstretched length L ;
at this instant, the acceleration of every point of the spring is
zero. You could extend the above analysis to find the velocity at
the end for any arbitrary length L' <L .

QUESTION:
What can I use to make a magnet repel? I can't find the magnets
with opposite ends so they attract and repel, is there anything else I
could use to show how this works?

ANSWER:
I would guess that you are probably trying to use
refrigerator magnets and they are manufactured in such a way that they
consist of many tiny bar magnets but there are both north and south
poles on the surface. You can find a detailed discussion at MadSci.net .
So what can you use? You need to find some simple bar
magnets. When I was a kid we had little toys which were dogs
(terriers) sitting on little bar magnets. These would attract or
repel each other and lots of kids had them so I guess they were readily
available in toy stores. I haven't seen them in years, but a
quick internet search revealed that they are still available and may be
bought at the Restless
Mouse Shop for $2.49 a pair (I don't usually do commercials
here!) Good luck and have fun!

QUESTION:
When calculating the moment of inertia(I) of a meter stick that is
pivoted about its end, the parallel axis theorem would be used. If a
weight is then clamped (tack welded or whatever) to the stick's
midpoint, would the new total (meter stick + new mass) mass be treated
as if the meter stick had gained mass, or would this additional mass be
treated separately when calculating I?

ANSWER:
The parallel axis should be used only if you already know the
moment of inertia about the center of mass of the stick. Of
course, the well-known result is ML ^{2} /3. Now,
attach a point mass m to the center. The moment of
inertia of this mass relative to the end is m (L /2)^{2} .
The net moment of inertia is L ^{2} [(M /3)+(m /4)].
If you assume that the effect is just increasing the mass of the stick,
then you would find (M+m )L ^{2} /3, clearly not the
same. The reason is that m does not have the same spatial
distribution that M does and I of anything depends on
not just how much mass there is but how it is distributed around in
space.

QUESTION:
An electromagnet is turned on and left on for one hour. At the instant
the magnet is turned on, I assume the magnet becomes surrounded by a
static magnetic field which grows larger at light-speed? - with
electromagnetic radiation present somewhere? If so, what's the "shape"
of the electromagnetic radiation? I mean, does the EM radiation have
the "shape" of a "thin shell" which is located at the boundary of the
magnetic field? After one hour the magnetic field has a radius of one
light-hour? Now the electromagnet is turned off. Does the entire
magnetic field instantly disappear when the magnet is turned off? While
EM radiation, somewhere, keeps propagating?

ANSWER:
Let us assume that there is a very short time during which
the field is changing when you turn it on and then a very short time
when it is changing again when you turn it off. During the first
time, there will be a pulse of electromagnetic radiation which moves
out with speed c and the space between it and the magnet will
be filled by a static magnetic field; the static field will become
smaller farther away from the magnet. Then when you turn it off
there will be another pulse of radiation and the space between the
second pulse and the magnet will have zero field. As time goes
on, this shell which consists of two shells of radiation and the
intervening space with static field all propagating away from the
magnet. However, the strength of the static field falls off much
more quickly as you get farther away than the strength of the radiation
fields does, so after not too long you would only see two pulses of
electromagnetic radiation.

QUESTION:
An operating pendulum is dropped from a great height above the earth.
It would be in freefall until it hit the surface.

ANSWER:
The pendulum bob would move in uniform circular motion about
the support point with the speed it had at the instant you released it.

QUESTION:
What is a "matter wave"? For example, if it's said that a rock is both
a material thing and also a wave, how is the rock a wave?

ANSWER:
DeBroglie's hypothesis is that any particle also has the
properties of a wave. The wavelength of the particle is given by h /p
where h is Planck's constant and has a value of 6.6 x 10^{-34}
J s and p=mv is the linear momentum of the particle.
Suppose that you had a rock of mass 1 kg which was moving with a speed
of 100 m/s. Its wavelength would be 6.6 x 10^{-36}
m. In order to observe this you would have to diffract the wave
around some obstacle which was about this size. But, the smallest
thing we know is an elementary particle like a proton which has a size
of approximately 10^{-15} m, far too big. However, the
wavelike nature of particles is well established by looking at much
smaller particles than your rock, say an electron which has a mass of
about 9 x 10^{-31} kg and so an electron with a velocity of 10^{4}
m/s would have a wavelength of about 7 x 10^{-8} m. Since the
spacing of atoms in a crystal is about 10^{-10} m, on may
easily see electron diffraction. This is the famous Davisson-Germer
experiment.

QUESTION:
A permanent magnet is travelling East at a constant speed - it bounces
off something and is now travelling West - did the magnet accelerate
when it changed directions? If so, I know two other things that cause
acceleration are rotation and a change of speed - anything else?

ANSWER:
Acceleration occurs whenever there is a change in velocity;
don't forget that velocity is a vector so that it can change by either
its magnitude (speed) or its direction changing (or both). To
actually compute the acceleration, you would have to know something
about the details of the collision. The average acceleration in
your example is change in velocity divided by the time it took to to
reverse the velocity. For example, if the speed in is 20 m/s and
the speed out is 20 m/s, the change in velocity is 40 m/s and it points
in the direction of the outgoing magnet. If the collision took
.01 seconds, the average acceleration would be 4000 m/s^{2} .
If the magnet had a mass of 1/2 kg, then Newton's second law tells you
that a force of 2000 N (about 450 lb) had to be exerted on the magnet
during the collision to cause the acceleration.

QUESTION:
I've read that even "non-magnetic" things, like the human body, a tree,
a piece of plastic, and anything else, all have extremely weak magnetic
fields surrounding them. If all things do have magnetic fields, what
causes that?

ANSWER:
Magnetic fields will exist only if there is an electric
current distribution of some sort. That is a true statement if
you think of magnetic moments of electrons, atoms, molecules, etc .
as resulting from local current distributions. In the human body
there are many electric currents — all
nerve impulses are primarily electrical, the heart beats in response to
electric currents, the blood moves through the body and doubtless
carries many ions and therefore constitutes an electric current, etc .
Similarly, any living thing is likely to be a source of magnetic
fields.

Virtually
any atom which has an intrinsic angular momentum also has a magnetic
moment. As examples, any atom which has an odd number of
electrons will have a magnetic moment; any nucleus which has an odd
number of protons or neutrons will have a magnetic moment.
Therefore, many atoms have magnetic fields around them because they
are, in effect, tiny bar magnets. The catch is, however, that a
chunk of stuff is made up of a very large number of atoms and almost
always the magnetic moments of the individual atoms are randomly
aligned so that the net effect is that there is zero field. The
few "ferromagnetic" materials in nature (like iron) have magnetic
fields around them because atomic magnetic moments have a tendency to
align with their neighbors. On the other hand, all materials are
"magnetic" to some extent — this
means that the atomic moments will have a tendency to align with an
external magnetic field. Therefore, I would suggest that the
piece of plastic you ask about would have a field of its own only if it
were in an external field (which it is since the earth has a magnetic
field), but it will be almost immeasurably small unless the external
field is very strong.

QUESTION:
If a permanent magnet is lying on a table, is it accelerating because
of the orbital and rotational motions of the planet? If so, is it
constantly emitting electromagnetic radiation? If so, it seems like
there wouldn't be any frequency or wavelength involved? Would there
just be a single magnetic field, and orthogonal to that a single
electric field - both fields constantly expanding larger and larger at
the speed of light? If all that is true, then are all things constantly
emitting that type of electromagnetic radiation, since all things are
made of atoms that are composed of electrically charged
particles?

ANSWER:
Technically, yes. Any accelerating magnet or electric
charge will radiate electromagnetic radiation. However, the
intensity of this radiation would be immeasurably small. Since
the accelerations you consider are not linear but centripetal, there
would be frequencies of 1 cycle/day and 1 cycle/year associated with
the radiation. There is an exception to this rule, however.
For very small systems of electric charges such as atoms, where quantum
mechanics is applicable, the system in its normal (ground) state will
not radiate even though my might think of atoms in terms of the Bohr
model in which we envision electrons as orbiting the nucleus; in fact,
the Bohr atom is a naive model which is not really an accurate
description of nature.

QUESTION:
what is edi current?

ANSWER:
I presume that you mean "eddy current". If you expose a
conductor to time varying magnetic fields, electric currents will run
around inside the conductor; these are called "eddy currents" because
they will often take the form similar to little whirlpools which is
what the word eddy originally means. You can find a more detailed
answer at physlink.com
. A nice example is "magnetic braking" of which there is a video
at http://www.pa.msu.edu/educ/lectdemo/index_e/3/E3_01.htm

QUESTION:
is electron particle or wave? is there experiment in this
matter?

ANSWER:
This is the fundamental question which leads to quantum
mechanics. The answer, surprisingly, is that it is both and it
will display the properties of either, particle-like or wave-like,
depending on what you look for. There are many experiments which
show that the electron behaves like a wave; these are essentially the
same type of experiments you use to determine that light is a wave — diffraction.
I suggest you do a Google search
on electron diffraction to learn more. You might also search on
wave particle duality which is the term for the fact that all things
(not just electrons) are both particles and waves.

QUESTION:
A permanent magnet has been lying on a table for one hour. Now, it is
rotated by 45 degrees. How large of an area do the new field lines
occupy? Is there anything left of the original field lines, beyond the
area occupied by the new field lines?

ANSWER:
First of all, the field occupies the whole volume around the
magnet, not some area as you imply. If the magnet is static (not
moving), the field is called magnetostatic; so when you finish moving
the magnet the field around it will be exactly the same (relative to
the magnet itself) as before. The real issue then is what happens
when the magnet is actually moving and how long does it take for the
field to transform from the old to the new? During the time it is
moving, the magnetic field is changing and therefore, according to
Faraday's law, there is an induced electric field also; the moving
magnet, since it is accelerating, will actually radiate electromagnetic
radiation (i.e. it is an transmitting antenna). However,
the information about the magnetic field progogates at the speed of
light, so the field, as accurately as you would be able to measure
unless you make the change extraordinarily rapidly, is essentially
smoothly and continously changing.

QUESTION:
If two magnets are brought near one another, do their fields completely
blend together to form one field?

ANSWER:
Yes. Magnetic fields obey the superposition
principle which states that if several different sources of field
individually cause fields at some point in space, the net field there
is the vector sum of all the individual fields. Since it
is a vector sum, it is possible that the sum will be zero even thought
the individual fields are not. For example, if two identical
magnets are positioned so that their north poles are some distance
apart, the field will be zero halfway between them.

QUESTION:
Does the speed of terminal velocity change on a planet with the same
ratio of gravitational pull and atmospheric gases as earth, assuming
that the air is made of the same gases in the preportions as on earth?

ANSWER:
Terminal velocity is when the velocity is just right that the
downward force of the weight of an object is just cancelled out by the
upward force of the air resistance force. Normally we compute the
air resistance force by assuming that it is proportional to the
velocity F = bv.
(Note, however, that
this is also an approximation and the dependence on velocity is much
more complicated.) Since we want the weight W to equal F
when v=v _{term} , then v _{term} =W /b ;
therefore, v _{term} depends on what the constant b
is. This proportionality constant depends on lots of things, the
nature of the fluid through which the object is falling and the
geometry of the object being the most important. I do not know
what you mean by you mean by "...the same ratio of gravitational pull
and atmospheric gases as earth..."; it is clear that what must be the
same is the ration W /b , so if b increases
proportionally to W , then the terminal velocity will be the
same. Also note that, since W =mg where g
is the acceleration due to gravity, g /b needs to be
constant. If we keep the geometry the same and go to a different
planet, then b will be primarily determined by the nature of
the atmosphere. If g is bigger on the new planet, then
the atmosphere must be more dense if v _{term} is to be
the same.

QUESTION:
An object that appears white (like a piece of paper) appears white
because it reflects all the colors of light that are incident upon
it. If that is true, then what is the difference between a white
piece of paper and a mirror? Doesn't a mirror also reflect all
the light upon it?

ANSWER:
The answer has to do with how the light is
reflected. Reflection from the paper is called diffuse
reflection ; a ray of light which comes in making some angle
relative to the normal to the surface is reflected in a random
direction. Reflection from the mirror is called specular
reflection ; a ray of light which comes in making some angle
relative to the normal to the surface is reflected in a direction at
the same angle relative to the normal (angle of reflection = angle of
incidence, "the law of reflection"). The result is that your
brain, interpreting the light it sees, can tell, by observing the rays
reflected by the mirror, where the rays came from but you cannot do
that with the paper. The net result, as anyone who has looked
into a mirror can tell you, is that an image of the "real world"
appears behind the mirror.

QUESTION:
Do electrons flow on the surface of wire/conductor or through it?

ANSWER:
It depends on the nature of the current. For DC
currents, the current is uniformly distributed across the whole volume
of the conductor. For AC currents, however, there is a tendency
for the current to be more concentrated toward the surface of the wire
so that, for very high frequencies, the current is mainly at the
surface; this is called the "skin effect". This effect tends to
make the resistance of a wire a function of frequency of the current
and it is therefore something hi-fi enthusiasts worry about; it is not
a big effect at audio frequencies, though, and becomes important at
radio frequencies (MHz). Very short pulses of current have high
frequency components. For example, you are safe inside a (metal)
car during a thunderstorm, even in the event of a direct hit, because
the current will flow only on the outer surface of the metal.

QUESTION:
Are magnetic lines of force literaly "strings" or "filaments" of energy
or are the lines shown in illustrations only there to show the size
& shape of a uniform field? i.e. is the force stronger directly on
the lines & weaker between them or is it a uniform field?

ANSWER:
The magnetic field lines are not "filaments of energy".
The field is a representation of the force which would be felt by the
north pole of a magnet (if you could separate it from its south pole
partner). The field lines you see drawn represent a uniform field
if the lines are all straight, parallel, and evenly spaced. In a
uniform field, a north pole would experience the same force, magnitude
and direction, no matter where you put it. If the field is not
uniform, the field lines either curve or have varying "packedness" (i.e.
some lines are closer to their neighbors than others) or
both; the more closely the lines are packed, the stronger the
field. Electromagnetic fields do contain energy, but you should
not think of the field lines as somehow representing this property.

QUESTION:
When working with projectile motion problems, the range (R) =
[(2*Vo^2)*(sin theta)*(cos theta)] / g, but when the launch angle is
horizontial (theta = 0) the sin theta is equal to 0, hence the range
would = 0. Since this is not the case, how do you calculate the range
when theta = 0?

ANSWER:
Ah, the dangers of using formulae instead of using your
head! You need to know the applicability of the formula and
understand precisely what it means. The range equation tells you
how far the projectile travels horizontally before it returns to the
same vertical position from which it was launched. Clearly, since
the projectile starts falling the instant it is launched, it never
comes back to that same height again.

QUESTION:
Do gravitons exist?Or are they only a theory?

ANSWER:
There is no working theory of quantum gravity and so a
graviton is a purely hypothetical particle. A graviton has never
been detected experimentally. Trying to unify general relativity
(the working theory of gravity) with quantum mechanics (the working
theory of everything else) is one of the holy grails of theoretical
physics.

QUESTION:
electrons are emiited from a conductor when the conductor is ?I think
the answer is cooled rapidly, but I AM NOT SURE.

ANSWER:
When: the conductor is heated up, when it is placed in a
strong electric field, when it is bombarded by photons of sufficient
energy, but I have never heard of rapid cooling causing emission.
In essence, it takes energy to remove an electron from an object and
cooling is energy being taken away. You can get more detail here .

QUESTION:
What exactly is the work function in the photoelectric effect? I know
it is in einstien's equations because electrons use energy in order to
get to the surface of a metal and thus be emmitted, but what exactly is
it measuring and determining?

ANSWER:
Suppose you have a perfect conductor from which you wish to
remove an electron. When you pull it out a little way, there will
be an induced charge distribution on the surface of the conductor
because the electron will push nearby other electrons away. This
problem is a classic problem in electrostatics and may be solved by
imagining an "image charge" inside the conductor which is oppositely
charged (positive). So, you see, work will be required to pull
the electron to where it is far away. This is essentially what
the work function is, the work necessary to remove an electron from a
conductor. There is one little catch —if you press this image
charge thing too far and try to calculate the work done it will be
infinite because eventually the two charges are on top of each other
(when the electron is exactly at the surface); this occurs because the
conductor is not really perfect and homogeneous but made up of atoms,
so when you get very close to the surface you see the individual atoms
which are separated by approximately 10^{-10} m. The
method to calculate the work function should then require you to
calculate the amount of work necessary to move an electron from r _{1} =10^{-10}
m to infinity; you will find that this gives you a ballpark estimate of
the work function, on the order of a couple of volts.

QUESTION:
Friction and air resistance cause kinetic energy of an object to
convert into heat energy. Why does this happen? Also, what kind of
"heat energy" is this? Photons of radiation or movement of particles?

ANSWER:
The detailed nature of friction at a microscopic level is not
very well understood. But the mechanism for heating must surely
be collisions between atoms resulting in increased kinetic
energy. For example, if a block slides across a table it loses
its kinetic energy but many individual atoms pick this energy up and
the block and table heat up.

QUESTION:
With the photoelectric effect are you hitting the free electrons or the
electrons in the atimic shells with photons? What would the difference
if you struck the different ones?

ANSWER:
The photoelectric effect is a phenomenon which occurs in
metals and the photoelectrons are the result of the conduction
electrons interacting with photons. In a conductor, to an
excellent approximation, the conduction electrons move around in the
material much as a gas moves around in a box and these electrons may be
thought of as free within the material.

QUESTION:
Whe you heat up an object by radiation (ie - hit it with infrared
photons) it gets hotter. Therefore, its atoms/ molecules must be
vibrating faster. However, I have also been told that when an electron
absorbs a photon it moves to a new evergy level and then releases the
photon again as it returns to its ground state. So, when do these two
different phenominum occur? ie - what must the photon in the 1st senari
strike, etc

ANSWER:
The photons scatter many times, elastically, from the atoms
but, since linear momentum must be conserved, each collision reduces
the energy of the photon a little while that atoms absorb that kinetic
energy and eventually share it with all their neighbors.

QUESTION:
Atoms go up in discrete energy levels. But what happens if a photon
hits an electron but does not posess the correct energy for it to move
up any level?

ANSWER:
If the photon has too little energy to excite the atom to any
level, the photon will just bounce off elastically from the atom,
leaving the atom in its original state but giving the whole atom
kinetic energy and slightly changing the energy of the photon (energy
and linear momentum must be conserved). If it has enough energy
to completely knock the electron out of the atom (ionize the atom), you
will have essentially Compton scattering. Things get more
complicated in a solid where the atoms are bound in a lattice; do some
research on the Mössbauer effect.

QUESTION:
Voltage or potential difference is "the difference in the energy per
unit of charge between two points". eg - If the pd across a resistor is
6V, then each Coulomb of electrons has lost 6J of energy and converted
this into other forms, such as heat. However, why do we then say that
if static electricity builds up on an object that the potential
difference between it and the earth or an uncharged object increases?
Where is the energy, voltage, etc?

ANSWER:
Perhaps a more fundamental way to think about potential
difference between two points is to think about the work it takes to
move a charge from one point to the other (if it takes one Joule of
work to move one Coulomb of charge from one point to another, then the
potential difference is one volt). It should then be obvious why
an object being charged is having its potential relative to some
reference point constantly increasing: it clearly will take more work
to move a charge from the reference point to the charged object as the
charge increases.

QUESTION:
"energy conservation requires that any energy entering the system via
the primary coil must equal the energy leaving the system via the
secondary coil." - But as far as I'm aware no energy is entering the
system. As electrons flow through the coil, they do not give electrons
to the core of a transformer. Therefore, how is energy entering the
system and consequently why does the conservation of energy apply
between the two coils? (ie - why does power in = power out? )

ANSWER:
See answer to the following question.

QUESTION:
Refering to others question on power and transformers - but surely
energy is not actually entering the system via the primary coil. After
all, the coil is not giving its energy to the core, is it?

ANSWER:
Of course energy is entering via the primary
coil. Do you think the transformer would operate if you did not
plug it into some power source? There is a time-varying current
in the coil which means that there is a time-varying magnetic
field. If it were just a coil it would be a time varying magnetic
moment which is a kind of antenna; electromagnetic waves would be
emitted from it resulting in energy leaving via those
waves. If you make a transformer, there will be a time-varying
field through the secondary coil which will cause a current to flow
through any load which is connected across it. Energy will thus
leave the system via the secondary coil and it comes from the power
supply which is driving the primary coil. The two will be equal
if you neglect losses due to heating of the transformer or the energy
which will be radiated away.

QUESTION:
Why is a transormer said to obey the equation Pprimary=Psecondary?
Surely, the two coils are completely seperate apart from their magnetic
fields, so why should the power (ie - V*I ) of one equal the power of
the other? If you take the concept of conservation of energy it
makes sense to say that the power in the primary coil of a transformer
is equal to the power of the secondry since energy cannot be created or
destroyed. But where does the energy change occur?

ANSWER:
I have consolidated two questions from the same person
here. Essentially, he/she answers the first question with his/her
second question: energy conservation requires that any energy entering
the system via the primary coil must equal the energy leaving
the system via the secondary coil. Of course, in the real
world the two will not be equal because of energy losses in the
transformer; for example, note that a transformer is always warm or hot
as some of the energy in is converted to heat.

QUESTION:
With a transformer, we say that all of the magnetic field of the first
(primary ) coil is trapped within the core. But if we look at a
solonoid containing an iron core, there is clearly a magnetic field
around the core. ( hence the familiar diagrams) How come that with a
transformer core we only get a magentic field within the core, and not
one surrounding it or around the coil?

ANSWER:
Well, the "familiar diagrams" show the core and a coil around
it, not a field. But a solenoid does have a field outside the
core, in fact the field looks, on the outside, just like the field of a
permanent bar magnet. Imagine now taking a bar magnet and bending
it around so that the two poles meet to make a closed loop of iron; now
nearly all the field will be confined to the inside of the iron.
This is the way the core of a transformer looks, closing on itself.

QUESTION:
The air conditioner in my apartment is terrible and I live on the top
floor of an a building with a BLACK roof. It's really hot.
I need an alternative method of cooling the apartment. My idea:
Fill metal pipes with dry ice, seal both ends, attach them near the
ceiling. I figure the pressure should keep the dry ice solid, the hot
air near the ceiling should cool near the pipes and fall creating minor
circulation and cool the room. Might it work? Could the pipes
explode?

ANSWER:
This is a sure recipe for disaster. Do not do it. Get a better AC
unit. Also, if you think about it a little, your idea attempts to
get something for nothing. If your pipes absorb heat from the
room, where is it going to go?

QUESTION:
if the psi is known along with the diameter of the tube the gas
is flowing through can the linear velocity of the gas be
determined? In my case the psi is 40 psi and the diameter is 1/8 in.

ANSWER:
No. You do not have enough information. If you
think about it, it is obvious that one possible solution is zero
velocity. Just add gas until the pressure is 40 psi.

QUESTION:
Why when an object gains an amount of charge ( eg - when charged with
static electricity) does its voltage increase?

ANSWER:
That is just the definition of electric potential.
Voltage is work done per unit of charge moved. So if it takes you
2 Joules of energy to charge something to 10^{-6} Coulombs, the
voltage is 2,000,000 volts relative to the place where the charge
originally resided.

QUESTION:
With capacitence, why is it that different bodies need to gain
different amounts of charge to gain a different voltage? ie - C=Q/V.
Surely, it would make more sense to think that since a body is gaining
xC of charge (ie - Q) and that each electron forming xC has a certain
amount of energy, that by gaining this xC any object's electrons would
possess this same extra amount of energy compared to earth. (ie - all
the objects would have the same potential difference or voltage)
Also, what does it mean when we say that a capacitor has a certain
value? What does this say about the particular capacitor?

ANSWER:
As you charge a capacitor, the amount of work to move each
electron from one plate to the other is constantly changing as the
capacitor charges up. It takes much less work to move the first
few electrons from one plate to the other than the last few. The
equation C=Q/V tells exactly what capacitance means —it is the
amount of charge moved from one of the two elements of the capacitor to
the other per volt applied. A capacitor is always composed of two
conducting elements which are separated by some nonconducting medium.

QUESTION:
1) Why can a capacitor only gain a voltage equivolent to the supply
voltage across it - no more?
2) What do you actually change about a capacitor if you increase the
value of it? Why does this mean it takes longer to charge?

ANSWER:
1) The voltage across a capacitor depends on many things
about the circuit in which it is placed. If you have in mind that
you just connect a battery across an uncharged capacitor, then it is
just the definition of potential difference which is the reason why the
two are equal when all current ceases to flow —the work it takes to
move a charge all the way around the circuit (including across the
battery and the capacitor) is zero. However, if you began with a
capacitor which already had a charge such that the voltage across it
was, V _{1} and you connected it across a battery which
was V _{2} , then the voltage across the capacitor would
be either V _{1} +V _{2} or V _{1} -V _{2}
depending on how you hooked it up.
2) To change the capacitance you need to change the geometry or put a
dielectric material in the intervening medium. For example, if it
is a parallel plate capacitor, you can increase the size of the plates
or decrease the distance between them to increase capacitance. In
a simple circuit with resistance of R and a capacitance or C
in series, the time it takes to charge depends on the product RC .
So, if you keep the resistance and battery constant and vary the
capacitance, the time is proportional to C . This should
make sense to you because as C increases, the charge which it
can hold at a particular voltage increases and it would take longer to
acquire that larger charge. Actually it takes infinite time in
principle to charge a capacitor completely, RC is the time it
takes to acquire 1/e =1/2.718 of its total final charge, but in
practice, wait a few times RC and it will have, for all intents
and purposes be totally charged.

QUESTION:
As I've understood it, Quantum entanglement of particles seems to mean
that information can be transmitted instantaneously between an
entangled pair, across any distance... Would you be able to set up an
instant link with the mars rovers - or a distant spacecraft - in this
way?

ANSWER:
No, because you have no control over the information being
"transmitted". You make a measurement on one of two particles
which puts that particle in a particular state and then the other
particle is "instantaneously" required to be in a particular state
depending on how the pair was originally prepared. But you cannot
send any information that way because you cannot predict in which state
you will find the first particle.

QUESTION:
Does work being done always require a force moving an object a
distance? eg - Work = force*distance, but suppose the transfer in
energy did not involve something being made to move a distance due to a
force, such as the conversion of chemical energy to light energry - is
work still being done? If so, what would it be equal to?

ANSWER:
The word "work" in physics generally is intended to mean
mechanical work, that is force*distance as you say. But work need
not be done in order for energy to be transferred. A simple
example is to add heat to a container of gas which has a fixed volume;
then no work is done but the energy of the gas increases, that is the
average kinetic energy of the molecules increases, that is the
temperature increases.

QUESTI ON:
what is a formula on how fast a roller coaster goes.

ANSWER:
Well, there is no formula because it is a complicated
problem. However, if friction is negligible and the roller
coaster starts from rest at a distance h above the bottom, you
can estimate the speed at the bottom using energy conservation, mv ^{2} /2=mgh
where m is the mass and g is the acceleration
due to gravity. Thus, v =[2gh ]^{1/2} .

QUESTION:
What is the volume differance between Diatomic Hydrogen(H2) and
Monoatomic Hydrogen(H1)?

ANSWER:
Assume that each is an ideal gas, so PV=NRT. (I
assume you know the ideal gas law.) If the two gasses have equal
pressure (P ) and temperature (T ) then V _{1} /V _{2} =N _{1} /N _{2}
where N is equal to the number of moles of each gas. If
we have one (1 gram approximately) mole of H^{1} (N _{1} =1)
and 1/2 mole (1 gram approximately) of H^{2} (N _{2} =1/2),
then V _{1} =2V _{2} ; so equal masses will
have the H^{1} occupy twice the volume as the H^{2} .
What you should realize, however, is that H^{1} never exists at
normal pressures and temperatures because it wants so badly to be H^{2} .

QUESTION:
If I could temporarily deviate the gravity lines (shield it against
gravitons) from a body on the surface of the earth and move it to a
higher position than I would obtain 'free' potential energy?

ANSWER:
Yes. However, there is nothing profound here.
Your question is essentially "if I could suddenly turn on a force field
would something which feels that force suddenly acquire potential
energy?" Yes, it would.

QUESTION:
Planck said that the energy of a photon is h / f. What are the smaller
and bigger frequencies a photon may have?

ANSWER:
First of all, energy is Planck's constant times
frequency, not divided by. There is, in principle, no limit on
the possible frequency a photon may have. Of course it could not
have zero or infinite frequency because zero energy would mean it does
not exist and infinite frequency would mean it had infinite energy and
that much is not available in the universe!

QUESTION:
Lets say I have a 60 watt light bulb and a 40 watt light bulb. If
the 40 watt light bulb is the one with the highest resistance, and the
composition of both the filaments are the same. Then what would
be the difference between the two filaments to make the 40 watt have
higher resistance?

ANSWER:
Most likely the length of the filament of the 40 watt bulb is
longer to make its resistance higher. Resistance of a wire is
proportional to its length and inversely proportional to its cross
sectional area. Therefore, you could also increase the resistance
by decreasing the cross sectional area, i.e. use a thinner wire.

QUESTION:
If a gun is fired from a speeding car in the direction of travel, the
bullet takes on the speed of the car in addition to its own muzzle
velocity, correct? If the bullet is fired backwards at a bystander, the
velocity of the bullet is reduced by the speed of the car, correct? So,
if the car could travel as fast as the bullet, then fired backwards,
the bullet would never hit a stationary observer (absent gravity, the
bullet would be stationary?). So, if light is emitted from a star as it
is propelled outwards within the expanding universe, is the light
travelling at 200,000 miles per second if the star is already
travelling at 14,000 miles per second (relative to a stationary
object)? ie. light would be travelling faster than the speed of light.
Conversely, is the speed of light reduced by the speed of the star's
retreat, from the point of view of a stationary observer? If this is
true, how does that affect E=Mc^{2} ?

ANSWER:
The principal postulate of the theory of special relativity
is that the speed of light in vacuum is a universal constant. A
beam of light, regardless of who views it, travels at the speed of c =186,000
mi/s=3 x 10^{8} m/s. So if you are in a car which is
approaching me with half the speed of light and you are shining a
flashlight at me, I will measure the speed of light to be c and
you will also measure the speed of the light to be c and a
third observer moving in the opposite direction as you with speed c /2
will also measure the speed of the light to be c , and so
forth. As paradoxical as this may seem, it is true and has been
verified experimentally numerous times!

QUESTION:
We know that in the case of earthquake waves, P longitudenal waves can
pass through the liquid core but S transverse waves cannot - why is
that?

ANSWER:
For a longitudinal wave to traverse a medium, the medium must
be compressible. This is how sound travels through air or water
or steel. The medium is displaced in a direction parallel to the
direction the wave is moving and does that by compressing closer to
gether. For a transverse wave to traverse a medium, the medium
must have a shear elasticity; I am not sure that
this is the right wording but if the medium is displaced in a direction
perpendicular to the direction the wave moves, it must "spring
back". A solid does this, a liquid or a gas does
not. Representations of the two types of waves are shown at
the right. The S wave could not exist if there were not a
restoring force in a vertical direction in the figure. The P
wave, however, has a natural restoring force because the pressure
increase in the regions of compression push the material back.

QUESTION:
i've noticed that sometimes if a rock falls off the bed of a truck, it
might, at some point, reach a height which is higher than the height of
the previous bounce. What energy is being transferred to
potential to allow this to happen? (ie. rotational, kinetic due to the
speed of the truck, etc.) and how can it be described
mathematically?

ANSWER:
When it leaves the truck, it has a velocity about equal to
the truck's velocity so it has a significant amount of kinetic
energy. Now, if the road were perfectly smooth and elastic, it
would keep bouncing always returning to the same height. However,
suppose that the road had an imperfection in it such that the collision
with the rock sent the rock straight up (or at least more up than its
downward velocity vector before it hit). It is easiest to think
of it going straight up: suppose the stone hit the ground with a
speed of 88 ft/s (60 mi/hr) and bounced straight up; also, to
make it more realistic, suppose it lost 1/2 of its speed in the
collision so it bounced up with a speed of 44 ft/s; I calculate that it
would bounce about 30 feet up which is pretty high. So, it is
easy to believe that a stone will frequently bounce higher than the bed
of the truck.

QUESTION:
Let's say I have a thermonuclear rocket, and I want to take my friends
on an extended tour of the solar system. My plan is to
continually accelerate at 1 G to keep us all healthy and
comfortable. Will I have to keep adding to my thrust to maintain
this if we approach relativistic velocities? For the return trip,
the plan is to cut the motor, turn around, and decelerate our way back
like they did in Tintin: Explorers on the Moon. But lets say
there is a malfuntion of the motor cut-off, so the throttle is
stuck open. Would I still be able to turn around and decelerate
toward home?

ANSWER:
Acceleration is not a useful quantity to talk about in
relativity. However, it is not possible to have a constant
acceleration of g for the following reason. Suppose that you
had accelerated to within 8 m/s of the speed of light; if your
accereration is and remains 9.8 m/s^{2} then you will exceed
the speed of light before the next second passes which is a
no-no. Anyhow, until you get very close to the speed of
light, you may have an acceleration of g but it will cost you
increasingly more energy as you go faster (finally requiring infinite
energy to achieve c ). The details are a little technical,
but if you know a little special relativity and calculus, they are not
hard. I have done a brief explanation which you can link to here . I
really don't see what your last question has to do with
physics —whether you can turn around with the motor running is a
function of how the craft is designed.

QUESTION:
Can you explain to me how to find the mass in grams, and volume in
grams of a rock? I have a project in school to do involving many
steps and the teacher is only giving us a week to do it. Please give me
hints!!!

ANSWER:
If you weigh the rock, it will probably be in pounds, so you
need to know how many grams in a pound: 1 lb = 453.6 gm. There is
a nifty little program called Convert which you can download from http://www.joshmadison.com/software/convert/
which will let you look up conversion factors like this one. Your
second question is nonsensical because you cannot measure volume in
grams. Volume must be measured in either a length cubed (e.g.
cubic centimenters, cm^{3} ) or in a unit defined as a volume,
like a gallon or a liter. The volume of a rock will not be real
easy to measure if it is not some regular shape like a cube or a sphere
whose volume you can calculate if you can measure the length of a side
or the diameter, respectively. What you will probably want to do
is to see what volume of water is displaced by your rock. Then
that is the volume of rock. For example, suppose that you find
the rock displaces 2/3 cups of water and you want the volume in cm^{3} .
Then since 1 c = 236.6 cm^{3} , your rock has a volume of
157.7 cm^{3} .

QUESTION:
Why in inelastic collisions is momentum conserved but kinetic energy
not? Energy is lost as heat, sound, etc, but why is momentum not lost
too?

ANSWER:
In a collision, if there are any forces which do work then
these forces change the total energy of the system. For example,
internal frictional forces will do negative work during the collision
taking kinetic energy away and changing it to heat energy. On the
other hand, linear momentum changes only if there is a net force on a
system. The easiest way to see why is to express Newton's second
law (usually force equals mass times acceleration) as net force equals
the time rate of change of linear momentum. For colliding
particles we generally regard the system as isolated, that is the only
forces which the constituents feel are forces internal to the system
itself. But Newton's third law states that all the internal
forces on a system must add up to zero (the force which particle 1
exerts on particle 2 is equal and opposite to the force which particle
2 exerts on particle 1). Thus the time rate of change of linear
momentum is zero, i.e. linear momentum is conserved.

QUESTION:
My question regards friction and determining reaction forces. If we do
the simple ladder problem, to determine the angle of slip, the typical
assumption is frictionless wall. What if I do have friction between the
ladder and the wall? If I use the static equilibrium equations, (Fx=0,
Fy=0, Mz=0), I get three equations for only two unknowns.
F1x - mu F2y = 0 mu F1x + F2y - mg = 0 mg cos(theta) x s = F1x
cos(theta) x L + F1y sin(theta) x L
where s is the length to CG and L is the length of the ladder, theta is
the angle the ladder makes with the ground, F1 reaction force on the
wall, F2 reaction force on the ground. I could solve the moment
equation for tan(theta) as a function of mu, s, L, and mg but when I
put a ladder up, it will stay up for a variety of thetas. Should this
be an equality relationship? I seem to be having a mental block
grasping it.

ANSWER:
The reason you are getting into difficulty is that for static
friction, the relation f= m N is not true. The
static friction force is whatever it has to be in order for the system
to be in equilibrium but may not exceed m N
(that
is when slipping will occur). The problem you describe is
thus an underdetermined rather than overdetermined one; there are four
unknowns and and three equations. There is no way to determine
them without additional information. Incidentally, the way you
describe the problem is valid if the ladder is just about to start
slipping. Then the angle q
is the third unknown
and you can predict the minimum angle the ladder can make with the
floor without slipping.

QUESTION:
If one had 2 droplets of water (1 with twice the radius of the other or
8x the volume) and one were then to put them on a flat surface, what
happens to the perceived radius from above? Does the larger
become more than double that of the smaller or not? Why?
Where can I read about this? Does it depend on the surface?

ANSWER:
It does depend on the surface. If the surface is
"wettable" then each droplet will spread out making a very thin film
and the thickness of each will be about the same. Therefore, each
flattened drop will have an area proportional to its volume, so the
larger will have an area 8 times that of the smaller, so the ratio of
the radii will be square root of 8 or about 2.83. If the surface
is not wettable, the shape which each drop takes on will depend on the
surface tension of water which keeps it from flattening out
completely. The surface tension depends on the temperature so,
using hot water with smaller surface tension will result in flatter
droplets. The flatter the droplet, the closer the radius ratio
(from above) will be to 2.83. The rounder the droplet, the closer
the ratio will be to 2. To learn more about this, you obviously
need to research surface tension.

QUESTION:
When perfect materials are specified in regards to a collisions, why is
there a "loss" of energy during inelastic collisions and not elastic
ones?

ANSWER:
Elastic and inelastic are just words. In usual
classical mechanics, an elastic collision is defined to be one in which
the kinetic energy of the colliding particles is the same after the
collision as it was before. But, the total energy of any isolated
system never changes, this is the principle of the conservation of
energy. So an inelastic collision, in which kinetic energy is
either lost or gained, must have kinetic energy changing into some
other kind of energy or vice versa . Examples are:

A bomb
at rest has no kinetic energy. Now the bomb explodes and pieces
of the bomb fly in all directions. Where did the kinetic energy
come from? It came from the chemical energy which was stored in
the explosive until it was released by a chemical reaction.
Two
putty balls collide and stick together (a so-called perfectly inelastic
collision). It is easy to show (using momentum conservation) that
kinetic energy is lost. Where did it go? It takes work to
squash a putty ball; this work is like friction and the kinetic energy
lost ends up as heat, i.e. if you carefully measured the putty
balls after the collision, they would be a tiny amount warmer.
For
macroscopic particles, one never sees a perfectly elastic collision
because there will always be friction-like forces to take energy
away. However, for microscopic particles elastic collisions are
the dominant process. For example, if you shoot protons at
protons, since protons are very elementary particles (with no internal
degrees of freedom which would let them "heat up" as it were) they
bounce off each other elastically. (If you give them a huge
amount of energy, it is possible to cause the proton to "heat up".)

QUESTION:
Ok. Well my question is, if the earth was suddenly to lose it's
atmosphere, where would the water go? Would it evaporate into space?
And then what? Will it seek a place with less density? What can
be less filled with atoms than space?

ANSWER:
No, it would not evaporate into space. It would
evaporate and form an atmosphere of water vapor. This process
would continue until the pressure of the atmosphere was just right so
that the water and the water vapor were in equilibrium at the
prevailing temperature. Atmospheres evaporate into space if the
atoms in the atmosphere have velocities larger than the escape velocity
from the surface of the earth. (The escape velocity is how fast
something must move in order to completely leave the earth, the
velocity you would have to give a rocket to send it to Mars, for
example.) At the temperatures present in the earth's atmosphere,
heavier molecules (like O_{2} , N_{2} , CO_{2} , etc .)
do not have velocities large enough to escape but lighter molecules and
atoms, most importantly hydrogen and helium, do and that is why there
is no hydrogen or helium in the earth's atmosphere; the only way to get
helium is to get it from inside the earth where it is trapped, for
example in natural gas wells. If the earth were to get a lot
hotter, the atmosphere would have larger velocities and would escape
into space.

FOLLOWUP
QUESTION:
But if there isn't an atmosphere, then does the temperature
get hotter or how then will the planet's temperature be? And what
I meant in the other question is, doesn't molecules travel from crowded
places to more open places?

ANSWER:
What determines temperature of a gas is the average kinetic
energy of the molecules (essentially the hotter the gas, the faster the
molecules are moving). What determines what the temperature will
be is a large variety of things like what energy is coming in (e.g. from
the sun), what energy is going out, etc. But any given
system will tend toward some equilibrium where the energy coming in
will equal the energy going out and some equilibrium temperature will
be reached. Regarding your other question, yes, there is a
tendency for the gas to spread out as far as it can moving to fill the
surrounding vacuum. But in the case of an atmosphere, this is not
the only thing to take into consideration. The atmosphere is
being pulled on by the earth's gravity which has the opposite effect of
the tendency of the atmosphere to spread out in space. If the
earth wins this tug of battle, it has an atmosphere; if not, it does
not (like the moon has no atmosphere).

QUESTION:
Let's say there is a stationary observer A and a stationary flashlight
B some distance away. Beyond B there is another flashlight C moving
toward B at a constant velocity. At the point when flashlights B and C
are equadistant to A, they are switched on. Which flashlight will A see
first?
A
B < — — —- C

ANSWER:
I presume that you mean that B and C are at the same location
when they turn on. Then A will see both flashlights
simultaneously. The speed of light is independent of either the
motion of the source or the motion of the observer; this is a basic
postulate of the theory of special relativity. The light from C,
however, will be blue shifted because of the motion of the source.

QUESTION:
Why is the formula for a moment "Moment = force * perpendicular
distance"? WHy does the perp.dist. have an effect?

ANSWER:
The easiest way to think about this is to define Moment=distance
* perpendicular component of force
where distance is the distance from the axis about which moment is to
be calculated to where the force is actually applied. This makes
sense because if the force has no component perpendicular to the
distance it will not tend to rotate the object as moments do.
(For example, think of trying to close a door by pushing on the edge
straight in toward the hinges.) Now, if you know a little
trigonometry, drawing a few sketches should convince you that the exact
same moment is achieved if you define moment as Moment=force
* perpendicular component of distance .
In the figure I have shown the vectors d and
F and their components
perpendicular (red) and parallel (blue) to the other. In one
case, moment is d (F sin q ) and in the other it
is F (d sin q ).

QUESTION:
Why does increasing the number of turns a coil has increase its
magnetic field? Also, why does increasing the current and adding an
iron core do the same as this?

ANSWER:
One coil of wire carring a current I will cause a
field B, so two will cause a field 2B , three will cause
a field 3B , etc.; and similarly, the same coil carrying
a current 2I will cause a field 2B , so two will cause a
field 4B , three will cause a field 6B , etc. The
question about iron has been previously answered .

QUESTION:
In an ac generator (ie - an alternator) you apparently get the ac wave
pattern because when the coil is in the horizontal position it is
cutting the magnetic field at the fastset rate, etc. Why is it cutting
it at a faster rate? Surely it is being turned at a constant speed,
meaning it shoud be a constant rate?

ANSWER:
The induced EMF in a loop rotating in a magnetic field is
proportional to the rate of change of magnetic flux through the
loop. The flux is defined as the area of the loop times the
component of the field perpendicular to the loop. If the
field is B and the area is A , then the flux is BA if
the field is perpendicular to the loop, 0 if the field is parallel to
the loop, and BA cos q
if
the normal to the loop makes an angle q with the field. So as the loop rotates with
constant angular velocity the flux through it will look like a cosine
function.

QUESTION:
In radioactivity, what causes the difference in the absorption of
alpha, beta and gamma? (ie - alpha is stopped by paper, beta by
aluminium, etc) Is it the size of the particle/ wave, its ionizing
power, or what?

ANSWER:
Radiation is stopped by losing energy and energy is lost by
interacting with matter. If we have one of each type of particle,
an a, b, g of the same
energy. The a
has a charge of +2e
(it is the nucleus of a helium atom) and a mass about 8000 times bigger
than the b which has a charge of +e
or -e (it is either an electron or a positron); the g has no mass and no
charge (it is a photon). They lose energy by interacting with
atoms and the slow, heavy, doubly charged a
loses it fastest while the massless, chargeless g loses it most
slowly. Learn more at HyperPhysics .

QUESTION:
How does a capacitor actually "store charge" and make a current flow?
Also, why does it take longer for the capacitor to charge and discharge
if the resistance is high or the capacitors value is greater?

ANSWER:
Basically, a capacitor is just two pieces of metal (I will
call them 'plates' below, but they need not be plates), not touching
each other. If you cause there to be an electric potential
difference between the plates, for example by connecting them together
with a battery and wires, then the electrons will flow off the plate at
the lower potential and go to the plate with the higher potential
(positive charges go from higher to lower potential). So, that is
how to "store charge"; actually you see that it is not storing some net
charge but has charge moved from one to the other. If you now
remove the battery and wires, a charge of +Q will be on one
plate and a charge of -Q on the other; a potential difference
will exist between the two plates, the one with +Q being at the
higher potential. So now, if you connect them back together, the
electrons will again move to higher potential, that is they will go
back to where they came from. If there is a resistance in the
wire, then the current will be inversely proportional to the resistance
and proportional to the potential difference. So a big resistance
means a small current which means a smaller rate of transfer of charge
(which is what current is) which means a longer time. Learn more
about electricity and magnetism and about capacitors at HyperPhysics .

QUESTION:
If you charge a piece of polythene through rubbing it, can you
discharge it by connecting it to earth?

ANSWER:
This question has been previously
answered .

QUESTION:
Does the physical hand positioning on a hockey stick affect the speed
of the hockey puck?

ANSWER:
The speed of the puck will be determined by the speed of the
stick where it touches the puck. This speed will depend on how
much acceleration you can give to that end of the stick. There
are two ways you can accelerate the end of the stick: either accelerate
the whole stick forward by pushing on it or give the stick a rotational
acceleration by rotating it. I believe that second way is much
more important and that it plays the dominant role in getting the puck
going very fast. Now, in order to impart a large angular
acceleration to the stick (get it rotating fast) you have to exert a
large torque on it. Imagine that your left hand holds still and
that your right hand exerts the force which causes the torque which
causes the stick to rotate; since torque is force times distance, and
since the maximum force you can exert is limited by your strength, you
can exert a bigger torque by having your hands as far apart as
possible. Of course, it isn't really quite that simple since if
your hands are at each end of the stick (as far apart as possible),
they are awkwardly placed and you would not be able to exert as much
force as if they were somewhat closer together. As I recall, a
powerful slap shot has the player's hands about half the stick length
apart.

QUESTION:
what does it mean to "trickle charge" a battery?

ANSWER:
Many types of rechargeable batteries cannot handle being
charged up too fast; trickle charge is charging them up at the right
(slower) rate.

QUESTION:
Logic Gates - what exactly are they? Ive been told the truth tables and
circuit symbols for AND, OR and NOT by my teacher but he didnt actually
say what they are, how you make one, what they look like in real life,
etc? Why for logic gates is it called a "Truth Table?

ANSWER:
This is really way too extensive a question to answer on this
site. You need to do research. I would suggest you start at
Hyperphysics .

QUESTION:
Is there a type of light which makes you see in black and white? If so,
how?

ANSWER:
No, not really. However, under very low intensity
situations, like at night, we cannot see colors very well because the
color sensors in our eyes are not sufficiently sensitive so we see
mainly in b&w in very low light.

QUESTION:
In terms of electronics - why would you ever need to turn an ac supply
into dc? (ie - why do you sometimes need to rectify it?) Also, when you
half wave rectify it using a diode and place a lamp in the circuit does
it go on then off, on then off, etc? is that the same if you full wave
rectify it using 4 diodes?

ANSWER:
Well, almost all electronics, your computer, tv, radio, etc .
use mainly dc. There is something in every one of these things
called a power supply whose job it is to supply dc to the
electronics. You should know that most electronics operate on dc
because many of them are supplied by batteries which are dc
devices. The rectifiers you refer to are seldom the last word in
making a dc source but are only the first step in conversion to
dc. The power from a dc power source also comes in ups and downs
and the light does not flicker because the change is too rapid and the
filament does not have time to cool down.

QUESTION:
What is the "intensity" of a wave? Is it the same for light waves as
other waves?

ANSWER:
Intensity is normally defined as energy per second per square
meter, the rate of energy flow per unit area. Often, for periodic
waves, this is defined as the average over one period of the energy
flow. For most waves the intensity is proporional to the square
of the amplitude of the wave.

QUESTION:
Firefighting test questions...PLEASE help

Question
1
Which wagon is easier to pull a heavy load

1. a
wagon with 2 big wheels in the front and 2 small wheels in the back

2. a
wagon with 2 small wheels in the front and 2 big wheels in the back

3. a
wagon with 4 small wheels

4. a
wagon with 4 big wheels

Question
2

Which
wagon is the easiest to pull considering the following angles of the
arm that is attached to the wagon

1, a
wagon with the arm parrallel to the wagon

2. a
wagon with the arm at 20 degrees

3 a
wagon with the arm at 45 degrees

4. a
wagon with the arm at 60 degrees

5 a
wagon with the arm at 90 degrees

ANSWER:
I think these questions might require more than physics, some
practical common sense.

Problem 2
is pretty straightforward: pull parallel to the ground and all of your
force will be applied to the wagon in the direction it has to go. Pull
at other angles and part of (the vertical component of) the force will
be wasted.

The first
problem is a little trickier. If you are on a hard, flat surface, it
should not make any difference at all. However, if the surface is
somewhat soft, the smaller wheels would sink in more deeply because the
force would spread out over a smaller area (like a needle goes into
your finger a lot more easily than, say, an unsharpened pencil). Also,
if there were bumps on the surface like rocks or ruts or the like, big
wheels would go over them more easily. I can think of no advantage to
having different wheel sizes front and back.

QUESTION:
What is the effect of increasing the amplitude of a light wave, since
this usually shows the amount of energy of the wave but for light this
is instead affected by the frequency?

ANSWER:
This question has been previously
answered .

QUESTION:
In reading Stephen Hawking's "The Universe in a Nutshell," I came
across a section about the spin of particles. It states that a particle
with a spin of 1 will look the same only after a full rotation of 360
degrees. A particle of spin 2 will look the same after a rotation of
180 degrees. The example given for this was the face of a playing card,
as it looks the same after turning it upside down. The section
continues to say that some particles have a spin of 1/2. This means
that the particles must complete 2 rotations of 360 degrees before it
looks the same. This is where I am lost. How could it be that an object
could rotate 360 degrees and look different but rotate another 360
degrees and look the same.

ANSWER:
The easiest way I can think of to see this is to think of a
mathematical function which has the properties you seek. Consider
the function sin(x /2); when you get to x =360^{0}
you have only gone through one half the cycle for the function, you
must go to x =720^{0} before the function came back to
where it started from. Incidentally, spin 1/2 particles are the
most important ones, and include protons, neutrons, and electrons from
which matter is composed. I cannot think of a good geometrical
example from everyday life but I can invent one: imagine that you had a
playing card which, when you turned it, changed suit from hearts to
spades and then back to hearts and it took 360^{0} for each
transformation to occur.

QUESTION:
I was wondering, is there a linear relationship between the steepness
of a slope and the rate at which an object falls down that slope?
For example, suppose I were timing the descent of a marble down a slope
with an angle of 10 degrees. If I doubled the number of degrees
of slope, how would the time change? I know it will take less
time, but is it an exact amount of time?

ANSWER:
If we denote the angle of the incline as A , then the
acceleration (a ) of an object sliding or rolling down the
incline (and which loses no energy to friction) is proportional to sin(A ).
Since the distance traveled if the object starts from rest is given by x =at ^{2} /2,
it follows that if the distance x traveled by the object is the
same, the time to go x must be proportional to the square root
of 1/sin(A ). For the example you give, sin(10^{0} )=0.174
and sin(20^{0} )=0.342, so t (20^{0} )/t (10^{0} )=[sin(10^{0} )/sin(20^{0} )]^{1/2} =0.713

QUESTION:
Can a sound wave in air bend light? Since sound waves are
compressions of air, and compressed air has a different index of
refraction than less dense air, I think the answer is yes. But is
it something noticeable without sensitive equipment or earth shattering
loudness?

ANSWER:
You are right, the answer is yes. The most important
examples of sound refraction result from sound speed differences in air
due to temperature changes. A nice example is given at HyperPhysics

FOLLOWUP
QUESTION:
thanks for the quick reply, but I was wondering if the sound wave can
*refract light* passing through it, not sound itself. My
assumption is that the sound will change the index of refraction of air
momentarily, and the light will bend just as it would in the case of a
mirage on a hot summer day?

ANSWER:
Again, the answer is yes but now I must say that I believe it
would be an extraordinarily small and difficult-to-observe
effect. Here is why:

If the
density variation in air is due to temperature differences, let's
estimate the effect on the density: treat the air like an ideal gas (PV=NRT ),
imagine the temperature (T ) to increase by 30^{0} , the
pressure (P ) to remain constant, the amount of gas (N ) to
remain constant, and the volume (V ) is allowed to
increase. (R is just a constant.) So if the
temperature were to increase from 300^{0} to 330^{0} ,
the volume would increase by 10% so there would be a 10% decrease in
density. As you note, this is how mirages happen and mirages are
easily observable.
If the
density variation in air is due to sound waves, let's estimate the
effect on the density: If the intensity of sound is at the level of the
"threshold of pain" (120 dB), the maximum change in pressure is about
30 N/m^{2} . But atmospheric pressure is about 10^{5}
N/m^{2} . Hence the pressure variation is only about
0.03%; hence the gas will compress by about 0.03% so the density
will increase by 0.03%.
So the
bottom line is: sound of normal intensities has very little effect on
the air.

QUESTION:
In terms of static electricity 1) Why do tv screens become very dusty?
2) Why are floors tiles in operating theaters made from conductiong
meterials?

ANSWER:
There is a lengthy explanation of dust on a tv screen at MadSci
Network .

The floor
tiles conduct electricity because if someone were to become
electrostatically charged, a spark could ignite flammable gasses which
might be present because of the anesthetics used. The floor provides a
path to ground to keep the personnel from acquiring an electrostatic
charge.

QUESTION:
Why is this formula true for a transformer: Vs/Vp = Ns/Np ?

ANSWER:
This requires too much space to explain. Look in any
elementary physics text for details. It is basically because of
Lenz's law and Faraday's law. There is a nice discussion at HyperPhysics
.

QUESTION:
If you have a solenoid coil (ie - which has a current passing through
it and thus a magnetic field around it similar to that of a bar
magnet), why if you place some iron in the middle does that increase
the magnetic field strngth?

ANSWER:
Because the iron itself becomes magnetized when placed in a
magnetic field. See the answer to another
recent question .

QUESTION:
Why when a wire is moved in a magnetic field does this induce a
current/ voltage? Also, why does a current in a magnetic field cause
motion?

ANSWER:
A moving charge in a magnetic field experiences a force
(unless that charge moves along the field lines). Atoms are made
up of positive and negative charges. In a nonconducting material
all the charges are tied down to their location in the material, but in
a conducting material, there are electrons free to move so they do and
those moving electrons are the current. Conversely, if you put a
current (moving charges) in a magnetic field, they will experience a
force which will cause there to be a force on the current carrying wire
which will then move if free to do so.

QUESTION:
Is there a four dimensional space-time coordinate system? If yes,
please explain this.Does not time remain a constant state of reference?

ANSWER:
The most elegant way to do the theory of relativity is to
introduce a fourth dimension which is, essentially, time. I say
"essentially" because the fourth dimension should be measured in the
same units as the other three, that is the fourth dimension should be
in units of length. The way this is done is to define the fourth
dimension to be x _{4} =ct where c is the
speed of light. It is not absolutely necessary to do this, it
just is a very sensible way to do it. Regarding your second
question, the one of the main results of the theory of special
relativity is that time is relative, there is no absolute time in the
universe and clocks of different observers (those in motion with
respect to the others) run at different rates.

QUESTION:
How can you measure the diameter of the sun using a pin hole?

