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QUESTION:
Fermilab's YouTube channel has an
enlightening
clarification of the phenomenon of
refraction: In it Dr. Lincoln explains that
charges in a material contribute their own
electric waves, which combine with the incident
light to form a new wave, which travels through
the material at a lower speed. However, I am
left with more questions. I believe that for
electromagnetic radiation to interact with
something, a photon must be absorbed, but his
explanation expressly mentions that this does
not need to happen for refraction to take place.
What am I misunderstanding?
ANSWER:
This is a case where he is doing
purely classical electromagnetism (EM). No
photons are needed. He is giving a (qualitative)
microscopic explanation about how v can
be less than c in materials. It was
well known in the 19th century, before quantum
electrodynamics was known, the nature of
electromagnetic waves. Everything you can know
about EM is contained in four equations,
Maxwell's equations. From these equations in a
vacuum you can find a wave equation describing
oscillating electric and magnetic fields and
traveling with a speed c =1/√(ε 0 μ 0 )
where ε 0 is the permitivity
of free space and μ 0 is the
permeability of free space. Anything not a
vacuum is not described by these two constants
but by corresponding constants ε and
μ which are larger than the vacuum
counterparts. Therefore the wave equation
describing a EM wave has a speed v =1/√(εμ )
which is always smaller than c .
Incidentally, the constants are simply constants
of proportionality dependent on how you define
magnitude of electric charge (usually Coulomb)
or electric current (usually Ampere). For
example, the force between two equal charges
Q separated by a distance r can be
determined experimentally to be proportional to
the square of Q /r , F∝Q 2 /r 2 ,
so F=CQ 2 /r 2 .
where C is a proportionality constant
which contains ε 0 , C =1/(4πε 0 )
if charge is in Coulombs.
QUESTION:
How massive does an object have to be in
order for its mass to change the rate of
acceleration due to gravity? And how does this
fit in with F=ma and a=F/m? I understand that
for small objects, acceleration doesn't change
due to mass, a 1kg object and a 2kg object will
fall at the same rate due to the Earth's gravity
in the absence of friction. But surely there
must be some change in acceleration due to mass
albeit miniscule for smaller masses? If we
consider 2 objects in a vacuum, in the absence
of any other objects they will exert a
gravitational pull on each other and fall
towards each other. If one is the size of the
Earth, and the other object is 1kg, that
acceleration is 9.8ms-2 , but if both
objects are as massive as the Earth do they
accelerate towards each other at 19.6ms-2 ?
What if one object is the size of the Earth and
the other the mass of the moon, or the Earth and
the Sun?
ANSWER:
This "rule" that you are trying to apply, that
two masses with different mass fall with the
same acceleration, technically applies only to
point masses in a uniform gravitational field.
Since the gravitational field of a large sphere
of radius R is approximately uniform at
altitudes h«R and most objects are much
smaller than the size of the earth, this rule
works approximately. But if the field is not
uniform (it actually falls of like 1/r 2
where r is the distance from the center of the
earth) and/or if the objects are not small
compared to the size of the earth, all bets are
off.
I
want to keep the algebra and integrations
simple, so as an example I will look at two thin
sticks whose masses are uniformly distributed;
one has mass M and length L ,
the other has mass M /2 and length L /2.
The gravitational field points in the downward
(negative y ) direction everywhere, but
decreases linearly as you move upward; so the
acceleration due to this weird gravity is
g=g 0 (1-δ(y /L ))
with g=g 0 when y =0
and g=g 0 (1-δ) at y=L
where δ is just a dimensionless constant which
specifies how rapidly g decreases as
you go upward. Note also that I have aligned the
centers of mass of the two sticks so that it is
a "fair race". The math is not hard but is a bit
tedious, so I will just give you the final
results. I find that the acceleration right now
is A=-g 0 (1-δ) for M
and a=-g 0 (1-(15/16)δ) for
M /2. If δ is very small compared to 1,
A≈a and if δ=0, A=a .
Your last
couple of questions are meaningless because you
have assumed that the acceleration due to
gravity is g everwhere, but it isn't as
I noted at the beginning of my answer—only very
close to the surface of the earth.
QUESTION:
If I set a timer to shine a light at my
space ship in 5 minutes. Then I get in my space
ship and fly away at 99% the speed of light, how
much time will I experience before the light
reaches me? And how much time would it take for
the light to reach me from the perspective of an
individual location at the light source?
ANSWER:
It is more convenient to answer your second
question first. You move with speed v =0.99
light minutes/minute (lm/m)* and the light moves
with speed 1 lm/m, so the ship is at a position
xs =0.99x5=4.95 lm when the
light turns on. Calling the time when the light
turns on t =0, the equations for the
ship and light front are, respectively, xs =4.95+0.99t
and xl =t ; if we set
xl =xs ,
it is easy to show that t =495 m and
then x =495 lm. Now we look from the
your perspective (your first question). You see
the 495 lm distance moving backward at a speed
0.99 lm/m so, because of length contraction, its
length is x' =495√(1-.992 )=69.8
lm and the corresponding time is t'=x'/v =69.8/0.99=70.5
m.
*Sorry for
the lousy notation here. The unit m here denotes
minute, not meter!
QUESTION:
In a Carl Sagan
video , "boy on bike with waiting brother"
provides a simplistic layman explanation of the
effects of light speed travel. IS IT POSSIBLE
for time dilation to occur when all "subjects"
(the boy, his high-speed vehicle and his writing
brother) are all earthbound?
ANSWER:
First of all, it is not "light speed travel"
which is impossible, speeds are close to the
speed of light. As to your question, being
earth-bound has nothing to do with time dilation
or length contraction; only speed matters. True,
the classic "twin
paradox " problem has one brother flying
straight out into space and straight back
because it is easy to calculate analytically,
but there is no reason the traveling twin
couldn't just go round and round in circles
before stopping.
QUESTION:
I have what I hope is a quick question that
you could help me with. I'm trying to find
equation(s) to chart communication across space
to a spacecraft. My hypothetical spacecraft is
traveling at the velocity of .1c towards a ship
that is relatively static and how long
communications would take as the distance
between them decrease starting at .25ly.
ANSWER:
Since the gamma factor at 0.1c is 1.005, I will
ignore relativistic effects. I am not sure what
you mean by communications, so I assume that you
mean the length of time for a message sent from
the moving spacecraft to (M) to arrive at the
stationary spacecraft (S) as a function of where
M is. The position d of M relative to S
is d =0.25-0.1t ly; the time
for light to travel that distance is T=d /c
where c =1 (ly/y). Therefore T =0.25-0.1t
y. The graph shows this.
FOLLOWUP
QUESTION:
What
I meant is to find out the length between each
communication between the the traveling ship and
the point in which it's traveling to. For
example communication with earth to ship to Mars
communication basically starts at no delay and
eventually time between the two is delayed as it
goes to mars and decreases as it returns. How
would I graph that and find out how long the
delay is as the ship is moving towards i.e.
Earth? [I was a little confused still about what
he meant. He sent the picture which cleared it
up.]
ANSWER:
What is wanted is the time, on earth (E),
between when a signal is sent to and a reply
received from the ship (S) as a function of the
ship's distance from the earth (ESE); his sketch
implies that he is also interested in the
reverse, the time (SES), so I will do both. The
zig-zag paths of the signals shown in the
questioner's sketch form many triangles which
are similar triangles; hence, if we do the
geometry for one triangle it will be correct for
all. The two graphs show one of these triangles
for each. Three important lines are labeled by
their linear equations. The quantity d
is simply any arbitrary distance between S and E
at the time when the communication is initiated.
It now
becomes a problem of analytic geometry, in
particular where linear functions intersect. In
the case of ESE, d -0.1t=t so
t=d /1.1; since this is an isosoles
triangle, the elapsed time T for the
communication is twice this, Δt =2d /1.1.
It is a little trickier if communication is
initiated by the ship because it isn't an
isosoles triangle; still, it is easy to find the
intersection of y=t-d and y=d -0.1t
to be Δt =2d /1.1, exactly the
same as for ESE. Finally we find that, whichever
initiates the communication when S is at y=d,
the total time is Δt =2d /1.1.
In your case, the first communication would be
when d= 0.25 ly, so Δt =0.455
y=41.5 days. A communication initiated when the
ship is halfway home, d =0.125 ly, would
take Δt =0.227 y=83 days.
[My wife
said to ask you if you are writing a sci-fi
novel!]
QUESTION:
My Question is related to Fluids. When a
washing machine is started(which is filled with
water), water inside it, form a shape of like a
tornado(which revolves around a common centre)
which can be explained by effective gravity
which has components- artificial gravity and
normal gravity (taking that revolving water as
our non-inertial frame of reference). But how
can the same phenomena be explained by inertial
frame of reference?
ANSWER:
I have never seen the centrifugal force called
artificial gravity before, but I guess that is
what you are referring to. The explanation in
the inertial frame is illustrated in the
diagram. The shape of the surface is a
parabaloid of revolution. I will focus my
attention on a tiny piece of the water on the
surface. There are two forces on it, its own
weight W and the force
F which the
surrounding water exerts on it. The horizontal
and vertical components of F
are shown. The vertical component must equal
W and the horizontal component must be the
centripetal force mRω 2 where
ω is the angular velocity and R
is the distance to the axis of rotation.
QUESTION:
Struggling with length contraction. Is it
just a matter of perspective and not actually
contraction? If I am on a rocket at high speed,
standing and facing forward, I experience no
difference in the depth of my thorax. From the
perspective of a person at rest, relative to me,
they see my thorax depth to have contracted. How
is that physically possible? My heart and lungs
cannot be infinitely compressed. Yes, I know the
math can be worked out for both perspectives,
but is it actually happening? It is just how the
rest frame "sees" me?
ANSWER:
T
here are two important issues
here:
I have
discussed these at great length in an
earlier answer . When looking at a
three-dimentional moving object the
results can be surprising.
QUESTION:
I have a simple camera - a 'VP Twin' - fixed
focus positive meniscus lens, aperture, shutter
speed. basically 'point and click' in bright
sunlight. I have become curious how it can be in
focus from 8 feet to infinity. On close
examination it has a curious feature - the
distance from the lens to the film plane is 47
mm, but the distance from the lens to the focus
of infinity is 50 mm. This is apparently
impossible. So two questions, Does 1/F = 1/U +
1/V hold for positive meniscus lenses? and is
the focus of infinity distance for a positive
meniscus lens the focal length?
ANSWER:
The equation you have written is called the thin
lens equation. You should be aware that it is
only an approximation but a pretty good one. I
prefer the notation 1/f= 1/i +1/o
where f is the focal length, i
the image distance, and o the object
distance. If o =∞ then clearly i=f
so clearly f =50 mm. So the image for an
object infinitely far away is not on the film,
and that object is not perfectly in focus. So
what is perfectly in focus? 1/50=1/47+1/o
and solving, o =-783 mm. What does this
mean?
It means that there is a virtual object located
783 mm behind the lens! This is obviously not
the way any simple camera works because to have
a virtual object you need another lens to create
it. You can easily show that for a simple
converging thin lens that all images of real
objects are farther behind the lens than the
image of an object at infinity. So you have
something certainly wrong with your numbers. If
you had said that the image of the sun (focal
length) is 47 mm behind the lens and the film is
50 mm behind the lens, it would have been
possible: o =783 mm, a real object
7.83 cm in front of the lens. I am not going to
analyze in any more detail because when I
googled this camera I found that the focal
length of the lens is 35 mm. I have included a
plot of i as a function of o
out to o =100 m (100,000 mm) for a 35 mm
lens. As you can see, beyond 10 m the image is
only 0.1 mm or less off the focal plane; so if
you set the film a distance f from the
lens, most of what you want is quite well in
focus. I have read that this is what is normally
done for fixed-focus cameras.
QUESTION:
If I was in free fall from a skydiving alt.
And I was falling on a platform that let's say
for sake of argument didn't flip over and over
and instead stayed flat while falling to the
earth.(let's say it was designed to fall flat on
the top side) It would obviously absorb some
amount of wind resistance, therefore slowing it
down ever so slightly. Me as a skydiving human
being standing on top of the flat platform would
not have any wind resistance because it would be
broke by the platform. Therefore my falling
velocity would be fast than the platform.
Therefore making it possible that I could walk
around on said platform (again, assuming it
couldn't flip over). Is that correct? Could I
walk around on a falling platform because of the
differential in velocity due to the air
resistance the platform would undergo and not my
body?
ANSWER:
Well, aerodynamics can be very
complicated, so there is no guarantee that you
will feel on effect from the wind, but that is
not the real crux of your question so I will
assume that it is true and that the platform
(absent you) will maintain its horizontal
orientation as it falls. To make this easier to
generalize, I will have the platform be a
uniform solid cylinder.
What
are all the forces on the platform (without
you)? There is its own weight W
which may be thought to act all acting at its
center of mass, and there is the air drag
D which would likely
be distributed uniformly over its area and
therefore you could treat it as a single force
acting up also at the center of mass. If the
mass of the disc were not uniformly distributed,
the center of mass would not be at the
geometrical center; the disc would not be in
rotational equilibrium, there would be a net
torque about the center of mass and the platform
could not be as you wish it to be—remaining in a
horizontal plane. This is shown in my diagram.
Now, let's
add you to the mix. If you stand at the center
of the disc all is well. But, if you wander off
on the platform, the force you exert down will
exert a torque and the disc will commence
rotatating.
You refer
to the effect of the drag to be "ever so slight"
but the drag would be quite strong for a
platform large enough for you to imagine
wandering around on it. That is how a parachute
works.
QUESTION:
if two objects fall at the same time with
different masses, neglecting air resistance,
will they fall at the same time? keep in mind
that their masses are different for example one
has 100kg of mass and one has 1kg of mass
ANSWER:
This is the old
Galileo experiment , famous even though it
probably never happened! If there is no air
drag, any object will fall with the same
acceleration and therefore both will always be
falling side by side; see the
experiment performed with a feather and a
hammer by an astronaut on the moon where there
is negligible air. And, if you know Newton's
first law it is not hard to understand:
The only
force on a falling object with no air drag is
its own weight which is its mass times the
constant g which is a constant which
tells you how strong the gravity is; on the
surface of the earth, g =9.8 m/s2 .
Newton's second law says that the force is equal
to the mass times the acceleration. So,
F=ma=mg and a=g ; no matter what
the mass is, it acceleration is always equal to
g.
You might
be interested the case when there is drag. The
drag force depends roughly on the size and shape
of the object and the square of its speed. So,
for example, if you drop two spheres of
the same radius but one has a mass m
and the other has a larger mass M , at
any given speed v each will experience
the same force because the force does not depend
on the mass. Newton's second law will be
F=ma=mg-cv 2 where c is
just some constant which carries all the
information about size, shape, gravity strength,
air density, etc . (The minus sign is
because the weight is a force down but the drag
is a force up. So a=g- [cv 2 /m ]
and similarly A=g- [cv 2 /M ]
for the more massive ball. In your example,
M =100 kg and m =1 kg, a=g-cv 2
whereas A=g- 0.01cv 2
for the 100 kg mass, much bigger. The heavier
wins.
QUESTION:
What grants authority to biological
materials to alter matter through DNA? If all
matter was created at the emergence of the
universe, how does DNA, plant and animal life,
allow for growth and create structures that did
not previously exist?
ANSWER:
YThe first thing you should
understand is that all energy was
created when the universe began, not all matter;
matter is just a form of energy. In elementary
chemistry we are taught a principle called
conservation of mass . In fact, this "law"
is only approximately true; what is true is
conservation of energy, that mass can be created
or destroyed as long as the total energy is
conserved. For example, when methane is burned
by combining it with oxygen, we get carbon
dioxide and water; energy is released in the
form of heat so the mass after the burning is
actually less than before. Chenmistry, however,
is a very inefficient way to get energy because
the change in mass is extremely tiny which is
why the law of mass conservation was, for
centuries, believed to be true because the
instruments were not accurate enough to detect
such small changes. Biology is just a
collections of natural machines which do
chemistry. For example, a plant takes energy
from the sun to to create sugar and growth by
chemically manipulating water, air, and
minerals.
QUESTION:
My twin brother and I are here on earth in
the same reference frame. We are both 20 years
old. He then travels in a rocket at 0.87c for
the next ten years. He then returns to earth. I
am now 30 years old. IS HE ONLY 25 TEARS OLD?
ANSWER:
You should read my earlier answer
about the
twin paradox ; if you looked on the faq page
you could have linked to it. But I will work it
out for you. If I understand your question, your
brother travels to some distance d for
five years and then back in five years. Since
his speed is 0.87c, the distance to his
destination is d=vt =0.87*5=4.35 light
years (ly). Now, because of length contraction,
he sees the distance he travels out to be d' =4.35√(1-0.872 )=
2.14 ly. Therefore, according to his clocks, he
has traveled not 10 years but 4.28 years and is
thus 24.28 years old.
QUESTION:
Why are particles waves and not spirals? I
feel like I can't imagine a particle going in a
zigzag motion as opposed to it rotating around.
ANSWER:
You never try to talk about what
the "particle part" is doing when you are
talking about the "wave part". Take light, for
example. We have know since the 18th cetury that
light is a wave. In the 20th century we found
that it is also a particle which we call a
photon. You, as you seem to feel, would not
imagine photons zig-zagging over the crests of
the wave. For the electron, for example, the
wave we envision is a wave which specifies the
probability of finding the electron particle at
some spot in space, not a physical wave in
space.
QUESTION:
When/where can matter and anti-matter pairs
form? Can this only happen in a vacuum? Also,
since matter and anti-matter both have mass, how
can the Law of Conservation of Mass not be
violated when these pairs are created?
ANSWER:
I
will talk only about photon
creation of electron-positron pairs. It can only
happen if the photon energy is equal to or
greater than twice the rest mass energy of an
electron; otherwise energy could not be
conserved. It cannot happen in a perfect vacuum
because the participation of another particle,
usually a nucleus, is required for the reaction
to be able to conserve linear momentum. Other
particle-antiparticle pairs would have similar
restrictions. There is no such thing as
conservation of mass, just conservation of
energy; mass is, afterall, just a form of
energy. Conservation of mass is a useful
approximation for elementary chemistry, but it
is false in detail; e.g ., the mass of a
water molecule is not equal to the mass of an
oxygen atom plus the mass of a hydrogen atom
plus the mass of a hydrogen atom. The difference
is small but not zero.
QUESTION:
A person is standing on a tower say three
meters above a swimming pool. The pool is on the
deck of a moving boat that is moving in the same
direction as I am facing. The speed although not
really important is say five knots. Now, let's
say I jump off of the tower and successfully
enter the water. Is there a point or height of
the tower where the length of travel or time of
fall that would be such that the pool would
eventually move out from underneath me and I
would hit the tower? I would be in straight free
fall and the platform is moving. I assume that
gravity would eventually stop all forward
momentum causing the fall to plumb to the earth.
If the target is moving to an unattached
platform underneath me, it would eventually have
to move out from underneath me, correct? Given a
long enough fall/time?
ANSWER:
For starters, the speed actually may be
important, but let me come back to that. First,
let me neglect air drag. When you step off the
tower you, like the boat, are moving forward at
5 knots. Now, the only force on you when you are
falling is your own weight which accelerates you
vertically; there is no force horizontally so
you continue to move forward, right along with
the boat so, no matter how high the tower is,
you will land directly on the spot in the pool
you were directly over when you jumped.
Now, if
air drag is considered, it is as if you were
falling but a v w =5 knot=2.57
wind were blowing you horizontally so you would
be accelerated backwards from the direction of
the boat. So, I have worked out the answer to
how far, given the numbers you have specified:
you will fall* from 3 m in 0.78 s and you will be
pushed backward about 7 mm and have acquired a
speed of about 2 cm/s. Not much effect.
Finally,
for anyone interested I will give an outline of
the calculation I did but not all the detailed
steps. To estimate the air drag at sea level I
like to use the approximation F =¼Av 2
where A is the area pushed on by the
moving air and v is the speed of the
air when it hits the object; in this case, v
should be replaced by (v w -v )
where v w is the speed of the
wind and v is the horizontal speed of the falling
person, so
F =¼A (vw -v )2 =ma=m (dv /dt ).
(It is
important to note that this approximation only
works if you use SI units.) This can now be
integrated to find the horizontal velocity of
the jumper as a function of time,
v=tv w /(k+t )= dx /dt
where k =4m /v w A .
This is
plotted for v w =2.57 m/s (5
knots), m =68 kg (150 lb), A =1
m2 in the first two figures above.
Note that for very large times (first figure)
the velocity approaches the wind speed as you
would expect. The second figure shows very small
times; the time to fall is about 0.78 s (see
footnote*). Finally, integrate the v
equation to find the distance traveled
horizontally as a function of time:
x =vw (t+k log(k /(k+t )))
where log is the natural logarithm.
The
position x is shown in the third
figure. For very large t this curve will
increase linearly with a slope of 2.57 m/s. I
haven't worked it out for other conditions, but
the distance traveled horizontally could be made
larger for larger v w , larger
h or smaller m .
*Assuming
that the person is jumping or diving straight
down, there will be little air drag going down.
The required time to drop to the ground from a
height h can be shown to be v =√(2g /h )
which is 0.78 s for h =3 m.
QUESTION:
I am curious to why mechanical advantage
works. I understand how pulleys offer mechanical
advantage. I understand the math. I do not
understand why there appears to be a mechanism
in physics that allows the translation of weight
into length of rope.
ANSWER:
I
am not sure what you mean by
"translation of weight into length of rope". To
understand what is happening you need to know
Newton's three laws. Let's consider the simplest
example I can think of with a pulley, weight,
and rope. The first of the three figures shows
the situation: a rope is attached to the
ceiling, passes over a pulley, the other end
pulled up by you, and a weight W is
supported by a rope attached to the axle of the
pulley; everything is at rest. If the force
which you pull up with is less than the weight,
then there is mechanical advantage. The physics
concepts you need are Newton's first and third
laws. The first law says that if an object is at
rest, all the forces on it must add up to zero;
the third law says that if one object exerts a
force on another, then the other exerts an equal
and opposite on the the first.
To solve
this problem it is best to focus on one object
at a time; first, focus on the weight as shown
in the second figure. There are two forces on
the weight, its own weight W which
points straight and the rope which exerts an
upward force whose magnitude I will call T 1 .
(I think we can agree that a rope is capable of
exerting a force and this force is usually
called the tension. There is no weight being
"translated" to the rope, rather the rope is
just exerting a force on the weight.) So now the
first law tells us that T 1 -W =0
or T 1 =W . You are
not holding up this weight and the
pulley is not holding up this weight,
this piece of rope is. Next, focus on the
pulley. There are three forces on the pulley the
tensions on the left, right, and certer
as shown in the third figure. You are not
pulling the pulley, the rope is; the ceiling is
not pulling the pulley, the rope is;
the weight is not pulling down on the
pulley, the rope is. Now we need to use the
third law: T 1 pulls up on
the weight so the weight pulls down on the rope
with a force of magnitude T 1 ;
the rope is at rest so the pulley must be
pulling up on it with a force of magnitude T 1 ;
therefore the rope is pulling down on the pulley
with a force of magnitude T 1
which just happens to be equal to the weight;
the weight is not pulling down on the
pulley. The other two forces on the pulley,
labelled T 2 and T 3 ,
must be equal in magnitude or the torque they
exerted would make the pulley rotate. So, now,
we can apply the first law to the pulley and
find 2T 2 -T 1 =0
or T 2 =W /2. Finally
we can use third laws arguments like we did for
the hanging rope to deduce that you must use a
force W /2 to hold up a weight W.
I know this must seem pedantic to you, but when
you are first learning this kind of problem it
is important to understand which forces are
exerted on something to avoid fuzzy thinking
like the weight being translated to the rope.
QUESTION:
I know the equation of intensity follows as
I = S / 4πr2 How is this equation
derived and what are the base formulas in terms
of the inverse square law.
ANSWER:
Technically, this is the intensity for a point
source or sources which are spheres. So let's
assume that a point radiator is emitting a total
power S which is energy per unit time.
A point source will radiate energy
isotropically, i.e. , uniformly in all
directions. Intensity is the power per unit area
passing throught an area A . The area
through which the energy passes is A =4πr 2
where r is the radius of any sphere
with the source at its center, so I=S /(4πr 2 ).
There is your derivation.
As I
said, this is only exactly true outside of
spherically symmetric sources, but most sources
look like a point source if you are far enough
away. The units of intensity are
Joules/second/meter squared (J/s/m2 )
and J/s is a Watt (W).
QUESTION:
In regards to the effect of dark matter on
galaxy rotation, would get the same results by
assuming a larger gravity effect exerted by the
galaxy's central black hole?
ANSWER:
I
normally do not answer questions
which are astronomy/astrophysics/cosmology. In
this case though, I think I can answer. There is
no way that the folks who do this kind of
calculation would not have included the effects
of supermassive black holes.
QUESTION:
I have been thinking about Einstein's
formula E equals MC squared. I can understand
where Einstein got his notion of mass and energy
what escapes me is the speed of light squared.
What clue did Einstein have That indicated the
speed of light should be squared why not just
the speed of light or why not the speed of light
to the power of three or 10. What indicated that
the speed of light squared should be used and
what pointed Einstein in this direction.
ANSWER:
I have gotten this question
before and it shows that you do not understand
the way we specify units in science. To take an
example you can understand, suppose that you are
traveling in a car with a speed of 57 miles per
hour; if I asked you how fast you were going and
you said "57 pounds" we would both know it was
nonsense. In the SI system of units scientists
usually use, energy has the units of kilograms
per meter squared per second squared (kg/m2 /s2 )
and this is called a Joule (J). As you see,
mc 2 has those units and mc 3
for example, does not.
(A more
familiar unit to you may be the unit for power,
the rate at which energy is created or consumed,
is the Watt (W) which is one Joule per second
(J/s)). This is not the only misconception in
your question, but also understanding how
theoretical physics is done. A new theory
generally starts with a very few suppositions
and then the theorist works out what the
implications of those suppositions are on
physics. In the case of special relativity the
suppositions are that the laws of physics are
the same in any frame of reference and that the
speed of light is the same for all observers.
What these two postulates lead to are many new
understandings of physics and one of them is
that E=mc 2 . You may be sure
that Einstein didn't sit around thinking and
suddenly a lightbulb flashed over his head and
he went "Eureka, E=mc 2 !"
QUESTION:
My dad was driving in a highway, when we
leave the highway there is a banked road. When
car turning right, my body is leaning to the
left. I thought my body will lean to the center
due to centripetal force. Why??
ANSWER:
Newton's laws do not work in
accelerating systems. When you are turning the
direction of your velocity is changing and so
you have an acceleration even if the speed is
constant; this acceleration, as you probably
know, points toward the center of the circle you
are turning and has a magnitude v 2 /R
where R is the radius of the circle and
v is your speed. But in your frame you
are at rest, not accelerating at all. If you are
leaning on a door on the outside of the turn,
you feel the door pushing you. That is a force
but, since your frame is at rest, you have an
unbalanced force but your acceleration is zero.
Your brain, however, feels that push and
interprets as you are being pushed by some
mysterious force against the door making you
feel like you are being pushed to the left. That
force does not really exist even though you feel
like it does. But this is all understood if
viewed by a pedestrian observing you: she sees
the door pushing you toward the center of the
circle providing the observed acceleration. The
real force is the force (centripetal) of the
door on you, the force (centrifugal) you think
you feel does not exist and is usually referred
to as a fictitious force.
QUESTION:
I am writing a fiction piece that includes a
carrousel space station (a la 2001) and have two
questions about the positioning and orientation
of said station. This would be a 200 meter
diameter I think. First - I intend that the
space station will not be geostationary (35000km
out) but want it to be semi geosynchronous. That
is, I would have it circle the earth once every
180 min as opposed to the ISS 90 min). I presume
it would higher than the 400km to slow the
revolution. At what height would that be?
Second, I was wondering what would the optimum
axis or rotation for a 100 to 200 meter diameter
station? Axis parallel to the line of gravity?
(I.E. the if it were on the surface it would
spin like a top) Axis in line with the orbit?
(the station disc is aligned more or less N-S )
Axis perpendicular to the orbit and the line of
gravity (so if it were on the surface it would
roll like a wheel)
ANSWER:
Kepler's third law states that the square of the
period T of a circular orbit is proportional to
the cube of the radius r of that orbit,
T 2 ∝ r 3 .
Keep in mind that you have specified the
altitudes h of orbits, not their
radii where r=RE +h
and RE =6378 km is the radius
of the earth. So now you can write (T 90 /T 180 )2 =(r 1 /r 2 )3 =¼.
You can now solve for r 2 ,
the radius of the orbit for the 180 minute
period, r 2 =1.59r 1 =1.59(6378+400)=10,778
km. You can express this as an altitude, h 2 =4399
km. For your second question you need to realize
that the spinning station has an angular
momentum (which points along its axis of
rotation) which remains the same if there is no
net torque on the station which there should not
be, at least to an excellent approximation.
Therefore if you try to orient the axis either
radially or tangentially it will soon not be
pointing in that direction since the angular
momentum vector would have to change but it
won't; e.g ., if you set it radially or
tangentially it will be pointing tangentially or
radially after having gone a quarter of the way
around its orbit. If normal to the plane of the
orbit, it would always be that. On the other
hand, I can think of no good reason why the
orientation would make any difference. You might
be interested in an
earlier answer and links in it which discuss
these rotating space stations in detail.
FOLLOWUP
QUESTION: I get your visualization
of the station starting off radially aligned
becoming tangential after a quarter revolution,
but am thinking that would only happen if the
station itself was spinning like a flipped coin
counter to the revolution. If the station were
aligned axis tangential and canted slightly
might it possibly give itself angular momentum
and propel itself around the orbit - like the
spinning bicycle wheel in the hands of a person
standing on a lazy susan.?
ANSWER:
Your space station is rotating as shown in the
picture. Its angular momentum points along the
axis of rotation. Now, there is an external
force on this station, the gravitational field,
but this force exerts no net torque; gravity
exerts a force which causes a torque due to each
piece of the station, but for each piece there
is a piece on the other side which exerts an
equal but opposite torque. Newton's first law
says that if there is no external torque, the
angular momentum will be unchanged no matter how
the station moves around. There is no angular
momentum corresponding to a "flipped coin"
rotation unless you exert a torque to get it
rotating like that, and you would never do such
a thing. The second figure shows what I was
talking about in the first answer: with no
external torque angular momentum direction does
not change as you go around the orbit.
Your
speculation about the satellite "propelling
itself" is totally wrong. No propulsion is ever
required for a satellite because it is in free
fall and just continues falling. Also the bike
wheel-stool-person example is irrevelent because
it is not analagous. It is also an angular
momentum conservation example but because an
internal torque (by the person) changes the
angular momentum of the wheel, the person
acquires an angular momentum of her own to keep
the total angular momentum the same.
QUESTION:
Why doesn't an irregular or polygon
shaped hole work like a pinhole camera when
held at or close to the screen? It requires
some distance between the hole and the
screen to make an image. When a card with a
hole of any shape is held close to the
screen, the lighted area on screen is of the
same shape as the hole. To make an image of
light source, say Sun, the card has to be
moved away from the screen to get an image.
Why does it not show the image in the first
case? I want to understand the path of light
rays for the two situations and the
transition. The explanations for the simple
pinhole with rays going through the hole to
make an image are shown for small holes; but
any size hole works as camera lens at
sufficient distance. I'm unable to use that
explanation and ray diagram for the above
question of large hole near and far away
from screen. If I think of a larger hole as
made from several small pinholes, even then
how do i explain that larger hole close to
screen gives a sharp image/shadow the same
shape as hole and makes the image of light
source as distance to screen is increased?
ANSWER:
I
haved shown above the photographs taken by
the questioner showing the results for the
pinholes both near to and far from the
screen; The object supposedly creating the
image is the sun. A pinhole is imagined as a
hole with infintesmally small size; of
course, such a hole would let in no light at
all, so we have to have a hole which is very
small compared to the distance to the
screen. No pinhole camera is perfect, but we
can fabricate one which is very small
indeed. So now I show in the next diagram
ray diagrams for two pinholes and the
resulting images they form of the arrow.
The
one on the left is so small that only one
ray from each point on the object can get
through the hole thereby forming a nice
sharp image. On the right the hole is much
bigger so that now the rays which can get
through are many, those between the two rays
shown coming from each end. I have shown
images made by these extreme rays but there
will be a continuum of such images and the
net effect will be an image which is only
barely recognizable as the the image of the
object, a total blur. You cannot even call
my second example a pinhole camera because
its size is not small compared to the
distance to the screen. Your reason for
expecting a sharp image from a big hole,
because a lens is a big hole, is totally
incorrect. The reason is that the purpose of
the lens is to bend the light rays so that
all the rays from any point on the object
end up at the same point on the screen.
QUESTION:
My friend thinks the reason a spaceship
would reach infinite mass if it achieved the
speed of light is because an infinite amount
of fuel would be needed to supply the
energy. I disagree. Is he correct? I hope
not because it would mean I don't have the
grasp on the psychics of this situation that
I think I do, which is one have notch above
the pure layman.
ANSWER:
Your friend has the right idea. Any object
with mass which is traveling at the speed of
light has an infinite total energy.
Therefore, to get the mass to that speed you
need to supply an infinite amount of energy.
But our entire universe does not contain an
infinite amount of energy.
QUESTION:
This is is a question relating to music,
specifically piano music. There is a lot of
misunderstanding among pianists about the
physics of how sound is produced. Basically,
the volume of sound produced by striking a
piano key is a function of the speed of the
hammer as it strikes the string. But,
assuming you strike the key three times, and
in each case the hammer is traveling at the
same speed on impact, is the volume of sound
the same, whether the hammer is
accelerating, decelerating, or at a constant
speed on impact?