ANSWER:
The ratio of the diameter of the sun over the distance to the
sun is equal to the ratio of the diameter of the pinhole image of the
sun over the distance to the image.

QUESTION:
I am in the 6th grade in Denver, Colorado. I made a machine out
of 2 by 4s that will kick a ball. I want to see if air pressure
in a ball has an effect on the distance the ball will travel. I
did some experimenting and found that the lower the amount of air
pressure in the ball, the farther the ball went when kicked with my
machine. I thought that the opposite would happen. Why did
the ball with the lower psi go farther??

ANSWER:
I would have expected it to go farther too. This is
kinda tough since I do not know the details of your machine. What
I would do first is determine the coefficient
of restitution of the ball at different pressures; do this by
dropping the ball from some height and measuring the height of
rebound. This tells you the fraction of energy lost when the ball
collides with something which is the most important thing.
Another thing I would do is carefully weigh the ball at different
pressures and see if there is a significant change in mass which would
certainly be an important factor; of course, the mass increases when
you increase pressure (add more air) but I just haven't the time to
compute it. Also, I would use a soccer ball, not a football
because of the odd shape of the football which could mess up your
experiment. Also, worry about your machine —does it impart impact
to the ball because of its own inertia or is it designed to always
deliver a constant force (a spring loaded device, for example). I
have not answered your question, but I have given you some things to
think about.

QUESTION:
If time dilation is measurable to an observer by its effects on the
actual physical properties of a very fast spaceship (eg people would
actually age more slowly), does increase in relativistic mass also
change the actual physical properties of an object? I mean, if a bullet
were travelling down a gun barrel at .99c, would the bullet swell up
and get stuck in the barrel?

ANSWER:
In a sophisticated formulation of special relativity, the
only mass that one talks about is the rest mass of an object.
Dynamics is done in terms of the four-momentum which, in any particular
frame, consists of the three components of the linear momentum plus the
energy of the particle. The catch is, however, that the linear
momentum is not p =m_{0} v like in
Newtonian physics, but p =m_{0} v /[1-(v ^{2} /c ^{2} )]^{1/2} .
One possible way to interpret this is to say that the mass has
increased to m =m _{0} /[1-(v ^{2} /c ^{2} )]^{1/2}
so that we can write p =mv but that is only a way
to think about it and the important quantity is momentum, not
mass. Regarding your second question, increasing mass does not
imply that all the spatial dimensions increase somehow. The
bullet would shrink along the direction of motion, the usual length
contraction, only; this is regardless of what happened to the
mass. Essentially, the mass would increase by the factor 1/[1-(v ^{2} /c ^{2} )]^{1/2}
and the length (and therefore the volume) would decrease by the factor
[1-(v ^{2} /c ^{2} )]^{1/2} so the
density would increase by a factor 1/[1-(v ^{2} /c ^{2} )].

QUESTION:
Why the color of sky is blue?

ANSWER:
The reason is that (white) light from the sun strikes the
molecules in the earth's atmosphere and is scattered via a process
known as Rayleigh scattering. Rayleigh scattering is most likely
to occur for light of short wavelengths and blue is at the short
wavelength end of the spectrum of light we can see. You can read
a little more detail in the answer to an earlier
question I answered. You can read a much more detailed
discussion at http://www.sciencemadesimple.net/sky_blue.html

QUESTION:
When people talk about magnetism they often talk about domains - what
exactly are the domains? ie - are they refering to particles, etc?

ANSWER:
First, read the answer to the question just after
yours. Electrons in iron tend to line up their magnetic moments
with their nearest neighbors. However, when solid iron is
forming, cooling down after being very hot, electrons in different
locations tend to line up in different directions. So there will
be many little permanent magnets in a piece of iron all pointed in
random directions so that the iron, as a whole, does not look like a
magnet; these little locally aligned regions are called domains.
When you take this piece of unmagnetized iron and place it in a
magnetic field caused by something external, the electrons have an
additional bias to also want to line up with that field; therefore,
domains aligned with the field tend to grow at the expense of other
domains and, when you take the external magnetic field away, the iron
tends to stay with a net magnetization. That is why you can take
an iron nail, for example, and make it into a little magnet using
another magnet.

QUESTION:
Magnetism occurs when an electric current flows. ( ie - there is an
electrostatic field around a still charged particle and a magnetic
field when it moves.) So, when is iron often magnetic?

ANSWER:
First, a moving electric charge also has an electric
field. You are usually not aware of this in a current carrying
wire because the net charge of the wire is zero even though the
electrons are moving. To answer your question, you must first
understand that an electron itself, at rest, has a magnetic field; this
is essentially because it looks like a little spinning ball of negative
charge and therefore looks like a tiny bar magnet (called a magnetic
moment). In most materials, the jillians of electrons line up
their magnetic moments in random directions so the magnetic properties
of the material are not easy to see. In a very few materials,
some miracle of nature causes all of the outer electrons in atoms to
align their magnetic moments with those of their neighbors so that the
whole macroscopic thing becomes a magnet. Such materials are
called ferromagnetic and iron is the most dramatic and best
known.

QUESTION:
If you place a book on a table, what actually is (ie- causes) the
reaction force of the table?

ANSWER:
You are still saying this wrong. I cannot tell for sure
what you want to know. Is it the force which the table exerts up
on the book or the force which the book exerts down on the table?
If it is the former, then the origin of the force is essentially
electromagnetic: when the atoms in the table get very close to the
atoms in the book, then the outer electrons in the atoms on the surface
of the book feel a repulsive force due to the presence of the outer
electrons in the atoms on the surface of the table. If it is the
latter, just replace the word "book" by the word "table" and vice
versa in the preceding sentence.

QUESTION:
Will an identical sound travel FURTHER in air or water?

ANSWER:
Two things determine the distance sound will travel.
One is that the sound spreads as it moves away from the source (unless
the source has been designed to send sound in only one direction);
hence it gets quieter and quieter as the energy available is spread out
over a larger and larger area, until the intensity gets below our
threshold of hearing. The other is that there is absorption in
the medium through which the sound is moving; here the energy is
actually removed from the sound by "frictional" effects in the medium
which results in heating of the medium. For normal sound, this
heating is imperceptibly small but it is possible for the heating to be
significant —ultrasound may be used to destroy tumors by heating.
It turns out that the energy loss from absorption in air (1.2 dB/100 m)
is much larger than in water (0.008 db/100 m). Therefore sound
will travel much farther in water. It is interesting that this
absorption also depends on the frequency of the source with low
frequency sound being much less absorbed than high frequency
sound. For this reason whales transmit low frequency sound and
are thought to communicate over tens if not hundreds of miles.

QUESTION:
Newton's 3rd law of motion says that there is alway "action/re-action",
but that these do not cancel out because the 2 forces are acting on
different objects. So, how come we say that eg a book on a table does
not move or accelerate up or down because the weight is the same and
cancels out the reaction force?

ANSWER:
Your first sentence is essentially the answer to the question
you are trying to ask in the second! And your second sentence is
all jumbled. Here is the ticket:

focus
your attention on the book and forget all about Newton's third law ;
there
are two things which can exert a force on the book, the earth and the
table;
the
earth exerts a force straight down called the weight (which we know)
and the table exerts a force, call it N ;
this
book is at rest, so Newton's first law stipulates that the sum
of all the forces on it must add to zero;
therefore
N must point straight up and have a magnitude
equal to the weight.
The
solution of this problem has absolutely nothing to do with Newton's
third law . Do you want to talk about Newton's third
law? If the table exerts a force N on the book,
then the book must exert a force -N on the table.
If the earth exerts a force W on the book, then the book
must exert a force -W on the earth. You might want
to read an earlier answer to a similar
question.

QUESTION:
How do we get the formula F= m(v*v)/r for centripetal force? I can
understand that it must come from F=ma, but why is a = (v*v)/r ?

ANSWER:
The centripetal acceleration of an object with speed v
moving in a circular path of radius r is indeed v ^{2} /r.
I am not going to derive it here because it is a little
involved and may be found in any introductory physics textbook.
There is a concise derivation here .

QUESTION:
How can we measure the half life of radioactive meterial when the decay
is completely random?

ANSWER:
The decay of any individual nucleus is indeed a random event,
but it will occur with a probablilty. For example, one nucleus
may have a 50% probability of decaying in the next hour whereas another
may have a 50% probability of decaying in the next century. You
learn this by watching a bunch of them and counting the number of
decays you see in some time. Now if you have a very large number
of radioactive nuclei (a typical macroscopic sample, say the head of a
pin, would have something like 10^{23} atoms in it), you can
compare the rate of decays you see with the number of nuclei and
thereby deduce the probability that any nucleus will decay in some
particular time. The half life is the time it takes N
nuclei to decay away to N /2 and, if you think about it, you can
convince yourself that this is essentially a probability.

QUESTION:
How come when we talk about the differing wavelengths of EM waves we
say that "the higher the energy of the wave, the smaller the
wavelength/higher the frequency", yet we also say with Sound waves that
the "more energy the Sound wave has, the higher the Amplitude"? Which
one is affected by more energy - Wavelength/frequency or Amplitude?

ANSWER:
For any wave, the intensity (which is essentially energy
delivered per second averaged over a complete cycle of the wave) is
determined by the amplitude of the wave —the bigger the amplitude, the
greater the intensity. Electromagnetic radiation, as you may or
may not have learned, is also a quantized wave. This means that
the smallest intensity possible is not zero but some tiny bundle of
energy called a photon. The energy of a photon is proportional to
its frequency, E=hf , where h is a (very tiny) constant
of nature called Planck's constant. Thus, the energy per photon
in an electromagnetic wave is what is being talked about when it is
said that "the higher the energy of the wave, the smaller the
wavelength/higher the frequency". (Incidentally, wavelength
l and frequency f
are related by l f=c where c is
the speed of light.)

QUESTION:
I wanted to ask about the famous Galileo experiment to do with gravity
and the leaning tower of piza. It has come to my notice that in this
"legendary" experiment, both the lighter and more massive ball fell at
exactly the same acceleration and hit the ground at exactly the same
time. Now, what I find curious is surely would this not be impossible?
Air resistance would occur between the ground and top of the tower and
mean that the acceleration of the lower mass would be reduce quicker,
since the air resistance would be greater in comparison to its weight
than the larger mass. Therefore, they could not have hit the ground at
the same time. If so, then what was the actual outcome of the
experiment? Or am I wrong?

ANSWER:
We cannot report the results actual experiment because the
story may be apocryphal —maybe it never happened. When describing
this experiment, the proviso "if air friction is neglected" should
always be stated. You are absolutely correct, the experiment is
very likely to fail unless the air friction is negligible (or
comparable); if you drop a marble and a feather, there is certainly no
argument about who would win the race. You do err, however, in
assuming that the lower mass will always lose the race since the air
friction depends also on the geometry. For example, a man with a
parachute races a marble to the ground —the marble (less mass) will
certainly win. The experiment reputedly was between cannonballs
of differing weights, and for these the air friction would be fairly
negligible for drops of this height.

QUESTION:
I'm experimenting with a small solar furnace (made with mirrors) for a
science fair project. I want to do the experiment inside so I can
control the environment. What type of light bulb would best
simulate the solar radiation that would reach the earth's
surface? I found one site that mentioned a 100 watt bulb at a
distance of 10cm but that won't work very well with my design. My
dad has a 200 watt spotlight, and I'm wondering if I could use it from
a farther distance.

ANSWER:
A rough number for the flux of solar energy hitting the earth
is 200 watts/m^{2} . Of course, this depends on your
latitude, time of day, time of year, cloudiness, etc. but,
let's just take 200 to get the idea. Suppose you have an
energy source which is radiating W watts (joules/s) equally in
all directions (like a light bulb). Then, if you go a distance r
away from it the flux will be W /(4 p r ^{2} ).
The catch is that an incandescent light bulb of wattage W is
not radiating W watts of light energy because much of the
energy is being wasted as heat. You can probably find the exact
numbers, but I would guess that a 100 W light bulb only radiates
something like 30 W of light. (Incidentally, that is why it is
energy efficient to use fluorescent instead of incandescent
lights.) So, to get 200 W/m^{2} what distance from a 100
W bulb would you have to go? Plugging in, 200=30/(4p r ^{2} );
solving, r =0.109 m which agrees (surprisingly!) well with what
you told me in your question. So, with the 200 W spotlight, let's
estimate that it radiates 60 W of light. But, unlike the light
bulb, the spotlight does not radiate in all directions equally but puts
its light in a forward direction (that is what the reflector
does). You need to estimate the flux at a distance r from
the light. Here is what I would do:

place
the light a distance s from the wall;
there
should be a rough circle of light on the wall of radius R, so
the area is A= p R ^{2} ;
assuming
that the whole 60 W falls on A , the flux at distance s
is F =60/A W/m^{2} ;
now
you should repeat the whole thing several times with different values
of s ;
now
graph F vs. s . From your graph you should be able
to approximate the distance s at which F is
approximately 200 watts/m^{2} .
This is
what is called an empirical approach.

QUESTION:
In a college level classical physics text, there's a section on
"relativistic mechanics" that I had a general question about as
follows. If an electron is moving along a path with v = 0.67c (c=speed
of light) and is acted on by a forceand the electron is given is given
an acceleration
(a) = 1 X 10^10 cm per sec^2 , what is the force IF: a) force is @
right angles to path (ie it would then be moving in circle)
(b) force is in same direction ( ie is parallel)

Since the
F = m(dv/dt) + v(dm/dt)......in general, the part a) is easy because
v(dm/dt) = 0 .........moving in circular path & equation reduces to
F = m(dv/dt) = ma & since the electron's rest mass ( 9.1 X 10^-28)
is known the force can be calculated using the relativistic mass. The
force is then directed toward the center of the circle & is the
centripetal force.

My
question is part b) & how I should think about the v(dm/dt) term?
The force & acceleration are in the same direction ( i.e. parallel
). Appartently, I've not recognized something that would allow me deal
with the (dm/dt) term. Can you give me some coaching on how to think
about this? The answer is not important, only the thought process.

ANSWER:
You are right, you have to worry about the changing mass, not
just the changing speed. In order to calculate dm /dt you
have to know how m depends on v and, thus how m
depends on t (since you know how v depends on t because
you know a ). What you need is m=m _{0} (1- b ^{2} )^{-1/2 } where
m _{0} is the rest mass and b =v /c .
Hence dm /dt =m _{0} b (1- b ^{2} )^{-3/2} (dv /dt )/c .
Perhaps part (a) is not as easy as you think: the answer is not 9.1 x 10^{-18}
gm cm/s^{2} .

QUESTION:
Assuming we compare samples of the four phases (excluding the plasma
phase) of any one given chemical element - i.e. Bose-Einstein
condensate, solid, liquid, vapour - will the size and energy of the
electron shells be identical in each phase or radically different?

ANSWER:
Not absolutely identical because the atoms interact with each
other, but not radically different either. The gas phase will
have atoms which are essentially isolated and may be considered on
their own. But, if the average spacing is on the order of the
size of the isolated atoms, the orbitals of one atom will be influenced
by the presence of neighboring atoms. Normally this is not more
than a few percent effect on what an atom might "look like".

QUESTION:
I was ask "How many cubic inches of water I would
displace?" I weigh 291 lbs and am 5' 9".
I would think it mattered what size of body of water you where in, if
it was salt water, how much you floated etc. or am I on the wrong
track?

ANSWER:
It does not depend on the size of the body of water unless
the depth is such that you rest on the bottom. If you are
floating, Archimedes'
principle says that there will be an upward force on you which is
equal to the weight of the displaced fluid (fresh or salt). Since
you are in equilibrium when just floating, this force (called the
buoyant force) must equal your weight; therefore, you will displace
exactly 291 lbs of whatever you are floating in. If you now look
up the density of whatever you are floating in, the volume is weight
divided by density (because density is weight divided by volume).
For example, the density of fresh water is about 0.036 lb/in^{3} ,
so the volume of 291 lb is about 8,083 in^{3} ; since salt water
has a different density, a different volume of it would weigh 291
lb. As you can see, your height is irrelevant.

QUESTION:
I working for an attorney on the cause of a car accident and have a
question on light. I know that our vision depends on the amount
of light returned to our eyes, so my question is if you double the
light on the object, two sets of headlights instead of one, does the
object reflect double the light back because double the light is
falling on it? Or would it be more inverse square law? And
as a side note, why don't a set of headlights on a car interfere with
each other?

ANSWER:
If you illuminate something with twice the light, it will
reflect twice the light, i.e. it will be itself twice as
bright. Inverse square law has to do with the distance from a
point source of light; for example, if the object in question is small
compared with the distance you are from it (so that you may approximate
it as a point), then if you get twice as far away it will be four times
less bright. But you must be careful about the inverse square law
because the object must be able to be approximated as a point source
radiating in all directions uniformly; the headlight of a car is
designed to defeat the inverse square law by sending all the light from
the source in a single direction (see figure).

QUESTION:
You've said in your previous answers that with a conductor the surface
its charged whilst the insulators charge is throughout - shouldn't it
be the other way round since the electrons in a insultator cannot move
so must stay on the surface? Also, why does all charge go to earth? As
well as this, would you be able to discharge a +ve insulator by
connecting it to earth, since the electrons from the earth can move?
Finally, how can you discharge an insulator? - and if tyhe charges
can't move how does the insulator discharge over time in air?

ANSWER:
The first thing we need to do is establish which electrons we
are talking about. There are the electrons which "belong" to the
conductor or insulator and there are those which have been added (or
subtracted if there is a net positive charge). For simplicity, I
will only talk about electrons since they are really the ones moving
around even if the net charge is positive. In a conductor there
are some electrons (on the order of one per atom), even if there is no
net charge on it, which are essentially free to move around; it is
these which carry current when a voltage is applied across the ends of
a conductor. An isolated, uncharged conductor will have all these
conduction electrons uniformly distributed to keep the total electric
field inside the conductor zero. (If the field inside were not
zero, the conduction electrons would move in response to that
field.) But, the conductor can also have a net charge, for
example we could to add a bunch of electrons to it. Both these
and the conduction electrons are free to move; now, the extra electrons
must arrange themselves in such a way that there is no electric field
inside the conductor because if there were a field inside the
conductor, any electron inside the volume would experience a force and,
being free to move, would move! It turns out that the extra
electrons must end up on the surface if this is to be true. An
easy way to see this is to imagine two electrons added to a conducting
sphere; since they are going to get as far as they can from each other,
they will move out to the surface. (This simple example is too
simplistic, actually, because the presence of the two electrons will
cause other conducting electrons on the surface to rearrange themselves
to make the field inside zero, but it gives you the flavor.) In
an insulator, there is nothing equivalent to a conduction
electron —each atom holds on tightly to all its
electrons. And, it turns out, any electrons which you add to the
insulator also are quite immobile. Hence, they can be put
anywhere because, even if they feel the electric fields of other
electrons, they are held in place by the surrounding atoms.
(Again, this is simplistic since it is certainly possible, with strong
enough fields, to move the electrons around.) Hence, connecting
the insulator to "earth" with a wire does nothing because the electrons
cannot move to it to exit. The reason that electric charges ever
move is to reach a point of lower electric potential energy (which is
just a fancy way of saying that they feel a force). That is what they
are doing when they move to "earth". If you have a charged object
which has an excess of electrons, the electrons would "like to" move
away from the object because of the field they cause; they do that if
you give them a path they can follow. Electrons can leak off
where they are if the air they contact can accept electrons; for
example, many molecules are such that they can add an electron easily
and become a negative ion.

QUESTION:
You can discharge a metal conductor which has been charged by static
electricity by "connecting it to the ground with a metal strip" - can
this also be done a work for charged insulators? If so or not, why?

ANSWER:
No. On a perfect insulator the charges are not free to
move, so even if they have a path to a place with lower electric
potential, they are not free to move. Of course, there is no such
thing as a perfect insulator and charge will slowly leak off. For
a conductor, excess electric charge is perfectly free to move; that is
why all excess charge on a conductor is always located at the
surface.

QUESTION:
What exactly is a spark? ie - is it a flow of electrons jumping the gap
or is it due to an electron avalanche? Could you please go through the
exact mechanism of a spark for me? Is it the same mechanism for
lightning?

ANSWER:
Lightening is just a big spark as you speculate. The
following question and answer will point you toward a complete
explanation.

QUESTION:
Someone said to me that a +ve "streamer" meets the -ve leader during
lightning and said that this "streamer" came from the ground - how can
this be since the +ve protons are fixed in the nuclei of the ground
atoms? What actually is the streamer made up of and what causes
it? It would really help if you could give a step by step
explanation of lightning if you would please - there are SO many
different versions that people give.

ANSWER:
There is not a concise answer to your question. The
most complete explanation I have found is at the Howstuffworks
site.

QUESTION:
We always talk about charging up "insulators" via static electricty.
(ie - through friction) Does it work for conductors? Does it still
remain a surface phenominum? Does the metal remain charged afterwards?
Does it dissapate naturally?

ANSWER:
Yes, conductors can become charged via static
electricity. For a conductor, all excess charge must reside on
the surface whereas, for an insulator, you may have charge
inside. The metal, in the real world, will discharge by leaking
its charge either to the air around it or through whatever (imperfect)
insulators may be supporting it. Even if it is perfectly isolated
from its environment, a metal will leak its charge away from any place
on its surface where the electric field (due to the excess charge) is
strong enough (this is called corona discharge); for example, a very
sharp needle will have a very strong electric field at the point and
electrons will stream away (or toward) this point.

QUESTION:
I'm having some trouble understanding densities and liquids, and I was
thinking about this problem. I think it could help me understand
how liquids work a little better, but I'm struggling to understand
it. If I suspend a rock from a spring scale and it has a weight
of W, then I take the same rock and, keeping it attached to the scale,
I completely submerge it in a liquid of density p and it now has a
weight of w, how do I find the ensity of the rock? I know
density=m/V, but this doesn't seem to be enough.

ANSWER:
You need to know Archimedes'
Principle . An object submerged in a fluid experiences an
upward force called the buoyant force which is equal to the weight of
the fluid which was displaced when the object came in, that is the
weight of an equal volume of water. If the buoyant force is less
than the weight of the object, it will sink and if it is greater than
the weight of the object, it will float. For your problem, the
forces on the object are the weight W= r Vg
down (r is the density
of the object and g is the acceleration due to gravity), the
force that the scale exerts up, w , and the buoyant force up, B= r _{w} Vg
( r _{w }
is the density of
water). All these forces must add up to zero because of Newton's
first law: B+w-W= 0= r _{w} Vg+w-r Vg. But ,
r Vg=W=Mg where M
is the mass of the object, and since you measured W
you know M. So r _{w} Vg+w-W= 0
and solving, V= (W-w )/(r _{w} g ).
Finally, r =M /V=W /(Vg )=r _{w} W/(W-w ).

QUESTION:
In physics class we performed an experiment called projectiles launshed
at an angle (to the horizontal). Now we made a setup where we let a
marble shoot away with the help of an elastic band. We then measured
the distance the marble travelled when letting it shoot from different
angles, 20, 30, 40,45, 50, 60, 70 degrees. I would appretiate some
theory concerning this experiment because I thaught that the marble
would reach the longest distance from the angle 45 but the distance of
the angles 40, 45 and 50 were very similar....would you please explain
to me a little about what a "fair test" is and what factors do
actually affect the range of this elastic band.

ANSWER:
Well, an elastic band is not a terribly good energy source
because it tends to get "fatigued" with repeated use so it will not
produce reproducible results. But, even if you were using a
better launcher, say a good steel spring, your results would not be
surprising. The expression for the range of a projectile is R=v ^{2} sin(2 q )/g where v
is the speed at launch, g is the acceleration due to gravity,
and q is the angle of
launch. To the right I have plotted the range as a function of
the angle for an experiment where the maximum range is 5 m. The
maximum is at 45^{0} . But the thing to note is how slowly
the range is changing when you are at angles close to 45^{0} ;
the right-most graph is the same as the other graph except that it is
plotted only over the range 40-50^{0} . As you can see,
any angle in this range will give about the same value for the range —5
m.

QUESTION:
what are some ways that atomic energy can affect the environment?

ANSWER:
The actual production of atomic energy is much less damaging
to the environment than traditional energy generation sources (coal,
oil). However, after the fuel is "burnt" in a nuclear power
plant, it is radioactive and will not become nonradioactive for
thousands of years. Still, this is not a danger to the
environment provided it can be reliably stored without leaking into the
environment. That, however, is a big "if". Imagine trying
to contain something with absolute certainty for thousands of
years. The containers would have to be impervious to corrosion,
earthquakes, fires, etc . People have thought about
sending this radioactive waste into outer space, but the small chance
of an accident at launch is too big to take. Some thought has
also been given to "transmuting" this waste into either radioactive
stuff which would decay away faster or not be radioactive at all; this
does not presently appear to be practical, though. The other
danger, obviously, is the possibility of a catatrophic accident at the
power plant causing radioactivity to leak into the environment.
There was one bad accident in the US in 1979 at the Three Mile Island power
plant in Pennsylvania; fortunately, most of the radioactive waste there
was contained. In 1986 there was a bad accident at the Chernobyl power plant in
Russia; this was a much worse accident since about 30 people were
killed immediately and bad radioactivity spread over a 20 mile radius
exposing many people to dangerous levels of radioactivity and many
cases of cancer were caused by this. Lower levels of
radioactivity were recorded all over Europe.

QUESTION:
what's the worst kind of radiation?

ANSWER:
This is an impossible question to answer. It depends on
the intensity of the energy and the damage that it will do to a
biological object (like you). Let me give you a couple of
examples:

Neutrinos
are particles which are released in certain radioactive reactions.
There is a huge number which come from the sun; about a million billion
per second strike you! But they do not interact strongly with
matter and just zip right through you.
Electromagnetic
radiation can interact strongly with your cells. For example, if
I were to put you at the doorway of a blast furnace, the radiation
would literally cook you within seconds. This is just plain old
fashioned infrared radiation which we use to broil foods and it can be
very bad for you.
The
various types of nuclear radiation, alpha, beta, and gamma, can do
damage if intense enough. Alpha radiation (helium nuclei)
interact very strongly but that is good in one sense since they are
easy to shield against: most alpha radiation can be shielded from with
a sheet of paper. On the other hand, alpha emitters can be
dangerous if in the environment because if ingested or inhaled; then
the source is right inside you so you can't shield against it.
Beta radiation is just high energy electrons and can be dangerous if
intense enough. Gamma radiation is just very high energy
electromagnetic energy (like light, radio, infrared, x-rays, etc .)
and it can do serious damage if intense enough. Sometimes that is
good because this is the kind of radiation usually used to irradiate
cancer tumors.
Another
kind of electromagnetic radiation which can be very bad is ultraviolet
radiation. This is the radiation responsible for sunburn and skin
cancer. There is evidence that the layer of ozone which is high
in the earth's atmosphere is being depleted by pollution. This
will lead to greatly increased rates of skin cancers.
QUESTION:
How exactly can one reconcile or relate the two different descriptions
of Heat as infrared EMR on the one hand and Heat as high frequency
phonons (quantised molecular vibrations) on the other?

ANSWER:
Heat is just another word for energy and is normally used
only for energy related to thermodynamics. Infrared radiation certainly
carries energy and, when absorbed by something, usually results in its
temperature increasing. The kinetic energy of atoms in solids,
liquids, or gases is also a form of energy we generally associate with
temperature; in an ideal gas, for example, temperature is a measure of
the average kinetic energy per molecule.

QUESTION:
what chemicals are in coke-a-cola

ANSWER:
Don't you know that the recipe for coke is a deep, dark
secret?! All you need to know is that it contains enough sugar to
make it bad for your health and teeth if consumed more than a couple of
times a week. One of the flavorings comes from the cola nut and
carbonated water has carbolic acid in it. Also, although the
original recipe 100 years ago included cocaine (hence the name), it
certainly does not today.

QUESTION:
What is electrochemistry?

ANSWER:
Passing electric current through something may cause a
chemical reaction to occur; an example is the electrolysis of water
where electric current causes water to split into oxygen and
hydrogen. Sometimes when a chemical reaction occurs, an electric
current is caused; and example is a simple battery where chemistry is
the source of the electricity. The scientific study of such
phenomena is called electrochemistry.

QUESTION:
As is commonly known in science, things float according to differences
in boiling points. Helium rises upwards because it is less dense than
air. But why upward. Why not downward since the natural course for
anything to take is with the flow of gravity. why then does helium
resist gravity and is propelled upwards. What propells it in this
manner. even if it is less dense than air i do not see why it would
ascend up.

ANSWER:
Well, I do not understand your "commonly known" assumption
that boiling points have anything to do with floating! But then
you go on to say the real thing which determines
floating —density. The reason that a helium balloon will go up in
air is exactly the same reason that a block of wood goes up in water;
it is because the density of the object is lower than the density of
the fluid in which it is immersed. The "why" of how this happens
has been known for thousands of years and is called Archimedes
principle. A freely floating/falling object in a fluid has two
forces on it —its own weight and whatever force the fluid might exert
on it. The force which the fluid exerts on it, called the buoyant
force , may be shown to be (see any elementary physics textbook)
upward in direction and equal in magnitude to the weight of the fluid
which has been displaced by the object; thus, if the object is less
dense than the fluid, the buoyant force up will be greater than the
weight down and the object will accelerate upwards.

QUESTION:
Say the orbits of two celestial bodies intersect...we'll say at two
different points I guess. How do you calculate the point and time at
which these two bodies will collide if we know the starting position of
both objects and assume that they must collide at some point. If the
role of gravity makes the calculation too complex we can ignore it.

[Because
of our groundrule
requiring "single, concise, well-focused questions", I have
considerably edited your question.]

ANSWER:
Well, for starters, the role of gravity certainly cannot be
neglected since it is gravity which determines the orbits in the first
place! What you probably mean is that maybe the gravitational
force between the two objects might be neglected. Of course, your
problem is way too complex to answer in detail on a site like
this. But, it is easy to do in a general qualitative way.
The objective of classical mechanics is, for a particular particle, to
specify its position as a function of time. For example, the
vector r _{1} (t ) tells us where a particle
is in some coordinate system for any time t assuming that we
have defined some particular time to be t =0 and that we know
the particle's velocity and position at that time (initial
conditions). This position vector is determined by solving the
fundamental law of classical mechanics, Newton's second law, for the
particle. This requires, of course, knowing what the forces
experienced by this particle are. Any intermediate-level
classical mechanics course includes the study in detail of the Kepler
problem in which we assume that the sun is infinitely massive and that
a planet only experiences the sun's gravitational force; this leads to
Kepler's three laws, in particular, of interest to you, elliptical
orbits for objects orbiting the sun. So, let's let r _{1} (t )
be the position of the earth as an example; and, as the rest of the
example, let's let r _{2} (t ) be the
position of some comet which we fear might strike the earth some
day. Assuming we have done the physics to get these two position
vectors, we simply set r _{1} (t )=r _{2} (t )
which gives us one equation with one unknown, the time(s) when the two
particles are at the same position. Problem solved!
However, much easier said than done! First of all, assuming that
the earth and the comet never interact with anything but the sun is
simply wrong; it will only take a little tweak of the comet's orbit
when passing close to, say, Jupiter, to cause it to be at a very
different place later in its orbit. And, when we start making the
problem a many-body problem rather than a two-body problem, chaos
(unpredictability) becomes an issue. Also, predictions using
computers, the only practical way to calculate, will rely on accuracies
of initial conditions which are far beyond our ability to measure; a
tiny change in initial conditions can make a huge change in the orbits
a long time in the future. Thus, our ability to do such
computations are very restricted —we might be able to make some
predictions reasonably accurately, even centuries into the future (e.g.,
I can be pretty sure that Pluto and earth will not collide in
the next 100,000 years even without doing a calculation) but others
predictions (e.g. for an asteroid which is known to cross our
orbit) can only be reliably done for maybe a year or two.

QUESTION:
I saw a static generator (millions of volts) charging a sphere. As the
great ball charged, the belt slowed as the motor labored to push the
tiny electrons on the belt into the sphere. This contradicted the
explanations I was hearing about how the voltage was created at the
contact between the different materials of the belt and drive rotor. I
think now only the electrons are obtained this way and the voltage is
inconsequential beyond what is needed for the comb at the sphere to
collect them in. As the electrons were forces toward the charged ball,
some kind of work was done on the character of the electrons stuck to
the belt, raising their potential and slowing the drive motor as the
charge rose.

[Because
of our groundrule
requiring "single, concise, well-focused questions", I have
considerably edited your question.]

ANSWER:
You have a basic misconception about how the device
works. The sphere actually becomes positively charged by giving
electrons to the positively-charged arriving belt. There is a
very detailed and easy-to-read explanation at howstuffworks.com .
Regarding your question of the belt slowing down, the half of the belt
which is moving toward the positively-charged sphere is positively
charged and therefore feels a repulsive force; this force will increase
as the charge on the sphere increases.

QUESTION:
I know that if you pump up a soccer ball to 9psi at sea level and bring
it to Denver, the pressure on the ball will change. Conversely, if you
pump it up to 9psi in Denver, and take it to sea level, it will also
change. (I think that it will pop going from sea level to high
altitude, and go flat when going to sea level, but that doesn't really
affect my question). My question is: Isn't 9psi at sea level the
same as 9psi in Denver? It might require a different amount of air
molecules placed inside the soccer ball to achieve 9psi at the 2
different locations, but aren't they both 9psi? If I use a pressure
gauge (like a tire gauge with a needle) to measure the psi, and it says
9, regardless of where I am, isn't it 9? Is the firmness the same
at both altitudes at 9? The flight of the ball and the bounce of the
ball might be different because of higher altitude, but the firmness of
both balls would be the same at 9psi, regardless of the altitude?

ANSWER:
Most pressure guages measure what is called "guage pressure"
which is the pressure above atmospheric pressure. Atmospheric pressure
is about 14.7 psi but varies fairly dramatically with altitude and also
somewhat with the weather. So a pressure of 9 psi is an absolute
pressure of 24 psi if the atmospheric pressure is 15 psi but only 20 if
the atmospheric pressure is 11 psi. But the absolute pressure is
not what matters; the guage pressure is what will determine the
"hardness" or "bounciness" of the ball because the properties of the
ball are determined by all forces on it, not just the forces on the
inside. A 9 psi ball in Denver should behave essentially
identically to a 9 psi ball in New York (with regard to "hardness" or
"bounciness").

QUESTION:
say i have a satellite that has a ridiculously low orbit of 6.4x10^6 m
and it passes at a high speed, i know the mass of the earth 6x10^24 kg.
and i need to know what the velocity of the satellite is?

ANSWER:
This is called a near-earth orbit and applies for orbits near
the surface of the earth. You do not even need to know the mass
of the earth, just that the acceleration due to gravity is g =9.8
m/s^{2} and that the weight of a mass m is mg .
Then the force (mg ) must equal the mass times the centripetal
acceleration (mv ^{2} /R ). Solving, v =[gR ]^{1/2} =7920
m/s which is about 17,700 mi/hr. This is only true near to the
earth since the weight, the force of gravitational attraction, gets
smaller as you get farther away.

QUESTION:
Why can't the flight of an airplane be fully explained using
Bernoulli's equation alone?

ANSWER:
I have previously answered this
question.

QUESTION:
I hope you have not already answered this question already, but I was
wondering what the basic concepts of string theory are.

ANSWER:
No I haven't. In fact, I do not know enough about
string theory to give a cogent answer and nobody in the department here
works in the area. I can tell you that a lot of physicists are
somewhat hostile toward a theory which has been fashionable for over a
decade yet has really few results or successes to date. Still, it
represents a fascinating glimpse of cutting edge yet controversial
research. There has recently been a series of three Nova shows devoted to string
theory which, I confess, I have not seen but which I have read is
really a marvelous production. The entire three hours is viewable on
the web .

QUESTION:
If the positive charge on the proton and the negative charge on the
electron cloud of an atom have an attractive force on each other, why
does the electron cloud not collapse onto the proton and stop "moving"?
What force prevents the electron cloud from being annihilated or
absorbed by the proton? If there is an attractive force between them,
why is the electron's momentum conserved? Why don't the two "particles"
merge or fuse together?

ANSWER:
I have answered this question before and you can find that
answer here . And your question about
momentum being conserved —it is not conserved which should be clear
since the electron's direction is continually changing and momentum is
a vector quantity; angular momentum, however, is conserved because the
proton does not exert a torque on the electron.

QUESTION:
If action & reaction are always equal in magitude & opposite in
direction, why don't they always cancel one another & leave no net
force to accelerating a body?

ANSWER:
Newton's third law states that if one object exerts a force
on a second, the second exerts an equal and opposite force on the
first. Therefore, the "action/reaction" forces are never exerted
on one body. If you select a body to study, its motion is
determined only by the forces exerted on it, not by forces exerted by
it. Students often make mistakes with this "action/reaction"
thing because they tend to identify any pair of equal and opposite
forces as being an "action/reaction" pair. For example, a 1 lb
book sitting on a horizontal table has two forces on it, its 1 lb.
weight pointing down and a force of 1 lb which the table exerts up on
it (usually called the normal force); these have nothing to do with
Newton's third law but are equal and opposite because the book is in
equilibrium and the force the table exerts is therefore required to be
1 lb up. If we now look at the table, the book exerts a 1 lb
force down on it because of Newton's third law; the "action/reaction"
pair is the force the table exerts on the book and the force which the
book exerts on the table. Lots of novice physics students want to
say that the weight of the book is the 1 lb force down on the
table —this is totally false since this is a force on the book, not the
table.

QUESTION:
Pardon me if this sounds a little far out, but I was just wondering if
time series analysis is ever used in trying to determine the origins of
the universe —that is, by studying the movement or motion of the
planets and other celestial bodies and trying to predict backward
toward the big bang or the creation or whatever got the ball
rolling. And if so, could you direct me to any books on the
subject?

ANSWER:
This is more than a little far out! Even if the laws of
nature which govern the motions of everything were really time
reversible, there are simply too many variables to be able to do the
calculations you suggest. And too many unknowns from the
past. Think of the complications —the solar system came into
existence long after the big bang and before that it was a cloud of
stuff which was the remnant of older stars which had lived out their
lives and exploded. Also, much information is lost as time goes
on. For example, suppose we had a big puddle of water on the
ground which, yesterday, was an ice sculpture of a swan. Can you
imagine trying to compute backwards from the puddle to the swan?

QUESTION:
I am having trouble understanding torque. What is it? A
force? A tendency? Something else? Also, how does it differ from
inertia? Does it not resist circular motion? I am very
mathematical, so maybe an explanation and/or derivation of the formula
t=Fd could help me.

ANSWER:
It is difficult to come up with an answer since you ask a
question, "...how does it differ from inertia?" which indicates that
you know very little about rotational physics! I will try, but
you must first understand translational physics, where everything may
be treated as a point mass. In this arena, you might very well
ask "...what is force and how does it differ from inertia?"

Inertia,
which is called mass (m ), is the property which measures how
resistant something is to being accelerated (a ) if it is pushed
or pulled; the amount of push or pull, called force (F ),
measures how effective the push or pull is at causing
acceleration. All this is stated, of course, by Newton's second
law, F=ma .
In
rotational physics the whole scenario sounds very much the same:

Inertia,
which is called moment of inertia (I ), is the property which
measures how resistant something is to having an angular acceleration ( a ) about some axis; that
"thing" which causes the angular acceleration, called torque ( t ), measures how
effective the "thing" is at causing acceleration. All this is
stated, of course, by Newton's second law in rotational form,
t =I a .
For
translational physics, the only thing which determines inertia is how
much stuff (mass) there is; for rotational physics, inertia is
determined by how much stuff there is but also how it is distributed:
an object with most of its mass near its axel is much easier to get
spinning than is an object with equal mass but much of it far from its
axel. For translational physics, the force is determined only by
hard we push; for rotational physics, torque is determined by how hard
we push but also where we push and the direction in which we
push. For example, suppose you want to open a door and you push
as hard as you can but you push at the hinged end of the door; the door
doesn't open, does it? This is because the torque increases as
you increase the "moment arm", the distance from the rotation axis to
where the force is applied (your t =Fd equation).
Actually, your equation is a bit too simplified to completely
understand torque. If you have an open door and push at the edge
of the door which is opposite the hinges but you push straight toward
the hinges, you do not close the door even though F and d
are both big. It is really only the component of the force F _{p}
which is perpendicular to d which contributes to the
torque, so t =F _{p} d.

I hope
this has been helpful.

QUESTION:
Okay, I have a stick of mass M and length L held so that it makes and
angle theta with respect to the floor. The stick is not hinged on
the floor, and the contact between the end of the stick and the floor
is frictionless. I release the stick and it falls to the
floor. How do I find the horizontal distance the left end (the
end on the floor) travels during the fall? And with what speed
does the right end (the end initially in the air) hit the floor?

ANSWER:
This is easier than you think. Since there are no
forces which have horizontal components, the center of mass of the
stick must fall straight down. The distance the end will move is (L /2)(1-cos q ). The second
question (not quite so easy!) should be addressed by energy
conservation. At the time the stick is released, it has only
potential energy E _{1} =Mg (L /2)sin q . Just before it
hits, the speed of the end will be v and the speed of the
center of mass will be v /2. The energy, purely kinetic,
will be the kinetic energy of the center of mass plus the kinetic
energy about the center of mass: E _{2} =(1/2)M (v /2)^{2} +(1/2)I w^{2} =(1/8)Mv ^{2} +(1/2)(ML ^{2} /12)[(v /2)/(L /2)]^{2} =Mv ^{2} /6.
Equating E _{1} and E _{2} now gives v =[3gL sin q ]^{1/2} .

QUESTION:
I have been curious about this problem for some time. If a solid
ball of mass m and radius r is resting on a block of mass M and a
horizontal force is applied to the block, how would I find the maximum
value of the force F that moves the block without causing the ball to
slide. This assumes there is friction between the block and the
ball, but not between the ground and the block. I think F acting
on the ball has to equal the force due to friction between the two
objects, but I'm not sure if, when finding this frictional force, you
consider the mass to be m+M or just m. And I feel like I'm
leaving something out. Thanks!

ANSWER:
This is a tricky problem. We know the ball has a
frictional force forward on it given by f= m mg
if the ball is about to
slip. Therefore, the center of mass of the ball has a forward
acceleration of a _{m} = m g.
However,
this is not the forward acceleration of the block. So now
consider the sum of torques about the center of mass of the
ball. This must equal the angular acceleration of the ball about
its center of mass times the moment of inertia of the ball about its
center of mass: t= fr= m mgr= (2mr (2mr ^{2} /5)(a _{t} /r )
where a _{t} is the acceleration of the rim of the ball with
respect to the center of mass of the ball. Therefore a _{t} =5 m g /2. Now, since
the center of mass itself has an acceleration, then the acceleration of
the point of contact between the ball and the block as measured in the
laboratory is a _{t} -a _{m} =a _{M} =3 m g /2.; I have
called this a _{M} because it is also the acceleration
of M since the ball is not slipping. So, finally, F-f=F- m mg=M (3 m g /2), so F= m g (m +3M /2).
Note that the ball exerts a force on the block of magnitude f
backwards because of Newton's third law.

Note that
I have used a theorem of classical mechanics: The sum of torques about
the center of mass of a rigid body is equal to I a
even if the center
of mass is accelerating.

(PS: In
future I cannot answer a question without an email address.)

QUESTION:
Does the force of gravity influence the movements in ice skating ?

ANSWER:
Ice skating would not be possible without the force of
gravity because there would be no force to push your skates down
against the ice. Apart from that minor detail, any movement which
involves motion in a vertical direction would be entirely
different. For example, a simple jump would become a launch from
which you would never come down.

QUESTION:
Does a sunset/sunrise over a large body of water appear differently
than a sunset/sunrise over a large body of land? If so, why?

ANSWER:
There are lots of possible answers. I asked the faculty
here if anybody had a good answer; I got several replies:

Yes,
it's much more romantic over water!
Maybe,
if the land is desert. Then there is less aerosol, a less hazy
sunset. If this is a 'green flash' question and was
rephrased I might have a more complete answer.
Yes,
all our 1112 and 1212 instructors know that there will be a reflected
beam into your eye that reflected from the water surface, and inteferes
with light that traveled directly from sun to eye. The angle of
incidence was so close to 90 degrees, that the path length distance for
the reflected ray is much less than one wave length more than the path
length for the direct ray. Since there is a phase change of pi
for the reflected ray, it interferes destructively with the other ray.
Consequently there are horizontal dark lines in the preceived image of
the sun. Of course variations in the refractive index of air can
also play a part. For sunsets/rises over land, there is no
appreciable reflected ray, because the land surface is not smooth
enough. If it were, such as flat ice, the same effect would show
up. (This is known as "Lloyd's mirror" and is analogous to the
Young's double slit experiment except for the phase change since there
is an image of the edge of the almost-set sun under the water surface.)
QUESTION:
i have read somewhere that according to general relativity, only mass
undergoing acceleration will produce graviton. so if an elementary
particle does not undergo acceleration, does it still have a
gravitational field? according to general relativity, only a mass
undergoing acceleration will produce gravitons. then why does a
stationary ball have a gravitational field around it?

ANSWER:
First of all, there is no working theory of quantum gravity
and so a graviton is a purely hypothetical particle. It is best if we
do not involve gravitons in the answer to your questions. Where
your confusion comes from is that general relativity predicts that an
accelerating mass will produce gravitational waves . All
masses produce gravitational fields. A mass which is not
accelerating will produce a field but not a wave. A gravitational
wave propogates through space, most likely at the speed of light much
as electromagnetic waves do. There is a lot of analogy between
gravity and electromagnetism here: any electric charge produces an
electromagnetic field but it will only radiate electromagnetic waves if
it accelerates.

QUESTION:
what is the spin of an elementary particle?

ANSWER:
Elementary particles have an intrinsic angular momentum; this
is called spin. A classical example would be the earth which has
an orbital angular momentum because of its revolution about the sun but
also an intrinsic angular momentum because of the rotation about its
own axis.

QUESTION:
does a stationary elementary particle have a gravitational field?

ANSWER:
Any object which has mass has a gravitational field.

QUESTION:
If a celestial object has Surface Gravity = 43054.5 cm/s^2 and Central
Pressure = 2.116347946 * 10^16 dynes/cm^2 i.e. 21.16347946 Gbars, how
do we go about converting these parameters given in CGS units into
their MKS equivalents?

ANSWER:
I always recommend to my students to multiply by one so that
the units come out right:

1.0 cm/s^{2}
x (1 m/100 cm)=0.01 m/s^{2
} 1.0 dyne/cm^{2} =1.0 (gm cm/s^{2} )/cm^{2}
x (1 kg/1000 gm) x (1 m/100 cm) x (100 cm/1 m)^{2} =0.1 (kg m/s^{2} )/m^{2} =0.1
N/m^{2} .

Incidentally,
there is a free software utility which is helpful for conversions of
units called, appropriately, CONVERT. You can download it
at http://www.joshmadison.com/software/convert/
. However, for reasons not clear to me, pressure units of N/m^{2}
are not included in the pressure options. However, you can use it
to convert dynes to newtons and cm^{2} to m^{2} .
Still, as seen in my little computation, unit conversions are generally
pretty easy from CGS to MKS. The CONVERT software is
useful for more arcane conversions, e.g. how many m^{2}
in an acre?

QUESTION:
how do scientists decide whether an element particle has mass or not?
how do they measure the mass and with what kind of equipment do they
achieve that?

ANSWER:
Elementary particles are created by observing byproducts of
collisions of known particles. For example, one can smash a very
energetic proton into another proton and many things will come
out. In special relativity there is a simple relationship which
connects the mass of a particle, its energy, and its linear
momentum. Furthermore, energy and linear momentum are always
conserved in a collision. Therefore one can measure or deduce the
energy and momentum of each particle that comes out of the collision
and if you know the energy (E ) and momentum (p ) of a
particle you can deduce its mass. For a massless particle, E=pc
where c is the speed of light. If the particle has mass, E ^{2} =m ^{2} c ^{4} +p ^{2} c ^{2} .

QUESTION:
If you are in a sealed train car (no way to see outside, soundproof,
vibration-proof etc.), can you devise an experiment to determine if the
train is moving? I'm only a high school physics student, but this
sounds pretty near impossible to me. Since speed is relative and there
is no way of comparison in the sealed car, I doubt there is an answer.
Any input would be greatly appreciated. Thanks!

ANSWER:
One of our basic postulates about nature is that the laws of
physics must be the same in all inertial frames of reference. In
classical physics an inertial frame is defined to be one in which
Newton's first law is true (e.g. a particle at rest will remain
at rest if there are no net forces on it). Once you have
found one inertial frame, any other frame with constant velocity
relative to the first will also be an inertial frame —find one and you
have found them all! If we approximate the earth as an inertial
frame (it really isn't because of its rotation on its axis and its
revolution around the sun), then if you are in a train going with
constant velocity , there is no experiment you can do which will
determine that the train is moving. On the other hand, you merely
stipulate that the train "is moving"; if the train were accelerating,
speeding up for example, you would certainly be able to detect that
(because you would feel like you were being forced into the back of
your seat). Frames of reference which are accelerating relative
to inertial frames are not inertial frames. The theory of special
relativity has the same postulate, that the laws of physics are the
same in all inertial frames, but an inertial frame is defined somewhat
differently and Newtonian mechanics must be reformulated; essentially,
the test for an inertial frame is that Maxwell's equations be the
correct laws of electromagnetism.

QUESTION:
Since the Earth is already travelling through space at something like
30 miles per second, isn't it already a bit shortened along its length,
and isn't time already a bit slower, and isn't earth's mass already a
bit greater than it "should" be? In short, isn't our perception of time
and space already mired in some degree of relativistic effects?

ANSWER:
The mistake which you make is that you are assuming that the
earth has some absolute velocity relative to some "special" frame of
reference. But the whole idea of relativity is that it is not
possible to distinguish between inertial frames of reference, so saying
that we move with speed of 30 miles/second is meaningless. All
that matters is that we be in an inertial frame; we are at rest
relative to our own frame so everything is as it "should" be.
Incidentally, an inertial frame is one in which all the usual laws of
physics are correct. Once you find one inertial frame, any other
frame moving with a constant velocity with respect to the first is also
an inertial frame; any frame accelerating with respect to the first is
not an inertial frame. Obviously, the earth is only approximately
an inertial frame.

QUESTION:
Ok this is a relativity problem.. Lets suppose that there is a
cylindrical hole in the ground (3m deep) with a bug trapped in the
whole that is (.5 metters high). There is a cylindrical plunger
traveling at .99c that is (2m in length). The traveler on the
plunger will see the hole shrink and he will squash the bug, but the
bug see's the plunger shrink so he has no worries of getting
squashed?? Who is right? They have to see the same thing
right?

ANSWER:
I think the issue here is the position of the end of the
plunger. The bug sees it approaching with speed 0.99c and
the moving observer sees the bug approaching the end of the plunger
with speed 0.99c . So both have to agree that the bug is
doomed.

QUESTION:
Just a simple one that i can't seem to find the answer to anywhere. Is
Atomic oxygen naturally present in the atmosphere at ground level and
if so, in what quantity.

ANSWER:
I have talked to several people and all agree that atomic
oxygen at sea level will be rare and transient since it is so
reactive. However, because oxygen is so reactive, O_{2}
molecules can dissociate on the surface of transition metals (which is
the principle behind a catalytic converter) and there might be
transient but detectable levels near such surfaces. It is
interesting, however, that there is a significant amount of atomic
oxygen at the very highest limits of the atmosphere where incoming
radiation from the sun dissociates O_{2} and the O has
difficulty finding other molecules with which to react. It turns
out that the reactivity of O is one of the most important issues in
corrosion of satellites and the shuttle in orbit.

QUESTION:
it is postulated that every object with non-zero mass produces a
gravitational field. now i know that photon has no stationary mass.
however, we can never observe a stationary photon since light travels
at the same speed to different frames. can we say that moving photon
has mass (since it has momentum) so there is a gravitational field
around every photon?

ANSWER:
One of the important findings of the theory of special
relativity is that a massless particle carries both momentum and
kinetic energy. The notion that momentum or kinetic energy
necessarily implies the presence of mass is a classical idea and not
correct.

QUESTION:
We are studying forces and their agents. The author of the text
states, "If you can't name an agent, the force doesn't exist!" I
have been unable to come up with possible examples where there appears
to be a force but no agent can be identifed. Any examples would
be helpful.

ANSWER:
Since Newton's laws are valid only in inertial frames of
reference, one often finds in accelerated frames of reference
(accelerated relative to an inertial frame) the appearance of
forces. For example, when you accelerate in your car you would
think that there is a force pushing you back in your seat, but there is
not. Such forces are called "fictitious forces" and the best
known is the centrifugal force; if you are in a spinning drum, you
think that there is a force pushing you into the wall of the drum
(centrifugal) but there is not. Always rememeber —there is no
such thing as a centrifugal force!

QUESTION:
i read from some book that the natural corollary from special
relativity is that every motion is relative, which implies that if you
are in a bus undergoing a uniform straightline motion, you cannot tell
your velocity or whether you are moving unless you look outside.
however, i think this phenomenon is also true under newton's law. if
special relativity does not exist, are there any ways, or by any
experiments that we can find whether we are moving or not in the bus
without looking outside?

ANSWER:
What you are referring to is what is called an inertial
frame of reference . In classical physics (Newtonian, as you
say), an inertial frame of reference is one in which Newton's first law
is observed to be true. Then it is an experimental fact that,
once you find one inertial frame, any other frame which moves with
constant velocity with respect to the first is also an inertial frame;
frames which accelerate with respect to inertial frames are not
inertial frames. Einstein extended this idea to all laws of
physics. He postulated, in formulating the theory of special
relativity, that the laws of physics must be the same in all inertial
frames of reference. Classical physics predicted that the laws of
electricity and magnetism would not be the same in all inertial frames,
and fixing that led to special relativity. The theory of general
relativity, the relativistic theory of gravity, postulates that the
laws of physics be the same in all frames of reference,
inertial or not; this results in your not being able to distinguish
between an accelerating frame and a nonaccelerating frame in a
gravitational field.

QUESTION:
As I drove to work this morning I noticed frost on the rooftops of
homes (it was 32def F). I remembered a friend of mine mentioning
that the frost forms due to black body radiation if the sky was clear
that night and the temperature is close to freezing. I'm still
wondering what that means. I understand that if it is cloudy,
then the clouds will reflect the escaping heat. But, isn't the
heat escaping just thermal heat. Why would the clear black
sky at night cool a rooftop or grass due to losing radiation?

ANSWER:
I have previously answered this question
in another context and in considerably more detail than I will answer
yours, so you might want to look there. In a nutshell, all things
radiate. Often everything in a particular environment is in
equilibrium with everything else which means that the energy it
radiates into its environment is exactly the same as the energy it
absorbs from the other things which are radiating. This is not
always the case though: E.g. something hot will cool down
because it radiates more energy than it absorbs. Similarly, the
roof will be radiating energy and if there is nothing to give it energy
back it will cool. In my previous example I
estimated that energy will be radiated from a blackbody at a rate of
about 30 W/cm^{2} . That is pretty rapid and can result in
cooling below the freezing point provided that the air (which will give
energy to the roof by convection) is not too warm. For this
reason (and also because of evapoarative cooling), frost can form when
the temperature is above freezing.

QUESTION:
My physics teacher assigned us a problem to solve and i NEED
HELP. The question is : Why does it take longer for a plane
to fly east to west traveling on the same route and going from the same
two places? What should I do? How should I start it?

ANSWER:
The
groundrules of this site clearly state: "do not use 'the Physicist'
to do your homework"! I can only give you a hint: the
"speedometer" in an airplane measures airspeed, not groundspeed.

QUESTION:
Is 2.116347946 * 10^16 dynes/cm^2 = 21163.47946 Mbars =
211.6347946 Gbars?

ANSWER:
Since 1.0 bar is 10^{6} dynes/cm^{2} , and 1
Mbar=10^{6} bar, the conversion from dynes/cm^{2} is
correct. However, since M means 10^{6} and G means 10^{9} ,
21163.47946 Mbar=21163.47946 x 10^{-3} Gbar=21.16347946
Gbar. Incidentally, there is a free software utility which is
helpful for conversions of units called, appropriately, CONVERT.
You can download it at http://www.joshmadison.com/software/convert/
. It would not have helped you with your problem, however, since
Mbar and Gbar are not listed (but it did help me check how a bar is
defined). You need to know the standard
prefixes for powers of 10

QUESTION:
I have spent a number of hours on the Internet over the last week
trying to find some information. I have been unable to find a
satisfactory answer to my question. My question relates to the
Bernoulli principal and conservation of energy. I would like to
understand conservation of energy at a molecular level, and how this
relates to a reduction in static pressure as air accelerates through a
venturi tube. (how it accelerates the flow). Perhaps, along with your
explanation you could give me some keywords that I can put into the
computer to help with my understanding in this area.

ANSWER:
You are right, Bernoulli's equation is a statement of
conservation of energy. However, it is not derived from
velocities of individual molecules but rather the velocity of the bulk
fluid. The velocity distribution of the molecules is determined
by the temperature of the fluid and has no relation to the velocity of
the fluid (which is added as a small drift velocity to each molecular
velocity). The things you are trying to understand
microscopically are best left as macroscopic.

QUESTION:
Is there an
advantage to having polarized lenses on sunglasses? I've heard
they are better; they are definitely more expensive. I have
polarized lenses now. The only difference I see is with these lenses I
see many different color patterns on car windows. Sometimes shiny
objects on the ground seem to "glow". What's going on?

ANSWER:
The advantage is that polarized glasses can reduce glare
reflected from surfaces. Reflected light is rather highly
polarized as shown by the figure at the right. At just the right
angle, the light will be totally polarized. The picture shows
unpolarized light before reflection becoming polarized after
reflection. Think of the surface as the hood of your car; then if
you are wearing polarized glasses which filter out horizontally
polarized light you will not see much of the directly reflected
sunlight (glare). Light from a clear sky (blue) can also be
highly polarized depending on the location of the sun and so
photographers use polaroid filters to get striking effects in black and
white photos.

QUESTION:
I'm
interested in a Career at NASA as a research scientist. I was
informed that you should have a good physics background during your
undergraduate years. What major would you recommend for this
career and what major in Graduate School would you recommend?

ANSWER:
This is an impossible question to answer. A "research
scientist" is simply too broad a thing to be aiming at. I am sure
NASA has scores of different types of research scientists —physicists,
astronomers, biologists, computer scientists, chemists, etc .
When you decide what kind of science you want to do, then you can plan
your college career more sensibly. Also, why restrict yourself to
NASA? There are many kinds of employers who carry on fascinating
science research.

QUESTION:
Could you
please explain to me the thermodynamics/heat transfer of: a) toasting a
marshmallow on an open fire.....b) and heating a can containing pea
cream soup

ANSWER:
a) Properly toasting a marshmallow requires that it be
exposed to the radiant energy from a bed of hot coals. (I
am a marshmallow-toasting gourmet!) The radiant energy then heats
and melts the sugar inside (change of state) and toasts the sugar
outside (a chemical change).

b) Heat
from the fire is conducted through the metal of the can making
the surface in contact with the soup become hot. Then the heat on
the inside surface of the can heats the soup in contact with it and
this is mainly transmitted to the rest of the soup by convection
(the soup moving around because of its heating). It is useful to
stir the soup occasionally (particularly cream soups which do not
convect so well) so that if the convection doesn't move it away from
the can a chemical change won't occur (scortching to the can).

QUESTION:
how can I
simulate invasion percolation procees?

ANSWER: (provided
by M. R.
Geller)
There is a nice site at http://www.ucls.uchicago.edu/People/02/Beckett_Sterner/
. There is even some shareware software you can download (I think
it is just for Mac.)

QUESTION:
What would
the mode be for a group of numbers that only appear once? such as 3,4,5

ANSWER:
This is really math, not physics. But check out http://mathworld.wolfram.com/Mode.html
. I think the answer to your question is that 3 4 5 would be trimodal
with each as a mode. This site http://scienceworld.wolfram.com/
is really helpful in getting facts about math, physics, astronomy,
chemistry, or biographical information about scientists.

QUESTION:
Have
isolated quarks ever been observed? I am getting conflicting
information from texts about whether they have actually been discovered.

ANSWER:
An isolated quark has never been observed; we believe that
the reason is that the force which binds quarks together is such that
the further the quarks get from each other, the stronger the force
gets. But the theory based on quarks accounts for such a rich
variety of experimental results that most particle physicists believe
in them wholeheartedly. You might ask if something unobservable
in the sense of seeing it all by itself is "real"; I think this is a
question for philosophers, not scientists.

QUESTION:
This a not a
brain teaser in any way at all. Me and a friend got in to an
argument the other day. We were eating pizza and it was too hot,
so he said that we should put it in the freezer for a little
while. well i don't remember how the argument got started but he
said that hot pizza would freeze in the freezer faster than pizza that
was room temp. I think he's nuts. What do you think?

And one more
question. Do "states of matter" (solid, liquid, and gas) have to
do with physics or chemistry?
ANSWER:
No, the hot pizza will not freeze faster. The classic
question along these lines is will hot water or cold water freeze
faster. For a lengthy discussion of this problem, see an earlier answer .

States of
matter are concepts used in both physics and chemistry.

QUESTION:
I have read
that the speed of light cannot be broken. I have also read that all
motion is relative. Suppose point A is a stationary reference point.
Point's B and C are start near point A and travel away from
each other at 51% the speed of light relative to point A. Does this not
mean that relative to each other, points B and C are traveling apart
from each other at 102%? Do you not add their speeds together as if two
automobiles are each traveling away from each other at 60 mph are
traveling at 120 mph relative to each other?

ANSWER:
No, your intuitive way of adding velocities does not work in
the theory of special relativity. Because of the fundamental
postulate of special relativity, that the speed of light is the same in
all frames of reference, the correct formula for velocity addition is v= (v _{1} +v _{2} )/(1+v _{1} v _{2} /c ^{2} )
where c is the speed of light. So, for your example, v _{1} =v _{2} =0.51c
so v =0.81c , 81% the speed of light. Note that the
formula reduces, to an excellent approximation, to your v= (v _{1} +v _{2} )
if the speeds involved are very small compared to the speed of light
because the term v _{1} v _{2} /c ^{2
} is extremely small compared to 1.

QUESTION:
I'd like to
know what charge is. Not the descriptive stuff like coulomb's law
and opposite charges attracting, but what is it? And how does it
work over a distance?

ANSWER:
This is a profound question. It is like asking what
gravitational mass is and how gravity works. Electric charge is
simply a property that some objects in nature have which allow them to
exert and feel electric and magnetic forces. Theories which try
to predict charges and masses of elementary particles are very complex
and only partly successful. It is known that quarks, from which
protons are made, have electric charge which is in multiples of one
third the elementary charge e . Regarding how the forces
act over a distance, classical electricity and magnetism postulates
what are called fields; and electric charge "distorts" the space around
it by placing a field there and then other charges feel this
"distortion". In quantum electrodynamics, a more complete theory
of electromagnetism, the picture is that charges exchange virtual
photons which are the mediators of the field.

QUESTION:
what is the
present device used in measuring acceleration due to gravity?

ANSWER:
Actually, the acceleration due to gravity is not considered a
fundamental constant because it is so variable. It depends on
altitude and where you are on the earth. The more fundamental
quantity which physicists are interested in measuring is the universal
gravitational constant G which appears in Newton's law of
gravitation, F=Gm _{1} m _{2} /r ^{2}
where m _{1 } and m _{2 } are two masses
and r is the distance between them. There is a very
readable article in the journal Physics Today in the November
2002 issue; the title is "Beam Balance Helps Settle Down Measurement of
the Gravitational Constant".

QUESTION:
What do you
think of the classical EM theory and why do we still use it when a lot
of it has been proven wrong? Tom Bearden - PhD, MS (nuclear
engineering), BS (mathematics - minor electronic engineering) claims
there are 34 flaws in the clasical EM theory which most people are
taught that it is still real ( http://www.cheniere.org/misc/flaws.htm
)

ANSWER:
Looking over the website you cite, many of these are bogus,
many are situations which simply point to the fact that classical
E&M is not properly quantized, and many I simply have no idea what
he is talking about. It is true that classical E&M is not
valid if quantum effects are important and that there are lots of
questions (self energy, virtual photon exchange, vacuum polarization, etc .)
which are not addressed. But quantum electrodynamics (of Feynman,
Dyson, Schwinger, Tomonaga, et al .) is a fully developed and
understood theory. To suggest that classical electromagnetism, a
beautifully complete, compact, rigorous, and precise theory is somehow
not "real" because it does not address quantum effects is
ludicrous. The first major breakthrough of modern physics,
special relativity, was discovered because classical E&M is a
completely correct relativistic theory. Surely you would not
throw out classical mechanics, absolutely imperative for computation of
orbits etc . in celestial mechanics, because it does not contain
relativity or quantum mechanics.

QUESTION:
I would like
to know if someone has conducted or tried to the following experiment:
with an isolated hydrogen atom, to confine the electron gradually
towards the nucleus restricting its orbital so that finally the
electron colapses onto the nucleus? What would be the main experimetal
difficulties with this and where is described such experiment?

ANSWER:
There is no way to do this and have the H atom retain its
identity. The ground state is the lowest one possible in such a
quantum system. Under extreme conditions (as in a neutron star)
it is possible to compress electrons and protons together to create a
neutron and a neutrino. You can also scatter very energetic
electrons from protons (this is done routinely at electron accelerator
laboratories) and the electron can probe the interior of the proton,
but this would certainly not be called a hydrogen atom.

QUESTION:
For a
science experiment, I decided to build three Newton's Cradles, each
with different ball materials, and observe the amount of times that
each would bounce. One would be rubber, one wooden, and one
steel. I concluded that rubber bounced the least amount of
times. I figured that more energy is absorbed into the balls with
each cycle, and so there would be less energy each time. I also
figured that the steel balls, being heavier, would have more potential
energy when lifted up. However, I'm specifically curious as to
why more energy is absorbed by rubber, and how that works
entirely.

ANSWER:
When you compress a ball, as happens when it collides with
another ball, you do work. Part of this work goes into the
compression like a spring where the work you do is stored as potential
energy; and part of the work is work against internal friction and that
work is lost, not stored, and shows up as heating the ball up. An
extreme example would be if you made your balls out of putty and all of
the work of compression would be lost. In your cases, the steel
balls are most elastic and energy comes closest to being conserved; the
rubber balls lose a considerable amount of the energy because it does
not come back out when the ball decompresses.

QUESTION:
I’ve read
that there is a maximum density limit for matter. If so, this would
mean an object made from the densest possible matter could not be
compressed any further. Isn't this a rigid body?

ANSWER:
As you probably know, nearly all the mass of an atom is
concentrated in the nucleus, an extremely small volume in the center of
the atom. Under extreme conditions it is possible to compress the
atom down to the size of a nucleus and compress all nuclei in the
object together. The mass density you come up with is called the
density of nuclear matter; this density is approximately 2x10^{17}
kg/m^{3} . Compare this with water which has a density of
1000 kg/m^{3} and you will appreciate that the density of
nuclear matter is almost unimaginably huge! But, does this ever
happen? Yes, there are special stars which, late in their
lifetimes, collapse under gravitational attraction to this density;
they are called neutron stars. Some stars collapse even further
such that their density becomes essentially infinite; they are called
black holes. So there is no maximum density limit.

QUESTION:
Can lines of
magnetic force be reflected or focused like rays of light?

ANSWER:
A beam of light is something which moves through space.
A magnetic field (or electric field for that matter) should not be
thought of as equivalent, so in some sense, reflection really has
little meaning. You can of course, distort things so that you
bend the field lines. If you want to call that reflection or
focusing, that is fine, but usually physicists don't use such
terminology.

QUESTION:
What are the
advantages/applications of a stem and leaf graph?

ANSWER:
Since you don't ask what it is, I assume that you know.
I didn't when I received your question! Actually, it has nothing
to do with physics, it is a reputed to be a convenient way to display
modest data sets (<100 points) which makes it easy to see some of
the qualitative statistical properties of the set like modality,
median, etc . It is more or less just like a histogram,
the only advantage that I can see being that you can actually pick out
each element of the set, not just see the lump sum of all the numbers
between 70 and 79, for example. Since I have not heard of it
before, it is safe to assume that phyiscists rarely if ever employ
it. Given modern computerized spreadsheets with automated
statistical analysis of virtually any kind you can think of, this kind
of plot looks real 19^{th} century to me!

QUESTION:
How much
stronger is the force of Earth's gravity over the magnetic force of the
Earths poles?

ANSWER:
This question cannot be answered. It is the old
"comparing apples to oranges" thing. The force of gravity depends
on the mass of the object which is feeling it whereas the magnetic
force depends on the magnitude and geometry of an electric current
which will feel it. Since the mass and current distributions have
no relation to each other, your question has no meaning. One may
make the general statement, however, that the gravitational force is
nature's weakest force. The only reason you think that it is a
strong force is that you happen to be in the vicinity of a very huge
mass (mass being the source of gravitational forces).

QUESTION:
I would like
to know your explaination of HOW magentic fields in Space are being
created. EARTHS MAGENTIC FIELD, SUNS MAGENTIC FIELD, MY FRIDGES
MAGNETS MAGENTIC FIELD, ELECTROMAGNETIC FIELD, ACTUALLY EVERY "BODY"
THAT IS IN SPACE. HE WHO KNOWS THE ANSWER, UNDERSTANDS THE
"WORKINGS" OF THE UNIVERSE. HE WHO KNOWS THE ANSWER WILL EXPLAIN
T.O.E.

ANSWER:
There are only two ways to make a magnetic field: have moving
electrical charges (electric currents) or have time varying electric
fields. Understanding the origin of some magnetic field may help
you to understand some specific thing, but believe me that one will not
understand everything by understanding magnetic fields.

QUESTION:
If I hold a
device that measures the speed of a passing light beam as C, isn't it
more reasonable to assume that the light is not moving, and that I am
the one moving at "light speed". Since such a device only measures the
difference in speed between me and the light beam, and we know that I
am wizzing through space on earth, isn't it more accurate to say that
light is standing still, and everythng else moves at light speed other
than the other way around?

ANSWER:
What you are suggesting is fallacious because c is a
universal constant, so somebody moving by you with a speed of half c
will also measure the speed of the light to be c . Now,
applying your logic, each of you is moving with speed c while
the light stands still, so the two of you are at rest with respect to
each other. You can't be both at rest and not at rest with
respect to the other fellow.

QUESTION:
A box
sliding across the floor is soon stopped by friction and its motion
(velocity) soon ceases. The frictional retarding force is an external
force acting on the "box-floor system" and momentum is not
conserved. However, I can redefine the system to include the
frictional force by redrawing the boundry.The force is now an internal
force in my new system, but the sliding box still stops. Is momentum
now conserved?

ANSWER:
You are all muddled. I have no idea what you mean by
boundary. You need to choose an object or objects to focus your
attention upon. Say it is the box. What are the forces
acting on it? Its weight, a force the floor exerts up on it to
keep it from accelerating through the floor, and a frictional force in
a direction opposite the velocity. As I told you last time you
wrote me, if the system you have chosen to focus upon has a net force
acting on it, then momentum will not be conserved. There is
therefore no mystery as to why the momentum of this box
disappears. To find a system in which momentum is conserved it is
easier to understand if we imagine the box sliding on a very big
(massive) block like the floor and this block sliding frictionlessly on
a horizontal surface. The big block (surrogate floor) is at rest
at the beginning of the problem. What are the forces on this
system? Just the weights of the big block and the box and the
normal force up on the big block which add up to zero; therefore
momentum will be conserved. The momentum starts as mv and
ends as (M+m )V. Conserving momentum, V= [m /(m+M )]v ,
so you see that if M >>m, then the final velocity
of the your floor and box will be zero. Of course, the floor is
not sliding frictionlessly on something, it is attached to the house
which is attached to the whole earth. Therefore M is the
mass of the whole earth plus the house plus the floor. I think
you would agree that that is a whole lot bigger than the mass of the
box!

QUESTION:
What is the
basic illusion with “movement” of any object? If I move one end of a
metal stick, the other end can not possibly move instantaneously
together, for that would be a faster-than-light phenomenon. With a
sufficiently long, hypothetically undeformable stick, a movement at one
end would reach the other end just after some time, therefore needing
some kind of “movement wave” or “wave of deformation” running along the
object. If I suppose a very long object, and a very large initial
movement, shouldn’t this “wave of deformation” be so strong as to
disrupt any kind of material? Where lies the problem here? Are objects
in some degree always deformed with any movement? How does the
relativistic question of simultaneity of events apply here?

ANSWER:
You are right, the information which tells the end of the
stick to have an angular acceleration cannot get there
instantaneously. The way the end of the stick gets accelerated is
by the force exerted on it by atoms attached to the part of the stick
you are looking at. This would be like a "shear" force.
This shear force must propogate from atom to atom along the length of
the stick at some speed which cannot exceed the speed of light since it
is carrying information. If the required shear force gets too big
it will be bigger than the material can bear and the stick will break
or undergo some other nonelastic deformation. There is no such
thing as an undeformable object, so you need not worry yourself with
that. Your question is not unrelated to the rotating disk problem
which has been answered earlier and you might want to look at and to
check out the links given.

QUESTION:
The extremes
of "perfectly elastic" and "perfectly inelastic" collisions are
somewhat confusing as most books show something between the
extremes. For instance, a billard ball that hits the billard
table's side bumper and moves away with the same constant velocity that
it approached with would have 0 change in the ball's momentum, because:
mass of ball * (+velocity) = (mass of ball) *(-velocity), hence net =
0, assuming only x component direction. Kinetic energy is conserved,
but momentum is not? s the way to think about this collision, or
think of a small object (the ball) colliding with a much larger object
(the table). Just seems "fuzzy"?

ANSWER:
Well, perfectly elastic simply means that kinetic energy is
conserved. Perfectly inelastic is a little more subtle since it
is defined as all the kinetic energy in the center of mass system as
disappearing in the collision; it is easy to recognize, though, since
it corresponds to the interacting particles being stuck together
afterwards.

The
billiard ball collision you cite you have incorrectly analyzed.
There is not zero change in linear momentum of the ball because linear
momentum is a vector. The change is a vector away from the bumper
with magnitude 2mv . The momentum of the ball is not
conserved because there is an external agent which delivers an impulse
(force x time) to it; that agent, of course, is the bumper. If
you want to see momentum and energy both conserved, you must look at
some isolated system: imagine the pool table to rest on a frictionless
floor so that the system is the ball plus (the much more massive)
table; then the table would recoil a tiny amount and the ball would
have slightly less speed after the collision such that energy and
momentum would both be conserved. In the limit that the mass of
the table approaches infinity, the ball would have the same speed after
the collision as before but the momentum of table would be nonzero
after the collision even though it would be at rest! This is
because zero x infinity need not be either zero or infinity. This
assures momentum is conserved. Of course, there is no such thing
as infinite mass and the real world has the table essentially at rest
after the collision.

QUESTION:
Can Landua
damping occur for frequencies that are lower than 1000Hz?

ANSWER: (Provided by
M. R.
Geller)
Most likely not. Landau damping does not occur until the
plasmon dispersion relation crosses the single-particle excitation
continuum. The crossing point depends on Fermi momentum of the system,
but is usually at higher frequencies (above, say, a THz). See "Theory
of the Normal Fermi Liquid" by Pines and Nozieres.

QUESTION:
There is a question I have that's twisting my brain. Although it's in a
chemistry text, the question is really more about physics: Work is
defined : w=fd, or force applied over a distance. One set of simple
units for this would be kg.M, right? ((I am using capital M's for
meters so as to distinguish from the variable mass as in f=ma.)) Ok
now, The Joule, which is often used to measure work has units of
(kg.M^2)/(s^2)..((sorry I can't use superscripts in this e mail
format)) Now I understand that this is really mass x acceleration x
distance that (f.d=m.a.d = kg x M/s^2 x M), but my question is: how can
it be both? How can it be kg.M^2/(s^2) and kg.M, or I guess the real
question is this: My understanding that a mass unit IS a force unit.
The kg is a unit of force, right? so, how can force have units of kg
(only) in one case and kg.M/(s^2) in another? Acceleration is not
unitless, so how can you have both sets of units for the same quantity?
My understanding of the above assumption is based on the fact that I
have seen the units for work expressed both as kg.M and
(kg.M/(s.s). I hope you can help me, and I really appreciate it!!

ANSWER:
You make the classic beginning physics student mistake of
thinking that mass and weight (a force) are the same thing. In
fact, they are totally different things (although intimately
related). A force is a push or a pull and the result of exerting
an unbalanced force on something is an acceleration of that
thing. Mass, on the other hand, measures the resistance which an
object has to being accelerated (inertia). Newton's second law
related the two. Newton's second law is two things. First, it is
a statement of an experimental fact: acceleration is proportional to
the force and inversely proportional to the mass. Second, it is a
definition of the unit of force which we agree to use, viz. one
unit of force is that force which causes a 1.0 kg mass to have an
acceleration of 1.0 m/s; the unit is called the Newton (N) and has the
units of kg m/s^{2} . Force can never have the units of kg
alone. So now, force time distance is N m = kg m^{2} /s^{2}
which is called a joule (J). I think that takes care of your question.

QUESTION:
If a bullet were fired into a wooden block and the bullet went thru and
exited the other side, would this be thought of as "elastic" collision
in linear momentum terms?

The block
would then side along the horizontial surface until frictional force
consumed all of the available kinetic energy and the block would come
to a stop? If so, then the work of friction could be equated to the
block KE and the coeficient of sliding friction calculated. Correct?

ANSWER:
Elastic has a very specific meaning —energy is
conserved. Most collisions, like the one you describe, are not
elastic. But momentum is conserved; there is no analogous word
for this conservation, but it isn't really necessary because in an
isolated system the total linear momentum is a constant of the motion.

What you
are saying regarding a specific problem with a sliding block will work
but it is not quite the correct way to frame it. What you should
say is that the work done by all external forces (friction in this
case) is equal to the change in kinetic energy. That is not
exactly what you said because you implied that the friction does some
amount of positive work when in fact the friction, taking energy away
from the system, does negative work. Of course, the change in
kinetic energy (final minus intitial) is also negative, the final being
zero.

QUESTION:
Am I correct in assuming the gravity of an object is dependant on its
mass and density? A singularity in a black hole is small yet dense and
has infinetely powerful gravity. My question is, is it possible to
increase the density of an object and change its gravity? The reason I
ask this question is because of a situation that I was trying to
imagine. Suppose an experiment could be done where a relatively small
sphere who's density could be increased could produce more
gravity than the earth itself and thus pull nearby objects towards
itself. I'm assuming that theres no way to increase the density of an
object to perform the experiment but I wondered wether the actual
concept was valid.

ANSWER:
Well, first of all the gravitational force due to a black
hole is not infinite. If it were a true singularity, a point
mass, then the force would approach infinity as the distance from it
approached zero. In order to answer your question, I am going to
make a simplifying assumption: I will talk only about mass
distributions which have spherical symmetry ; the shape of the
object is always spherical and the density depends only on the distance
from the center of the sphere. So, for example, a basketball has
spherical symmetry since all the mass is on the outer surface but the
density does not depend on where you are on the surface.
Spherically symmetric objects are, at least approximately,
representative of many objects of interest in the universe. A galaxy is
an example of something which normally does not have approximate
spherical symmetry since they tend to be disk shaped spirals. If
you have a mass, say a star, the gravity outside all the mass is
exactly the same as it would be if all the mass were a point mass in
the center. Therefore, if this star were compressed down to the
size of a pea (corresponding to a huge increase in density), the
gravity outside the original surface of the star would be
unchanged. However, the gravity halfway to the center of the star
would be smaller for the star than the same place for the pea-sized
star with the same mass. Generally speaking, the gravity will be
due only to the mass inside where you are measuring it. The most
dramatic example is a spherical shell, like a basketball. Imagine
a star with all its mass on its surface; if you were inside it there
would be no gravity whatever.

QUESTION:
I have a question that i really want to know the answer to. I'm
hoping you can help. Sometime when you're watching tv or driving
on the road and you look at cars it appears that the rims of the cars
are actually rotating in a backwards motion. Why does this happen?

ANSWER :
If it happens on the road, I don't really know why; I have
never seen that. I can tell you why it happens in movies and
TV. The classic example is a wagon wheel in a cowboy movie.
The key to understanding is to recognize that a movie is a series of
still photographs in a sequence of time. Imagine a spinning wheel
with two of the spokes labeled A and B such that if you watch one place
you will see A there first and then, some time t _{0}
later you see B at the same place; so t _{0} is the time
it takes the wheel to rotate the angle equal to the angle between two
spokes. Now suppose that you are taking a movie of this spinning
wheel and the time between frames is some time t ; then, if t=t _{0}
each frame would have all the spokes in the same position so the wheel
in the movie would appear to not be rotating! [An example is
shown to the right: here the wheel rotates as you can see from the red
dot on the rim of the wheel, but it does not appear to be if you don't
watch the red dot.] If t were a little more than t _{0} ,
the wheel in the movie would appear to be going in the correct
direction, but slowly; if t were a little less than t _{0} ,
the wheel in the movie would appear to be going backwards. There
you have it.

You can
get a similar effect in "real life" if you view something with a strobe
light; some light sources have a normally imperceptible flicker (often
60 cycles per second like the line voltage powering them) which can
cause the source to be a strobe.

QUESTION:
I am researching the Hall effect for a possible innovention. I
understand the concepts well, but I have not been able to find any
information on the time sensitivity of a Hall effect measuring
device. I assume that once a B-field and voltage are applied,
that electric potential at the sides increases and then begins to level
off, approaching the final voltage. And that after some amount of
time, the measurement made would be close enough to the final voltage
to be accurate. But what kind of time delay are we talking
about? Milliseconds? Seconds? Minutes?
Hours? Could I alter ther B-field and make a new measurement
right away? Or do I have to sit on my thumbs and wait for my
little electrons to crawl to one side of the plate?

ANSWER:
Intrinsically the device should be very prompt,
microseconds. What will be much more important is the circuitry
in which the device is installed, often it will be an RC or RL
circuit which will have its own time constant. The probe, for
example, will have some capacitance which depends on its size.
So, if you are interested in superfast response, you have to worry
about all these things. Also, the response time of the device
should be listed in the specs of the probe when you buy it. I
have used Hall probes to map magnetic fields in magnetic spectrometers
and have never had to worry about how long to leave the probe before
making a reading.

QUESTION:
Is it possible to make a wind-up pendulum movement that can swing in an
arc of at least three feet and has a bob that is exposed to air as well
as water friction? I have an invention in mind and do not know if the
type of large pendulum movement I am envisioning is even possible.

ANSWER:
This is an engineering question, not physics really. In
any case, it is likely that it would be possible to do what you want
depending mainly on what you mean by water friction. It would
probably be hardest if the water the pendulum had to go through was not
a regular periodic event. But, if each swing of the pendulum
encountered identical friction, it would be straightforward to devise a
mechanism to put lost energy back into the system.

QUESTION:
I am doing self-review in electromagnetism. I was wondering how I
would set up a problem. A micron of oil with specific gravity =
0.9 and a net charge of 100 electrons is placed between two parallel
plates that are 5mm apart. I need to find the voltage required to
keep the oil suspended. I was thinking of relating the force to
the electromagnetic field. But I am confused about units.
How do I find the net downward force of the oil drop using specific
gravity? Would the charge simply be 100 times the charge of an
electron?

ANSWER:
The specific gravity of something is its density relative to
water, so this particular oil has a density of 0.9 that of water (which
is 1000 kg/m^{3} ), or 900 kg/m^{3} . So the mass
is the volume times the density and I presume that 1 micron drops of
oil mean spheres of radius half a micron. The the weight of a
drop is its mass times the acceleration of gravity; I compute the
weight to be W =4.62 x 10^{-15} N. The magnitude
of the charge on an electron is 1.6 x 10^{-19} C, so the
(magnitude of) net charge on a drop is q =1.6 x 10^{-17}
C. The force of a charge q in an electric field E
is qE , and this force must be equal in magnitude to the weight,
qE=W . I compute E =289 N/C. Since the
charge on the electron is negative, the electric field must point down
for the electrostatic force on the drop to be up.

QUESTION:
I am 45, a wee bit bright but scientifically and numerically challenged
in the traditional sense. Other than independent study, my last science
course was biology, 1973 and math, Geometry or Algebra 2, 1974. I'd
like to return to school for a degree in physics as it fascinates me.
Is there hope? As I am a believer in the school of thought "we are only
limited in our potential by our desires," I think it's possible. Yet,
it seems a bit far-reaching. What do you think? By the way, I am so
right-brained, I tilt slightly in my gait. Hmm. - Sunny

ANSWER:
Take it one step at a time. First of all, you cannot
really do much with physics without quite a bit of higher-level
math. Take a course at a local college in analytic geometry
(essentially the prelude to calculus); this course might even be
labelled "precalculus". Enlist some good advisor to make sure you
get into the right course. Then, if you like this and do well,
take two terms of calculus where you learn both differential and
integral calculus. If this goes well, take an introductory
physics course (it should be referred to as "calculus-based
physics"). You are probably two years into your program now, and
if you are doing well and loving it more than just about anything you
have done before, you are off and running! Good luck!

QUESTION:
Several questions regards SHM:

1) In
using the SHM equations to find the time required for a body to move
from one position to another in y = A*sin(w*t+theta), is it always
permissible to set theta = 0? This allows the easy solution to "t" via
the inverse sin assuming the amplitude & w (omega0 are known?

2) The
net force acting at either extreme of motion can be found from F = m*a
= m(w^2*A) where A = amplitude (or Xmax). This where the acceleration
is max & velocity is 0; however, when the body passes thru the
equilibrium position (say a vertical spring & on the way up) the
velocity is max, but the accel = 0, hence the net force = 0.
There is still a force acting (as it passes the midpoint) just to
support the weight of the body at the end of the spring..is this
correct?

3) Is the
sign convention (+ or - ), how you would know whether a body in SHM is
moving up or down? Example, when y = - 5 the acceleration becomes
= -w^2*y = (-w^2)*(-5) = positive number & the body is moving up,
while a negative would indicate that its moving down. Is this correct?

ANSWER:
1) The choice of when t =0 is always up to the person
solving the problem, just like the choice of coordinate system.
However, it may not always be the best choice for any particular
problem.

2) The
net force is (by definition) zero at the equilibrium position of the
spring. The equilibrium position of a vertical spring is
determined by the weight of the mass such that when the spring is
stretched by an amount s , ks=mg.

3) The
signs take care of themselves if you do the problem carefully.
For the example you give, it should always be such that the
acceleration is opposite the displacement; this is the basic
characteristic of a restoring force like SHM. For a given
position you cannot determine the direction of the velocity because it
is inherently ambigous. At any particular time, however, the
velocity is well determined where v< 0 means moving in the
negative coordinate direction and vice versa for v >0.

QUESTION:
I'd like to ask for some help regards how to evaluate time
requirements when working with shm equations. Here's a problem that has
a time component asked for: "A body of mass = 100 grams hangs from a
long spiral spring. When pulled down 10 cm. below its equilibrum
position, it vibrates with period(T) = 2 sec. The problem then
asks for velocity as it passes thru equilbrum position & what is
the acceleration when 5 cm above the equilbrum position. These are easy
to solve with the basic equations. However, the next step is to find "
when the body is moving upwards, how long a time is required for it to
move from a point 5 cm below its equilibrium position to a point 5 cm
above its equilibrium position. How do I get the time out of the
wave equation y=A*sin(WT+theta)? I know thru differentiation the
velocity & acceleration will fall out, but getting the time isn't
as straightforward, but I think it's asking for the body to evaluated @
90 degrees & @ 270 degrees. Is this correct or how would I proceed?
Thxs.

ANSWER:
It is all a matter of what you are solving for. You
just have to be able to invert the equation x =A sin[w t +f ] to get t =[sin^{-1} (x /A )- f ]/ w . One
important tip is to always work in radians, not degrees. For your
particular problem, I would choose f = 0 which means
that the mass is at x =0 at t =0 and moving in the +x
direction. In your problem, w = p and so
the time when the mass will first reach 5 cm is t =sin^{-1} (5/10)/p
= ( p
/6)/ p
= 1/6 s. So the
time it will take to go from -5 cm to +5 cm is clearly twice this, 1/3
s.

QUESTION:
This question is on the brink between biology and physics. When
we feel the sensations of "hot" or "cold", are we feeling heat tranfer,
or temperature? I ask this because temperature doesn't seem right
(bathroom tile feels cooler than the rug at the same temperature do to
their different heat capacities). On the other hand, it would
make sense if there were certain neural receptors that expanded as
their temperature increased. I cannot think of a mechanism by
which heat transfer could be transformed into a neural sensation.
I have had trouble getting a good explanation of this.

In short,
how do the heat receptors in our body work on a molecular level, or
where could I go to find this out?
ANSWER:
I have no idea! However, I did a Google search on heat receptors and
many sites which looked very helpful regarding details of
heat/cold/pain receptors were returned.

QUESTION:
I have been pondering a question that I have no idea how to put into
mathmatical terms. I have postulate a very large disk, in a vacuum and
no gravity, with a slot running the entire diameter of the disk. In the
center of this disk you place a ball bearing so that if the disk is
spun, the two ball bearings would move apart. Now if the disk is spun
fast enough, the ball bearing should come close to the speed of light
provided the disk is big enough and the rotation (rpm) is great enough.
Would the velocity at which these 2 ball bearing are moving away from
each other not be greater than the speed of light? If not, is there
some relationship between the two ball bearings other than the spinning
disk that prevents them from accelerating past the speed of light? What
I mean is there some relationship that exceeds the normal relationship
of two ball bearings simply because they are close to light speed? Does
this state exist if the two bearings are not at light speed?

ANSWER:
Your question is essentially the spinning disk paradox.
You do not need to introduce your ball bearings, just ask if it is
possible for a point far enough out on a spinning disk to exceed the
speed of light. The answer, of course, is no. See the
answer to a previous question .

QUESTION:
Since light vibrates when it moves it makes a zig-zag line when it
travels through spacetime right?
EXample/\/\/\/\/\/\/\/\/\/\/\/\\/\/\/\/\/\/\/\\/\/\//\/\\/\/\//\/\/\/\/\//\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
and the distance from the begining of that zigzag to the end is the
distance that it has traveled in a certain time right? But
shouldnt the actual distance be the distance of a stretched zigzag?
since it is the actual path that an atom makes through spacetime?
The zigizag I drew above if stretched could be twice as long.
Example — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —
— — — — — — — — — — — — — — — — — — — — — —-
Therefore the actual distance light travels in a certain time is longer
so the speed of light should also be longer.

THis is just
an idea and its probably been thought of already and proved wrong or
something but it has been bugging me for months and would really
appreciate an answer.
ANSWER:
Although the electric and magnetic fields are oscillating
sinesoidally, the wave itself is traveling in a straight line.
This is really no different from a water wave, for example, where the
water is oscillating up and down but the wave travels the straight line
distance between two points on the surface.

QUESTION:
I am 14 years old and my best subject is science. The question is
would I have to major in physics to get into UGA?

ANSWER:
I don't know what you mean. Getting into UGA is
determined by numerous things including SAT scores, high school GPA,
letters of recommendation, extracurricular activity etc . and
what you want to major in is relatively unimportant. Some
programs (but not physics) have strict limitations on how many majors
they will accept, but that is something which is more likely to be
enforced when you are already here than to affect your admission.
(By the way, you should include an email address with your question so
that I can communicate with you more effectively; you did not fill in
that box in the form.)

QUESTION:
If light (electromagnetic radiation) is known to be heavy,
why are the constituent photons they are composed of considered to be
massless (zero rest energy)?

ANSWER:
I do not know what you mean "known to be heavy". Viewed
either as electromagnetic waves or photons, light has no mass.
Perhaps your confusion is that electromagnetic radiation is known to
carry energy and momentum even though it has no mass. In the
(well verified) theory of special relativity, massless particles may
have kinetic energy and mass, and photons are the prototypical
relativistic particles. Furthermore, in the classical theory of
electromagnetism the electromagnetic field has both an energy density
and a linear momentum density.

QUESTION:
You mount a laser on a rigid tripod and shine the beam out a
few meters to a mirror (again, rigidly fixed) so that the beam reflects
back to a spot, X, near the laser. Will the beam continue to shine
exactly on X throughout the day (as the earth turns)?

ANSWER:
It seems to me that the only things which will affect the
path of the beam of the light are the gravitational field and the
acceleration of the reference frame (which are in some sense, the same
thing in the view of general relativity). The most important
gravitational field is that of the earth which does not change with
time and the most important acceleration is that due to the earth's
rotation which does not change. However, there are other
gravitational fields, in particular due to the sun and the moon which
do change over the course of a day because of the earth's
rotation. For example, suppose the laser were at the equator and
pointing straight down (toward the center of the earth); at dawn the
sun's gravitational field would point east and at dusk it would point
west, so the deflection of the beam of light would differ. So,
technically, the answer to your question is no. However, these
effects are so incredibly small that you could never hope to observe
them.

QUESTION:
I am a proffesional engineer who designs ASICS for the
telecommunication market. I hold a Ph.D. in Computer Engineer. I
have always enjoyed physics and am interested in finding like minded
people. Do you know of any group/club that meets on a regular basis to
discuss physics?

ANSWER:
I must say that I have never heard of a club which meets to
discuss physics; there are, of course, many clubs of amateur
astronomers which certainly will talk about many physics-related
things, so you might look in your vicinity for astronomy clubs.
Another possible avenue for what you want would be at local colleges
and universities. Two things you could do are:

Check
out the web sites of physics departments and check their program of
colloquia; these are generally intended for nonspecialists (unlike
seminars which tend to be more for specialists in subdisciplines).
Look,
again in physics department web sites, for Society of Physics Students
(SPS) chapters at local colleges and universities; these often have
regular meetings and programs at which I am sure you would be welcome.
QUESTION:
If an object hanging on a spring has its equation of motion
as y = A*sin W*t where A & W(omega) are constants & t = time,
how can I express the velocity as a function of the coordinate?
The velocity & acceleration are easy to express as functions of
time by taking the first and second derivatives of the equation of
motion. This results in v = dy/dt = WAcosW*t and accel = - W^2 A sinW*t.
It seems easy to find the accel as as a function of coordinate by
substiuting as follows: since accel = - W^2 (A sinW*t) & with Y =
A*sin W*t (original given equation of motion), then accel = -W^2*(y).
Applying the chain rule to get the velocity : a(accel) = v(dv/dy) hence
a (-W^2*y*dy) = v*dv. If this is intergrated over the limits (y= 0 to y
and t= 0 to v), I get v^2 = -(W^2)*(y^2). I now have velocity as a
function of its coordinate? Am I correct?
How could I determine the maximum velocity? Usually, this when
something is set = zero?
I've plotted the curve on an Excel spreadsheet and it is sinusodial
& by assuming different values for A & W (omega), I can easily
make the curve take on different shapes.

ANSWER:
If you are interested in velocity as a function of position
for the harmonic oscillator, it is far simpler to never get time
involved in the calculation. Just use energy conservation: kA ^{2} /2=mv ^{2} /2+kx ^{2} /2;
solving, v =[k /m ]^{1/2} [A ^{2} -x ^{2} ]^{1/2} =w [A ^{2} -x ^{2} ]^{1/2
} (this will be either positive or negative). The
maximum speed will be when x =0 and will be w A . You could
also get it from your answer, it being the coefficient of sin[w t ] above,
w A again; this is
because the maximum magnitude of the sine function is unity.

QUESTION:
Does general relativity explain an apple falling from a tree?
Even if space-time is curved, does that somehow result in an apple
falling? It seems like, for the apple to fall, a force grabs the apple
and pulls it to the ground - does general relativity explain that?
Thanks!

ANSWER:
If you are happy with the empirical explanation that the
earth has a mass and therefore causes a gravitational force which can
be felt by other masses, then there is no need for general
relativity. This is essentially what Newton did —he stated his
universal law of gravitation as an experimental fact and it then
becomes one of nature's laws if it describes nature accurately.
However, if you want to ask "why does the earth cause a
gravitational field?", that is what general relativity addresses.
And the answer to your question is that yes, general relativity does
explain why there is a gravitational field around an object with
mass —mass warps the space around it.

QUESTION:
Consider two charged particles traveling in equal but
opposite directions as shown on the website at www.hypercomplex.us/Questions/bforce.htm .
In a reference frame in which both charges are moving, the second
charge experiences a magnetic force with non-zero magnitude resulting
from the magnetic field produced by the first. The magnitude of this
force, however, becomes zero when the velocities are transformed to a
reference frame in which the velocity of either charge is zero. Why
does the force disappear given this transformation of coordinates?

ANSWER:
There are a couple of problems with this web page.
First of all, the magnetic field given is incorrect. This is not
a magnetostatic situation which means that the magnetic field is not
constant with time at every point in space. The actual field must
be computed at the 'retarded time', which takes into account the time
it takes the field information to propogate to the second charge (at
the speed of light); however, the given field is approximately true if
the speeds are small compared to the speed of light (which is called a
'quasistatic' situation). The second problem is that the Coulomb
force between the two particles is totally ignored. I would guess
that part of your problem conceptually here is that you are finding a
particle with no force in one frame and with force in another. If
you remember that there is also an electrical force, then you will not
have this problem. Because the magnetic force depends on the
velocity of the charge, then it should be obvious and not disturbing
that there will be some frame where the magnetic force is zero.
The correct transformation of magnetic and electric fields from one
frame to another is a topic in electromagnetism which is quite advanced
and the details are beyond the scope of what 'Ask the Physicist' is
supposed to be about.

Here is a
similar example of magnetic forces going away. In this case it is
the field itself which I can make disappear. Imagine a long line
of electric charge at rest. Obviously, there is no magnetic field
but there is an electric field. Now imagine making the whole line
move with some speed v ; you now have a long straight
current and therefore a magnetic field has appeared. The electric
field will increase because the charge density on the line will
increase because of length contraction.

QUESTION:
Suppose there's a lamp next to a railroad track, and a train
begins accelerating away from the lamp, and someone on the train
measures the speed of the light from the lamp. My understanding is that
if the train were in uniform motion then C would be unchanged, but if
the train is accelerating away from the lamp, will C be found to be
less than usual? If the train is accelerating towards the lamp will C
be greater than usual?

If an
observer is standing next to the track, will the accelerating train
appear shorter to the observer, and will the people on the train appear
to be moving slower than usual? If so, will those effects be more
pronounced than if the train were in uniform motion?

Did
Einstein have any theories about those things, and have experiments
been done for them? Maybe it's extremely difficult to prove length
contraction and time dilation? But maybe there's been some pretty
convincing experiments to show if C is affected when measured from an
accelerating frame of reference?

Are
experimental findings in agreement with Einstein's theories? - if he
had any theories about those things. Thanks

ANSWER:
The answer to your first question is that the speed of light is always c
as measured by any observer. The fact that the train is
accelerating is of no importance. If the train were accelerating
perpendicular to the direction of the light ray, the light would travel
in a curved path (not constant velocity) but it would move with
constant speed.

Lengths
along the direction of motion are shorter than if at rest; therefore an
accelerating train is shorter than its "normal" length and getting
shorter as it accelerates to faster speeds. I do not know what
you mean about the people on the train appearing to be moving slower
than usual; what is usual?

Length
contraction and time dilation are completely verified experimentally,
but obviously not with trains and such. Rather, elementary
particles moving with speeds comparable to c are used.
Any good book about special relativity will discuss this.
Regarding the accelerated frame of reference, the theory of general
relativity says that an accelerating frame is indistinguishable from a
gravitational field. Therefore, a beam of light bending when it
passes a massive object like the sun or another star, is identical to
what would happen if a beam of light entered the side of an
accelerating elevator and was observed to bend down as it crossed the
elevator.

FOLLOW UP QUESTION:
If a train is moving past an observer who's standing on the
ground next to the track, from the point of view of the observer, the
train would be subject to time dilation? - and from the point of view
of the observer time would be running slower on the train, and clocks
on the train would be running slower, and the people on the train would
appear to be moving in slow motion? Maybe I should be asking about at
rocket, because maybe then the speed of the rocket would be closer to
what is required for a noticeable effect?

ANSWER:
You should avoid using the word 'appear' because relativity
is not about how things look , it is about how things are .
Measurement of velocity involves measurement of both length and time
both of which depend on the frame of reference, so you have to be
careful. In your one-dimensional example (the velocity of someone
on the train is in the same or opposite direction as the velocity of
the train itself, i.e. walking up and down the aisle), there
is a simple equation which tells the velocity of the person as measured
by your 'stationary' observer. It is called the velocity addition
formula: u'= (u-v )(1-(uv )/c ^{2} )
where v is the velocity of the train, u is the
velocity of a person as measured by somebody on the train, and u'
is the velocity of the person as measured by somebody on the ground.

QUESTION:
I assume that the Surface Tension of Water is caused by
Electro-Magnetic Forces between Water Molecules? So it seems like Water
should be Attracted pretty strongly by a Magnet? - but it doesn't seem
to be Attracted very strongly, if at all?

ANSWER:
The word 'electromagnetic' encompasses all forces which
result from either electric or magnetic fields. The
electromagnetic force which is responsible for surface tension is
purely electrostatic, that is it has nothing whatever to do with
magnetism. The water molecule is a polar molecule, meaning it has
an electric dipole moment, and it is the linking of these dipoles which
results in surface tension. An electric dipole (or an electric
charge for that matter), when at rest, is unaffected by magnetic
fields. To ask what you would expect to happen if you put a
magnet nearby, you must ask what the magnetic properties of water
molecules are. I recently answer ed a
question about how water is affected by a magnet.

QUESTION:
If we put a 2 masses in the space (vacuum where there is not
any kind of gravity field) Is this 2 masses attract to each other? If
yes, Why?

ANSWER:
This is basic Newtonian physics. Newton's universal law
of gravitation states that any two masses exert gravitational forces on
each other which are proportional to the product of the masses and
inversely proportional to the square of the distance between them, F=Gm _{1} m _{2} /r ^{2}
. However, unless at least one of the masses is very large, the
force will be very small because gravity is nature's weakest
force. This is reflected in the constant G being very
small, G =6.67 x 10^{-11 } N m^{2} /kg^{2} .
So if you put two 1 kg masses a distance of 1 m apart, each would
experience a force of 6.67 x 10^{-11 } N and so each would have
an acceleration near the other of 6.67 x 10^{-11 } m/s^{2} .
To put this in perspective, it takes something with an acceleration of
6.67 x 10^{-11 } m/s^{2 } (and starting from rest) a
time of about an hour and a half to travel 1 mm!

QUESTION:
Here's a question about the Length Contraction Aspect of the
Special Theory of Relativity - say there's a Railway Car sitting
Motionless on a Track, with a Lamp on the East End of the Car - the
Lamp is Switched On and the Time is Measured for the Light to Reach the
West End of the Car - then the Car begins Rolling Eastwards, and again
the Light is Switched On and the Time Measured again - this time it
Takes Less Time for the Light to Reach the West End of the Car, so the
Special Theory would say that, since C is Constant, the Length of the
Car has Changed - the Car's Motion caused the Car to become Shorter?

Is all
that right? - if so, then what about if the Car begins Rolling West,
and again the Light on the East End of the Car is Switched On - this
time it would take Longer for the Light to reach the West End, so by
the same Reasoning, it seems like the Special Theory would say that now
the Car's Length has Increased!

Is that a
Flaw in the Special Theory? - maybe the Wording should be something
like "Sometimes an Object in Motion becomes Shorter, but Sometimes it
becomes Longer"

Do I
deserve a Nobel Prize? - or am I Overlooking Something?

ANSWER
I will answer your last questions first —no, you do not
deserve a Nobel prize! And it is not so much that you have
overlooked something, but that you have computed the length of the car
incorrectly. In order to measure the length of something, you
need to measure the positions of each end at the same time, not
different times. So first you must use special relativity to get
length contraction so that you know what the length of the moving car
is and then you can analyze your experiment.

You
cannot deduce the length of something by simply measuring the time it
takes light to travel between its ends (which is what you have
discovered here!). Let's do your problem but do it
correctly. Suppose we call the length of the car, in its own rest
frame L _{0} . Then, if it has a speed v relative
to some observer, its length as measured by that observer is L=L _{0} [1-v ^{2} /c^{2} ]^{1/2} .
For your first experiment (car at rest) the time is t =L _{0} /c .
For your second experiment, (car moving east) let us call the time t _{1} .
Looking at the first picture (dashed lines show where the car has moved
to when the light, red, strikes the other end), it is clear that L=L _{0} [1-v ^{2} /c^{2} ]^{1/2} =vt _{1} +ct _{1} ,
so t _{1} =t [1-v /c]^{1/2} (you will
have to do a little algebra to get here). t _{1}
is, as you state, indeed less than t but not because L
is shorter than L _{0} , although it is. In your
third experiment, refer to the second picture and call the time t _{2} .
Now we have that L=L _{0} [1-v ^{2} /c^{2} ]^{1/2} =ct _{2} -vt _{2} ,
so t _{2} =t [1+v /c]^{1/2} . So
now, t _{2} >t but not because the car is any
different length from experiment #2 (it has exactly the same length);
rather because the light had to travel farther than the length of the
car.

QUESTION:
How will light behave when entering a cone shaped
prizm? Assuming the light enters through the bottom of the
prizm. Will the light exit the prizm through the sides or just
bounce around and refract until it exits through the bottom of the
cone? If it does exit through the sides of the cone will it exit
with an even distribtion from the cones sides?

ANSWER
I will assume that you mean that the light enters the base of
the cone with the rays perpendicular to the base. The answer,
essentially, depends on two things —the angle of the cone and the index
of refraction of the material from which the cone is fabricated.
There are additional details which would depend on the polarization of
the light, so let us assume that the light is unpolarized; I won't go
into the details which depend on the polarization. A ray strikes
the inner surface of the cone with an angle of incidence
q _{i} which you can see is
determined by the angle of the cone a , q _{i} =90^{0} - a . Part of the light will
be reflected (blue in the figure) at an angle equal to the angle of
incidence; this ray, it may be shown, will be (partially) reflected,
possibly bounce once more on the inner surface, but eventually reemerge
back out the bottom exactly opposite the direction in which the
incident rays are entering. But also, part of the incident ray
will emerge (red in the figure) from the cone at an angle
q _{r} which is larger than
q _{i} . This is called
refraction and the value of the angle depends on the index of
refraction of the cone, n : sin q _{r} =n sin q _{i} . The relative
amounts reflected and refracted depends on q _{i }
and n and on
the polarization. Also, if the incident ray is unpolarized as we
have stipulated, the refracted and reflected rays will not be, so the
fractions on the next bounce will not be the same. So, you see,
the details will be very complicated. There is one very important
special case. Suppose that n sin q _{i} >1; then, since sin q _{r} cannot be bigger than
1, there will be no refracted ray. This is called total internal
reflection and all the light will be reflected back in the direction it
came from.

QUESTION:
How does absorption occur in paper towels?

ANSWER
The short answer is capillary action, the tendency of water
to creep along very narrow passageways. This is how sap rises in
very tall trees and how blood moves through very narrow veins and
arteries in your body (called capillaries). You can find out lots
more about your question by doing a google
search on "paper towels" AND capillary .

QUESTION:
My question is rather simple, but has created much
controversey among my friends and I. When I am at an amusement park, I
often wonder... Which part of the rollercoaster goes faster? The front,
or the back? I immediately come to the conclusion that the
rollercoaster MUST be going the same speed, in the front and the back.
My friends disagree with me, and say that the back is going faster. I'm
pretty sure that this is untrue. Could you explain the reason our
confusion, and give a brief answer.

ANSWER
For a roller coaster, every car moves with the same
speed. The reason is that all the cars are constrained to move on
the track. If they were not constrained to move on the track,
however, one part of the train could easily move with a different speed
from another. Imagine, for example, that the train rotated around
the front car which was standing still (except for its rotational
motion); then each car, as you moved toward the other end, would be
moving faster than the one before. When I was a kid we used to
play "crack the whip" when ice skating: a long line of kids would skate
across a pond and the kid at one end would essentially stop and the
kids near the other end would get a very exciting and fast ride!

QUESTION:
If you bring a compass to the right side of a bar magnet (the
north is on the right, and the south is on the left of the bar magnet),
which way should the compass needle point? To the left (away from the
north pole of the magnet), or to the right (so it touches the north
pole of the magnet)

ANSWER
A compass is itself a bar magnet and the end which points
north is, by convention, called the north pole of the compass
(magnet). Now, as you probably know, like magnetic poles repel
and opposite magnetic poles attract. Therefore the north end of
the compass will be repelled by the north pole of the magnet so the
compass will point away from the magnet. Incidentally, note that
the convention means that the magnetic north pole of the earth is
actually a magnetic south pole if the north pole of the compass points
to (is attracted to) it.

QUESTION:
When an atom releases energy as a photon, does the photon
travel along only one direction? I am thinking that since light
can also be interpreted as wave, then the emitted light may be a
spherical wave front in 3D space. This is my dilemma.

ANSWER
The description of the properties of the photon depend upon
the quantum state in which it is put by some measurement. For
example, if you know nothing about the direction in which the photon is
traveling, it is equally probable that you will find it going in any
direction so it is like a perfect spherical wave. On the other
hand, if you know the direction perfectly then you will, according to
the Heisenberg uncertainty principle (HUP), be totally ignorant of the
position so it is like a perfect plane wave. The classic example
is to put a slit in the way of a beam of photons to try to pin down
where they are and in fact you destroy that information because of the
HUP. This is because a slit of width D x
means that the x
position of a photon which passes through is uncertain by an amount
D x
so the momentum in the x
direction will be uncertain by an amount approximately equal to
Planck's constant divided by D x. Therefore, the
smaller D x
is the more uncertain
you are of the direction in which the photon is going after passing
through, it is more spherical (actually cylindrical for a slit) like in
wave language.

QUESTION:
What are the physics behind the fun of an air hockey table?

ANSWER:
Well, I do not know much about air hockey, but the most
important feature of the game, physics-wise, is the fact that the pucks
slide with very little friction. This is achieved by having many
small holes on the board through which air is squirting. The puck
then slides on a very thin cushion of air which causes there to be
negligible friction. Thus, as the puck slides across the board it
moves with approximately constant velocity until something exerts a
force on it —a wall or a paddle.

QUESTION:
My name is Sam and I am 11 years old. Where can I find
someone to help me with my science questions? My mom told me to find
some web-sites and e-mail to see if I could get any suggestions. I'm
researching photons. My science book and most of my research states
that photons do not have mass. This interested me since photons have
energy, are affected by gravity, and (I'm still researching this) can
be split. If you have any suggestions for my research please
reply.