ANSWER:
I started off answering your question by
acknowledging that a piano is a very
complicated system, but still trying to do
some simple analysis of your question making
simplifying assumptions. As I was plowing
forward, I was simultaneously doing research
since I knew little about the mechanisms of
a piano. What I discovered is that your
question involves situations which cannot
occur in real piano. The hammer comes up to
the strings from below so that it will fall
back down when the strings are struck. But,
an essential component of the hammer
mechanism is that the hammer is released
from the mechanism connecting it with the
key just before it strikes the strings; the
mechanism which causes this to happen is
called the escapement. Therefore, your
question is moot because the hammer is
moving freely when it hits the strings and
because the only force on it as
it approaches the strings is its own weight,
it is slowing down. If you want an
excellent introduction to the physics of the
piano, a
talk given by Professor Nick Giordano of
Purdue University at Georgia Tech.
(Unfortunately, this video is slightly out
of focus but still worth watching.)
QUESTION:
Can a bolt/Arc of electricity be
affected by a wave of kinetic energy such as
an explosion’s blast wave. I’m not looking
for any real world application this is for a
fiction book I am writing. The only reason I
refer to an explosion is because I can’t
think of another example close enough to
what I mean.
ANSWER:
A spark is just the break down of a gas
which occurs if there is a strong enough
electric field. Free electrons in the gas
are accelerated by the field and they
collide with molecules which results in more
electrons and the positive ions they leave
behind. So a current will be created through
the previously insulating gas. So, if you
disturb the air, it will certainly affect
the way this spark forms.
QUESTION:
I've been under the impression that if 2
objects of the same shape and size but
different weight are dropped they will fall
at the same speed. I've seen videos and
articles also proving this. However my
father and I conducted a test with a piece
of steel 3x1x1 inches and a piece of wood
3x1x1 inches. Neither weigh that much
probably less than a pound each. When
dropped the steel always hits the groumd
first. I know it must deal with air
resistance some how but they have the same
surface area so I can't figure out why the
steel falls first.
ANSWER:
You are certainly right that it is air drag
which causes this to happen. And you also
note that they have the same area but you
don't say what that has to do with anything.
Let's just write down a general expression
for the drag force, f =½ρCAv 2
where ρ is the air density, C
is a constant determined only by the shape
and orientation of the object, A is
the area presented to the onrushing air, and
v is the speed of the object. So,
you see, the drag force on the steel is
identical to the drag force on the wood. But
what matters is not so much the force on the
object but rather the effect this force has.
The net force on an object with mass m
is f-mg and, using Newton's second
law, ma=f-mg ; so a=f /m-g .
So, you see, the larger the mass, the
smaller the effect in slowing down the
falling object. There would have been a much
more dramatic difference if you had
fashioned a 1x1x3 box out of paper.
QUESTION:
Do rotating bodies have to be a finite
distance from the center of rotation to
move? Consider a circular pad of sand
grains. Any line drawn from the edge to the
center will have interlocking sand grains.
If there is one grain at the exact center it
must rotate to make the whole pad move. But
the grain has molecules and atoms and etc.
There has to be an exact center (R=0). The
rotation formulas fall apart with zero
radius. Rotation can occur if there is no
rigidity.
ANSWER:
To be clear here, we are talking about
classical Newtonian physics where such a
thing as a point or a point mask are assumed
to exist, even though they don't (except
maybe black holes which are not part of
classical physics). A point, having no size
at all, cannot rotate about an axis which
passes through it. So a point mass at the
axis of rotation of some rotating rigid body
has neither a velocity nor rotational
velocity. You mention a "grain"; thinking of
a spherical grain of radius r for
simplicity, if its center is located on the
axis of rotation neither it nor any other
point on the axis is moving, but all other
points are. If you think of the the central
point as an atom or radius r , the
same arguments apply, but classical physics
is not applicable to objects as small as an
atom. I do not know what you are talking
about that "The rotation formulas fall apart
with zero radius"— v=rω so v =0
if r =0.
QUESTION:
If you have a light source that is
emitting a constant number of visible red
photons per unit time as seen by an observer
that is stationary relative to the light
source and then the observer starts moving
towards the light source at, say, 99% the
speed of light, what exactly would the
observer experience? I presume one would see
the photons very strongly blue shifted,
maybe even to the point of becoming gamma
rays?? Maybe killing the observer because of
the ionizing radiation?? Also since the
observer is moving into a continuous stream
of photons traveling away from the source at
the speed of light, would brightness (the
number of photons per unit area per second)
of the source increase? To the point of
vaporizing the observer? Or what if the
observer was moving perpendicular to the
light source at a very high speed, would it
intercept a much larger quantity of photons
moving away from the light source and thus
make the source somehow appear much brighter
although not be blue shifted?
ANSWER:
The first part of your question is called
the relativistic longitudinal Doppler shift
and is easy to calculate: λ'=λ √[(1-β )/(1+β )]
where β =v /c . In
your case, β =0.99 so the blue shift
is λ' =0.071λ . So, suppose
that we take λ =700 nm; then λ' =49.7
nm, at the low-energy end of vacuum
ultraviolet, probably not energetic enough
to kill the observer and pretty easy to
shield against. For the case of light coming
from a source not directly ahead, you can
see the math in the
Wikepedia article on relativistic
Doppler shift.
More
interesting to me though is the effect on
all the stars all around me when I am going
at a very high speed. Their apparent
positions all tend to collapse into a very
small-angle cone directly in front of me.
Hence, I would find it almost impossible to
navagate and would have a very intense of
high-intensity, high-energy radiation coming
directly at me. See an earlier
answer .
QUESTION:
Scientists are saying a nova 1286 LY
away will nova, if so won't it be 1286 years
before we can see it on earth do telescopes
shorten the time. The reports I see act like
when the nova occurs we will see it at that
time
ANSWER:
If it is 1286 light years away, anything we
see of it will have happened 1286 years ago,
telescope or not. So if those scientists say
that it will happen soon, that is what our
clock will say, but the actual event
happened a little more than 1286 years ago.
QUESTION:
My mother says one radiator is better
than another at heating our house. But I
think it's only insulation that makes a
difference. Because hot air can't just
"disappear" because your radiator is made of
one thing rather than another or has a shelf
above it.
ANSWER:
If the two radiators are identical, their
effectiveness depends only on their
environments. One aspect would be the rate
at which heat escapes the room; it is not
only insulation, but also windows, doors,
etc. Also, the radiator in a small closet
would keep you toasty warm but not in the
astrodome, so volume of the room being
heated is also important.
QUESTION:
if you replace a 100 watt incandescent
light with a led light using much less
wattage that has the same amount of lumens
will the output on a solar cell (on your
calculator) be the same?
ANSWER:
I once
answered a question which delved into
the definition of a lumen and its relation
to other units of electromagnetic wave
energy, power, etc . which you might
be interested in.
To
answer your question, though, you have to
look into the spectrum of emitted light of
your two bulbs and the spectral response of
the solar cell. I have included two figures
above, one showing the the spectra of the
sun, a typical led, an incandescent bulb,
and a cfl lamp. The visible spectrum is
about 350-750 nm (0.35-0.75 μm). Note that
the led is taylored to having nearly all of
its radiation in the visible range, whereas
the incandescent bulb is most intense in the
red and infrared range (long wavelengths).
But, as the spectral response of the solar
cell figure shows, the cell is most
responsive in the long-wavelength regime.
Therefore, your calculator battery will
charge faster with an incandescent bulb than
an led bulb.
QUESTION:
I am 72 years old and not a crank, yet.
Cannot get answer to simple query: How much
energy is in ONE electromagnetic wave of any
particular wave length, say one radio wave?
I know the usual equations such as
energy-h*frequency. I cannot see how to make
them work with one wave alone. I am good at
algebra and OK at calculus. I know waves
rarely come alone but usually come as part
of photons. Still, waves do have a distinct
identity when traveling, at least ideally. I
have self-studied physics.
ANSWER:
Unless you are talking about a single
photon, there is no such thing as "…ONE
electromagnetic wave". In a classical
electromagnetic wave there are rays and wave
fronts. The way to specify the energy
depends on the particular wave. Electric and
magnetic fields have energy densities (the
amount of energy in a volume divided by the
volume) associated with them. But an
electromagnetic wave has constanty changing
fields so the energy density will be
constantly changing and we are generally
more interested in the time averaged energy
density. Also, though, a wave has a flow of
energy away from the source; This flow is
generally expressed by a quantity called the
Poynting vector S ,
the units of which are watts per square
meter. An example is a plane wave in a
vacuum, a wave in which wave fronts are
planes. Here S =|E |2 /Z 0
where Z 0 =√(μ 0 /ε 0 )
is the impedance of free space and E
is the electric field. The time average
value of the poynting vector is <S >=<|E |2 >/Z 0 =E 0 2 /(2Z 0 )
where E 0 is the maximum
magnitude of the electric field. So, you
see, it is a complicated situation.
QUESTION:
Is building a time machine physically
possible
ANSWER:
It is possible to travel to the future but
not the past. We already have time machines;
your car, for example, moves you forward in
time when you make a round trip but since
the speed of a car is so small compared to
the speed of light you never notice it. For
example, suppose that you travel at 60 mph
for an hour and then return at the same
speed.
QUESTION:
What will happen to the momentum of
raindrops as it falls yo the ground
ANSWER:
Any object falling through air will
experience a downward force (its own weight)
which will speed it up as it falls. But,
when it starts moving it also experiences an
upward force due to air drag which increases
as speed increases. Eventually the drag
force up becomes equal to the weight force
down, so it then falls with constant speed
until it hits the ground. The drag force
depends on the size (hence on its mass) and
shape of the raindrop as it falls. Because
it is deformable, it will change its shape
as it falls, so to compute its speed as a
function of time would be a very complicated
problem to do in detail.
QUESTION:
Can megnetic fields exist independently
of electric charges?
ANSWER:
All electromagnetic fields ultimately depend
the source of the fields, electric charges.
Magnetic fields may be caused by electric
currents in a wire which has no net charge,
but the electrons in the wire are moving
which is the cause. Magnetic fields are also
caused by an electric field which is
changing, but you ultimately had to have
electric charges to cause that changing
electric field.
QUESTION:
what happens when sound waves are
produced in a chamber with diminished air
molecules. I’m a musician and am considering
experimenting with vacuum chambers and hand
held percussion instruments while recording
to try to capture some unique sounds. I know
sound will not travel through a vacuum, but
I am wondering about the effect of a space
with a diminished level of oxygen molecules.
ANSWER:
The speed of sound c in a gas,
which is what is going to determine how a
musical instrument will sound, is given by
the Newton-LaPlace equation, c= √(K /ρ )
where K is the bulk modulus of air
and ρ is the density. If you remove
air from some volume you are reducing the
density, so it would appear that the speed
gets larger as you remove air; however,
K depends on temperature and pressure,
so you need to dig deeper to find out what
c is because if you take air out
the pressure is going to decrease. You are
probably not interested in all the details*,
so I will just give you the final results:
in dry air at a temperature T in
°C, the speed is approximately c =20.05√(T +273.14).
Since you will probably do your
experimentations at constant temperature,
the speed of sound and therefore the musical
behavior will likely not change
significantly.
*If
you want to see details, see the Wikepedia
article on the
speed of sound .
QUESTION:
Does light impart any kinetic energy?
Put another way, with enough light can you
move an object?
ANSWER:
Light carries linear momentum and can
therefore exert a force on whatever it
strikes. So, yes, light can move an object.
Using the sun's light to propel spacecraft
has been
suggested .
QUESTION:
I am a junior high school student and I
would like to know how a black body will
develop if it is given a continuous
injection of power? Does the mass get
bigger?
ANSWER:
There is a simple answer here and it does
not require that we are talking about a
black body. Everything radiates energy at
some rate which depends on its temperature,
size, shape, composition, etc . If
you add energy at a rate greater than the
rate it is emitting radiation, it will gain
energy (most likely by increasing
temperature); if the input energy is at a
smaller rate, the object will lose energy
but at a rate smaller than it would have if
you hadn't injected energy. The mass of an
object depends on its total energy (E=mc 2 ),
so increasing energy means increasing mass.
In most cases, the mass increase would be
unmeasurably small.
QUESTION:
Diatomic hydrogen becomes monatomic upon
absorption into palladium-from whence comes
the energy to separate the atoms?
ANSWER:
You are right, it takes energy to separate
the atoms in molecular hydrogen. But when
palladium absorbs hydrogen, a single
hydrogen atom is bound in the crystaline
structure of the metal. Since the H atom in
the palladium is more strongly bound than it
would be if bound to another H atom, the
system will reduce its total energy by
breaking the H2 bond and forming
the Pd-H bond. The answer is that the Pd is
the source of the energy to separate the H2 .
QUESTION:
This question is about strobe lights and
smoke/fire. I lit a small clump (1"x1") of
pine sap on fire and put my strobe
flashlight up to it to see if there was any
visual effect. Instead, I heard a buzzing
sound coming from the smoke. The closer I
got to where the smoke was forming the
louder the sound. If I increased the
brightness of the flashlight the sound also
got louder. If I brought the strobe to
around 10 flashes a second I could hear 10
small blips of sound a second. If I covered
the light the sound went away, so the sound
was not coming from my flashlight.
ANSWER:
At the outset I will admit that I really do
not know for sure. I did a little research
and came up with some information and some
ideas. The ideas I have are based on
speculation, but I think they are pretty
sound. I find that a typical pulse power is
several kilowatts. Now, that pulse has a
significant infrared component. Going
through air, this does not have much heating
effect because the air is pretty
transparent. When the pulse hits the smoke
the infrared is pretty strongly absorbed and
superheats the gas which expands rapidly
making a pop.
QUESTION:
How can the electrons in an antenna
simultaneously oscillate an many different
frequencies and modulations presented to it
by electromagnetic spectrum.
ANSWER:
An electron oscillates because it is
responding to a force due to an oscillating
electric field. Because of the superposition
principle, the fact that if two different
oscillating electric fields with different
frequencies are applied at one point in
space, the new electric field there is
simply the sum of the two at that point and
time. So the electrons will behave in the
same way. The figure shows this for two sine
waves, one with frequency 10 times larger
and amplitude 5 times smaller than the
other, added together. Now, an antenna will
see hundreds of frequencies falling on it;
in order to be useful, e.g. for
tuning a radio, you need to have an
electronic circuit which filters out all but
the desired frequency.
QUESTION:
This Autumn here in the UK (like many
other countries) the clocks went back an
hour. Everyone talks about the clocks going
back and 'gaining' an hour. But surely,
didn't time move forward? I clearly
understand time in this sense doesn't
actually move forward or backward as it is
just our measurement of it. But.... the
clocks may have gone back, but time has been
brought forward not backward surely? For
instance. On the 30th of October the Sun
will have set here at around 17:30. But the
31st it now sets at 16:30. Time has not gone
'back' as it were, we have in fact moved
time 'forward'. The sunset has been brought
forward in this sense. The clocks go
backward....but time is brought forward?
ANSWER:
In any frame of reference (e.g.
the UK), time proceeds forward at a constant
rate. So we need some operational definition
of how to measure time intervals, the time
elapsed between two events. Pretty
universally used is the second (s). (I will
not go into how the second is measured, but
you know what it is. Now, when we do a
physics problem which involves time, the
first thing we do is to choose some time
where the time has a value of zero; all
later times are positive and earlier times
are negative for this choice of t =0.
In your case, t =0 could be
midnight. Then 1:00 AM would be t =3600
s. Imagine that you go to sleep at 1:00
according to this clock. Now, suppose the
British government decrees that midnight
will henceforth occur when your clock reads
11:00 PM. According to the government's
choice of t =0, you go to bed at
t =0 s, midnight. But, just as in your
sunset example, your going to bed is an
event that happens at a particular time
regardless of how you label it (according to
your choice of t =0).
QUESTION:
Because of drag, a human-made satellite
needs occasional rocket thrusts to stay in
its orbit, otherwise it will fall to earth.
Why does not the earth experience drag from
its OWN atmosphere, and fall to the sun?
Similarly, why does not the moon experience
drag from its OWN atmosphere, and fall to
earth?
ANSWER:
Not all satellites experience
significant drag because if their orbits are
far from the surface of the earth there is
virtually no air. The earth's atmosphere
rotates with the earth and therefore there
is no drag. You might argue that if there is
a wind it is rotating at a different speed;
but I would guess that you looked at all the
winds on the earth there would be just as
many speeding up the earth's rotation as
slowing it down. In any case, even if the
atmosphere did cause a drag it would change
the rotational speed, not the orbital speed.
The moon has no atmosphere and is well
outside the earth's atmosphere, so it
experiences no drag. There are drag forces
on the moon and earth due to the tides which
the moon causes; see an earlier answer .
QUESTION:
I'm attempting to calculate the velocity
of an object over time given air resistance
and am not finding an equation online. The
force from air resistance is 1/2 v^2 x d x A
x cp where d is the density of air, A is the
area of the object and cp is a constant
given the geometry of the object's surface
area. If an object weighs 10 kg and is
traveling at 10 m/s in a 0 gravity
environment with an area and cp both set to
a value of 1 then the air resistance or drag
on the object is 61.25 N. Dividing by the
mass of the object and multiplying by an
incremental change in time of 1 second is a
change in velocity of 6.125 m/s. The
object's velocity after 1 second is 3.875.
Each calculation for velocity requires the
previously changed velocity. After 2 seconds
the velocity is 2.955 m/s and 3 seconds is
2.42 m/s. If I draw a graph with time on the
x axis and velocity on the y axis is there a
function F(x) to calculate velocity given
time and the initial velocity?
ANSWER:
Yes, there certainly is. You should
realize that what you did is a reasonable
calculation but that since the velocity is
constantly changing, it is not very accurate
to assume that v stays the same over 1
second. What you are doing is the first step
in calculating the integral of the quantity
but you need to find how it improves as you
make the time interval smaller and smaller.
Since you ask this question, I assume that
you have had no calculus; therefore you will
probably not understand the following
derivation. Do not lose heart, the function
you seek will emerge at the
end!
First, I usually use the approximation that,
at sea level, the drag equation may be
approximated
F =-¼Av 2 =Ma=M (dv /dt )
where
M is the mass of the object, A
is the area it presents to the onrushing
air, and a is its acceleration; the
minus sign signifies that the drag force is
opposite the direction of the velocity. [It
is important that this equation is not
dimensionally correct and will only work
using SI units.] So, we can write that
dt= (-¼A /M )-1 v- 2 dv.= (-4M /A )v- 2 dv
Integrating from t=0 to t, v (0)=v 0
to v (t )=v and
then doing a little algegbra, I find
v=v 0 /(1+kt )
where k =¼Av0 / M.
For
example, the graph for v 0 =20
m/s and k =1 s-1 is
shown. The constant k is just
whatever the coefficient of v 2
in the force is divided by the mass. You
could use your value, for example, k= ½dAC p v 0 /M .
QUESTION:
I was thinking about light sails.. if I
take a photon, with energy = hf , obviously
that is the energy of the photon. Then we
have this photon momentum : p= hλ. If I have
1 photon of light and I make it incident up
on a free floating stationary mass with no
external forces upon it that is mirrored
such that the photon is reflected and a
certain proportion of its energy is
transferred to kinetic energy on the mass it
is incident upon.. and then I repeat the
experiment but this time I make this photon
incident upon a metal first with a very
small work-function such that it liberates
an electron, fully transferring all its
photon energy into both liberation and the
kinetic energy of that election.. and then
that electron collides with our original
free floating stationary target mentioned
earlier.. would either the first case or the
second case impart more momentum to the
object in question (I know some energy would
be used to liberate the electron, but would
more of the photon energy be transferred to
kinetic energy)?
ANSWER:
(The momentum of a photon is h /λ ,
not hλ.) Let's look at the first
situation first. Before the collision the
energy of the photon is e=hf , and
its momentum is p=h /λ=hf /c ;
the sail has E=P= 0. Assuming an
elastic collision, after the collision the
photon has a momentum (approximately) of
p'=-hf /c so momentum
conservation gives us that p=p'+P'
or P'= 2hf /c. But
the mass (M ) is macroscopic and
moving very slowly, so P'=Mv where
v is its speed; so v =2hf /(Mc)
and E' =½Mv 2 =(2/M )(hf /c )2 .
Next
you propose that we use the photoelectron
created by the photon to collide with sail.
The energy of that electron will be
E=hf-W where W is the work
function. If we assume that the electron
bounces back just like the photon of energy
hf did, it will have less momentum
to transfer so the recoil and hence the
resulting kinetic energy of the sail will be
smaller.
QUESTION:
I’m a retired saxophone player who likes
to play bongos on a paddleboard, on Lake
Michigan. It was my understanding that any
for any resonance chamber, the pitch is
higher for a shorter tube, and lower for a
longer tube. But when I set the bongos on
the paddleboard, resting on their lug nuts,
there is about 3 cm space between the bottom
of the drum shell and the surface of the
paddleboard, where water can wash underneath
the drum during a big wave. And the pitch
goes down. Isn’t the water underneath
effectively making the drum smaller?
Shouldn’t the pitch go up?
ANSWER:
Actually, the main thing that is happening
is that you are changing an open pipe to a
closed pipe; the shortening of the length is
a negligible effect on the pitch compared to
the closing off of the end. In the figure
above I have sketched in the vibrating drum
head in red for the two kinds of situations.
The right end of the drum must be an
antinode where the air is vibrating its
maximum amount because that is where the
drum is beating on the air. The "open ended
pipe" is the situation when the bottom of
your drum is open to the outside air;
because the air there is totally free to
move, that is also where there must be an
antinode of the resonant wave. The
fundamental tone is then the lowest-pitched
wave which can fit in there and, as you can
see, it is one half of a full wave. When the
water comes up it closes the previously open
end and keeps the air from vibrating there,
thus becoming a node. Now the lowest pitch
which can fit is a quarter wave, a longer
wavelength and therefore a lower pitch than
the open ended pipe.
QUESTION:
Hi, regarding superposition of nuclear
particles, it is said the electron (for
example) exists in a superposition of states
(all states at once) until it is measured
(momentum, spin etc) and this measurement
colapses the "probability wave". The
probability wave as i understand it, is not
a real thing, it's a graph showing where the
electron is most likely to be in space. Now
my ripe with it, isn't it rich to assume
that an object exists in all possible states
before being measured ? My nighbour, i
haven't seen him today, does that mean he's
everywhere else on the planet until i go
check his house? I understand he's not a
quantum object but still, this superpostion
assumption is to me really far fetched
unless i can be show actual evidence of
something being in two places at once.
ANSWER:
You are confusing "superposition" with
"probability wave"; they are actually
different things. For example, an electron
can have two possible spin states, either up
or down; sometimes the electron will be have
a little amount of each, like 60% up and 40%
down and a measurement of the spin of that
electron would be more likely to find the
spin up than down. Although not really a
good example, your neighbor's superposition
is that his state has zero probability
anywhere than in his house. A probability
wave is essentially the square of the wave
function of a particle or system of
particles; the wave function has no direct
physical meaning until it is squared. The
resulting probability distribution does have
physical meaning in that it tells you that
if you look in one particular place you will
have some particular probability of finding
it. If you look at a cloud in the sky, do
you say that it is in more places than one
just because it is spread over some volume?
Think of your electron as a cloud; it is not
in many places at the same time, rather it
is just spread out. The electron cloud is
different from the cloud-in-the-sky cloud
because if you make a measurement of some
place inside the sky cloud, the whole cloud
is not suddenly there.
QUESTION:
I understand that a light year is a
measurement of distance, but aren’t the time
and distance interchangeable? For example,
if I’m looking at a star that is 10 million
light years away that means that I’m
effectively seeing that start 10 million
years in the past, correct?
ANSWER:
I would not say that time and distance
are interchangeable, but the distance light
travels in a certain time is a good measure
of distance because the speed of light is a
universal constant. But you wouldn't say
that the time it took for a car to go from
New York to Miami was interchangeable with
the distance between the two cities. Indeed,
as you suggest, when you look at the stars
you are looking into the past because of the
time it takes their light to arrive.
QUESTION:
Do sub atomic particles exist or just a
theory?
ANSWER:
Yes, they exist. Very few are stable but
those which are unstable play an important
role in the interactions between particles.
Nearly all have been observed directly in
experiments. A few, quarks in particular,
are unobservable directly but predictions
assuming that they exist have been so
successful that they either exist as
particles or else they provide a very
accurate description of how nature works.
QUESTION:
Since soundwaves can create heat, and
the louder the soundwave is, more heat is
created, does pitch affect how much heat is
created as well?
ANSWER:
The energy carried by a wave is
proportional to the square of its amplitude
but does not depend on its frequency.
QUESTION:
Has Heisenberg's uncertainty principle
being proven either right or wrong? or is it
still under experimentation?
ANSWER:
The HUP is one of the keystone
components of the theory of quantum
mechanics. It posits that the observable
quantities, position and linear momentum,
are not independent things but actually
coupled; if you change one of them it has an
effect on the other. (Mathematically, in the
theory, each is the Fourier transform of the
other). These are called conjugate variables
and the result is that the more accurately
you measure one, the less accurately you can
know the other. The proof that it
is correct is that quantum mechanics has
never failed to correctly describe the
properties of very small systems like atoms.
So everytime you find that some physical
system (e.g. the structure of some
atom) is well-described by quantum
mechanics, you are proving the HUP. See a
recent question very much
like yours.
QUESTION:
Can a force cause
a
moving object to accelerate,
a
moving object to rest, and
a
resting object to move.
I just want to know
the science it was a test question and I got
it wrong and I kind of want to know.I
ANSWER:
Force (F , a
vector) causes a mass m to
accelerate (a , a
vector) as indicated by Newton's second law,
F =ma .
Acceleration is the rate of change of
velocity (v , a
vector). A vector can change in two ways, it
can either change its magnitude (speed up or
slow down) or change its direction; all
three of your situations are the result of
changing the magnitude of the velocity. We
could have had a fourth situation:
If an
object moves in a circle with constant speed
(the speed is the magnitude of the velocity
vector), there must be a force which is
always perpendicular to the direction of the
velocity vector. An example is a planet
moving in a circular orbit around a star
where the force is the gravitational
attraction to the star.
QUESTION:
I was recently working through some
practice problems for a mechanics course
that I am enrolled in. The current unit we
are covering is work and power. One question
asked us to find the power required for a
mass (.119 kg) to accelerate from 0 m/s to
27 m/s in 0.36 sec. I thought that you could
find the acceleration using the kinematics
formula 'V = Vo + at' to solve for
acceleration; and then use the formula 'P =
m*a*v' to solve for power. When I looked at
the answer key, this was not the way that
they solved the problem. They used 'P = W/t'
where 'W' was the change in KE. I can
understand why this approach is correct,
however, I was puzzled as to why the answer
that I got solving it my way was exactly 2x
greater than the one found using their
method? I have been thinking about this for
a while now, but unfortunately am having no
luck.
ANSWER:
v (t )=v 0 +at
gives you the velocity at some time
t if the acceleration (and hence the
force) is constant; but you have no idea
whether that is the case. The equation you
used, P=Fv is not applicable if
v is changing which it surely is.
However, it will work if you use the average
velocity over the time interval. Since you
have assumed that a is constant,
v is increasing linearly and so
v average =½v (t=T )
where T is 0.36 s in your case.
There is your factor of 2!
QUESTION:
Is quantum uncertainty intrinsic to
physical reality or a measure of our
ignorance?
ANSWER:
Both. If the uncertainty principle is
correctly stated, it is a correct physical
principle which indicates the degree to how
well you can possibly know something. It
does not say that no quantity can be known
to arbitrary accuracy. But if you have two
quantities and they are what is called
conjugate variables , then it is not
possible to know both of them to arbitrary
precision. For example, position and linear
momentum of an electron are conjugate
variables. If you form the product of the
uncertainty of each, it cannot be smaller
than approximately the rationalized Planck
constant, ℏ. So the more accurately you
try to determine the position of the
particle, the less you can know about the
momentum; conversely, the more accurately
you try to measure the momentum of the
particle, the less you can know about its
position. If you were to know exactly what
the momentum is, you would know absolutely
nothing about where that particle is, could
be anywhere in the universe! A second
example of a conjugate pair is time and
energy. So suppose you have a stable atom of
mass M in its ground state; you may
measure its energy E=Mc 2
to any degree of accuracy you wish because
you have an inifinte amount of time to do
so. But if the atom is excited to some
excited state of energy E+E' , it
will decay back to the ground state and so
you only have a short time to measure its
energy. Thus the energy of that state is not
exactly E' but actually broadened
by some amount determined by the half life
of the decay; the shorter the half life, the
more uncertain the energy of that state is.
QUESTION:
My step dad is adamant that you could
put a wind turbine on a car that will help
charge the battery on the car to make it go
further or help recharge the battery. Afaik
this doesn’t work due to perpetual motion
and this is similar to that drawing of a fan
trying to blow its own sail. That
thermodynamics wouldn’t allow enough of a
wind turbine or a “fan” he says that soaks
up the wind that the car produces(which I
think is just air resistance so not enough
energy) to make this idea viable. Am I
correct it wouldn’t work or is there someway
we could use a fan to produce electricity on
a car to recharge the battery or help
battery last longer etc?
ANSWER:
Well, your stepfather is right. Driving
through still air the turbine will be able
to charge the battery, but dead wrong that
this will make the car go farther. Where
does the energy sent to the battery come
from? The forward motion of the car lets
there be a wind which will spin the
propeller. But the car needs the energy to
keep it moving with a constant speed. Since
the turbine is removing energy from the
moving car, the car needs to provide more
energy to keep moving. So the energy that
goes into the battery is supplied by the
engine. In the real world there is friction
and other energy losses in the turbine, so
the net effect is for the whole car to lose
energy.
QUESTION:
Gravity waves propagate at the speed of
light. We know this because of very
sensitive lasers. How can any phenomenon
that is hypothetically faster than light be
measured by something that is not, for
example photons? Wouldn't everything
measured always be at the speed of light or
less?
ANSWER:
Actually, the speed of gravity has not
been measured. In general relativity it is
assumed that the speed is the speed of light
and nothing has indicated that it isn't. The
most convincing evidence that I am aware of
does come from the laser-interference
measurements. Although most of the events
detected are from black hole collisions, one
was a collision between neutron stars. A
collision between black holes should be
unobservable with conventional telescopes
because no light, xrays, gamma rays etc.can
escape. However the neutron stars collision
was observed by telescopes at just about the
same time; given the enormous distance away
of the event, the speeds of light and
gravity must be very close.
If you
look at my site ground rules, you will see
that I do not answer questions about faster
than light speeds because
physics-as-we-know-it forbids that
happening.
QUESTION:
How does speed affect our aging? I have
read on the internet that the faster you go,
the slower you age. I'm just wondering how
can time affect the biological aging of a
person. I also watched the twin paradox
video wherein the guy who traveled near the
speed of time looks younger than his twin
who stayed on earth. Does that mean our
biological process of aging that happens
inside and outside our body gets affected by
our speed?
ANSWER:
You should first read
my explanation of the twin paradox. For
the traveling twin the time proceeds for him
just as it would if he had stayed at home.
But the distance he has to travel before
turning around becomes shorter. The
earth-bound traveler observes his brother's
clock running more slowly. Regarding aging,
if you think about it, the body is
essentially a clock; average age about 80
years, average time between heartbeats about
1 second, for example. The earth-bound twin
observes all clocks on his brother's space
ship slow down, including his brother's body
clock.
QUESTION:
I recently thought about why I can see a
smaller object than reality reflected when a
hold an pen in front of a spoon? Normally I
think the object should look bigger because
the surfaces of the spoon is bended inwards
and some of the areas outside the real pen
in the spoon should also reflect the pen. Is
it something wrong with my spoon?
ANSWER:
I think from your question that you have
not studied geometric optics. So I will give
a very brief tutorial here. First of all,
the thing which you call "reality", the pen,
is referred to as the object. The reflection
from the mirror which you refer to as an
object is referred to as the image. A spool
is approximately a bowl shaped like a part
of a sphere and the kind of image you will
get will depend whether you look into the
inside of this bowl (called a concave
spherical mirror) or into the backside
(convex).
A
spherical mirror has what is called a
principle axis which passes through the
center of the mirror perpendicular to the
surface. The sphere has a radius R
and halfway from the surface to the center
of the sphere on the axis is a point called
the focal point (F ). Light coming
in parallel to the principle axis is
reflected out through the focal point; Light
coming in through the focal point is
reflected out parallel to the principle
axis. Above are diagrams for the concave
mirror showing the geometry and a ray
diagram for an object placed farther away
than R from the spoon which will
certainly be the case for the spoon you are
looking at. Note that the image is smaller
than the object (just as you found) and
inverted. The image is called a real image
because it is on the side as the object and
could be seen on a screen placed there.
If you happen to be
looking at the back of the spoon you have a
convex mirror. Above are shown diagrams of
the geometry and ray diagram for this
situation. Here rays coming in parallel are
reflected directly away from F and
rays coming in directly at F are
reflected parallel to the principal axis.
Now the image is again smaller but now
upright. This is called a virtual image
because it is not really where you see it to
be; if you put a screen behind the spoon you
would not see the image there.
I have used some
images from the Wikepedia
article on curved mirrors. I have added
(in red) rays which come in toward F .
You can learn a lot more detail there.
QUESTION:
The hydrogen material in a thermonuclear
device is supposedly compressed into helium
by the force of the X-rays produced by the
detonation of the fission component of the
weapon. X-rays are photons. Particle physics
shows that photons are massless. How can
something that has no mass compress
anything?
ANSWER:
Even though a photon has no mass, it does have
linear momentum. The reason for this is that
at high velocities (the speed of light is
the highest possible), linear momentum is
not mass times velocity. If it has linear
momentum, it can exert a force.