Thanks,
Sam and Sam's mom (who won't let Sam into cyberspace unchaperoned)

ANSWER:
Hi Sam! Answering questions of hungry young minds like
you is the primary reason for this web site! Here is a brief
overview of photons:

In the
beginning there was light! You know what light is,
electromagnetic radiation which includes visible light, radio waves,
x-rays, etc . The study of the physics of light is one of
the oldest areas of physics and is called optics. Optics existed
long before anybody actually knew what light is (electromagnetic
radiation). But the study of electricity and magnetism (E&M)
yielded the wonderful four Maxwell equations (in the last half of the 19^{th}
century) which summarized everything there is to know about
E&M. Now, here is the remarkable thing: when one manipulates
Maxwell's equations there emerges equations which describe waves of
oscillating electric and magnetic fields (in sound waves, air
oscillates; in water waves, water oscillates, etc .). And
guess how fast these waves travel —exactly with the speed of
light. It was, before now, known that light is a wave, but nobody
knew what was "waving". Now it turns out that the speed of light
is a universal constant —no matter who measures the speed of a beam of
light, everybody gets the same answer. This is amazing, if you
think about it; for example, suppose that you were going 50 mi/hr down
a road and somebody else going the other direction was going 100 mi/hr
(the speed somebody standing by the road sees) toward you. Then,
you see that other car approaching you with a speed of 150 mi/hr,
right? That would not be right if 100 mi/hr were the speed of
light: both you and the guy by the roadside would see the same
speed. I know it is hard to swallow, but it is true. This
gave rise to the theory of special relativity (by Einstein) which
revolutionized the way we think about physics (around 1905).
Around this same time another revolution in physics was happening, the
discovery of quantum physics (by Planck) which found that nature is not
a smooth continuous thing but rather discretized or kind of
grainy. For example, a system of particles which is bound
together (like an atom or a solid or the solar system or a galaxy)
cannot exist in any old energy state it wants but only is certain
states. You never see anything like this when you look, e.g.
at the solar system because the states are very close together and we
can't see anything but smoothness; but when you start to look at tiny
things like atoms, you start to see these "quantizations".
Anyhow, it turns out that one of the consequences of relativity and
quantum physics is that light is both waves and particles
(called duality by physicists). If you look for a wave you will
find one; but if you look for a particle you will find a
particle! In your research you should find out about the photoelectric
effect which was the first instance in which the particle nature of
light was actually apparent. (Again, it was Einstein who
explained the photoelectric effect. It was this achievement, not
relativity, for which he won the Nobel prize.) Particles of light
are called photons. Now, by definition, photons must travel with
the speed of light (they are light!). But relativity says that
the only things which can travel with the speed of light are things
with zero mass. Therefore, photons must have zero mass.
There is no doubt whatever of this fact! Yes, photons do have
energy and momentum even though they have no mass; this is allowed in
relativity. And photons do feel gravity even without mass; this
is a consequence of the theory of general relativity which Einstein
formulated about ten years after the 1905 theory of special
relativity. When you are doing your research, also look up the Compton
effect (the change in wavelength of photons when they scatter from
electrons) which is one of the seminal confirmations of the notion of
photons. The energy of a photon is hf where h is
a constant (called Planck's constant) and f is the frequency of
the corresponding electromagnetic waves; you may think of this as the
smallest amount of light of a particular frequency which is
possible. Yes, a photon can be split in two but since energy must
be conserved, the "colors" of the two new photons will be "redder" than
the original. A recent reference for this is http://focus.aps.org/story/v10/st3 .

I realize
that this is sort of a rambling "stream of consciousness" answer to
your question. Hope it is of some help!

QUESTION:
I have worked out enough "projectile motion" problems such
that I think I understand the principles, but there is one type that I
can't nail. It's where a ball rolls off the upper floor and then you're
asked to tell which step the ball will land on? For instance, a
ball rolls off upper landing at 5 ft/sec in the horizontial direction,
and the steps are 8" high by 8" wide. It asks which step the ball will
land on?

The
problem tells us that initial velocity is horizontal & hence the
launch angle (theta) = 0 degrees so sin (theta) = 0 & cos (theta) =
1. We also know that the initial velocity is constant along the X
(horizontial) axis. We know that the Y axis velocity component = 0
initially, but begins to fall at rate(v) = -g*t (gravity x time) and
the distance (y) it falls = -1/2gt^2.

We don't
know time required & I can't make it fall out, and without the time
I can't find the X distance... which is what I want to know, as this
will tell which step the ball will land on.

What am I
missing? I've not connected with something, but what?

ANSWER:
Imagine a ramp running down the stairs (just imagine a board
laid on the stairs). Its equation is y =-hx /w
where h is the height of a step and w its width.
The equations of motion of the ball are

x=v _{0} t
y=-gt^{2} /2.

Now the
ball "hits" the ramp when -gt^{2} /2=-hx /w.
But, since x=v _{0} t, we can get t =2hv _{0} /(gw ).
Now, knowing t you can get x : x= 2hv _{0} ^{2} /(gw ).
Once the ball crosses the ramp it must hit the next available step; x /w
will be the number of steps when it crosses but it will (except for
very special cases) not be an integer since it will have passed a
fractional step. You just need to go up to the next integer from x /w= 2hv _{0} ^{2} /(gw ^{2} )
to get the number of the step which is hit. For your particular
problem, v _{0} =5 ft/s, h=w= 0.75 ft, g =32
ft/s^{2} , so x /w= 2x0.75x5^{2} /(32x0.75^{2} )=2.08,
so the ball lands on the third step.

QUESTION:
How does a magnet interact with water?

ANSWER:
Water is diamagnetic which means that it is repelled by a
magnet. However, the force is extraordinarily small and therefore
difficult to observe.

QUESTION:
Police agents flying at a constant speed of 200 km/h
horizontally in a low-flying airplane wish to drop an explosive onto
master criminal’s automobile traveling 130 km/h on a level highway 78.0
m below. At what angle (with the horizontal) should the car be in their
sights when the bomb is released?

ANSWER:
This sounds like a homework problem to me! Are you
telling me you just happened to wonder about this situation?
Anyhow, dropping means that the initial conditions for the velocity of
the explosive are an x -component of its velocity of 55.6 m/s
(200 km/hr) and a y -component of zero. I choose a
coordinate system with the origin at ground level and directly below
the airplane when the explosive is dropped, so the initial conditions
for the position are zero for x and 78 m for y .
We can now write the kinematic equations of motion (I presume you know
how to do that) for the explosive:

x= 55.6t
y= 78-4.9t .

Now, we
are interested in where the explosive hits the ground because that is
where the car is. So set y= 0 to get the time.
Solving, t= 4 s. Knowing t , we can say where the
car must be: x =55.6 x 4=222 m. Now, what is the equation
of motion for the car? Since we do not know where it is (that is
the objective of the problem) when the explosive is dropped, let's just
say that it is x _{0} . Then the equation of motion
is

x =x _{0} +36.1t

where
36.1 m/s=130 km/hr. Now, we want x =222 m when t =4s;
solving, x _{0} =76.4 m. There is probably some
roundoff error, but this is close. Finally, to express the answer
in terms of your question, the tangent of the angle of the line of
sight to the car is 78/76.4 below horizontal; pretty close to 45^{o} .

QUESTION:
I'm curious (and you may have been asked this recently,
but...) In martial arts, what are the components which influence
the damage done by a strike, what are thier relationships, and most
importantly, a laymans explanation as to why (sounds like a thesis
level question, don't it? :-) ?

ANSWER:
I haven't been asked this question in this context, but if
you carefully read the q&a just below you will get all the
information you need to answer your question. Just as in that
question, you want to deliver the maximum possible force over the
minimum area to do the maximum damage. [Below I only briefly
touched on the 'minimum area' aspect, but it is pressure, force per
unit area, that counts. For more on this see my discussion
on the 'penny falling from the skyscraper' question.] So, using
the edge of hand, as in karate, is more effective than using the flat
of your hand or than using your fist (other things being equal).
Also, to deliver the maximum force you need to maximize the momentum
transfer as described below (for example, it is better if your hand
'bounces off' than if it doesn't). And, the time of the blow
should be minimized to maximize the force. A good example of this
is using a hammer to drive in a nail: if you watch a master carpenter,
you will notice that the hammer bounces back with each blow and (of
course) each blow is of extremely short duration; someone less skilled
tends to have the hammer stop. Have you ever tried to push a nail
into a board with your hand?

My
discussion has focused on the force which the hand or foot delivers to
whatever it is attacking. However, a very important principle of
physics is Newton's third law which states that if one thing exerts a
force on another, the other exerts an equal and opposite force on the
first. Therefore, the hand or foot experiences exactly the same
force as it delivers. Which is why, of course, sometimes boxers
break their fingers. So I guess that one of the skills in martial
arts is knowing which areas of the body are most vulnerable so that a
given force there will do a lot of damage but an equal force on the
hand would not.

QUESTION:
A physics teacher asked a class this question, but I wasn't
around for the answer (not my class).
Start with two objects
of mass m, travelling at velocity v, neither of which can penetrate a
particular target. Now, double the mass of one object, and the velocity
of the other, while leaving all other characteristics unchanged.
Which one is more likely to penetrate the target, or are they equally
likely? The class seemed to be divided into two categories, those
who thought that kinetic energy was the key, and that therefore
doubling mass doubled the impact, but doubling speed quadrupled the
impact force, and those who thought that the two were equal.

ANSWER:
What is important in breaking through the target? The
most important thing is the force which is applied to the target.
Of course, also of importance is the area over which the force is
applied, so a sharply pointed object which exerts the same force as a
blunt object is more likely to break through; but I assume that the two
objects have identical geometries and compositions. So we need to
focus on the force which a projectile exerts on something during the
time of impact. Newton's second law (N2) is usually thought of,
particularly by students just learning physics, as force equals mass
times acceleration, F=ma ; however, when collisions are being
studied, it is much more useful to think of N2 as force equals time
rate of change of linear momentum (which is mass times velocity, p=mv ),
or the force (F _{avg} ) averaged over the time of the
collision (D t ) is change in
momentum of the projectile (D p ) divided by
D t, F _{avg} =D p/ D t .
[Incidentally, it should be clear to you that the two forms of N2 are
the same since, if the mass does not change, the rate of change of mass
times velocity is mass times the rate of change of velocity and rate of
change of velocity is acceleration.] So, if we are interested in
force we have to think about both the change in momentum and the time
it takes for that momentum to change; big D p
or small D t means big
force.

My guess
is that your physics teacher wants to teach the lesson here that the
important thing is momentum, not kinetic energy, so let me talk about
momentum and then I will add a little bit about time. What is the
range of possible changes of linear momentum if the projectile does not
break through (the first experiment)? The least it can change is
if the projectile comes to rest after the collision, in which case
D p=mv ; the most
it can change is if the projectile bounces back elastically, in which
case D p= 2mv (don't
forget that momentum is a vector quantity). For simplicity's
sake, let's just assume that both projectiles stop in the first
experiment (because if they bounce back they will never, by definition,
go through!). Then, in the second experiment we will double the
momentum of both projectiles so they are equally likely to break
through if both collisions take the same amount of time .

But, are
they likely to take the same amount of time? No, because the
projectile with the doubled speed is likely to take longer to transfer
the same amount of momentum as the projectile with the doubled mass and
thus will exert less force . Therefore, I believe that the
likeliest one to break through is the projectile with the doubled mass.

QUESTION:
Two capacitors in series have a neutral middle. The right
plate of the left capacitor is connected to the left plate of the right
capacitor. This section is isolated from the battery. Charges in this
section move because the plates connected to the battery are charged by
it. Free elctrons in the neutral middle are effected by the generated
field. The neutral middle becomes polarised and the charge on each
capacitor becomes equal.
How long does this polarization take given that drift velocities are
very low for electons in wires? Would the time be significant? The RC
constant in a series circuit seems to be reduced because C equivalent
is less)not increase becasue of drift through the middle???

ANSWER:
First of all, it is really inaccurate to say that the
"middle" is isolated from the battery. As soon as the plates
connected to the battery begin to charge up, they give rise to an
electric field which the middle plates feel. This field is what
causes charge to flow from one of these middle plates to the
other. I would say that the battery is causing the fields so the
middle plates are "seeing" the battery. The time constant has
nothing at all to do with drift velocity of the electrons. You
could just as easily have asked why the charges get to the plates from
the battery so quickly. The fields in the wires set up by the EMF
of the battery cause all the electrons in the wires to begin moving
instantaneously (actually, these fields propogate with the speed of
light, so it isn't quite instantaneous); you don't have to wait for the
electrons to travel the distance from the battery to a plate or, for
the "middle" plates from one plate to the other.

QUESTION:
what do we mean by friction? i.e-friction is defined as a
force that apposes the force applied by a person on a body, Many books
agree that when ever 2 bodies are in contact and if they rub together,
the junctures or the uneven surfaces, get smoother and even, so we can
say that friction is proportional to force applied if and only if 2
bodies are in contact, then how are we able to make precise
calculations about it.

ANSWER:
When two objects exert contact forces on each other there may
always be two components of the forces, one perpendicular to the
surface of contact and one parallel to the surface of contact.
The parallel component is usually called friction and the perpendicular
component is usually called the normal force. In many cases of
interest, the magnitude of the frictional force is proportional to the
magnitude of the normal force; the proportionality constant is
determined by the nature of the surfaces and is called the coefficient
of friction. This proportionality is only approximate, certainly
not a law of physics or anything like that. To understand
friction microscopically, at the atomic level or even at the level of
the roughness of the surfaces, is a very difficult problem and it is,
in fact, not well understood at all. I previously
answered a question about friction which you might want to look at.

QUESTION:
In physics we have classified all the forces into 4
categories as said in feynman's lectures-nuclear forces, gravitational
forces, electromagnetic and weak interaction..
But where do we categories forces like friction and push or pull. They
don’t satisfy any category.

ANSWER:
These forces are what are normally referred to as contact
forces, forces which depend on the contact between two objects which
are exerting forces on each other. For example, when your hand
pushes on a ball the atoms in your hand are interacting with the atoms
on the ball and the fundamental force by which atoms interact with each
other is the electromagnetic force; essentially the electrons in your
hand repel those in the ball.

QUESTION:
I wish to discuss a technicality in a sound wave problem. The
problem states: "Suppose that a sound wave is passing through a gas.
Does a change in the temperature of the gas affect the wavelength of
the sound wave if the frequency remains constant?" I understand that
wave speed=wavelength*frequency but changes in the temperature in a gas
must affect the speed at which sound travels through it and the
frequency and wavelength must change too right???? Is the problem
stated falsely?? Thank you for your time and consideration. I greatly
appreciate your help.

ANSWER:
The frequency of sound is determined by the source and is
completely independent of the medium through which the sound
passes. Even if there were no medium and therefore no sound, the
frequency of a 440 Hz tuning fork would still be 440 Hz. If the
frequency if f , the source causes the medium to compress every
1/f seconds. Now, it is well-known that the speed of sound
in a gas depends on the temperature of the gas. Since, as you
state, v= l f,
if v changes
and f does not, then l must
change.

QUESTION:
I know that a rocket has to have enough thrust to accelerate
itself to about 25,000 mph to escape the Earth, but why that speed?
Let's say that in some futuristic vessel, it had 100 times more power
than the fuel-propelled rocket and had an almost unlimited source of
fuel. Couldn't this vessel just push it's way up gradually out of the
Earth's atmosphere at , let's say, 1000 mph?

ANSWER:
What escape velocity means is the speed a projectile would
have to have at the surface of the earth such that when it gets
infinitely far away it will just come to rest. It is understood
that "projectile" means something with no propulsion. However, if
you have propulsion, velocity is irrelevant. If you had adequate
fuel you could drive off to infinity going 1 mile per hour; it would
just take a lot longer than if you were going faster. If you got
going faster than escape velocity at the start, you would still have
some velocity (i.e. some kinetic energy) when you were
infinitely far away.

Incidentally,
infinitely far is a "mathematical" construct here. What it really
means is very far away, far enough that you could not detect the
gravitational force of the earth with any reasonable experiment.

QUESTION:
At science centers, you can usually find the activity with
the bicycle wheel and a spinning chair? Can you explain the gyroscopic
precession principles involved with that activity and what makes you
spin when you change the angle of the bicycle wheel? Also, can you
please explain that using simple language...most websites I have
visited about this question are difficult for a layman to understand.

ANSWER:
Precession is not an issue here; rather, conservation of
angular momentum must
be appreciated. Imagine that you are holding a spinning wheel
with its axel pointing vertically and that you are standing on a
turntable which can rotate without significant friction. Although
there are external forces on you (your weight, the weight of the wheel,
and the force the table exerts up), none of these forces exerts any
torque on you and the wheel. The total angular momentum of a system
which has no external torques on it must never change which is what it
means to say that the angular momentum is conserved. Angular
momentum is a vector quantity. If you look down on the spinning
wheel and it is spinning counterclockwise, it has an angular momentum
vector which points up and has a magnitude which I will call L.
You are not spinning and have an angular momentum, therefore,
of zero. Therefore the total angular momentum of you + the wheel
is L upward and this is what it must always be. Now you
flip the wheel over so that the angular momentum of the wheel is L
downward , so the angular momentum of the wheel has changed by an
amount of 2L downward. But, the total angular momentum
cannot change so an angular momentum of 2L upward must pop up
somewhere; the only part of the system which can get angular momentum
is you, so you must start spinning such that your angular momentum is 2L
upward which means, since up means counterclockwise as seen from
above, you will spin counterclockwise as seen from above. Here is
a movie .

Incidentally,
if you care, conservation of angular momentum is simply a consequence
of Newton's first law in its rotational form. In translational
form, Newton's first law says that, if there are no external forces on
a system, the linear momentum will be conserved; in rotational form it
is that, if there are no external torques on a system, the angular
momentum will be conserved.

QUESTION:
Xrays should be called VEXrays!
An electron is accelerated under the influence of an electrostatic
potential, generally in the neighborhood of 60-80 kilovolts under a
hard vacuum. The electron collides into a target atom, causing an
orbital electron to momentarily absorb this kinetic energy.This
energised electron relaxes back to its ground state producing a quantum
x ray photon.X ray photons have great energies compared to electons
accelerated to speeds of 60 kilovolts. Adding to this quandry,
the target electrode heats up, seemingly stealing energy, from this
process, generating Xrays. I try to envision this phenomenon using
billiard balls. Not a chance! Never hit an object ball to see it roll
away faster or spin faster than the ball I hit it with.
Sincerly! VEXED!

ANSWER:
Actually, you are wrong. If you send in 60 keV
electrons you will see all photons coming out as having less than or
equal to 60 keV. Energy must be and is conserved when electrons
interact with atoms. Also, classical physics has no problem with
the particles coming out having a greater velocities than particles
coming in: for example, imagine bowling balls with a speed of 5 m/s
colliding with bb's at rest. After the collision the bb's will
have speeds near 10 m/s.

QUESTION:
What is energy? A usual definition (the ability to do work)
only tells me what it can do.

ANSWER:
This is a nice succinct question. It certainly does not
have a succinct answer, though! Books have been written to answer
this question. I will hit a few high points:

Perhaps
you would like your "usual definition" better if it were turned around:
energy is what you get if you do work. For example, if you do
work by pushing on some block resting on a smooth horizontal surface
you end up with energy in the form of kinetic energy plus any
temperature increase caused by your doing work against friction plus
the energy associated with any sound that came out while you were
pushing. Actually, all this energy is kinetic energy because
temperature is a measure of the average kinetic energy of the molecules
and sound is wave motion of air molecules. Kinetic energy is K=mv ^{2} /2
(m is mass and v is speed) and, if friction and sound
are negligible, the kinetic energy you end up with is numerically equal
to the work you do.
Potential
energy is energy which you get when you do work but may not convert all
that work into kinetic energy. For example, if you lift a block
from the floor to a table which is a height h above the floor
you have done work but not increased K. The potential
energy in this case is U=mgh (g is the acceleration due
to gravity) and if other effects are negligible, the potential energy
you end up with is numerically equal to the work you do.
Since
the discovery of the theory of special relativity nearly 100 years ago,
we now understand that mass itself is a form of energy. That is,
you may destroy a little bit of mass and that amount of energy will
appear (usually in the form of kinetic energy). For example, in a
nuclear power plant uranium nuclei split and convert a bit of mass into
the kinetic energy of the steam used to power the turbines used to
generate electrical energy. The relationship is the famous E=mc ^{2
} (c is the speed of light, 3 x 10^{8}
m/s), so, if you create a mass m , you must do an amount of work
equal to mc ^{2} .
Perhaps
it is reasonable to ask: why are physicists so enamored with the
concept of energy? The reason is that it is a conserved
quantity . In an isolated system, the total amount of energy
never changes, it just changes from one form to another. This
turns out to be very powerful in understanding the physics.
I hope
this is a little enlightening for you.

QUESTION:
I don't knwo if anyone if farmiliar with "A Brief History In
Time," but it has leaft me confused. Can you please explain the 4
dimension rule. How Time and Distance are related. Also how a clock on
the ground of earth would be slower than one on the top of a mountain.
I understand the slower frequency rule due to gravity, but i don't
understand how that affects how the clocks tell time. Their mechanisms
which keep time shouldn't be related to waves.

ANSWER:
You need to learn the theory of special relativity. It
requires only algebra to understand its basic concepts.
Essentially, if we demand that the speed of light in vacuum is the same
for all observers (by now a well-documented experimental fact), the
inescapable conclusion is that space and time are not separate entities
but are "entangled". The "entanglement" is much like that of the
three spatial dimensions. An example is the rotation of a
coordinate system shown at the right. Note that the components of
the vector in the black coordinate system (x,y ) are not the same
as in the rotated (red) coordinate system (x',y' ).
In fact, (x',y' ) depend on what (x,y )
are:

x'=x cosq +y cos q
y'=-x sin q +y sin q .

The
rotation "mixes up" the spatial dimensions. It turns out that in
special relativity, the transformation between space and time if one
system is moving with speed v with respect to the other looks
very much like a rotation in "four-dimensional space" because x
(the direction of the relative velocity) and t get mixed up:

x'= g
(x-vt )
t'= g
(t-vx/c ^{2} )^{
} g
=(1-(v/c )^{2} )^{-1/2}

The fact
that this transformation mixes up these two dimensions leads us to
recognize that space and time are no longer separate unrelated
concepts. Instead, we should think of "space-time" which includes
both time and the three dimensions of space, thereby suggestive of a
four-dimensional world.

Regarding
your second question, I found an interesting
example on the web which might be instructive for you.

QUESTION:
Quantum physics is my weak spot so treat me like I have never
even heard the word quantum before. How is it that in the two slit
experiment (assuming you are farmiliar with it) there can be more than
two lines on the screen behind the two slits if there is only one white
light source? Also Can you embelish the uncertainty principle for me?
Thank you.

ANSWER:
First, you don't want to consider a white light source since
it contains all wavelengths which considerably complicates the
problem. Understanding the double slit experiment hinges first on
something called Huygen's principle: every point on a wave front acts
like a new point source. Hence, when you put a narrow slit in
front of a source of light, it behaves like a line source and emits
cylindrical wavefronts. If there is another slit parallel to it,
the other slit is a source of waves in the same way. Now you have
to accept that when waves from the two sources hit some place in space (e.g.
a screen in front of them), the net result will be the simple sum of
the two waves. Hence, if the crests (wavefronts) of these two
waves arrive at that point simultaneously, the resulting waves will be
twice the amplitude of one source (bright), whereas if the crest of one
arrives at the same time as the trough of the other, they will cancel
and there will be no wave there (dark). A nice simulation may be
seen by clicking
here . Of course, with light this has nothing to do with
quantum mechanics —it is called the Young's double slit experiment and
has been known for 200 years. See also the answer to a previous question. When
quantum mechanics was beginning to be developed in the early 20^{th}
century, the notion that things we traditionally think of as particles
might behave, under the right circumstances, like waves, was verified
using exactly the same ideas as the double-slit experiment.

"Embellish"
implies that I already know what you know and add to it! The
uncertainty principle states that, as an inescapable consequence of
quantum mechanics, there are certain pairs of observables (called
conjugate variables) which cannot both be known to arbitrary
precision. The best known pair is the momentum and position of a
particle. One may get the idea from the ideas of the slit
described above. Perhaps a small hole would be more instructive;
here Huygen's principle says that the hole, if very small, acts like a
point source so there are spherical wavefronts coming out of the
hole. But what does the hole try to do? The incoming waves
come straight into the mask but with a completely undetermined
position. The hole localizes the position of the wave but the
result is that the wave spreads out spherically so now we have almost
no determination of its direction (momentum is essentially equivalent
to velocity for our purposes here). So we start with no position
information and perfect direction information and end up the other way
around! If you make the hole larger, more of the light will go in
the forward direction coming through it (you will get a fuzzy image),
so there is always a compromise between knowledge of position and
velocity. This is a result of the wave nature of what you are
looking at and, when you start thinking of particles as waves, they
take on this uncertainty property as well. We never see particles
doing this in everyday life because the product of the uncertainties of
momentum and position is on the order of 10^{-34} kg m^{2} /s.
So, if you have a 1 kg ball whose uncertainty in position is 1 mm=10^{-3}
m, you could (according to the uncertainty principle) measure its speed
as accurately as 10^{-31} m/s. Since it is absolutely
impossible to make such an accurate measurement, you would never notice
your inability to measure it perfectly.

QUESTION:
Why is it that if hypothetically, two twins were placed in
different situations, where one is put on a space ship traveling at the
speed of light, and the other stays on earth, that the twin in space
wouldn't age as fast? Wouldn't the body clocks remain the same?

ANSWER:
First of all, if you believe special relativity, and all
physicists certainly do and it has been irrefutably proven many times,
this is not hypothetical —the twin who went on the trip would age more
slowly (never mind that we do not have the technology to accelerate
large objects like you to speeds comparable to the speed of
light.) Also, nothing can move with the speed of light except
massless things (like light itself), so the traveling twin would, as I
said, travel with a speed comparable to but smaller than the speed of
light. You seem to think that the body clock is somehow different
from all other clocks. It is different only in that many
different bodies age at somewhat different rates so that two 60 year
old people can appear to have aged very differently. But, on
the average , our clocks run down somewhere around 80 years. So, if you
have a person and he has a clock with him in his spaceship, and the
clock ticks off 20 years while he is gone, he certainly will not appear
to be 80 when he returns. His body will run at the same rate as a
mechanical clock he carries with him. Regarding the details of
your question, I have answered it before —click here .

QUESTION:
I have always been fascinated with the interrelationship of
matter and energy. In the excellent book entitled, "The Physics of Star
Trek" by Lawerence Krauss, he discusses about the transporter and how
it works. It basically converts crew members into energy and once that
energy is sent to it's appointed destination, the energy is reconverted
back into matter.

Granted
Star Trek creator Gene Roddenberry created the transporter as a means
of speeding up storytelling, but let's focuse on the matter-energy
conversion process itself for the sake of argument. Krauss stated that
in order to convert matter into energy, one must heat it up to 1000
billion degrees!!!

1) If we
were to accept the premise of the show, why would heating the crew
members up to 1000 billion degrees be necessary in order to convert
them into energy?

2) Is
there another way to theorically convert them into energy without
heating them up?

ANSWER:
As Krauss explains, if something is heated to this energy
everything would become unbound, including the quarks in the nucleons,
and when this happens, he says, all matter becomes radiation. I
can think of one other way to convert your mass to energy: Make an
anti-you, which is a copy of you created with antimatter, which when
brought in contact with you would annihilate all your mass (and its
mass as well) and radiation would result to carry off the energy.

QUESTION:
Is it possible for a metal to be both magnetic and have the
properties of an insulator (heat) ?

ANSWER:
By magnetic I presume you mean ferromagnetic. All
materials are magnetic, the classifications being paramagnetic,
diamagnetic, ferromagnetic, and antiferromagnetic. The most
dramatic magnetic material is iron which is ferromagnetic. Other
metals are also ferromagnetic, notably nickel and cobalt. All
metals are conductors, not insulators. However, there are many
compounds and alloys which are ferromagnetic, notably many ceramics,
and many are insulators. I don't know what you are referring to
with your parenthetic 'heat'.

QUESTION:
1.)Are the masses of protons of all the elements same?
2.)where does the energy come from in a nuclear reaction?

ANSWER:

This
is primarily a matter of interpretation. If you mean is the rest
mass of a proton in any nucleus the same, the answer is yes. If
you mean the total energy of a proton (including its kinetic and
potential energies in the nucleus), the answer is no.
There
are uncountable things which may be labeled as "nuclear
reactions". Some release energy, some require energy. You
are probably interested in fission and fusion reactions which are
energy sources in reactors and bombs. In fission a heavy nucleus
like uranium splits into two lighter nuclei; the lighter nuclei are
more tightly bound than the heavy one which means that the mass after
the fission is smaller than before. Thus energy is released, the
source being the nuclear binding energy. Similarly, two light
nuclei like deuterium may fuse together to make a heavier nucleus,
helium in this case. Helium is more tightly bound than deuterium,
so again mass decreases releasing energy.
QUESTION:
I have a question concerning rotating electromagnetic fields
such as might be found in a photon with angular momentum. Do you think
that the circulation of the EM fields, which are described as being
perpendicular to the velocity, would give rise to magnetic and electric
dipole moments oriented parallel (and antiparallel) to the velocity?
I've not seen this described. Reciprocally, it would seem that
motion of a dipole would generate a field, no?

ANSWER:
Fields do not give rise to dipole moments, charge
distributions do. A photon has no charge distribution so it has
no moments.

QUESTION:
Can we spin a disk fast enough so that the tangential linear
velocity of a point near the edge of the disk approaches the speed of
light? Can we continue to dump energy into the disk, even if a point
near the edge of the disk cannot exceed the speed of light? Finally,
how would the Lorentz contraction phenomenon effect such a disk?

ANSWER:
I am sorry that I have not had the time to research and fully
formulate an answer to your question. It turns out that the question
you ask is fairly difficult and well beyond the scope of the "Ask the
Physicist" site. According to one site I found, the "...rotating disk
paradox (it turned out to be Ehrenfest's paradox) has been extensively
analyzed by many people (including Einstein himself, who developed
general relativity to answer this problem, as one author
speculates...)" Let me comment first on your first two
questions:

As
long as nothing goes as fast as the speed of light (c ), you can
(in principle) do it. However, you would never find a material strong
enough to withstand that kind of centrifugal force. The disk would fly
apart long before anything got close to c .
Again
in principle, there is no limit to how much energy you can dump into a
system (but again, no material could withstand such speeds). What
you would be doing (in the common interpretation) is increasing the
mass of the disk, not so much its velocity. A simpler illustration of
this is a very energetic electron having a speed 0.999c ; if you
double its energy it only increases its speed to 0.99975c .
Your
third question is the most subtle and difficult of the batch. Here I
refer you to two good web sites which discuss the problem in some
detail and give some good references:
http://galileo.phys.virginia.edu/~vc/743/answer/disk/disk.html
http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

QUESTION:
I have come across conflicting information on two items
regarding the production of C14 (radiocarbon) in the Earth’s
atmosphere. These two items are:
(1) The process of changing N14 to C14
One source I have seen states that the change from N14 to C14 is caused
by an incoming neutron experiencing “charge exchange” with a proton in
the nucleus of N14 whereby the incoming neutron changes into a proton
and continues on its way while the proton in the nucleus stays in the
nucleus after changing to a neutron. The other source I saw states that
the incoming neutron replaces or “knocks out” the proton in the nucleus
so that the two nucleons do not change their charge but actually trade
places in the nucleus.
Questions: Which of these two explanations is the correct one? And, if
the “charge exchange” explanation is the correct one, what is the
mechanism of the charge exchange, i.e. what particle mediates the
transfer of charge?
(2) The source of the neutrons
One source I have seen states that the original comic rays are mostly
in the form of protons which strike atmospheric atoms and produce
secondary cosmic rays in the form of neutrons which then strike the N14
atoms. The other source states that the neutrons are the original
cosmic rays, thus implying that they are not secondary rays that are
produced by the initial proton cosmic rays.
Question: Which of these two explanations is the correct one.

ANSWER:
Since you state that you are interested in the creation of ^{14} C
on earth, let me address your second question first: a neutron is not a
stable particle, it has a half life of something like a half hour, so
there will not be many free neutrons found floating around the
universe. The neutrons responsible for the ^{14} N(n,p)^{14} C
reaction which creates ^{14} C on earth are created as
secondary cosmic rays when the primary cosmic rays interact with the
atmosphere. Because cosmic rays are so energetic, one collision
with a nucleus is likely to cause more or less complete disintegration
sending many neutrons out.

Regarding
your first question, the two are, in some sense, the same. Not
really, but they are entangled because neutrons in the nucleus and the
incoming neutron are indistinguishable particles and, when the incoming
neutron becomes part of the system containing the other neutrons, you
can't put a label on the incoming neutron to keep track of it.
When you are doing the calculation you have to do something called
antisymetization to account for this. That said, we can talk
about the two processes as being approximately different and I believe
that charge exchange dominates.

Charge
exchange, where electric charge is transferred from a proton in the
target to the transient neutron, is mediated by the weak
interaction. In effect, it is exactly equivalent to nuclear beta
decay (the classic example of a weak interaction). Imagine having
a beta decay of ^{14} N: ^{14} N would emit a positron
(which has the same mass as an electron but positive charge) and a
neutrino and ends up as ^{14} C. The only catch is that
if you add up all the energy of such a decay there is more rest mass
energy (mc ^{2} type energy) after the decay than before
so it can't happen. But it "can happen" if you put energy in and
that is essentially what you are doing with the (n,p) reaction.
"Knock
on exchange", requires that there be good matching between the incoming
n and the target p. As you can imagine, if the projectile is
enormously faster than a typical target proton, the likelihood of the
neutron almost coming to a stop in so short a distance is small.
However, it is possible because neutrons and protons have about the
same mass so a pure head-on collision results in the incoming particle
being almost at rest (think of colliding billiard balls).
QUESTION:
The electrostatic, magnetic, and gravitational forces all
seem to imply action at a distance. What is the underlying mechanism
that is responsible for these attractive and repulsive forces?

ANSWER:
Forces which are "action at a distance" are common for the
fundamental forces of nature. Two of the three you mention,
electrostatic and magnetic, are actually one, the electromagnetic
force. There are generally two ways physicists think about such
forces: the classical way where the space is thought to be filled by a
field created by sources of the force (electric charge in the case of
electromagnetism); then if there are objects in the field which can
feel it (also electric charge), the force will be felt via the
field. The other way to look at it is to quantize the field; in
the case of electromagnetism, then, there are "quanta of the field"
which transmit the force. For electromagnetism these quanta are
photons, little bundles of light, which electric charges "throw back
and forth among themselves" and thereby feel the forces. This is,
of course, a pictorial representation of what requires considerable
mathematical prowess to describe quantitatively. Electromagnetism
is a theory which we physicists believe we understand completely.

Gravity
is in principal just the same but in practice there are glitches.
The field is called the gravitational field and the source is
gravitational mass. In fact, the theory of general relativity
even allows us to get a picture of why gravitational mass sets up a
gravitational field —it is because it "warps the space" around
it. However, there is no successful theory of quantum gravity but
many theorists are working hard on it. There should be quanta of
the field, which have a name, gravitons, but there is no experimental
evidence of their existence.

Note
added later: The idea of photon exchange must not be taken too
literally. It is meant as a "cartoon" picture to convey the qualitative
ideas of field quantization. For example, you cannot just think of the
recoil of the emitting particle which you normally would in classical
physics and seems like an easy way to see a repulsive force. The reason
is that these photons are necessarily what are called virtual photons,
in some sense they do not exist. You cannot observe a virtual photon.
It is easy to see that they cannot be real photons, because if the
charged particle suddenly spit out a photon, where did the energy
needed for the creation of that photon come from? Energy conservation
has been violated! However, in quantum physics, it is ok to violate
energy conservation by an amount D E and that violation
can last as long as D t as long as D E D t < h /(8 p ) where h is
Planck's constant, about 6.6x10^{-34} J-s; this is the famous
Heisenberg uncertainty principle. It is good if budding young minds can
be convinced that simple, qualitative models are good for getting
insight but not necessarily good for a detailed theory. Another example
is the Bohr model of the atom: we know that it describes a lot of data
very nicely but we also know that the idea of electrons running around
in little circular classical orbits is total nonsense in a rigorous
model of the atom.

QUESTION:
If there is a battery with two wires connected, and end at
the end of each wire there is a light bulb, one a 40 watt bulb, and the
other a 100 watt bulb, when the battery is turned on, which bulb will
glow brighter?

ANSWER:
For starters, if you hook up one wire to a light bulb, it
will not work at all because electric current must flow through the
filament for it to work. So, the answer to your question is that
neither bulb will glow. However, if you connect a wire between
the two bulbs to complete the circuit (called putting the bulbs in
series), the bulbs will light up. Now the ratings of the bulbs
which you gave me were incomplete, because you need to tell the energy
consumed (watts) for a particular potential difference (voltage) across
the bulb. You are probably thinking of bulbs you use at home
where the voltage is typically 120 volts. But, the 40 watt bulb
could be much brighter if the voltage across it were much larger than
across the 100 watt bulb. And in the envisioned scenario (the
bulbs in series), the voltages are different across the two bulbs, so
we have to figure it out. The power consumption is given by P=VI
where V is the voltage and I is the electric current;
we will also need Ohm's law, V=IR where R is the
resistance of the bulb. To make things simple, let's assume that
the voltage at which the bulbs have their power rated is 100
volts. Then for the 100 watt bulb, I= 1 ampere so R= 100
ohms; for the 40 watt bulb, I =0.4 amperes so R =250
ohms. In general, the resistance of a 40 watt bulb is 2.5 times
larger than the resistance of a 100 watt bulb. Now imagine
connecting them across a 100 volt battery: the net resistance of the
two is 350 ohms so a net current of 100/350=0.286 amperes flows through
each. Therefore, the voltage across the 100 watt bulb is 28.6
volts and the voltage across the 40 watt bulb is 71.4 volts.
Therefore, the power consumption of the 100 watt bulb is 81.8 watts and
that of the 40 watt bulb is 20.4 watts, so the 100 watt bulb will be
brighter. This assumes that both are equally efficient at
converting the electrical energy to light; an incandescent bulb
converts quite a large fraction of the energy to heat instead of
light. This could have been easily proven in general, but I
thought that a numerical example would be more enlightening (pun
intended!).

QUESTION:
A number of websites describe the red shift that occurs in
the light received from distant type 1a supernovas as being dependant
on the speed at which they are travelling away from us and the space
between us. I thought that red shift only depended on the speed, yes
the distance would affect the ammount of light that reached us but not
the frequency shift. For example there is a type 1a supernova 9 billion
light years from earth, an observer on earth and an observer only 1 LY
away from the supernova(but stationary with repect to the earth) would
observe the same red shift. Of course the observer not on earth would
get a much brighter view! Does the space between the observers add more
of a red shift?

ANSWER:
You are right, red shift (Doppler effect) is determined by
speed. However, it is indirectly dependent also on the distance
because it has been empirically determined that the speed of an object
increases as the distance to it (from earth) increases. In fact,
this is so well established that red shift is usually not thought by
astronomers as a measurement you do to determine its speed but rather
its distance. There is also a gravitational red shift but it is
not important except for the most massive of sources.

QUESTION:
My refrigerator has its interior white in color, but
shouldn't it be better if it was painted black? I know that white
color reflects heat, so now the heat inside will be bouncing back and
forth inside the fridge but not out of the walls. If the walls were
painted black, then heat will be absorbed and transferred to the
cooling agent, so why are the fridges all have their interior white?

ANSWER:
Well, first of all I believe that the way a refrigerator
works is to cool the air by moving it across the coils, not by the
contact the air has with the walls. Therefore, depositing some of
the heat first in the walls would not be advantageous. The other
thing I can tell you is that an efficient absorber is also an efficient
radiator, and therefore black walls at a given temperature would
radiate more energy into the air than white walls. Finally, my
guess is that color difference of radiation and absorption by the walls
is a trivially small consideration and that you would be hard-pressed
to see any difference in the operation of white- or black-walled
refrigerator. I would guess that the choice of white is largely
aesthetic —people associate white with cleanliness.

QUESTION:
My question is unfortuneately on the subject of time, which
confuses and irritates me intensely. I would like to know more about
the notion of "blocktime" or how some physicists describe the past and
the future and the present coexisting. Surely they dont mean it
literally as that.

ANSWER:
I have no idea what you are talking about, have never heard
of blocktime. Is that anything like block scheduling, the idiotic
scheduling scheme in my daughter's high school? Kidding aside,
let me just give you a sort of rough sketch of how physicists think
about time. The theory of special relativity, devised by Einstein
in 1915, showed that space and time are not unrelated, unchangeable
entities. There is no longer such a thing as universal time and
universal space. Rather, we get unexpected (based on everyday
experience) things happening: for example, not all observers will agree
on the rate at which some clock is running and not all observers will
agree on how long a certain stick is. The implication is that
past, present, and future no longer have universality —your past might
be my present but somebody else's future. You do not see such
bizarre things in everyday life because such things happen only when
observers have relative speeds which are comparable to the speed of
light.

QUESTION:
A classical physics problem is the block that's released from
rest & falls a specified vertical distance (h) onto a spring that
is then compressed a distance (y). By knowing the block's weight (or
mass), the block's velocity just before impact with the spring can be
calculated. The KE of the block is then transferred into elastic
potential of the vertical spring. All this is given by the relationship:
1/2*m*v^2 = 1/2*k*y^2 & hence solve for y since the block comes to
rest at maximum compression. This usually results in the solution of a
quadratic equation, et al.
Several questions that aren't clear to me are:
1) isn't the block continuing to lose PE as it compresses the spring?
Where/how is this accounted for? Usually the compression is small, but
nevertheless, shouldn't it be in the calculation?
2) does the block continue to accelerate downward after initial
contact, or does it begin to slow down? What I'd like to understand is
where (velocity as a function of y) does "maximum velocity" occur? I
suspect that the max or min position would result from taking the 1st
derivative of velocity in terms of y? Where could I find such a problem
worked out ? Thxs.

ANSWER:

You
are absolutely right, the solution to the problem is incorrect if you
include spring potential energy but not gravitational potential
energy. The solution you described is correct only for a
horizontal spring. If we choose y =0 to be where the mass
strikes the spring, then mv ^{2} /2=mgy +ky ^{2} /2
is the equation to solve. Be certain to choose the solution with y <0
for maximum compression |y |. The solution y> 0
corresponds to how high the mass would rebound if it remained attached
to the spring.
The
mass accelerates down as long as the weight is bigger than the force of
the spring. Eventually (when |ky |=mg ) the spring
force becomes bigger than the weight and the mass starts slowing
down. I have given you all the information you need to work it
out for yourself.
QUESTION:
Do raindrops spherical or of the classical raindrop shape?
And what make it shape that way? How can somebody see the shape of
raindrops? Is there a simple experiment I can do to verify it? Will
raindrops from clouds higher in the sky larger/smaller than those from
lower clouds?

ANSWER:
Shown to the right is a picture I stole from a site
which explains this all very nicely. Raindrops smaller than about
1 mm in radius remain spherical and, as the size gets bigger, they tend
toward a shape like a hamburger bun (called "oblate"). (The
numbers shown are the radii of spherical drops which have the same
masses as the ones shown.) To imagine the shape for each case
drawn here, imagine rotating the the shape about a vertical axis
passing through the center. What is happening, of course, is that
the drop sees air rushing "up" which tends to flatten it out. All
these are assuming that the drops are falling in still air. It is
interesting that for very large drops, which would be on the order of
one centimeter across if spherical, a little parachute actually forms
which then breaks like a bubble and results in smaller drops. I
believe that there is no easy way to correlate size with altitude since
other factors will determine it. In any case, raindrops never
have the classic teardrop shape. Verification would probably
require high speed photography; the shapes are equilibrium shapes after
terminal velocity has been reached, so you couldn't just look at drops
falling a few feet.

QUESTION:
I'm wondering if I have enough information to answer this
question, and if so, could you explain it? If there's a metal rod with
a metal ring on the end, and the ring is held over an open flame, when
the temperature of the ring increases, will the hole become larger,
smaller, or stay the same?

ANSWER:
The hole becomes larger. The metal expands when its
temperature increases so it becomes thicker, taller, and bigger
around. However, do not forget that the metal rod will expand
to. The relative expansions will depend on what the two metals
are. This could be used as a method to remove a tightly-fitting
ring provided that the rod expanded less or you were able to keep the
rod from getting as hot. You can find discussions of thermal
expansion in just about any introductory physics book.

QUESTION:
Using Friction, how do we walk? I know that we couldn't walk
without friction and stuff but I want it all explained to me.

ANSWER:
Suppose that you are standing still and you wish to move in
some direction. Then you must accelerate in the direction you
want to move and Newton's second law (F=ma ) tells us that there
must be a force in the direction you want to accelerate. How do
we get that force? What we do is push back on the ground with our
feet using muscles in our legs. If there were no frictional force
between us and the ground, there would be no way we could push back on
the ground. Now, the force we exert on the ground is of no
particular interest to us if we are interested in what happens to us;
rather, we are interested in forces on us —those are what determine how
we move. But, Newton's third law tells us that if our foot exerts
a force on the ground, the ground exerts an equal and opposite force on
our foot. So the ground pushes forward on us and it is that force
that causes us to accelerate forward. The little walking man
shown above has, during most of the time that a foot is in contact with
the ground, a forward force on him. Of course, when your next
foot hits the ground it is moving forward and must come to rest to
begin the process over again, so there is a brief time just after your
foot touches the ground when the ground must exert a backward force on
your foot to provide the acceleration to stop it.

A similar
example is a car which must accelerate forward. If there were no
friction, putting your foot on the gas would be futile —your wheels
would spin and the car would stand still. The ground must exert a
force on the wheels which is pointing forward in order to keep them
from doing what they want to do —spin. It is the friction which
accelerates your car forward.

There is
a similar question below with some links which
address mainly other aspects of the physics of walking besides friction.

QUESTION:
Why physicist needs to breakdown a resulting velocity to 2
components such as x and y components? For example, rolling a ball
diagonally on the rectugular table.

ANSWER:
We don't "need" to do it, it is just convenient in many
cases. The main reason is that it is often easier mathematically
to work with two or three scalar equations rather than the more compact
single vector equation simply because vector arithmetic is more
difficult than scalar arithmetic. The usual example, of course,
is projectile motion without air friction. Here the motion in the
horizontal direction is uniform (constant speed) and in the vertical
direction has uniform acceleration. In your example, a ball
rolling with constant speed across a table, there is probably no good
reason to break the velocity into its components: you should simply
choose a coordinate direction along the velocity vector.

QUESTION :
How do humans walk? I know this could have many parts to it
and I'd like to know about them all. From the friction side of things
to the pendulum side of things. Thank you.

ANSWER:
This question, as you note, is really too broad and extensive
to have a concise answer as is the function of this site. I
recommend that you do a google
search on "physics
of walking". I found several informative sites including:
http://www.discover.com/july_01/featphysics.html
http://www.discover.com/archive/index.html
(search for an article by Zimmer, August 1995)
http://amstel.wins.uva.nl/~heck/research/walking/walking.pdf
Also, the physics of walking is important to robot design and
you can find information by searching on "robots and walking"; one
useful site I found is http://www.fzi.de/ids/WMC/walking_machines_katalog/walking_machines_katalog.html

QUESTION:
I have a question concerning simple harmonic motion. I had an
experiment using spring. I found the amount of time it takes to
complete one cycle in seconds for various masses. When I drew the
graph, and manipulated it into linear graph, I got a simplified
equation of T=K x M(square rooted) I know the formula for the SHM
is T=2pie x m/k squretooed. what I like to know is the meaning of the
slope value. what is it that is being measured by the slope value in
the equation I got?

ANSWER:
I am a little unsure about your notation. I will use
the notation [expression]^{1/2} for the square root of
"expression". The analytic result that T =2p [m/k ]^{1/2}
is based on the assumption that the force which the spring exerts on
the mass is given by F=-kx where x is the amount that
the spring is compressed or stretched; the minus sign means that the
force always tends to push or pull the mass back to the equilibrium
position (x =0). The constant k is called the
spring constant and, as you can see, is a measure of how stiff the
spring is. For example, if k =100 N/m, this means that it
would take a 1 N force to stretch the spring by one cm (10^{-2}
m). Now, if you plot your data for T as a function of [m ]^{1/2} ,
if you get a straight line its equation will be T =S [m ]^{1/2} ,
where S is the slope. Comparing your experiment with the
theory, your slope is 2 p
[1/k ]^{1/2} ,
so you are determining the spring constant of the spring: k =(2 p /S) ^{2} .

You could
do one additional experiment to provide an independent measurement of
the spring constant k : Hang several different weights (W )
from your spring and measure the amount of stretch (x ) for
each. Now, if you plot W vs. x, you should get a
straight line with slope equal to the spring constant k .

QUESTION:
I have a problem I thought you might be able to help me out
with. I am trying to figure out how much force is holding a piece of
film that is laminated to a large piece of glass, covered in an index
matching solution, and finally sandwiched with another large piece of
glass. This force seems to be important in keeping the film from
expanding while exposing, and I'd like to try other means of applying
the same force (maybe with static or something else). The only problem
is I am having a hard time finding a way to calculate the amount of
force on the film sandwiched between two pieces of glass.?. Would you
know how to tackle such a problem? Thanks for any help,

ANSWER:
I doubt that I will be of much quantitative help to you, but
I may be able to get you thinking correctly about the problem. I think
the essentials of the problem depend only on the film and its contact
with something on its surfaces, so I will forget about the
index-matching film between the glass and the film. So my
simplified picture is shown to the right. Now, what are all the
forces acting on the film? That is easy: its own weight (W )
points down, the bottom plate exerts an upward force (B ), and
the top plate exerts a downward force (T ).
T and B
are spread more or less uniformly over their respective surfaces of
contact and W is spread uniformly over the volume of the film,
but that does not really matter. Newton's first law now tells us
that, since the film is at rest, all the forces must add up to zero, so
B=T+W. The fact is that the weight of the piece of film is very
small compared to the weights of the pieces of glass, so for simplicity
I will assume that W=0 which means that the upper piece of glass
presses down on the film with a force equal to its own weight and the
lower piece of glass pushes up on the film with a force also equal to
the weight of the upper piece of glass because B=T if W=0.

Now
suppose that we exert some force (S ) which tries to make the
film slide; this could represent whatever causes your film to
shrink when you expose it. This single force is an
oversimplification, of course, but it will illustrate the important
features. So when you first start to push with S , nothing
happens; why not? Because the pieces of glass will exert forces
to keep the film in equilibrium; these forces (F_{T} and
F_{B} ) are called static frictional forces, are
more or less evenly distributed over the surfaces of contact, and are
roughly equal because B and T are about equal. If the film does
not start to slide, Newton's first law applies and so F_{T} +F_{B} =S.
This is the kind of thing which happens when you try to move a very
heavy box on the floor by pushing on it —you push but the friction
pushes back and nothing happens. But eventually, if you push hard
enough, the box "pops away" from being at rest and starts
sliding. So there is some maximum amount of force which you can
exert and have the box remain at rest; that equals the greatest static
friction you can get in a particular circumstance. If I
understand your question, your film is slipping and you want to keep
that from happening. So the appropriate question to ask is how
can we get more static friction?

There are
two things which determine the maximum amount of static friction you
can get. First is the nature of the surfaces which are in
contact. So in the box example above, if we grease the floor the
box is easier to get sliding and if we glue the box to the floor it is
harder to get sliding. The other thing we can do is to press the
surfaces together harder. For the box, one way you would make it
harder to get sliding is to put a 500 lb piano on top of the box; it is
not directly the extra weight which makes it harder to move, it is the
fact that the presence of the piano results in the floor and the box
being pressed together harder. So, finally the only ways to keep
your film from shrinking are to either make it stick better to the
glass (say with epoxy) which you probably do not want to do, or to
squeeze the glass to the film harder (for example clamp the pieces of
glass really tightly, which might be hard to do given that glass wants
to break!) At any rate, those are your options.

QUESTION:
I'm confused on low orbital mechanics. An object in low orbit
has two forces acting on it: (1) gravity=mg; (2) centripetal force=mv^{2} /r.
Both pulling it towards the earth. The object is in "freefall". When
computing the velocity of the object, it seems "practice" is to equate
the two forces (as if one was opposing the other): mg=mv^{2} /r.
v=sqrtgr. There is no state of equilibrium in this system, so why is
this invalid assumption made? I can understand analyzing this problem
from an energy point: P.E.= K.E.(translation)+K.E.(rotation). (mgr=1/2mv^{2} +1/2Iw^{2} ;
mgr=1/2mv^{2} +1/2mr^{2} (v/r)^{2} . This reduces
to v=sqrtgr. This assumes the object is a point source compared to
Earth. Where is my thinking going wrong?

ANSWER:
Here is what students just learning about circular motion
often do:

What you
do early in solving a mechanics problem is to identify all the forces
on some body (the satellite in your case). The student will say:
"there is such and such force due to such and such, and then there is
such and such force due to such and such, etc ., and then
there is the centripetal force ." But, the centripetal force
is not a new force that magically appears when something moves in a
circle; instead, the centripetal force is simply a name for the sum of
all the components of the real forces which point toward the
center of the circle. Then as usual, F _{c} =ma
where a is the component of the acceleration toward the center
of the circle which is, of course, simply a=v ^{2} /r .
In the case of a near earth orbit, there is (neglecting air friction)
only one force on the satellite, its own weight mg.
Therefore F _{c} =ma => mg=mv^{2} /r
where r is the radius of the orbit (approximately
equal to the radius of the earth).

Incidentally,
your analysis using energy is totally incorrect. There is no law
that says potential energy = kinetic energy. And, if you do
analyze it that way, r does not mean the radius of the earth,
it would mean the height above some prechosen zero of potential energy,
probably the ground. Furthermore, the kinetic energy is either
the translational one or the rotational one, not their sum —that is
double counting.

An answer
to an earlier question may also be helpful to
you.

QUESTION:
How does the chemical content and density of chemicals
influence the color of a planet's sky? I have attempted to locate the
answer to this question in various textbooks and have asked a few
astronomers and the closest I get to an answer is a reference to
Raleigh scattering in Earth's atmosphere tends towards blue in general
and red due to the extra depth of atmosphere and angle of the sun at
sunset. This still doesn't really answer my question, so I am hoping
you can help.
Forgetting for a moment that life as we know it could not exist on
earth without our particular atmosphere, if the Earth had a different
atmosphere composed primarily of some other chemicals as ammonia, or
cyanide, or methane, etc. instead of the Nitrogen-Oxygen mix we have,
how would one determine or predict what color would be visible at
ground level on a sunny day? Is there a way to calculate this based on
particle size or a chart one could consult comparing scattering to
particle size?
I realize that each element absorbs and reflects different wavelengths
of light, but I am assuming that this does not influence Raleigh
scattering so much as the size of the particles in the atmosphere.
Given that our atmosphere is predominantly N2 and O2 (with formula
weights of 28 and 32), I would expect much less scattering in an
atmosphere composed primarily of NH3 or CH4 (with formula weights of 17
and 16), but no appreciably difference in an atmosphere composed of HCN
(with a formula weight of 27). Is this correct? Would an atmosphere of
HCN have a blue sky? How would I determine the sky color of an
atmosphere with NH3 or CH4 or any other chemicals?

ANSWER:
It is my understanding that Rayleigh scattering, from
anything smaller than the wavelength of light, including atoms or
molecules, always preferentially scatters the shortest wavelengths (the
distribution of scattered intensities being something 1/l^{4}
where l^{ }
is the
wavelength). Therefore, regardless of the chemical content of the
atmosphere, the sky will be blue. Actually, since violet is a
shorter wavelength than blue, you would expect the sky to be
violet. Actually, it is, but our eyes are much more sensitive to
blue than to violet so the sky "looks" blue.

However,
this is not the whole story. As the size of scatterers approaches
the wavelength of visible light or larger, the 1/ l^{4}
law ceases to be
correct. A new type of scattering begins to dominate called Mie
scattering. Aerosols such as water droplets or ice crystals tend
to scatter light of different wave lengths with equal
efficiency —that's why clouds are white. And, of course, if there
were a significant number of "large" dust particles suspended in the
air which had a specific color themselves (for example, a sandstorm on
Mars which has a very red dusty surface), the sky would take on that
color.

QUESTION:
First, given a vacuum with a single atom present, say
hydrogen, what keeps the electron from colliding into the
nucleus? Second, is the isolation of one atom a valid situation?
Given an
environment where there are multiple atoms present, I hypothesized that
the nucleus of each atom exerts a force over neighboring electrons of
other nuclei, thereby keeping the electrons from collapsing into any
one nucleus. However, I don’t think this is completely right. I find it
hard to believe that an atom’s existence is dependent upon the
existence of other atoms. Are individual atoms in a particle
accelerator isolated? Can’t more stable atoms, i.e., helium, in space
exist alone? If so, while electromagnetism is responsible for the
attraction of the electron to its nucleus, what is the repulsive force
keeping it so far away?

ANSWER:
Your hypothesis is, in fact, completely wrong. I
recently answered a very similar question
which, I believe, should address your question. One atom is a
very stable thing with no tendency whatever to collapse any more than
the solar system has a tendency to collapse.

QUESTION:
I have a question for you. My physics teacher gave this to me
just for fun, and I don't know how to go about it. There is a block of
wood floating at the top of a body of water. Is the density of the
block of wood floating in the water, greater that the density of the
water, equal to the water density, less than 1/2 the density of water,
more then 1/2 the density of water, or is there not enough info? How
would I go about finding the answer? Thanks very much!