QUESTION:
Say I have one cannon that it is free to
roll backwards due to recoil when I fire it,
and I have another that is secured very
firmly, and cannot freely roll backwards due
to recoil. Will the more secured cannon that
cannot freely roll back and forth give the
cannonball more energy than the one that can
freely roll if given the same amount of
powder and the same mass of ball?
ANSWER:
I will discuss this assuming that the cannon
fires horizonally and that the ground is
horizontal. You can also do the problem if
the cannon ball is fired with some vertical
component of its velocity, but the details
obscure understanding the concepts. The
total energy created by the explosion (call
it E ) must be divided up between
the kinetic energies (Km =½mv 2
for the cannonball and KM =½MV 2
for the cannon) of the cannonball and the
cannon; this is called energy conservation.
The amount of energy each gets is determined
by the velocity each has after firing; this
is found by applying momentum conservation
which says the amount of momentum before
firing must equal the momentum after firing,
linear momentum being defined as mass times
velocity. Since momentum is zero before
(both with zero velocity), 0=mv-MV
or v=MV /m . (The minus sign
is because the cannon recoils in the
negative direction.) For example, suppose
the masses are equal; then v=V and
each will then have half the energy. Or if
the cannon were 100 times heavier than the
cannonball, v =100V , so the
ratio of energies is (½(0.01M) (100V )2 /½MV 2 )=100.
Finally, lock the cannon to the ground; now,
M is the mass of the whole earth.
So, since the mass of the earth is so much
larger than m, let's just call it infinity
and the cannonball gets all the energy.
QUESTION:
"the physics definition recognizes that
sound exists independently of an
individual’s reception." I don't agree with
this. surely they should be referred to as
compression waves? They only become sound
when they are detected by the ear
ANSWER:
This is purely a case of semantics. It is
just fine to call sound waves compression
waves, but in physics sound waves are
defined to be longitudinal compression waves
in some material. If you don't want to call
it sound unless it is detected by the human
ear, be my guest, but that is not physics.
In addition, you definition of sound wave
should probably include the compression
waves being interpreted by the brain to be
sound?
QUESTION:
if the earth is rotating at roughly 792
mph at the equator why dont cars need to
travel faster than that in the same
direction to go any distance?
ANSWER:
Because the car is already going 792 mph
when it is still relative to the earth. So
is the air if there is no wind.
QUESTION:
Is a photon's size less than a planck
length?
ANSWER:
A photon has no well-defined size. Because
of the uncertainty principle, the "size"
depends on how well you know its "color".
For example, suppose you have a laser whose
color you know rather precisely; then the
size of a photon would probably be bigger
than a house. On the other hand, if you know
quite precisely where the photon is, you
will be able to know almost nothing about
its color.
QUESTION:
Good day, this question is posed from
the view of physics as a branch of science
which has it's own structure, a structure
that can be built on and adjusted if need
be. I have looked for an obvious start to
physics as a whole structure but I haven't
found one that has a "start to finish"
approach. It always looks as though there is
underlying disorder and there is no obvious
way to go about studying physics
structurally. If you can, please suggest a
particular structure in which physics should
be studied such that no context is left
haphazardly intertwined with another and
every context can be derived from the
previous without having to dwell at all on
other related contexts.
ANSWER:
Your question assumes that the structure of
physics is linear, each step a new step and
based solely on what went before. But this
is not how physics is actually structured.
Relations among the principles of physics
are complex; in fact, understanding how all
of physics is constructed from its numerous
components is a key goal, "unification".
Your need to have no intertwining among
various concepts of physics is impossible to
satisfy. In my opinion, any discipline is
pretty uninteresting if it can be explained
using the constraints you seek. An
additional problem with understanding
physics is that it is not just a bunch of
equations, but it is two-pronged, both
theory and experiment; in fact most
physicists consider observation as the
fundamental source of knowledge, theory
being the attempt to systematize what we
observe.
In my
opinion, the best way to study physics is
from a historical perspective, the way the
discipline actually progressed. An excellent
example is the study of electromagnitism.
Ancients had observed electric and magnetic
forces in nature. Mainly in the 18th and
19th centuries many investigators quantified
these forces (Coulomb's law and Ampere's
law, e.g. ) but then, in 1820,
Ørsted discovered that the two forces were
"intertwined" as you wish not to be. By the
end of the 19th century Maxwell compressed
all that was known about classical
electromagnetism to four equations. And
Maxwell's equations also allowed classical
physics to understand a long-debated
question—what is light composed of—dragging
optics into this intertwinedness (just made
that word up!) But all this then led
Einstein to discover relativity which
resulted in a revision of Newtonian
mechanics, dragging in even more
intertwinedness! But then, in the early 20th
century, quantum physics started to be
discovered and developed which led to the
issue that light also has particle-like
properties which means that quantum
mechanics and electromagnetism had to become
intertwined! It was not until the mid 20th
century when Feynman and others developed
quantum electrodynamics (QED), intertwining
the whole set of theories into a single
theory.
QUESTION:
Does negative energy exist? I have
looked online and found that one article
said no, then another from less than a year
later says yes. I can not seem to be able to
find a absolute answer.
ANSWER:
First of all, you need to know that the
absolute amount of energy in any closed
system is arbritrary to an additive
constant; if you say some system has 6
joules of energy and I say it has 5 joules,
we are both right and would arrive at the
same conclusions based on laws of physics.
The laws of physics do not care about the
absolute value of the energy. Of course,
some forms of energy you would probably want
to be positive definite; kinetic energy, the
energy by virtue of its motion of an object
with speed v and mass m , ½mv 2 ,
and rest mass energy, mc 2 ,are
always positive because m , v 2 ,
and c 2 are. Negative
energy is used all the time in various field
theories. Potential energy is the best
example. A planet of mass m is
orbiting a star of mass M with a
circular orbit of radius R ; its
potential energy is -MmG /R +C
and we choose the constant C
to be zero so the potential energy will be
zero at R =∞.
Similarly, the mass energy mc 2
of a nucleus comprised of N
neutrons and Z protons but it is
smaller than the total mass energies of its
constiuents: mc 2 <Nm neutron c 2 +Zm proton c 2 .
The difference is negative and called the
binding energy.
QUESTION:
Is the human body a form of energy? We
have heat and electrical impulses. If so,
would this not be breaking the first law of
thermodynamics in that energy cannot be
created or destroyed? A person is created in
the womb, or is this just a transfer of
energy from the mother?
ANSWER:
The first law of thermodynamics is simply
the principle of energy conservation. This
law states that the total energy of an
isolated system cannot change. Any body
with mass has lots of energy, E =mc 2 ;
and there is a lot of chemistry going on in
your body, and lots electrical impulses, so
if we put the body into a box filled with
air where nothing
can come into or go out of the box, all the
energy in the box will constantly be
shifting from one form to another, but if
you measure the total energy at any time it
will always be the same. The baby in the
womb gets the energy from the mother who
would usually get it from eating food, no
longer available inside the box. The woman
would pretty soon use up all the air,
converting into mostly carbon dioxide and
die. But the total energy in the box will
always be the same. You know, that is the
problem with conservation principles—you
have to know the conditions under which the
principle is obeyed.
QUESTION:
Hi, how do I calculate the force
(joules) required to maintain an object in
motion's speed with earth's gravity? ex.:
how much force do I need to apply to a
tennis ball to keep it going at a constant
speed?
ANSWER:
(In the diagrams below, red vectors are
forces you do not exert, green vectors are
forces you exert, and blue indicates the
velocity (not a force) at this instant.)
The
Joule is a unit of energy, not force. Force
is measured in Newtons* in SI units or
pounds in Imperial units. Newton's first law
tells that if the sum of all forces on an
object is zero that it will move in a
straight line with constant speed. The
forces on an object moving in a uniform
gravitational field with some velocity
v at some particular
time is (neglecting air drag) is simply its
weight
W , a force
straight down. So if you wish to exert a
force
F so that the
object continues with that same velocity,
you must exert a force straight up but with
the same magnitude as the weight. This is
shown in the first diagram and the baseball
will continue moving in the direction as the
blue arrow and the speed will not change; if
you did not exert
F ,
the ball would follow the arc we are used to
seeing baseballs move in. If there is an air
drag
D (which always
points in the direction opposite the
velocity) as shown in the second figure, you
must also exert a force
d
which is equal and opposite to
D ;
also shown is the net force you could exert
which is just the sum
F + d .
*a
Joule is a Newton·meter
QUESTION:
I’ll form my question as a statement:
correct or not correct. I am trying to get a
grip on the centripetal, centrifugal, and
Coriolis forces, particularly as regards the
creation of effective gravity by rotation.
An object in a rotating drum on earth
experiences an apparent force (centrifugal)
as the surface of the drum moves
continuously into the object’s inertial path
of motion. A stationary object placed inside
a rotating drum in space (that is to say,
outside any significant gravitation) would
sit in place and not move. This would be
true whatever the distance is between the
object and the drum (excepting some amount
of electromagnetic effect depending on the
composition of the two things). If however,
the object contacted the drum, the object
would gradually begin moving with the drum
as the apparent force between the two
increases. Is this correct? This was
motivated by an article in Medium magazine
which has been circulated widely by fellow
nerds, asserting that (what was in the
article misattributed to centripetal force)
the coriolis force on people in this
situation would make it difficult to to
move, their blood to flow properly, their
inner ear…etc. I.e. it can’t work. Clearly
this would have something to do with the
relationship of the size of the drum to the
person. Ok, an additional question. Look for
a June 2 article at allanmlees59.medium.com
ANSWER:
I have previously answered questions like
yours in great detail. See the
faq page.
QUESTION:
I've googled several variations of the
question: What would happen if neutrinos
suddenly disappeared? The most frequent
Google answers pertain research
possibilities that physicists would be
deprived of. But of course, what I really
want to know is how it would affect physical
interactions in the universe and life on
earth. In short: in what ways are we
dependent on neutrinos to the extent that,
if they were to suddenly disappear, but with
no negative effects to solar nuclear fusion,
how would humans, earth, and the universe be
adversely impacted? What good do neutrinos
do for us now?
ANSWER:
I can think of two possible ways to
interpret your question:
what if all the current neutrinos were
to suddenly disappear? and
what if neutrinos did not exist at all?
I
believe that the answer to the first
question is that it would not change much at
all. Neutrinos interact extremely weakly
with matter; it was decades after neutrinos
were hypothesized before they were directly
observed. And, although they were originally
thought to have zero mass, they are now
known to have a small mass; it would take
about
six million neutrinos to make the mass
of one electron. Still, the universe has
huge numbers of neutrinos flying around and
their mass could have significant effects on
the dynamics of galaxies, for example. As
you probably know, the universe seems to
have about six times more mass than we can
see; this unobserved mass is called dark
matter . Could dark matter be neutrinos?
Estimates indicate that only about
1% of dark matter could be neutrinos.
The
answer to the second question is that
neutrinos play a crucial role in β-decay and
β-decay plays a crucial role in the workings
of stars; so without neutrinos the universe
as we know it could not exist at all.
QUESTION:
I would like to know if present physical
science has the answer to, ,'why does the
attractive force between two bodies diminish
in the order one by d square?' The answer,
'according to Newton's equation...' is not
the expected answer. Why should it decrease
if space is absolutely vacuum? If there is
no answer I have, based on a new space
concept. If satisfactory answer is already
there I had better stop my philosophy.
ANSWER:
It is helpful to envision the light from a
very small (point) source. In some time
t , the light carries a certain amount
energy (E ) away from the source
with a power of P=E /t
which is usually measured in Watts. Now, the
total power at some distance d from
the source must remain the same because of
energy conservation. Now, intensity of light
is essentially the power per unit area,
Watts per square meter; since the area of a
sphere is of radius d is 4πd 2 ,
the intensity of the light falls off
proportionally to 1/d 2 .
The behavior of a force field (gravity in
your case) follows exactly the same
arguments except we talk about the flux
(field stength/area/time) and, again, the
area a sphere increases quadratically with
the radius. In a nutshell, I guess you could
say that the 1/d 2
behavior is because clasically the geometry
of the universe is Euclidian.
QUESTION:
My question if that a hand-sized metal
ball (assuming it would not disintegrate)
was propelled to the limit of the area of
Earth's gravitational pull, how long would
it take for this object to come back to
earth. This question was inspired by the
idea that if an indestructible baseball
could be hit this far how long would it take
to reach the Earth again.
ANSWER:
The
gravitational force which an object with
mass m experiences a distance r
from the earth's center is F=MmG /r 2
where G is the universal
gravitational constant and M is the
mass of the earth. Note
that the force never becomes zero so "… the
area of Earth's gravitational pull…" is
everywhere. You have to get infinitely far
away from the earth for there to be no
force. There is a speed, ve
called the
escape velocity , which is the minimum
speed which will project an object from the
surface of the earth so that it will never
return: ve =√(2GM /R )=11.2
km/s=25,020 mph; R is the radius of
the earth. So the answer to your question is
"forever".
QUESTION:
Sir in our earlier classes we are told
that gravity comes from the centre of masses
but it's the Newtons theory right, but
einstein's general theory of relativity that
is proved to be correct, is that gravity is
due to the curvature of space - time fabric
and therefore masses get attracted towards
each other due to slop in the curvature of
warped space - time then what is true if
Albert Einstein's theory is correct than
gravity emerging from the centre of masses
is wrong, because both theories are way
different!! Even though the curvature is
indused due to masses there's no point in
gravitational force from the centre So why
were we told in our 9th class textbooks that
gravity comes from the centre??
ANSWER:
First of all, gravity does not come
"from the centre of masses", it comes from
everywhere in that mass. If the object is
spherically symmetric, the net
force always points directly to the
geometrical center of the mass which is also
the center of mass; this is not true for a
mass with an arbitrary shape and mass
distribution. I guess your 9th grade
textbooks only talked about spherical
objects like the earth and moon and sun for
simplicity. Newton's theory of gravity is
mainly empirical; he showed that Kepler's
three laws for the motions of the solar
system could be perfectly understood if the
force of gravity falls off like 1/r 2
where r is the distance from the
center of the sun to the center of a planet;
this was a monumental achievement, based
mainly on experimental observations of the
planets. One of the main reasons that Newton
had to invent calculus was to prove that the
gravitational force outside a spherical mass
was the same as if the whole sphere were a
point mass with the same mass as the sphere.
This is a wonderful and astonishingly
accurate theory which is easy to calculate
with. Nearly all calculations for celestial
mechanics can be done perfectly accurately
and are used on a daly basis by astronomers.
So now
Einstein comes along and says to himself
that this is how gravity works, but why
is the force the way it is? He invents
general relativity and says that this
wonderful theory of Newton's is correct
because the presence of mass warps
space-time. Admitedly, general relativity
gives some new insights as well, things like
light being bent by gravity, black holes,
gravitational waves, etc ., but that
doesn't mean Newtonian gravity is incorrect.
Calculations using general relativity are,
unlike for Newtonian gravity, generally
difficult to do because the calculations
involve extremely advanced mathematics like
Riemannian geomtry. It would be foolhearty
to simply dump the classical gravity theory
in favor of general relativity because then
nobody but advanced graduate students would
be able to understand, in any quantitative
way, how the solar system works. You were
taught about gravity the right way.
QUESTION:
When I fill up water jugs in the tub I
put a wet towel around a bodywash or shampoo
bottle so I can set the jug under the faucet
with a funnel and it won't slide as the jug
fills. Is there a specific law or laws to
that whole process? I took physics in high
school and got an A but I'm 40 now and 20
years of bartending in lol...
ANSWER:
The
thing which keeps anything from sliding is
friction. All you are doing when you insert
the cloth between the bottle and the tub is
increasing the friction. The equation for
frictional forces is f=μN where
f is the fictional force, N is
the normal force between the object and the
surface it rests on, and μ is the
coefficient of static friction. If you long
to harken back to high school physics, here
is some more detail. If an object is on an
incline of angle θ , its weight
W is doing two things as shown
by its two components: N
is trying to press the object through the
surface of the incline and D
is trying to push the object down the
incline. So, to hold it on the incline there
must be a force the same magnitude as
D which points up
the incline; that would be the friction.
Trigonometry tells us that D=W sinθ
and N=W cosθ . Therefore
W sinθ=Wμcosθ or μ =tanθ
; notice that μ does not
depend on the weight. For example, suppose
that θ =50 ; then μ= 0.09.
Suppose that μ= 0.05 for your bottle
sitting on the tub; then it will slide. But
suppose the towel has μ= 0.13 when
in contact with the tub; the bottle, if
sitting on towel, will not slide.
QUESTION:
So Muons have a tiny mount of mass
right? And we accelerate them to near the
speed of light in particle accelerators
right? And the faster something goes the
more mass it has, until it hits infinite
which is why we can't get it all the way to
the seed of light right? Assuming I am
correct above, doesn't that mean if we
wanted to say create artificial gravity on a
space ship, we could accelerate a single
Muon until it's moving so fast that it has
the gravity equivalent of earth? Could this
also be used to warp space and create a warp
drive essentially? Wouldn't it create like a
spiral gravity field that moves around the
acceleration track with the Muon? Wouldn't
this create some kind of gyroscopic gravity
effect? If you could stop it and start the
acceleration it seems like it would start
causing near by objects to accelerate
towards it and then past it. Likewise if the
Muon keep being accelerated closer to the
speed of light maybe it would enter it's own
gravity wake which would be warped, and thus
allowing it to accelerate faster then the
speed of light?
ANSWER:
Your idea will not work for many reasons.
First, look at an
answer to an earlier question very much
like yours; the muon is so tiny, even at
huge speeds it still only gets more massive
by like a few hundred thousand times, and
ends up still tiny. Also, the muon is not
stable and although time dilation would
cause it to live much longer than if at
rest, it would still only live a few
seconds. Also, how are you going to
accelerate it to such great speeds. Also, if
you could get it massive enough so that the
field would be equal to earth's at say 20 m
away, the field at 10 m would be 4 times
bigger.
FOLLOWUP
QUESTION: So You mentioned the
field would be bigger, does that mean that
the particle would get larger as it moves
faster? I'm not sure I understand what you
mean by that. I guess I assumed the particle
would stay the same size, but act
gravitationally as if it was larger. I guess
if it actually has to grow to the size of
earth it would not be useful. Then the other
challenge is energy, I agree that's a
limiting factor, if energy wasn't a barrier
could it still be used to produce 1G, or
would the particle become earth sized?
ANSWER:
The field being larger means the force you
would feel is larger, not the size of the
particle is larger.
QUESTION:
Cool, thanks! So If I can get for all
the energy needed I got artificial gravity
figured out ??
ANSWER:
You have obviously not read my answers
carefully. The most important reason that
your idea for artificial gravity will not
work is that the source is a point mass and
the field falls of quadratically This means
that force you will feel will only be what
it would be on earth if you were one
distance from the particle; move closer and
it gets bigger fast, move farther and it
gets smaller fast. What you need is to have
a uniform gravitational field, which means
that the force you feel should be the same
everywhere like it is on the earth.
Furthermore, how are you going to keep this
particle inside your ship if it is moving,
for all intents and purposes, at the speed
of light? Just for the heck of it, let's
figure out how big the mass M of
your particle would have to be for the force
on a mass m a distance 10 meters
from M is equal to mg
where g =9.8 m/s2 is the
acceleration due to gravity on earth.
mg=MmG /r 2 , so,
M=gr 2 /G =9.8X102 /6.67x10-11 =1.5x1014
kg
Here
G is Newton's universal gravitation
constant. If you were 1 m away from this
mass, the force on you would 100 times your
weight. The very presence of a mass this
large would likely result in the collapse of
the entire ship. The energy you would have
to give a proton would be about 1031
Joules; it would take all the power plants
on earth about 100 billion years (much
longer than the current age of the universe)
to deliver this much energy. Please do not
tell me that you have "artificial gravity
figured out"!
QUESTION:
Hello! I’m coming without a question
about Mariana Trench. Since it’s so deep and
pressure is so strong, would it be possible
for a £2 coin to fall to the very bottom of
it? And would it be transformed in any way?
ANSWER:
A coin and water are both essentially
incompressible; in other words, the coin and
the water both will have the same density at
the bottom as at the surface. The only
forces (ignoring drag if it is moving) on
the coin are its own weight and the buoyant
force which is equal the the weight of the
water with the same volume as the coin. So
the motion of the coin does not depend on
the depth and it will fall to the bottom
because its weight is greater than the
weight of the displaced water.
QUESTION:
The weak force is stronger than gravity.
Bosons reportedly mediate or generate the
weak force.
Wiki says that "the high masses of
bosons limit the range of the weak
interaction". Perhaps the reason this seems
odd to me is because I'm conceiving the weak
force to function similarly the way gravity
functions. Looking at it from that
perspective, the greater the mass of an
object, the greater its gravitational pull,
right? So, if the weak force is stronger
than gravity, why would the greater mass of
bosons limit the range of the weak force? Or
perhaps what I should be asking is: how does
the function of the weak force differ from
that of gravity?
ANSWER:
I should first warn you that it is very
difficult to compare gravity with any of the
other three fundamental forces because it is
the "odd guy out"; our best theory of
gravity is general relativity and the
associated field has never been quantized
while the others have. Therefore, it is not
really possible talk about the gravity in
any quantum-mechanical detail because we do
not have any gravitational boson mediator to
compare. However, it is generally agreed
that we should call this (unobserved,
unpredicted) boson a graviton. I will answer
your question assuming gravitons do exist,
so you should be wary.
Let's
first talk about the electromagnetic force.
The boson which mediates this force is the
photon and it has zero mass. The
electrostatic force is proportional to 1/r 2
where r is the distance between a
point electric charge and where the force is
felt. All indications are that gravity is
also a 1/r 2 force and
that gravitational waves travel at the speed
of light which would mean that the graviton.
It is also thought that the gluon has
approximately zero mass whereas, as you
note, the bosons associated with the weak
interaction, as you note, has massive
bosons. So you see, the mass of the boson
has nothing to do with the relative strength
of the force.
The
range of the force is not the same as the
strength of the force. The mechanism in
field theory for force is that "virtual
bosons" act as mediators. But of course,
these bosons have energy so if one of them
suddenly pops into existence, doesn't this
violate energy conservation? Yes it does!
The way this can happen is via the
uncertainty principle which permits a change
in energy ΔE as long as it only
does so for a limited time Δt where
ΔE Δt ≈ℏ where ℏ
is the rationalized Planck constant.
Finally, the only reason gravity, the
weakest of all is the only one we are
normally aware of is that there just happen
to be huge chunks of mass, the source of
gravity, around us.
QUESTION:
I'm trying to get a handle on particle
"spin". Most of my sources say that
scientists long ago gave up the notion that
particles actually spin on an axis the way
the earth does. Wiki depicts it as a 720
degree rotation.
https://en.wikipedia.org/wiki/Spin-%C2%BD
Then other sources say that the spins of
bosons, electrons, nuclei (and the quarks
contained therein) are not the same spin.
And then there's the 1/2 spin, etc. So what
makes spin a spin?
ANSWER:
I suggest you look at three earlier answers, 1 ,
2 ,
3 .
QUESTION:
i am trying to invent something. a tank
of water, six inches deep, 20 feet in
diameter. 157 cu ft. Tank rotates 60 rpm, 62
fps. What psi does outer wall have on it?
ANSWER:
There is no single answer because the
pressure in a fluid in a gravitational field
depends on the depth. Also, note that it
will take some time for the whole volume to
come to the same angular velocity, i.e .
rotate as a rigid body. Now, there is an
important effect which will occur when you
start the tank spinning: a centrifugal force
will push the water away from the axis and
the surface will become parabolic in shape.
Unless the tank has walls higher than the
depth of the water, the water will spill out
over the top. Your proposed angular velocity
is pretty large (1 revolution/second) and
this will make the problem extremely
difficult to do. This is illustrated by
using an equation which I
found which expresses the the height
h above the fluid in the center and
upper edge of the water at the wall of the
tank as a function of the angular frequency
ω and the radius R of the
cylinder, h =ω 2 R 2 /(2g )
where g is the acceleration due to
gravity. I prefer to work in SI units and
will switch back to imperial units in the
end:
h =[((2π)2
s-2 )(3.052 m2 )/(2x9.8
m/s2 )=18.7 m=61.4 ft!
What
this means is that virtually all the water
in the tank would spin out to the walls and
spill out. I must admit that this surprised
me. To make this a doable problem we must
make the surface of the water stay flat; we
can do this by putting a flat lid right at
the surface of the water. That is what I
will now do to answer your question; note
that the pressure at the outer surface will
depend on the depth.
I will
assume that you are interested in the gauge
pressure, the pressure in addition to the
atmospheric pressure; if you want absolute
pressure, add about 14.7 psi. First,
assuming the tank is not rotating at all,
write the pressure of the water (P w )
a depth d below the surface; P w =ρgd
where ρ =103 kg/m3
is the density of water. So, at the surface
(d =0) the pressure is zero and at
d =6 in=0.152 m, P w =1490
N/m2 =0.216 psi. The pressure at
any depth d is P (d )=(0.216/6)d =0.036d
where d is in inches. That was
gravity without rotation; next do rotation
with no gravity. (I am going to do this
using calculus, so if you don't know it you
can just jump to the end.) All infinitesimal
volumes of water (dV =r dφ dz dr )
will have a centrifugal force dF c =(v 2 /r )dm
where r is the distance from
the axis of rotation, v=rω is the
speed of the tiny volume, and dm =ρ dV=ρr dφ dz dr
is its mass. You will note that the pressure
depends only on r and if you do all
the integration you will find that P c (r )=½rω 2 ρ .
I find that P c (r )=6.02x104
N/m2 =8.73 psi. Finally you can
simply add P c and P w
to get P (d )=(8.73+0.036d )
psi.
QUESTION:
I am in the middle writing what I
believe will be a series of speculative
fiction novels which take place in the far
future. I've been working on them for over
20 years while I was working my normal job.
Now that I'm disabled, I've become ENABLED
to focus 100% of my energies on completing
this first book. I'm fairly decent at basic
math but I was never taught how to utilize
mathematical formulae - I'm forever sorry
about it. As such, I was wondering if it's
possible to put a basic math series of steps
together for me to be able to fill in the
slots to figure out how fast a large
rotating ring would need to spin to achieve
the equivalent of 1G (9.81 m/s2) for an
orbiting ring with a radius of 132,500,000
kilometers? I have all the statistics of my
habitat ready to go, I just need to know
where to put which number and then a.s.m.d.
which to which to get to the needed number.
ANSWER:
The acceleration a of an object
moving with speed v on a circle of
radius R is a=v 2 /R
so v =√(Ra )=√(1.325x1011 x9.81)=1.14x106
m/s; the acceleration is a vector directed
at the center of the circle. Suppose that
object was a person of mass m . The
ring would have to exert a force F=ma
on her to keep her moving in the circle; she
would interpret the force she felt from the
ring as a force force pushing her against
the ring (centrifugal force), and she would
feel like she felt on earth. With a speed of
1.14x106 m/s, the rotational
velocity would be ω=v /(2πR )=1.3x10-5
revolutions/second=425 revolutions/year.
Have
you ever read
Ringworld by Larry Niven? Your
world sounds a lot like his.
QUESTION:
I have a piece of metal that is designed
to pass an electromagnet at very high speed.
The set up is that the electromagnet is
designed to manipulate this piece of metal
as it passes it. I need to know how quickly
an elctormagnet can be turned on and off and
on and off again in order to affect the
peice of metal - the important fact is it is
travelling very quickly indeed. For example,
if this were to be posed as a circlar
movement of the metal as the hand of a clock
and the electromagent sitting at one side
outside of the perimeter of the clockface,
say at 8 oclock, this metal piece could
easily be achieving speeds akin to a high
powered motor. Yet at the same time the
magnet musty be able to give full effect,
turn off and given this thought example turn
back on again at the next pass. Can an
electromagnet react this quickly?
ANSWER:
I don't know the details of your device, but
I can tell you that very rapid changes of
magnetic fields from electromagnets are very
difficult to achieve. The physics reason is
that electromagnets are always coils of wire
which have an inductance which, because of
Farraday's and Lenz's laws, resists changing
the current in the wire. Even a current in a
simple circuit cannot be instantaneously
turned on or off because of the inductance
of the single loop of current-carrying wire.
I would encourage you to study up on
inductance .
QUESTION:
When we
throw light on opposite wall in an elevator
by a torch, the light bends . Is is it
because the photons leaving the torch move
with constant velocity but the opposite wall
of the elevator accelerates and we see it as
bend?
ANSWER:
The light does not "bend" if the elevator is
moving with a constant speed, only if it is
accelerating. Also, you must imagine the
elevator in the middle of empty space, no
gravitational field. This is going to be one
of my long-winded answers because it is a
good learning opportunity on several levels.
I will change your question slightly because
it is more conventional to consider this
problem if the light source is outside the
elevator and enters through a small hole in
the side; the light moves perpendicular to
the velocity of the elevator as seen by an
observer at rest outside.
First
let's look at the case where you are in the
elevator moves with constant velocity v
and and there is also an observer at rest
watching you go by. You, in the elevator see
the light follow a path shown by the solid
red line. The length of this line is d =√(L 2 +(vt' )2 )
where t' is the time your clock
measures from when it entered until it hit
the opposite wall. The time it takes
t'=d /c where c is the
speed of light. It is a straight line
because v is constant and, for
example, the elevator has moved d /2
at t' /2. Meanwhile, the stationary
observer sees the light move straight across
(dashed red line) and strike the wall at the
same place because the wall has moved up in
the time it takes to go across which is
t=L /c where t is the
time measured using his clock. Therefore,
the clocks of the two observers do not run
at the same rate. Now, the time the observer
measures is t' =√(L 2 +d 2 )/c
where d=vt' is the distance the
elevator has traveled. Now, putting it all
together and doing some alegbra, (t' /t )2 =(1+β 2 (t' /t )2 )
which leads to t'=t /√(1-β 2 )
where β =v /c . This
is the famous time dilation formula.
Finally, let's consider the case of the
accelerating elevator. I will assume that
you, in the elevator, have measured the
acceleration to be a and the the
velocity at the instant that the light pulse
enters the elevator is zero; therefore,
v'=at' . The outside observer sees
exactly the same as he saw with constant
velocity: the light goes straight across and
then strikes the wall wherever you saw it
strike the wall. You, however, see the light
follow a parabolic path (it would seem*)
since the position of the elevator is z' =½at' 2 .
This was the thought experiment by Einstein
where he realized that there is no
experiment you can perform in an
accelerating system in empty space which
would not have the same result in a uniform
gravitational field, the equivalance
principle, a cornerstone of general
relativity.
*The
previous part of my answer (accelerating) is
technically wrong but conceptually correct.
The actual mathematics and physics become
quite technical. The main reason it is not
easy is that acceleration does not play the
important role it does in Newtonian
mechanics because it is not the same for
observers. Also contributing to the fact
that the path is not exactly a parabola is
that the speed everywhere on the path is
c whereas the speed increases as it
falls for an object with mass. Also
emphasizing the difficulty here is that,
when Einstein first calculated the
deflection of light by a star, he made an
error of a factor of two.
ADDED
THOUGHT: If you want to learn more
about how acceleration and force are treated
in relativity, see the links on
this subject on the FAQ page.
QUESTION:
I have had this theory for about three
years now and I was wondering if you could
see if this is valid or not. My question is,
if everyone person except the pilot on a
passenger plane were to all stand up and
jump at the same time, would the plane drop
or be disturbed at all? I feel like it would
be disturbed as average mass it at 60-70kg
each therefore surely it should move.
ANSWER:
This is not a "theory", it is
straightforward Newtonian phyics. For an
isolated system (which we can consider the
plane plus passengers to be, approximately)
the linear momentum must be conserved.
Suppose that there are 200 passengers each
with mass 80 kg in a Boeing 737-100 with
mass 28,000 kg. If each jumps so that they
go up 40 cm, they need to have a speed of
v=√(2gh)≈2.8 m/s. Their momentum is p =80x200x2.8=4.48x104 .
The plane must recoil downward with the same
magnitude of momentum, P=28,000V =4.48x104
so V =1.6 m/s=3.6 mph. The plane is
flying with an approximate speed of 600 mph,
so this would not be very violent, certainly
not nearly as much as bumpiness due to
turbulance; and when everyone returned to
the floor there would be no vertical speed
at all again.
QUESTION:
I understand that the speed at which
time passes varies with altitude, or more
accurately, with the strength of the
surrounding gravitational field. I also
understand that the speed at which time
passes changes for an observer as their
velocity changes (I.e. time dilation effect
as you approach the speed of light). So...
is the a difference in the speed time passes
between a point at sea level at the equator
versus at sea level near one of the poles?
ANSWER:
The first part of your question alludes to
gravitational time dilation. However, your
question asks only for the comparison of
clocks where the gravitational field is the
same. So my answer will discuss only the
normal time dilation. Suppose that you have
a clock on the north pole. You observe a
clock at the equator which has a speed of
about 464 m/s. When if you observe your
clock to tick off 1 s, the clock at the
equator will click off a time t =1/√(1-(464/3x108 )2 )≈1+1.2x10-12
s, a very tiny bit slower than your clock.
QUESTION:
I'm researching issues of physics and
wanted just a clarification, specifically
with regards to gravity. I've heard it said
what actually keeps gravity from crushing us
is the electromagnetic force - that the
electrons in our bodies are pushing against
the other atoms, which keeps us from being
crushed. Is this true?
ANSWER:
Well, interactions between atoms are
primarily electromagnetic and interactions
betweeen atoms can bind them together to
make a solid. If there are to be living
beings, that binding must be strong enough
so that it is not broken when the gravity
tries to push each atom in a solid down; any
evolution will certainly be driven by the
environment and gravity is a very important
part of the environment. An instructive
example might be an ice cube. It sits on the
table top not being crushed. But as it melts
the bonds break and the water spreads out on
the table as gravity "crushes" it. If you
did not have bones you would probably end up
pretty flattened on the ground.
QUESTION:
Can gravity be used to create energy,
and if so, can dark matter be used to create
energy?
ANSWER:
Drop a ball and it gains energy; gravity is
giving it that energy. Water from a
reservoir runs down and spins a turbine to
generate electricity; gravity is causing
that energy to be generated. You have to be
careful with the word "create" though. The
total amount of the energy in the universe
always remains the same and our
manipulations simply convert one kind of
energy to another. I wouldn't speculate
about dark matter, though, because nobody
actually knows what it is in any detail at
all.
QUESTION:
I keep reading/hearing on videos that
there have been experiments to prove that
the speed of light is the same for all
observers. A sample claim is "Since
Maxwell's work, numerous experiments have
been performed to test the prediction that
electromagnetic radiation travels at the
same speed for all observers - and none have
failed." The articles or videos never go
into any additional details on the
experiments.
ANSWER:
Let's first talk about the origins of the
idea that light in a vacuum has the same
speed regardless of the motion of the source
or the observer. Maxwell's equations, the
equations of electricity and magnetism,
unambiguously predict this behavior. When
Einstein developed his theory of special
relativity, perhaps the most important
postulate was the constancy of the speed of
light, c . The predictions made by
special relativity were revolutionary and
would be false if c were not a
constant for all observers. So, certainly
the most important and accurate experiments
verifying constant c were experiments
checking the predictions of special
relativity. No experiment was ever performed
which contradicted the predicitions.
If you are not familiar with special
relativity, look up
time dilation and
length contraction . Incidentally, no
prediction of Einstein's theory of general
relativity (our best theory of gravity) has
ever been found to not agree with
experiments.
QUESTION:
I only went through Physics 2 in college
so I am a relative rookie, albeit a long in
the tooth rookie. My Question is related to
Quantum Mechanics. If a particle, electron
or whatever is spinning one way; and we send
it through a magnetic field of one direction
or the other; and the object goes up or down
or right or left and supposedly we know it's
spin. I understand that sometimes, however,
we send it through again and the spin is
different. The question may be stupid but
for a rookie we think of simple things like
English on a pool ball, it can put the
opposite spin on an object it collides with.
So the question is do physicists consider
friction and it's effect on spinning objects
when measuring these things?
ANSWER:
This is a really hard question to answer.
Electrons do indeed behave like they have an
angular momentum. We generally refer to this
angular momentum as intrinsic spin or just
spin. We are therefore led to believe that
they are actually tiny spinning balls, just
like the English on a billiard ball. But, as
is almost always the case, we generally run
into trouble if we try to apply our
classical physics to quantum physics. For
example, consider a proton whose spin
quantum number is, like the electron, ½; its
actual angular momentum S is S =ℏ√[½
(1+½)]=0.87ℏ where ℏ is the
rationalized Planck constant. Suppose now
you put in reasonable numbers for the mass
and radius of a proton, assume that it is a
uniform sphere, and calculate the rate of
rotation. You find a totally nonphysical
answer; the surface of the sphere would have
to move faster than the speed of light!
Certainly when particles collide with other
their spins will be affected depending on
how they interact; to find out the details
you would have to do quantum mechanical
calculations. Friction is a wholly
macroscopic concept and has no existence in
a microscopic world. Also, the way we
observe spin is to observe the requisite
magnetic moment of the particle; recall that
a current loop has a field just like a tiny
bar magnet and a charged particle with spin
will be a tiny current loop. A bar magnet in
a uniform magnetic field will experience a
torque but no net force; you must make the
field non uniform for the particle to see a
force. It seems that you are a little hazy
on how this observation is done and you
should study the
Stern-Gehrlach experiment to sharpen up
that understanding.
QUESTION:
Is refraction an actual physical
phenomenon, or is it a strictly biological
perception, like with touch? To elaborate,
does light actually bend from its path when
its refractive medium is changed, or do we
only see it a such?
ANSWER:
What a strange question! Of course it is an
actual thing. If refraction didn't really
result in bending the path of light, your
eye would not work; your eye captures light
from some object and bends that light (using
a lens) to form an image on your retina
which is then interpreted by the brain.
Likewise with cameras and telescopes. I have
attached a photograph of light rays being
bent to a focus by a convex lens; the camera
which took this photo would not be able to
see a "perception". Maybe you are thinking
of whether a stick stuck into water is
really bent like it appears to be; no, but
it looks like it is because the refraction
is bending light rays.
QUESTION:
Newtons cradle: When you release 2 balls
and they send their kinetic energy down the
line,2 balls on opposite side move. But why
doesn't it just send one ball much higher?
ANSWER:
Newton's cradle is usually explained
assuming all collisions are perfectly
elastic. Therefore, both kinetic energy and
linear momentum must be conserved. Suppose
that two balls, each of mass m ,
move together with speed v 1
when they hit the other balls. Then the
initial linear momentum is p 1 =2mv 1
and the initial kinetic energy is K 1 =½(2m )v 1 2 =mv 1 2 .
Now, suppose a single ball exits the other
side with speed v 2 ; its
linear momentum must be equal to p 1 =2mv 1 =p 2 =mv 2 ,
so v 2 =2v 1 .
Now, what is the final kinetic enregy? K 2 =½mv 2 2 =2mv 1 2 ,
twice the initial energy. Therefore energy
is not conserved so this cannot happen.
QUESTION:
If a singularity is a small point inside
of a black hole where there’s infinite
density, does that mean if we take an iota
of that singularity (presuming its tangible
or transportable in its ways) would that be
able to sustain gravity within a spacecraft
or would that iota of singularity overpower
everything in its way
ANSWER:
You are
asking if you could use some small black
hole to provide a gravitational field in a
spaceship. Suppose that we put the black
hole a distance R =1000 m from the
captain's chair. What mass M would
be required to make the gravitational
acceleration the same as on the surface of
earth (9.8 m/s2 )? So, a=F /m =GM /r 2 =6.67x10-11 M /106 =9.8
or M =1.5x1017 kg. But
the field would decrease rapidly away from
the captains chair and increase rapidly
toward the black hole; e.g ., 10 m
away from the captains chair there would be
an acceleration 9.8(100/110)2 =8.1
m/s2 and 10 m closer to the black
hole 9.8(100/90)2 =12.1 m/s2 .
Also, how would you hold it there? Anything
close to the black hole would would
experience huge gravitational forces tending
to obliterate whatever was holding it.
QUESTION:
For example I'm going to home with my
car at 40mph and I open my car's lights.
These photons from my car won't go over the
speed of light and my 40mph wont gonna add
to this speed so what happend to this speed
to this energy. Where is it go?
ANSWER:
You are right, the photons emitted forward
from a source moving forward have more
energy than if the source were at rest. You
are also right that either photons will be
going the same speed. So, how could their
energies differ? The energy of a photon is E=hf=hc /λ where h
is Planck's constant, c is the
speed of light, f is the frequency
and λ is the wavelength. So the
wavelength of the gets shorter if the energy
gets larger. This is just the Doppler effect
for light and in this case it is referred to
as a blue shift. No energy has been lost.
(By the way, since 40 mph is so small
compared to c , it would be really
difficult to observe any effect at this
speed.)
QUESTION:
I have a question that is driving me
crazy. I'm a children's author; my stories
have appeared in Highlights magazine. I'm
working on a middle-grade novel in which the
protagonist plays lacrosse. The boy's father
is an engineer who is fond of explaining how
things work, and at one point he and his son
have a discussion about what keeps a
lacrosse ball in its pocket when a player is
cradling (i.e. the rocking motion a player
makes with a lacrosse stick to prevent the
ball from falling out of the pocket). I've
researched this endlessly online and most
explanations claim centrifugal force is what
keeps the ball in the pocket. But a few
claim that it's actually centripetal force.
I tend to think that is the correct answer:
the ball wants to fly off in the same path
as the head of the lacrosse stick as the
player swings it back and forth, but the
pocket's mesh netting gets in the way and
keeps the ball in the pocket. It would be
the same force that stops water in a bucket
from following the path of inertia and
spilling out and when you swing it over your
head in a big circle. BUT ... I'm simply not
sure.
ANSWER:
It all depends on who is observing the ball.
Let us, to keep the situation simple but
still illustrate the basic principles,
assume that the ball is moving in a
horizontal circle of radius R with
a speed v . We focus our attention
on the ball as we observe it watching the
game. (I have drawn three Cartesian
coordinates shown in green to help
visualize.) First, find all the forces
on the ball .
(Ignore the dashed vector labeled
C for now.) Two things
exert forces on the ball, the earth
(gravity) and the netting. The earth force
is called the weight as W
in the diagram; the netting force I have
broken into its vertical (F V )
and horizontal (F H )
components. The vertical component is equal
in magnitude and opposite in direction to
W to ensure that
the ball moves horizontally rather than fall
vertically as the weight would have it do;
the horizontal component (centripetal force)
keeps the ball moving in a circle and
provides the acceleration of a=v 2 /R
which points toward the center of the
circular path. You can now calculate the
force F H
from Newton's second law, F H =mv 2 /R
where m is the mass of the ball.
So, there is your answer: the ball continues
to be in contact with the net because the
force the net exerts on the ball causes the
ball to accelerate; as long as the player
keeps changing the direction and/or speed of
the ball, it can be kept in the net. This is
the centripetal argument.
But,
what about the centrifugal force which seems
to be the majority opinion? Is it wrong? Not
if you understand what a centrifugal force
is. Imagine that you are the ball and you
wish to use your coordinate system to
understand the physics. In your coordinate
system you are at rest so all the forces on
you, according to Newton's laws, must add up
to zero. What are all those forces? Exactly
the same as when you observed the ball from
the stands. There is nothing else but the
earth and the net which can exert forces on
you. What this means is that Newton's laws
are false if the observer is accelerating
relative to some coordinate system in which
they are true. So what do we do? We invent
and add in a force which compels the system
to obey Newton's laws. That force is called
a fictitious force and in this case it is
the so-called centrifugal force. So we add
the vector C =-F H
to make the sum of all vectors equal to zero
even though it does not exist! So this is
the centrifugal argument since clearly
C is holding the ball
to the net in this picture. Is it "cheating"
to just add a nonexistent force? Not if you
do it for a good reason. Newton's laws are
great and greatly help understanding what is
happening. As long as you know what you are
doing, if you can force Newton's laws to
work, you can often more easily solve the
problem.
Finally, just a little linguistics! Fugo
is Latin for I flee (the root of
fugitive); centrifugal is center-fleeing.
Peto is Latin for I seek ;
centripetal is center-seeking.
QUESTION:
I am 17 years old and I want to know why
like charges repel and unlike charges
attract. I know this is due to the exchange
of virtual photons but I am interested to
know the deep facts in simplified way. I
have read
this , but it is still complicated to me.
I would be grateful if you simplified it.
ANSWER:
Gravity has only one kind of force,
attractive, because there is only one kind
of mass. However, electrostatics has two
kinds of force, attractive and repulsive, so
there must be two different kinds of
electric charge. Measurements are able to
determine the forces quantitatively
(Coulomb's law). One can build up everything
we knew before the early 20th century about
classical electromagnetism by taking the
attractive/repulsive nature of electric
charge as axiomatic and based on
measurements; physics, after all, is an
experimental science at its heart. But just
like Einstein sought an answer to "why"
masses attracted each other, physicists (e.g.
Feynman, Dyson, Schwinger, Tomonaga, et
al .) developed quantum electrodynamics
(QED) which quantized the electromagnetic
field and therefore provided a theory for
the interaction between electric charges.
Unless you are an extraordinary genius, at
age 17 you will not have the mathematical
toolbox needed to understand the theory
beyond the qualitative understanding you
have (exchange of virtual photons); if you
are that extraordinary genius, you should
not need me to explain it to you!
QUESTION:
It's a simple question, but I can't find
an explanation. What is the total energy of
a block which is at rest, at ground level on
a flat, frictionless surface? Intuitively it
should be zero, since it is not moving (no
KE) and it's "height" above the ground is
effectively zero (so V=mgh=0) However, I
thought an object with mass must have
energy.
ANSWER:
More specifically, we would say that
(provided that we have chosen the ground to
be zero potential energy) the total
mechanical
energy of the block is zero.
Still, it has
rest mass energy mc 2 .
QUESTION:
I have a question for you. In your book
“Atoms and Photons and Quanta, Oh My!”
appendix B page B-2 Energy in special
relativity, you derive E=mc^2 using an
approach I’ve not seen before. What I don’t
get is the 2nd line where F = m
(d[v/(1-v^2/c^2)]/dt), where m is the rest
mass. Can you show the steps you took that
aren’t showing (to save space)? I don’t see
how you arranged this expression to include
the differential operator “d”. I’ve
seen/used several derivations using
differentials & integration by parts & so
forth from old physics texts & on-line.
ANSWER:
Here is a screen shot of the page the
questioner is referring to:
So,
keep in mind here that m is rest
mass, F is force, p is
linear momentum, K is kinetic
energy, W is work, v is
velocity, c is speed of light. In
classical mechanics, p=mv . But, if
v is not very small compared to
c , p must be redefined or else
it will not be a quantity which is conserved
if F =0. The correct definition in
relativity is p=mv /√(1-(v 2 /c 2 )).
So the second line (after F =dp /dt
) simply inserts the expression for p .
The third line multiplies and divides by dx
and rearranges so as to have v =dx /dt
appear as a factor. The fourth line
rearranges and replaces dx /dt
by v. Finally, the fifth line
is the result of evaluating the derivative
with respect to x :
m (d/dx )[v (1-(v 2 /c 2 ))-1/2]
=m (dv /dx )(1-(v 2 /c 2 )-1/2 +(-½)(v )(1- (v 2 /c 2 ))-3/2 (-2v /c 2 )(dv /dx )
=m (dv /dx )(1-(v 2 /c 2 )-1/2 [1+(v 2 /c 2 )/(1- (v 2 /c 2 )]
=m (dv /dx )(1-(v 2 /c 2 ))-1/2 [(1- (v 2 /c 2 ))-1 ]
=m (dv /dx )(1-(v 2 /c 2 ))-3/2
To do
this last step you need to know your
elementary differential calculus, (d/dz )(f (z )g (z ))=f (dg /dz )+g (df /dz )
[product rule] and (d/dz )f (g (y )=(dg /dy )(dy /dz )
[chain rule]; and, of course, a bit of
algebra.
QUESTION:
If a raindrop is affected by the earth's
gravity and it should accelerate at a rate
of g(9.8m/s^2) downward. Why does when it
hits the ground it doesn't result in a big
impact?
ANSWER:
Yes, a raindrop experiences the downward
force of its weight; but it also experiences
an upward force due to air drag. The drag
force depends on the drops size and speed
and increases like the square of the speed.
Eventually the speed reaches a point where
the upward force of the drag is equal the
the downward force of the weight and the
drop stops accelerating. This speed, called
the terminal velocity, is usually about
15-25 mph depending on the size of the drop.
QUESTION:
Can gravity efect electron velocity?
ANSWER:
Yes, see a very recent answer . But
you would have trouble seeing it. An
electron which is accelerated over 1 volt
has a speed of about 6x105
m/s=1.3 million mph!
QUESTION:
So I know that gravity is caused by mass
curving spacetime and all mass following
those curves. That would explain how
something is pulling something without
actually touching it. But a question arises.
If gravity is a curve in spacetime, how do
magnets attract each other? Yes, I know that
magnets create magnetic fields but what are
those fields exactly? They just feel like
some magic arrows that go in and out of
magnets. I haven't really seen anyone
explain this so if you can help me
understand this I'd be grateful.
ANSWER:
There are four fundamental forces in
physics, gravity, electromagnetism, the
strong nuclear force, and the weak nuclear
force. To answer your question, I only need
to discuss gravity and electromagnetism.
Electric forces and magnetic forces are two
faces of one force, the electromagnetic
force. There are two kinds of electric
charges which are characterized by assigning
a sign, plus or minus. Two electric charges
will repel (attract) each other if they are
of the same (opposite) sign; they do so
because of the electric fields they produce.
But an electric charge can also cause
another type of field if they are moving,
called a magnetic field, which we can
associate with electric currents. The
simplest electric current is simply a tiny
current loop as shown in the figure. Notice
that the magnetic field points toward the
loop at the bottom and away from the loop at
the top; the field of this loop looks just
like the field of a tiny bar magnet as
suggested by labeling below the loop S
(south) and above N (north). Many atoms look
like a current loop and hence like a tiny
bar magnet. In most materials all these
atomic loops are randomly oriented such that
they cancel each other out. In a few very
special materials the fields of the atoms
tend to line up with the fields of their
neighbors and the whole object is said to be
magnetized. The electromagnetic field has
also been quantized which means that we
understand microscopically the details of
how the forces are communicated among
currents and charges; the field quanta are
called photons (tiny parcels of light,
really) and these are the "messengers".
Gravity may be
interpreted, as you note, as geometry, a
warping of spacetime. However, general
relativity (Einstein's theory of gravity) is
also a field theory; nobody has yet been
able to quantize the field. When (if?)
gravity is successfully quantized, the field
quanta would be called gravitons./font>
One additional detail:
electromagnetic and weak nuclear forces have
been unified, shown to be different faces of
a single force, the electroweak interaction.
QUESTION:
Can gravity slow down the atom ?
ANSWER:
You are asking, essentially, if atoms feel
gravitational forces. Anything which has
mass will feel the gravitational force and
atoms have mass; so, yes, atoms experience
the gravitational force and thus can be made
to slow down. For example, if you shoot an
atom straight upwards, it will slow down
just like a baseball thrown upward with the
same speed. The very fact that we have an
atmosphere is proof that gravity is acting,
keeping all those atoms from speeding off
into space.
QUESTION:
What is the smallest change in frequency
within the visible light spectrum that the
human eye/brain can discern? My new TV says
it can show over a billion colors. Can
humans even see 1 billion colors!
ANSWER:
The way a tv works is by mixing the three
primary colors, red, green, and blue, in
varying amounts. There are an infinite
number of possible ratios. All the colors
you can see are not all the possible
wavelengths you can see (also an infinite
number) but mixtures of many many
wavelengths. When the manufacture refers to
"billions" he is probably referring to how
accurately the electronics can mix the
three. For example, if the relative amounts
of the three were 10, 7, 3, the electronics
would have an uncertainity in how accurately
they could set these numbers; maybe the
numbers could only be set to a 0.01%
accuracy, the ten would be 10±0.0001. Then
you could estimate the number of
possibilities you could have over the whole
visible spectrum.
QUESTION:
I have been thinking about this for
countless hours and though not a physics
student, and haven’t taken any classes of
the sort I have wanted to know this
question: we know black holes are brought
down to a point of singularity. So could and
atom being so small do the same? And if this
is true would that be the last atom we see
on the periodic table?
ANSWER:
The force which holds a star together is
gravity; a normal star does not contract to
a black hole because it has collapsed to the
point where burning hydrogen (fusion) in the
star heats it up so that the pressure
balances the gravity trying to push it
smaller. As the fuel runs out, the star
starts getting smaller. At this stage,
various things can happen, depending on
various conditions, mainly the total mass of
the star; one possible thing is eventually
collapsing all the way to a black hole. An
atom is a completely different kettle of
fish. We should really look at the nucleus
since that is where nearly all the mass is.
Unlike a star, energy is not being produced
in the nucleus, its energy remains constant.
An atomic nucleus is held together not by
gravity but by the strong nuclear force.
Unlike gravity which is a very long-range
force, the strong interaction is very short
range, so each proton and neutron see only
their near neighbors, not all the other
particles in the nucleus. The result is that
the heavier a nucleus becomes, on average,
the less bound it comes so its tendency is
to break apart, not collapse. By the way,
the nucleus does not collapse because the
nuclear force at extremely short distances
becomes repulsive, not attractive.
QUESTION:
How much does a thought weigh? And how
much does one molecule of carbon dioxide
weigh? I want to compare the two.
ANSWER:
A molecule of CO2 has a mass of
about 7.3x10-26 kg (about 1.6x10-25
lb). A "thought" is a collection of billions
of electrical impulses and chemical reations
in the brain. Even if these could be
assigned a "weight", which they can't, all
"thoughts" are not the same and would have
different weights.
QUESTION:
Franklin first coined the term negative
and positive in 1750 and Volta invented the
battery in 1800. Did Volta assign a positive
and negative end to the battery? Or did he
think the negative and positive applied to
electrostatics and not "flowing"
electricity. You see where I am going with
this. I need to go up the chain to J.J.
Thompson to see where the goof up was on
adoped circuit diagrams. The explanations I
have read are non sense. They do not
actually describe why the convention had to
be reversed. All the examples I read use
circular reasoning and thus far no smoking
gun of how and when the goof up happened. I
thought if I could start with Volta I might
be able to unravel it.
ANSWER:
I really don't understand your perception of
"goof up". There are two different "kinds"of
electric charge. Coulomb's law shows that
the magnitude of the force F
which a point charge q feels due to
the presence of another point charge Q
is proportional to the product of the
charges Qq and inversely to the
distance r between them, F ∝Qq /r 2 .
Also, if the charges are of the same "kind"
they repel and of different "kind" they
attract; since attract/repel result in
exactly different directions for the vector
force if we randomly assign signs to the two
"kinds", we can write F ∝1 r Qq /r 2
where 1 r
is a unit vector in the direction of the
vector r from
Q to q , 1 r =r /r.
This proportionality does not
depend on which "kind" we call positive and
which we call negative, there is no right
and wrong choice for the sign for an
electron. Franklin made a random choice
which resulted in electrons as having
negative charge; he might have made a
different choice if he knew electrons
existed and were usually the charge carriers
in currents (nobody knew then), but it was
not a wrong choice. Later, when currents and
magnetic fields were studied, all of
electromagnetism simply continued the
original choice with no problem whatever.
All this means that positive electric
currents in wires actually are almost
electrons moving in the opposite direction
which is endlessy confusing to students. I
really don't know anything about Thompson
correcting "a goof up" in circuit diagrams
other than pointing out that currents shown
moving in one direction (from + to - battery
terminals) is actually electrons moving in
the opposite directions. No physics was
changed.
QUESTION:
What will be the motion of the earth if
the sun disappears after t=0 sec?. I was
asked this question in an exam and the
answer is not tangential to the motion
ANSWER:
It is assumed, although never very
accurately measured, that the speed of
gravity is the same as the speed of light.
Therefore, the earth would continue in its
orbit until t =8.3 minutes, the time
it takes light from the sun to reach the
earth. At that time the earth would go dark
and proceed in an approximately straight
trajectory.
QUESTION:
How do gravitons that exit a certain
mass cause other objects to be attracted to
that mass. I can imagine Einstein's idea of
curved space being the cause for the seeming
attraction, but what does that have to do
with gravitons?
ANSWER:
There is no successful theory of quantum
gravity; therefore gravitons are
hypothetical particles, never observed, just
an idea. Gravity is the only one of the four
fundamental forces of nature which has not
been quantized. See an
earlier answer and the links in that
answer. You will see that the theory of
general relativity is not unambigously
described by the notion of warped space-time
but is also a field theory. Any field theory
should be able to be quantized; in a
quantized field theory the "messenger" of
the field would be called a graviton, just
as the messenger for the electromagnetic
field is the photon, and is the gluon for
the strong nuclear field.
QUESTION:
Why does a rounded pencil I put on a
flat ground make a small harmonic movement
before stay steady?
ANSWER:
One side of the pencil is heavier than the
other. For example, if you had a uniform
ball and glued a weight to its surface, it
would always rotate so that weight would go
to the ground. With some pencils the lead in
the middle is not exactly on the axis. The
"lead" is actually graphite (carbon) with a
density of 2.26 g/cm3 ; wood has a
density of about 0.6 g/cm3 . So
the lead will seek its lowest p
ossible
point it can possibly find. The figure
illustrates this situation. The weight of
the lead will cause the pencil to roll to
the right; but when it gets there it will be
moving and so the pencil will continuing to
roll until the lead stops and then begins
rolling to the right. It will oscillate
forever if there is no friction; but there
is so it will quickly damp out with the lead
directly below the center axis of the
pencil.
QUESTION:
I am a glass artist. When I fume gold
(vaporize in the torch flame) and it
collects onto the borosilicate glass it
appears pink/purple. If I encase this gold
with clear, it then looks green. I can
throughly manipulate the glass and maintain
the green color. I read that Faraday
obtained green gold in a thin film with
stress on the film. I also fumed gold onto a
rod and stretched it at a fairly cold glass
working temperature to stress the gold and
avoid annealing the gold. The gold took on a
green appearance. I have been doing this for
over 25 years and have not been able to find
out why gold turns green when encased (or
stressed like with the stretch) I am asking
a physicist because this has to do with
light and gold, plasmon resonance etc are
pretty heavy topics. If you can answer this
question I’d love to gift you one of my art
works! Why does a thin film of gold turn
green when encased in glass?
ANSWER:
I am not sure why it is pink/purple when you
first deposit it. One video suggested that
has to do with the amount of oxygen in the
film; it is also possible that it is a
thin-film interference effect where the
thickness of the film is an integral
multiple of wavelengths of the color you are
seeing and the light of that wavelength
which is reflected from the front and back
of the film add up to become the dominant
light you see (the same thing going on in
the pretty colors you see on an oil slick).
There such things which use thin-film
interference, for example the magenta
interference filter shown in the figure.
Now, to
the green issue. Why is gold the color we
see with just a block of gold? The reason is
that when white light strikes the surface,
red/yellow/orange are preferentially
reflected back making the gold look golden.
But some of the light will continue into the
bulk of the gold, and, coincidentally,
red/yellow/orange are preferentially
absorbed. This means that once you get to a
certain depth in the block the only light
surviving is the blue/green part of the
spectrum; eventually all the light gets
absorbed and you will see no light coming
from the back side which originated on the
other side. However, if you have a "block"
which is thin enough, smaller than the
distance where everything is absorbed but
thick enough that all the red/yellow/orange
has been absorbed, the film will look
blue/green. I am quite sure that this is
what is going on but, since I don't know the
geometry of your art, I do not have a
detailed explanation for your particular
situation; e.g ., does it look green
when seen from any direction? Maybe if you
sent a photo or two I could go farther.
I have
a question for you. I am a
stained-glass hobbiest and find that red
glass is often more expensive than other
colors. I have been told that the reason is
that gold is required to get a rich, pure
red glass. Do you know if that is true
and/or why?
QUESTION:
A thought occurred to me. From what I
know about particle accelerators, scientists
speed up accelerator particles of matter and
crash them into slower moving or stationary
particles of matter so that they can examine
the breaking apart of all into other
elementary particles. This then allows them
to find more and more basic particles of
matter. I have read that the speeds these
accelerator particles are hoped to reach is
the speed of light, hence the effort to
build more powerful particle accelerators.
So, why not simply have photons of light
(that are travelling at the speed of light
since they are the elementary particles of
light) crash into those other particles of
matter? Why use so much energy trying to
push accelerator particles to reach the
speed of light, when we have light in the
first place and photons already moving at
186,000 miles per second?
ANSWER:
It is not the speed of the incident
particles which matters, it is their energy.
There are no good available sources of
high-energy, high-intensity photons. Also,
photons interact only via the
electromagnetic force and we often want
projectiles which interact via the strong
nuclear interaction. So the accelerators are
actually poorly named because they do almost
no change in velocity when they increase
their energy by a large amount; I have
maintained that they would better named
energizers. For example, a proton with speed
99.999% the speed of light accelerated to
99.9999% the speed of light increases its
speed by 0.0009% but its energy increases by
216%.
QUESTION:
A photon hits an atom perpendicularly to
its speed v and it is absorbed as is known
immediately. So it can not act after the
time when it is not perpendicular (e.g. the
force is 0 after that). Certainly all its
energy goes to the atom when it is
perpendicular to v.
But the
impact (absorption) applies a force on the
atom and it is postulated that a
perpendicular force can not do work. So work
has not been done on the atom. Consequently
its energy can not change. If all photon
energy goes to increase internal energy of
an orbit electron from Eo to E1 how then is
the DIRECTION of velocity of the atom
changed? Isn’t this a contradiction?
ANSWER:
Your first paragraph really makes no sense.
Talking about "perpendicular" in a collision
really has no meaning; I think what you mean
is a head-on collision, one where the recoil
of the atom is in the same direction as the
incident photon. In a case like this you do
not want to think about force and work.
Rather you should ask if energy and momentum
are conserved; if so, you should solve your
problem that way. Suppose that the system we
are interested in is the photon (mass zero,
energy E=hf , and momentum p=hf /c )
plus the atom (rest mass M , energy
E =Mc 2 , and at
rest p =0). Because there are no
external forces on this system, its total
energy and momentum must remain constant.
After the absorption, the mass of the
excited atom is M' , its kinetic
energy is K , and its momentum is
P' ; there is no need to think about
what goes on microscopcally inside the atom.
So we have two equations:
E 1 =E 2
or
hf +Mc 2 =M'c 2 +K
p 1 =p 2
or
hf /c=P'.
The
notions of force and work are seldom useful
when looking at atomic sizes. In this case,
the photon carries energy into the atom so
the atom's energy has to increase.
QUESTION:
If l video an energy source such as a
man-made Ultraviolet Curing light, can
watching the video hurt your eyes? And is
there a way of capturing this light to make
it even though it is invisible?
ANSWER:
The video camera is not sensitive to UV
light and therefore records none. The
monitor is designed to emit almost entirely
visible, not UV light and therefore radiates
none. So no, the video will not hurt your
eyes. There are certainly detectors which do
detect UV radiation but it is not really
'capturing' it in the sense that you could
'make' it.
QUESTION:
This question has resulted from
considerable back-and-forth communications
between me and the questioner. The essence
of the question is this: Two cylinders are
rigidly attached to each other. The smaller
has a radius R and its center axis
(B in the figures below) is separated by a
distance d from the center axis (A)
of the larger. A torque is applied about the
axis of the smaller cylinder but exerts no
net force. Find the resulting torque about
the axis of the larger cylinder.
ANSWER:
The figure shows one possible way of
exerting a torque about point B. The four
tangential forces all have magnitude F /4
so that the torque is τ B =RF.
There is a net force of zero. The
torques about the point A are:
τ 1 =RF /4
τ 2 =(d+R )F /4
τ 3 =RF /4
τ 4 =-(d-R )F /4
When
all four torques are added,
τ A =RF=τ B .
A
more general approach to the problem would
be to have a uniform tangential force all
around the perimeter of the smaller cylinder
as shown in the second figure. If we write
the magnitude of the vector dF
as dF =[F /(2π )]dθ ,
the integral ∫dF from 0 to 2π
will be F and τ B =RF.
Now the torque due to dF
about A will be dτ A =|r × dF | dθ=[RF /(2π )][(d sinθ /R )+1]dθ.
Integrating this from 0 to 2π
yields, again, τ A =RF
because ∫sinθ dθ= 0 over one
whole cycle.
There are two
important provisos here: there must be no
other forces present which exert torques
about axis A and the forces exerting the
torque about axis B must sum to zero net
force.
Sometimes I like to
interject examples of how science is done
and sometimes progress hits snags. Most
often preconceptions are the biggest
stumbling blocks. When I first started to
work on this problem I assumed that the
answer had to depend on d . As you
can see, this is not a particularly
difficult problem. I quickly tried the
general approach and got the answer as I did
above; but I was so sure that d
should be in there somewhere that I assumed
that I had done something wrong. So I did
the calculation for each quadrant
individually; these are tedious
calculations, not difficult but certainly
prone to algebra and arithmetic errors. I
spent hours! Finally, when I was sure that
everything was right I found
τ A =RF.
That was when I tried the simplified
torque with four forces (above) to convince
myself that it was correct.
QUESTION:
I am trying to clarify in my head an
idea regarding electromagnetic force. It is
usually explained in very superficial terms.
One way to explain my question is this: I
understand that two positively charged
particles will repel each other, for example
two protons. They both have a positive
charge, but if they are at rest there will
be no magnetic field. Do electrical charges
repel like magnetic charges? Are they the
same thing? We measure magnetic strength in
one set of units such as Teslas but
electrical potential in another set of
units. I understand the strength of a
permanent magnet to be somehow related to
the arrangement of its electrical charges.
Is that correct? Is a proton at rest a
permanent magnet? Is there some simple
relationship between a static electrical
charge and a permanent magnet? I’m sorry the
question is so long but it is all part of
one big question.
ANSWER:
Your question violates site ground rules for
"single, concise, well-focused questions" in
spite of your saying it is "one big
question"; even so, it is neither concise
nor well-focused. I am sorry, but I cannot
give you an entire disquisition on
electromagnetic theory in a single concise
answer. You have a number of misconceptions.
First, there is no such thing as a magnetic
charge. I always like to like to refer to an
answer to a question I answered long ago:
Answer: The laws of
electromagnetism are perfectly symmetric: a
changing magnetic field causes an electric
field and a changing electric field causes a
magnetic field. The first of these is called
Faraday’s law and the second is part of
Ampere’s law. You seem to think that only a
permanent magnet is magnetism. In fact, any
moving electric charge causes a magnetic
field. The most common source of magnetic
fields is simply an electric current. Here
are some facts about electric and magnetic
fields: • electric charges cause
electric fields, • electric currents
(moving charges) cause magnetic fields,
• changing magnetic fields cause electric
fields, and • changing electric fields
cause magnetic fields.