ANSWER:
There is enough information given to determine that the wood
has a density less than that of the water. There is not enough
information given to determine the density of the block. Archimedes'
principle states that the buoyant force is equal to the weight of the
displaced water. Since the buoyant force must be equal to the weight of
the block for it to float, it should be clear that the density of the
block is less than that of water (its mass occupies a larger volume
than an equal mass of water). If you were told what fraction of the
volume of the block were under the water, you could determine its
density exactly. If exactly half the block were submerged, its density
would be 500 kg/m^{3} , half that of water.

QUESTION:
Our SUN is the source of all energy in our SOLAR
system.Therefore it has great stored quantities of potential energy in
the form of unfused HYDROGEN. I see the SUN having all the properties
of a bomb.It's ignited and it has lots of unexploded fuel.What prevents
it from going off all at once? OR! Is it more plausible to think of the
SUN, much like a smouldering log, burning slowly.The logs' rate of
conflagration governed by it's access to oxygen. If the log was
chemically converted to cellulose nitrate and ignited it would release
all its' potential energy in an instant. Why doesn't the SUN do this
similarly? Is it possible, the SUN gleans its' energy synthesising
helium from hydrogen via gravity, from its' surrounding solar system?
In this manner, gravity and its magnitude would seem to be a reasonable
explanation for the long lived thermodynamics of the SUN. A radical
thought! I know!

ANSWER:
The sun is, as you say, a huge ball of hydrogen. It is
very hot, so the sun is a plasma with many electrons and protons
running around very rapidly. However, the process is not just a
simple one-step process as in your burning log where we essentially
have carbon combining with oxygen to produce CO_{2} . What
happens is first two protons (^{1} H) fuse and beta decay to
heavy hydrogen ^{2} H; then the ^{2} H fuses with a ^{1} H
to make ^{3} H; then the ^{3} H beta decays to ^{3} He;
then ^{3} He either fuses with a ^{1} H and beta decays
to make ^{4} He or else two ^{3} He fuse to make one ^{4} He
and two ^{1} H. But for each step to happen you have to
have the right parts come together which happens with a probability
determined by many things; and even if the parts come together, there
is only some probability that the desired reaction will happen.
So it doesn't all happen at once because of the limiting probabilities
governing things. A nice place to read about this so called
proton-proton cycle is here .
Incidentally, the H-bomb does not work like the sun (but of course does
use fusion) but rather works as a single-step process of fusing two ^{2} H
to make ^{4} He.

QUESTION:
When elements of low atomic weight, under fusion, form
progressively heavier atomic weight elements, energy is released!Is it
correct to assume that each stepwise, atomic condensation, in the
pathway to forming such heavy unstable elements such as Uranium or
Thorium, has given off energy. If this is so! Why do elements such as
Uranium now give off energy when they decompose into elements of lesser
atomic weight,under fission?

ANSWER:
No, once you get heavier than around iron you must add energy
to fuse nuclei.

QUESTION:
Near the end of the "old history" questions there is an
interesting one about the boy & swing
(Seraway text ?) that's moving up @ constant velocity (1st law) with
the boy pulling down on the rope. I wondered how I'd apply F=ma to this
same problem, if the boy & swing were to rise with a uniform upward
acceleration = 1 m/s^{2}
What would the FBD look like under acceleration? What's not clear to me
is how to show the downward force (due to the boy's pulling) & the
rope tension. Since I can't can't get the the FBD correct, I can't
apply F=ma sensibily.
Can you show the FBD? Is it one one or two diagrams?

ANSWER:
The free-body diagram has nothing to do with whether it is a
Newton's first or second law problem. So the FBD is of the boy
plus swing. The force down is the weight, 180 lb, and there are
two forces up, both of magnitude F just as before. But
now, 2F- 180=ma rather than zero. Let us suppose
that the acceleration is 2 ft/s^{2} (instead of 2 m/s^{2} )
since the weight is given in pounds. Now the mass of the body is
weight divided by the acceleration due to gravity which is 32 ft/s^{s} ;
so m= 180/32=5.625 lb-s^{2} /ft. Solving the above
equation, I find F= 95.625 lb.

QUESTION:
Given the laws of attraction, what (force?) is it that keeps
electrons from "falling" into or colliding with the atomic nucleus?

ANSWER:
You are actually asking the wrong question. An
identical question in many ways is 'what keeps the moon from falling
into or colliding with the earth?' In both cases we have a light
object orbiting a heavy object; they are able to do this because the
attractive force (gravity in one case, electrical force in the other)
keeps them in orbit. Newton understood this orbiting phenomenon
qualitatively by imagining throwing a rock off the top of a mountain:

Just
throw it and it falls to the valley below.
Throw
it harder and it goes much farther before it hits the ground.
Throw
it harder and it passes over the horizon before it hits the ground, so
you never see it hit.
Throw
it harder and it keeps "falling" but never hits the ground and goes all
the way around the world and comes back and hits you in the back!
Try
it yourself!

So, what
do I mean that you have asked the wrong question? An electron
which accelerates (as a circling electron does because the direction of
its velocity is always changing) radiates energy away. This is
how the antenna at a radio station works —electrons are accelerated
back and forth in the antenna. Therefore, an electron in a
circular orbit should radiate its energy away and spiral into the
nucleus as it loses energy. This does not happen and that is the
puzzle. It was solved by the discovery of quantum mechanics back
in the early 20^{th} century; it was found that certain special
orbits (quantum states of the system) are stable and do not radiate
their energy away.

QUESTION:
Could it be that the "dark matter" is NOT a dark matter but
simply regular light, which is acting as particles? Hense, the gravity
of that elusive "dark matter"?

ANSWER:
Well, I have never heard a serious astrophysicist suggest
that photons might have mass. You may be aware that the evidence
is now becoming quite convincing that neutrinos, previously thought to
be massless, have a tiny amount of mass which contributes to the total
mass of the universe (and is certainly "dark"). Electromagnetism
and quantum electrodynamics, however, are perhaps the best understood
topics of all of physics, and there is no reason to suspect that light
carries any mass (although it certainly does carry energy and
momentum). If photons were not massless, light would not travel
at the speed of light; that would be sort of paradoxical, wouldn't it?

QUESTION:
My wife and I were discussing how our feet feel much colder
in our living room as the outside temperature has dropped. We are
standing on carpet and the sensation is not on the bottoms of the feet,
but rather on the tops of the feet and the ankles. So it is the air
temperature inside the house that is causing the sensation. My
contention is that 1) there is a gradient of temperatures from the
ceiling to the floor and 2) that the gradient is larger as the outside
temperature decreases. Here is a description of the problem as I
see it:

Assumptions:
The house is ranch house on a slab. All rooms are the same
size. All ceilings are 9ft. and no vaulted ceilings. All heating
ducts are in the ceilings. There is nothing to disturb the air
except the heating unit itself. The heating unit could be gas or
electric, it does not make a difference. The thermostat is
approximately 5 ½ feet from the floor. The thermostat is
calibrated to turn the heating unit off once the temperature in the
room at the height of the thermostat reaches +1 degrees above the
desired temperature set on the thermostat. The thermostat is set
at 72 degrees F. The air coming from the ducts is heated to some
unknown temperature well above 72 degrees.

The
heated air flows into the room from the ducts with some amount of
pressure. Because the air is
being pushed out of the vents towards the floor, the hot air will move
towards the floor, but will very quickly be pushed towards the ceiling
by the colder air on and near the floor. My contention is that
the temperature at the ceiling is higher than at the floor and that
there is a gradient of temperatures between the floor and the ceiling.
Further, I contend that the temperature at the floor will
decrease as the outside temperature decreases. This is the crux
of the discussion my wife and I had. Is my contention correct?

ANSWER:
Well, most of the information you have given me is not really
necessary! Here is what is important:

When
the outside temperature is lower, the temperature of the slab and
therefore your floor itself is lower.
On the
average, the temperature at the thermostat is the same regardless of
the outside temperature.
The
temperature at the thermostat will be higher than the temperature of
the floor (warm air rises).
Therefore,
as you note there will be a gradient of temperature with temperature
increasing, in probably a more or less linear way, going up. (The
picture at the right shows a possible scenerio assuming a linear
temperature dependence.)
Therefore
the temperature a few inches above the floor will be lower when the
temperature of the floor is colder (as the picture clearly shows).
Finally,
I would be remiss as a scientist if I did not point out that it is
extraordianarily easy to prove all these conjectures by actually
measuring the temperature a few inches above the floor on a cold day
and and on a warm day! (Of course, this is essentially what your
feet are doing for you!)

QUESTION:
I have a question that has been bothering me. It came about
when my teaher at my high school was lecturing about Albert Michelson
and his speed of light experiment. How did he control the rate of
rotation of his eight-sided mirrors. Without the help of computers what
machine did he use to keep the speed constant as well as measuring the
time it took for the mirrors to make one rotation.

ANSWER:
I was not able to find anything about the details of
Michelson's experiments. He apparently did it once around 1870
and again with greater accuracy around 1920. What I can tell you,
however, is that regulating the speed of a rotating motor was within
the realm of 19^{th} century technology; there are things
called "governors" which mechanically sense changes in speed and feed
back to the motor to keep it going at a constant rate. Governors
were used way back in the era of steam engines running mills in
England. Regarding measuring the rate, that is a really simple
matter if you are certain that the speed remains constant: just count
the number of revolutions in a minute; at the high rates probably
involved in these experiments, this would still be relatively
simple —just have a cam on the axel which actuates a counter. So
the short answer to your question is that computers aren't needed for
this level of technology.

QUESTION:
Where can I find information on current patways in seawater?
In response to a DARPA request for submarine detection, I thought I
might be able to detect a submarine in shallow water by the change in
conductivity when a submarine passed between two electrodes, about 200
feet apart. But if it was that easy, the Navy would have done it a
century ago.
I read up on it a bit and tried the Feynman Lectures on Physics, which
I find to difficult on my own. But I read that conductivity is
proportional to the area of the conductor, and inversely proportional
to the length. A submarine passing between two electrodes effectively
changes the pathlength between them, which I had hoped to detect. But,
my guess is that the current passing between the two electrodes would
in fact remain constant, because because it is free to move through a
greater area to compensate for the change in pathlength.

If you
could tell me where to look for information in how to model this
mathematically, I'd appreciate it. I've read up on div, grad, and curl,
but don't see how to set up the equations to account for when an object
is present or absent in a field.

ANSWER:
This is an interesting question. The answer can be
found in a problem in the well-known textbook on electricity and
magnetism, Introduction to Electrodynamics by David J.
Griffiths (Prentice-Hall); it is problem #1 in chapter 7. Here
the student is first asked to compute the resistance between two
concentric metal spheres of radii a and b (b >a )
where the volume between the two spheres is filled with a conducting
material with conductivity s.
The
answer is R= [(1/a )-(1/b )]/(4 ps ). Why is this
interesting? Because if b >>a , it is, to a
superb approximation, given by R= 1/(4a ps ), completely
independent of the size of the outer sphere. So we can say that
the resistance between the inner sphere and a point infinitely far away
is essentially the same as the resistance between the inner sphere a
point 100a away; the resistance is nearly completely determined
by the water near the inner sphere. So the way you measure the
conductivity of seawater is to put two small spheres a large distance
apart, maintain a constant potential difference between them, and
measure the current which flows. Since the current which flows is
determined by the resistances (1/(4a ps ) for each or 1/(2a ps ) net) and the
resistances are due only to the water near the spheres, then unless the
submarine passes very close by one of the spheres (in which case you
could see or hear it so you wouldn't need your fancy detector!), it
will have no effect on the current.

Where you
need to do your research is in the computation of resistance. You
see, although the "length" you refer to above is just (b-a ), the
"area" you refer to is changing as you move out, so you have to
integrate, not just multiply area times length.

QUESTION:
A 2.8kg rectangular air mattress is 2.00m long, 0.500m wide,
and 0.100m thick. What mass can it support in water before sinking?

ANSWER:
A question this specific sounds like a homework
problem. So I will outline the solution conceptually. When
the matress is about the sink it will be completely submerged so there
will be an upward buoyant force equal to the weight of the water
displaced (Archimedes' principle). There will be downward forces
of the weight of the matress and the weight of the load on top.
They must all end up being equal to zero. The only unknown is the
weight of the load.

QUESTION:
If a solid symetrical block was lying on the bottom of a
lake, swimming pool, or whatever and the block's bottom had no exposure
to the water would it or wouldn't it have a buoyant force acting
upward? I don't think it does because there's no fluid pressure
acting underneath, but it still occupies a volume of water causing
displacement. I'd agree there's total force acting downward due
the block's weight + fluid pressure acting over the top surface area.
The horizontial forces would cancel out. There is a normal force
acting upwards, but no buoyant force. What do you think?

ANSWER:
The buoyant force results from the pressure difference at the
top and bottom of the block. But, with a few exceptions (weight,
electrostatic forces, magnetic forces, etc. ), all forces in
mechanics are contact forces. Therefore, the only things which
can exert a force on the block are things that touch it —the water on
the top and sides and the bottom of the lake. Of course, the
forces on the sides cancel, so the net force on the block is zero and
it is the normal force - weight - pressure x area at the top. I
agree with you —no buoyant force. There is an important proviso,
though: do not add the atmospheric pressure to the pressure on the top,
just the pressure due to the weight of the water. Basically, the
bottom must hold up the weight of the block and the weight of the water
on top of the block.

QUESTION:
I want to ask a few questions about "coefficient of friction"
1.Apart from experimentally determine the coefficient of friction
between two surfaces, is there any theotical way to do it? For example,
if we know the structure of the two materials, can we then proceed to
calculate the coefficent of friction, and if we can, how? What factors
(temperature, charge on the sufaces, etc.) are involved in the
calculation?
2.If we know the coefficient of friction between, say, surface A and
surface B and name if x, that between surface B and surface C be y, can
we then say that the coefficient of friction between A and C would be
at maximum x+y? Similarly can we say it be at least the minimum of
(x,y,abolute(x-y))?
3.In my textbook it tells me that coefficient of static friction
between two clean steel surfaces is approximately 0.6 and the
coresponding value of kinetic friction is still 0.6, which means the
value differ very little, the same is true for telflon, which in my
book it says both value are 0.04, however for lubricated steel
surfaces, the values for static and kinetic friction are 0.09 and 0.05,
which is nearly a double, the coefficient of static and kinetic
friction between two glass surfaces is 0.9-1.0 and 0.4, which is once
again nearly a double. What makes the difference between them? Is there
any pair of surface that has a great difference between coefficient of
static and kinetic friction?
Lasly is there any reference you would recommend me about this
topic?
Thanks very much for answering my questions.

ANSWER:
Ah, such an innocent-sounding question! Every freshman
physics student learns about the very simple idea of a coeffiecient of
friction, the proportionality constant between the magnitude of a
frictional force and the normal force (how hard the surfaces are
pressed together). And that there are two types, static and
kinetic, and that they are so easy to understand. But here is the
catch: f= μ N
is a statement of an
approximate experimental fact, not a law of physics. You cannot
derive it from known laws, you cannot deduce it from first
principles. Surprisingly, friction is fairly poorly
understood. I have gone to the website
for the Tribology (study of interacting surfaces) Laboratory at the
University of Florida and find the following statement: "When two
bodies are in contact and relative motion, a finite force is required
to maintain this motion, the friction force . A coefficient of friction
is calculated by dividing the friction force by the normal force. To
date, despite considerable efforts, there is no model capable of
predicting friction coefficients from first principles. Thus, careful
and proven experimental techniques represent the most sophisticated and
reliable technique for investigating, designing, and assessing the
tribological worthiness of new materials." The most famous
example of a large difference between static and kinetic friction is
the violin bow; here the instrument is excited by the bow repeatedly slipping/sticking
on the string. The two links I have given here would be good
places to start your research. Do a google search on "theory of
friction"; much of what you find will be quite technical.

QUESTION:
I'm trying to understand a concept that has been in the media
lately regarding the measurement of the speed of gravitational waves.
Last fall Jupiter passed near a Quasar close enough to measure the
effects of its mass on the Quasar's light. I found an article online
that explains part of the experiment:
"Retardation in propagation of gravity is a central part of the
Einstein theory of general relativity which has not been tested
directly so far. The idea of the proposed gravitational experiment is
based on the fact that gravity in general relativity propagates with
finite speed so that the deflection of light caused by the body must be
sensitive to the ratio of the body's velocity to the speed of gravity.
The interferometric experiment can be performed, for example, during
the very close angular passage of a quasar by Jupiter. Due to the
finite speed of gravity and orbital motion of Jupiter, the variation in
its gravitational field reaches observer on Earth not instantaneously
but at the retarded instant of time and should appear as a
velocity-dependent excess time delay in addition to the well-known
Shapiro delay, caused by the static part of the Jupiter's gravitational
field." -
http://arxiv.org/abs/gr-qc/0206022
How do they actually measure the gravitational field speed? I don't
understand the statement that "deflection of light caused by the body
must be sensitive to the ratio of the body's velocity to the speed of
gravity" — wouldn't the light deflection of the quasar only be
sensitive to the velocity and mass of Jupiter?

ANSWER:
Well, I looked up this article. This is a preprint and
has not, as far as I can tell, been published yet so it is not gospel
as they say. Nevertheless, it appears to be done by competent
researchers. General relativity is an extremely specialized and
mathematically intensive topic in physics, and frankly I cannot read
and understand the details of this article in any reasonable amount of
time. You should note that, as best as I can tell, gravity waves
are not at issue here, rather the propagation speed of the
gravitational field; in electromagnetism, light and fields propagate
with the same speed but until you know this, measuring the speed of
light is not necessarily equivalent to measuring the speed at which the
field propagates. To the best of my knowledge, gravitational
waves have never been unambigously observed directly and the speed at
which the gravitational field propagates has never been measured (this
paper proposes a way).

Your
question is beyond the scope of what this web page is intended to
do —communicate with laymen and young folks some of the basic exciting
ideas of physics, not discuss serious current research issues.
May I suggest that you post your question to the sci.physics.research
moderated newsgroup? It would better fit your needs, I think.

QUESTION:
I have a question regarding an experiment I read about on the
net at http://www.sumeria.net/phys/gravity.html .
In effect it says that the earths center of gravity is above the earth
surface. Was this test valid? Was there ever a test that disputed the
results?

ANSWER:
This article has, in my opinion, zero credibility. The
author is unknown, there are no references, and no specific data are
given. The premise that the pendulum will point straight at the
center of gravity of the earth is wrong (except at the poles and
equator) because of the earth's rotation. The very notion that
the center of gravity is somewhere out in space is utterly ludicrous,
because if it were, why don't I fall there? Don't believe
everything you read on the web!

QUESTION:
This question is purely theorical. If you somehow got a
ship moving at 4/5 the speed of light in one particular direction. Then
the same ship is accelerated to the same speed in a direction
perpendicular to the original direction. So given that, wouldn't the
resultant speed be greater than the speed of light? And if so wouldn't
that go against the theory that faster than light travel is impossible?

ANSWER:
To answer your question you must know the Lorentz
transformation for velocities. I shall call the direction of the
velocity of the ship relative to the ground, before the acceleration in
the perpendicular direction, the x -axis; so the velocity of the
ship in its own rest frame is, of course, u'_{x} = 0 and
the velocity of the ship in the ground frame is u_{x} =0.8c
(where c is the speed of light). Now the ship
is given a velocity in the primed frame (the one moving in the +x
direction with speed 0.8c ) of u'_{y} = 0.8c .
(It may be easier to visualize the ship firing a projectile in the y'
direction with speed 0.8c and asking what the speed of the
projectile seen from the ground is; it is equivalent to your
question.) The inverse Lorentz transformation for components of
the velocity perpendicular to the direction of relative motion (i.e.
the transformation which tells you u_{y} if you know u'_{y} )
is

u_{y} =u'_{y } /[ g
(1+(u'_{x} v /c ^{2} ))]

where
g =1/[1-(v /c )^{2} ]^{1/2}
and v is the speed of the moving frame (in the x
direction), 0.8c in your question. It is now a simple
matter to compute u_{y} =0.48c . Therefore
the speed of the ship in the ground frame is u =[u_{x} ^{2} +u_{y} ^{2} ]^{1/2} =0.933c
which is, as it must be, less than the speed of light.

Let me
speculate where you may have gone astray in your reasoning here.
You were probably assuming that u_{y} =u'_{y }
because the Lorentz transformation for lengths in the y direction
is y=y'. As shown above, however, this is not the case
because of the fact that the two observers use clocks which run at
different rates to measure the velocities in their own frames.

QUESTION:
How does aluminum foil work? If you wrap food in
aluminim foil, does it keep it warm? Does aluminum foill conduct
heat? Why doesn't it get hot when you put it in the oven?

ANSWER:
Aluminum is an excellent conductor of heat. Therefore,
it would be a poor choice for wrapping something to keep it hot.
And, it does get hot when you put it in the oven. It soon will
reach the temperature of the air in the oven. So, you ask, why
don't you get burned when you pick up a piece of foil you have just
removed from the oven? The internal energy which something at a
particular temperature has is proportional to its volume and the rate
at which the energy is radiated away is proportional to its surface
area; so hot foil has little energy (because of its small volume) and
it loses it very quickly (because of its large surface area). If
you took a block of aluminum from a hot oven it would be unwise to
touch it for several minutes.

QUESTION:
How much more work (as a percentage) is required by a person
to walk up an incline of 5 degrees or 10 degrees versus walking on
level ground with all other factors being equal? Thank you.

ANSWER:
The only thing which determines the amount of work done
(against gravity) is the altitude to which an object is lifted.
Therefore, you do no work walking on level ground and the work you do
walking up an incline is independent of the slope, it just depends on
how high the hill is.

Of
course, you and I know that someone walking on level ground does work,
but this is not the work done against gravity. Rather, it is work
done against internal frictional forces (for example, bending and
unbending your knees repeatedly) and the friction of moving through the
air; these are very hard to calculate using simple physics and would be
more or less comparable (I would think) regardless of the slope of your
walk.

QUESTION:
what is the name of the curly E used to represent emf??
something so simple, yet so undocumented leaves me pondering....

ANSWER:
I don't think it has a name. Just "capital script e".

QUESTION:
Given the level of educational and laboratory material
available and access to technology today, is there a future for a
self-taught physicist?

ANSWER:
Well, I guess like anything else you would have to sell
yourself and prove yourself. It would certainly be harder to get
your foot inside the door in the first place. And, like anything
else, having a mentor (teacher) and a peer group of fellow students can
greatly enhance the learning experience. Even the world's most
brilliant individual can obtain irreplaceable benefits from dialog
because that is part of what being human is.

QUESTION:
When we measure the speed of light and conclude that it is
186,000 miles/sec does that mean we the observers are without motion
e.g. (0 miles/sec) in reference to this velocity.Since we don't move
through space in a zero mile/sec situation. What then does that imply
on our calculations on the speed of light?

ANSWER:
The main cornerstone of the theory of special relativity is
that the speed of light in a vacuum is a universal constant, it does
not depend on the motion of the observer. As counterintuitive as
this may seem, it is a fact which has been experimentally verified to
extraordinary precision. So if you measured the speed of light
coming from a lightbulb in your room and then got in a rocket ship
which was moving past your room with a huge velocity and remeasured the
speed, you would get exactly the same result!

QUESTION:
I have a question that deals with relativity. A space ship
approaches a space station at a speed of .5c. In the space stations
rest frame the distance from the station to the ship is 10^6m. Then the
ship fires a laser beam and a rocket at the station, the rocket has a
speed of .3c. I am trying to find the times the laser and the
rocket hit the station. I need help organizing each frame for the
station and the ship. Could you help me oragnize the information
I am given to achieve my desired results?

ANSWER:
The answer to your question may be found here .

QUESTION: (follow-up
on previous question )
Thanks, and I wondered : isn't the pressure at the top (at
the surface) really = zero? Hence, the only downward force is the
weight of the fluid, while the force at the bottom acts upwards (really
is acting in all directions @ bottom). The net force (F) = (pressure at
bottom)*area - (fluid weight ). why is mass of container not used, as
it must also be accelerated up? how could this this have been worked if
the container were an inverted cone, where area @ bottom was ~
zero? Had this been a glass of water is accelerating to the right
along a horizontal surface, what is the "origin of the force" that
produces the acceleration on a small element of water in the middle of
the glass? We all know that the fluid would move to the left side of
the glass as the glass was accelerated to the right. Also, this is an
incompressible fluid, and would assume the shape of its
container. Just curious!

ANSWER:
No, the pressure at the top is atmospheric pressure unless
you have contrived to do your experiment in a vacuum. And this
pressure is very large (about a ton/ft^{2} ). The mass of
the container has no relation to the pressure gradient in the fluid;
there are forces on the container, including its own weight, but those
forces are not relevant to fluid. Variation of pressure with
depth is independent of the shape of the container. The origins
of the forces on a small element of fluid are its own weight and any
forces which the surrounding fluid exerts on it; if it happens to be at
the surface, it also experiences a force due to the atmosphere.
Generally speaking, the only forces an object may experience (apart
from its own weight, the force the earth exerts on it) are contact
forces. (I am assuming we have no "exotic" forces, e.g.
electric or magnetic, other than gravity.) If the glass is
accelerated horizontally there would have to be a horizontal gradient
of pressure as well. This is why the surface of the fluid would
not itself be horizontal.

QUESTION:
I was sitting in physics one day and thinking about car
racing, and bicycle racing. So I started to compare the two, and one
thing jumped out at me both of them race counterclockwise. I started
thinking that if both racers constantly turn left that it was probably
easyer to turn left. I would like to know if there is an explanation
for this.

ANSWER:
I was not aware that there was a preferred direction for
racing. However, I can assure you that there is no physics reason
for this —turning left is no easier than turning right.

QUESTION:
I keep hearing about particles having intrinsic properties
known as "spin". I looked it up, and all I got from it is that it
devides particles into different groups (bosons and fermions) and that
it is closely related to angular momentum. I think I understand what
angular momentum is, but I don't see how it is measured in numbers.
Could you explain that for me? Also, what IS spin exactly? I understan
it is also related to some sort of rotation symmetry. How's a 1 spin
particle different from a 2 spin particle? why are particles with half
integral spin placed in a seperate group?

ANSWER:
Let's think of a classical particle —think of a bowling
ball. What intrinsic properties can it have? Well, it has
mass, so it also has rest mass energy. If this particle retains
its identity, then its mass is something which you cannot change.
If it is moving it has linear momentum and kinetic energy. These
can be changed by changing the velocity of the particle; in fact we
could put the particle at rest and it would have no kinetic energy or
linear momentum. It might be spinning about its center of mass,
so it would have angular momentum and rotational kinetic energy.
These can be changed by changing the angular velocity of the particle;
in fact we could stop the particle's spinning and it would have no
rotational kinetic energy or angular momentum. These are the main
things; we could subject it to some force field (like gravity) and talk
about its potential energy as an intrinsic property, but it is easier
to just restrict ourselves to talk about bowling balls in the middle of
empty space.

But now
let's think about a particle like an electron or an atom or a
proton —something in the microscopic realm. What makes it any
different from a bowling ball apart from the fact that it has less
mass? Well, if you were a 19^{th} century scientist, you
would see no reason why there would be anything at all different.
But when detailed measurements on such small systems started to be
made, it was found that expectations of their properties based on "old
fashioned" classical ideas turned out not to work. Quantum
mechanics was born to explain the strange new phenomena. A
particle still has a mass and a rest mass energy. A particle
still has linear momentum and kinetic energy, but they behave
strangely. If it is a free particle, its momentum cannot be
precisely known unless we are totally ignorant of its position and its
kinetic energy cannot be precisely known unless we have forever to
measure it; these are the results of the famous Heisenberg uncertainty
principle. What this means, essentially, is that a particle may
not be put perfectly at rest somewhere. Finally we come to what
you are interested in: can a tiny particle, say an electron, be
spinning? The answer is yes, but only with the discrete amount of
angular momentum which nature gives it. For particles in quantum
mechanics, the spin angular momentum has a magnitude of ħ √[s (s +1)]
where ħ is a physical constant called Planck's constant, and
where s is the spin quantum number for the particle or state of
the particle. s is called the spin of the particle and
may only take values of 0, 1/2, 1, 3/2, ..., n /2 where n
is an integer. And it is unchangeable : an electron (which
has s =1/2), for example, may not be stopped from spinning, it
always has a spin angular momentum of ħ √(3/4). This is
just the way that nature is! Why do particles have spin? It
turns out to be a relativistic thing. If you do quantum mechanics
relativistically, spin just comes out of the theory.

Finally,
what can you tell about a particle by knowing its spin? It turns
out that spin determines the "statistics" of the particle. What
"statistics" means here is that nature seems to have divided its
particles into two types:

particles
which, if in some quantum system with other identical particles (e.g.
a bunch of electrons in an atom) may not be in exactly
the same state as any other particle in the system. These
particles are called fermions. (Incidentally, "state"
means all the things which are needed to characterize a
particle —energy, orbital angular momentum, spin angular momentum, etc .)
particles
which are not subject to this restriction. These are called bosons.
The restriction is called the Pauli Exclusion Principle.
What
determines whether something is a fermion or a boson is its spin.
If the spin is 0, 1, 2,... it is a boson; if it is 1/2, 3/2, 5/2,... it
is a fermion. It is a lucky thing that electrons are not bosons
since all atoms would have essentially the same chemistry as hydrogen
and there would be no you.

QUESTION:
Subject: A fundamental physical peculiarity.

I am
extremely sorry to bother you. Unfortunately my college physics'
teachers are not able to answer a very interesting question that I have
been pondering for quite some time. It has to do with Planck's time
constant (5,54.10^-43 s) and Einstein's theory of Relativity (in this
case - the Special one).

I would
enormously appreciate, if you could respond with a clarification on the
problem that I've been thinking over recently. You would put to rest an
uneasy mind. :-) *** It goes like this:

I.
According to Einstein's special theory of relativity (TR) it turns out
that in a moving inertial system time is slowing down in respect to a
static observer.

II.
According to Quantum Mechanics (QM), Planck's time (~ 5,54.10^-43) is
the smallest possible time interval. ...

If time
slows down in a moving system, would that mean that Planck's time
constant "increases" also? If looked from the point of view of the
moving system itself, Planck's constant will be measured as
"decreasing", which would violate Quantum Mechanic's assumption that
Planck's time constant is the smallest possible time interval. In any
case Planck's time constant must be changing, which is in violation
with the initial assumptions by QM. If this is the case that would
mean, that either QM or GR is wrong. Both theories are cancelling each
other. Could you please clarify the issue for me? Thank you very much
in advance.

ANSWER:
Suppose we made a clock which "ticked" once every 5.54x10^{-43}
s. This clock, as seen by any other observer, would have ticks
greater or equal to this time (time dilation). No observer
would see this clock having a tick interval shorter than your Planck
time.

However,
I believe that the Planck time is based upon the Planck length —it is
the time it takes a photon to travel one Planck length (about 1.6x10^{-35
} m). So, if you had a "stick" equal to one Planck
length and had it move by, it would become shorter (length contraction)
so the time it took a photon (which all observers see to be moving with
the same speed) to move its length would be less than the Planck
time.

So, yes,
paradoxes do exist. However, it is very important for you to
appreciate that the theories of quantum mechanics and relativity are
not compatible on many fronts. For example, photons really have
no place in general relativity. This incompatibility is the
impetus for research into seeking a theory of "quantum gravity".

QUESTION:
In 1998, researchers at Stanford University's Linear
Accelerator Center successfully converted energy into matter. This feat
was accomplished by using lasers and incredibly strong electromagnetic
fields to change ordinary light into matter. The results of this
experiment may allow for the development of variety of technological
gadgets. One such development could be matter/energy transporters or
food replicators that are commonly seen in some of our favorite science
fiction programs.

My
question is, if we were to build a transporter that worked by
converting people into energy, would they be able to theorically
survive the procedure once their energy has been reconverted back into
matter, since both are interconvertible?

ANSWER:
Changing electromagnetic energy into mass has been seen for
decades. It is called pair production and happens when a very
high energy photon spontaneously creates an electron-positron pair.
Conversely, any time a particle meets its antiparticle the mass usually
totally disappears and electromagnetic energy appears. Regarding your
question, it has no short answer. But, there is enormously more
to the transporter idea than simply destroying and creating mass —you
have to put it back together the way it was. I should also note
that it has never been possible to take some macroscopic thing like a
person and totally annihilate all the mass. Anyway, if you want
to read a good discussion of the physics of why it is extremely
unlikely that a transporter will ever work, I would recommend a book
called The Physics of Star Trek by Lawrence M. Krauss.

QUESTION:
If I have a container of fluid subjected to an upward
acceleration "a" and I want to show that the pressure variation with
the depth is given by: pressure = (density of the
fluid)*(height)*(g+a). Fluids in a container, under acceleration,
act differently than say an elevator being accelerated upwards. The
force must act in the same direction as the acceleration. Hence, I
would expect the net acceleration term to be = (g-a)? On a "free body
diagram" the accelerating force is upward, while the weight acts
downwards. This says the term = (g-a). What am I missing? If the
acceleration were downwards, what would the term become? What
part of the concept am I missing? Is it that that the fluid and
container are not in the same inertial system?

ANSWER:
Consider a volume of the fluid which has a height h
and a cross sectional area at its top and bottom ends of A ; the
volume of fluid is V =hA and so its weight is W= r ghA
where
r
is the mass density of
the fluid and g is the acceleration due to gravity. Now,
there will be a pressure at the level of the bottom surface of P _{bottom}
and a pressure at the level of the top surface of P _{top} ;
since the force on an area A is PA , the net force on
the volume of fluid will be F _{net} =P _{bottom} A -P _{top} A- r ghA.
But Newton's second law
says that F _{net} =ma= r hAa . Therefore,
setting these equal and doing a little algebra, P _{top} =P _{bottom} - r (g+a )h. Note that
if a =-g , that is the container is simply freely falling,
the pressure is the same at all levels.

QUESTION:
I wish to know if it is possible for a small car #1 weighing
+/- 3200 lbs to travel at a speed of 30 mph, crash into a car #2
weighing +/- 3200 lbs, and push it a distance of 45 feet (neglecting
the distance car #2 was pushed into a house). Please assume typical
coefficients, if applicable, for drag, friction, et al. Also assume a
flat ground surface with grass cover (some gravel and asphalt areas,
but should be negligible). And assume a level ground surface (a very
slight upward slope, but should also be negligible). If it is NOT
possible, please suggest to me how to determine at which speed such a
collision and subsequent reaction could indeed occur, at the stated 30
mph speed.
Thank you for your answer!

ANSWER:
What you have described is two cars of approximately equal
mass which stick together after colliding with one of them initially at
rest. This is called a perfectly inelastic collision and it may
easily be shown, using the principle of conservation of linear
momentum, that the two cars have, immediately after the collision, a
speed half that of the incoming car, so 15 mph in your question.
Let's convert that to feet/second, about 22 ft/s. (I have no idea
what you are talking about when you refer to the house, but you say to
neglect it so I will.) I will not give you the details (write back if you want
them), but I compute that, for the cars to come to rest in 45 feet the
coefficient of sliding friction would be about 0.17. This number
is quite low which would mean that the surface over which the cars were
sliding was fairly slippery. Consider, for example, the
coefficient of friction of wood sliding on wood for reference: it is
about 0.2. So, you have to judge: was the "grass cover with some
gravel and asphalt areas" as slippery, on average, as wood sliding on
wood? If yes, then the cars could go 45 feet, if no then they
would not.

QUESTION:
i have three questions that i cant find definite answers to...
1. does the speed of light ever change? i know that the time it takes
light to travel through a given space can change, ( when passing
through glass or water for instance) but can the actual speed of light
be slowed down? why does it take longer for light to pass through glass?
2. do photons have mass? my physics teacher would like me to think
so..but then how do black holes affect light? i thought the basis of
einstien's equation E=mc2 was to show that E=m...
3. how is heat and kinetic energy different than electromagnetic
energy? are either passed in photons?
if you could answer these questions i would be very very happy. thank
you.

ANSWER:
Well now, that is not really a "single,
concise, well-focused question" is it? Since they are at
least loosely related, I will go ahead and answer:

The
speed of light in a vacuum is a universal physical constant
which never varies. What is really cool is that the speed of a
beam of light is independent of the motion of the observer or the
source of light. This is the main foundation of the theory of
special relativity and very well verified experimentally. The
speed of light depends on two constants which characterize the medium
in which the light is traveling, ε
which characterizes electrical properties and μ which characterizes
magnetic properties: v =1/√(με ). Since με for
any material is greater than με for a vacuum, light travels
fastest in a vacuum.
Photons
definitely do not have mass. They do, however, carry both energy
and momentum. You cannot understand this using Newtonian physics,
of course, because both kinetic energy and linear momentum are
proportional to mass. Although we generally write E=mc ^{2} ,
the more general equation for the energy of something with mass m
and momentum p is E ^{2} =p ^{2} c ^{2} +m ^{2} c ^{4
} (note that this reduces to E=mc ^{2} , if
p =0). So, you see that a massless particle may
have energy if it has momentum, even if it has no mass. The
energy of a photon is hf where h is a physical constant
called Planck's constant and f is the frequency of the
corresponding light. The reason black holes (or any massive
object for that matter) affect light is because of the theory of
general relativity which shows that the effect of gravitational mass is
to "warp" the space around it. Therefore, light bends when
traveling past a massive object but it is really traveling in a
straight line in the warped space.
Energy
is the ability to do work, so essentially all forms of energy are
equivalent. Heat is the same as kinetic energy; when an object is
heated, the average kinetic energy of the atoms increases. So the
molecules in a hot room are moving, on average, faster than in a cool
room. Electromagnetic energy is the energy contained in the
electric and magnetic fields in some region where they exist.
Clearly, fields have the ability to do work because they can push
electric charges around. You may think of light as either photons
or electromagnetic waves. If you describe it as waves, the energy
is computed as electromagnetic; if you describe it as photons, then the
energy is purely kinetic energy, hf per photon. Both
descriptions will give the same amount of energy being carried by the
light.
QUESTION:
I am doing a science fair project on sonoluminescence. Are
there any other liquids other than water in which imploding bubbles
will emit light? If so, can any of these substances enhance the light
emitted from imploding bubbles?

ANSWER:
I do not know much about sonoluminescence but I do know that
it is currently a pretty hot topic in physics and therefore there are
abundant references to be found on the web. Do a Google search on
sonoluminescence. Just in the first couple of dozen returned
sites there were a several with suggestions on how to enhance the
intensity:

One suggestion was to
add small amounts of a noble gas to the air;
another
suggested that glycerin be added to the water for stability;
a third
detailed how to enhance intensity by carefully tailoring the
characteristics of the sound.
As a sort
of "base camp" for your research I would recommend the page http://www.aip.org/physnews/preview/1996/chain/ ;
this site is authored by the American Institute of Physics , the
parent body of the American Physical Society which is the most
prestigious professional society of physicists in the US and therefore
a reputable source of "noncrackpot" information.

QUESTION:
Hi, I was wodering if you knew whether when a plane passing
through the origin intersests an ellipsoid centred on the origin, the
cross section of the ellipsoid is an ellipse? If this is the case,
where could I find the proof?

ANSWER:
Well, now, this isn't really physics, is it? Also, it
sounds suspiciously like a homework assignment! Just for
curiosity, I worked it out for myself and found that I could prove that
the intersection of the plane with the ellipsoid is indeed an
ellipse. The proof is fairly straightforward; I will sketch
it but not give all the details. I am assuming that by an
ellipsoid we mean an ellipsoid of revolution, i.e. a surface
generated by rotating an ellipse about either its semimajor (prolate)
or seminor (oblate) axis. Let the axis of revolution be the z- axis.
Now write the equation for the ellipsoid and solve for z .
The plane will make some arbitrary angle q
with
the x-y plane and it will intersect the x-y plane in
a straight line passing through the origin; no generality is lost by
choosing this line to be the x -axis and considerable
simplification in algebra results. You may now write the equation
for this plane simply as z =y tan q .
Now set the z
for the plane equal to the z for the ellipsoid and you get an
equation involving only x and y which you can show is
the equation of an ellipse. However, this is not yet the solution to
the problem because this ellipse is the projection of the intersection
curve onto the x-y plane, not the equation of the curve
itself. You must now transform to a coordinate system x'-y'
on the plane, the transformation being x=x' , y=y'
cos q . Now you will find
that the equation in the x'-y' plane is also an ellipse.
Finally you can check that everything is right by looking at the
special cases q =0 (should be a circle)
and q = p /2 (should be the
ellipse which generates the ellipsoid).

NOTE ADDED:
I looked up the proper terminology. An ellipsoid may
have three unequal axes in the three orthogonal Cartesian directions
and is described by (x ^{2} /a ^{2} )+(y ^{2} /b ^{2} )+(z ^{2} /c ^{2} )=1.
If a=b , the ellipsoid is called an ellipsoid of revolution or a
spheroid; if c<a, it is an oblate spheroid, if c >a,
it is a prolate spheroid. Of course, if a=b=c ,
it is a sphere.

QUESTION:
1. Can you tell me how and why did the special theory of
relativity lead to the atomic bomb and revealed the secret of stars
?
2. How did general relativity give us space warps , the big bang and
black holes ?

ANSWER:
The answer to your first question is that one of the
consequences of the theory of special relativity is that there is a
connection, an equivalence, between mass and energy. That is,
mass may be converted to energy and vice versa. As a
result, if you cause a very heavy atomic nucleus to split (nuclear
fission) energy is released because the mass after the fission is less
than that before; this is the principle behind a nuclear reactor or a
conventional atomic bomb. But if you take two very light nuclei
and fuse them together (nuclear fusion) you also get energy released
because the mass after the fusion is less than that before; this is the
principle behind how stars generate energy and also behind the hydrogen
bomb.

The
answer to your second question is that general relativity is the theory
of how gravity works. In essence, in order to satisfy the
postulates of the theory, the structure of space and time themselves
are altered by the presence of gravitational mass; this alteration is
often described as a warping of the space around massive objects which
then explains why massive objects attract each other (they simply fall
along "straight-line" paths in this warped space). I don't think
that I would say that general relativity "gave us" the big bang or
black holes. They just happened. Of course, to properly
describe a black hole, you need to use general relativity. But
the reason a black hole forms is that when a star is all "burnt out"
the force of gravity (which is always trying to compress any large
object) has no force opposing it which is strong enough to keep the
object from collapsing.

QUESTION:
Do you know where I can find background theory about Lambert
diffusers. I can't find anything.

ANSWER: (with
some help from R. S. Meltzer)
This question really stumped me for a while because I have
never heard of Lambert diffusers. A little bit of research
reveals that Lambert scattering refers to the idealized case of a
surface reflecting light equally in all directions. The precise
definition of what this means can get a bit confusing because it
involves what things like 'equally' mean. You must define a
quantity called radiance and then the radiance of the surface is
independent of the direction from which it is viewed. This does
not necessarily mean that it looks the same from all directions.
When I did Google searches on
"lambert diffusers", "lambert scattering", and "lambert diffusion" I
found that there are apparently many uses of this approximation in
astronomy (analysis of planetary images, etc. ), acoustics,
radar imaging, sonar, etc . But, as you have apparently
also found, all these sites talked about the Lambert scattering without
getting quantitative; they assumed that the reader knew all about it
(you can tell that I am not an expert in any of these areas!).
Then I found that it is apparently more common to refer to these
idealized surfaces as "Lambertian surfaces" and a search on Google revealed many sites which
quantitatively talk about and define what you want, I believe.
Two which seemed particularly good to me were http://www.cs.berkeley.edu/~efros/cs280/notes/lec16.ps
and http://www.intl-light.com/handbook/ch06.html .
To see others, do a Google search
on "lambertian surface".

QUESTION:
How are hyperbolic functions (sinh, cosh, tanh, etc...)
related to free fall and are there any web sites or books where I can
read more on this?

ANSWER:
When the problem of a falling object is studied without
neglecting air friction (which is usually neglected in introductory
discussions of the "free fall problem"), the solution to the problem
involves exponential functions and exponential functions may always be
expressed in terms of hyperbolic functions. I found a web site
where this problem is worked out: http://www.owlnet.rice.edu/~phys111/Lab/expt02.pdf
(which is the writeup for an undergraduate laboratory exercise).

QUESTION:
Regards the classical mechanics of a rocket launch, at t=0
there is no motion, hence velocity & acceleration both = 0.
However, at ignition (t+dt), there is vertical motion and hence
velocity, but is there instant acceleration? Is there
always an acceleration of an object that was initially at rest? For
example, a car starting its motion at stop light when "green" comes on,
is it accelerating? I never really understood this & thxs in
advance.

ANSWER:
Acceleration is the rate of change of velocity; therefore,
what the velocity is does not matter. An object can be at rest
and be accelerating (that is it will not be at rest afterwards) or it
may be moving with a speed of 99% the speed of light and not be
accelerating. To further help you understand this, think of
Newton's second law F=ma which reveals the cause of
acceleration to be force. If an object has a net force on it, it
necessarily has an acceleration of F/m regardless of its state
of motion (velocity). Another good example is stone thrown
straight up in the air: when it is at its highest point it is at rest
but has an acceleration because it was moving just before and will be
moving just afterward.

QUESTION:
Do you know of any videos that illustrate clearly why the sun
appears to move in different paths across the sky at different times of
the year and how the paths change with latitude.

ANSWER:
Sorry, I don't keep up with videos. I can tell you,
though, that you should be able to convince someone of the origins of
these different paths pretty easily with a globe and a penlight.
The whole key, of course, is that the earth's axis tilts and the axis
always points the same way (toward the north star). Therefore we
(in the northern hemisphere) are tilted toward the sun in the summer
and away in the winter.

QUESTION:
I need help showing my 7^{th} grade class how to
figure the potential energy and kinetic energy from a point on a model
roller coaster. They will be making a coaster that is no higher than
one meter, and no longer than 1 meter excluding the loops,etc. They are
familiar with the formula for kinetic energy KE=mv ^{2} /2.
They will need help figuring velocity so they can complete the formula.
They have never figured potential energy (not in their text). However,
I would like them to try it.

ANSWER:
Conveying the ideas of potential energy is not easy in a
college class, so conveying it to your 7^{th} grade class will
not be easy! I will try to give you a few tips:

Suppose
we have never heard of potential energy (which is evidently the way
your text author would like it!) How does the kinetic energy of
something get changed? It gets changed by doing work on it.
Work is force x distance (as long as the force is pushing along the
distance direction) and the change in kinetic energy is numerically
equal to the work done on the object. For example, suppose that
there is a 10 kg box at rest on the floor and we push with a force of
10 N and the frictional force is negligible. If we push for 2 m,
then 20 J of work have been done so the kinetic energy has increased by
20 J. But, the initial kinetic energy was zero, so 10xv ^{2} /2=20
=> v ^{2} = 20/5 => v =2 m/s.
Suppose
that instead of pushing the box across the floor we drop it from 2 m
above the floor. It also gains kinetic energy, but what is the
force which is doing the work which increases the kinetic energy?
The force on the box which does this is the box's own weight which is mg =10x9.8=98
N (g =9.8 m/s^{2} is the acceleration due to
gravity). This force acts over a distance of 2 m as in the
previous problem and the initial kinetic energy was zero as in the
previous problem, so the work is 98x2=196 J which must equal the change
in kinetic energy, so 10xv ^{2} /2=196 => v ^{2} = 196/5
=> v =6.3 m/s.
But
there is another way to look at the box-dropping problem. Suppose
we ask how the box got 2 m above the floor. Well, of course, I
put it there by lifting it there. I did this by pushing up with a
force equal to the weight of the box over the distance of 2 m, so I did
an amount of work on the box which was 10x9.8x2=196 J. (Why did
the kinetic energy not change? Because as I was doing 196 J of
work on the box, its own weight was doing -196 J of work so no net work
was done.) What we could say is that I gave the box a new kind
energy, energy by virtue of where it is, which we call potential
energy. The potential energy is easily calculated as the amount
of work you need to do to lift it from where y =0 (you would
probably choose the lowest point on the roller coaster) to some height y ,
so PE =mgy. The clever thing is that if we now say that
the total energy of something is the sum of its kinetic and potential
energies, then we never have to compute the work done by gravity
(weight) again; I often tell my students that PE is just a real clever
bookkeeping device to automatically calculate the work done by
gravity.
To the
extent that friction is negligible, the total energy of your roller
coaster car will be constant and equal to mgy+mv ^{2} /2.
The constant to which it is equal is mgh where h is the
height of the first hill (assuming the car starts there at rest).
Then v may be easily calculated for any y : v ^{2} =2g (h-y ).
Of course, in the real world there will be friction and it will do
negative work on the car and eventually take all the energy away.
If the car starts from rest at the top of the first hill, all other
hills must be the same or lesser height because the energy at the top
of another higher hill would be greater than the energy the system
started with.
I hope
that this is not too detailed and will be helpful to you. Your
being a teacher specializing in 7^{th} graders should make it
possible to translate my discussion to the middle-school level, I hope.

QUESTION:
About that "moving molecules" question —let's say we had a
glass of water. The water molecules would have the "tendency" to form
hydrogen bonds with each other. Wouldn't we need a constant input of
energy to break these bonds and keep the water in a liquid state? If
so, where does this energy come from? Is it, for the most part,
radiation?

ANSWER:
The answer is pretty much the same as the one just
below. Yes, the water molecules want to bond to other water
molecules and that is exactly what happens when the water freezes to
ice. But they do not bond to each other if you have liquid water
above 0^{0} C because the kinetic energy which an average
molecule has is greater than the energy which is necessary to break the
bond. It is kind of like saying: suppose we have two heavy balls
which each are moving very fast in opposite directions and we decide to
tie them together with a little rubber band; the rubber band would just
break. However, if you take the kinetic energy away from the water
molecules (i.e. cool the water down) then eventually they will
bond together. Just as below, there is no need to "keep"
everything moving; Newton's first law takes care of that.
Incidentally, radiation has absolutely nothing to do with anything here.

QUESTION:
Can you explain to me what exactly KEEPS molecules moving?
With no energy being added, they should just eventually stop, shouldn't
they? Where does this energy that keeps them moving come from? In the
end, does it all come down to radiation from the sun?

ANSWER:
You have fallen into one of the most common traps regarding
misunderstanding how the universe works. Newton's first law
states that an object which experiences no net force will continue to
move with constant speed in a straight line. What this means is
that if something is moving and nothing is pushing or pulling on it,
then you do not have to do anything to keep it moving. In terms
of energy, if something has a certain amount of energy, then it will
retain that energy until some external agent changes it; this is called
conservation of energy. I am not sure what you have in mind with
your question, but probably the molecules moving around in a gas.
As you probably know, the temperature of a gas is a measure of the
average kinetic energy per molecule. If the gas is in thermal
equilibrium with the walls, then when a molecule hits the wall it
rebounds (on the average) with the same kinetic energy it had
beforehand. You don't have to do anything to keep it
moving. Incidentally, if Newton's first law were not true we
would never have sent probes to the distant planets like Saturn and
Jupiter or even the close ones like Mars and Venus. The reason is
that if we had to KEEP the probe moving by burning an engine the whole
way we could never carry enough fuel. What actually happens is
that we burn up almost all the fuel escaping the earth and acquiring a
high speed and then we just turn off the engines and coast the rest of
the way.

QUESTION:
I am trying to find out approximately what the speed of an
electron around a hydrogen atom is? A teacher at my son's school said
that an electron around an atom moves close to the speed of light? I
found this difficult to swallow since according to my understanding of
relativity, the mass of the electron would become very large at that
speed and this would be the opposite of its known properties, ie small
mass. I've done some research on the web and according to Bohr's
theory, the speed of the electron should be approximately 1% that of
light. But since his theory is not the prevailing theory and the
Quantum Theory is, I was wondering if their is a probability
distribution for the speed of electrons around an atom. It seems to me
that if Quantum Theory can give us a probability of the location of
electrons around a nucleus, then it can give us a probability of its
speed around a nucleus.

ANSWER:
The teacher who claims the speed to be comparable to the
speed of light is wrong; the Bohr theory gives the correct
approximation of the electron speed to be 2.2x10^{6} m/s=0.7% c
(c is the speed of light). However, as you
note, the Bohr theory is not correct and quantum mechanics gives a
better picture of what is actually going on. However, the
essential details are the same —the most likely place the electron will
be is near the Bohr orbit and the most likely speed is still around 1% c .
Since the ground state of the H atom has a well defined energy, the sum
of the potential and kinetic energies of the electron must always be
the same. If the electron is at a smaller distance from the
nucleus than the most likely, the potential energy will be smaller
(more negative) so the kinetic energy will be larger (implying higher
speed); similarly, if the electron is further away it will be moving
more slowly. To give some perspective, if the electron were 1/100
of its most likely value (a very unlikely place to be found), its speed
would still be less than 10% c . Again, using
nonrelativistic quantum mechanics (Schrödinger equation) is not
exactly right because of the neglect of the (small) relativistic
effects, and the problem should be done relativistically.

QUESTION:
I know metals conduct as the valence band is not filled,
allowing a potential difference to raise an electron's energy level.
But why does this make an electron move, so a metal can conduct?

ANSWER:
The general idea, in very simplified terms, is this: when the
material forms into a solid, the valence electrons of adjacent atoms
interact in order to hold the solid together. In some solids,
called conductors, the interaction is such that all atoms share their
valence electrons in a way so democratic that it is no longer possible
to say that a particular electron belongs to a particular atom.
In fact, to a very excellent approximation, these "conduction
electrons" behave like a gas of electrons. But, of course, in a
gas there is no net wind in a box of the gas because at any instant
there are always just as many particles going in one direction as in
the opposite direction. Similarly, there is no current flowing in
a conductor just because it is a conductor. However, in the box
of gas, if there is something akin to a fan which forces there to be a
net flow of the gas then we will have a "wind". Similarly, if we
cause there to be a force on the electrons by putting in an electric
field created by a voltage across the conductor,
then we will get a net
flow of electrons which is called a "current".

QUESTION:
I am confused by some questions concerned with the difference
among nuclear physics,particle physics,high energy physics etc.I'd love
to study nuclear physics in UGA.But is it true that the particle
physics and high energy physics are also related?

ANSWER:
Well, all of physics is ultimately one science. That is
the beauty of physics —that the connections between all the various sub
areas are pretty well understood (with a few rare exceptions).
Nuclear physics, particle physics, and high energy physics are among
the most closely related branches of physics.

Particle
physics is the study of elementary particles, their properties, their
constituents, and the theories which describe them.
High-energy
physics is sort of ill-defined because there is no clear line where
energy is high above it! Usually it refers to physics done at
particle accelerators around the world which have the highest
energies. In the 1950's 1 GeV (particles with energies equivalent
to accelerating an electron across a potential difference of a billion
volts) was high energy. Today it is more like a TeV (a thousand
GeV) for high energy. Usually high energy physics is synonymous
with particle physics but need not be. Sometimes high energy
accelerators are used to study solid state physics or nuclear physics.
Nuclear
physics was traditionally separate from particle physics, but if you
think about it, that is kind of artificial, because what is a nucleus
made up of anyway? From the 1930's through the 1980's the study
of nuclear physics was just that —attempting to understand the
structure of the nucleus. The two main means of achieving this
are to look at the radioactive decay of nuclei and to look at particles
scattered from nuclei (particles which have been accelerated in an
accelerator).
By the
1980's and 1990's, great progress had been made in understanding
nuclear structure in terms of a collection of protons and neutrons
interacting via phenomenological forces. So much so that
traditional nuclear physics is now thought by many to be an essentially
solved problem. There are still lots of details and loose ends,
and these can be interesting problems, but that is not the real essence
of forefront science. Today, much of the field of nuclear physics
has merged with particle physics/high energy physics. Now nuclear
physicists are attempting to understand nuclear properties in terms of
the underlying constituents of the protons and neutrons —quarks and
gluons. In addition, inside the nucleus is being used as a place,
a "laboratory" if you like, to do particle physics. Nuclear
physics is still a vibrant and exciting branch of physics —it just
isn't what it was 20 years ago and that is good!

QUESTION:
The north pole of a bar magnet is moved towards a conductive
loop of metal, from one side, perpendicular to its plane. Within the
loop, which direction does the current flow, from the magnet's point of
view? Which way does the current flow if the magnet approaches from the
opposite of the loop, again as seen from the magnet?

ANSWER:
The figure (plagiarized from Serway and Beichner) illustrates
the answer to your question: counterclockwise as seen by the
magnet. The reason that it goes that way is because Faraday's law
states that the induced current will create a field which tends to
oppose the change of field causing it. If, as shown on the left,
the N is moving toward the loop, the magnetic field through the loop is
increasing and pointing to the right. Hence the induced current
should have a field pointing to the left as shown by the right-hand
figure. If the N were approaching from the other side, the
current would go the opposite way but it would still go
counterclockwise as seen by the magnet.

QUESTION:
Hello. I hope my question isn't too unfocused.

There is
a theory I've been trying to remember, but I can't get the details
right. I just remember it in a vague way. Hopefully you can help.

It's
about two particles (atoms?) with similar charges (?) (or identical in
some way, or perhaps connected and then broken into two pieces), so
that if something affects one of the particles, it will affect the
other particle. The theory is that this will be true no matter the
distance between the two particles: if one is here and the other is on
the other side of the galaxy, affecting the particle here will
INSTANTANEOUSLY affect the particle on the other side of the
galaxy.

Naturally,
this seems to contradict the limits of the speed of light, but it had
something to do with bending space and the seemingly instantaneous
affects of gravity. I think.

Anyway,
hopefully this sounds familiar to you. I'd appreciate any help you can
give on details about this theory (briefly is fine). Thanks very much.

ANSWER:
What you are thinking about is often referred to as the EPR
(Einstein, Padolsky, Rosen) paradox or, in modern parlance, quantum
entanglement. The idea, roughly speaking is the following.
Attributes (things which can be measured) of physical systems in
quantum mechanics are often not definitive until you make a measurement
at which time your measurement actually places the object in the state
which has the attribute which you observe. I am going to make up
an example which, although frivolous, illustrates the paradox.
Suppose there is some particle and that particle can be either red or
green. Furthermore, suppose that if you have a pair of those
particles, one must be red and the other green —you can never have two
red or two green. So the attribute, that which the so-called
"state" of the system would describe, would be "red/green-ness".
The fundamental truth about quantum mechanics is that the state of a
system can be a mixture, that is a particle could be say 50% red and
50% green (strange as that may seem); so, for a pair of particles, if
one is 50% red and 50% green, so is the other. In practice what
this means is that if you have many such pairs and measured the color
of one of them, half the time you would see green, half red. You
are said to be, by virtue of the measurement (looking at it),
"collapsing the wave function", actually placing it in a pure red or
green state; but, since the pair must be one red and one green, you
must simultaneously place the other particle in the opposite
state. So the "faster than the speed of light" part comes in when
we let one of the particles go very far away before we look at the one
we keep at home. Then, instantaneously when we measure one the
other is determined even if it is on the other side of the galaxy!

There is
a very nice discussion of all this on the web at http://fergusmurray.members.beeb.net/Causality.html

QUESTION:
What are the wavelength and frequency of red, orange, yellow,
green, blue, and violet lights?

ANSWER:
W ell, these are subjective things, not some well-defined
standard. The picture to the right shows somebody's qualitative
idea of where the colors fall on a wavelength scale. The units,
nm, are nanometers, 10^{-9} or one billionth of a meter.
(Incidentally, the size on a single atom is about one tenth of a
nanometer.) You can roughly read off the center of each of the
colors you ask about:

red:
680 nm
orange:
640 nm
yellow:
610 nm
green:
540 nm
blue:
460 nm
violet:
420 nm
To get
the frequencies you divide the wavelengths into the speed of light, 3x10^{8}
m/s. For example, the frequency of 680 nm red light would be 3x10^{8} /680x10^{-9}
s^{-1} =4.41x10^{14} cycles/second.

QUESTION:
If you placed a one gallon pot of water on a six inch
electric burner that had a temperature of 500 f the water will soon
boil. How much water can this six inch 500 f burner boil. five gallons,
10 gallons, 100 gallons, or more. Do we have a formular to solve this
problem. What would be the maximum gallons of water the burner will
boil and how long will it take to come to a boil.

ANSWER:
W ell, since you say an electric burner, I presume you mean that
this thing is plugged in. In that case, energy is entering the
system continuously and there is no limit to the amount of water you
could boil if you are willing to wait long enough. In the real
world, of course, you would have to worry also about energy leaving the
system; for example if you stuck your burner into a lake then heat
(energy) would enter the lake from the burner but it would leave via
the surface and the bottom of the lake, etc. and would
therefore certainly never boil the lake because equilibrium would be
reached where heat was leaving as fast as it was entering. So it
is always a question of rates of heat flow in and out. On the
other hand, many physics problems in textbooks assume that you are
dealing with an isolated system, so if you could isolate the whole lake
then eventually the lake would boil.

On the
other hand, perhaps you are interested in what happens if the burner is
not plugged in but is very hot. A simplified analogous example
would be dropping a 1000^{0} chunk of iron into a pot of water
and asking what is the maximum amount of water that it would
boil. This is a simple example of a class of physics problems
called calorimetry. Calorimetry is based on the equation Q =mC D T where Q is the
amount of heat necessary to increase the temperature of a mass m
of something by an amount D T=T _{final} -T _{initial
} and where C is called the specific heat of the
material. Notice that if the initial temperature is larger than
the final temperature then Q will be negative indicating that
heat was given up by the object. For isolated systems, problems
usually take the form

Q _{net} =Q _{1} +Q _{2} +Q _{3} +...=m _{1} C _{1} D T _{1} + _{1} +m _{2} C _{2} D T _{2} + _{2} +m _{3} C _{3} D T _{3} +...= _{3} +...= 0

which
says the total heat entering the system is the sum of all the heats
entering and leaving the individual pieces and that must equal zero
because the system is isolated. One quick example: suppose
that you have a 2 kg chunk of iron at 1000^{0} C and an unknown
amount of water m at 20^{0} C in a 1 kg aluminum pot
also at 20^{0} C. Find m such that everything ends
up at 100^{0} C. Then D T _{iron} =- 900,
D T _{water} = D T _{aluminum} = 80.
The specific heats of iron, aluminum, and water are 448, 900, and 4186
respectively (the units are not important for this problem, but they
are J/kg^{0} C). Putting all this information into my
equation above I find:

2x448x(-900)+1x900x80+m x4186x80=0
=> m =2.19 kg.

So, you
could boil 2.19 kg of water. There is a complicating feature of
calorimetry which I have not discussed here: nothing must evaporate or
freeze or melt or condense since these do not absorb or reject heat
according to Q =mC D T .

QUESTION:
If 2 combined materials that are carbon dated to be from the
year 1300 AD, and it is found that 60% of the material was from the
year 1500 AD, can the date of the other 40% of the material be
approximately calculated?

ANSWER:
Well, I am certainly not an expert but I understand how
carbon dating is done. I can give you an answer in theory.
Someone who actually does it would have to judge whether it would be
practical in practice. Also I am going to take a somewhat
different perspective (from your actual question) because it lends
itself to a clearer explanation: rather than consider what I would
consider an erroneous dating, the details of which I do not know, I
will assume that we have two materials, one of known age and one of
unknown age whose relative amounts are also known, and that we simply
take a measurement of the radioactivity of the source. I am not
going to go through the derivation of the formula I derived to answer
this question for myself. If you are really interested, send
another question. What I find is:

t _{B} =t _{A} -(1/l )ln[(N _{0A} /N _{0B} )((R
exp{ l t _{A} })/(l N _{0A} )-1)]

where t _{B}
and t _{A} are the ages of the two materials, N _{0A}
and N _{0B} are the number of ^{14} C atoms at
the "creation" of each of the two materials, l
is the decay constant of ^{14} C (about 1.2x10^{-4} yr^{-1} ,
corresponding to a half life of about 5700 yr), and R is the
current rate of radioactivity coming from the source. (ln[] is
natural logarithm, exp{} is exponential function.) Everything on
the right side of the equation should be known:

If we
know the total amount of the sample and the relative amounts of the two
materials then we can infer N _{0A} and N _{0B}
(assuming, of course, that carbon dating works).
We
know t _{A} .
We
measure R .
Here is
an example: t _{A} =1000 yr, N _{0A} =10^{11} ,
N _{0B} =2x10^{11} , R =3.07x10^{7}
yr^{-1} (corresponding to about one decay per second), then t _{B} =2023
yr. That does not explicitly answer your question, but it shows
that the answer to your question is a definite yes in theory. If
you know abundances of both and age of one then you can infer age of
the other from the observed radioactivity.

QUESTION:
what is "energy" really? heat, motion, capacity to do
work.....all of these?! i am very confused about the concept of
energy....any info where it is looked into philosophically?

ANSWER:
Yes, all of these and more! When I teach the concepts
of energy to students in introductory physics, I always emphasize that,
in many respects, energy is just a clever trick to either make problem
solving easier or to allow us to solve harder problems. But it is
not a new fundamental thing; it is just casting Newton's laws (which
are the fundamental things in classical mechanics) into a new
form. In more advanced courses, energy arises naturally from
mathematical manipulation of the differential equation which is
Newton's second law. The "elemental" form of energy, the one most
easily grasped, is kinetic energy, the energy something has by virtue
of its motion. The way that kinetic energy can be changed is to
do work on the system. Kinetic energy is useful in the
kinetic theory of gasses and solids —temperature is nothing more than a
measure of the average kinetic energy per molecule. The next
thing we introduce in mechanics is the potential energy which
attributes energy to something by virtue of its position; I always tell
my introductory students that potential energy is basically a
bookkeeping device —it automatically keeps track of work done by some
force which is always present. What you gain with energy
compared to forces is that energy is a scalar which leads to easier
problem solving; the price you pay is loss of information about the
time development of the system. An important thing we learn from
studying energy in classical mechanics is the beauty and power of
conservation principles —some things, like the total energy of an
isolated system, are constant with respect to time.

Physics
at more advanced levels attributes more importance to energy than a
cute trick or a bookkeeping device. In the theory of special
relativity, the energy associated with the mass of an object emerges as
an unexpected important property of material objects (E=mc ^{2} ).
In electromagnetism we are often interested in the energy contained in
electromagnetic fields. In quantum mechanics, energy is one of
those variables in nature which you can not know to arbitrary precision
(that is, a Heisenberg uncertainty principle applies to energy).
Even empty space contains energy (called vacuum fluctuations) whereby
the vacuum is viewed as particle/antiparticle pairs popping into and
out of existence (since the uncertainty principle allows conservation
of energy to be violated if for a short enough time).

I do not
know much about philosophy of science (although I do a lot of
philosophizing on my own!), so I cannot point you to a philosophical
reference on energy. Surely there must be good ones, though.

My answer to an earlier question might also be of
interest to you.

QUESTION:
Please solve the following problem for both teachers and
students and explain your rationale. We are confused or we are thinking
too hard. Thank you, Judy

A curious
kitten pushes a ball of yarn at rest with its nose, displacing the ball
of yarn 17.5 cm in 2.00 s. What is the acceleration of the ball of yarn?

ANSWER:
There is technically not enough information given to answer
the question, but I am sure the intent was for you to assume uniform
(constant) acceleration. So, you may apply the "equations of
motion" for uniform acceleration,
x =x _{0} +v _{0} t +at ^{2} /2,
v =v _{0} +at
where x _{0} and v _{0} are the position
and velocity when t =0. Choosing t =0 when the
kitten starts pushing, then v _{0} =0 and we may choose x _{0} =0.
So,
17.5=2^{2} a /2=2a
=> a= 8.75 cm/s^{2} .
Incidentally, you can also find the speed the ball has at this time:
v =8.75x2=17.5 cm/s.

QUESTION:
WHAT IS YTTRIUM?

ANSWER:
Yttrium is an element with atomic number 39 and atomic weight
of about 89. You can get some detailed information from webElements .

QUESTION:
I know a little bit about Physics but have always wondered;
Suppose a Spacecraft zooms by me at a velocity just below c. A
scientist is onboard conducting Young's Double slit experiment with the
screen lying parallel to me with the Photon emitter between me and the
screen.

=> v~c
SCREEN
_______________________________________________________________
____________________ ____________________
_____________________
SLITS
^
Emitter
o
-[]- ME!

If, say, at the very moment the craft passes me, a photon is
fired at the screen and I start a timer to measure from the time the
photon is fired to the moment it strikes the screen.
I know I'm not supposed to know which slit the photon passes through ,
but wouldn't a photon passing straight through the slit on the right
take longer than if it passed straight through the slit on the left?

ANSWER:
Gosh, it's hard for me to believe that this is something you
have always wondered about! Anyhow, it is a question which simply
cannot be answered within the constraints you put on it —that we know
which slit it goes through and that we know that it goes from the
emitter straight through the slit and straight to the screen.
This is the whole idea of quantum mechanics, that these particles
(photons) will behave like waves (light) or vice versa
depending on what you look for. The guy in the craft will see a
diffraction pattern on the screen even if he only shoots one photon per
day and this is an indisputable fact. If he sees this wavelike
behavior, so will you.

So let us
agree to dispense with the double-slit part of your question. It
seems that what you really want to know is: If a photon takes a certain
straight line path (and time) between the emitter and the screen (as
seen by the on-board observer), what path and time are observed by
you? That is not a hard question to answer. Click here .

QUESTION:
What is the chance of a quantum atomic orbital (ex.1s,2p,2d)
transforming or transmuting into another kind of atomic orbital (ex. 1s
to 1p)?

ANSWER:
The example you state is impossible if you mean that the
orbit must "transform itself ". The 1p orbital will be
higher in energy than the 1s, so for this to happen spontaneously would
violate energy conservation. However, it can be induced to
happen. The most common way is to shine electromagnetic radiation
on the atom, for example a laser. Then, the chance that it will
happen will depend on things like the intensity and wavelength of the
light. However, transformations from a higher state to a lower
state happen all the time. For example you can get a transition
from the 1p to the 1s level and the excess energy comes out as
electromagnetic radiation. This is essentially where all light
comes from —deexcitation of excited atoms. The probability of
this happening must be calculated using quantum mechanics. Most
atomic radiations are what is called electric dipole radiation and the
way you calculate the probability is to calculate something called the
transition matrix element. {If you are familiar with quantum
mechanics, this is <y _{2} |P
|y _{1} >
where P
is the electric dipole
operator and y _{2} and y _{1} are the wave functions
of the final and initial states.}

QUESTION:
I heard speeding up (gradually stepping more on gas pedal)
while driving on a long, sharp curve was better (safer) than slowing
down (gently braking). Is this true?

ANSWER:
Let us assume that the main safety concern in turning a curve
is the possibility of skidding. My answer will address that
concern. There are lots of other issues that could affect
"safety" such as whether you have better control while accelerating a
front-wheel or rear-wheel drive car, or whether you have a four-wheel
drive car, or the fact that braking occurs on all four wheels but
acceleration uses (usually) only two, etc . These are
engineering issues which I don't want to get into. In the safe
use of a car, the force of static friction between the road and
the wheels provides the force to brake the car, accelerate the car, or
move a car around a curve. The fact that it is static friction
confuses some people since the car is obviously not static. However, if
all is well, the tires are at all times at rest where they touch the
road. An important feature of static friction is that there is a
limit to how much you can get; for example, if a heavy box is sitting
on the floor, you can push gently and it won't move, push harder and it
won't move, push harder still and it won't move...but eventually it
will "break away" and start sliding across the room. This is what
happens when you slam on your brakes and start to skid, "peel out" when
accelerating, or break into a skid when turning a curve too fast.
When you go around a curve, the static friction from your tires must
provide a force equal to mv ^{2} /R where m
and v are the mass and speed of the car (this could be smaller
for a banked curve) and R is the radius of the curve. Let
us assume that you are going through with the fastest speed possible so
that if you went any faster you would skid. This force is labeled
'turn' in the two pictures above which show the forces on the
"footprint" of a tire on the road; this 'turn' force is pointing toward
the center of the circle about which the car is turning. The two
pictures show what happens if you either accelerate or brake: an
additional force is required and the sum of the two, labeled 'net', is
bigger than the static friction can give and you will go into a
skid. Therefore, if you fear you might be approaching a curve
going too fast, you should do your slowing down before you
start the curve, don't try to do it while you are negotiating the
curve.

QUESTION:
I was wondering about the physics of swimming. Is the
friction created by moving hand through the water a linear dependence
on the velocity of the hand or some exponential. I was wanting to know
how the push one would recieve depends on how fast they pull their
hands. And any other physics that relates to swimming.

ANSWER:
Resistive forces like friction are very complicated.
There are lots of little "rules of thumb" in certain circumstances
which give one an approximate answer for what the frictional force
is. For example, for sliding friction the force is proportional
to the force pressing the surfaces together (normal force), for slow
velocities through fluids the force is roughly proportional to the
speed, and for faster velocities through fluids the force becomes
roughly proportional to the square of the velocity. So it looks
like the drag force would initially increase linearly with speed and as
you got faster and faster it would eventually increase quadratically
without bound! However, these rough guidelines do not work if the
motion through the fluid causes there to be any significant turbulence;
if there is turbulence, all bets are off and you are most likely to get
a drastic reduction in the drag.

One
example of this is a golf ball; did you ever wonder why it has
dimples? They are there to induce turbulent flow so the ball will
go farther.
Another
example, closer to your swimming problem, is the propeller in a
boat. You would think that the faster it spun the more thrust you
would get. In fact, turbulence causes something called cavitation
where the propeller makes a "hole" in the water so it just spins on its
own.
As is
often the case, there is no simple answer to your question.
Basically, you have to experiment to find the best arm speed for your
stroke.

QUESTION:
After using google to search countless web sites, I still
can't find the answers to the following questions.

How
permanent are permanent magnets?
If 2
permanent bar magnets are place in a way that their "north" pole face
each other (causing them to repell each other), will their polarity
change?
How do
you calculate (or measure) the strength of a permanent magnet and
electromagnets.
ANSWER:
Well, of course, nothing is truly permanent.

What
makes a permanent magnet (PM) permanent? In a normal piece of
material from which the PM is made, there are regions small compared to
the whole piece but large compared to atomic scales which are called
domains. In these domains, the magnetic moments of many electrons
are all aligned in one direction so the domain is, itself, a tiny
permanent magnet. However, these domains are all randomly
oriented relative to each other so the piece of stuff is not itself a
permanent magnet. The trick is to get all the domains to align
with other. The only way I know to accomplish this is by brute
force —put it in a very strong magnetic field. Some materials,
like just pure iron, will become magnetized but when the magnetizing
field is turned off the domains will tend to migrate back
toward what they were so the iron will be a weak PM which will, over
time become weaker. Many other materials are more suitable as PMs
because, after being magnetized they tend to stay magnetized.
Nevertheless, over a long time they will tend to become
weaker.
A second way that a PM can lose its magnetization is if it is heated
up. When a PM reaches a temperature called the Curie temperature
(CT), all magnetization is lost and it will cool back down to an
unmagnetized state. In fact, even heating a PM up to a
temperature which is lower than the CT will lead a more rapid decay of
its magnetization than would happen at cooler temperatures.
If a
magnetic field can cause magnetization, then it certainly can change it
also. Putting these two PMs together like this would cause both
to become more weakly magnetized but, because the fields of individual
PMs are considerably smaller than the fields which magnetized them in
the first place, the effect would likely be small.
The
strength of a magnetic field depends on lots of things —not just how
magnetized the magnet is but on the shape of its poles, how far you are
away from it, etc . I presume that you are just interested
in how to measure a magnetic field. You can buy commerical
gaussmeters or build
your own . You just stick the probe in the region where you
want to measure the field and read it off a meter. How do they
work? Several ways which I will not give details of how they work
since you could look them up: (a) the Hall effect where a voltage is
generated when a known current passes through the probe in a magnetic
field; (b) a rotating coil in which there is an induced current when it
is in a magnetic field (like how a generator works); (c) nuclear
magnetic resonance in which magnetic moments of nuclei flip when
exposed to electromagnetic waves of just the right frequency.
Another way would be to measure the force on a current carrying wire or
the torque on a tiny bar magnet (little compass).
QUESTION:
A mass m on a string of length R _{0}
is rotating with angular velocity w _{0} . The
string breaks and a force, with no radial component, is exerted on the
mass; the effect of this force is to keep the angular velocity of the
mass unchanged. Describe the motion of the mass.

ANSWER :

QUESTION:
What is the speed of gravity? If you don't have a exact
speed calculated, is it faster than the speed of light?

ANSWER:
No one has ever measured the speed at which the gravitational
force propagates. In Newtonian physics we calculate orbits by
assuming that the forces are instantaneously transmitted, but nobody
believes this —it just works pretty well because realistic propagation
times are bound to be very short over the distances of interest.
A better theory of gravity is general relativity. Here the theory
predicts that gravity is due to distortions of spacetime caused by the
presence of mass and these distortions are predicted to propagate with
the speed of light. In general, all the physics we know forbids
any information propagating faster than light speed. Finally, we
believe that a theory of quantum gravity will someday be found in which
the quanta which transmit the force will be massless particles called
gravitons (similar to photons which transmit the electromagnetic
force). Massless gravitons, like all massless particles, would
have a speed equal to the speed of light so the force which they
transmit would move with that speed. There is a very enlightening
essay at http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

QUESTION:
Assume for the moment that I can make a giant Bose-Einstein
condensate, as big as a building. (Obviously, I would need a big
refrigerator!!) (See Cornell & Weiman) Anyway, if I could make such
a construct would the strong force and weak force display their effects
on a macroscopic level. Optional bonus question: If the universe is
going to be like super duper cold in a trillion gazillion years and all
matter reduced to its basic component parts, (See Fred Adams, Five
Universal Stages or something like that) is it possible that phase
transitions in that super cold environment would create super biggie
Bose-Einstein condensates? i.e much much bigger than buildings!!

ANSWER (#1):
(provided by M. R. Geller)
No. The atoms in these experiments behave essentially as
elementary particles, even though they are really composite objects.
The only thing that matters is their statistics (Bose or Fermi) and
their inter-atomic interaction forces. Things happening inside the
nucleus are not important at these cold temperatures.

ANSWER (#2):
I am no cosmologist and there is no cosmologist in our
department; but I can give a general answer based on what I have
read. (You might want to ask elsewhere to see what other answers
you can get.) The universe will not end up as just some big cold
chunk of stuff. Although nobody knows for sure what the detailed
picture of the end of the universe will be, the likely scenarios are
either a collection of black holes flying away from each other or one
black hole "containing" everything. I have difficulty imagining a
black hole making a "phase transition" into a BE condensate.

QUESTION:
If there is a ball that is traveling on a vertical axis at
2m/s and hits a 45 degree angle, it will then travel at 2m/s to the
left. Now lets say that before the collision the ball and the angle
were traveling to the left at 3m/s relative to the planet Venus (or
whatever). After the ball collides with the angle it's total velocity,
relative to venus will be 5m/s to the left. Before the collision,
relative to Venus, the ball was traveling at 3.8m (I used the
pythagorean theorm with 2m/s and 3m/s). How is it that the ball has
increased velocity relative to Venus without any energy being added to
the system. The angle should only redirect the velocity. If the ball
hit any kind of angle that was fixed in space relative to Venus there
would be no way to change the velocity from 3.8 to 5. How is this
possible?

ANSWER:
I can answer your question, but you have made it far harder
than it need be. Furthermore, I am not sure that you have gotten
the velocity of the ball after the collision right for the second
scenario, but it doesn't really matter since we can look at a much
easier example (easier because it is a one-dimensional collision,
whereas yours is two-dimensional) and learn where this mysterious added
energy comes from. Suppose we have a ball which collides
elastically with a stationary wall; if the ball comes in with speed v
to the right, then it rebounds (to the left) with speed v
also —energy is conserved. Now, if the
wall is moving toward the ball (to the left) with speed u , the
velocity of the ball after the collision will be v'=v +2u to
the left and the velocity of the wall will still be u'=u to
the left. How did I get these speeds after the
collision? Well, these are the speeds only if the wall has an
infinite mass. We get the results by conserving linear momentum
and kinetic energy for a collision of a ball of mass m and a
wall of mass M and then let M approach infinity.
So they are approximate solutions to the real problem if M is
enormously bigger than m which is what you are presumably
assuming. But these solutions cannot be exactly true because no
wall is really infinitely massive. But, if it were, it would have
an infinite amount of energy before the collision if it were moving; it
then loses the amount of energy which the ball gains, but if you
subtract a finite number from infinity you still get infinity. So
it sort of looks like energy materialized from nowhere but it really
comes from the infinite energy which the wall has! So, basically,
any energy which the ball gains must come from the wall.

Incidentally,
the general solutions to the problem if M is not infinite are
v' =[(m-M )v -2Mu ]/(m+M )
u' =[(m-M )u+ 2mv ]/(m+M )
which you can show are approximately v'=- (v+ 2u )
and u' =-u for very large M , that is M>>m.
(The minus signs denote that they are both traveling to the left.)

QUESTION:
When the sun is in the sky, it is too bright to look directly
at, but at sunset we are able to look directly at it. Why is this so?

ANSWER:
When light passes through our atmosphere, molecules of air
and particles of dust and smoke scatter the light. The picture at
the right shows light coming straight down through the atmosphere at
midday and light coming through the atmosphere at sunset. As you
can see, the light at sunset has much more atmosphere to pass
through. Therefore, a much larger amount of the light from the
sun is scattered out of the original beam of light at sunset so it is
much less intense and therefore easier to look at. Incidentally,
this is also the reason the sky is blue. As you know, the sunset
is very red. The reason for this is that the atmosphere scatters
blue light much more efficiently than red light so more blue gets
removed by scattering. When you look at a point in the sky you
are really seeing light from the sun which has scattered from that
point and you see blue predominantly. If there were no
atmosphere, the sky would be black even in the middle of the day; you
may have seen pictures taken on the moon where the sky is black when
the sun is in the lunar sky.

QUESTION:
I have a question regarding Kinetic Frictional Force. The
magnitude of the kinetic frictional force is equal to the coefficient
of kinetic friction multiplied by the value of the normal force;
therefore, there is no dependence on surface area. Based on that, I was
wondering if it were fair to say that tire companies that advertise
that wider tires "grip the road" better than average size tires (made
of identicle material), are deceiving consumers? Since wider tires cost
considerably more, I was just wondering if they were misleading the
public or if there were other forces/laws that apply to this scenario
that do involve surface area.

ANSWER:
This is not an easy question you have asked! I did a
little research and here a few particularly enlightening answers to
this question which I have found (stolen!) on the web:

"In
general, most of the responses were speculative, and centered on
factors such as the nature of the tire-road surface interface, heat
dissipation, tire deformation and elasticity, stability, and a myriad
of other environmental and engineering constraints. The 'best' answer
suggested that the tire problem did not fall under the standard Coulomb
(dry) friction parameters, thus using tires as an example was comparing
apples to oranges.
"Indeed, further digging supports the latter statement. According to
Engineering Mechanics: Vol. 1, Statics (2nd ed), JL Meriam and LG
Kraige, Wiley and Sons: New York, 1986, the coefficient of ROLLING
RESISTANCE, while analogous to the coefficient of static or kinetic
friction, is really an entirely different beast. It would be most
difficult to describe fully without a free body diagram, but is a
function of many factors, including, but not limited to: road and tire
deformation and the resultant pressure over the area of contact,
elastic and plastic properties of the mating materials, wheel radius,
speed of travel, and roughness of the surfaces. Meriam and Kraige
state, '... depends on many factors which are difficult to quantify, so
that a comprehensive theory of rolling resistance is not
available.'" (Plagiarized from Biomch-L Newsgroup )
"This
is a good question and one which is commonly asked by students when
friction is discussed. It is true that wider tires commonly have better
traction. The main reason why this is so does not relate to contact
patch, however, but to composition. Soft compound tires are required to
be wider in order for the side-wall to support the weight of the car.
softer tires have a larger coefficient of friction, therefore better
traction. A narrow, soft tire would not be strong enough, nor would it
last very long. Wear in a tire is related to contact patch. Harder
compound tires wear much longer, and can be narrower. They do, however
have a lower coefficient of friction, therefore less traction. Among
tires of the same type and composition, here is no appreciable
difference in 'traction' with different widths. Wider tires, assuming
all other factors are equal, commonly have stiffer side-walls and
experience less roll. This gives better cornering performance."
(Plagiarized from physlink.com )
"Now
let's look at what happens at the tire-to-road interface. In a perfect
physic's laboratory world, one usually asserts that the friction force
(i.e., the braking force) between two objects is a function of two
things: the coefficient of friction at the interface, and the normal
force. (Muddying the waters more than a little bit is the fact that
this is only true for two "smooth" surfaces sliding against each other.
It doesn't take into account that rubber is relatively malleable, and
tends to fill the small nooks and crannies in a roadway, thereby
changing the nature of the "braking" problem into one containing both
frictional and shearing-type forces. This turns out to be a VERY
difficult complication. Worse, the vulcanized forms of rubber used in
most automotive tires has rather weird coefficient properties that
don't behave nice and linearly. But I digress...)" (Plagiarized
from datsuns.com )
You can
find a lot more by going to Google
and searching on friction AND "surface area" AND tires .

QUESTION:
I understand that if negative and positive matter touch you
get E=0, m=0, or in other words: Nullification. But what happens just
before they touch? Since E=mc^2 can never be violated, and since
everything in the universe ultimately comes from one field or another,
when negative and positive matter 'almost' touch could local normal
space/time be modified, creating a new E, m, and c? Relativity would
still apply, but with a different c, which may be many times faster
than our 'regular' c.

ANSWER:
When you refer to "negative and positive matter", I assume
you mean "matter and antimatter". Positive and negative are words
which usually refer to electric charge. "Nullification" is
certainly not what usually happens when a positive and negative charge
get together; for example, if you put an electron (negative) and proton
(positive) together you get a hydrogen atom. A particle and its
antiparticle usually have opposite electric charge (exceptions being
neutral particles which have neutral antiparticles). When a
particle and its antiparticle "touch", they annihilate, that is all
their mass disappears (m =0). However, where you are
incorrect is in saying that you get zero energy (E =0) because
energy is conserved. E =mc ^{2} doesn't have
to mean that if you have no mass then you have no energy. For
example, radio waves carry energy but no mass. What normally
happens when an electron and its antiparticle (a positron) annihilate
is that two "photons", quanta of electromagnetic energy, are created
which have all the energy which the electron and positron had before
the annihilation, but none of the mass.

QUESTION:
Have you heard about "red mercury" fusion bombs (they use
mercury in place of plutonium) or would you know if sensors could be
made to detect such a bomb?

ANSWER:
No I haven't heard of them. I can tell you, however, that the
only plutonium in a fusion bomb would be that used in the fission bomb
used to trigger the fusion reaction. I can also tell you that mercury
is not a fissionable material, at least not to the extent that you
could sustain a chain reaction with any reasonable amount of it. It
sounds to me like what we have here is an urban legend!

QUESTION:
I understand the basics of photoelectric effect, in that a
photon hits a metal and if the photon brings sufficient energy an
electron is emitted with an energy equal to (incoming photon frequency
times Planck constant) minus work function - I think.
OK, so the problem is this. What happens if the emitter and the target
are approaching at some high velocity - say a good proportion of the
speed of light. The incoming radiation will be blue-shifted by Doppler
effect, so each electron will have higher energy when it hits the
target due to its higher frequency. However, does the intensity of the
radiation (and thus the number of emitted electrons) change due to
relativistic time dilation effects, or are there any other effects to
be considered?

ANSWER:
For simplicity of illustration, I will assume that each
photon results in one photoelectron; this is completely false, but if
there is some fraction f which represents the number electrons
per photon, this can just be applied to any answer. Also, there
may be some slight dependence of this fraction f on the energy
of the photon; this does not seem to be what you are interested in, so
I will assume that there isn't. Of course, there is because if
you moved S and T apart so that you redshifted the photons below the
work function you would get nothing. Although I am not certain, I
believe that, provided the energy is greater than the work function,
then the probability of getting a photoelectron is pretty much
independent of the energy.

Now, to
your question: it all depends on what the relative velocities of the
source (S) and the target (T) are, relative to the observer. Let
us think of the source as a clock which emits photons at the rate of R= 1/t
photons per second in its own rest frame where t
is the time between photons. Let's enumerate the possibilities:

If S
and T are both at rest relative to you, you see R electrons per
second.
Suppose
S moves toward T with speed v _{S} , T stays at
rest. Now you have a moving clock and, because of time dilation,
you will see not R but rather R (1-(v_{S} ^{2} /c ^{2} ))^{1/2}
photons (and thus electrons) per second, a lower rate because moving
clocks run slower.
Suppose
T moves toward S with speed v _{T} , S stays at
rest. Now you see photons at the rate R so the distance
between these photons is ct (=c/R ). Imagine
T encountering a photon; if it encounters the next one at a time t'
later then it has moved a distance v _{T} t' and
the next photon has moved a distance ct' to meet it. So ct=ct' +v _{T} t'.
Solving, t'=t/ (1+(v _{T} /c)).
Therefore, you see not R but rather R (1+(v _{T} /c))
electrons per second, a higher rate because the target intercepts the
photons more quickly than they are coming.
If
both are moving, the rate will be the product of both effects, R (1-(v_{S} ^{2} /c ^{2} ))^{1/2} (1+(v _{T} /c)).
You
could also consider motion apart rather than toward each other.
This would have no effect for S (because V _{S} is
squared) but would make the rate at which T intercepted photons smaller
instead of larger. In that case, the rate would be R (1-(v_{S} ^{2} /c ^{2} ))^{1/2} (1-(v _{T} /c)).
QUESTION:
I have a question about "work". Many of the basic science
texts discuss "work" in the context of a human pushing on some object.
However, I'm curious as to whether stars can do "work". For example, is
the sun's converting hydrogen to helium or some such other activity
considered "work".

ANSWER:
Without getting into the technical details of how to compute
the work which some force does on some system, we can generally define
work as that which changes the energy of some system. If positive
(negative) work is done, the energy of the system increases
(decreases). So, essentially, what the system is defined to be
determines whether any work is done. For example, if you define
the energy of a falling stone to be its kinetic energy, then the work
which gravity does on the stone changes its kinetic energy; on the
other hand, if you introduce a potential energy for the stone and
define the energy to be its kinetic plus potential energy, then the
energy doesn't change so there are no external agents doing work on the
stone. In the example which you cite, fusion in the sun, we could
say that the energy of the sun is the kinetic energy of all its
components plus the "rest mass energy" of all its components (this is
the mc ^{2} part); when fusion happens mc ^{2}
gets smaller, so kinetic energy gets bigger and this added kinetic
energy is eventually radiated away into space. No work was done
by an external agent. One could, however, define the energy of
the sun to be its total kinetic energy; in this case we would have to
say that the process which caused the energy to increase (fusion of
hydrogen) did work on the sun.

Most
people think of science of being rigid and absolute, but answers to
questions like yours point out that different precise definitions of
what we mean can lead to different answers to the same question.
But the answers are all correct!

QUESTION:
1. Does gravity bend light or the space it's moving through?
2. If all motions and forces could be subracted from the Universe, what
would remain?

ANSWER:
1. It all depends on semantics. If we look at the world
as having a Euclidean "flat" geometry and watch a ray of light pass a
very massive object, we see the light bend. But, the way that a
general relativity guru would have us see it is that the light follows
a straight line but that the geometry of the space in the vicinity of a
massive object is changed from the flat Euclidean space we envision.
2. This question doesn't really make any sense to me! I don't see
what you are after.

QUESTION:
I'm writing a science fiction story in which humans go to
another planet where everything on the planet is seen by them as black,
white and gray. Is there a scientifically plausable way that this could
happen?

ANSWER:
Your question has been previously
answered in a different context. There the question was
whether you could make a "filter" which would take light from real
objects with colors and then pass an image which is only "black and
white and gray". You will see that, essentially, the answer is
no. The "scientifically" plausible ways it could happen would
be

if
everything on this planet really was black, white, or gray;
something
caused the human brains to intepret the information from the eyes
without the color information; or
something
induced total color blindness in the humans.
The last
might be of most interest to you for your story line. There are
detectors in your eye called rods and cones. The cones are
responsible for color vision whereas the rods detect only the intensity
or brightness of the light. If, for some reason, none of the
cones (there are three kinds, for red, green, and blue light) worked,
you would see in black and white. The rods tend to be more
sensitive than the cones which is why you have trouble seeing colors in
very low light conditions. So you could introduce some "cone
destroying or cone debilitating force field"! Not really very
scientific, but hey, this is science fiction, right?

QUESTION:
Given that the neutron has a magnetic moment - is this why
neutron stars have very strong magnetic fields? Are all of the neutrons
within the extreme density of the star aligned along their magnetic
fields?

ANSWER:
The simple answer to your question is no! The reason is not
simple but is related to the fact that a neutron star has lots of
neutrons but it also has lots of other stuff. An astronomer here
in the department is going to consult with friends of his who are
neutron star experts and then we can get back to you with a more
complete answer.

Have a
look at
http://www.astro.umd.edu/~miller/nstar.html#internal .

ANOTHER ANSWER: (Provided
by M. C. Miller )
(I referred the questioner to the above site. He
emailed the author of that site, Professor Cole Miller of the
University of Maryland who then provided the following answer.)

"The
origin of neutron star magnetic fields is not, by any means, well
understood. The existing field is supported by normal currents in
normal fluids (superconductors exclude magnetic fields), but why it got
so strong is an evolutionary question. Let me start by giving you the
pat answer you'll find in many textbooks, then I'll explain why it's at
best misleading and tell you a bit about current thinking.
It's often "explained" like this. The Sun has a magnetic field, with an
average strength of about 100 Gauss. If you were to compress the Sun to
the size of a neutron star and conserve magnetic flux, the strength of
the field would be about a trillion Gauss, which is about the strength
inferred for young pulsars. Voila, that's why you expect such a strong
field.
The problem is that the Sun will not become a neutron star. Stars that
do become neutron stars (massive O and B stars) have relatively little
convection in their interiors, meaning that they don't generate a field
via a dynamo. If you were to take their central cores and compress them
to the size of a neutron star, you'd get a rather weak field.
So, people think that the field is generated during the collapse of the
core that initiates the supernova. Somehow, with all the associated
turbulence, you stir things up and produce a dynamo that generates the
observed field. Again, the field itself isn't especially mysterious —
it's standard currents — but how it got there is not well known. One
problem is why it stops at a trillion Gauss for many or most neutron
stars. In principle, the dynamo could generate something a good hundred
thousand times stronger, so why doesn't it? Maybe there's something
preventing such growth, or maybe it's just that such a star would spin
down so fast we usually wouldn't see it. There is a class of objects
thought to have magnetic fields up to 1000 times stronger, so perhaps
such stars are out there."

QUESTION:
Why does the neutron, a neutrally charged particle, have a
magnetic moment? Isn't the magnetic moment defined as the direction of
the magnetic field due to a rotational charge?

ANSWER:
Magnetic fields (which are how you detect magnetic moments)
are caused by electric currents which are moving electric
charges. The source of the field need not have a net electric
charge to have a net electric current. The most obvious example
of this statement is a simple current-carrying wire: although its net
charge is zero, electrons move in the wire while the positively charged
remainder of the wire does not; there is a current but no charge and
there is a magnetic field. A neutron, as you know, has no net
charge but experiments have shown that it has a charge density.
Roughly speaking, the neutron has a positively charged core and a
negatively charged surface. So if the whole neutron were thought
of to be spinning as a rigid body (which is a naive but useful picture)
there would be a magnetic moment opposite the angular momentum of the
particle (in agreement with what is known about the neutron magnetic
moment). This is because there would be more negative current than
positive current since the negative charge is farther away from the
spin axis than the positive charge so it moves faster.

QUESTION:
Consider a universe limited to one massive object, nearly the
minimum size of a black hole. A photon is emitted. The gravity of the
object causes the photon to red shift. Over time, the shift
increases and increases. (1) If the
photon starts as visible light, what does it shift through over the
millenium? (2) Since gravity can
bend light, is it possible that the photon eventually stops and
reverses direction to return to the object of origin? (3) Is it possible to describe the properties
of the photon over the entire cycle? [numbering
by "The Physicist"]

ANSWER: (with
helpful comments from C. R. Johnson )
This is a pretty complicated question. I have talked
with the relativist in the department and he has given me some ideas to
convey. I will keep things pretty general and qualitative.

The first
thing he advises is that we should not talk about photons when we are
investigating general relativity which is what your question is all
about. A photon is a quantum concept and quantum mechanics and
general relativity are theories which have not been reconciled.
Nevertheless, I will ignore his advice but not try to push the idea of
a photon too far. Indeed, the photons will have a red shift when
compared to what you expect them to be, for example relative to some
atomic line. The theory of general relativity predicts what this
red shift will be at infinity for photons moving radially out from the
object; in fact, the red shift is well understood and there exists a
formula which gives the frequency of the radiation as a function of the
distance from the star: D f/f =-(GM/c ^{2} )((1/r _{o} )-(1/r ))
where D f/f is the
fractional frequency change, G is the universal gravitational
constant, M is the mass of the object, c is the speed
of light, r _{o } is the position of the source,
and r is the position of the observer. Some things to
note:

Since
D f/f is
negative, the frequency for r>r _{o} is smaller, so the
wavelength is longer —red shift.
If the
object is not a black hole, there is no point where the photon will
turn around and come back or even disappear.
This
equation is correct even if the photon is not traveling radially, that
is, the red shift depends only on the radial position; so you can
think, qualitatively at least, about the photon having a potential
energy.
If the
photon is not emitted radially, it will travel on a curved path but
will escape to infinity.
So, that
answers all three of your questions.

It is
interesting to ask what happens if the object is a black hole.
There is a critical radius which, if a photon is inside it, it cannot
escape. Any photon originating outside the critical radius will
escape if it is moving radially outward but, if it is not, its fate
depends upon the direction it is moving. For a particular
distance from the center of the black hole, there will be a cone; if
the photon is emitted traveling inside that cone, it will escape,
otherwise it will not.

For an
interesting discussion of this whole issue (and which illustrates how
naive my answers above are in the context of general relativity!), see
the Sci.Physics.Research
Archive .

QUESTION:
If I were a photon travelling along an x-axis (of a
three-dimensional universe), would my universe become two-dimensional,
since everything along the x-axis would collapse to zero length?
Alternatively, would I be everywhere along the x-axis at the same
instant (of time)?

ANSWER:
Well, let's clearly state what we mean by "If I were a
photon"! Nobody can be a photon because a photon is a massless
thing and nobody and no thing can make that claim! But let us
assume that I am going really, really fast, say 99.9999999999999% of
the speed of light and along the x -axis of some coordinate
frame. What would the universe look like to me? I would
judge that by looking at things; for example, I would see the light
from some star coming from somewhere. If I were at rest in the
coordinate frame, the light would come to me from some direction which
would make some angle with the x -axis. However, it may be
shown, the light that I see when moving very fast is shifted to the
very forward direction. Interestingly, even light coming from
"behind" me (if I were at rest), will come from in front of me if I am
moving fast enough. So, as you approach the speed of light,
everything collapses to a single point right in front of me! So,
the answer to your question is that your universe becomes
one-dimensional, lying along a line in front of you; it would look like
a point to you, though, a zero-dimensional universe.

This kind
of question was what inspired Einstein to formulate the theory of
special relativity. He said that as a boy he pondered what would
happen if he rode on a wave of light. These are provocative
questions but not ones which can be definitively discussed since they
cannot happen (we conscious beings traveling at the speed of light with
respect to some frame of reference).

If you
want to do further research, it is the "relativistic velocity addition
formula" you need to learn about. Measuring velocities requires
that both length contraction and time dilation be invoked.

QUESTION:
In the case of photoelectric effect, do the electrons come
out in a preferred direction or do they come out at random. Furthermore
do they come out from the same side of the metal (the side where the
radiation is casted ) or they come out from the other side. Do we need
to consider the conservation of momentum when analysing photoelectric ?
Thank you

ANSWER:
Linear momentum is always conserved in an isolated system .
In the case of the photoelectric effect, the photon carries both linear
momentum and energy into the interaction. At first glance, one
would say that the electron must carry that energy and momentum
out. However, the electron and the photon are not the only
participants in this process; in fact, the photoelectric effect will
not happen if we shoot photons at isolated electrons. The whole
atom, in which the electron initially resides, participates in what
goes on and can carry a large amount of momentum but very little
energy. This is because the atom has a very large mass compared
to the electron. Therefore, with a very little velocity the atom
can carry a relatively large amount of momentum (mv ) but,
because the energy is mv ^{2} /2, and if v is
small then v ^{2} is very small, it will have
very little energy. Therefore, the electron can go in whatever
direction it likes and it will carry off (most of) the energy while the
atom carries off (most of) the momentum. In an actual experiment,
any electron which starts off into the metal will quickly stop as it
plows through the material. I don't really know if there is a
preferred direction for those emerging from the metal, but I would
guess that it is pretty random, i.e. isotropic.

QUESTION:
How does it work when you can't see the bottom of the pool? A
boy drowned a couple of weeks ago in LA and the police couldn't find
him. However, the boy turned up in the pool a couple of days later. The
police claim that the pool was dirty and when they looked in the pool,
it looked like they were looking at the bottom when in fact, they
really couldn't see the bottom where the boy was. How does this work???

ANSWER:
This doesn't seem to be a physics question, rather just
common sense. If a pool is totally dirty, the water is not
transparent and you could see nothing under the surface. It seems
possible to me that there would be a layer of muddy water near the
bottom of a pool and that the water above that layer could be cleaner;
so you could see anything above that layer but not below. There
is also the question of refraction which alters where something appears
to be because of light bending when it goes from the water into the
air. This phenomenon could make it more difficult but not
impossible to see something on the bottom of a swimming pool even if
clean. This has been fully discussed in the answer to an earlier question .

QUESTION:
Could you explain how the ZPE(Zero point energy) comes about
when we talk about a particle in a box?

ANSWER:
In classical physics, suppose we have a particle of mass m
inside a box and there are no forces on the particle except
when and if it collides with the walls and those collisions are elastic
(no energy is lost in a collision). What is the lowest possible
energy state of this particle? Well, of course, it would
correspond to the particle being at rest somewhere in the box and we
would define that state as zero energy. All other states of the
system have positive energy mv ^{2} /2 where v is
the speed of the particle. So the spectrum of all allowed states
of the system is zero or any positive energy.

These
ideas (the ideas of classical physics) are in accord with our everyday
experience, but it was discovered around the turn of the 20^{th}
century that they are wrong. Classical physics is only an
approximation of the "truth" which is quantum physics. One of the
most important results of quantum physics is called the Heisenberg
Uncertainty Principle (HUP) which states, in essence, that you cannot
simultaneously know both the position and the velocity of a particle to
arbitrary precision. Therefore the notion that the particle is
perfectly at rest would imply a perfect knowledge of its velocity which
in turn, according to the HUP, implies total ignorance about the
position of the particle. However, we know for certain that the
particle is in the box so we are not totally ignorant of its
position. Therefore, the particle must not have zero energy but
must instead have some minimum energy>0. One can estimate this
energy using the HUP or one may solve for all energy states (which
happen now to be not any energy above your ZPE, but only discrete
"quantized" energies) by solving an equation called Schrödinger's
equation.

So why
don't we see this effect if we put a 1 kg ball in a 1 m^{3}
box? It turns out that quantum effects only become apparent for
very microscopic systems (like atoms). If you calculated the
minimum energy allowed by quantum physics of the 1 kg ball, the
corresponding speed would be so slow that you would have to wait for
something like the age of the universe for the ball to move from one
wall to the other! No chance of measuring a velocity that small!

QUESTION:
I have been a serious student of science for many years,
having created a hypertexted science knowledgebase with over a thousand
interrelated links. Yet one quandary still puzzles me: if gravity is a
phenomenon of the geometry of space, then why do all accounts of the
forces of nature, include the notions of an "attractive force",
moderating "gravitons", and "action at a distance", etc.?

ANSWER:
Well, for one thing the theory of general relativity is very
subtle and based on novel ideas of the structure of space and
time. Because we live in a world where the ideas of force and
mass and motion —Newtonian ideas —tend to be intuitive, it is easier to
understand many aspects of everyday physics using these notions rather
than trying to understand gravitational force as resulting from a
distortion of space which is not intuitive. So, it is simply
useful to be able to quantify gravitational effects in a Newtonian way
and then, when we are pretty good at that and understand things well,
we can finally ask the question "...But, why is there a
large gravitational force in the vicinity of a large chunk of mass? ";
it is at that stage where we must learn the theory of general
relativity.

Another
thing is that gravity is the only one of nature's forces which is
understood in terms of geometry alone. Other forces in nature are
understood in terms of quantum theories. There is no satisfactory
quantum theory of gravity and therefore a graviton is a purely
speculative thing. One of the most important current areas of
theoretical physics research is the attempt to unify our understanding
of all of nature's forces and to develop a satisfactory theory of
quantum gravity.

QUESTION:
I have heard through books that I read and other sources that
objects cannot excede the speed of light. Why is this? If you are
travelling at any speed and you turn on a flashlight, how fast is the
light moving? Is it moving at C + the speed you are moving? Or is it
just moving at C? Please send I response, I would be thankful.

ANSWER:
The cornerstone postulate of the theory of special relativity
is that the speed of light is a universal constant. Therefore, no
matter who measures the speed of light coming out of the flashlight you
allude to, the measured speed will be the same, 3x10^{8}
m/s. This is hard to swallow because it conflicts so seriously
with our "Newtonian-Galilean" intuition, but it has been shown with
great accuracy to be true. With this postulate under our belts,
we can reformulate mechanics. One of the consequences of this
reformulation is that the amount of energy it takes to accelerate a
mass m to a speed v is given by

E=mc ^{2} / (1-v ^{2} /c ^{2} )^{1/2}
- mc ^{2 }

where c
is the speed of light. Notice that if we want to accelerate the
object up to v=c, an infinite amount of energy needs to be
supplied. Sorry, but nobody has an infinite amount of
energy! You need to find a popularized book on special relativity
(there are many); it does not take very sophisticated mathematical
ability to understand the theory.

QUESTION:
Contrary to the usual description of the twin paradox, where
one of them stays on earth and the other is travelling at relativistic
speeds, consider the situation where both of them ( say A & B ) are
travelling at 'identical' speeds but away from the same point ( some
arbitrary point in space, away from the influence of any gravitational
field so as to neglect the effects of General theory of relativity ) in
diametrically opposite directions and then returning to the same point.
Then, according to a third 'neutral' observer at rest with respect to
the 'point' from where A & B started off, both of them would have
aged equally. But in A's Frame Of Reference (FOR in short ) B has been
constantly travelling ( except at the moment where they stop and start
retreating ) and hence B has aged according to A; the same applies even
to A from B's FOR. Here, though we ( the 'neutral observers' ) know
that they are of the same age how is one going to resolve the paradox
keeping in mind the views of A & B ?

ANSWER:
My favorite way to understand the twin paradox is to suppose
that each twin, using his own clock, sends a light pulse to his brother
once a year. Each brother receives all the pulses from the other
but the moving brother sends fewer since, because of length contraction
of the distance to the object he travels to, he has less far to travel
in his frame than his brother observes. The diagram below shows
how this works for a brother who travels to a star which is 8 light
years away and back with a speed of 80% the speed of light. Since
the traveling brother sees, because of length contraction, a distance
to the star of only 4.8 light years, he sends out six light pulses on the
way out and six on the way back, and all are received by the earthbound
brother, so both agree that he has aged 12 years. The earthbound
brother, of course, sends out 20 pulses and the traveling brother
receives them all so both agree that he has aged 20 years. It is
interesting to note that this graph illustrates that the rate at which
clocks run is not the same thing as the rate at which they appear
to run. The earthbound brother sees his twin age only 6
years in his first 18 years and then age 6 more years in only 2
years. The traveling brother sees his twin age only 2 years in
his first 6 years and then 18 more years in his remaining six years of
travel.

So now we
come to your question. How would we analyze the situation if one
twin went on the original trip and the other twin went to a star 8
light years in the opposite direction and back? The graph below
illustrates this scenario. It is clear that each twin sends and
receives 12 pulses so each sees both himself and his twin age 12 years
over the duration of the trips. Again, note the interesting
disparities between actual aging and the appearance of aging.
Neither twin receives any pulses until the return trip.

QUESTION:
A 1934 text states:

"It has
never been possible naturally to assemble over one hundred orbital
electrons in one atomic system. When one hundred and one have been
artificially introduced into the orbital field, the result has always
been the instantaneous disruption of the central proton with the wild
dispersion of the electrons and other liberated energies."

I
understand this to be saying that elements with 101 or greater
electrons do not naturally occur in nature, and when they are
artificially created, they have very short half-lives. The term
"instantaneous disruption" is used in the text relative to time scales
of billions of years. This sounds consistent with the behavior of the
transuranium elements (although I am not considering the "island of
stability") and their relatively short (sometimes referred to as
fleeting) half-lives. It is inconsistent with the number of naturally
occurring elements, scientists noting there are 92, while the text says
there are 100. To the best of your understanding, would these comments
be accurate?

ANSWER:
The statement from the old text says nothing about the
nucleus. Therefore, your understanding is incorrect because the
instability of heavier elements has nothing whatever to do with the
electrons in the atom but rather is determined by the numbers of
neutrons and protons in the nucleus. The statement from the text
would seem to me to be completely archaic. If you want to
understand which elements occur naturally and why, I strongly urge you
to consult more up-to-date references.

QUESTION:
What is the relationship —if any —beween physics and
artificial intelligence?

ANSWER:
Physics, being the most fundamental of all sciences, is
related to everything by definition! Seriously, though, AI is
simply the attempt to program computers to have "behaviors" similar to
human behavior for certain circumstances. This, in itself,
doesn't have a lot to do with physics.

QUESTION:
What would be the best microwave-resistant material that I
could buy at Home Depot or other such store?

ANSWER:
Shielding from electromagnetic radiation is usually done by
putting a conducting material in the way. This works because when
the radiation enters the material it quickly loses its energy by
inducing currents and causing the material to heat up. The depth
to which the waves will penetrate is determined by the properties of
the conductor and the frequency of the waves. This property is
parameterized by a quantity called the "skin depth" which tells how far
a wave will penetrate before losing 1/e (about 1/3) of its
amplitude; it is given by

d= (p m s f )^{-1/2}

where
p =3.14159, m is a magnetic property
of materials called permeability, s
is a property of
materials called conductivity, and f is the frequency of the
wave. This formula, incidentally, is valid only for good
conductors. Microwaves have wavelengths in the range of 0.3-30
cm, and therefore frequencies in the range 10^{9} -10^{11}
Hz. For aluminum I compute a skin depth of about 3x10^{-6}
m; any good conductor will be in this ballpark. The whole point
of this discussion is to illustrate that microwaves do not penetrate
very far into metals. By the time you go five skin depths into
the metal you will find only about (1/3)^{5} of the amplitude
of the original wave, about a half percent. That means you don't
even have to go to Home Depot —just go to the grocery store and get
some aluminum foil.

QUESTION:
I am researching the anomaly known as spook lights. Can you
give me a detailed explanation of how headlight refraction works, since
this is one of the theorys proposed to explain this.

ANSWER:
Basically, I think you simply want to know about
refraction. Light, when it goes from one medium to another,
changes its velocity and as a result the direction of a beam of light
will change unless the direction is normal to the surface between the
two media. For example, light coming out of water appears to come
from somewhere other than from where it is actually coming. If
you want to spear a fish from a boat, you need to understand this so
you can adjust your aim or you will miss every time the fish is not
directly below you! More complicated situations occur in gases
like air. Here, if the density of the air changes, then the
direction of a light ray will change due to refraction because the
speed of light in air depends on the temperature. The most
familiar example of this kind of refraction is a mirage. If a car
with its headlights on is approaching you when the air directly above
the road is hotter than the prevailing temperature, you will see both
the headlights of the car and what looks like another set of headlights
under the road; it looks like the car is driving on a mirror. You
might say that there is a "spook car" under the road! So, the
bottom line is that the phenomenon of refraction can cause light to
look like it comes from somewhere it isn't.