Magnetic
fields are not caused by a charge like
electric fields are, they are caused either
by electric currents (moving charges) or
changing electric fields. So then, what is a
permanent magnet? On the atomic level, think
of electron orbits as little currents and
electrons as tiny spinning charges, also
currents. In a material where all these tiny
magnets align with each other results in a
sum of all their magnetic fields. Regarding
your writing “but if they are at rest there
will be no magnetic field”, this has nothing
to do with any force they exert on each
other by virtual of their charges. Indeed
protons do have a tiny magnetic moment
because they can be roughly be thought of as
spinning, but those moments are very weak.
If you want to really understand all this,
you need to seriously study
electromagnetism.
QUESTION:
As solar energy is becoming a major,
maybe eventually the main source of
electricity, I wonder what the effects of
capturing and retaining significantly more
of the Sun's energy in the system of our
planet could be. Would the overall effect be
comparable to the warming effect of
greenhouse gases, where less energy is
radiated back to Space? What about other
forms of renewable energy, that ultimately
just harness the Sun's energy?
ANSWER:
In greenhouse warming, radiation is absorbed
by the earth and then radiated back into the
atmosphere; some of that energy goes back
into space and some bounces around in the
atmosphere, heating it. Factors, including
the amount of CO2 in the
atmosphere, determine the fraction of
reradiated energy which is trapped. Now, if
we capture this energy and store it in
batteries, it stores the energy in a form
which is not heat; then this energy is used
to drive cars, run our utilities, lift
material to build buildings, etc .,
all of which will convert a fraction of the
energy to heat, but a much smaller amount
than if all that energy were absorbed by the
ground and then was reradiated into the
atmosphere. And, using this energy instead
of fossil-fuel burning will further help the
situation by not adding to the CO2
in the atmosphere.
QUESTION:
I am trying to calculate the energy
consumed by an electric bike when it is
charging at home. It has a 48V, 14Ah battery
and takes about 6 hours to charge. Is the
electricity in my power bill going to be 672
watt hours (48 x 14) or ~4,000 watt hours
(48 x 14 x ~6)?
ANSWER:
I got exactly the same answer as you but
took a different path to get there. Since an
ampere is a Coulomb/second (C/s), 1 A·h=3600
C, so 14 A·h=5.04x104 C. The
energy to move that charge across a
potential difference of 48 V is E=QV =48x5.04x104 =2.4192x106
J=2.4192x106 W·s(1h/3600 s)=672
W·h.
QUESTION:
I learnt about the Rutherford experiment
where they disproved the plum pudding model
for atoms where they used a gold leaf which
was apparently 100 atoms thick. I have since
read about a gold leaf that was 2 atoms
thick. If atoms are mostly empty space why
wouldn't you be able to see right through
said gold leaf.
ANSWER:
Gold looks golden because it reflects
yellow, orange, and red if illuminated with
white light. Interestingly, if an object
reflects a particular color, it is also a
very good absorber of that same color. What
that means is that, for gold, yellow,
orange, and red will be exceedingly unlikely
to get through gold leaf; if any light at
all gets through it will be other colors.
Indeed, light transmitted through a gold
foil will be bluish in color. Typical gold
leaf is more than ten atoms thick and two
atoms is a very recent achievement, I
believe. I would think that the thinner the
foil, the more of the gold colors would also
transmit through.
QUESTION:
The Laws of physics say nothing can
travel faster than the speed of light. Given
the universe is expanding, two stars on
opposite sides of the universe would be
travelling away from each other at much
faster than the speed of light. From the
perspective of planet a, planet b is
travelling faster than the speed of light;
how is that possible.
ANSWER:
I presume that you are using Galilean
relativity to deduce the speeds of two
objects relative to each other if you know
their speeds relative to something which you
will call "at rest". Let's do an example:
suppose that, relative to earth planet a is
moving away from you with a velocity V ay =0.8c
and, on the other side of the universe,
planet b is moving away from you with a
velocity V by =-0.8c
where c is the speed of light.
You conclude that the velocity of a relative
to b is V ab =1.6c
because the velocity addition formula for
Galilean relativity is V ab =V ay +V yb
and V yb =-V by
(or, you might say, just by "common sense").
However, for objects moving with speeds
comparable to the speed of light, Galilean
relativity, and indeed, all of Newtonian
mechanics, are not valid laws of physics.
The correct
velocity addition formula is V ab =(V ay +V yb
)/[1+|V ay V yb |/c 2 ]=1.6c /1.64=0.976c .
QUESTION:
I am emailing you with a specific
question my 5 year daughter asked my husband
and I about rainbows. I’m hoping you might
be able to help us, or point us in a
direction that would help us with an
explanation. We found you online, and we
need a professional. We showed our daughter
a prism refracting the light just like a
raindrop would. She asked, if we hung 100
prisms from the ceiling that should make 100
little rainbows, right? I said correct.
BUT!!!! she asked "If each raindrop can
refract the sunlight into the colors of a
rainbow. Then why aren't a million rainbows
in the sky? Why do we only see one big
rainbow? How do the reds stay together, the
orange together etc...Please help. I haven't
been able to find any information on the
web.
ANSWER:
The prism demonstrates dispersion ,
the fact that different colors of light
travel with different speeds in any medium.
The prism is made of glass and the rain drop
is water; both demonstrate dispersion but
the details are different because the rain
drop is a sphere. All the light rays from
the sun come in essentially parallel. In my
figure, several rays are shown and one is
followed as it refracts when it enters the
water, disperses into the colors, and
strikes the back of the drop; when it
reaches the back, part of it goes back into
the air (not drawn here), and part of it is
reflected back through the water as shown;
then it reaches the surface again and part
of it refracts back into the air as shown,
part of it reflects into the water again
(not shown). For this particular ray, and
all others from millions and millions of
other raindrops for
which
light enters at the same place on the
surface, the emerging light constitutes the
rainbow you see. But, note that the angle
between the direction of this incoming ray
and the exiting "rainbow rays" is 42°. But,
what if I had chosen one of the other rays?
I would get a different path for the ray
bouncing around inside the drop and the
rainbow would emerge in a different
direction. So, the whole sky would be
covered in overlapping rainbows! And, it
actually is. But not all of these infinite
rainbows have the same brightness; it turns
out that at 42° the emerging light is the
brightest. You can actually derive this
angle of maximum brightness if you know
calculus and trigonometry; see the Wikepedia
article on
Rainbow . So, if you are standing in just
the right place to see the colors from this
drop, then all other drops which sent their
rainbow in your direction will also be seen
but as coming from different points in the
sky. The locus of those points will be a
perfect circle. If you are in an airplane it
is possible to see the full circle. One last
issue: it turns out that we often see a
second, dimmer rainbow above the bright one.
The reason for this is that some of the
incident rays will reflect twice instead of
once inside the drop before they come out;
these form the second rainbow. I hope this
is not too complicated for a five-year old,
but hopefully you can digest it and explain
it in terms she would understand.
QUESTION:
It says in a lot of places that
wavelength is inversely proportional to
frequency. Would it not be possible to
increase the speed of the wave, (and
therefore the frequency) so more waves pass
through a point per second, without
increasing the wavelength?
ANSWER:
In general, v=fλ and you can hold
any one of the three constant and ask what
happens to the other two. Here is a simple
example: A guitar string of length L
vibrating with its fundamental frequency is
a standing wave with λ =2L.
You can change the speed of the wave by
changing the tension in the string (which is
what the tuning pegs do). But the wavelength
cannot change so the frequency must.
QUESTION:
Does the gravity of earth comes from
molten core.and if so can we create
artifital gravity by making molten core
simillar to earth by rotating it at high
speed.
ANSWER:
Gravity is caused by mass. The earth's core
constitutes approximately 1/3 of its total
mass and rotational speed has nothing to do
with it. Neither does the "moltenness" have
anything to do with gravity. So if you had a
molten core alone, rapidly rotating, its
gravitational field would be just the same
as if you had a solid sphere of the same
mass; there would be nothing "artificial"
about it.
QUESTION:
what force are acting on a ping pong
ball at the top of its bounce is there just
gravity or is the kinetic energy from the
ball still at play.
ANSWER:
The first thing to note is that kinetic
energy is not a force. And I don't know if
when you say "top of its bounce" whether if
is in purely vertical motion or is in a
trajectory during which it will still have a
speed at the top. If it is the former, it is
at rest and the only force on it is its
weight (gravity). If it is the latter there
will also be an air drag force opposite its
direction of motion. (Air drag is quite
important for a ping pong ball.)
QUESTION:
My question is about the actual energy
consumption that is required to produce a
certain amount of electrical output: Framing
the setup for the question: If I have an
electrical generator that can produce 1kw of
electrical energy and I spin the armature
with NO load, the amount of work I am
producing to spin the generator is fairly
nominal. It's simply the frictional losses
of the bearings, the amount of energy to
accelerate the mass of the armature to a
particular rpm (lets say 1000 rpm), and
perhaps some wind resistance. This will have
some numerical value, I am assuming in
joules or horsepower. (please correct me if
I am wrong).
Now, if
I throw a switch that connects 10, 100 watt
lightbulbs. The back EMF is tremendous.
Anyone who has tried to spin a generator by
hand can attest to how difficult it is to
try and light just 1 100 watt light bulb.
Assuming the generator is 98% efficient; how
much work (or energy) is required to produce
the 1kw of electricity to light the bulbs? I
would like to compare the "NO LOAD" power
requirement to spin the armature to the
"FULL LOAD" requirement to spin the
armature.
ANSWER:
I don't think there is much of a mystery
here. Let's forget about your first
paragraph because when you say 98%
efficient, that presumably includes all the
losses you enumerate. This generater can
generate 1 kW of power at 1000 rpm. Now you
are asking it to give you 1 kW if you attach
10 100 W bulbs. You ask how much energy you
have to put in but that is not the right
thing to ask; you should ask how much energy
per second (Watts) you need to supply,
power. Since the efficiency is 98%, you need
to put in 1 kW/0.98=1.02 kW of power.
QUESTION:
Hello, i was curious why science hasn't
utilized the physics of "slipping" into
generating energy... I understand there are
road blocks, but is the potential of
slipping to generate energy not extreme
enough to warrant overcoming them? Can you
not generate a lot of energy with slipping?
or is it just that in comparison to other
forms of generating energy, it just doesn't
compare and it's not as exponentially
capable of generating energy as im assuming.
Like when i imagine moving a large object at
high speeds once you've propelled it to the
speed you want it wouldn't the physics of
slip, give it the continuous push it needs
to maintain or even exceed the level you had
put it too..and would not the physics of
slipping then take over the propelling of
the object...and if we could create or
find..a suitable force that creates slip and
resists wear and tear long enough to warrant
using would that not enable us to
drastically improve our capacity to generate
energy??
ANSWER:
I presume you mean friction? Friction
generally takes energy away from a system.
For example, a box with kinetic energy
sliding across a floor stops because
friction takes the energy away from it. That
energy which the box had does not really
disappear but shows up as heat (the floor
and box will get a little warmer) and sound
(while sliding you can hear it). Are you
suggesting that we could use the heat and
sound energy somehow? Why not just use the
energy the box already had? Usually we try
to eliminate friction as much as we can to
maximize other sources. One example is the
brakes of a convential car which transform
all the kinetic energy of the car into heat;
electric and hybrid cars use magnetic
induction to put that kinetic energy into
the batteries rather than into heat in the
drums or disks of brakes.
QUESTION:
Hi, today I read an article on Armstrong
Limit and I have a question about it, if
saliva/tears boil in 36 degrees Celsius in
the Armstrong Limit why is it a problem,
does it cause any health issues?
ANSWER:
Your body is about 60% water. How could you
ask if it would be any health issues if all
that water started boiling?
QUESTION:
If you could dissassemble a human atom
by atom, and then store the human in his or
her dissassembled atom state, could you
reassemble said human in the future without
them aging essentially creating some form of
atomical time travel? Also if it were
achievable, would they retain their
conciousness, memories and all that?
ANSWER:
The answer is no. To read a discussion of
the physics of the impossibility of the
"beam me up Scotty" problem, I suggest
The Physics of Star Trek by Lawrence
Krauss, HarperCollins Publishers, 1996.
QUESTION:
I am wondering the following; 1.
Will the electron of atoms always apear when
looked for in its cloud, or can it sometimes
be "missed"? 2. Will it actually only be
in two places at once or could it be where
ever it is looked for simultaneously? 3.
Is the "quality" or "strength" of an
electron compromised when seen
simultaneously in its different locations?
ANSWER:
The problem with your questions is that you
have the idea that you can even talk about
the electron being at any particular
position. All you can know, until you make a
measurment, is the probablity of the
electron being in any particular arbitrarily
small volume. A measurement means that you
look at a particular small volume and if the
probability of finding it there is 1% you
will only see it once every 100 times you
look there. It is never in two places at the
same time, it is really smeared over all
places until you observe it and then it is
where you observe it to be. It can never be
observed at the same time in two places;
otherwise there would mysteriously be two
electrons even if you know there can only be
one.
QUESTION:
I have a potentially unusual question to
which I am currently unable to find an
answer and I thought that you may be able to
help. I was riding my motorcycle today and
turned a corner and the front tyre came into
contact with a large patch of gravel on the
surface of the road. I had touched the front
brake immediately prior to this and, as the
front tyre hit the gravel the bike began to
slide and to pitch sideways. I put both feet
on the ground and steadied the bike,
effectively holding it upright and avoiding
a spill. Now for the question: My body is
very sore after around 5 hours after the
event and this has made me wonder what kind
of energy did I absorb in my attempt to
avoid the accident? The bike weighs 220kg, I
weigh 125kg, I was travelling at 20mph
(after I had released the brakes )on
relatively flat ground and stopped the
momentum of the bike within 2.5 yards (after
I had released the brake). I am completely
unable to fathom a calculation for this, I’m
afraid . Can you help me at all?
ANSWER:
So, we have m =345 kg moving with a
speed v =20 mph=8.94 m/s and
stopping in d =2.5 yd=2.3 m. The
initial kinetic energy was E 1 =½mv 2 =2.73x104
J and the initial linear momentum was p 1 =mv =3.08x103
kg·m/s; the final energy and momentum are
zero. Change in energy is equal to work
done, so W=-E 1 =-Fd= -2.3F
where F is the average force
exerted by the ground on you (frictional
drag); so F =1.19x104 N.
You could also say that the change in
momentum equals the Ft where t
is the time to stop, so t =1.19x104 /3.08x103 =3.86
s. The average acceleration over the stop
was F /m =34.5 m/s2 ,
about 3.5 times the acceleration due to
gravity, 9.8 m/s2 . So why were
you so sore? Because Newton's third law says
if the ground exerts a force on you, then
you exerted an equal but opposite force on
the ground. So you were exerting a force on
the order of 10,000 N, about 2000 pounds,
for about 4 seconds. That sounds like an
awful lot to me. You said the brakes were
not engated as you traveled 2.5 yards but
was the bike skidding sideways? That would
have been like having your brakes on and
have lessened the amount of force you needed
to apply to stop the ride.
QUESTION:
My question is about Newton's 3rd law. I
understand it pretty good, so I'm making
myself questions trying to understand the
limits. Here is my question. Let's suppose
I'm in a target range, shooting at the
target. Obviously the bullet go through it.
In the very precise instant that the bullet
hit the card, how can that be explained by
the 3rd law. Is there an opposite force,
equal in magnitude, opposite in direction,
even though the bullet go through the
target?
ANSWER:
During the time the bullet and target are in
contact with each other, the bullet exerts a
force on the target; this force is evidently
stronger than the target can withstand and
that is why it tears and the bullet passes
through. Now when that is happening, Newton's
third law states that the target exerts an
equal but opposite force on the bullet. It
is not very hard to punch a hole through a
sheet of paper, so the force is not very big
in this situation; nevertheless, the bullet
feels that force and as a result emerges on
the other side with a slightly smaller
speed. If you had 100 sheets of paper to
make your target, the bullet would be going
with a considerably smaller speed when it
came out the back; with a thick or strong
enough target the bullet would just stop in
the target.
QUESTION:
In this
video , starting at 2:00, a pair of
2-liter bottles suspended over 2
loudspeakers are made to orbit about a
fulcrum on which they are balanced when
excited at their resonant frequency purely
by sound waves. But such wave motion is back
and forth, so how can it impart momentum in
just the forward direction? I must know how
this works!
ANSWER:
Yes, the pressure at any point in the
bottle, including the open end, will
fluxuate with the frequency of the air in
the bottle. When the pressure is low, air
from the outside is sucked into the bottle;
when the pressure is high, the air is pushed
out. But the two air motions are not
analogous. As shown in the figures below,
when air flows in it comes from many
directions so this gives the bottle only a
small thrust to the right. But when it flows
out, the neck tends to align the air motion
causing a larger thrust to the left.
A cavity with a
narrow, open neck at one end is called a
Helmholz resonator.
QUESTION:
Einstein’s theory claims that a large
object in space will create a gravitational
field that will bend Space-Time. My question
is - if indeed a large body like the Earth
does create a bend in Space-Time do we exist
in another aspect of time than that of the
surrounding space which is empty and thus
does not have a gravitational effect on that
around it?
ANSWER:
The rate at which clocks run depends on the
intensity of the gravitational field. A
clock 100 m above you runs at a different
rate as yours. So I guess you do "exist in
another aspect of time" (whatever that
means!) from any region where the magnitude
of the gravitational field is different than
yours. This effect is often referred to as
the gravitational red shift. It turns out
that gps systems must correct for this
effects since timing between you and
satellites at high altitudes must be
extraordinarily accurate. I should note that
the differences in time are extremely small,
even for something as large as the sun which
is incredibly massive.
QUESTION:
If at the start of the universe all the
matter was within a cubic meter how didn’t
it create a black hole
ANSWER:
I state clearly on the site that I do not
normally do
astronomy/astrophysics/cosmology, so take
with a grain of salt that I am no expert. I
would guess that there is so much energy in
this cubic meter that whatever forces might
be acting are simply inadequate to reverse
the expansion. Also, I used the word "other"
because we do not know what the laws of
physics were in the very young universe.
QUESTION:
got a discusion with a friend, about the
riding wind versus side wind. when driving,
the air thats in front of the care needs to
move, and puts presure on the car. does that
increase when you drive harder, and will a
side wind have more or less effect when you
drive harder?
ANSWER:
If you are driving into a headwind there is
a force back on you due to the speed of the
wind. You would therefore have to press
harder on the accelerator to keep going with
the same speed you would go in still air.
But that does not have a handling the car,
merely holds you back causing you to use
more gasoline. For example, if your car has
a speed of 60 km/hr into a 20 km/hr
headwind, it would be the same as driving 80
km/hr in still air. In a crosswind, however,
handling is more of an issue. There is a
tendency to push the car in the same
direction as the wind is blowing; if the
road is wet or icy and the wind is strong
enough, the friction of the tires could be
inadequate to keep you from sliding across
the road. Even if the road is dry, the
tendency for the car to turn with the wind
and you need to slightly steer into the
wind. Also keep in mind that a strong
headwind will not necessarily keep blowing
exactly opposite your direction or else you
may need to do a maneuver where the wind now
is partly blowing across your path. So,
always be extra careful on a windy day.
QUESTION:
I respect that Heisenberg's Uncertainty
Principle can't be violated. . But a daily
lab event seems to say "no you can". Here's
the situation. Take the cathode ray tube in
an oscilloscope. An electron is ejected from
the cathode, deflected by coils around the
tube's neck and then impacted at a precise
screen location. The UP says I can never
accurately know the electron's position and
momentum at T. But I do accurately know P,
it's relativistic mass is constant and so is
it's velocity. In order to place that
electron at the precise point on the tube
face, the circuit must know its exact
location to know when to increase voltages
to the neck coils for proper deflection. Why
is this scenario not a violation of UP?
ANSWER:
Well, how accurately can you actually know P ? Certainly not better than maybe
0.1%. And position? Maybe to a micron or so.
But, the P you are talking about
and the position you are talking about are
not the proper quantities to be talking
about in terms of the UP. Suppose we call
the line between the electron gun to the
center of the screen the z -direction.
Then the UP is Δz Δpz ≥ℏ
but the position on the screen would be
perpendicular to z , could be either
x or y ,and Δx Δpz
has no uncertainty principle associated with
it.
QUESTION:
So I'm in high school and I'm in
physics. I'm also a musician (piano,
singing, etc.). I was wondering about
pendulums, but specifically metronomes. If I
wanted to use a metronome as a model in an
energy conservation project, would I simply
apply the same rules as normal pendulum
situations? I guess I'm just saying that I'm
having trouble with the difference between
metronomes and other pendulums. I tried to
find the answer online, but whenever a
metronome is used in a physics example is
always is talking about simple harmonic
motion or synchronizing metronomes, which
isn't really what I'm asking. I'm not
advanced enough yet to figure out the
physics behind a metronome by myself,
either. Sorry about the long paragraph for
what's probably a pretty simple question!
I'm basically just asking if, when
discussing energy transfer/conservation,
work, and force in physics does a metronome
abide by the same rules as a normal
pendulum, and if not, how does it differ?
ANSWER:
A metronome is certainly not a simple
pendulum which is basically a weightless
stick of length L
frictionlessly pivoted at one end and
attached to a point mass m at the
other end; the point mass oscillates below
the pivot. As you have probably learned,
this seemingly simple problem cannot be
solved analytically but can be approximately
solved is the angle is small. The result is
that the frequency f (cycles per
second) is approximately f ≈[√(g /L )]/(2π )
where g is the acceleration due to
gravity. Note that the mass does not matter,
a bit of a surprise perhaps.
Now the
metronome is clearly not a simple pendulum
because the mass is above the pivot point.
But, how can that be? If you just stuck a
mass on the end of a stick and rotated the
mass to a point above the pivot, would it
oscillate about the very top? Of course not,
it would still oscillate about the very
bottom but with an amplitude which was very
big. So, how does the metronome do this?
Well, I have earlier
worked out how a pendulum works. You may
want to have a look at this but, being in
high school you probably are not ready for
the math there. And maybe your physics class
has not even gotten to rotational motion yet
so you would not know about moment of
inertia, angular acceleration, etc .
But you find out the nature of the pendulum
in a metronome: there is a bigger mass below
and out of site. So this is really a "double
pendulum" and as long as the hidden mass is
bigger, it (the bigger) will oscillate about
the bottom while the smaller oscillates
about the top. The picture here of the
metronome made of plexiglas shows that
bottom mass. You can also get the final
answer if the "stick" has negligible mass
compared to the two masses, f=ω /(2π )=[√(g (MLM -mLm )/(MLM 2 +mLm 2 )]/(2π )
where M (m ) and LM
(Lm ) are the mass and
distance from the pivot of the lower (upper)
weight.
Now to
your question, whether energy conservation
applies to a metronome. No real-world
pendulum as its mechanical energy conserved
because friction of some sort is always
present. However, if you put it in a box
from which no energy can come in or go out,
the total energy will be conserved, even if
the pendulum stops, the air in the box will
heat up a little bit and if you were to
measure all the energy contained in thermal
energy you would find all the energy which
the pendulum originally had. So the answer
to the "basically
just asking" is yes, the two pendula "abide"
by the same physics rules. Also note that
there is a little spring-driven motor you
can see inside; when you wind it up you do
work to give it potential energy and it then
gives this energy to the pendulum to replace
energy lost to friction.
QUESTION:
Let's consider crushing some item with
scissors blade. It is easier to do it when
the item is positioned closer to the
pivot(assuming the force of squeezing the
scissors is constant). I am asking to check
the following explanation: Is it so because
to avoid being crashed the item must
generate reaction force impacting the blades
of such magnitude that the blades are not
moving(theirs torques have to equal
zero).Let's assume that force momentum
coming from squeezing the scissors is
constant(we don't change force of our
fingers nor the distance from the pivot) and
potential of generating reaction force by
the item also. Then the only thing we can do
to decrease this torque coming from reaction
force(and therefore making the item less
"resilient") is to shorten the distance
between item and pivot. My wonder is if it
all comes down to considering movement of
scissors blades around axis, meaning: blades
are moving the item gets crushed, blades are
still(torque equals 0) the item is in one
piece.
ANSWER:
You can understand the principle by just
considering one half or the scissors; then
the other does just the same but with
opposite forces. All your talk about
"reaction force" is wrong; look at my
diagram—the force labelled F 1
is the force your thumb exerts on the
scissor and the reaction force is the force
which the scissor exerts on your thumb which
is the same magnitude but oppisite
direction. I will assume that there is some
small object you want to "crush" located
somewhere along the blade. This gets a
little confusing, so read the next sentence
carefully. If you exert the force
F 1 on the
scissor and the scissor is not closing, the
sum of the torques exerted on the
scissor must equal zero; if the object
is located near the pivot, like where
F 2 is, the
object must exert a force opposite to
F 2 but
of the same magnitude since F 2
is the force the scissor exerts on the
object which is what you are interested in.
(Ignore F 3
for now.) To calculate the magnitudes of the
torques you need to multiply the force times
the moment arm (yellow lines). The moment
arm for F 1 is much
longer than for F 2 and
therefore F 2 should be
much larger than F 1 .
Now, consider placing the object out near
the end of the blade where F 3
is drawn; its moment arm is larger than for
F 1 so
its magnitude is smaller. I hope it is now
clear why placing the object closer to the
pivot results in a larger force on it than
farther away.
QUESTION:
I have a question about the weak nuclear
force. Why do we call weak force a force?
All what's said about the weak force is that
it cause decays by changed one particle into
another (or at a fundamental level, by
changing the flavour of quarks). But , "what
makes it a force?" Do we call it a force
only because it involves W/Z bosons or
there's something else.
ANSWER:
It is better named the weak interaction. The
concept of a force is generally not useful
in quantum physics, but, since we all have a
gut feeling what a force is in macroscopic
classical physics, it is natural to think of
interactions between objects as resulting in
forces. But, to treat interactions between
objects quantum mechanically we use the
concept of fields to describe those
interactions; fields are also used in
classical physics as well—think electric,
magnetic, and gravitational fields. At the
quantum level fields alone are useful, not
forces. And when you have a field it can be
quantized and the quanta are often viewed as
the "messengers" of the field; the field
quanta are photons for the electromagnetic
field, gluons for the strong interaction,
and W± and Z bosons (as you note)
for the weak interaction.
It is
interesting that the fourth "force" in
nature is problematical: no one has
successfully quantized the gravitational
field and achieving that is one of the holy
grails of physics.
Also
interesting is that the notion of force is
not useful in the theory of special
relativity because the usefulness of force
depends on the usefulness of acceleration.
At high speeds where special relativity is
important, different observers will see
different accelerations for the same object;
if we think of force in terms of Newton's
second law (F=ma ), different
observers would deduce different forces. If,
instead, we thought of Newton's second law
as force equals time rate of change of
linear momentum, force could be invisioned
but only if linear momentum (mass times
velocity) were differently defined. But I
digress...!
QUESTION:
I am a 180 lb man competing 180 lb
weight class five time world champion
powerlift. I bench 500 lb and I would like
to know how much force am I executing to
lift 500 lbs?
ANSWER:
To hold 500 lb at rest or to move 500 lb
straight up with constant speed requires a
force of 500 lb. Since the weight you lift
is initially at rest and momentarily at rest
when your arms are fully extended, you must
exert a force larger than 500 lb to get it
started moving upward and smaller than 500
lb for it to stop at the top. I have watched
videos of the bench press and it appears
that during most of the time the bar is
moving with a constant speed; so most of the
time the force you exert is about 500 lb. To
do a rough calculation of the force
necessary to get it moving, I will use a
total lift distance of about 1 m in a time
of about 1 s so the speed is roughly v =1
m/s; 500 lb has a mass of about m =227
kg. (I will use SI units, like scientists
prefer, and convert back to pounds in the
end.) Suppose that the time it takes to
accelerate the bar to 1 m/s is t =0.1
s; so the acceleration is a =(1
m/s)/(0.1 s)=10 m/s2 . So the
force F you must exert is
calculated using Newton's second law, ma =(F-W )
where W=mg is the weight of the bar
and g ≈10 m/s2 is the
acceleration due to gravity; F =227(10+10)=4540
N=1000 lb, twice the weight you are lifting.
Similarly, if it takes 0.1 s to stop the
bar, the force you would apply would be
F =227(-10+10)=0, the weight will stop
itself because of gravity. Keep in mind that
these are estimates using reasonable
numbers; the actual numbers depend on the
way the individual does the lift.
MY ANSWER
IS WRONG, OR AT LEAST INCOMPLETE. SEE BELOW
QUESTION:
Recently when cleaning, I opened the
cover of our ionisation smoke detector, as I
was opening the smoke detector cover, dust
fell from the inside of the smoke detector
into my eye. The whole inside of the cover
was covered in this dust-a brown colour
dust, not grey colour. I was concerned that
this dust has been given off by the
Americium due to the slats on the ionisation
chamber which allow particles to escape from
the radiation. Could this be so? I was
concerned as it said online that it was
dangerous to ingest or inhale the americium
from smoke detectors and this dust entered
my eyes?
ANSWER:
I have recently answered a question about
smoke detectors; you should read
that . The
americium is in a sealed container and
cannot get out. Also, the color of dust
depends on where it comes from. There is
probably some source of brown dust in your
house and you wouldn't notice its color
except when something like your smoke
detector accumulates a relatively large
amount over a long time.
CORRECTION:
I received an
email with additional information about
which I was not aware. Essentially it said
that the recoiling americium nuclei could
collide with and eventually damage the
casing around the source. Since your smoke
detector is 27 years old and the rule of
thumb is that the useful lifetime of
americium smoke detectors is about 10 years,
you should dispose of this one and get a new
one. (Use the recommendations of the
answer you included in your question to
dispose of it safely.) I still believe that
you need not obsess over the small exposure
you received when you were cleaning, given
the very small numbers I quoted in the
earlier answer about smoke detectors; the
unit quoted there,
rem (Roentgen equivalent man), is
specifically meant to indicate exposure to
ionizing radiation of human tissue and is
therefore more useful than Curies used by
the author of the
answer which just measures the amount of
radiation emitted without consideration of
their health effects.
QUESTION:
A piece of cork is thrown vertically
downwards from a sky scraper 300 meters
high. Its initial velocity is 2 m/s. Air
resistance produces a uniform acceleration
of 4 m/s2 until the cork reaches
a terminal velocity of 5 m/s. Why does the
cork reach terminal velocity?
ANSWER:
There is something terribly wrong about this
question. I see only two ways I can
interpret this question.
I
can interpret everything completely
literally as written. Of course, this
cannot be physical because the air
resistance F A must
depend on the speed v (usually
F A ∝v 2 )
of the cork and it will certainly not,
in the real world, result in an
acceleration which is constant. If the
mass of the cork is m , then
F A =4m ; there is
also the force of gravity F G =-mg
where g =9.8 m/s2 . So
the net force is F net =ma net =m (4-9.8)=-5.8m ,
so a net = -5.8
m/s2 . So the cork will fall
with a downward acceleration of
magnitude 5.8 m/s2 and never
have a terminal velocity since it just
keeps speeding up.
Maybe you didn't really mean to say it
has a "uniform acceleration" but that
F A =kv 2
at the instant you release it with speed
down of 2 m/s and the acceleration due
to that force is 4 m/s2 ; so
k (2)2 =4m or
k=m . Now we have F net =ma net = (mv 2 -mg );
when v=√g= 3.13 m/s, a net = 0,
so if this is the interpretation of the
problem, the terminal must be 3.13 m/s,
not 5 m/s.
I
cannot think of any interpretation of this
problem which would be self-consistent.
QUESTION:
How much energy in joules would it take
to propel an object with a mass of 14 grams
to a speed of 1.25 kilometers per second
over the course of 0.15 seconds? Further,
how much power would it take to create this
amount of energy? I know this sounds like a
homework question, but it isn't. this is a
question from someone who barely understands
any of these physics terms and who has
hardly any idea how to calculate the
simplest of formulas, yet is trying to get a
concrete idea of what it would physically
require to complete this hypothetical task.
ANSWER:
I think it is a homework question and you
are trying to pull the wool over my eyes;
nobody just wonders how much energy a 15 gm
object going 1.25 km/s has. I will outline
what you need to solve, but just this once
and not in complete detail.
The kinetic energy
E of an object with mass m
and speed v is E =½mv 2 .
If you want the energy to be expressed in
Joules, m must be in kilograms and
v must be in meters per second. If
the object takes a time of t to
achieve an energy E , the average
power expended to do so is P=E /t .
If you want the power to be in Watts, E
must be expressesed in Joules and t
must be expressed in seconds.
QUESTION:
If a full cart is pushed at the same
time as a empty cart with the same force
witch one will stop in motion first? And
why?
ANSWER:
I assume that the carts are identical except
for the loads; and that the force "does the
same thing" to both of them. The heavy cart
has mass M , the light cart m .
There are two ways you can apply the force
F :
Push
over the same distance d for
each. In that case you give the same
kinetic energy to each, ½MV 2 =½mv 2 ,
so V=v √(m /M )
where V (v )is the
starting speed of M (m ).
Note that, as you would expect, v>V .
Push for the same time for each. In that
case you give the same linear momentum
to each, MV=mv , so V=v (m /M ).
Again, v>M but by a different
factor.
Now,
what stops the carts? Friction f
which is proportional to the weight of each
car. For the loaded cart, fM =-μMg
and for the empty cart, fm =-μmg
; here the negative sign indicates that
the force is slowing the cart down, μ
is the coefficient of kinetic friction, and
g is the acceleration due to
gravity. Because of Newton's second law, the
accelerations of the two cars are A=fM /M =-μg
and a =fm /m =-μg.
Because the accelerations of the two
cars are the same, the one which started
fasted (empty) will go farther and take a
longer time to stop.
QUESTION:
My question is hypothetically ( or
exactly ) does or would an "alien" be able
to negotiate a 90 degree turn and not be
splattered on the inside of the vessel.??
ANSWER:
That would depend on two things. First how
long does it take the vessel to turn which
would determine the average acceleration?