QUESTION:
I want to generate electricity by using a slowly falling
weight. The weight is suspended 10 feet above the ground. The
weight is attached to a gear system by means of a rope. The gear system
is attached to generator. Two wires from the generator lead to a 100
watt light bulb. Disregarding friction, how heavy must the weight
be (in pounds) to sustain a 100 watt light bulb if the weight falls at
only one foot per hour? What is the formula through which this is
calculated?

ANSWER:
First of all, let's put things in metric units: the 10 ft is
not relevant and 1 ft/hr=8.5x10^{-5} m/s. The formula for
the potential energy of the mass is given by E=mgy where m
is the mass, g is the acceleration of gravity (about
10 m/s^{2} ), and y is the distance above where the
potential energy has been defined to be zero. So, for example, if
a 2 kg mass falls a distance of 3 m in 3 s, its potential energy
decreases by 60 J in 3 s so it is losing energy at a rate of 20 J/s= 20
W. In your case, the power, which is energy over time is given by
P=E/t=mg(y/t)=mgv , so 100 W=m x 10 m/s^{2}
x 8.5x10^{-5 } m/s, so m= 1.2x10^{5} kg (which
corresponds to about 60,000 lb). This does not seem very
practical to me!

QUESTION:
If X-rays are a form of EMR, then they have a changing
magnetic field and a changing electric field, yet one of they're
properties is that they are not deflected by magnetic or electric
fields. How can that work?

ANSWER:
Essentially, electric and magnetic fields affect only
electric charges or charge distributions. In some sense, this is
the definition of a field —a measure of the force felt by particular
electric charges. If there are two different electric fields at
some point of space, for example, the two fields simply add. But
one field does not "exert a force" on the other; the notion of fields
exerting forces on each other doesn't even mean anything.

QUESTION:
If you take a beam of high-energy electrons and slam it into
a metal plate, a certain amount of the electrons’ kinetic energy is
converted into heat, but a certain amount is converted directly into
photons. If you examine the outgoing photons in detail, you find that
they span a wide spectrum of frequencies – but there is a sharp
“cut-off ” in the spectrum at high frequencies, beyond which one never
sees any photons.

a)
Explain briefly in general terms why there is a specific high-frequency
cut-off.

b) Let’s
suppose the electrons in our beam are moving at 1% of lightspeed. What
is the cut-off frequency of the photon spectrum?

ANSWER:
a) It is an energy conservation thing. When an electron
interacts with matter, it may lose its kinetic energy in two main ways:

It
could give some of its kinetic energy to atoms in the matter.
Think of a moving ball striking another at rest; after the collision
the ball originally at rest will be moving. This is where the
heat comes from.
When
an electric charge suffers an acceleration (which means either that its
speed changes or the direction of its velocity changes or both) it
emits electromagnetic radiation called bremmstrahlung which
carries off some of its energy. This is where the photons come
from.
One thing
is for sure, though. You can't get more energy from the electron
than it has. Therefore the most energetic photon corresponds to
one having all the kinetic energy of the electron. This can be
associated with the frequency of the photon by using Planck's famous
equation E=hf where E is the photon energy, f
is the frequency, and h is Planck's constant (6.63x10^{-34}
J s). (Actually it was Einstein who hypothesized photons and
showed that their energies are given by this equation.)

b) Since
you have asked for the cutoff f for such a low electron speed
(which is usually not the case with electrons), I can, to an excellent
approximation, do the calculation without invoking relativity.
So, E=mv ^{2} /2=9.3x10^{-31} kg x (3x10^{6}
m/s)^{2} /2=4.2x10^{-18} J=(6.63x10^{-34} J s)f
and so f =6.3x10^{15} s^{-1} . This
corresponds to a wavelength of about 48 nm, ultraviolet.

QUESTION:
I want to know (among other things i consider wrong) why you
believe light is massless? Do you believe these to be true: light bends
around stars, light can't escape from black holes, light is able to
push light-craft such as futuristic light-sailships and spherical craft
that WAS able to be propelled 200 feet into the air by a high-powered
laser, or that we can see light at all? Well MY point is that anything
that acts on anything else must have mass, otherwise there is 0
momentum to transfer onto the next item. If light was to be attracted
by gravity, wouldn't it have to have some sort of mass that the gravity
acts on since the amount of attraction between two objects by gravity
is related to the mass of BOTH objects?

ANSWER:
None of the things you cite require light to have mass. The
problem is that you are thinking classically. In fact, even in
classical electromagnetic theory, light has momentum and energy but no
mass. In modern physics, the electromagnetic field is quantized, i.e.
the light behaves like photons in certain circumstances which are
massless particles with velocity c =3x10^{8} m/s, energy
E=hf (f is frequency, h is Planck's
constant) and momentum p=E/c. Light being affected by gravity
is a result of the principle of equivalence in general relativity. This
states that there is no experiment which you can perform to distinguish
between your being in a gravitational field or in an accelerated frame
of reference. Thus, for example, imagine that you are in an elevator
which accelerates upward; if light enters through a hole in the side of
the elevator it will clearly appear to fall like a projectile because
of the acceleration of the elevator. So, the same thing will appear to
happen in a gravitational field the acceleration due to which is
exactly the same as the acceleration of the elevator. Being able
to see something has nothing to do with mass; the detection only
requires that we have a detector sensitive to the electric or magnetic
fields associated with the light.

QUESTION:
Who defined density as mass divided by volume? Newton
did not define density.

ANSWER:
The notion of mass density was used even by the
ancients. Surely Archimedes used the notion of density to
understand what floats and what sinks. It is not some profound
idea which is credited to somebody. It simply characterizes some
material in terms of other properties of the material (mass and
volume). There are all sorts of densities, for example energy
density, charge density, and momentum density which specify the amount
of energy, electric charge, or momentum respectively per unit volume.

QUESTION:
i have two separate questions.
1. i was researching and came across a book that looked interesting
called Surface Phases on Silicon and i was wondering if you can tell me
what its about in a little more detail or guide me to resources that
can.
2. i was wondering what exactly layer-boundary theory was and what it
has to do with.
i am interested in quantum mechanics and have researched many different
theories and concepts, so i do understand some of it. but if you could
briefly help me out with these two concepts i would appreciate it.
thanks for your time

ANSWER to #2: (provided
by M. R.
Geller )
Boundary-layer theory is concerned with classical viscous
fluids in the presence of a surface (it has nothing to do with quantum
mechanics). When a fluid flows along a surface, viscosity causes the
fluid near the surface to partially stick to it, slowing it down
relative to the fluid far from the surface. The "layer" of fluid that
is most slowed down by the surface is called the boundary layer, and
boundary- layer theory deals with the properties of this layer (for
example, its thickness), as well as how the presence of the boundary
layer affects the rest of the fluid. Boundary layers are also important
in aviation, because the air flow right near the surface of a runway is
considerably different than higher up, and pilots have to know how to
land in the presence of these boundary layers.

ANSWER to #1: (provided
by S. P. Lewis )
The term "surface phases" can mean several things, but in
each case it has to do with a "phase" of some material at the surface
of either the same material or a different one. Clearly "Surface Phases
on Silicon" refers to the situation where the native surface of
interest is the material silicon.
The crux of the matter, then, is the meaning of "phases" in this
context. The concept of "phase of matter" is hard to define. In
general, it has to do with the degree of order that the material
possesses. Structural phases refer to the degree and type of order in
the structural arrangement of the atoms making up the material. So, for
example, atoms arranged in a crystalline solid are more ordered than
atoms making up a liquid, which are more ordered still than atoms in a
gas. However, even within one of these broad categories, there can be
different phases. For example, one can imagine a crystal structure in
which the atoms sit on the points of a simple cubic lattice. This
crystalline phase would be distinct from the case where the atoms sit
on the points of, say, an hexagonal lattice. These are two different
crystalline phases.
Another example of phases of matter are magnetic phases, which refers
to the ordering of atomic magnetic moments (i.e., polarity). Therefore,
even if we just consider atoms arranged on, say, a simple cubic lattice
of points, there can be several different magnetic phases. For example,
if the magnetic moments of all of the atoms are pointing in the same
direction, then the material is "ferromagnetic". However, if the
magnetic moments alternate in direction (e.g. ,
up-down-up-down-...) as you go from atom to atom, then the material is
"antiferromagnetic". Examples of both types of magnetic ordering (and
still others) are observed in nature in real materials.
A system can be driven from one phase into another by applying
"fields", for example temperature, mechanical pressure, or magnetic
field. These "phase transitions" exhibit fascinating physics and are
the subject of very active study.
So now we come to surface phases. Very often, the atoms at the surface
of a material will adopt a different geometrical arrangement than one
would expect from merely cleaving the surface along a plane of the bulk
crystal. The reduced coordination (i.e. , number of neighbors) of
the atoms at the surface often causes them to favor a structure of
different symmetry than the underlying material. When this happens it
is called a "surface reconstruction". Semiconductor materials, like
silicon, often exhibit a wide range and variety of reconstructed
surface phases, depending on the conditions. Another "surface phase"
issue is that the surface of a solid material can sometimes melt at a
different temperature than the underlying material. Finally, the
magnetic structure of a surface can also be entirely different from
that of the bulk material beneath. These issues undoubtedly form the
substance of a book called "Surface Phases on Silicon".

QUESTION:
If force is a property of motion, then it follows that forces
cannot exist independently of motion. Yet, the motions producing the
"fundamental forces" are not identified. Tis seems strange to me. My
question has two parts. 1) What is the motion producing each or the
four fundamental forces? 2) What is motion? (Note: I'm not asking what
terms to use to express the motion (speed) of an object, but rather
what is motion in general without respect to a material object). A good
example is the three dimensional motion of gravity or the three
dimensional motion of expanding space). Obviously, in a universe of
space and time devoid of objects motion still must exist, right?

ANSWER:
The questions you are asking do not have well defined answers
in physics. This is because much of modern physics involves
quantum mechanics, general relativity, or special relativity all of
which have introduced new concepts related to our ideas of force and
motion.

There is
no "law of physics" which states that "force is a property of
motion". Newton's second law tells us that if an object is
observed to accelerate, then there must be an unbalanced force on
it. However, Newton did use the word "motion" in his classic work
Principia ; today we call this quantity "linear momentum"
and it is the product of the mass times the velocity of an
object. The force is precisely equal to the time rate of change
of the linear momentum; but this fact absolutely does not imply that
"forces cannot exist independently of motion". Rather, if there
is no acceleration in Newtonian physics this means that the net
force is zero, not that there are no forces. Incidentally, do not
confuse motion in this context with velocity. An object with no
forces on it may move with constant velocity. Force does not
cause velocity, it causes velocity to change (accelerate).

In the
theory of special relativity, Newtonian mechanics is revealed to be
only an approximation of the true behavior of nature. Force
becomes defined again as the time rate of change of linear momentum but
linear momentum is different —it is the mass of an object times its
"proper velocity". In the theory of general relativity, the
gravitational force is understood as the warping of space-time due to
the presence of mass.

In
quantum mechanics the concept of force becomes not particularly useful
and the concept of potential energy, closely related to force, acquires
the position of importance. In fact, in sophisticated
formulations of classical mechanics (e.g. the formulations of
LaGrange and Hamilton) potential energy is the key player also.

Most
modern theories of the fundamental forces of nature become "field
theories" where the fundamental concept is a field rather than a
force. The most accessible field theory is that of electricity
and magnetism. Here it turns out that the fields themselves may
contain linear momentum or energy. This is how it is that
electromagnetic radiation, such as light and radio waves, carry both
linear momentum and energy even though there is absolutely no mass
involved.

QUESTION:
Exactly how does a megaphone "magnify" sound?

ANSWER:
It does not magnify the sound. Rather it focuses the
sound. It is very much like the reflector in the headlight of an
automobile where light which would have gone straight up is reflected
to go straight ahead instead. In the picture above, the star
represents the source of sound and the heavy black lines the sides of
the megaphone. All the sound "rays" which are red would not have
been there if the megaphone had not been there. Instead they
would have gone along the dashed black lines.

QUESTION:
Please, is everything basically relative in the universe?

ANSWER:
Hmmm...I really do not know what you mean! The central
postulate of the theory of relativity states that the laws of physics
are the same as seen by any observer, no matter what his relative
motion is. In that sense, everything is indeed relative because
the laws of physics ultimately describe everything.

QUESTION:
Who first measured the wavelength of one electromagnetic wave
and how?

ANSWER:
Credit is usually given to Dr. Thomas Young
(1773-1829). Most famous of his "measurements" is the so-called
double slit diffraction experiment in which light strikes two closely
spaced slits in an opaque sheet. Each slit acts like a source of
waves and these two waves interfere with each other when they strike a
screen some distance away. What interference means, essentially,
is that the two waves add up to give a net disturbance at that
point. So, if the two waves are both at a crest they add up to
look twice as big (therefore bright) but if one is at a crest and the
other is at a trough, they will cancel each other out and there will be
no light. Young explained many well-known optical phenomena (such
as the colored fringes you see from an oil slick) using the idea of
interference; he and his ideas were, as often happens with
revolutionaries, reviled by many of his contemporaries. Before
Young, Christiaan Huygens (1629-1695) was an advocate for a wave model
of light; he was able to show that light must slow down in a medium
like glass and thus explained refraction. Some 13 years after
Young had described interference, Augustin Jean Fresnel (1788-1827)
independently developed the notion of interference.

QUESTION:
My students swear to me that when they did their science fair
projects last year in middle school they "proved" that hot water will
freeze faster than cold water. This doesn't make sense to me so we are
experimenting tomorrow to analyze what factors may contribute to their
theory. In the meantime, is that true? If so, why?

ANSWER:
This is the perennial question for scientists from
laymen! Alas, it has no clear and simple answer since it depends
very much on the conditions under which an experiment is
performed. In a nutshell, here is the important concept: if
evaporation from the hot water plays an important role, it is possible
and even likely that the hot water will freeze faster. What is
well known is that evaporation results in cooling. Another way to
put this is that it takes energy to cause a transition from liquid
phase to a gas phase. An easy way to demonstrate this is to take
a damp rag and swing it vigorously around your head; it gets really
cold. So I will give you two scenarios, one in which the hot will
likely freeze first and one in which the cold will certainly freeze
first.

Suppose
we have two wooden buckets each containing a gallon of water, one hot
and one cold. The cold one must not be so cold that it is just
about to freeze; imagine one room temperature (~70^{0} F) and
the other hot tap water (~160^{0} F). The buckets, being
wood, are pretty good insulators, so most of the conductive/convective
cooling comes from contact with the very cold air by the
surfaces. But the hot water has an advantage: in addition to its
conductive/convective cooling it has (greater) evaporative cooling, so
it is likely to "catch up" with the cooler bucket before the cooler
bucket freezes. Now we have two buckets of water at the same
temperature, but there is now less water in the originally hot bucket
because some of it evaporated away. Starting with two buckets of
water one with a gallon and one with less, and having the rate of
energy leaving each bucket about the same, the one with less water will
cool faster and therefore freeze first!
Suppose
we have two one gallon sealed glass bottles filled with water, one hot
and one cold. Because they are sealed, evaporation cannot play a
significant role so each will cool at about the same rate, so the one
that started cold will certainly freeze first.
Because
of the subtleties here, this is not a good experiment for novice
scientists! There are too many variables in terms of how energy
gets in and out of the system and even in terms of parts of the system
escaping. The important things to teach your kids are

A
certain amount of heat (energy) added will result in a certain increase
of temperature in a certain amount of stuff (specific heat),
a
certain amount of heat is required to change a certain amount of a
solid into a liquid (latent heat of fusion), and
a
certain amount of heat is required to change a certain amount of a
liquid into a gas (latent heat of evaporation)
To read
more about this topic, go to the MadSci Network and
search on "freeze water hot" with the AND choice. That page, a
much bigger one than ours, has received at least 46 queries like yours!

QUESTION:
What would the infrared frequency be, in microns, for the
temperature 86 degrees fahrenheit.
What is the frequency
range, in microns, for low, middle, and far infrared.

ANSWER:
For starters, frequency is not measured in microns since
frequency is number of vibrations per second and a micron is a unit of
length. I presume that you want to know the answers to your
questions as wavelengths in microns (1 micron =10^{-6} m, a
millionth of a meter).

Now,
regarding your question of what the wavelength is for an object at 86^{0}
F is: the fact is that objects radiate at all wavelengths, so we need
to discuss how the energy is distributed over all the
wavelengths. The details of the radiation depends on the
particular object which is radiating, but we can get a good idea of
what is going on by assuming the object is a perfect radiator and that
is called a blackbody. Blackbody radiation is extremely well
understood in physics and it was, in fact, the theoretical study of
blackbody radiation which gave quantum physics its birth. So, let
us find the wavelength at which the most energy is being radiated by a
blackbody. There is a handy equation which lets you find out at
which wavelength this maximum occurs: l =2900/T where
l
is
the wavelength in microns and T is the temperature in degrees
Kelvin (the absolute temperature scale where zero degrees Kelvin,
absolute zero, is -273 degrees Celsius). 86^{0} F is about
303^{0} K, so the most energy is being radiated at a wavelength
of 9.6 microns. A picture of the distribution of energy as a
function of wavelength is shown to the right.

The near
IR band contains energy in the range of wavelengths closest to the
visible, from approximately 0.750 to 1.3
microns .
The intermediate
IR band (also called the middle IR band ) consists of
energy in the range 1.3 to 3.0 microns .
The far IR band
extends from 2 to 14 microns .

TWO RELATED QUESTIONS:

How is
an EEG related to physics, please explain
What
was the first EEG like and how did it relate to physics?
ANSWER:
Well, the details of an EEG are quite complex, but the ideas
are not too hard. Basically, the brain consists of cells called
neurons which constantly have electrical impulses (tiny currents) which
turn on and off. Therefore there is a large amount of low-level
electrical activity in the brain. So there are time varying
electric and magnetic fields in the vicinity of the brain. Time
varying fields are particularly convenient to detect because of
something call induction. For example, if you have a coil of wire
which has a changing magnetic field passing through it, an induced
current will flow in the coil, that is there will be an induced voltage
around the coil. This is how lots of devices you are familiar
with work —electric motors, electric generators, automobile ignition
coils, to name a few. The electrodes which are placed on your
head during an EEG measurement are devices which, by using this
induction idea, create tiny voltages which vary the same way the
electrical activity in your brain is varying; these voltages are then
amplified and recorded. So, you see, the physics of the EEG is
primarily the physics of electricity and magnetism.

I found a
picture of the very first recorded EEG, done by Hans Berger in 1928:

QUESTION:
For a 7th grade science class could you suggest additional
mini experiments the students could do to demonstrate the relative
density of air, CO2, and helium gases. We have done ones with balloons
to see which rise. Class equipment includes an aquarium, vacuum chamber.

ANSWER:
Here is one that I can think of. I presume that you
have sources of CO_{2} and He (cylinders of the gases).
Your students will know that a candle will burn in air. But it
will not burn in CO_{2} or He. If you slowly leak CO_{2}
into your aquarium (keep the air around it still while you are doing
this), after a while the aquarium will fill up with CO_{2}
because it has a higher density than air. If you now put a
lighted candle down into the aquarium it should go out. Now empty
the CO_{2} out of the aquarium so that it is again full of
air. Now slowly leak He into the aquarium for about the same
amount of time. Now, the candle will not go out when you put it
into the aquarium because the He, having a lower density than the air,
will float out instead of settling in. If you can get a piece of
dry ice (solid CO_{2} ), and put it into the aquarium you will
be able to see the CO_{2} gas (which will look like a little
cloud) laying on the bottom because of its higher density.

QUESTION:
I was wondering why magnetism worked. I mean at the
atomic/molecular level. And does this have anything to do with the
"Spin" of a particle.

ANSWER:
Basically, magnetism is a force which has its source in moving
electric charges. The most common laboratory source of
a magnetic field is an electric current. And, curiously, a
magnetic force is felt only by a moving electric charge. But I
suspect what you are really interested in in why ferromagnetism works;
this is the magnetism associated with permanent magnets and materials
like iron. To elucidate this, let us first talk about a small
circle of wire which carries an electric field. This causes a
magnetic field which will exert a force on another current; in
particular, if you place two little loops coaxially they will repel
each other if their currents are in the same direction and attract each
other if their currents are in opposite directions. Now, it turns
out that most elementary particles behave like tiny current
loops. In particular, an electron gives rise to a magnetic field
which may be qualitatively thought of by arising from the
field of a spinning sphere of electric charge which is essentially just
like a current loop (a stack of loops, really). This is ok to get
a qualitative idea of what electron spin is, but if you try to push it
too far (like try to calculate how fast it would have to be spinning to
explain the observed magnetic field) it gets ridiculous; spin is a
result of relativity.

Now, back
to ferromagnetism. In most materials, the spins of all the
electrons align in random directions and cancel each other out so a
piece of aluminum, for example, does not have any magnetic field.
For a very few materials, however, it turns out that the solid forms
such that the electrons shared by neighboring atoms will align their
spins with each other. There are therefore huge numbers of
electrons all with their spins (that is their magnetic fields) all
pointing in the same direction and these cause the whole piece of iron
to have a strong magnetic field.

Finally,
the question naturally arises as to why every piece of iron isn't a
strong magnet. When the solid piece of iron originally forms
(cooling down from molten iron, then hot solid iron), a group of
electrons in one location will align in one direction but groups in
different locations align in different directions. Thus we end up
with many domains each of which is pointing in a random
direction and the net magnetization of the whole piece of iron is
zero. But, when an external magnetic field is applied to the
whole piece of iron, the electrons want to point in the direction of
the field so the domains already in that direction grow at the expense
of the domains not pointing in that direction. When the field is
turned off, the domains relax back down but not all the way to where
they began and a net magnetization remains. That is how you can
make an ordinary nail become a permanent magnet —just put it in the
vicinity of another magnet. That also explains why a permanent
magnet can pick up a piece of iron —the originally unmagnetized piece
of iron becomes magnetized and so the two attract each other.

QUESTION:
i was wondering something about time dilation. If motion
causes particles to resonate at a slower speed, and that causes atomic
clocks to slow down, which results in time dilation, then wouldnt
decreasing temperature also be time dilation? And wouldn't time stand
still not only at the speed of light, but also at absolute zero? Thanks
for your time.

ANSWER:
I think the problem here is that you have associated the
vibrational energy of molecules with the temperature of
molecules. This will be easier to explain if we suppose that the
molecules are in a gas form. Now, what do we mean by
temperature? Essentially, the temperature of a gas is a measure
of the average translational kinetic energy of the gas molecules; by
'translational' we mean that this is the energy due to the motion of
the molecules moving around in their container, not including the
energies associated with internal energy of the molecule such as
vibrational energy and rotational energy. So as the gas gets
hotter, the average molecule moves faster so a clock located on that
molecule will run slower. So it would seem to me that an
infinitely hot gas would have, as you say, 'time stand still', not a
gas at absolute zero. At absolute zero, all the molecules would
be at rest so their clocks would run at the same rate as ours.

Incidentally,
the fact that temperature is dependent only on the translational motion
has an impact on what are called specific heats of materials. If
you have a monatomic gas (like helium), it cannot either rotate or
vibrate so when you add heat to such a gas all that heat shows up as an
increase in temperature. When you have a molecule, it can accept
energy by vibrating or rotating on its own axis, so less of the energy
goes into increasing the temperature. Specific heat is a measure
of the increase in temperature for a given amount of heat added, so a
monatomic gas has a bigger specific heat than a gas made up of
molecules.

QUESTION:
Saw your "Ask a physicist" site. I have been trying to work
out the problem below for MONTHS! I work at a science museum with a
Foucault Pendulum, but few people realize how it demonstrates rotation
of the Earth. I have a very effective partial demonstration:

To
demonstrate the most basic concept of the pendulum, I have a child
stick out her finger, attach a yo-yo, start it swinging, and have her
twist her body 90 degrees. We note that the pendulum continues to swing
in the same direction after twisting.

To
demonstrate the Coreolis Effect, we form a line of people, with me at
the end. I have the kids move around in a circle, with me at the
center, trying to maintain a straight line of bodies as the line
rotates (kind of like professional figure skaters). The kid on the end
opposite me can't keep up because he is travelling the farthest. I have
no problem. Since I'm in the center I only twist, and don't travel any
distance.

But, HOW
DO I GET ACROSS THE CONCEPT OF PERIOD OF PRECESSION???? People realize
that at the poles, the period is 24 hours (actually 23 hours, 56
minutes), and some even suggest that the period of precession is
infinite at the equator. But, is there a simple way of getting across
that the period in between the poles and the equator varies
specifically as 23:56 / sin(latitude)?

Personally,
I would like to have a SIMPLE explanation for the origin of the
sin(latitude) denominator, which I don't understand yet, even though a
full derivation of the equations of motion describing the pendulum
motion is provided in texts and on the web. At an absolute minimum, I
need to provide to a science fearing public an easy to understand
reason for why the period of procession increases (instead of remains
equal to) ~24 hrs. as you get farther away from the poles. A
qualitative explanation, avoiding the sin(latitude) specifics, is much
better than nothing.

Demonstrations
that physically involve our guests (such as holding a yo-yo or running
around in a line) are preferred.

On behalf
of our museum guests, thanks very much in advance for considering this
problem!

ANSWER:
I have reviewed my classical mechanics texts and I cannot
find a simple intuitive way to explain why the period of precession
varies like 1/sin(latitude); this factor arises from the cross product
of two vectors |a xb |=ab sinq _{ab}
and I believe that this will not provide an intuitive route for your
guests! However, I think you have the explanation at your
fingertips without knowing it. If you call the period of
precession T and the period at the north pole (q= 90^{o} ) T _{N} ,
then you could propose that T =T _{N} /sin( q ). Of course this
works at the north pole since sin(90^{o} )=1. Now, you
could easily convince everybody that the period at the equator ( q= 0^{o} ) should
be infinite and, since sin(0^{o} )=0, the proposal works.
Now you could consider the south pole ( q=- 90^{o} );
the proposal now says, since sin(-90^{o} )=-1, that T _{S} =-T _{N} .
What does this mean? What it means is that the period is -24
hours, i.e. a period of one day but the precession is in the
opposite direction as at the north pole . I think that you
could convince them that a pendulum at the north pole precesses
clockwise and one at the south pole precesses counterclockwise.
This is not a rigorous proof of the proposal, but it at least works for
three points on the earth.

Incidentally,
I think you could do more with the Coriolis force. Have a look at
the video at http://bake2.physast.uga.edu/x4101/coriolis.mpeg .
Note that if you watch the ball from the ground, it goes in straight
lines. But if you watch it from the merry-go-round, it does not;
it seems to behave as if there were some mysterious force on
it. If you could rig up a merry-go-round you could demonstrate
the Coriolis force very nicely.

One more
thing is that the period of the pendulum at the poles will be exactly
24 hours, not a little less. 24 hours is the time it takes the
earth to rotate once on its axis, not the time between the two times
when the earth is equally oriented toward the sun.

QUESTION:
I've read the definition of a perpetual motion machine but
I'm still a little unclear. Would a system that captured latent heat
(and ran forever) be considered a perpetual motion machine? Does
"perpetual motion machine" refer only to isolated systems or could it
also be used to describe a system that ran forever exchanging energy
but not matter with its environment ? I know its not possible, I'm just
interested... Thanks

ANSWER: (provided
by S. P.
Lewis )
An elementary consequence of the 2nd Law of Thermodynamics is
that it is not possible to convert random thermal energy (heat)
entirely into usable energy (work). Suppose there is some source of
heat, usually called a "heat reservoir", and a system (an engine)
designed to extract some heat and convert it to work. However, simply
extracting heat from a reservoir causes entropy to decrease. Entropy is
a fairly abstract concept that physicists use to quantify the degree of
randomness in the Universe. The 2nd Law of Thermodynamics tells us that
no physical process can cause the entropy of the Universe to decrease.
Therefore, since extracting heat from a reservoir and converting it all
to work would decrease entropy, some of the extracted heat has to be
expelled as waste to make a net increase in entropy. For this to work,
it turns out, the expelled waste heat has to be dumped into another
heat reservoir at a colder temperature than the first. That's why you
may sometimes hear about engines operating betweeh "hot" and "cold"
reservoirs.
So what does all this have to do with perpetual motion machines? Well,
over the years, some people have suggested machines that extract heat
from a reservoir, convert it all to work to do something useful, and
then at the end return the energy back to the reservoir, replenishing
it for the next user. One famous example describes an engine designed
to extract heat from the ocean to run a ship. The fundamental problem
here is that this engine would make entropy continually decrease,
violating the 2nd Law of Thermodynamics. At best, some heat would have
to be expelled to a lower temperature reservoir, removing it forever
from the ocean. Some external agent (e.g., the Sun) would then have to
supply energy to replenish the heat lost from the ocean. Thus this
scheme does not define a perpetual motion machine, because even the
lifetime of the Sun is finite.
Another aspect of your question hints at the possibility of supplying
energy to an object that has absolutely no dissipation, like the
physicist's idealization of a block sliding on a frictionless table.
However, real macroscopic systems can never eliminate all dissipation
channels. There will always be some dissipation, causing the initially
pushed object to slow down and come to a stop. Things are different in
the microscopic world, but that is a topic for a different discussion.
For a nice discussion of perpetual motion machines that is very
accessible to non-scientists, go to http://www.kilty.com/pmotion.htm .

QUESTION:
I'm a junior in high school and currently taking Classical
Mechanics at Michigan State University. In learning about planetary
motion, I remembered a question on the Michigan Science Olympiad Reach
for the Stars event from last year and it goes as follows: what happens
to the orbit of the Earth if the sun increased its mass 4 times?
I didn't know how to answer it, but, this year, having learned about
eccentricity, Kepler's laws, and planetary orbits in depth...my
professor and I still can't work it out! According to our equations
(with alpha and a using reduced mass, angular momentum, etc), the orbit
must be an ellipse because the eccentricity is between 0 and 1. But
when we work backwords, it doesn't work! I'd greatly appreciate it if
you could provide some insight to this problem! Thanks!

ANSWER:
(For simplicity and clarity, I am going to assume that the
mass of the earth is very small compared to the mass of the sun, i.e.
I will ignore reduced mass effects.)

Qualitatively,
this problem is easy to understand. The earth is in an
approximately circular orbit with the sun being its current mass.
If the sun mass suddenly quadruples, then so does the force, so the
orbit will no longer be circular because the velocity will be wrong: F =mv ^{2} /R =k /R ^{2} ,
so v =[k /(mR )]^{1/2} .
Here, m
is the mass of the earth, R is the radius of the earth's
current orbit, and k =mMG , M being the mass of
the sun. If the mass M became 4M , the velocity
would have to double for the earth to stay in a circular orbit.
Since the earth, when the mass quadruples, has no radial component of
its velocity, it would be at the slowest point of its new orbit, that
is the earth would be at the apogee (farthest point from the sun) of
its new orbit.

Quantitatively,
the equation you probably used for eccentricity, is e= [1+(2EL ^{2} /(mk ^{2} ))]^{1/2}
where E is the total energy and L is the angular
momentum. However, you must slightly revise this equation
(probably your error here) since the mass of the sun is now 4 times
larger, you must replace k ^{2} by 16k ^{2} :
e= [1+(EL ^{2} /(8mk ^{2} ))]^{1/2} .
Here k still means mMG where M is the "true"
mass of the sun. It is now straightforward to calculate E
and L using the expression for v from above, v =[k /(mR )]^{1/2} .
I find
E =mv ^{2} /2-4k /R =-7k /2R ,
L ^{2} =(mvR )^{2} =kmR ,
(EL ^{2} /(8mk ^{2} ))=-7/16.
So, finally, e= [1-7/16]^{1/2} =3/4.

QUESTION:
what is the laboratory prove for light doppler effect?

ANSWER:
There are innumerable experimental verifications of the
Doppler effect for electromagnetic radiation. I will just give a
couple of examples:

The
light which atoms emit are always at very specific wavelengths.
However, when we look at light coming from distant stars the
wavelengths are almost never exactly what they would be here on
earth. The reason is that, because the universe is expanding,
distant stars and galaxies are moving away from us with large
velocities and all light (and other electromagnetic radiation) is
Doppler shifted because of this relative motion.
When a
policeman measures your speed using radar he is using the Doppler
effect to do so. Suppose that the policeman is at rest and you
are speeding by. The radar leaves the police car and you receive
a Doppler shifted frequency because you are a moving observer.
Then you reflect the radar back to him so you become a moving source
and there is a Doppler effect again. So this reflection becomes a
double Doppler effect. The policeman now compares the wavelengths
of the original radar and that reflected back from you to determine
your speed.
QUESTION:
Would it be possible to detect nuclear material (using some
sort of sensor) used in nuclear (including neutron) bombs or other
nuclear material coming in to the country in trucks, cars, luggage, etc?

ANSWER:
The material used in bombs is the fissionable material,
usually uranium or plutonium. Both of these are unstable
elements, that is they decay into other elements by emitting some kind
of radiaocativity. Basically, there are three kinds of
radioactivity that are important —alpha (just helium nuclei ejected
from the nucleus), beta (high speed electrons or positrons ejected from
the nucleus), and gamma (energetic electromagnetic radiation, like very
penetrating x-rays ejected from the nucleus). So, if you can
detect any of these (for which many kinds of detector certainly exist),
you can infer the presence of the radioactive nucleus itself.
But, you can forget about trying to detect the alpha radiation because
alpha particles are stopped very easily and if you put your contraband
in a cardboard box, no alpha radiation would escape. To a lesser
extent, this is also true of beta radiation, so it would be easy to
hide the beta radiation by light shielding. Gamma radiation, the
most penetrating of the three, would be your best bet. However,
it is also possible to shield the gamma rays but it would take a fair
amount of lead, the weight of which might trigger suspicions. I
guess the bottom line is that, if the smuggler knows what he is doing,
it would be very hard to detect the presence of the material; if he
doesn't, then detecting telltale gamma radiation would be relatively
easy.

QUESTION:
I am in the 8th grade and I am doing a project on what
affects the period of a pendulum. I ran into a differential equation
that you might be able to help me with.
d^{2} q /d^{2} t+(g/L) q =0

ANSWER:
Wow, you are in 8th grade and you know what a differential
equation is?! I suspect it is at least a little obscure to you,
so I will keep my answer as qualitative as I can. The
differential equation you have given me is actually a statement of
Newton's second law (basically, F=ma) for a simple pendulum. A
"simple pendulum" is one which has a mass on the end of a string and
the string has negligible mass and the mass has negligible size.
In your equation, q
is the angle the
pendulum makes with the vertical at any instant, g is the acceleration
due to gravity, and L is the length of the pendulum. But the
catch is, that this is only approximately correct. The correct
equation which describes the simple pendulum is d^{2} q /d^{2} t+(g/L)sin( q) =0. (I hope that
you know a little trigonometry; you really have to to be able to
quantitatively understand the pendulum.) The trouble is, you
cannot solve the correct equation exactly, only by, for example, using
a computer to solve it numerically. However, it turns out that, if
q
is a small angle ,
the sine of the angle (sin(q))
is approximately equal
to the angle q
itself. This is
called the small angle approximation, and you must measure
q in radians (obviously
you can't use degrees because the sine of 2^{0} can't possibly
be close to 2 because the biggest the sine can ever be is 1.)
Radian measure has 90^{0} = p
radians=1.571
radians. A graph which illustrates the small angle approximation
is shown to the right; notice that the small angle approximation is excellent up to about 20^{0}
and quite good up to 40^{0} or 50^{0} . Now,
what is the solution to your differential equation, now that we know
where it came from? The solution will be q and it will depend on
time, so first we have to choose a time when we call time zero.
Imagine the pendulum swinging back and forth such that its maximum
angle is Q
(which is called the
amplitude of the motion) and that it is exactly vertical at the time
which we call t=0. The solution to the differential equation is
q=Q sin(2 p t/T) (see the picture
at the left). The constant T is called the period of the pendulum
and is the time that it takes the pendulum to make one complete cycle
of its motion. For example, if you release it from rest from the
amplitude position q=Q
, then T is the time
it takes for the mass to come back. T, of course, must depend on
what the other constants in the differential equation are. The
solution is T^{2} =4 p ^{2} L/g.
So, you see, the longer the pendulum is, the longer its period
is. One of the most interesting features of a pendulum is that
the period is completely independent of what the mass is; to see this,
go to a park and watch a little kid and his dad swinging side by side
on swings of the same length —they will have the same periods.
Also, the period is independent of the amplitude; this is what makes a
pendulum useful as a regulator for a clock. Don't forget,
though —all this is only approximately true and then only for angles
less than about 40^{0} .

An earlier question I answered might also be of
interest to you.

QUESTION:
My physics students and I are debating which principle better
explains why airplanes and helicopters are able to fly - is it Newton's
3rd law or Bernoulli's principle? Also, if a helicopter is stationary
and you have a scale with enough capacity, can you measure its weight
while it levitates above the scale? Can an airplane do this if it were
able to levitate?

ANSWER:
Indeed, both are important in the design of a wing or a
helicopter propeller, but Bernoulli's principle is the more important
one (that is, normally the pressure difference due to the difference in
velocity of air over the top and bottom of a wing provides most of the
lift). Newton's third law (N-3) also plays a role for the
following reason: Because of the shape of the wing, the air which came
straight in before it encounters the wing leaves the wing with a
downward component of its velocity; the wing must therefore have
exerted a downward force on the air so, because of N-3, the air must
have exerted an upward force on the wing. The balance between the
two types of lift can be altered by changing the "angle of attack", the
angle which the wing makes with the direction of flight. If the
angle of attack becomes bigger, the air is thrust more in the downward
direction than otherwise, so N-3 becomes relatively more
important. (Although, changing the angle of attack also increases
the Bernoulli effect because the speed of the air over the top will be
greater relative to over the bottom when the wing is tilted.) I
would guess that sustained level flight with the airplane upside down
(which is possible) would depend primarily on N-3 and therefore require
a rather large angle of attack. Incidentally, there is a limit to
how much lift you can gain by increasing angle of attack; eventually,
air on the upper of the surface of the wing will be unable to flow
smoothly, will become turbulent, and all the Bernoulli lift will be
lost. This is called a stall. I have had some flying
lessons where stalls were purposely caused and I can assure you, the
plane begins falling like a rock which shows that N-3 alone will not
keep the plane in the air!

The
answer to your second question is no. If you are standing under a
helicopter which is hovering, you experience the downward force of the
air which is thrust down. However, this force is nowhere near the
downward force of the whole weight of the helicopter. If you were
able to "gather" all this downward wind and if the wind lost no energy
coming down from the helicopter, the net force down would still be much
smaller than the weight of the helicopter. This again emphasizes
that the role played by N-3 is relatively small.

QUESTION:
I was wondering if I could ask another question that is
fairly related to the first one. Why do electrons "quantasize" (as in
they "jump" to the next quantum electron level) and what proof is there
that electrones don't raise the circumference of their orbit smoothly?
I was thinking that if when electrons gained more heat, they just went
a little faster making their orbit a little higher which would make the
atoms be more spread out so as to allow flow, then when the electrons
move fast enough, they hit other atoms hard enough for them to start
flying around and being more separated from other atoms than their
magnetic field strength accounts for. Why isn't this the case?

ANSWER:
In some sense, the answer to your question is that bound
systems are quantized because they are! The history of how this
was discovered is too long to go into here, but I will give you a
couple of examples which you can check out in the literature:

Planck
found that the spectrum of radiation which is emitted by an object
called a "black body" can only be understood if one assumes that the
oscillating electrons in the black body which give rise to the
radiation (each like a tiny antenna) have only integral multiples
of hf where h =6.6x10^{-34} J-s is called
Planck's constant and f is the frequency of the oscillator.
Bohr
found that the spectrum of a hydrogen atom could be explained by
assuming that the only allowed orbits are those for which the angular
momentum is an integral multiple of Planck's constant divided by 2 p .
Quantizing the angular
momentum results in also quantizing the energy of the atom.
So,
although it may seem to you like an atom could accept any amount of
energy, it simply is not the case. The technical reason for this
is now well understood. Underlying is the fact that particles are
not really particles but, in certain instances, behave like
waves. For example, an alternative picture of the Bohr atom is
that the electron in its orbit is a wave and, if the wavelength is just
right, the wave will exactly fit as a standing wave on the orbit; if it
is not just right, the wave will interfere with itself and end up as
not existing.

Any
system which is bound together (like an atom, or a molecule, or a
nucleus) has only certain quantized states in which nature allows it to
exist. This is also true for macroscopic systems like our solar
system or a 1.0 kg mass attached to a spring, but it turns out that in
those cases the allowed energies are so incredibly close to each other
that you could never tell that there was not a continuous distribution
of possible energies (which is why you have to look at microscopic
systems to discover nature's quantum properties).

QUESTION:
I hope you can help settle a dispute. We are looking at a
high school physics text book which states that because of Snell's Law,
total internal reflection and the critical angle, someone could find a
location underwater in a swimming pool in order to not be
visible to an observer outside the pool. We found a convincing similar
problem and solution in a college text which asks for the location a
shark would need to be in order to not be visible to an observer in a
boat. However, in the second example, the solution states that you
assume that the observer's eyes are very close to the surface of the
water. Some of us maintain that if the observer's eyes are at any angle
above the surface, there is no hidden region under the water (assuming
there is nothing floating on the surface or anything like that). Can an
object be hidden underwater so as to not be visible to an observer on
the surface?

ANSWER:
The critical angle for water is about 49^{0} . Referring to the picture, note that light which has an angle
of incidence more than 49^{0} (black ray) will not leave the
water. Light with angles less than 49^{0} (red rays) will
leave the water. Of course, the picture you should have in your
mind is that the black ray defines a cone and only rays within that
cone will leave the water. Now, if you think about it a little,
you will see that there is really no place above the water which a red
ray cannot reach (except right along the surface). This is why
the solution you found said that you must be close to the
surface. Therefore, there is really no place the source can be
and be truly invisible to anybody above the water. There is one
additional catch —the red rays also reflect (I haven't show that) some
of the light back into the water so a , b , and c are all
less intense than the light coming from the source; furthermore, the
intensity of the reflected ray gets bigger as the angle gets bigger, so
the intensity of transmitted ray c is
much less intense (bright) than a .
Therefore, if you need ray c to see the
object, you will have trouble seeing it because it is so dim. It
looks to me like your high school textbook failed to take into account
the refraction of the rays leaving the water.

QUESTION:
My question is, why do particles in a solid stick together?
Also, what happens on the atomic scale to make the solid "break" apart
when the substance is melted?

ANSWER:
Think of an atom as a tiny positively charged nucleus
surrounded by a "cloud" of electrons. When these two atoms
approach each other there is, when they are farther apart than their
sizes, no force because each is electrically neutral. As the
clouds begin to overlap, there is a repulsion due to the force between
electrons, but there is also an attraction because the electrons in one
atom are attracted to the nucleus of the other. In most cases
there is a region, that is a range of separations, for which the
attractive force is larger than the repulsive force and the result is
that the two atoms may bind together. This is what causes
molecules to form and what causes solids to form. However, this
sticking together depends upon the kinetic energy of the individual
atoms; and, the average kinetic energy of the atoms is simply a measure
of the temperature of the collection of atoms. For example,
consider a hydrogen molecule H_{2} : think of the binding as a
tiny spring connecting the two atoms and, when the H_{2} gas is
heated up, some of the kinetic energy is taken by the oscillating pairs
of atoms such that the amplitude of oscillation may be large enough to
"break" the spring. This same thing happens in a solid where we
may think of all the atoms as being rigidly bound to their
neighbors. As the solid is heated up, the bonds break and the
solid is no longer a solid. At first it is a liquid where the
atoms or molecules move around in close proximity to their neighbors
with bonds being constantly in a flux of bonding and breaking
again. As more energy is added to the liquid, individual atoms or
molecules acquire enough kinetic energy to escape entirely forming a
gas.

QUESTION:
We recently had an unfortunate incident in Chicago where
scaffolding fell from the John Hancock Building and killed three people
who were in their vehicles. This led to a barroom discussion on what
could kill a pedestrian if it was dropped off the Empire State
Building. Some are of the opinion that something as small as a penny
could do the job. Others say it would have to be something the size of
a rock.

ANSWER:
This is a great question, but it does not (which is the case
for most "real-world" questions) have a simple answer. I think
that there are essentially four things you need to understand in order
to discuss the problem —force, pressure, acceleration due to gravity,
and terminal velocity.

You
probably know all about acceleration due to gravity. If there
were no air friction then all objects would have a constant
acceleration down of approximately 10 m/s/s. What this means is
that when you drop something from rest, it will have a speed down of 10
m/s after 1 second, 20 m/s after 2 seconds, 30 m/s after 3 seconds, etc .
I will not burden you with any equations but will assume that you will
trust me to calculate for myself and tell you the answer of how fast
something will be going after having fallen some distance or some
time! Obviously, this will be necessary to us later.
In the
real world, there is air friction and this can seriously affect the
acceleration due to gravity. Because the air friction depends on
how fast something is moving (for example, imagine the force of the
wind on your hand sticking out the window of a car going 20 and a car
going 80), it turns out that instead of accelerating forever, a falling
object eventually reaches some terminal velocity and then falls with
constant speed after that. How big that terminal velocity is
depends on lots of things, but very much on the size and shape of the
object. Some examples: a parachute causes the terminal velocity
to be so low (but only when open!) that you can safely jump out of an
airplane with it; an ant could jump off the empire state building and
experience no injury; in fact, cats have fallen out of apartment
windows 40 stories up and survived; a dust particle has a terminal
velocity so close to zero that it will stay in the air for years
because of drafts pushing it back up every time it starts to fall.
Of
course, you know what force is qualitatively. Quantitatively
force is mass times acceleration. So, for example, if you want to
stop a 1000 kg (about 2000 lb) car going 10 m/s (about 20 mi/hr) in one
second, its acceleration would have to be 10 m/s/s; the force you would
have to exert is 1000x10=10,000 newtons (about 2500 lb).
I'll come
back to pressure later. So, let's drop a penny off the empire
state building and neglect air friction. I would guess that the
height is something like 100 stories x 3 m/story=300 m. The speed
it would acquire would be about 80 m/s. Now, it hits your head
which must stop it before it enters the brain, maybe 5 mm inside.
So, I figure that it must stop in about .0001 seconds which means that
it must have an acceleration of 800,000 m/s/s! If we estimate the
mass of the penny to be 1 gram=1/1000 kg, them mass times acceleration
is about 800 newtons (200 lb). So your head must exert an upward
force of 200 lb on the penny and the penny, therefore, will exert a
force of 200 lb down on your head! So the question is, could your
head withstand such a force? Yes and no. Imagine a penny
laying on your head and a 200 lb man standing on it. This would
quite likely not cause your skull to be crushed. On the other
hand, imagine a penny standing on its edge and a 200 lb man
standing on it. In all likelihood, his weight would drive the
penny through your skull. So, if the penny hit you flat side, you
would probably survive. If it hit you edgewise, you might
not. If you now factor in air friction, the penny would be going
slower, maybe only half as fast; so in the final analysis, I would say
you have a fair chance of surviving a penny strike. Something
much more massive than a penny, say baseball, would almost certainly do
you in since the force is proportional to the mass and the final speed
would be about the same.

Finally,
why is the penny more dangerous on its edge than on its side?
Because in the former case the whole force is spread over a much
smaller area of your head than in the latter. Force per unit area
is called pressure, and pressure on your skull is more indicative of
strength against breaking than force is. A sharp dagger dropped
from the second story of a building would have no problem getting
through your skull because the force is all applied over essentially a
point so the pressure is almost infinite.
QUESTION:
How can one measure the molecular vibration of a body?
Is it using the electromagnetic radiation? If this is the way,
how would one measure the frequency of elec. waves, how would measure
something so fast in a terahertz or more? Molecular vibrations
are in terahertz, aren't these molecules (it's so fast) supposed to
produce so much heat, that the matter would simply melt?

ANSWER:
The answer to this question requires that you understand a
little about quantum mechanics. When models of atoms and
molecules were first being devised, back around the end of the 19^{th}
century, one of the most vexing problems was how to reconcile classical
electrodynamics with the possibility of having charges move around
inside an atom or molecule. The problem is that classical
electrodynamics says that an accelerating charge radiates its energy
away. As you probably know, the simplest approximately correct
model of an atom has electrons circling the nucleus of the atom.
But, something moving in a circle is constantly accelerating and so it
should quickly radiate away all its energy. However, this does
not happen for certain special states (which are said to be
quantized). That is just an unexpected fact of nature.
Where radiation from atoms comes from is when the atom in one allowed
state makes a change into an atom of a lower allowed state.
Electromagnetic radiation is "spit out" in order to conserve the energy
of the system. The same thing is true for molecules. If you
have a molecule which is vibrating, the energy of those vibrational
states are quantized and the molecule only radiates when it drops from
one state (having a large vibrational amplitude) to another state
(having a smaller vibrational amplitude). Therefore, if a
molecule is in its lowest possible vibrational state as would be normal
in, for example, O_{2} molecules moving around in the air, it
is vibrating with a large frequency but it does not radiate. The
only way to make it radiate is to excite the molecule to a higher state
and let it relax back down to the lowest state by emitting
electromagnetic radiation.

To answer
the question of how to measure the vibrations we must get a little more
quantitative. It turns out that the energy of a quantized
vibrator (visualize a diatomic molecule with a little spring attaching
one atom to the other) is given by E_{n} =(n +(1/2))hf
where n is an integer (0,1,2,3,...), h= 6.6x10^{-34}
joule-s (known as Planck's constant), and f is the vibrational
frequency of the molecule. Now what you have to do is excite the
molecule from its lowest energy state (E_{0} =hf /2)
to its first excited state (E_{1} =3hf /2) and
observe the electromagnetic energy coming out and measure its frequency
which would be the same as the frequency of the oscillator. As a
concrete example, the frequency of radiation emitted by an excited CO
(carbon monoxide) molecule is 6.42x10^{13} Hz. The
corresponding radiation is in the infrared region of the spectrum and
has a wavelength of 4.7x10^{-6} m.

QUESTION:
My brother and I are currently having a discussion
(argument) about whether or not the water in a solar collector can
freeze if the air temperature does not go below 32 degrees Fahrenheit.
He argues, and I have read on websites, that this is possible because
solar collectors are near perfect blackbody radiators and are
transferring their heat into a cold dark night sky. I argue that this
would be true if there was no intervening atmosphere and the heat
radiated into cold dark space. But no matter how ‘perfect’ a blackbody
absorber/radiator the solar collector is it is radiating its heat into
an atmosphere at some temperature. If the collector did get cooler than
the atmosphere then heat would transfer from the atmosphere back into
the collector to establish equilibrium.

ANSWER:
I do not know much about solar collectors. But, what I do know
for certain is that water freezes at 32F! Regardless of what the
other conditions are, if the water gets down to 32F it will freeze.
I presume that the water is contained in some pipe, maybe a copper tube
painted black? Then, I presume the question is whether the copper
can be colder than the surrounding air by virtue of radiating energy
away. Of course it can —but , it depends
on how fast the energy is being radiated away. If the tube
becomes colder than the air, then energy will flow from the air to the
tube via conduction and convection. So it is all a balance
problem —the warmer the tube the greater the rate of radiation;
however, the greater the temperature difference between the tube and
the air around it, the greater the rate of conduction from the air back
to the tube. I strongly suspect that you are right for most
conditions you might have —that is, I expect the air to carry
energy back to the tube just as fast as it radiates away.

In
thinking about this problem, I have found it is instructive to do a few
rough calculations to get a feeling for the numbers we are talking
about. I emphasize that these are just rough estimates to get the
idea of what we are dealing with.

First,
let's calculate a typical energy radiation rate for a black body.
To do this you need to know Stephan's law which states that, for a
black body, the energy radiated per unit time per unit area of an
object with temperature (in degrees Kelvin) T is
s T ^{4 } where
s = 5.67 x 10^{-8}
joules/m^{2} /s/(^{0} K)^{4} . (A joule/s is
a watt.) So, suppose we have a cube 1 cm on a side whose surface
area is thus 6 cm^{2} near the freezing point of water of 273^{0} K.
Then I calculate that the rate of energy radiation is about 180
joules/s=180 watts. Frankly, I was a little surprised at how
large this number is. What it means, as I will show next, is that
if this were all that was going on, the water in your collector would
freeze very quickly.
So,
let's now estimate energy change it takes to change the
temperature. To do this, I will calculate how much energy it
takes to change the temperature of 1 cm^{3} of water (which has
a mass of 1 gram=10^{-3} kg) by one degree Kelvin (same as one
degree Celsius). The heat (in joules) you must remove in order to
lower the temperature of a mass m by an amount
D T
is given by mC D T where C
is a constant called the specific heat. For water C =4186
joules/kg/^{o} K, so I calculate that the heat needed to cool
the gram of water by 1 degree is about 4.2 joules. It loses (at
the rate of 180 watts) this amount of energy in about 0.02
seconds! This is why frost can form at air temperatures above
freezing and why your brother is worried about your solar
collector. Even if we are dealing with something far from being a
perfect black body, the rate of energy radiation would still be on the
order of a few watts so the time to lower the temperature to freezing
would be on the order of no more than a few seconds.
Assuming
that radiation is the only thing happening, the water will quickly go
down to freezing temperature. Now, as more heat is removed, the
water will start turning into ice. About 300 joules must be
removed to freeze a gram of water, so another few seconds should do it.
So,
why doesn't this just happen all the time? For one thing, there
are things all over the place which are also radiating at about the
same rate and the black body absorbs any of this radiation which falls
on it. Furthermore, the air, by conduction and convection, will,
if warmer than the water, carry heat to it. This is a complicated
problem (because air generally transfers heat mainly by convection
which is very complicated) which does not lend itself well to rough
calculations. However, you can get an idea of expected rate of
energy transfer by conduction by imagining the water to be enclosed in
copper container. So, imagine our cubic centimeter of water
enclosed in a copper box 1 mm thick and suppose that the temperature of
the water (and inside of box) is one degree colder than the temperature
of the air (and outside of box). The rate at which energy is
conducted is given by kA D T/ D x where A
is the area (6 cm^{2} ), D T
is the temperature
difference, D x
is the thickness (1
mm), and k is the thermal conductivity (397 for copper).
Doing this, I find that about 240 watts of power flow into the water,
more than it is radiating but close. Now the problem reduces to
whether the air can carry heat to the outside of the copper fast enough
to maintain this temperature difference. It certainly cannot do
it by conduction because the thermal conductivity k for air is
very small (0.0234). But, I believe that under normal
circumstances convection, along with other radiation being absorbed by
the box, will be able to warm the outside of the box fast enough to
keep the energy loss by radiation balanced.
So, where
has all this led us? Your brother is certainly right —this could
happen. The proof is that it is well known that frost can form
when air temperatures are above freezing. (Actually, I believe
that evaporative cooling also contributes to this phenomenon.)
However, I earnestly believe that you are right in all but particularly
extreme situations. If the air temperature is within a degree or
two of the freezing temperature, you should probably consider the
possibility that it might freeze.

QUESTION:
In my Physics class we are doing a project on making our own
hot air balloon which only flies from 2 ordinary birthday candles. How
would I make something like this?

ANSWER:
Since the net force up (i.e. the buoyant force minus
the weight) will be depend on the buoyant force on the balloon, you
want to make the balloon as big as you can to make the lift big.
On the other hand, as you make the balloon bigger, the total weight
will inevitibly get bigger (the material in the balloon weighs
something), so this says make the balloon smaller! So, what is
the ratio of the volume to the area? Assuming that it is
something like a sphere, this ratio is (4p r^{3} /3)/(4 p r^{2} )=r/3.
So you have a net gain in overall force up as you increase the size of
the balloon. However, the last variable is the heating device, in
your case pretty puny! So, there is obviously no way you could
keep hot a huge balloon full of air because heat would escape as fast
as you could put it in. So, here is my advice to you: make the
payload (which is just what you need to hold the candles) as light as
you possibly can. Then find a material from which to construct
your balloon (essentially just a bag open at the bottom) from the
lightest material you can. Then experiment with different size
bags until the whole thing lifts off the ground. Good luck!