Second, what is the physiology of the alien,
in particular how much acceleration can her
body endure?
QUESTION:
Can you please explain why Magnetic
force is Non-Central when the
Electromagnetic forces are Central forces?
On the same note, Why is the friction non
conservative when the EM forces are
considered conservative in nature?
ANSWER:
You have some misstatements here. For
"central" I think you mean conservative
because, for example, the electric field for
a uniformly charged wire does not all come
from a single point (which is the definition
of central fields). Also, when you say
"electromagnetic" I think you mean electric.
Now, the force due to a static electric
field is conservative, but not all electric
fields are conservative. For example, the
electric fields induced by changing magnetic
fields are not conservative. You ask why the
magnetic force is like it is—because that is
the way nature is. Be aware that electric
and magnetic fields are manifestations of
one field, the electromagnetic field, and
not separate fields but intimately linked.
Regarding why friction, admitedly due to
electromagnetic forces, is not conservative,
I refer to the first part of this answer
where I emphasized that there is no reason
to assume that electric or magnetic forces
are conservative.
QUESTION:
If black is the absence of light, and
thus of color, how is it one can mix primary
colors together to get black paint?
ANSWER:
Because mixing paint is not the same thing
as mixing light. It makes sense
simplistically: red paint absorbs everything
but red, blue paint absorbs everything but
blue, so red+blue paint absorbs
everything—black. Like I said, this is
simplistic because they are not purely red
nor purely blue, but it gives you an idea of
why mixing paints would look black.
QUESTION:
Can something rotate if it is not
symmetrical?
ANSWER:
Yes. It is easiest to visualize an object of
any shape you like which is in empty space
at rest. You now give it a kick somewhere on
its surface. Unless the direction of the
force is directed directly at the center of
mass, the kick will cause the object to move
away from you and be also rotating as it
moves. The rotation will be about an axis
which passes through the center of mass. You
could make the object rotate about any axis
you wanted but you would have to hang on to
the "axle".
QUESTION:
Do we know how a nucleus of an atom is
structured with its protons and neutrons? Is
it merely a discombobulated mess, or is
there actual structure to it? Are they in
movement within that space between itself
and the electrons (spinning, rotating, doing
a disco dance, etc)? This question has been
on my mind for a while and I cannot find any
resources on it, and graphical
'representations' just show a mass of random
protons/neutrons.
ANSWER:
I will note at the outset that this question
is in violation of site ground rules:
"…single, concise, well-focused questions…"
Nuclear structure is an entire subsection of
physics and it would take a whole book to
give you even an overview. I spent my entire
research career, more than 40 years,
studying nuclear structure. The answer is
that it is not "merely a discombobulated
mess" but pretty well understood. I will
give you a few examples.
The
force between nucleons (neutrons and
protons) is very strong and results in
the nucleus being extremely small
compared to the the atom (nucleons and
electrons). The size on an atom is on
the order of Angstroms (10-10
m) where the nucleus is on the order of
femtometers (10-15 m). If the
atom were about the size of a football
field, the nucleus would be about the
size of a golf ball.
One
of the first successful models of the
nucleus was the shell model. Because of
the average force due to all the other
nucleons, each nucleon moves in orbitals
like electrons do in atoms under the
influence of the Coulomb (electric)
force.
A
later, also successful, model is the
liquid-drop model where the particles
all move collectively. Imagine a liquid
drop which, when bumped will oscillate
in some way. Or imagine a nucleus which
is deformed like an American football.
It could rotate about its center of
mass. Many heavier nuclei have
rotational band structures. In these
cases the nucleons all move in
choreographed unison, maybe more like a
line dance than a disco dance.
Hope
this gives you the idea that structure of
nuclei is fairly well understood. The
examples I give are over simplified because,
among other things, when inside a nucleus,
nucleons lose their individuality.
QUESTION:
I was recently doing an OCD Exposure
homework and now I am unsure of what I did.
I opened the smoke alarm cover and touched
around all the inside parts of it. I touched
all the sides of the outside of the
ionisation chamber with the radiation symbol
on it for around 5-6 minutes. I did not open
the ionisation chamber. I want to find out
1. Did this expose me to a lot of
radiation? 2. Will this increase my risk
of taking cancer in the future. Can you
please tell me honestly, even if it is not
what I want to hear, I just need the facts.
ANSWER:
The radiation from smoke detectors is
trivially small. Even if you had removed the
source and kept it around it would have
given you negligible radiation exposure. A
study by the Nuclear Regulation Commission
showed that "…a teacher who removed the
source from a smoke detector could receive a
dose of 0.009 millirems per year from
storing it in the classroom. The teacher
would get another 0.001 mrem from handling
it for 10 hours each year for classroom
demonstrations, and 600 mrem if he or she
were to swallow it…" To put this in
perspective, you receive approximately 2.2
millirems per year from natural sources
(from space above, ground below) if you are
at sea level, even more if you are above sea
level. Don't swallow it, though!
QUESTION:
I have a question about the operation in
drift chambers. Charged particles enter the
chamber medium to ionize the gas atoms. The
electric field is applied to drift resulting
electrons and ions. Also, there is a
magnetic field applied to measure the
particle momentum. However, all the papers
use the Lorentz force on the charged
particle to calculate its momentum in this
form F= qvb and don't include the force due
to the electric field. In other words, why
the electric field doesn't have an effect on
charged particles entering the drift
chamber?
ANSWER:
The figure above shows a schematic sketch of
one layer of a wire chamber. The particle
ionizes atoms near wires it passes close to.
The wires carry a charge which creates an
electric field near them which then collects
charged particles and sends a pulse to a
computer. The electric field is fairly
strong near the wires but quickly becomes
nearly zero in a very short distance
distance from the wires. Shown in the second
figure are equipotential lines due to the
voltage on the wires; note that at distances
about equal to the wire spacing the
equipotentials are constantly spaced
indicating nearly zero fields. A second
consideration is that the particle being
detected has an extremely large energy which
makes electric fields of the magnitude near
the wires practically invisible to the
detected particle.
QUESTION:
We know that our universe has fine tuned
gravity constant so my question is can a
universe born with a different gravity
constants other than fine tuning or there is
no possibility other than fine tuning
constant
QUERY:
This is a very deep question, and you
greatly underestimate the situation. In
fact, the properties of a universe are
extremely sensitive to the values of many
fundamental constants, not just G .
So you cannot just think of how things would
change if G were changed. Usually
the question is not about whether a universe
could be "born" if the fundamental constants
were different (partly because we really do
not know the mechanism of the big bang);
rather, the question is usually "how much
could we change the fundamental constants
and still have a universe where life is
possible?" Here is an interesting example I
read about of the consequences of changing
fundamental constants: If the strong nuclear
force were increased by 2%, the diproton
(two protons bound together) would be
possible. The consequence would be that
stars could not exist because the
proton-proton cycle which produces the
energy in stars would not occur if two
protons could just stick together.
QUESTION:
We know from Maxwell equations that c =
1/(e0m0)1/2 which indicates that
electromagnetic fields propagate in vacuum
with the speed of light. Thus, light is an
electromagnetic wave as proved later by
Hertz. Consequently, Can induced
electromagnetic fields change the energy
carried by light waves if they were exposed
-in a certain way- to them?
QUERY:
It is not clear what you are asking. "...in
a certain way..."?
REPLY:
Would light interact with electromagnetic
fields or not?
ANSWER:
Yes, light will interact with any electric
or magnetic field. First, there is the
superposition principle; at any time and
place the net electric or magnitic field is
the sum of all fields from all sources. Here
is another example how "light" can interact
with an electric field: a photon with
sufficient energy, if it passes close to a
nucleus where the electric field is
particularly intense, may convert into an
electron/positron pair.
QUESTION: I ride motorcycles
and do have BS degree in Biology and am
familiar with simple physics. There is a new
airbag technology that has been recently
released to consumers. This technology was
initially developed for motorcycle racers
close to 15 years ago. The early tech used a
cord attached to bike and rider, when the
rider and bike were separated the airbag
deployed. Due to advancements in electronics
new system use a small computer with
appropriate algorithms sensors to determine
when a crash is happening and activate the
airbag. This technology is now available to
consumers like myself at reasonable cost
$700-1,100 + for a system. The European
Union has developed standards for impact
protection pads for motorcyclist's. EN
1621-2:2014 defines the exact spinal pad
impact reduction capabilities, standards and
testing requirement's. I have found it
difficult to find technical data on the new
airbag system with regard to exact impact
reduction capabilities. What I have found is
that rider report being in a crash with the
airbag system and it registered a IMPACT
force of 18 G-forces. The rider was not
injured. Assuming that the rider suffered a
TRANSIMTED force of the no more than 12kN
(as per the EN 1621-2:2014). How much energy
in Kn did the airbag system absorb ? What is
the energy absorption capability difference
between the airbag and a EN 1621-2:2014 pad
with 12 kN transmitted energy ? Problems
1) I have been unable to determine the EN
1621-2:2014 IMPACT test force, I only know
the TRANSMITTED test force is no greater
than 12kN 2) Rider crash data uses
G-force to measure IMPACT force, and
provided no transmitted force data, due to
there being no injury, I feel it safe to use
a transmitted force of 12 kN . 3) Some
forces are given in kN and some G-forces, I
do not understand the difference of these
force measurement systems and I do not know
how to correctly convert from one
measurement system to another to solve my
problem
ANSWER:
I am sorry, but your question violates one
of the rules of the cite, "...single,
concise, well-focused questions..." I can,
however, help you with the single question
of confusion between kN and g -force.
The g -force is technically not a
force at all, it is an acceleration. As you
likely know, the acceleration due to gravity
is g =9.8 m/s2 . The g -force
is usually expressed in g s; for
example, because of some force you
experience an acceleration of 10 g s,
so that means that if your mass is 100 kg,
the force you experience is F=ma =100x10x9.8=9800
N=9.8 kN.
QUESTION:
I was wondering at sea level, given
variable air densities. What is the rate at
which a human would slow in relation to the
earths rotation if suspended in the air for
an extended length of time. If you were able
to nullify the air density acting on your
body could you theoretically travel the
globe at 1000 mph opposite the earths
rotation. Am I forgetting variables other
than loss of contact with the earth and air
density? I know that inertia would keep you
moving with the earth however you would
begin to shed that inertia quickly correct?
ANSWER:
Any way you could be "…suspended in the
air…" would rely on something attached to
the earth. Even a
hovering helicopter is being held up by
the atmosphere which rotates with the earth.
No, you cannot travel around the earth by
staying still.
QUESTION:
Why does a volume of any random-motion
particles begin spinning when it is
compressed, either by outside force, or by
gravity? Why they heck do things go into a
spin? Is this some form of conservation of
momentum?
ANSWER:
The things do not "go into a spin", they
were already spinning but more slowly,
perhaps too slow to notice. It is, indeed,
because of a conservation principle, the
conservation of angular momentum. The
principle states that if there are no
external torques acting on a system of
particles, the angular momentum will remain
constant. Let us take a very simple case, a
single point mass moving with speed v
in a circle of radius R . It is
attached to string which passes through a
hole in the table; you are holding it in its
orbit by pulling down with just the right
force F (which happens to be
F=mv 2 /R , but we
don't need to know that now). Its angular
momentum is equal to L 1 =mvR .
Now you pull down with a bigger force so
that the new radius is R /2; as you
did this, no forces on m exerted
any torque so the angular momentum was
conserved. L 2 =mv'R /2=L 1 =mvR ,
so v' =2v ., spinning faster
than before your force pulled it in. But,
you argue, if there are many particles with
random velocities there is, surely, for each
particle a second particle which is rotating
in the opposite direction so the net
rotation is zero. But, this argument will
only work exactly if there are an infinite
number of particles. In a very large
assembly of particles even a very small
difference between a pair of particles can
cause a large angular momentum if they are
far away from the center of mass of the
assembly.
QUESTION:
What is the terminal velocity of a 5
inch hail stone, and how many psi would be
necessary to launch that roughly 1.1 pound
ball of ice down a barrel style tube? I’m
looking to do some stress testing on some
roofing material simulating the most severe
weather.
ANSWER:
I have a good way to estimate the air drag
force at atmospheric pressure: F =¼Av 2
where F is the force in Newtons,
A is the cross sectional area in m2
of the object, and v is the
velocity in m/s; you must work in SI units
because the factor ¼ contains constants like
drag coefficient, density of air, etc .
Now, F=ma= ¼mAv2
where a is the acceleration. The
hailstone has a downward force on it, its
weight mg where g =9.8 m/s2
is the acceleration due to gravity; when
v gets big enough that the drag force
(upwards) equals the the weight down, the
acceleration becomes zero. This leads to the
terminal velocity v =2√(mg /A ).
I calculated the mass to be about 0.74
kg=1.6 lb and the area about 0.0167 m2 ,
so v =48 m/s=107 mph.
Regarding the cannon you want to make, you
cannot just know the psi but the distance
over which the force from that pressure will
act. If you want to push it with a force
F over a distance d with a
pressure P , you will (ignoring any
friction) give it a speed v =√(2Fd /m )=√[2PAd /m ]
where m is the mass of the
projectile and A is the cross
sectional area of the tube. So, you can
either push it with a big pressure over a
short distance or a small pressure over a
long distance. Let's use the number I used
above, m=0.74 kg, A=0.0167 m2 ,
and use a d =2 m long tube; then
v= 48 m/s, 48=√(2xP x2x0.0167/0.76).
Solving, I find P =2.83x105
N/m2 =41 psi. Don't forget that
there is atmospheric pressure, 14.8 psi, on
the front of the ball so a rough estimate
would be 56 psi. The actual required
pressure would be quite a bit larger than
this because friction (including air drag)
would not be negligible.
QUESTION:
When actors are shot off of roof tops,
why do they fall forward when the bullet
should propel them backwards?
ANSWER:
I have already
discussed this question in some detail.
The bottom line is that when a bullet hits a
man the recoil is negligibly small.
QUESTION:
i have come up empty in trying to figure
out a certain issue .. I am doing a project
with air cylinders and need a calculation
based on how much compressed air would be in
a cylinder thats 2 x 6 inches and yes i
searched google and YouTube and there is
mention on these but it appears different
folks have different formulas and different
conclusions... i am startled because of the
lack of information on this, i wouldn't
assume this would be THAT elusive.. most
answers I got was - P1 x v1 = p2 x v2
formula and that seems something i will
learn at one point in time after i learn the
basics but here is what i got so far if i
may- Pressure times volume divided by
atmospheric or : P x V / 14.7 .. seem ok at
first when i calculator a pressure of 40 psi
times the volume of my cylinder being 18.85
Cu inch then divided by 14.7 atm which comes
to 51.3 Cu inch … but if i change 40 psi to
10 psi things get hairy... 10 (psi) x 18.85
(volume) / 14.7 (atm) = 12.82 Cu inch.. but
this is less air than regular atmospheric
pressure .. so the formula must be faulty
ANSWER:
The relation P 1 V 1 =P 2 V2
(called Boyle's law) which you state is
correct provided that the temperature and
the amount of gas inside the cylinder do not
change. But then you do not apply it
correctly. I take it that you are not adding
air to the cylinder but are compressing the
air in the cylinder. I am also assuming that
the cylinder has a height of 6 in and a
radius of 1 in and that so the volume is
V 1 =6x3.14x12 =18.8
in3 . I also assume that when the
cylinder is at 18.8 in3 there is
1 atmosphere of pressure, P 1 =1
atm=14.7 psi, so P 1 V 1 =276
in·lb. We now have that V 2 =276/P 2 ;
if P 2 =40 psi, V 2 =6.90
in3 ; if P 2 =10
psi, V 2 =27.6 in3 .
Notice that for the 10 psi case the volume
is bigger than the the original volume
because the pressure is smaller than
atmospheric pressure. If you can't make the
volume any smaller than 18.8 in3 ,
the only way to reduce the pressure is to
remove some gas or cool it. If you are
interested, the most general expression for
an ideal gas is PV /(NT )=constant
where T is the absolute temperature
and N is some measure of how much
gas you have.
ADDED
THOUGHT: Rereading your question I am
thinking that it is not a cylinder with a
piston but maybe of constant volume 18.8 in3 .
In that case, keeping V and T constant, the
appropriate relation would be P 1 /N 1 =P 2 /N 2
or N 2 =P 2 N 1 /P 1 .
Suppose we call the amount of gas you can
put in the cylinder at atmospheric pressure
1 gas unit. Then if you fill the tank with
10 times atmospheric pressure, you will
store 10 gas units.
QUESTION:
Imagine a 10 m2 plate of
aluminum (thermal conductivity 225.94
W/mK)that is 0.1 m thick. In the center of
the aluminum plate protrudes a long skinny
aluminum bar that is 20 m high, 0.1 m wide,
0.1 m long. This would look like a dirt
tamper. The temperature of the large plate
is at 100°C and the top of the long skinny
pole is 10°C. How can I calculate heat flow
in this system? I certainly appreciate your
help.
QUERY:
Is the plate maintained at 100
and the top of the bar maintained at 10 and
you want the rate of heat flowing through
the bar? Or do you want to know the final
temperature is allowed to come to
equilibrium insulated from the environment?
And if you actually want an analytic
expression for any point in the system as it
is coming to equilibrium, it would require
that I have the shape of the plate; this
problem probably cannot solved analytically
and would require a numerical solution which
I am not able to do.
REPLY:
Plate is maintained at 100; top heats up but
I understand that is hard to model because
the rate would decrease as the top increases
in temperature (eventually becoming zero
when the top reaches 100). I guess I'm
really interested in the heat flow if the
top is at 10 (so we could imagine it is held
at 10)
ANSWER:
I am going to
assume that there is no heat leakage from
the sides of the bar. Initially I will
assume that the top end of the bar is kept
at a constant 10°C, so heat flows through
the bar at a constant rate. We can figure
out the heat rate as a function of the
temperature difference, ΔT =100-T .
I will neglect any edge or geometry effects
in the bar, in other words I will treat it
as a one-dimensional problem where the heat
flow vector in the bar is in the direction
of the bar and uniformly distributed across
the cross-sectional area. In that case the
equation for the rate R is
R =ΔQ /Δt=kA ΔT /L
where
A =0.01 m2 is the cross
sectional area, k =226 W/(m·K) is
the thermal conductivity, and L =20
m is the length. So R =0.113ΔT
W. For ΔT =90, R =10.2
W. Now, if the top remains at 10°C, that
means that energy at the top is being taken
away at the rate of 10.2 W; since ΔT ∝L ,
the temperature must increase linearly along
the bar when equilibrium has been achieved.
TIME
DEPENDENCE: The questioner indicated
interest in the more difficult problem of no
heat flowing out the top end of the bar
which started out 10°C. What I did was to
just give the steady-state solution like you
would learn in any elementary physics
course. So I dug into the transient case; I
learned a lot! Below I find the general
solution to the heat-flow problem and apply
it to the case in question, the bar starting
out at 10° and ending up at 100°. I have
left out a lot of details but have provided
links to derivation of the 1-d heat equation
and its general solution. Algebraic steps I
have omitted in applying boundary conditions
could be filled in by anybody interested in
these details.
The
one-dimensional heat equation is
∂T /∂t =c 2 ∂2 T /∂x 2
where c 2 =k /(C p ρ )
where k is thermal
conductivity, C p is
specific heat at constant pressure, and
ρ is the mass density. (You may find a
derivation
here . ) The general solution of this
equation is
T (x,t )=exp(-c 2 s 2 t )[A sin(sx )+B cos(sx )]+Cx+D
where A ,
B , C , D , and
s are constants to be determined for
the specific problem. (For a derivation, go
here . Go to Section 2.2)
Suppose the boundary
conditions are that (1) the temperature of
the bar is at some constant value T 1 ,
(3) one of the bar is held at a temperature
T 2 , and that (2) the
other end of the bar (and the sides) are
insulated:
T (0,t )=T 0
∂T (x ,t )/∂x|x= L =0
T (x ,0)=T 1
These boundary
conditions lead to the following:
B exp(-c 2 s 2 t)+D=T 0
B =0, D =T 0
exp(-c 2 s 2 t )[sA cos(sL )]+C= 0
C =s (exp(-c 2 s 2 t) [A cos(sL )])
s ≠0 so C =0A n cos(s n L )=0⇒s n = ½nπ /L,
n odd
∫(T 0 -T 1 )sin(s m x )dx =-Σ∫{A n sin(s n x )sin(s m x )}dx =(L /2)A n δmn
An =- (2/L )(T 0 -T 1 )∫ sin(½nπx /L )dx
=- 8(T 0 -T 1 )sin2 (nπ /4)/(nπ )
So,
finally,
T (x,t )=T 0 -Σ{exp(-c 2 s n 2 t )A n sin(s n x )}
c 2 =k /(C p ρ ),
An =8(T 0 -T 1 )sin2 (nπ /4)/(nπ )=4(T 0 -T 1 )/(nπ ),
sn = ½nπ/L , n
odd
We should tabulate
the constants to be used:
T (x,t )=100-Σ{(115/n)exp(-0.00206n2 t )sin(nπx /40)}
(t is in hours here.)
The plot of this
function including just the first three
terms of the series (1,3,5) is shown in the
first figure. Note that because the bar is
so long it takes the system several hundred
hours to fully equilibrate. The second
figure shows the calculation for the first
hour; this is clearly wrong since the whole
bar is at 10° at the beginning. The reason
for this is that so few terms have been
included in the infinite series. However, to
understand the long-term behavior, n>1 plays
almost no role because the n2
behavior in the exponential damps out higher
n contributions, particularly at large t .
One more
calculation, for the initial temperature
profile linearly decreasing over the length
of the bar may be seen
here .
QUESTION:
I'm writing a novel and want to make
sure I'm describing a scene correctly. A
spaceship built in a "tower" formation
(rocket at the bottom, levels stacked on top
of each other to the bridge at the
top/front) is forced to land on Earth in an
emergency. It initially enters at a steep
angle nose first, causing rapid deceleration
in the atmosphere, and then does a flip and
burn, the engines faced to the ground and
firing to slow them down. My question is
around the gravitational forces an occupant
would endure, sitting in a crash-couch/seat
that cushions them and rotates to ensure
they're always being pressed into it. Which
way would these forces push/pull on them
from start to finish, particularly around
the flip and firing of the engine? I can't
seem to wrap my head around it.
ANSWER:
Here is the trick you need to
understand: Newton's laws, by which we
usually do classical physics problems like
what you are describing, are not valid in
systems which are accelerating. When your
astronaut is in empty space and turns on her
engines, she thinks she feels a force
pushing her into her chair; but she cannot
understand this because she is at rest in
her frame and Newton's first law says that
if she is at rest all the forces on her
should be zero. If we look at it from the
outside we see the chair pushing her, that
being the force which is accelerating her in
accordance with Newton's second law.
However, she can do Newtonian mechanics in
her frame if she invents a force which is
equal in magnitude to her mass times the
acceleration of the system (rocket) but in
the opposite direction as the acceleration;
this is the "force" pushing her into the
chair and in physics we call it a fictitious
force. You have probably heard of a
centrifugal force,
the force which tries to throw you from a
merry-go-round; that is a fictitious force
because rotation is a kind of acceleration.
In the
following examples, the green dotted line
vector indicates the orientation of the
chair desired for most comfort for the
astronaut. Note that in each instance this
vector points exactly opposite the net
acceleration vector.
First the initial reentry, entering the atmosphere
at a steep angle. The rocket has some
velocity v . When
it encounters the atmosphere the result is a
retarding force opposite to the the
direction of the velocity which causes an
acceleration a of the rocket (and
astronaut). There will also be an
acceleration g due
to the gravitational force (weight); if
there were no air, this would be the only
acceleration. The net acceleration is the
sum of the two and is labeled a net .
The chair will orient with the green aligned
with the net acceleration as shown.
Next
we have the situation where the rocket is
rotating to be tail down for landing. It is
still falling with some velocity
v so there
is still an upward acceleration due to the
air drag, labeled a v
here. There is still the acceleration due to
gravity g . But now
there is also a rotation about the center of
mass of the rocket so there is a centripetal
acceleration a c experienced by the astronaut
which points toward the axis of rotation.
Add all three to
get the net acceleration a net .
Now the chair will be aligned as shown.
Finally, the landing position which is the
easiest. There is still the gravitational
acceleration g .
There is now an upward acceleration due to
both the engines and the air drag but the
engines will be the main contributer as the
speed decreases. The net acceleration now
points straight up and the chair orients as
shown.
Keep in
mind that I have not made any attempt to
have any particular relative values of the
various accelerations but they are not
unreasonable. They will vary considerably
depending on the specific conditions at
any particular time.
QUESTION:
So from the movie space cowboys. Two
pilots ejected from a plane and deployed
their parachutes at different times. Assume
they deployed their parachute at terminal
velocity. They started at 100,000 feet.
Assume same size parachute. If pilot A
weighs 230lbs and deployed their chute 5
seconds later than pilot B who weighs
250lbs. How possible is it that pilot B gets
to the ground first?
ANSWER:
Well, this is kinda tricky for a couple of
reasons. First, "terminal velocity" is not
some constant number; it depends on geometry
of the object (which you try to keep the
same by having "same size parachute"), the
density of the air, and the mass of the
object. So, falling from 100,000 ft they
will experience very significant change in
the density of the air. The second reason is
that, because of the first reason, your
phrase "deployed…at terminal velocity"
doesn't really tell me much. Also, terminal
velocity is not attained at a particular
time so you would have to specify something
like 'at 99% of terminal velocity'.
All
that aside, let's do the calculation
assuming that the density of the air is
independent of altitude, the same as at sea
level. The air drag is proportional to v 2
where v is the speed and I will
call the proportionality constant k ;
k is where both the density of the
air and the geometry of the falling object
are 'hiding'. Then Newton's second law may
be written as ma=mg-kv 2
where m is the mass, g is
the acceleration of gravity, and a
is the acceleration. Terminal velocity v t
is when a= 0 or v t =√(mg /k ).
Since we have stipulated that k is
the same for both A and B and is altitude
independent, only their masses would affect
v t . If they both jumped
and never opened their parachutes, B would
clearly be the winner (loser?!) since his
terminal velocity is larger than A's as
would be his speed at all times. Similarly,
if each deployed his parachute immediately,
B would get to the ground first. And,
certainly, if A deploys first he will reach
the ground after B since he was already
behind and will end up falling more slowly
than B. So, finally we come to your scenario
where A deploys after B. What will happen
depends on the altitudes where each deploy
and how far ahead B is when A deploys. If A
does not pass B during the 5 second free
fall, B will get to ground first. If A does
pass B he will end up below B but going more
slowly; if they fall for a very long time B
will pass A after some time. If that time is
less than the fall time left for A to hit
the ground, A will win.
Now,
falling from 100,000 ft will be quite
different because terminal velocities at
very high altitude will be much greater than
at atmospheric pressures. That means that
the distanced between the two will be much
greater when they have both deployed. If
they wait until they are fairly close to the
surface, it seems to me quite likely B will
hit first.
QUESTION:
So, when earth moves faster around its
axis (=one full rotation=1 day), does that
mean we have a "faster time"? If time is
correlated to biological age, does that mean
we "age faster"?
ANSWER:
Time is not measured relative to any
astronomical motions. If the earth were to
spin twice as fast, that does not mean that
time is running at twice the rate; rather it
means that a day would now be 12 hours long.
Any clock would run at the speed it did
before the earth sped up. That includes
biological clocks. (I have neglected
relativistic effects which are exceedingly
tiny at the speeds which the earth rotates.)
QUESTION:
In beta plus decay proton gets converted
to neutron— where does the extra mass come
from?
ANSWER:
A free proton does not (as far as we know)
decay for the reason you have inferred from
the masses; it must be inside a nucleus to
β -decay. The nucleus begins with
some mass M . It emits a β + ,
mass mβ and kinetic
energy Kβ ; and a
neutrino, mass mν and
kinetic energy Kν . The
nucleus now has mass M' and kinetic
energy K' . The energy of this
isolated system must be conserved,
Mc 2 =(M'+mβ + mν )c 2 +K'+Kβ +Kν .
Essentially, the energy came from the mass
of the nucleus; you would find the new
nucleus more tightly bound than the original
nucleus which you could interpret as being
why it decayed in the first place.
Therefore,
β+ -decay
does not occur if the new nucleus is less
tightly bound than the original. (By the
way, the neutrino mass and kinetic energy of
the nucleus are negligible in most cases.)
QUESTION:
Just listened to a Mark Parker Youtube
where is was stated (obviously) that at the
top of a sphere (the earth), say near the
north pole, you are travelling slower, i
guess in some reference frame, than someone
at the equator. ie. you travel a shorter
'distance' in a a 24 hour period than
someone at the equatòr (larger
circumference, same 24 hr period). This is
obvious with hindsight but not something
I've ever tried to think about. My question
is where does the additional energy come
from if I travel south of the north pole (or
north from the south pole) that enables me
to travel faster (along the direction of
rotation). I expect I'm misunderstanding the
problem in some way. At the equator I travel
40000km in 24hrs or 1666.7km/hr (I'm
assuming a spinning sphere and can ignore
anything external to the earth). I've
randomly picked a latitude of 60degrees
where the circumference is 20000km, which in
24 hrs = 833.3 km/hr The degrees don't
matter except to show a smaller
circumference travelled in the same time
(slower). So, as I travel north of the
equator my rotational speed decreases. where
does that energy go ? is the question valid.
If I travel south of some arbitrary northern
latitude towards the equator my rotational
speed increases. Again, I'm grossly missing
some key aspects of physics here. Am I
actually (by travelling towards the equator)
benefiting from the rotational energy of the
earth, and when travelling away from the
equator......actually is this a conservation
of energy (or angular momentum) thing
similar to pulling my arms in while spinning
on a chair.
ANSWER:
You hit the nail on the head when you
mentioned angular momentum conservation. You
plus the earth are an "isolated system" with
no external torques acting on you (forget
the moon, etc .) and the angular
momentum L of this system must
remain constant regardless of how you
move around. Suppose that you are at the
north pole; then the angular momentum of the
system is L 1 =Iω 1
where I is the moment of the earth
and ω 1 is its angular
velocity. Now you walk down to the equator.
The angular momentum is now L 2 = (I+mR 2) ω 2
where m is your mass and R
is the radius of the earth. But, L 2 =
L1 so if you do the algebra
you will find that ω 2 =ω 1 /(1+ (mR 2 /I )).
So the earth slows down by the tiniest
amount; I estimate (ω 1 -ω 2 )/ω 1 ≈10-29 =10-27 %!
Regarding the energy, the initial energy is
E 1 =½Iω 1 2
and the
final energy is
E 2 =½(I+mR 2 )ω 2 2 =½Iω 1 2 /(1+ (mR 2 /I ))
and so
E 2 /E 1 =(1+ (mR 2 /I ))-1 .
So, the
energy has not been conserved but is a tiny
bit smaller even though your energy has
increased by
½mR 2 ω 2 2 =½mv 2 .
If you
work it out you will find that the energy of
the earth alone is
½Iω 1 2 /(1+ (mR 2 /I ))2 .
So
we conclude that energy is lost by the
system although you have gained energy, so
the earth lost more energy than you gained.
Energy is changed by forces doing work on
the system, so what force is doing negative
work here? Since we are in a rotating
coordinate system, we will introduce the
appropriate fictional forces so that
Newton's laws can be used. If you are at
some latitude θ , one force you will
experience is the centrifugal force shown in
the figure, C . It
has two components, a normal component
N which would
reduce your apparent weight and a tangential
component T which
is trying to drag you toward the equator. In
order to keep T
from accelerating you, something must exert
an equal but opposite force on you; this is
simply the frictional force f
between your feet and the ground. As you
move toward the equator f
does negative work thereby decreasing the
energy of the system. Finally, the reason
you gain energy is that there is another
component to the frictional force which
points also tangentially but perpendicular
to f such that as
you move south it speeds you up to keep the
earth from sliding away from you.
QUESTION:
I'm trying to explain to my gf that if
the car in front of us is moving 60mph we
would have to be moving 60mph also in order
to maintain a 25 ft distance behind it. I
know it
sounds
dumb but I can't seem to make her believe it
ANSWER:
Ask her to imagine the car in front of you
is going 60 mph and is towing you with a 25
ft rope. Ask her to imagine looking at your
speedometer.
QUESTION:
This may be super simple but I keep
wondering about it. If a vehicle in the void
of outer space fires its boosters, why does
it move anywhere? Here’s why I ask: movement
on Earth requires friction and a medium
through which to move. If I’m in a swimming
pool and kick off from the edge with my
legs, then I move far. But if I’m in the
middle of the pool and make the same kicking
motion without contacting the edge, I
probably won’t move. Well, not unless I move
a lot of the medium around me (the water).
If there’s no medium in outer space, why
does the rocket booster have any effect?
Here’s my theory- is it correct? There’s no
resistance behind my spaceship, but there’s
no resistance in front of it either. After,
say, the first second of the ship firing its
boosters, the next second’s worth of
focused, exploding fuel is contacting the
exhaust and energy of the first second. This
gives the later second’s worth of exploding
fuel something to push against and, with no
resistance in front of the ship, it moves
forward easily. Is this right?
ANSWER:
It is a common misconception that a rocket
needs a "medium" against which to push, but
it is wrong. Your attempt to think of a way
to "create" an atmosphere to push against is
therefore also incorrect. In order for
something to accelerate, i.e.
change its speed, all that is needed is for
the sum of all forces on an object to be not
zero. Even if you were on a surface with no
friction and you were in a vacuum, you could
start moving if someone behind you pushed
you. The best way to understand a rocket is
to consider the following example using the
above situation of you in a vacuum with no
friction on the floor.
You
have a ball in your hand;
you
throw the ball with some speed;
in
order to give it that speed, you need to
exert some force on it;
Newton's third law says that if you
exert a force on the ball, the ball
exerts an equal and opposite force on
you;
therefore you end up moving in the
opposite direction as the ball;
it
turns out that the speed you acquire is
smaller than the the ball's speed by the
ratio of the ball's mass to your mass;
for
example, if the ball has a mass of 1 kg
and your mass is 100 kg, your speed will
be 1/100 of the ball's speed.
The
rocket engine is essentially throwing
countless tiny "balls" (atoms, molecules,
ions) with very high speeds to propel the
rocket ship.
QUESTION:
If a system of two separate modules
connected together with a linear passage(A
spaceship carrying space traveler) is
suspended in space with no gravitational
influence and is rotating to create
artificial gravity, at which point in the
system will there be no force experienced by
the accommodates while transitioning from
one module to the other through the linear
passage? It is the center I suppose, but I
am unable to figure out the math.
ANSWER:
It depends on what the masses of the modules
are. It also depends if the masses M 1
and M 2 of the modules
themselves are much larger than the mass
m of the person moving from one to the
other. The pair will rotate about their
center of mass which, if M 2 =M 1 ,
is at the halfway point. If the person is
very light relative to the rest of the
system, she will experience no centrifugal
force at the center as you guessed; the
reason is that the force is proportional the
square of the speed she is traveling and her
speed will be zero at the axis of rotation.
But if m is not relatively small
the location of the center of mass will vary
when she moves from one module to the other.
The
following is probably more than you want,
but I like to do it anyway. I will denote
the masses of modules as
M 2
and M 1 , including
anybody who is in them, and will treat them
as point masses separated by a distance
R ; the traveling astronaut has mass
m and is a distance d from
M 1 and will also be treated
as a point mass and I will assume the
passage has negligible mass. The center of
mass is located a distance r cm
from M 1 . Calculating the
center of mass location, relative to M 1 , is
straightforward and results in
r cm =Σ(m i r i )/Σ(m i )= [(M 2 -m )R +md ]/(M 2 +M 1 )
For
example, if M 2 =M 1 =M ,
r cm =½R -(m /M )(R-d );
and if
m<<M ,
r cm ≈½R.
Now, you
are interested in when d=r cm :
d=R (M 2 -m )/(M 2 +M 1 -m ).
And
again, for a check,
if M 2 =M 1 =M ,
d=R (M-m )/(2M -m ),
which
is, if
m<<M ,
d ≈½R.
QUESTION:
Okay so lets say your free falling from some
odd feet in the sky in a car. Dont worry
about the reason why but that its happening.
If you put yourself in a position to jump
out of the car before it hits the ground and
explodes, will you be able to change your
momentum enough to not take so much damage
from hitting the ground so fast?
ANSWER:
It depends on how far the car has fallen. If
it falls from say 10 ft, you could do it
fine but wouldn't even need to jump. But if
the car fell from a high enough altitude to
reach its terminal velocity, which I
estimate to be about 200 mph, could you
survive it by jumping upwards at the last
minute? How high can you jump from the
ground? The highest humans can jump is about
6 feet and to do that requires that you can
jump up with a speed of about 13 mph. If you
jumped that fast relative to the falling
car, your speed relative to the gound would
be about 187 mph; you would be just as dead
as you would have been if you hadn't
bothered to jump.
QUESTION:
Hello. If a 200 pound man runs at 5
miles per hour, hits another man going same
speed same weight, opposite. What is the
force of impact? Fyi. I'm 56. Past homework!
ANSWER:
There is no way to answer this question
because it depends on the details of the
collision. What matters most is the time the
collision lasts (or, equivalently, the
distance over which each moves during the
collision) and whether the two essentially
stop or each rebounds backward. I can give
you a couple of suggestions. I will work in
SI units (v =5 mph=2.25 m/s, m =200
lb=90.7 kg) which is usual for physicists
and will convert back to imperial units
(pounds) at the end. I will assume that the
collision is perfectly inelastic, that is
both runners are at rest following the
collision; in that case all the kinetic
energy the two had (each has ½mv 2 =230
J) is lost in the collision.
Suppose,
first that the two have lowered their heads
much like bighorn sheep do when fighting.
Since there is not much flesh on the
forehead, they will stop in a very short
distance, let's say d=5 mm=0.2 inches. The
work done by the average force each man
feels must equal the energy he lost, Fd /2=230
J so F =460/0.005=92,000 N=20,700
lb; of course, this force will be spread out
over an area of several square centemeters.
Still it is pretty darned big.
Suppose
the two men have big beer bellies and that
is where they collide. Then the distance
over which the collision occurs will be more
like 5 cm and the resulting force more like
2000 lb. And, the force will be spread over
a much larger area.
In the
real world, such a collision would occur
over a large amount of the body so the total
energy, converted as a force, would be
spread out over maybe a square meter. places
which are hard (like your head) would be
hurt more badly than places which are softer
(like your torso). Also, if the forces look
unbelievably big, keep in mind that they
last for a very short time.
QUESTION:
If energy and mass are interchangeable
or equivalent can E=mc2 be
written as M=ec2 and mean the
same thing? Thanks! I’m totally ignorant
maybe there’s is an obvious reason why not
here.
ANSWER:
Mass and energy are not "interchangeable or
equivalent". That would be like saying that
electricity and wind, both forms of energy,
are the same thing. Mass is a form of energy
and the total amount of energy which a mass
potentially has if it could be totally
converted into energy is given by E=mc 2 .
If you wanted to know how much mass could be
realized by converting an amount of energy
into mass, m=E /c 2 .
Another way to see why your M=ec 2
is an impossible equation is by doing
dimensional analysis: The units of mc 2
must be kg·m2 /s2 so
energy must have units of kg·m2 /s2 ;
but your equation would have the units of
energy being kg·m4 /s4
which is not what the units of energy are.
QUESTION:
I hope there really is no such thing as
a stupid question, if there is you are
really in for a treat. Okay, so if something
is being propelled from the rear it is being
pushed, and if something is being propelled
from the front it is being pulled, correct?
Is there a term for something being
propelled from the front AND the rear at the
same time?
ANSWER:
The words 'push' and 'pull' are qualitative
words, not physics words. Pushes and pulls
are both denoted a forces in physics. For
example, if you have a horse pulling on a
cart with a force of 100 lb and a man
pushing on the back of the cart with a force
of 10 lb, the net force on the cart due to
these two forces is 110 lb forward. If the
man is instead pulling on the back of the
cart, the net force is 90 lb forward.
QUESTION:
Light cannot escape a black hole. What
is the nature of light inside a black hole,
theoretically? Static photons? Can light be
"static" and exist in only one point in
space time? Perhaps it ricochets around
inside the event horizon rather than static?
Perhaps gets squeezed into a different
subatomic particle? Am an old guy who has to
much time and thinks of such things.
ANSWER:
Note that I usually do not answer questions
on astronomy/astrophysics/cosmology as I
state on my site. When light is captured by
a black hole, it ceases to exist as photons
and basically becomes mass, increasing the
overall mass of the black hole. When you
when you ask what is going on "inside a
black hole" it can mean two things: either
inside the Schwartzchild radius (from inside
of which no light can escape) or inside the
black hole itself. There is no answer for
the latter because at nearly infinite
density we have no idea at all as to what
the laws of physics are. For the former, you
can say that a photon loses energy as it
falls from the Schwartzchild radius to the
surface of the black hole, the lost energy
being converted to mass; the photon
continues moving at the speed of light but
its frequency decreases until the photon
disappears at the surface of the black hole,
totally converted to mass.
QUESTION:
Say you have a rocket ship in a vacuum away from all gravitational fields firing its engines to maintain uniform circular motion. Since the engines are burning fuel, it makes sense to me to calculate the power output in watts. But the Force in the direction of the Velocity is zero so the power is zero. I'm confused. And since the Kinetic energy is not increasing, where does the energy expended by the engine go?
ANSWER:
You are going to have a rocket engine which can fire
perpendicular to your velocity vector. None of the energy
generated by the engine will be given to the rocket ship
because the force the engine applies is perpendicular to
the displacement, hence no work is done. The energy
generated goes into the kinetic energy of the ejected
gases. It will be a little tricky to maintain uniform
circular motion because mass is being ejected, so as the
ship loses mass you need to reduce the force to keep
velocity and radius constant: F=mv 2 /R .
QUESTION:
How can universal gravitation and conservation of energy both exist together? If gravitation is universal there should be no where in the universe where a system is not acted on by an outside force (gravity).
ANSWER: Conservation of energy is
applicable to systems with no external forces doing
work on them. You have to be very exact when
applying conservation principles. Suppose you choose the
solar system as the system; if the solar system were in
the middle of empty space its energy would be conserved,
but it isn't an isolated system in real life and the rest
of the galaxy exerts forces on it which might change its
energy. Suppose you choose the milky way galaxy as the
system; if the galaxy were in the middle of empty space
its energy would be conserved, but it isn't an isolated
system in real life and the Andromeda galaxy, the
nearest major thing which exerts forces on it, would
change its energy. You can see where I am going with
this. Suppose the entire universe were the system; then
there are only internal forces acting on this system so
its energy will not change.
QUESTION:
I have some understanding issues with Current and Voltage regarding Induction. When there is a change in magnetic flux there must be either a change in the area of the e.g. wire or the magnetic field. But how is there a flow of current then induced by it without any poles where I can measure the voltage. I guess, it's just me thinking wrong of voltage, but I really didn't understand that one if you ask me. Connecting to that, how must I think of a coil that is getting an induced voltage by changing magnetic flux. How are the electrons flowing through the coil, because right now i always try to think, that in every nth part of that coil there is some sort of electric field that is created because of the electrons getting pushed. I really do think it's a question of understanding, but the mathematic definition and explainition is not really enough for me, i want it visualized for me.
ANSWER: When electricity and magnetism is
taught, there is usually a sequence:
First we talk about
electrostatics where all electric fields are constant and
conservative. A conservative
field is one for which, if you move an electric
charge from one point in the field to another, the
amount of work you do is independent of the path you
choose. This means that potential difference
(voltage) between two points is a meaningful number.
You are worried because you have been told that a wire loop with
no battery in it might have a current cannot flowing in
it.
Next we talk
about magnetic fields which are caused by electric
currents which are constant and presumably caused by
potential differences. This is called
magnetostatics .
Finally we come
to the full theory of electromagnetism where the
fields are no longer constrained to be constant.
All of
electromagnetism is described by Maxwell's four
equations. It gets mathematically dense so I have
previously
qualitatively described these equations without any math.
Electric charges cause electric fields.
Electric currents cause
magnetic fields.
Changing electric fields cause magnetic
fields.
Changing magnetic fields cause electric
fields.
So the answer to your
question is that you do not need charges to create the
electric field which will drive a current; a loop of wire
will have a current running in it if you cause the
magnetic field passing through its area to change. This
is called Faraday's Law.
QUESTION:
Okay, so i may dissagree with a master here, aka. Stephen Hawkings himself, but i have some questions that need to be answered, so here goes.
Hawking Radiation. It's basically a theory that says particles and anti-particles spontaneously materialize in space, seperating, joining, and annihilating each other. But then a black hole comes and sucks the a particle, leaving the other without a partner to annihilate with. This particle appears to be in the form of black hole radiation. Sooo, blackholes are not eternal. But what if all of the particles are sucked into the black hole? This is particle and anti-particle that we're speaking of, that has POSITIVE mass. It's not exotic particles that have NEGATIVE mass right? So does Mr. Hawkings law apply? Please answer this, i'ts kinda been bugging me for a while...
ANSWER:
First I will note that I do not normally do
astronomy/astrophysics/cosmology, as clearly stated on
the site. However, since I have answered this and related
questions many times before, I will accept your question.
I would say that you stop thinking about mass and simply
think of energy. A particle can have negative energy
without having negative mass. I think the best and most
detailed answer I have given is
here .
QUESTION:
why light cannot curve around object
ANSWER:
In fact, light does curve around an object with mass.
Originally it was known that light had no mass and
therefore would not be affected by gravity. But
Einstein's theory of general relativity predicts that
light passing near a massive object like a star or a
galaxy will be deflected. This has been observed to be
correct. There is a nice
article on gravitational lensing on Wikepedia.
However, for everyday life the bending is hardly
noticeable because it is small. Even the entire earth has
too little mass for this bending to be noticable (see a
recent answer ).
QUESTION:
Assuming no prior information is given, except the formula for the time period of a pendulum (T=2π√(L/g)), would the time period of a swing be changed at
all if someone went from sitting down on the swing to standing up on the swing? Eg. would the mass distribution affect the time period?
ANSWER: The formula you state is true
only for a point mass M attached to a massless
string of length L . Therefore you cannot solve
this problem using that formula. However, you could guess
that the period would be proportional to the square root
of the distance D from the suspension point to the center
of gravity of all the mass (the person, the swing, the
ropes). In that case, the period would get smaller when
the person stood up because the center of gravity would
be closer to the suspension point. The correct equation for the period
is T =2π √[I /(MgD )]
where I is the moment of inertia about the
suspension point.
QUESTION:
Why doesn't your brain explode in a PET scan? Doesn't antimatter meeting matter create a lot of energy?
ANSWER: Yes, matter/antimatter
annihilation releases the maximum amount of energy—when
a positron meets an electron, 100% of their mass is
converted into energy. But how much mass do they have and
how much energy is that? Each particle has a mass of
about m =10-30 kg, so their total mass
is about 2x10-30 kg. The energy is then
E=mc 2 =2x10-30 x(3x108 )2 =1.8x10-13
J where c =3x108 m/s is the speed of
light. To put this into perspective, this would be the
energy of a particle of dust which has a speed of about 2
inches per second. Furthermore, nearly all of the energy
(which is in the form of two x-rays) escapes without
leaving any energy in the brain.
QUESTION:
Under "miscellaneous" here on your site,
you answer the following
question : "If the earth is curved how is
it you can get a laser to hit a target at
same height at sea level more then 8 km
away? How is it that it's bent around the
earth?" Part of your answer states that the
laser is perfectly straight. But spacetime
is bent (curves) within the gravity well of
a massive object like the earth. Astronomers
have shown that it's possible to see stars
that are actually behind such massive
objects because the light from the star is
bent around the massive object as it
necessarily follows (somewhat) the curvature
of spacetime around the object. So how is
the laser beam in the question you answered
perfectly straight?
ANSWER:
It was clear to me when I was answering this
question that the questioner was interested
in a classical physics question, not one
taking general relativity into account. You
are right, any mass (or other energy
density) will cause a ray of light to bend.
But in this case the amount of bending is
very, very tiny. If you apply the angle of
deflection equation θ= 4GM /(rc 2
to a beam of light tangent to the surface of
the earth at the surface of the earth, you
will find that θ≈ 0.0006"=1.67x10-7 °.
I think you will agree that this is
negligible in the context of the question I
answered!
QUESTION:
It takes 375 joules of energy to break a
human bone. How high must a 60kg person fall
to break a bone?
ANSWER:
There is no answer to this question. And it
really does not make any sense because it is
force, not energy, which breaks a bone. If
you delivered 0.01 J/s over 37,500 s
(approximately 10 hours) would you break the
bone? So what matters is the time the
stopping collision takes to deliver the
energy impulse.
QUESTION:
Hi, hoping you can help settle a debate I’m having with a colleague. We know that when a solid object spins, radial and circumferential tension exists within it due to centripetal/centrifugal force. However, for a massive object such as a planet, whose gravitational acceleration far exceeds centrifugal force, does that tension still exist within the object? I say that the product of gravitational and centrifugal forces results in a net force towards the object’s centre of mass leading to a net compression, and an object under compression cannot also be under tension. He states that the tension that would have existed due to the centrifugal force only would still exist and that the gravitational force makes no difference to this. As you can probably tell we’re not physicists! Can you help answer whether tension (such as a hoop stress force) would still exist in such a massive object, or would the gravitational force
'overwhelming' centrifugal force prevent tension from forming in the first place?
ANSWER:
Whenever you want to understand something, you need to
include all the forces acting on it. Centrifugal force is
what we call a fictitious force, it doesn't really exist.
When you are experiencing an acceleration, Newton's laws
do not work. However, if you add fictitious forces
cleverly, you can do Newtonian physics. If we are in a
rotating system like the earth we are accelerating
because in physics acceleration does not just mean
speeding up or slowing down but also includes changes in
your direction which is constantly happening to you as
the earth spins (unless you are at a pole). To see how
this works, look at an
earlier answer . So, suppose that you are standing on
the equator; there are three forces acting on you, your
own weight (gravity) which points toward the center of
the earth, the centrifugal force which points away from
the center of the earth, and the force (which points up)
which whatever you are standing on exerts to keep you at
rest. Suppose that your mass is 100 kg and the
acceleration due to gravity is approximately 10 m/s2 ;
then your weight is approximately 1000 N. The centrifugal
force is your mass times your speed (464 m/s) squared
divided by the radius of the earth (6.4x106
m), C =100x(4642 )/(6.4x106 )=3.4
N. Suppose you are standing on a scale; since you are at
rest, the force which the scale exerts up on you is
1000-3.4=996.6 N, about 0.3% smaller than your weight.
This is a long-winded answer to your question: yes,
forces due to rotation apply to anything regardless of
its mass or size. (If you are not comfortable with metric
units, 1000 N=225 lb and 3.4 N=0.76 lb. If you weighed
yourself at the poles the scale would read 1000 N if the
earth were a perfect sphere.
Here is another example: the earth is not a
perfect sphere, it bulges at the equator.
The reason is that the earth, when it was
just forming billions of years ago, was very
hot, almost molten, and therefore more
"plastic"; so the centrifugal force caused
it to flatten as it rotated. That would mean
that your weight at the poles would really
be greater than what it is at the equator
because it is closer to the center of the
earth.
QUESTION:
Hey, i wanted to discuss about the MPEMBA EFFECT. Mostly i have read that it has no valid and accepted explanation but i think it should be taken as common sense like if we draw an analogy with electrodynamics we can say that a body with higher temperature should be at higher potential and a cold body should be at lower temperature. So as the potential difference gets bigger the rate of flow of charge(current) gets higher, similarly the rate of flow of heat charge should also get higher. Hence, it should be taken as obvious that a hot body will cool down much faster than a relatively cold body.
ANSWER: I first note that I have already
given a quite lengthy
answer to the "hot-water-freezes-faster" hypothesis.
As you will see, what happens depends a lot on the
conditions of any measurement or experiment you might try
to do. Your attempt to bring in "potential" is pretty
muddled, so let's review rate of flow in electrodynamics
and in thermodynamics. Materials have a property called
electrical conductivity; this property tells you how much
electric current you will get if there is a voltage
(potential difference) between two points in the material—the
larger the conductivity, the larger the current will be.
Similarly materials have a property called thermal
conductivity; this property tells you the rate at which
heat travels throught the material for a given
temperature difference—the larger the temperature
difference, the larger the energy flow will be. But, this
will have very little influence on how quickly the water
will freeze. First, the change in thermal conductivity of water
changes by only about 10% between say 20°C and 70°C;
second, water is not a very good thermal conductor; and
third conduction is not the primary way water cools
because the density of water, unlike most materials in
the molten state, gets larger as it cools, the cooled
water at the surface sinks so most of the transfer of
heat inside the water is by convection. Certainly the hot
water loses energy at the surface faster than the cold
water, but it will eventually catch up with the cold
water and then the two will be indistinguishable unless
the cold water freezes before the hot water catches up
with it. As I explained in the earlier answer,
evaporation cools the hot water more and if a significant
amount of the hot water evaporates before it catches up
to the cold water, it will win the race because there is
less of it; but that is really cheating, isn't it. Sorry,
I do not know anything about the mpemba effect.
QUESTION:
I just want to ask if, can we solve the force of attraction between two identical pendulum bobs with only mass(0.35kg) as given? Is it solvable? I am just curious about this, my teacher discussed about this topic but with enough components to solve it. But with this problem, I really can't think of something to solve it with only one given component.
ANSWER:
You cannot calculate the gravitational force without
knowing the distance between them. And you could not
easily calculate the force unless the bobs were spheres.
QUESTION:
I was wondering, in particle annihilation between say a electron and a positron, how long does it take to occur? I know it's refered to as being instantaneous, or happening in a immeasurable amount of time in clearer terms.
ANSWER: I believe there is no good answer
to this question because when would you start and stop
the clock? And even if you could specify some particular
times, they would inevitably depend sensitively on the
initial conditions like how fast each was moving
initially. You can, however, measure the lifetime of a
positronium atom (one electron bound to one positron). In
vacuum the singlet state (spin zero) atom has a lifetime
of about 0.125 ns and the triplet state (spin one) has a
lifetime greater than 0.5 ns. The times for the atom in
various materials would be longer.
QUESTION:
If a carousel is out of control, spinning at high speed, and suddenly it is stopped, there will be chaos and horses and riders flying out of the carousel structure. How do you explain this in terms of physics? Are they ejected due to the centrifugal force? Loss of centripetal force? And what happens to the kinetic energy? Is it transformed to what kind of energy?
ANSWER:
If the carousel is moving in a circle with constant
angular velocity, the only forces horizontally are forces
necessary to provide the centripetal acceleration. In the
photo, the girl holds the pole and presses on the side of
the horse with her leg to provide those forces; the horse
is held by the force of the pole to which it is attached.
All horizontal forces are toward the center of the
carousel because the speed of everything on the
carousel is constant. What happens if the carousel
suddenly stop. All those radial forces quickly drop to
zero. The tendency is for everything on the carousel to
continue moving with the velocity they had before, but
now in a straight line . But now, although the
horizontal radial forces are gone, each object
experiences horizontal forces opposite the direction of
their velocities. The girl would initially move forward
until she smashed into the pole; the horse would probably
be held in place by the pole although the force required
to stop the horse could very possibly bend the pole;
someone standing on the spinning floor would only have
the friction of the floor to stop her and would likely
keep moving in a tangential direction. Nothing is
"ejected", it just keeps going in the direction it was
going when the carousel stops, unless something stops it.
QUESTION:
I push on a wall and am accelerated backward. Per Newton’s third law, how does the acceleration of the wall manifest itself.
ANSWER: You push on the wall with some
force F and, as you note,
Newton's third law says that the wall pushes with a force
-F . So the magnitude of your
acceleration a is a=F /m where
m is your mass. The magnitude of the wall's
acceleration A is A=F /M where
M is the mass of the wall. For all intents and
purposes, the mass of the wall is infinite so its
acceleration is zero.
QUESTION:
Are electrons made of quark/anti-quark pairs? If yes, which pair?
ANSWER: No. To the best of our knowledge
electrons are elementary, i.e. they have no
components.
QUESTION:
What force pulls the train? I am doing a science project about this simple electromagnetic train. This project looks so simple yet so complicated, here is the
video
of the project that I am working on.
ANSWER:
The construction details and a brief description of the
physics are shown in another youtube
video . Essentially, when the magnets touch the copper
wire the battery causes a current to flow in the coil
between the two batteries which causes a magnetic field
in that section of the coil; each magnet, now being in
that field, experiences a force moving it forward.
Important things:
The same pole, north or south, must point away
from the battery.
Be sure to use bare copper wire. Often copper
wire has an insulating layer on its surface
and this would not allow current to flow
Good luck on your project. It is really not as
complicated as you thought.
QUESTION:
I can't grasp how two waves can pass each other on the same piece of string. For a wave to travel on a string each piece of string is, let's say, pulled up wards by the preceding piece of the string, and the wave propagates forwards. If a wave moving in the +x direction meets a reflected inverted wave in the - x direction, a node will be formed as one wave pulls that piece upwards and the other wave pulls it downwards. Therefore the piece of string doesn't move, so how can either wave travel past this point?
ANSWER: What you need to understand to
see why the two waves, moving in opposite directions are
able to do so is a little bit of the physics of waves on
a string. Any wave which moves through a medium is a
solution of a very famous equation, the wave equation:
d2 f (x ,t )/dt 2 =v 2 d2 f (x ,t )/dx 2 .
Here, x the position on the string, t
is time, v is the velocity of the wave, and
f (x ,t )
is the solution to the equation which will describe the
shape of the wave at a time t . You may not know
calculus and this equation is goobledy-gook to you, but
all you need to know is that, mathematically, any
function f is a solution as long as it of the form
f (x-vt ). The most commonly used example of
waves is sinesoidal, for example, f=A sin(kx-ωt )
where k is called the wave number and ω is
the angular frequency of the wave. Note that ω /k=v.
Also, to touch base with quantities you might be more familiar with,
k= 2π /λ and ω= 2πf where
f is the frequency of the wave and λ is the wavelength.
Now comes the the important part: because any f
will be a solution to the wave equation, if you have a
wave traveling to the right, f right =A sin(kx-ωt ),
and an identical shaped wave traveling to the left, f left =A sin(kx+ωt ),
their sum will also be a solution to the wave equation.
As you can see from the figure, waves traveling
simultaneously right (red) and left (green) add up to a
"standing wave" (brown) which does not appear to move but
is still oscillating*. You are correct, there are nodes
which do not move but both component waves go right on by
them nevertheless, it just isn't apparent when you look
at there sum. The fact that the net motion of the medium
(string) is just the sum of all individual waves is
called the superposition principle.
*If you're handy with trigonometry, you can
calculate f right +f left
which is apparently not moving even though
you know that before you did the calculation
you could certainly see that it was two
waves.
QUESTION:
Hi, I have been toying with the
concept of using gravity as a perpetual energy source,
and have had a lot of pushback in the process of trying
to ask questions. The conversation usually ends before I
can get an answer. The skepticism surrounding perpetual
motion is understandable. If I understand correctly the
issue is that closed systems lose energy to various
forces such as friction. Firstly, does it count as
perpetual motion if the energy is supplied constantly
from outside forces?Secondly, if not, is there a proper
term for what I've described?
ANSWER: I am afraid that you have it
exactly backwards. In a closed system, defined as one
which has no external forces acting on it, energy is
conserved. In a closed system where friction is present,
when kinetic energy is lost (e.g . a spinning
wheel slowing down or a box sliding to a stop on a
surface), the lost kinetic energy shows up mainly as
heat. And, if you do work on a closed system to keep it
moving forever, that is not what we mean by perpetual
motion. Proper term for what? If you mean motion that you
keep going by pushing on it, it has no particular name.
QUESTION:
First, I am way out of my field of understanding here so please keep it simple. I watched some videos on E=MC2 which led to how light reacts differently than matter at high speeds causing time to slow down when moving fast. My question is, if I was to shine a flashlight perpendicular (90 degrees) to the direction traveled am I correct to say if I was moving at half the speed of light the beam would actually be at a 45 degree angle and when travelling at the speed of light the beam would be horizontal (0 degrees). This would also be true whether the beam was inside or outside of the spacecraft, correct?
ANSWER: No, you have it wrong. The most
important thing to remember here is that the speed of
light, c , is the same for all observers. If you
are in the rocket and shoot a beam of light straight
across the ship, it will go straight across the ship in
some time ts and the distance it goes
will be cts ; if the width of the ship
is w , ts =w /c .
Now, someone on the ground will see the rocket moving by
with some speed v , so the point on the inside of the ship
where the light beam will have moved a distance vtg
when the light strikes it; but since this is light, the
distance it will travel is longer because the light has
to go farther. Now, to find the angle the light relative
to the perpendicular to the velocity v ,
we note that sinθ =v /c (tg
cancels out). In your question, since v=c /2, so
θ =sin-1 (0.5)=30°. This
example also demonstrates time dilation because the two
observers see different times of transit of the light. If
you work it out, tg =ts /√[1-(v 2 /c 2 )].
QUESTION:
The radius of earth is 6440 km . Suppose, the angle between Canada and USA inside the earth is 10 degrees , then what is the distance between them? I have been stuck with this problem . Can't i answer this problem using The Cosines Laws? If i do so....does it would go wrong? please answer my question...? It is not a homework question . i am thinking to solve this one with a different way....
ANSWER: This is a very strange question
because the USA and Canada are adjacent and therefore the
angle would be 0°. However, you can easily find the
distance between two points which subtend 10° just by
knowing the definition of radian measure of angles. The
angle θ subtends a distance s for a circle of radius
r . This angle, in radians, is defined as θ =s /r .
Since the circumference of a circle is 2πr and there are 360° in a circle, there are 2π radians in 360°.
So 10°=10x(2π /360)= 0.175=s /6440,
so s =1124 km.
QUESTION:
Hi! I would like to settle a bet with my father.
My questions is that if I was sitting in a chair that was tied to a rope, and the rope was divided by a pulley on the ceiling, would it be possible to lift yourself up just by pulling down on the other side of the rope.
ANSWER: The figure on the left shows all
the forces on you: T is the
force which the rope exerts up on you*, N is the force
which the seat of the chair exerts up on you, and
Mg is
the force of gravity (your weight) down on you. Choosing
+y up as indicated, Newton's second law in the
y direction is, for you, N+T-Mg=Ma
where a is your acceleration and M is
your mass. The figure on the right shows the forces on
the chair: -N is the force you
exert on the chair, the same magnitude but opposite
direction as the force the chair exerts up on you because of Newton's third law; T
is the force which the rope exerts up on the chair, the
same magnitude as the tension on you because the rope
and pulley are assumed to have negligible mass;
mg is the force of gravity (weight)
down on the chair. Newton's second law in the y
direction is, for the chair, -N+T-mg=ma where
a is the acceleration of the chair (the same as
yours) and m is the mass of the chair.
Let's first assume you are exerting just the right force
T on the rope so that you are at rest, so a =0.
In that case, if you solve the two equations you will
find that T =(M+m )g /2. In other
words, you need to be able to exert a force down on the
rope which is equal to half the weight of you and the
chair combined to hold the chair from falling. All you
need to do is be able to pull just a little harder to
move upwards. That answers your question but it interests
me to look at a case where a is not zero.
If you solve the two equations for a and N
you get
a =[(2T /(M+m ))-g ] and
N=T (M-m )/(M+m ).
A few observations about what these results tell us:
If you let go of
the rope, T =0, a=-g , N =0;
you and the chair are in free fall.
If m=M ,
N =0, a =(T /m )-g ;
the motion of you and the chair are identical even
though there is no interaction between you and the
chair.
You will
accelerate upwards if T >½(M+m )g ,
half the total weight. Your strength is the only
thing limiting how fast you can accelerate upwards.
If 0<T <½(M+m )g
you will accelerate downward with an
acceleration smaller in magnitude than g . Of
course T <0 is not possible for a rope.
*It is
important to note that, because of Newton's
third law, if the rope exerts a force up on
you, you exert an equal and opposite down on
the rope. T is a measure of how hard
you are pulling. How large T can be
depends only on how strong you are and how
strong the rope is.
QUESTION:
We define image as the intersection point of reflected rays and the focal point is the point at which light will meet after reflection.
So, why is it that we have different positions of image when object is placed at different distances from concave mirror and not always at focal point except when object is placed at infinity?
ANSWER:
Because only rays which come in parallel to the optic
axis are reflected through the focal point. This occurs
only (approximately) for objects very far from the
mirror.
QUESTION: Does radiation from your phone stay on your hands after you use it?
ANSWER: Absolutely not. Any radiation
from the phone is electromagnetic, essentially radio
waves. It is transient and does not "stay" anywhere. Even
when you hold the phone the radiation is not "on your
hands" but harmlessly passing through them just like the
waves from some radio station is passing through your
whole body.
QUESTION:
When a uranium or plutonium atom is fissioned, energy is released visa E=MC2. I believe this energy is in the form of photons. What I do not understand is how these photons (light) create such high temperatures in an uncontrolled nuclear explosion. Can you please explain?
ANSWER: Actually, emitted photons account
for only about 2.5% of the total fission energy. Emitted
neutrons account for 3.5%. Most of the energy is
contained in the kinetic enerngy of the fission
fragments, 85%. The other approximately 9% of the total
energy shows up later when the fission fragments, which
are not stable, decay radioactively, mainly from β -decay.
The high temperatures are due to the kinetic energy of
the fission products which have speeds typically 3% the
speed of light.
QUESTION:
I'm embarking on a horological project. I need to buy a very particular type of spiral spring. I guess it's a type of torsion spring. The only supplier I can find who sells these springs classifies them with two numbers. The overall diameter in mm
and the torque in gm/cm/100°.
(As noted below, I assume that this means the torque on the spring when
the angle is 100° and it should be gm·cm, not gm/cm.) I know the spring I need. But I only know in terms of the CGS System. I'm following an old book and I've ascertained that the spring I need has a
"CGS number" of 0.76.
So, my challenge is to find a way to convert between these two unfamiliar units of torque measurement. A
"CGS number" and gm/cm/angle.
There are plenty of calculators for converting between different units of torque. But I don't think it's as simple as that. I have two questions.
I know a bit about the CGS system and there is loads of info online. But I'm not clear what it is in the context of classifying a torsion spring. The book only says
"the CGS number indicates the restoring couple of the spring when its diameter is 1cm." So what is the CGS number exactly? Is it dyne-centimetres? If so, then that's easy to work with since that's a measure of torque I'm familiar with. So I'm half way there. If it's something else, then perhaps you can help explain.
Then there is the question of gm/cm/angle. The problem is that none of the calculators I've seen have a notion of angle. This is specific to torsion springs. The torque, in this case, is the working load when the spring bends 100°.
Will I ever find the correct spring? Sadly, I have spoken to the manufacturer, and they were unable to help me.