QUESTION:
Hi, I am not a physicist or anything, but i was wondering
about gyroscopes. Is it possible that "dark matter" could actually be
explained mathematically by gyroscopic effects? For instance, since
galaxies are spinning and are huge masses, is it possible that the spin
could actually affect gravity and cause the effects that we currently
believe to be dark matter? barring that, is it possible to explain dark
matter if the entire universe was spinning in a gyroscopic manner? or
am i just insane? I do not have the math background or the background
in physics to even attempt to answer this questions, but as they say,
sometimes wisdom comes from the mouth of babes! :> Please let me
know.

ANSWER:
Of course, there is really no answer I can give you to your
hypothesis. As you may know, almost all really great
breakthroughs in science occur with really "off the wall" hypotheses
rather than just the usual incremental advances on what we already
know. Examples are are tectonic plates (modern geology), diseases
being caused by microorganisms (modern microbiology), universal
constancy of the speed of light (special relativity), and wave-particle
duality (quantum physics). However, you can never make any
headway without some description of how the hypothesis might explain
the unexplained phenomenon and without experimental verification of the
veracity of the hypothesis. You refer to "gyroscopic effects"
which really has no meaning in physics. The behavior of
gyroscopes is related to a measurable property that a spinning object
has which is called angular momentum (AM). There is no doubt that
most objects in the universe have AM relative to some axis about which
their rotation occurs. The earth has AM about its own axis; it
also has AM about the sun. The sun has AM about the center of
mass of our galaxy and in fact the whole galaxy has AM around its
center. Does the whole universe have a nonzero AM? I don't
think anyone really knows. But now we come to the hard
part —what, if anything, does AM have to do with dark matter? You
suggest that maybe somehow AM will affect gravity. All I can tell
you is that nobody has ever observed any change in the gravitational
force due to AM. For example, we know that the force with which
the earth attracts us is exactly the same as it would be if it were not
rotating on its axis or revolving around the sun, and that force is
what gravity is. Of course you are not insane (at least not on
the basis of your question!) I think it great that someone thinks
about pivotal questions in physics whether or not they are physicists.

QUESTION:
Im a lower sixth student studying physics in the UK [2 yrs
before uni] and have a question:
Gravity pulls air particles towards it, and the further away from the
earth the particle is the weaker the pull of gravity, hence the higher
you go the thinner the air. Why then, is the ozone layer there? Just
because O3 is heavier than both O2 AND N2, the molecules making up the
greatest percentage or air, it should be pulled right down close to the
earth because by F=ma it will have a stronger force, or does the
pressure of the air at sea level hold it up.
Also, another point of argument this question stems from is am i right
in thinking that as you leave the atmosphere and go into space,
pressure drops off rather than there suddenly being no pressure as u
exit the outer layer of the atmnosphere because there are stary
particles etc.

ANSWER:
The first thing we need to discuss is why the atmosphere gets
thinner as you go up. It is most definitely not because, as you
believe, that gravity gets weaker as you go up. It is true that
gravity gets weaker, but by a very small amount over the distance where
there is any appreciable atmosphere (which could be anything from like
200 km to 1000 km depending on what you mean by "appreciable").
For example, if the acceleration of gravity is 9.8 m/s^{2} at
the earth's surface, it will be about 9.2 m/s^{2} at an
altitude of 200 km. The ozone layer is in the region 20-30 km of
altitude where the gravitational acceleration would be about 9.7 m/s^{2} .
As you probably know, in the ocean as you go deeper and deeper down the
pressure gets greater and greater. The reason for this is that
the weight of all the water above has to be held up. But the
water does not get "thinner" as you go up —why not? Because water
is, for all intents and purposes, an incompressible fluid —no matter
how big the pressure, the volume of 1 kg of water stays almost the
same. Air, which is also a fluid, is far from what we would call
incompressible. The air inside your bicycle tire is at a
relatively high pressure and if you let it out it will occupy a much
bigger volume. Therefore, if you view yourself as being at the
bottom of a 200 km deep sea of air, the atmospheric pressure you
experience is due to the weight of all the air above you. Just
like in the ocean, if you go up the pressure will decrease but this
will now result also in the air being less compressed, i.e.
less dense. So that is why the air gets thinner as you go up.

Now, why
are the ozone molecules in the atmosphere mostly up high? First
let's ask why they are formed up there. The density is much lower
high up in the atmosphere, but there are still a lot of O_{2}
molecules —just not as many per cubic meter as there are here at the
surface of the earth. But there is almost nothing above them, so
they see all the radiation which is striking the earth from outer
space, in particular they see pretty intense ultraviolet (UV) radiation
from the sun. When UV strikes an O_{2} molecule, the
energy is about right to dissociate it into two O atoms. These
float around until eventually the O atom will find an O_{2}
molecule and the two will combine to form your ozone molecule O_{3} .
That's how the ozone gets there and the reason the same thing doesn't
happen at lower altitudes is that much of the UV has been absorbed or
reduced as it plows down into the atmosphere; of course it does happen
at lower altitudes, but just not enough to create a significant amount
of ozone.

Finally,
why doesn't the ozone get "pulled down close to the earth" as you
suggest? Does the water at the surface of a lake fall down to the
bottom? This only happens by convection, for example if the water
at the surface is colder than the rest of the water in the lake it will
be more dense and therefore sink to the bottom. (Of course, that
is why the water at the bottom of a lake is generally colder.)
The same idea applies to the atmosphere —convection causes cold air to
sink and warm air to rise. This convection is the origin of all
our planet's weather. However, in the region where the ozone
exists (called the stratosphere), the air gets warmer as you get higher
so all the air is already "where it wants to be" so there is very
little convection and the ozone stays put.

Your last
question is about how the density falls off. Yes, it is smooth
and gradual and in fact there is no place in the universe where the
pressure is zero. In intergalactic space there are still a few
atoms per cubic meter.

I hope
that I have helped you understand this process. A very good web
site to learn more about the atmosphere is http://daac.gsfc.nasa.gov/CAMPAIGN_DOCS/ATM_CHEM/ozone_atmosphere.html .

QUESTION:
I am a photographer. I will tell you the facts. I take
pictures in a studio with a strobe that is off camera. When I move the
strobe closer to the subject (person I'm photographing) the exposure
increases and I can use a smaller aperture in my camrera (i.e .
the subject reflects more light into my camera, so I can 'stop down the
lens', which closes the diameter of the lens, so less light enters).
However, if I leave the strobe in place and move only the camera closer
to the subject the exposure remains the same. In fact, even if I move
the camera 100 feet away from the subject the exposure of the subject
(although he appears smaller in the viewfinder) remains the same. Now
why is that so? The light travels the same distance so shouldn't the
exposure remain the same?

ANSWER:
First, you need to understand what is called, in physics, the
intensity of the light. Intensity is defined as energy per second
per square meter carried by the light. In your case, since you
use a strobe, I will refer (incorrectly, but it serves our purpose) to
intensity as energy per square meter since the light all comes in a
very short time. The important thing to understand about
intensity is that it falls off like 1/d ^{2} where d
is the distance you are from your subject. The reason for this is
fairly easy to understand. The light goes away from your subject
as a spherical pulse and the total energy in the pulse must stay the
same and the area of the sphere when it reaches you is 4p d^{2} ; ; so
to keep to total energy constant it must be spread out more
thinly. Thus, if you are 50 m from your subject, the light
intensity will be 4 times greater than if you are 100 m from your
subject.

If you
move the strobe closer to the subject the light illuminating the
subject will be greater so the light coming from the subject will be
greater so you will have to stop down your camera. This, as you
note, is not surprising. However, if you move the camera farther
away, it does seem, at first analysis, that you should have to open the
camera up because as you get farther away the intensity gets
smaller. However, if you think about it a moment, you will
realize that although the intensity decreases like 1/d ^{2} ,
the size of the image of your subject also decreases like 1/d ^{2} .
Since the size of the image is smaller it takes less light to get a
proper exposure of the subject by exactly the same amount as the
intensity has decreased. If you use a telephoto lens to make the
image the same size you would have to open your lens up to get a proper
exposure. (Note added : I have been informed that a
telephoto lens, by its construction, lets in more light for the same
f-stop. So, I should say that using a device which enlarges the
image without increasing the amount of light available will result in
underexposure of the film. )

QUESTION:
We know that when a ball is thrown it is being affected by
air resistance, lighter balls will then have a short range (as a ballon
does), heavier balls, due to its mass, will have smaller initial
velocity and thus have a short range also. So I'm going to investigate
which texture (density) of ball will have the longest range. However, I
have encountered a few problem.
1. Should I assume my hand exert constant force or constant power?
2. In deducing the equation for the range, I came across with the
following equation:
1/R * LN(1-(bR/mvCOS(z)))=bTAN(z)/mg - (b^2)/(v(m^2)COS(z))
where R is the range, m is the mass of the ball, z is the angle of
projection, b is the resisive parameter (I assume air resistance to be
bv), and g the gravitational acceleration. But I'm not able to
solve for R, can you help me solve that?

ANSWER:
You have several serious misconceptions here which you should
surely straighten out by doing more reading and research. The
trajectory of a projectile is determined solely by its initial
velocity, the constant you refer to as b, and, if b is
not zero, the mass. Your statement that "heavier balls, due to
its mass, will have smaller initial velocity" is wrong. Initial
velocity is whatever you give it. So you make a mistake to think
of your hand as having anything at all to do with the problem. I
am not saying that mass has nothing to do with the problem, but it
enters only in a subtle way as you will see below; if there were no air
friction, the trajectories of all objects with identical initial
conditions are identical. But mass is not the only thing which
determines what the effects of air resistance will be. For
example, a parachute surely has much more mass than a BB but encounters
enormously more air resistance. If you want to do an experiment
comparing trajectories of different things, you will need to figure out
how to give those things all the same initial conditions (particularly
speed and direction at the beginning). Once your hand or whatever
is no longer in contact with the thing, your hand or whatever is
irrelevant.

So, you
found a formula and you want me to solve it! First of all, there
is an error in your formula (you are missing one minus sign). The
correct formula is ln(1-x )/x =-[1+(bv/ (mg
cos(z)))] where x =(bR )/(mv cos(z )).
I should tell you that this is a well studied problem and is standard in almost
all intermediate level classical mechanics textbooks, so you will find
this formula in just about any classical mechanics textbook. It
is, alas, what is called a transcendental equation, that is one which
cannot be solved in closed form. That doesn't mean you can't
solve it for a specific case (specific b, m, z, v ), you just
have to do it approximately or numerically. Since R is
the only thing you are varying for a specific case you could, for
example, program a computer to keep computing the left side of the
equation until it was close enough (say within 1%) to the right side of
the equation. Or, you can solve it graphically. Since your
formula is of the form

ln(1-x )=-Cx

where C
is just some constant and x is proportional to R .
So, if you plot both ln(1-x ) and -Cx vs. x , where they
cross will be the x corresponding to the R you
seek. An example, for C =2 is graphed to the
left.

Finally,
there is an elegant way to approach the problem if x is very
small. By small I mean small compared to one; notice that x
cannot be bigger than or equal to 1 because ln(0) is negative infinity
and logs of negative numbers do not exist (for your purposes). In
that case you may approximate ln(1-x ) as the first few terms of
a power series expansion, -x +x ^{2} /2-x ^{3} /3.
Using this you can show that you may approximate R =(2v ^{2}
sin(z ) cos(z )/g )[1-(4bv sin(z ))/(3mg )].
This is a formula which you can evaluate directly, but don't forget
that it is approximately correct only for (bR )/(mv
cos(z )) small.

An
excellent reference for this topic is Mechanics by Keith Symon
published by Addison-Wesley but many other texts will cover it also.

QUESTION:
I have recently being investigating, damping of a rigid
pendulum? I used the equation y_{1} =y_{0} e^{-kt}
(the -kt is to the power of e, y_{0} is the initial amplitude
and y_{1} is the final amplitude). I damped the pendulum using
different areas of card and timed how long 5 oscillations took. I would
like to know how I would justify using 5 swngs and not 10 for example?
I had to draw a graph of Area against k. I found that T(period) is
independent of area. However, I would like to find out if I could
expand this investigation to obtain more results. Also, is it true that
a simple pendulum is an approximation of simple harmonic motion? and
does s.h.m only occur when there is a small amplitude?
QUESTION TO THE
QUESTIONER:
Please clarify what you mean by area.
REPLY:
We used card to dampen the ruler, by place it at right angles
to the plane of oscillation. To vary the damping, I varied the
area(centimetres squared) of card.

ANSWER:
OK, so the picture I have is that you used a ruler as a
pendulum and you attached cards of varying areas to the end and
perpendicular to the direction of swing to induce varying air
friction. Let me first answer your last question: as you state
correctly, a pendulum is not an example of simple harmonic motion, one
characteristic of which is that the period is independent of the
amplitude. However, to a very good approximation it is SHM if the
amplitude is not too large. For a pendulum, amplitude is the
maximum angle through which it swings. How large is too
large? That depends on how accurately you measure things.
My guess is that you are using a stopwatch or maybe just the second
hand of a clock to measure times; you would probably find constant
periods for your pendulum up to maybe as large as 40^{0} .
In fact it would be an interesting extension of your study to measure
period as a function of amplitude: measure the period for amplitudes of
say 10^{0} to as big as you can get (if your ruler is mounted
so that you can have amplitudes bigger than 90^{0} you could go
all the way up to almost 180^{0} .) If you then plot your
data you should find a straight horizontal line up to some maximum
angle and then it should go up.
Now, we come to the crux of your
experiment which, as I see it, is to observe the effect of damping on
the motion. So you measured the time of five oscillations and
found that you got the same answer for all dampings. This is a
good start. What you have shown is that different small
oscillation pendula with different damping (that is different k because
that is what you vary by changing the card area) all have the same
periods. In fact, that is what you are supposed to find if you
study the theory of the damped oscillator which predicts that you get
exactly the same oscillatory motion except that the amplitude decreases
exponentially as a function of time. The graph to the left shows
what theory predicts. You see that the damped oscillator has
exactly the same period as the undamped one. So you have verified
that aspect of the theory. However, what you need to do now are
some measurements of the amplitude as a function of time and see if it
really is exponential. The way you would do that is to measure
the angle of the pendulum each time it reaches the amplitude (when you
release it, one half period later, one half period later, etc) .
This is a little trickier measurement since you would have to tape an
angle scale to the wall or something like that. What would be
really cool is if you could take a video of it and then make the
measurements on the tv or computer screen. If the theory is
right, then you should get a series of points which can be plotted vs.
time (one half period per point) which fall off like the blue lines
above. And, the data for different damping should now look different
with the larger cards falling faster than the smaller cards. Now,
finally, if you are good at math, you should be able to extract the
constants k for each card so you can see how k depends on area.
The easiest way to do this is to plot your data on semilog graph paper
(or plot the logarithm of amplitude vs . time); if the data are
truly exponential, you should get a straight line and k is essentially
the slope of that line.

Incidentally,
when measuring these amplitudes it would be a good idea to use a very
long pendulum so that the period becomes much bigger because it would
be easier to measure because it would go slowly longer near the end of
each swing; and you could tape a more accurate (that is bigger) angle
scale to the wall which would be easier to read.

QUESTION:
Suppose I have a pot of boiling water on a stove. If I turn up the
stove to a higher setting, what will happen to the temperature of the
boiling water?
Water boils at 100 degrees celcius. So the way I see it is that when
the setting is increased, the rate of boiling will increase while the
temperature of the water in the container will take the temperature of
the heating apparatus. I need know if this sounds correct.

ANSWER:
Water undergoes what is called a "phase transition" at 100
C. (We are talking here about what happens at "normal" conditions
of pressure, i.e. about one atmosphere. E.g.,
water boils at a lower temperature at high altitudes, and if you have
ever lived there, you know that you it takes longer to boil a potato
there.) Until you get to that temperature, heat you add (that is
energy you add) shows up as an increase in temperature. When the
phase transition temperature is reached, additional energy does not go
into increasing temperature any more but changes the water from a
liquid state to a vapor state (which requires energy to do). So
when you are adding energy at a certain rate you are converting water
to steam at a certain rate; now if you increase the rate that you are
adding energy you increase the rate of converting water to steam but
you no longer cause there to be any temperature increase. The
hottest the water will get is 100 C. That is why it is a waste of
energy to have the heat turned up any higher than you have to to keep
water boiling if you are boiling potatoes. The temperature of the
container is irrelevant since it is merely a means of conveying the
heat from the heater to the water; the hotter the container, the faster
you are converting water to steam.

QUESTION:
Am I correct in assuming that water at 4 degrees centigrade is more
dense that water at say 5 or 6 degres centigrade. The change in density
is negligible between the two, but yet theoretically existant?

ANSWER:
You are completely correct. Water, in fact, has its
largest density at 4 C. Most materials expand as you heat them up
and therefore, since the volume increases while the mass stays the
same, the density decreases. This is true for water only above 4
C. The bizarre thing is that water also expands as you cool it
below 4 C. This simple fact has profound effects on the nature of
our world which has so much water in it. In the winter the
environment causes the water in a lake to cool down, first near the
surface which is in contact with the cold air. The layer of water
at the surface cools and gets denser and therefore sinks to the
bottom. This continues happening until all the water in the lake
is 4 C. Now, as the water near the surface cools further it
becomes less dense so it floats (stays at the top) until it reaches 0 C
at which point it freezes and becomes ice. This is why ice floats
in water. For most materials (take solid iron in molten iron, for
example) the solid sinks. If water behaved like this, a lake
would freeze from the bottom up. You would not be able to ice
skate until the entire lake was solid! So you see, these tiny
changes in density are not, as you assert, negligible. Also, it
has nothing to do with "theoretically"; this is an experimental fact,
not a theoretical hypothesis.

QUESTION:
I was wanting to know the official mathematical equation for the Second
Law of Thermodynamics. I found a couple of sites, each with conflicting
information (argh!)

ANSWER: (Provided
by D. P. Landau)
The 2nd Law of Thermodynamics is the result of research by
multiple individuals who took slightly different approaches to the
study of thermodynamics. Consequently, there is no single "definition"
of the 2nd Law of Thermodynamics but rather several statements that
have equivalent consequences. The Clausius statement says: "No process
is possible whose sole result is a heat flow out of one system
at a given temperature and a heat flow of the same magnitude into a
second system at a higher temperature". The Kelvin-Planck (combined)
statement is: "No process is possible whose sole result is a
heat flow Q out of a reservoir at a single temperature, and the
performance of work W equal in magnitude to Q". Most books on
Thermodynamics (e.g. Thermodynamics, Kinetic Theory, and
Statistical Thermodyanmics by Sears and Salinger) will discuss both
of these.

QUESTION:
I had a CAT scan on Thursday because I have had a long-term
nose problem that causes hearing loss. In the second part of the test
they had me lie on my stomach, head down. I closed my eyes while the
test went on, and I "saw" the x-rays. I saw the image(s) separately in
each eye. What I saw: it looked like a solar eclipse with the "sun"
completely black as in an eclipse, but the "corona" of the "sun" was a
very pale blue color. But very bright. It appeared as multiple flashes
and it flashed 8 or 9 times, much in the same way as one of those photo
booth machines where you put in a dollar to get several photos.
I seem to remember that somebody connected with the discovery of x-rays
or radioactivity could "see" radioactivity with their eyes closed, too,
and that this is the result of some direct action on the brain, not
through normal eyesight.
I've checked the internet fruitlessly in search of any information. But
your site popped up on google as a response to search terms ["see
x-rays" brain spectrum] so I'm hopeful you can give me the answer or
direct me to where I can find it. I'm not a scientist, (majored in
history).

ANSWER:
I have asked around the department and nobody has much
knowledge here. What I can tell you is the following: Your eye is
filled with a fluid (mostly water). When ionizing radiation (which
includes x-rays, gamma-rays, and fast charged particles like electrons)
passes through matter the result is often that it causes there to be a
little flash of light. This process is called scintillation and is the
basis of an important type of detector of radiation (in fact,
scintillation detectors may very well be used to detect the x-rays in
the CAT machine). At any rate, the fluid in your eye scintillates and I
suspect that this is the origin of what you "saw" —real light but
coming from inside your eye rather than from outside as it usually
does. I very much doubt that the radiation is directly stimulating your
brain.

QUESTION:
How do I calculate torque of a vertical axis turn table drive
by an internal gear given the weight of the table (20 lbs) and the
pitch diameter of the gear (10.125 in)? I want to accelerate at .38
rad/s^{2} .

ANSWER:
You have given me mostly the wrong information to answer your
question. Newton's second law is given for rotational dynamics as
t=Ia
where t
is the torque, I
is the moment of inertia, and a is the angular
acceleration. If your turntable can be approximated as a uniform
solid disk, then taking its mass m to be 9.1 kg (which is about
the mass of 20 lb), its radius R to be 0.15 m (which would
correspond to about 6 in), its moment of inertia would be about I=mR^{2} /2= 0.1
kg m^{2} . Thus the required torque would be
t =0.1 x 0.38=0.038 N
m=0.336 in lb. To get this torque you would, for example, have to
exert a force (tangentially, that is tangent to the circumference) of
0.056 lb at 6 in from the axis (because torque is force times the
distance where force is applied relative to the axis). There is
one more complicating factor you will have to take into account.
Your turntable will not be frictionless, that is there is some
(negative) torque due to friction. Therefore you will have to
measure this frictional torque (by measuring the angular acceleration
of the turntable when not driven) and add this amount of torque to the
driving torque calculated above (which was 0.336 in lb for the numbers
I chose) to get the proper acceleration. So, for example, suppose
you measure the angular acceleration of the wheel which is slowing down
with the motor off to be -0.1 rad/s^{2} . Then the
frictional torque is -0.1 x 0.1=-0.01 N m=-0.089 in lb. In that
case you would have to exert a torque of 0.089+0.336=0.425 in lb to get
an angular acceleration of 0.38 rad/s^{2} .

QUESTION:
Is there any way to repel charged solute particles in an
aqueous solution using some kind of electromagnetic field? For example,
if I had a "U" shaped tube filled with water, could I somehow use a
magnetic field as a partition between the two sides so that if I put
some charged solute molecules in the water in one half of the tube,
they could not diffuse to the other side? Thanks.

ANSWER:
In principle, any field which would exert a force on the
these charged solute molecules could be used to achieve what you
want. However, a magnetic field would probably not be a very good
choice because the force that a magnetic field exerts on an electric
charge is proportional to the velocity of the charge and the charged
solute molecules move around in the solvent like a gas, that is with a
distribution of velocities some of which are very small.
Therefore the magnetic field might very well not be strong enough to
keep the slower molecules from "leaking through". A much better
choice would be a strong electric field which would exert an equal
force on any charged solute molecule. To keep them from leaking
through, you would simply have to have a strong enough field such that
you could stop the fastest of them which were heading straight toward
the other side.

QUESTION:
How would you go about calculating the autocorrelation of a
Gaussian laser pulse:
E(t) = [ (s/pi)^2 * e^-(s(t^2)) * e^(-iwt) ]
- I appreciate you taking the time out to help me with my question.

ANSWER: (provided
by W. M. Dennis)
Normally what is required for determining the pulse width of
a laser pulse is the intensity autocorrelation function. This is the
quantity that would be measured using a second harmonic crystal.

_{ }
The
upper integral can be evaluated by completing the square and the
resulting autocorrelation function is another Gaussian.

If we define the intensity as proportional to e^{_t} ^{2
} the autocorrelation function which is a function of
the delay, t , is given by e ^{_} ^{t} ^{2/2} . The two pulse widths
are related by

_{D} _{t} _{=Dt/1.414.}

For comparison, if we were to assume a pulse of the form I (t )~sech^{2} (t )
we would find

_{D} _{t} _{=Dt/1.543.}

QUESTION:
We know that during beta-decay, an neutron is converted to a
proton and an electron. The electron was emitted leaving the proton
inside the nucleus. So, it will increase the net charge of the atom by
1, changing it into an ion. As the decay goes on the net charge grows,
the attraction between the ions is reduced due to the net charge
produced. The radioactive material will explode eventually due to the
repulsion mentioned above. However, it does not agree with our common
experience that the material will not explode. How is it justified?

ANSWER:
This is more a question of scale and number than anything
else. In order that the premise of your question be satisfied,
let us assume that the radioactive source is in a vacuum and that it is
thin enough that most of the electrons actually escape the
source. A quite "hot" radioactive source would have one Curie (C)
of radiation coming out and 1 C corresponds to 3.7x10^{10} _{
} decays per second. That sounds like a lot, but
consider how many atoms there are in the whole sample, something like 10^{23} ,
enormously larger than the number actually decaying over some
time. So, on the average, the ions which form will be far
separated from others. As time goes on, of course, the source
will acquire a net positive electrical charge, so its electric
potential relative to its surroundings will increase. Eventually,
there will be a large enough potential difference that perhaps a spark
will jump to neutralize the charge. Or, if the source is in the
air, a large enough potential would lead to corona discharge to the
surrounding air. In the real world it is pretty difficult to
maintain a large static charge on something, and a static charge large
enough to "explode" something just doesn't happen.

Oh, yes,
one more issue: You have to consider what happens to the charge once it
is created on one atom inside the volume of the source. If the
material of the source is a conducting material, then any net charge
must reside on the surface, so this puts the excess charge on the
surface where it is easier for it to be neutralized by the environment.

QUESTION:
Why does the flame of a candle blow out in a wind or a
breath? What exactly is the explanation of this?

ANSWER:
I must admit that I stole this answer from another web site,
but I'm not into reinventing wheels! Fire is an autocatalytic
oxidation process, which cannot go on if you disperse the zone where
the reaction happens, i.e. the flame. In other words, when you
blow on the candle you rapidly dilute the contents of the fire, which
makes the amount of heat per unit of volume not sufficient for the
chain reaction of burning to continue. Consider also that the flame
"feeds" off vaporized paraffin, and if you blow it away the flame has
nothing to burn. Third factor is cooling, which is (again) associated
with dilution. As a result the flame goes out.

QUESTION:
Hi!
I'm involved in a "science problem of the week" contest. The idea is to
research each question, making contact with anyone who might be of
help. Perhaps you could assist me with this latest question?
Thanks!
Consider the following scenario...
In this situation, a person sits in a seat and pulls down on a rope. As
a result, the seat and the person in it rise up. If the seat and
a man together weigh 180 pounds, with how much force must the man pull
down on the rope in order to raise himself and the seat up with a
constant speed?

ANSWER:
This is a
classic problem often appearing in textbooks. Shown in the figure
above is the picture from Serway's text Physics for Scientists and
Engineers (I'm probably violating some copyright by reproducing
it!) So I will refer to the boy and swing instead of your man and
seat since that is the picture I found. Since the boy and swing
are to move up with constant velocity, this is a Newton's first law
problem, i.e. the sum of all forces on anything you choose to
look at must add up to zero. The easiest thing to look at is the
boy+swing. The forces on the boy and swing are its 180 lb. weight
(down ), the tension of the rope tied to the swing (up ),
and the tension of the rope the boy is pulling on (up ).
But the ropes tied to the swing and on which the boy is pulling are the
same rope so they have the same tension. Therefore, twice the
tension in the rope must equal 180 lb, so the tension in the rope is 90
lb. Technically, the problem has not yet been solved since what
is asked for is how hard the boy must pull down on the rope and we have
found how hard the rope pulls up on the boy; however, Newton's third
law tells us that these two forces must be equal and opposite, so the
final answer is 90 lb down.

QUESTION(1):
Today there is evidence that the big bang occurred. But do
scientists have any theories as to WHY it occurred? Or is it,
that they just don't know for sure and are working on it? Also, any
theories are not testable right?

ANSWER(1): (Provided
by L. A. Magnani)
Right now,
there is no consensus on why the Big Bang happened. Modern physics
breaks down in conditions typical of the Plank Time (10^{-43}
seconds after the Big Bang) so the safest thing to do is to discuss
what happens after the Plank Time. However, people who work on Grand
Unified Theories and string theory feel that when these theories are
fully developed we will be able to discuss what happens before t=10^{-43}
seconds. People talk about vacuum energy fluctuations and phase
transitions, but, at the moment this is just speculation backed up by
theories which are in the developmental and not the testing stage.
String theories and GUTs may be testable in the years to come, but they
are not testable now. However, the next few decades could lead to
breakthroughs in this area.

QUESTION(2):
Oops! I have another question. According to this article: http://www.space.com/scienceastronomy/generalscience/physicists_bigbang_000209_wg.html
Does this mean that at the most we can have quarks come out of this
"recreation" of big bang conditions? Or do scientists think they
can make other matter, like trees, rocks, gold, etc. through this
matter, or do they think that it only occurred through the big
bang?

ANSWER(2): (Provided
by L. A. Magnani)
I hadn't
heard of this result and I would like to know more about it. But,
I think the answer to the student's question is that the conditions in
the early Universe which led to quark production subsequently led —as
the Universe expanded and cooled —to the production of protons and
neutrons. Hydrogen, Helium, Lithium, and Beryllium followed. But by
then, the Universe had cooled to the point where nucleosynthesis could
no longer proceed and the abundances of those elements and their
isotopes were frozen in (some elemental and isotopic compositions
changed with time because of processing of original material in stars).
The rest of the elements, necessary for trees, rocks, gold, etc. were
synthesized in stars.

QUESTION(3):
Hello, I have one last question!!! Do scientists believe
that the big bang explosion came from a single atom, a glob of matter,
or they just don't know and are working on it?

ANSWER(3):
See ANSWER(1).

YET ANOTHER
FOLLOW-UP QUESTION:
Prof. Magnani had said that "elements, necessary for trees,
rocks, gold, etc . were synthesized in stars." What does
he mean by this? I mean, you never see gold, or a tree, or a rock or
plants, or a coral reef coming out of a star or inside a star? How
could these things come to earth through a star? Also, how are stars
and the Sun formed? I couldn't find this on the Web!!! I mean, stars
are balls of gas, and the sun is too right? So could we just put
hydrogen and helium together to make a mini-sun?" How do
scientists think the earth is formed? I have a feeling that the
professor will tell me that stars were created through the matter by
the big bang, but what about the earth? You'll never find the earth
coming out of those balls of gases, right?

ANSWER:
As Dr. Magnani said, after the big bang the universe
consisted entirely of the very lightest elements, mostly
hydrogen —nothing you could make a tree out of. The reason is that what
was created were the "elementary" particles (protons, neutrons,
electrons) and to make a heavy element like gold you have to put
together a bunch of protons which takes a lot of energy to push those
positively charged protons together which want to keep apart. Therefore
when things have settled down after the big bang you have what is
essentially a universe full of hydrogen. And that would be the end of
it if the universe had been homogenous (perfectly the same in all
directions). But there were lots of places where the density of H gas
was bigger than neighboring regions and the gas started to contract
because of the gravitational attraction. Eventually the gravitational
force has compressed this H gas so much that it gets real hot, that is
the atoms are moving around with a very large velocity. Now the H
starts "burning". It isn't really burning in the usual sense; rather it
is colliding with other H atoms and causing nuclear fusion to occur.
The main thing that happens in a normal star in the "prime of its life"
is that H burns to Helium and the result is the release of huge amounts
of energy. That is the energy that keeps us all alive. As the star gets
older and begins to exhaust its H, things get complicated and things
start happening relatively fast. Basically the He starts combining to
form heavier elements, the star starts cooling down, so that it starts
to contract under gravity again. In many cases the star at some point
just explodes (a supernova) throwing all its matter out into space. Now
we have in the instellar space material other than H. But there is
still a lot of H around. So now a new star will form but when it does
there is stuff other than H there. What apparently happens is that
there is an "accretion disk" which forms around the new young star
which has much of the heavier elements in it (sort of like Saturn's
rings which are sort of an accretion disk) and the "stuff" in the disk
condenses into lumps which become planets. So our sun must have formed
long after the big bang or else there wouldn't have been any of the
heavier elements around. A favorite poetic saying astronomers like to
use is that "we are all made of star dust".

The
answer to your question "So could we just put hydrogen and helium
together to make a mini-sun?" is yes! That is what a hydrogen
bomb is —a mini-sun. That is what scientists trying to harness
fusion energy are trying to do is to make a controlled H bomb as an
efficient power source. It is a very difficult job still far from
finished.

You need
to get yourself a good book on elementary astrophysics and one which
addresses stellar evolution. There were things called books and
libraries back before the internet and they actually still exist!
Seriously, though, you clearly have a very inquiring mind and you
should read a good book or take a course so that you can get the big
picture. It will be better for you than surfing the web!

QUESTION:
I need to know which element has a symbol which does not
relate to its name, eg: Copper = Cu = Derived from latin name Sulpher =
S = First letter

ANSWER:
To the best
of my knowledge, no element which has been officially named by the
international agency responsible for naming ( IUPAC ) has a symbol
which has no relation to either the actual name or to its name in
another (usually classical) language (e.g . Ag for argentum which
is Latin for silver). Many of the names are in honor of famous
scientists (Einsteinium, Curium, etc .) and many are named for
places where they were first discovered (Ytterbium, Erbium, Yttrium, e.g .,
are all named for a village in Sweden called Ytterby). There are a
number elements for which agreement upon their names has not been
reached, so perhaps these are what you are looking for. They have been
given temporary names and symbols. For more information you may go to
the website Weblements, http://chemserv.bc.edu/web_elements/web-elements-home.html ,
and links in that site. For information on the naming issues, see http://www.webelements.com/webelements/properties/text/definitions/name.html .

QUESTION:
I want to know about polylog. What is a polylog and How
polylog to use? And I want to know about photon gas...and the
difference between ideal gas and photon gas ..thank you.

ANSWER:
(provided by M. H. Lee)
1. Polylogs
Polylogs are transcendental functions much like the Bessel functions.
(A transcendental function is one that is not a polynomial, sometimes
called an algebraic function. It has an infinite number of roots.) They
are probably least well known of this class of functions. The simplest
is called dilogs, invented by the great mathematician Euler in 1768.
The next simplest is trilogs by an English mathematician Landen some 20
years after Euler.
These functions are basically integrals involving log terms. As a
result, most mathematicians probably shied away from them. But in
physics there are many log related quantities. Thus, whether they like
or not, physicists must use them. Recently these polylogs have come to
simplify certain problems in thermodynamics and statistical mechanics.
Polylogs are also useful in mathematics since Riemann's zeta functions
are special cases of polylogs. Perhaps one could look up "Polylogs and
Riemann's zeta function," by M. H. Lee, Physical Review E 56, 3909-3912
(1997).
2. Photon gas
If a solid is heated at some fixed temperature T, it will emit energy
or light (both visible and invisible.) If this process is done in a
solid carved into a cavity, the energy is collected inside the cavity.
This energy is called a photon gas (black-body radiation) —a gas made
up of photons, which are units of light energy. These photons have both
wave and particle behavior. If one thinks of them more in particle
sense, this gas of photons is rather like a gas of ordinary particles
like He. The important difference is that if the temperature T is
raised, there will be more photons. (It doesn't happen with a gas of He
atoms.) This photon gas is sometimes called thermal photons for this
reason.
3. Ideal gas
The term ideal gas is a general term, meaning a collection or gas of
any particles which do not mutually interact. A photon gas is a good
example since photons are not known to interact with each other.
Another example is a neutrino gas. A He gas is not an ideal gas since
He atoms will interact with each other. But at very high temperatures
the interaction appears very small, so that often it is regarded as an
ideal gas. This idea applies to almost all atomic gases. For a recent
reference, see "Carnot cycle for photon gas?" by M. H. Lee, American
Journal of Physics, August 2001.

QUESTION:
I am having some hard time trying to get some data and laws
on how metal conductivity diverts (decreases indeed) from bulk
resistivity when the metal sheet gets very thin ( i.e.
significantly smaller than the average electron free run in the metal).
Any idea ? Text books ? Articles ? Web sites ?
Thanks for your help.

ANSWER:
(provided by M. R. Geller)
The effects that lead to this change from bulk behavior are
usually called "size effects". They have many origins because there are
many length scales (e.g., Fermi wavelength, thermal wavelength, elastic
and inelastic mean free paths, phase coherence length, magnetic length,
screening length) associated with a metal, and as soon as the film
becomes thinner than one of these length scales the transport
properties change. Some good references are Chapter 8 of Abrikosov's
book (A. A. Abrikosov, Fundamentals of the Theory of Metals) and the
review article by Ando et al. (T. Ando, A. B. Fowler, and F. Stern,
Rev. Mod. Phys. vol. 54, p437, 1982).

QUESTION:
As I understand it, osmosis does not require any energy to
occur; it is passive. If a passive process can move water in an upward
direction against gravity....then the osmosed water would then have
potential energy...correct? If you can't create or destroy energy, you
can only change it from one form to another....and it took no energy to
give the water potential energy....what was the energy which had to be
converted to the potential energy the water now has? Also, where
is the entropy in that conversion? Thanks, Brett

ANSWER:
First a disclaimer: I don't have any expertise here! I
did a little research, though. First of all, let's define what
osmosis is: essentially a membrane lets water pass through but does not
allow cer tain
molecules dissolved in the water thro ugh. The classic
example illustrating what happens is a U-shaped tube with water in it
and a permeable membrane across the tube at the bottom. Water may
pass through the membrane unimpeded and therefore, if there is just
water in the tube, the level of fluid in each side is the same.
However, if you put, for example, some salt in one side, water will
flow through from the pure water side to the salty side raising the
level of the salty side and lowering the level of the pure water
side. This system, as you note, has more gravitational potential
energy than before the water flowed, so the question is: where did this
energy come from? The solute behaves like an ideal gas and moves
around in the solvent with some average velocity. The solute
molecules are therefore colliding with the membrane transferring
momentum and energy to it and the net result is a pressure differential
across it. This pressure difference is what causes the
water to flow preferentially in one direction and the flow will
continue until the pressure difference is the same as the hydrostatic
pressure due to the weight of the upper column of water. The work
(energy) to increase the potential energy came from the kinetic energy
of the solute molecules which must therefore have slowed down (that is,
this "solute gas" must have cooled down in the process).

You state
that "osmosis does not require any energy". What this means is
that no external agent does any work; that doesn't mean no work
was done. As you can see, the total energy of the system has
stayed the same because the potential energy increased and the kinetic
energy decreased.

Finally,
with regard to entropy, the second law of thermodynamics states, in one
incarnation, that "as two interacting macroscopic systems approach
equilibrium, the changes in the system variables will be such that the
entropy of the combined system increases." So, the entropy of the
system increases.

The
website where I learned about osmosis is http://members.tripod.com/~urila/

QUESTION:
When placing an hourglass on a scale, will its mass increase,
decrease, or stay the same as the sand is falling?

ANSWER:
Here is a question with just layers and layers of
answers! I will address all the possibilities I can think of:

First,
let's just think of the problem purely classically (relativity is an
issue as you will see below). Let us assume that the hourglass
has nothing in it except sand (in particular, there is no air).
During the time the sand is transferring from the upper to the lower
chamber there is, at any time, a fraction of the sand falling and
therefore not exerting any downward force on anything. Therefore,
during the time the sand is falling there will be some fraction, say
1%, of the sand not being weighed, so the scale will read less at this
time.
Next
let there be air in the hourglass. During the time when sand is
falling the air will exert two upward forces on the sand, a frictional
force and a buoyant force. If the air exerts an upward force on
the sand, then Newton's third law stipulates that the sand exerts an
equal but opposite force on the air which is part of the system being
weighed, so the amount by which the weight is less will be smaller than
if there is no air in the hourglass.
A
possibility related to part 2 of the answer is that, if the forces
upward on the sand are not small compared with the weight of each sand
particle (for example think of the particles of sand as being a very
fine powder or think of the hourglass being filled with molasses), then
the falling particles will quickly acquire a terminal velocity where
the upward forces are exactly equal to the weight of the particles; in
this case the hourglass system will always have the same weight as the
sand falls.
Another
small effect is that the earth's gravitational field is not really
uniform, that is the weight, which is just the force which the earth
exerts on the sand, gets bigger like 1/r ^{2} as r
decreases where r is the distance to the center of the earth;
it is weight, not mass, which the hourglass measures. Therefore
the total weight when all the sand is in the bottom will be larger than
when it was all in the top. This is a very small effect since the
radius of the earth is about 6,400,000 m and the distance the sand
falls is only something like 10 cm, so this effect will cause the
weight to increase by an amount (I calculate) of about 3x10^{-6} %!
Now we
come to relativistic effects. Einstein says E=mc ^{2 } where
E is energy, m is mass, and c is the
speed of light (about 3x10^{8} m/s). Let's think about
the energy of one grain of sand. At the top it has, relative to
the bottom, a gravitational potential energy of mgh (where m
is the mass of the grain of sand, h the distance it falls, and g
the acceleration due to gravity) and no kinetic energy. Let us
ignore effects like 2-4 above for purposes of clarity, so imagine an
evacuated hourglass in a uniform gravitational field. As the sand
falls its potential energy gets smaller and its kinetic energy gets
larger, all the while keeping the total energy the same. When it hits
the bottom, all of its energy is kinetic and equal in magnitude to mgh ;
but now it stops so this energy disappears. Or does it? No,
energy is still conserved because the kinetic energy will be converted
into heat and the sand+hourglass will warm up a bit. But, if we
suppose that the whole system will then cool back down to the same
temperature as its surroundings eventually, there has been a net loss
of energy Mgh where M is the mass of all the
sand. But a loss of total energy means a loss of total mass
because E=mc ^{2} , so the change in mass is
D M=Mgh/c ^{2} ;
if we again take the size of the hourglass (h ) to be about 10
cm, I find D M/M= 10^{-17}
or about 10^{-15} % change in mass!
In
summary,

#1 decreases
the mass measured by something on the order of 1% while the sand is
actually falling ;
#2 and
#3 cause the effect in #1 to be smaller or zero;
#4 increases
the weight measured (but not the mass) by something on the order of 10^{-6} %
after the sand has fallen ; and
#5 decreases
the mass by something on the order of 10^{-15} % after the
sand has fallen and cooled down .
This is
probably way more than you wanted, but the answer you wanted should be
in there somewhere! And, I had fun thinking about it!

QUESTION:
I'm a chemist who's having a tough time getting his
head around a little electric field problem. Here's the
deal - I've got a supramolecular system that orders itself into 3
nm "tubes". The surface of the tube is made up of the ionic ends
of molecules and form a positive and the corresponding negative
layer (i.e. a dielectric). I wanted to make the tubes line up
perpendicular to the substrate I've made them on, and I have, by
putting the film into a capacitor i.e. 2 copper plates 4 mm apart
with 3000 volts between them. The question is, why does this
work? I'm glad to give any further clarifications you may
require. I originally thought this would be a simple problem, but
have been stumped at each step.
QUESTION TO THE
QUESTIONER:
I'm not sure I understand the problem or the question! If I
had one of your tubes in empty space, describe the charge density
on it to me; e.g. positive charge distributed uniformly on the
outer surface and net negative charge inside the tube such that
the net charge is zero? Is there any net charge on the ends of
the tube? How do you know that they all stand up straight? Why do
you call it a dielectric? Maybe you mean dipole? What happens if you
reverse the field?
REPLY:
First, thanks for your speedy response. Second, perhaps more
detail is required. I have a micelle (the tube) which is made of many
long alkyl chains with amino groups (which is positively charged) at
the terminus. Charge is balanced with chloride ions (negative).
Surrounding the micelle is silicate (another different dielectric),
which is only partially polymerized, and still flexible. We apply the
field, the silicate continues to condense, we remove the field and the
shape is stuck. The ends of the tubes are the ends of the micelle,
which will have no net charge, we know they stand straight by TEM
images of the silicate after the procedure, there are dipoles in the
system, but I don't think they contribute much to these supramolecular
events, and reversing the field produces the same end product.

ANSWER:
My suspicion is that the answer to your question is, in fact,
totally independent of all the details you have given me. Imagine a
needle in a uniform electric field. If the needle is made of a
dielectric material, the needle will become polarized throughout its
volume but the ostensible effect will be for each end of the needle to
become electrically charged (the ends having opposite charges). The net
effect of this charge will be for there to be a net torque on the
needle which will tend to align it with the field; indeed, this will
happen regardless of the direction of the field (which is why I
asked!). This same thing would happen with a conducting needle except
the end charges would be due to real electrons migrating to one end and
leaving positive ions at the other end. I believe that this explains
your effect. Of course when you turn off the field, the polarization
goes away; it sounds like the silicate gives rigidity to your tubes so
that they remain in position after the field is turned off.

QUESTION:
Dear Physicist, can you please tell me why photo current is
independent of frequency if intensity is kept constant. Arthur Beiser
on page 55 of his text perspectives of modern physics makes this
statement. This seems impossible because if intensity is defined as
power/metre squared then there must be few electrons emitted by higher
frequency light. I have been told that Beiser may be defining intensity
as flux of photons but on page 54 he quotes intensity in units of Watts
per metre squared. Please Help!!!

ANSWER:
Suppose that the intensity I is energy/unit
time/unit area (the standard definition as you note). Then if dN/dt
is the rate at which photons are striking an area A
on the photocathode and hf is the energy/photon, we may
write
dN/dt=IA/(hf).
This just says photons/unit time/unit area=[energy/unit
time/unit area]/[energy/photon]. Now, clearly, the rate at which
electrons will be ejected/unit time from A will be proportional
to the number of photons striking A per unit time so it follows
that the current (which will be proportional to the number of electrons
ejected from A per unit time) will be proportional to dN/dt
and therefore inversely proportional to f and not independent
of it for a given I. Therefore you are right! The
photocurrent is independent of frequency if the photon flux remains
constant (some authors call this the photon intensity), but not if the
intensity remains constant.

When I first read the previous question I misread it and
therefore answered the wrong question! I leave my original answer
here in case anyone is interested in learning more about the
photoelectric effect
The question
you ask was one of the pivotal questions which led to the development
of quantum physics around the beginning of the 20^{th } century.
It is an excellent question and one which truly perplexed the best
minds in science at the time. It took Albert Einstein to come up
with the answer to this puzzle and his explanation later was recognized
with the Nobel Prize. Surprizingly, the citation for his Nobel
Prize did not mention the theory of relativity at all but did say
"especially for his discovery of the law of the photoelectric effect".

So your
question is basically "if the energy is present in that incident light
to knock out an electron, why do no electrons get knocked out?" I
will make the answer rather brief. This is a topic which you can
further research in just about any book on modern physics. The
answer is that light can only give up its energy in little increments
and those increments happen to be proporional to the frequency of the
light. The energy which light can give up in one little bundle is
E=hf where f is the frequency and h is called Planck's constant which
has a value of about 4x10^{-15} eV s (where s means seconds and
eV is the amount of energy an electron acquires if you let it
accelerate across 1 volt of potential difference). These little
"bundles of energy" are called photons. So suppose you have light
with a wavelength of 6000 Å (~orange light); its frequency is
about 0.5x10^{15} s^{-1} , so the energy of a photon
would be about 2 eV. Now, what do we need to knock out an
electron? In metals the electrons which are responsible for
electrical conduction are, for all intents and purposes, just freely
moving around inside the metal like a gas. But, when they come to
the surface, there is some barrier keeping them in like gas in a box;
if there weren't, then all the electrons would eventually just leave
the metal which we know does not happen. The amount of energy
necessary to remove one electron from the surface is called the work
function (a very poor name, by the way!) and the work functions for
most metals is around 2-5 eV. So, unless the photon has energy
greater than the work function the electron will not get removed from
the metal. You might wonder why two or more photons couldn't work
together to remove an electron. That is just a matter of
probability and, using even the most intense sources of radiation
available around 1900, there is a really small probability of two
photons interacting with the same electron at the "same time".

QUESTION:
My high school physics teacher told our class one day that
when we bounced a ball off of the floor, that eventually at some
instant the atoms of the ball could arrange themselves perfectly so
that they all miss the atoms of the floor and therefore fall straight
through. I wanted so badly to argue, but seeing as how I was puny 12th
grader and my grade depended on my mouth I decided otherwise. So my
question would be, is there any logic what-so-ever in that statement???

ANSWER:
I suspect
that your teacher was trying to provide a dramatic example of the
notion that, as you know, most of the volume of an atom is "empty
space". The positively charged nucleus of an atom has a radius on
the order of 10^{-15} meters while the atom itself is of the
order of 10^{-10} meters but almost all of the mass of the atom
is in the nucleus; the electrons, which occupy the "empty space" have a
mass which is about 4000 times smaller than the mass of the
nucleus. However, his statement that the ball could pass through
the floor because of an accidental alignment of the atoms is, in fact,
incorrect. The reason for this is that the "empty space" in
atoms, while containing almost no mass, carries half the electrical
charge of the atoms and the negative electric charge on the outside of
the atoms in the floor exerts a repulsive force on the negative charge
of the atoms in the ball.

To
understand what is going on, imagine one atom slowly approaching
another single atom. Although introductory courses in chemistry
or physics usually have you visualize the electrons in little
"planetary" orbits around the nucleus, a much more accurate picture is
that the electrons form a "cloud" of negative charge around the
nucleus. Thus the two clouds repel each other and the incident
atom "bounces back". If the incident atom has a high enough
velocity the clouds may penetrate each other and then the attractive
force of the nucleus may be seen and a molecule might form. If
the incident energy is higher still, the two may simply pass through
each other. The velocities where the two might pass through each
other (maybe something like a tenth the speed of light, 30,000,000
meters per second!) is far higher than any ball is likely to have and
at such energies neither the ball nor the floor would remain in
tact. So, you see, the force which makes the ball bounce back (or
simply holds up the ball's weight if the ball is just sitting on the
floor) is electrical in origin and that force would not go away if the
atoms happened to "line up" as your teacher suggests.

There is
one more subtle possibility which your teacher may have been trying to
illustrate; it is called "tunneling" and is an effect of quantum
physics. In classical physics if an object is confined by forces
to move in a region of space it is called a "bound" particle and cannot
escape. Even if there is some nearby region of space where it
would not be bound it still can't get there. In quantum
mechanics, however, there is a probability that it could be found
outside the bound region and so if you watch long enough it might just
magically pop out there! In fact George Gamow showed back in the
30's that tunneling is the proper explanation for radioactive
alpha-particle decay in heavy nuclei. I suspect that the
probability of the ball tunneling through the floor would be so small
that you could wait for several times the age of the universe to expect
it to happen!

QUESTION:
Is there a color filter that can absorb color but transmit
shades of gray?

ANSWER:
If you mean
is there a simple piece-of-glass kind-of-thing which turns what would
have been a color image into a black and white image, I believe that
the answer is no. All light is characterized by its wavelength
and lights of different wavelengths correspond to different
colors. (You might want to read the answer to the question below
for more detail.) "Shades of gray" are not colors and therefore
do not actually exist in the light which forms the image.

On the
other hand, there are certainly ways to create a black and white
(B&W) image from what would have been a color image if viewed by
your eye. The simplest to understand, of course, is B&W film
in a camera. Here we have a chemical which undergoes some
chemical reaction (turning to a different chemical) if light strikes
it. The brighter the light is, regardless of its color, the more
of the chemical reaction takes place so that you are actually forming
an image based on brightness of the light rather than color of the
light. This is an oversimplification because photographic film is
not "linear", that is the chemistry is not equally sensitive to all
colors; for example, most B&W films are relatively insensitive to
red light which is why someone's (bright) red shirt often looks black
on a B&W picture even though you might have thought it should look
lighter. This is also the reason dark room lights are red.
You can also make B&W images electronically; digital cameras often
have an option of taking B&W pictures. Scanners can also
create a B&W image from a colored object.

So, the
bottom line is that a color image is one where the image is formed by
using wavelength information whereas a B&W image uses brightness
information, and no "passive" device can create the brightness
information from the color information. Filters are passive
devices which block out certain colors and transmit others.

QUESTION:
I am a student at the University and am writing an article on
alternative therapies. This topic, while interesting, is very difficult
to scientifically ground. I have some questions that I think you may be
able to answer. It has been so long since I've taken a physics class,
that I just want to be absolutely sure that what I am writing is
factual. I would appreciate your time to look over and reply to these
questions:

Does
color literally vibrate? Is it just our eyes that vibrate when we look
at a color, or does that color actually vibrate? If so, do different
colors vibrate at different intensities?
Does
light vibrate?
Are
light and color both considered energy, or are they thought of as the
same thing? Are they physically interchangable?
Are
our bodies comprised of energy, including light and color?
ANSWER:
First, what
is light? Light is a very narrow region of what is called the
electromagnetic (EM) spectrum. It is that type of EM radiation to which
our eyes are sensitive. Other examples of EM radiation are radio waves
(far "above" what we can see), radiant heat, also called infrared (just
"above" what we can see), x-rays (somewhat "below" what we can see),
ultraviolet radiation (just "below" what we can see), and nuclear gamma
radiation (far "below" what we can see).
Now what do I mean by "above" or "below"? EM radiation is a type of
wave and all EM waves move through space with the same speed (186,000
miles per second). Being waves, they pass us or hit us with a
characteristic frequency and have a characteristic wavelength; think of
a water wave: if you are standing still on the shore looking at waves
coming in, frequency is, for example, the number of waves per minute
which hit the beach and wavelength is the distance between two adjacent
waves. If you think about it (this is subtle!), if all waves travel
with the same speed which you know, like light waves do, then if you
know the wavelength, you also know the frequency (or vice versa). Thus
waves can be characterized by either wavelength or frequency; I think
wavelength is a little easier to visualize, so in the examples above,
"above" means a longer wavelength than our eyes are sensitive to and
"below" means a shorter wavelength than our eyes are sensitive to. The
shortest wavelength which we can see is about 4000 angstroms (an
angstrom is 1/10,000,000,000 meters, so you see that an angstrom is a
very short distance) and the longest is about 7000 angstroms.
So what is different between 4000 and 7000 angstrom light? If 4000
strikes your eye you perceive violet; if 7000 strikes your eye, you
perceive red; If 10000 or 1000 strikes your eye you perceive nothing
(black or darkness). Thus, you see that what color is is simply a
qualitative description of the wavelength of the light. Incidentally,
another wave with which you are familiar is sound; the pitch of sound
is determined by the wavelength of the sound waves. So pitch is to
sound as color is to light.
Now, I haven't told you what is doing the "waving" in EM radiation. For
water waves, the water "wiggles"; for sound waves, the air "wiggles",
both vibrating. For EM radiation, it is more complicated and subtle
(light can propagate through totally empty space where nothing is
available to "wiggle") and I think you get the gist of the general idea
without going into details. If you want more details, let me know.
With that background, let's answer your questions (not necessarily in
the order you asked them):

"Does
light vibrate?" Sure, just like the waves hitting the beach with some
frequency, the light hits your eye (or anything else it hits) with some
frequency. But you never perceive this because the frequency is
amazingly large (for example 400,000,000,000,000 times per second for
red light). Nevertheless, your eye/brain evidently have a way of
measuring the color because you perceive it.
"Does
color literally vibrate? Is it just our eyes that vibrate when we look
at a color, or does that color actually vibrate?" Color is just a
qualitative measure of a length of the wave, as stated above, so it
does not vibrate. I am not an expert in vision, but I know that the eye
does not "vibrate" in response to the incident light but absorbs the
energy of the light in a time averaged way.
"If
so, do different colors vibrate at different intensities?" Well, 'not
so ', but you have raised another issue, intensity. Intensity has
nothing to do with color; rather it is a measure of another property of
waves which I did not discuss above —how big are they? E.g., water
waves an inch high have a different "intensity" than water waves a foot
high. (Actually, intensity is defined to depend on the square of the
height, so in the preceding example the intensities would differ by a
factor or 12x12=144.) Intensity is a measure of how bright light is, so
bright and dim green lights would have the same color but different
intensities.
"Are
light and color both considered energy, or are they thought of as the
same thing? Are they physically interchangeable?" Light carries energy
with it as it moves along. This is easily seen by feeling the warming
effect of light that strikes you; you probably, when you were a kid,
used a magnifying glass to focus sunlight on something and burn
it —energy. Again, be careful to realize that color is only a way of
characterizing the light, so of course light and color are not
interchangeable. What color does is determine how much energy the light
can carry. Purple light carries more energy than a comparable amount of
Red light because the shorter the wavelength, the more energy. That is
why x-rays can penetrate into your body but visible light does not.
"Are
our bodies comprised of energy, including light and color?" Einstein
showed that all matter is a form of energy. In that sense, our bodies
are comprised of energy. Just because light carries energy does not
mean that you can conclude that our bodies are comprised also of light.