ANSWER: I think the "number" of interest is not a torque, rather
the spring constant associated with the spring for
angular displacement. First, a review of linear springs
which when stretched stretch a distance x when a
force F is applied to the spring. It is
experimentally determined that, to an excellent
approximation for a good steel spring, is that the
distance stretched is proportional to the force applied,
x ∝F or F=kx . The
proportionality constant k is called the spring
constant and is measured, in SI units, as N/m (Newtons per
meter) [dynes/cm in CGS units, lb/ft in Imperial units].
k indicates the stiffness of the spring
and this equation is called Hooke's law. The form of
Hooke's law for angular motion (as in a torsion spring)
is that the angular displacement θ is proportional
to the applied torque τ or τ=κθ ;
I believe the quantity you want is κ
(kappa) and is measured, in SI units, N·m/radian
[dynes·cm/radian in CGS units, lb·ft/degree in Imperial
units]*. I have noted that κ is often called, by
manufacturers and vendors, the rate
of the spring since it is the rate of change of torque
per unit angle.
I am puzzled by your reference to a constant defined as
gm/cm/degree; 1 dyne=1 gm·cm/s2 , so κ
should have units gm·cm2 /s2 /degree;
sometimes the gram is treated as a unit of force (called
a gram force where 1 gram force=980 dynes), in which case
κ
would be gm·cm/degree, not gm/cm/degree. So, there
must be somewhere in the reference you are using a
definition of what the CGS number is. Suppose that it is
0.76 dynes·cm/radian (which would be the likeliest
units if purely CGS system of units is applied) but you want it in
lb·in/degree; then 0.76 (dyne·cm/radian)x(2.25x10-6
lb/dyne)x(0.394 in/cm)/(57.3 deg/radian)=1.18x10-8
lb·in/degree. Or, suppose you wanted it in (gm
force·cm/degree); then 0.76 (dyne·cm/radian)x(gram
force/980 dyne)/(57.3 deg/radian)=1.39x10-5 (gm force·cm)/degree).There are lots of unit converters
around, my favorite is
here .
The choice of units for κ which it seems most vendors use
is lb·in/deg. The best calculator I found to get
an idea of the magnitude of these springs can be found
here .
Until you are able to deduce the units of the so-called CGS value of
0.76 you cannot know what it is.
*Sometimes
κ is implicitly given by specifying the
torque at a particular angle.
ADDED
THOUGHTS: In the book by Hans Jendritzki,
Watch Adjustment , he says that in the CGS system
for characterizing the spring in question (which I will
call N CGS ), the unit of force is the
dyne and the diameter of the spring is 1 cm. He then
uses not the restoring torque to calculate N CGS
but rather the
restoring couple associated with the force; the
restoring couple is the restoring force times the
diameter whereas the torque is the restoring force
times the radius . Most vendors use the
restoring torque to characterize springs, so ½N CGS
should be used to convert to the vendor's units.
Although the units of the angle are not given in
Jendritkzi, I have found that the more natural radians
gives values much too small when converting to vendor
units; so N CGS =(restoring couple in
dyne·cm for 1 cm diameter)/degree.
One vendor charactorizes the spring by specifying a
torque given by N gm/cm/100°; this does not
appear to be dimensionally correct unless the gram is
gram force (I will denote it as gmf) where 1
gmf=981 dynes (the weight of 1 gm in dynes) and /cm
means that again this is for a 1 cm diameter spring. So,
assuming that the vendor's torque is indeed torque
(force times half the diameter), ½N CGS =N .
For example, suppose N CGS =2; then
N =½(2) (dyne·cm/1°)(1 gmf·cm/981
dyne·cm)/(100°/1°)=0.102 gmf·cm/100°; so the conversion
from the CGS value to this vendor's value is N =0.0491N CGS .
So, in the case of N CGS =0.76
dyne·cm/1°, N =0.037 gmf·cm/100°.
[Disclaimer: I am
not a clockmaker and I am not familiar with conventions
clockmakers might use or meanings they may attach to
certain words which are not in accordance with
definitions, units, or conventions used by physicists.
In the event that I have misunderstood or that either
Jendritkzi really meant torque or the vendor really
meant couple, the conversion factor would be 0.102
instead. I have done my best to determine conversions
between two numbers representing the same thing but in
different systems of units.]
QUESTION:
If you lined up frictionless gears over an exceptionally long distance, could you effectively communicate faster than the speed of light through one person moving the first gear and a second person reading/interpreting the resulting movement of the last gear?
ANSWER:
I have answered this question in many guises
many times . In your
variation, each gear will experience a force from the
previous gear in the line and exert a force on the next
gear. The time it takes for the force to travel from the
input force location to the output force location is
determined by the speed of sound in the gear. The message
you send would travel at the speed of sound, much smaller
than the speed of light.
QUESTION:
If you are driving side by side at 60mph with another car and throw a can of coke at them would the can be travelling at the same speed so hit the target as if you were parked next to each other or merely disappear behind your vehicle?
ANSWER:
When you throw it out the window it originally has a
forward component of its velocity of 60 mph. At this
speed, the can will experience a very significant force
slowing its forward velocity down because of the air
drag. So the can will not keep pace with the two cars and
will appear, from either car, to be accelerating
backwards. So, no, the can will not hit the other car
where it would have if the cars had not been moving.
QUESTION:
Under relative motion, it's experimented that A person sitting on a back seat in a moving bus has a speed with respect to the earth which is the same as that of the bus. But if he now walks towards the driver of the bus,he has a speed relative to the earth which is more than that of the bus. Suppose the bus is moving at 100km/h and the person walks at 5km/h towards the driver, his forward speed relative to the earth is 100+5= 105km/hr. But when he walks back from the driver to his back seat at the speed of 5km/hr his speed relative to the earth is now 100-5= 95km/hr.
WHY the increase speed of such person when moved forward and the decrease when moved backward
ANSWER:
The problem you are having is that you do not understand
the phrase "relative to the earth". Let's alter your
example by having the train moving at 5 km/hr. If you are
in the train walking 5 km/hr in the opposite direction,
someone standing by the side of the tracks will
see you standing still; the same as running on a
treadmill in a gym with the treadmill and you moving in
opposite directions with the same speed. If you walk in
the same direction as the train is going, you will be
seen by the trackside observer as moving with a
speed of 10 km/hr.
QUESTION:
IF WAVES need a medium to transmit energy but light can travel through a vacuum.
Does dark matter have Anything To Do With Lights Ability To Do This?
ANSWER:
There is absolutely no mystery about light waves;
electromagnetism is arguably the best-understood theory
in all of physics. You may be sure it has nothing to do
with dark matter. Dark matter is arguably the
least-understood theory in all of physics. There are many
features of the universe which do not fit into our
current understanding of physics and introducing some
kind of particle which interacts with the rest of
creation only via gravity would solve a lot of
them; all attempts to directly detect dark matter have
failed and until there is direct observation we
understand nothing. To my mind, failure to observe dark
matter may be indicative that we do not understand
gravity as well as we think we do.
QUESTION:
Bowling ball vs.
billiard ball
What would ultimately travel further, a 12 lb bowling ball or a standard
billiard ball: if thrown/rolled with the same amount of physical strength/athletic ability, at ideal trajectory for each, on a frozen lake (not without small imperfections but as close to perfect as could be found naturally), under normal atmospheric conditions (Wisconsin in the winter), with little to no wind? Would temperature play a roll in determining a winner? Would the small difference in friction coefficients be significant? Any other variables Etc...
My brother and I have been debating this for years while ice fishing.
ANSWER:
(You might not want to read this rather long
introductory paragraph if you are just interested in the
final answer. ) This problem was very interesting to me
because I initially thought it was pretty easy to
understand. The billiard ball starts out with much greater
speed than the bowling ball, jumps into the lead.
Neglecting air drag is usually what is done initially in
considering such problems and there has to be some
kinetic friction for the balls sliding on the ice; but
because the kinetic friction is proportional to the
weight, each ball experiences the same rate of change of
its speed as it slows. Therefore the billiard ball is the
obvious winner. When I initially solved the problem this
way using reasonable numbers, I found that the bowling and
billiard balls had initial speeds of 14 mph and 81 mph,
respectively; the bowling ball went 686 ft and the
billiard ball went 21,700 ft, a little more than 4 miles! When I
was writing the final paragraph I discussed the possible
effects of two things I had neglected, rolling of the
balls rather than sliding and the possibility of air
drag. Just for fun I estimated the effect of air drag and
discovered that at the beginning, particularly for the
billiard ball, the drag force was considerably larger than the
sliding friction. I realized that I had to redo the whole
problem, now much more difficult and mathematically
involved than originally. Because the air drag estimate I
used is valid only if you work in SI units, I will do
that throughout and convert to imperial units when needed.
The mathematics gets very complicated and I will only
sketch my calculations and whoever is interested can fill
in the gaps.
First I will tabulate quantities needed for the problem:
We need to
quantify what is meant by "…the same amount of
physical strength/athletic ability…" I will
assume that the thrower exerts a constant force F
over a distance d on both balls. Physics says
that the result is that a kinetic energy K =½mv 2
is acquired where m is the mass of the ball and
v is the speed it has when it leaves the
thrower's hand; this kinetic energy is equal the the work
W done which is W=Fd . that is, Fd =½mv 2 .
If you solve this equation for V you find the speed,
v =√(2Fd /m ). I will use
Fd =110 J (about 25 lb over 1 m) because this
value would give the billiard ball a speed of about 80 mph
which seemed reasonable to me.
The masses of the balls are
0.17 kg=6 oz for the billiard ball and 5.4 kg=12 lb for the bowling ball.
I will use
a coefficient of sliding
friction for each ball which is very small, μ =0.01.
Each ball experiences a frictional force of f=μmg=ma
(Newton's second law) where a is the
acceleration and g =9.8 m/s2 is the
acceleration due to gravity.
For the air drag
I will use the approximation D =¼Av 2 where
A
is the area presented to the oncoming air (πR 2 )
and v is the speed. The radii of the balls
are 0.0286 m and 0.108 m. As noted above, this
equation is only valid for SI units because the
factor ¼ includes things like air density, drag coefficients,
etc .
We are now ready to
start doing the physics. Start with Newton's second law
ma=m (dv /dt )=-¼A v 2 -μmg
dv /dt=-Cv 2 -μg
(C =¼A /m ).
This equation is
integrable. The result is
t =tan-1 [v √(C /(μg ))]/√(Cμg )|v(0) 0 .
Evaluating
this expression between the limits v=v (0)
to v =0 and inverting to solve for v (t )
is straightforward but messy. The essential
results are illustrated in the two graphs below.
The effect of air drag on the bowling ball is
modest but noticeable, reducing the time to stop
by about 10 s. The effect on the billiard ball,
though, is enormous; because of its high speed
at the start, it is slowed much more than due to
sliding friction alone. Still, it stayed in
motion about 20 s longer than the bowling ball
and was going faster most of the time, so
certainly went farther before stopping. Still,
it would be useful to find the position as a
function of time to determine how far each ball
went.
To find the position, x (t ), is
straightforward but again messy: integrate the
differential equation dx /dt =v (t )
where v is the function being plotted
in the graphs above. The results are shown in
the graphs below. The bowling ball stops at
t =54 s at a position x =158 m=518
ft; the billiard ball stops at t =77 s
at a position x =535 m=1755 ft. The
winner is the billiard ball which goes more than
three times farther than the bowling ball! I
believe that regardless of any details of the
friction and drag forces or how we choose the
quantity Fd , the billiard ball will
always go faster if the initial kinetic energies
are equal.
Finally, I want to discuss rolling vs .
sliding. It is really hard to get an object
rolling without slipping on a very slippery
surface. A sliding ball will not roll until it
is nearly ready to stop, and that would have
very little effect on my calculations. Also,
rolling friction is velocity-independent and
proportional to the normal force so it would
affect both balls the same.
QUESTION:
I have a question about gravity. In order to general relativity mass bends the space. I saw a simulation of gravity that a man put a mass on a surface like silk. that the mass bends the silk. and we put another mass they attract to each other. that 2D silk is an examinition of our #d world. I wanna know why masses attract to each other. is it becuase of bending space?
ANSWER:
That simulation is a good way to understand how space
bending can alter how objects move but keep in mind that
it is only a cartoon to illustrate the warping of
spacetime. See my
faq
page .
QUESTION:
If repulsion forces between protons is high why hasn't the nucleus exploded
ANSWER:
It is true that the Coulomb force experienced by protons
in a nucleus is very strong. However, the nuclear force
at very short distances is stronger and wins the
"tug-of-war" and holds them inside. However, this strong
force is also a very short-range force and if you pull a
proton just outside the nucleus, it will be repelled and
fly off. It is also interesting to ask why the nucleus
doesn't collapse if the nuclear force is so strong. The
nuclear force becomes repulsive if the protons get too
close together; this is called the saturation of nuclear
forces.
QUESTION:
What would be a ballpark estimate of all the mass converted to energy (lost) by all the fusion reactions in all the stars that are or have ever been in the last 13 billion years (lots of assumptions)?
Would this lost mass effect the overall gravitational attraction of the universe?
Would it be significant enough to be a factor in the current acceleration of the universe?
ANSWER:
I would not presume to make even the roughest estimate
you ask for. However, it is not relevant to the questions
you seem to be interested in. According to general
relativity, gravitation is caused by energy density, not
specifically mass. (Don't forget, mass is just a form of
energy, E=mc 2 .) So changes of mass
will affect the details of the gravitational field in
some volume of space but if you get very far from that
volume, the gravity you see originating from that volume
will not change.
QUESTION:
Our 5-year-old niece wants to know why it's so hard for her to ride her Big Wheel uphill (3% slope) on her family's driveway. My BSN-RN wife would like a friendly, informative way to break that down for her (Daphne), including whether her weight (50 pounds), her position on the Big Wheel, the coefficient of friction on said drive wheel, and other factors play a part. Said aunt would also like a kid-friendly way to calculate the horsepower necessary for her to ascend that grade on her Big Wheel for perhaps 100' (the driveway's length). No, this is in no way homework. Except it's at their home. And, well, it is work.
ANSWER:
A little tough for a
five-year old! Here, maybe, are some suggestions:
Get her to
understand Newton's first law: an object at rest or
moving with constant speed has all the forces on it
cancelling each other out.
Get her to
understand Newton's second law. If the forces are not
cancelling each other out, the object will start
moving and continue to speed up in the direction of
the uncancelled forces.
For example,
have her imagine sitting on the trike on the
incline with her feet off the pedals or the
ground. She will start moving down the incline.
What is happening is that her force, which is
still straight down, must have a little bit of
itself pushing her down the incline. (This is
basically a component of the weight but the
concept of vectors and their components is likely
too hard for a five-year old to comprehend.)
Finally, if, on
the incline, there is a piece of the weight which she
will have to compensate for in order to keep that
force from making her go down the hill: she does this
by pedaling to create a force up the incline: the
force she generates must be greater or equal to the
piece of the weight pointing down the incline. On
level ground there is no unbalanced weight so the
pedaling is much easier.
Friction, I
would opine, is an unnecessary complication.
If she were
heavier, the piece down the incline would be bigger
so it would be harder to pedal.
(This part is
for you, the little girl will not get it.) Your
question about the horsepower is incomplete because
to calculate the power you need to know the time it
takes her to go the 100 ft. Suppose she maintains a
speed of about 2 ft/s. Now, the component of a 50 lb
weight on
a 3% grade is about 50x0.03=1.5 lb so the work done
is 1.5x100=150 ft·lb. The time it takes her is
(100 ft)/(2 ft/s)=50 s, so the power is 150/50=3 ft·lb/s=0.0055
hp.
Friction has
little to do with it because the friction is not
significantly different on level ground vs .
the incline. Position on the trike plays no role
since any change would be the ease of pedaling which
would be the same in either case.
QUESTION:
what would a light/sound wave look like in 3 dimensions, all diagrams are in 2d and it just came to mind that it wouldn't be like that in real life
ANSWER:
Here is a cylindrical wave. The blue surfaces represent
wave fronts.
QUESTION:
I am currently doing an experiment on the metronome pendulum. As you know a metronome pendulum has two types of pendulum which is the inverted pendulum and simple pendulum. It is called as double-weighted pendulum. This type of pendulum has two weight which is one above the pivot and one below the pivot. In my experiment I want to find the relationship between the varying mass of the top mass of metronome pendulum and its angular frequency but I cant seem to find any formula that describes my metronome pendulum. What formula for angular frequency should I use because from my data, I will be looking for its period of oscillation. There are too many angular frequency formula for different types of pendulum so I am confused on which formula I should be using.
ANSWER:
I will show you how to get a general answer. Then you can
apply it to the conditions of your experiment. Newton's
second law for rotational motion is τ=Iα=I (d2 θ /dt 2 ) where
τ is the net torque about the pivot (the blue cross in the figure),
I is the moment of inertia about the pivot, α is the angular acceleration, and
θ is the angle of the system at some time
t . If the system is a pendulum with no forces
exerting torques except weights (vertically down), all torques are
proportional to sinθ where here
θ is the angle relative to the vertical.
Now our equation may be written as
(d2 θ /dt 2 )=-(|τ| /I )sinθ
where
|τ| is the torque about the pivot when
the pendulum is horizontal (θ=π /2). The
negative sign results from the fact that the angular
acceleration and the angle are always opposite, a
restoring torque. This equation is extremely difficult to
solve, but if the angle is small (much smaller than 1
radian), you can use the small angle approximation, sinθ≈θ ,
which gives us
(d2 θ /dt 2 )≈-(|τ| /I )θ
This is the simple harmonic oscillator equation for an oscillating
system for which we know the angular frequency, ω =√(|τ| /I ).
So, let's apply it to your situation as illustrated in my
figure. There are two point masses and the mass of the
rod itself which I will take as m . The rod has a center
of gravity (orange cross) a distance of L 3
from m 1 as shown; if your rod is
uniform, L 3 =L /2. (If the
mass of the rod is much smaller than both the point
masses, you can approximate m as zero.) Now,
since all weights are of the form W=mg , you can
write |τ| :
|τ|=g (m 1 L 1 -m(L 3 -L 1 )-m 2 L 2 ).
The moments of inertia for the two point masses are
I 1 =m 1 L 1 2
and I 2 =m 2 L 2 2 .
The moment of inertia for the rod is trickier, so to
avoid a lot of messy algebra I am going to assume a
uniform rod where L 3 =L /2. I
find
I rod =(m /(3L ))(L 3 -L 1 3 )
The total I is just the sum of the three moments
of inertia. So, finally,
ω =√[(g (m 1 L 1 -m (L /2-L 1 )-m 2 L 2 ))/(m 1 L 1 2 +m 2 L 2 2 +(m /(3L ))(L 3 -L 1 3 )]
And, if m can be neglected,
ω =√[g (m 1 L 1 -m 2 L 2 )/(m 1 L 1 2 +m 2 L 2 2 )]
QUESTION:
So I have just thought about this, let’s just say that hypothetically you were able to fall through the earth completely. I was wondering what the gravity would do to you if you went in one side and came out the other
ANSWER:
See the
faq page .
QUESTION:
If everything is expanding, and every galaxy is moving away from the Milky Way, why is the Milky Way and Andromeda predicted to colllide in 3.5
billion years?
ANSWER:
See an
earlier answer . See a simulation
video .
QUESTION:
this is for a horror book I am writing. how fast can a bat fly, if it weights 4500 pounds and is 25 feet 6 inches tall? and in 1 flap of its wings, can rise a rise a quarter, of a mile? also how wide would its wingspan be?
ANSWER:
The best I can do is to estimate how the wings would have
to be if the ratio of weight (W ) to wing area (A )
is the same as a real bat. I find that a typical value
for that ratio for bats is about R =W /A =20
N/m2 =0.42 lb/ft2 .
So, for your case, A=W /R =4500/0.42≈10,000
ft2 or one wing has an area of about 5000 ft2 .
If I model a wing to be a triangle with base B =25 ft and
a width H , 5000=½BH =12.5x25xH
or H =400 ft, a wingspan of about 800 ft.
Since it has the same R as for a real bat, it
should be able to fly as fast as a real bat which can be
as high as 100 mph.
QUESTION:
I am a part time instructor and one of the things I teach is rigging. An
important part of the class is calculating sling stress. I was asked a
question that I couldn't quite answer. I don't believe it was my high
school or college physics teachers that failed me rather the more than two
decades since I sat in a physics class. The question relates to vector
forces. If I am hoisting a load of a given weight (say 2000 lbs) and I have
two slings that are at a given angle (say 45 degrees) from the load. Each
sling would be carrying half the load or 1000 lbs straight up. However, the
sling in this example would have 1414 lbs of force (inverse sine if I
remember correctly). The question I was asked was regarding a crushing or
compression force. Since each sling is carrying 1000 lbs vertical load that
means it has 414 lbs of force perpendicular to the vertical. Does that mean
that there is 414 lbs of "crushing" force or is it 828 lbs since each sling
is pulling towards the center of gravity.
ANSWER:
After an email exchange with the questioner, I was able
to determine that "sling" is just a rope or a chain or a
string etc . In the diagram I have shown all the
forces on the load, the tensions in the slings,
T 1 and T 2
and the weight of the weight, W .
The magnitudes of the tensions are equal, T 1 =T 2
=T . Also shown are the horizontal (x )
and vertical (y ) components of the tensions;
they are T 1x =T cosθ ,
T 2x =-T cosθ ,
T 1y =T 1y =T sinθ.
So the equilibrium equation is 2T sinθ-W= 0;
therefore T=W /(2sinθ ).
So, for your example, sinθ =cosθ =1/√(2),
W =2000 lb, and T =1414 lb. Now, your
interest is in the horizontal components; each is T cosθ=W /(2tanθ )=1000
lb "crushing force" on each side. Your main error is that
you treated the vector forces as scalars, you assumed
T=Ty +Tx where in fact T =√(Ty 2 +Tx 2 ).
QUESTION:
Since everything in the universe is always moving, is the idea of being still/idle/stationary just an illusion?
ANSWER:
I would not put it like that. If you are in a system
where Newton's first law (if the net force on an object
is zero, it will remain at rest or moving with a constant
velocity) is correct, you may think of yourself at rest.
This is called an inertial frame of reference. But the
catch is that this is not the only frame of reference in
the universe where Newton's first law is true, any other
frame which moves with a constant velocity relative to
yours is also an inertial frame. In other words, there is
no such thing as absolute rest. A frame accelerating
relative to yours, however, is not an inertial frame.
QUESTION:
I am a cancer patient about to receive Y-90 treatments. If Y-90 has a half life of 64 hours: 1. When does that time clock begin running and 2. How can they ever have a full strength dose on hand at any given time?
ANSWER:
90 Y is the decay product of 90 Sr
which has a half life of about 29 years. 90Sr is an
abundant byproduct of the fission of uranium and
therefore available in the waste of reactors. When the
90 Y is needed it can be chemically separated
from the 90 Sr. I am sure that the hospital
cannot keep a supply on hand and will have it delivered
when it is needed for your treatment. It does not matter
how long it has been since it was separated, only that
the radiation level be correct for the prodedure. It
probably arrives with a level too high and they wait for
it to be at the needed level.
QUESTION:
I am struggling with the concept of relativity, as it relates to time. Not in terms of the underlying principle or the mathematics. These have been confirmed experimentally and are now in everyday use through technologies such as satellite navigation systems. My concern relates to the way that relativity treats time as a variable rather than an absolute quantity. It seems to me that we can only measure or experience time by reference to some physical process or change, whether that is the oscillation of a quartz crystal or the lifetime of a muon! Any experimental proof of time dilation as a result of the effects of either velocity or gravity really just relates to the way that we experience the passage of time and does not exclude the possibility that for the universe, as a whole, time is absolute and unchanging, passing in a constant fashion regardless of relativity and whether we measure it or not. Therefore, although relativity theory works in practice, this is only because it deals with our limited understanding of, and ability to measure and/or experience, the passage of time. Is this nonsense or is it a possibility? It follows from this reasoning that we can never prove experimentally that time either speeds up or slows down, only that the rate of how we measure or perceive time can vary. It also implies that time (and indeed space) are concepts that we will never fully explain as they ultimately cannot be qualified or quantified by measurement or mathematics.
ANSWER:
You are correct that "…time is absolute and unchanging,
passing in a constant fashion…" but only in your
frame of reference; any clock at rest relative to you
runs at a constant rate. But, what special relativity
tells us is that clocks not at rest relative to us do not
run at the same rate as ours. I have always felt that the
best way to convince someone that this is the case is the
light
clock . In order that this example is convincing to
you, you must accept that the
speed of light is a
universal constant in all frames of reference; but this
is an experimentally well-verified fact and also is a
consequence of the principle of relativity—that the laws
of physics are the same in all inertial frames of
reference.
QUESTION:
SINCE FREE FALL DONT NATURALLY OCCUR ON EARTH BECAUSE FLUID FRICTION IS PRESENT IF YOU
DROP A BOWLING BALL AND A BASEBALL OFF A TOWER OF PISA WHICH WOULD LAND FIRST
ANSWER:
The (upward) drag force on a falling sphere of radius R and speed
v in air
at sea level may be approximated as F =¼πR 2 v 2 =0.79R 2 v 2
(this is correct only in SI units); the (downard) force of
the weight W on an object of mass m is
W=mg . The net force is the mass times the
acceleration a (Newton's second law), ma =-mg +0.79R 2 v 2 .
So as the object falls it goes faster and faster until
v is large enough that it falls with a constant
value speed vt (a =0), vt =[√(mg /0.79)]/R .
For a baseball, m =0.144 kg and R =0.037
m, so vt baseball =9.28 m/s.
For a bowling ball, m =8 kg and R =0.12
m, so vt bowling ball =83.0
m/s, nearly ten times greater than the baseball. So the
baseball stops accelerating sooner than the bowling ball
and loses the race to the ground.
QUESTION:
My 8 year old asked me if the cycle of the universe expanding and contracting will ever end?
ANSWER:
What makes you and your son think that the universe is in
such a cycle? The ultimate fate of the universe is one of
the most important questions in cosmology and nobody
knows the answer. I believe most cosmologists believe
that the universe will continue expanding forever.
Eventually the supply of hydrogen and helium in the
universe will be exhausted (fused into heaviers elements)
and stars will no longer exist. The universe will be
dominated by black holes which will eventually decay away
by Hawking radiation, so the universe will be a very
dilute, cold, dark place filled with radiation. To get
more information, check out the
Wikepedia article on the fate of the universe.
QUESTION:
My son has a question.
A shadow is long or short, wide or narrow. He thinks this seems like area (taking up space). But the real question is... can you have a shadow without matter? Because if you can't have a shadow without matter, then doesn't that mean the shadow is actually matter? (I have to ask because I'm not sure how to answer this; I suppose the shadow is actually the part of the matter that doesn't have light reflecting off of it...? Or how do I answer this for him?)
ANSWER:
A shadow is not matter;
it is not anything, it is a region in which light
illuminating something is blocked by some obstruction.
You might say that a shadow is the lack of anything,
light which might have otherwise illuminated where it is.
And there are degrees of "shadowness". If all the light
illuminating an area is blocked by the obstruction, it is
called an umbra ; if only part of the light is
blocked it is called a penumbra . (Umbra is Latin
for shadow.)
QUESTION:
U recently answered a question regards Earth's precession = 26,000 yrs. Can u show what specific equation(s) used to yield this time? I doubt that a single equation was used.
ANSWER:
Just like the precession of a top is due to the torque
exerted by its own weight, the precession of the earth is
due to torques on the earth, mainly by the sun and the
moon. Your equations may be found in this
Wikepedia article .
QUESTION:
On an ordinary day as one goes upward from the surface of the ground the electric potential increases by 100 volts per meter, this means that outdoors the potential of your nose is 200 Volts higher than that at your feet. Then why is it that none of us get a shock when we go out in the street?
ANSWER:
As is often the case, the most lucid answer to your
question can be found in the
Feynman Lectures .
QUESTION:
Why doesn't the earth just stop spinning and orbiting don't you need energy for motion where does the earth gets it's energy and is the earth in perpetual motion?
ANSWER:
You are hitting one of Isaac Newton's most important
discoveries—inertia. It says that you do not need
to exert a force (torque) to keep an object moving in a
straight line (rotating about an axis) to keep it going;
this means, also that the energy it has by virtue of its
motion will not change. So the rotating earth keeps
rotating since there are no torques on it*. The earth is
in a circular orbit† around the sun and so its orbital
motion keeps going since, even though there is a force on
the earth, there is no torque.
*Actually, the moon does exert a torque on the earth
which is causing the spinning to get smaller, but the
change is so tiny that it takes millions of years to be
significant.
†The earth's orbit is
not exactly circular. When it moves closer to the sun it
speeds up, when it moves farther from the sun it slows
down by exactly the same amount, so the orbit just
continues forever.
QUESTION:
With the devastating tragedy that happened in Lebanon, there has been some videos surface of the explosion. One such
video is of some ladies in a shop, the pressure wave hits the shop and the lady outside the shop doesn’t get thrown back but inside the other lady does. What is the explanation here??
ANSWER:
Note that the glass doors are closed. Outside the front
of the pressure wave hits, but it takes some brief time
to reach its maximum. However, the pressure difference
between inside and outside is not large enough to break
the glass until it is near its maximum, so the pressure
difference changes much more quickly so the air rushes in
quickly. I will admit that this is just an educated
guess!
QUESTION:
If electricity always take the path of least resistance, why does a lightening discharge form a zigzag path to earth and not straight down which would be the shortest distance?
ANSWER:
Because air is not just a uniform homogeneous medium.
There are local fluctuations in temperature, humidity,
density, wind speed, and composition which makes the path
of least resistance at any instant, not a straight line.
QUESTION:
how momentum is conserved of both objects with same masses moving opposite direction to each other collide to a point and at same time both objects chemically combined to each other?
ANSWER:
In an isolated system, linear momentum is always
conserved; this is because the definition of linear
momentum is designed to be conserved in the absense of
any external force. This is also true in special
relativity where the linear momentum is not p=mv but
rather p=mv /√[1-(v 2 /c 2 )]
where c is the speed of light. In the event of
chemistry going on, there is either energy gained
(endotermic) or lost (exothermic). If lost, the momentum
of the radiation has to be taken into account. If gained,
the energy will be taken from the kinetic energies of the
incident objects but momentum will be conserved.
QUESTION:
Does precession and nutation ( or anything else ) affect the axis of the earth?
ANSWER:
Precession of the earth's axis is caused mainly by
torques exerted on the earth by the moon and the sun. The
period of precession is 26,000 years. But precession does
not cause the actual rotation axis to change. What causes
the axis to change (such that the geographical positions
of the poles would change) is thought to be three things:
glacial ice
melting and sending its water to the oceans,
glacial rebound
which is the raising up of the land previously
supporting the melting glacier, and
tectonic motion,
the drift of large land masses.
QUESTION:
The momentum transmitted to the golf ball comes from the club head and,to a lesser extent, from the club shaft.
Golfers are instructed to hold the club very loosely. Thus, it seems, no momentum is transmitted from the arm.
What if the golfer instead holds very tightly to the club? Will the momentum of the arm be transmitted to the golf ball?
ANSWER:
The physics of a golf swing is much more complicated than
you might think. From the little research I saw, your
"hold the club very loosely" is an oversimplification.
The club needs to accelerate very quickly on the way to
the ball and that acceleration is only going to come from
the torque you provide. What should happen is that, just
before impact, your wrists must relax. There is an
excellent discussion on the golf swing which, for the
first half is fairly conceptual and the second half goes
into the detailed physics analysis of the swing.
FOLLOWUP QUESTION:
I understand that the standard golf swing depends on uncocking the wrists to generate velocity at the moment of impact.
Moe Norman, a Canadian golfer now deceased, was the best ball striker in the history of golf. Moe used a completely different swing that he developed. Among other things, it involved holding the golf club tightly.
Thought experiment: The shaft of the golf club is fused to the golfer's arm and runs all the way up to the shoulder. That is, there is no wrist play at all. I understand that such a set-up will generate less velocity than a hinged wrist. Therefore, the 'v' in 'mv' will be lower. My question is whether the 'm' will be higher. Given that the golfer is now swinging a much more massive club, albeit from the shoulder, I posit that the mass striking the ball is greater.
Take the center of gravity of the club-arm, calculate its velocity, multiply by the mass of the club-arm, and one has the momentum that is transmitted to the ball (I think). This may or may not be as great as the momentum of the faster wrist-hinged swing.
I just want to know whether the 'm' will be greater with this fused club.
ANSWER:
I believe that there is no way to understand the club as
a point mass, so to talk about its mass in a collision is
meaningless. Think of it as a machine which imparts
momentum to a small mass by exerting a force on it over
some short time. I think it probably boils down to having
the machine achieve the greatest possible velocity when
it hits the ball; if you can achieve that by holding the
club more tightly, go for it! It was interesting to learn
a little about Moe Norman of whom I had never heard.
QUESTION:
When a golf club hits a golf ball, part of the momentum of the golf club transfers to the golf ball. The golf ball, being light, takes off at a velocity greater than the club head speed.
What if the golf club is accelerating at the point of impact? Will the ball take off faster (and go further) than when it is hit at the same impact velocity by a club that is not accelerating?
Note: amateur golfers tend to reach maximum velocity at the point of impact. Pros reach max velocity after impact. Pros do swing faster, which imparts more velocity to the ball. I would like to know whether their acceleration adds even more velocity.
ANSWER:
What will determine the exit speed of the ball is the
speed of the club at impact. The impact time is
approximately Δt =0.0005 s. How big would the acceleration
need to be in order to significantly change the speed
over this time? A typical speed of the club head is about
150 mph, about v= 70 m/s. Suppose that the speed increased
over the impact time by Δv =1 m/s. Then the acceleration
would have to be a =Δv /Δt= 1/0.0005=2000 m/s2 ;
this is about 200 times greater than the acceleration due
to gravity. We could estimate the acceleration by
guessing that the club took about 0.1 s to go from zero
to 70 m/s, a =700 m/s2 . So the speed
of the club might increase by 700x0.0005=0.35 m/s and the
average speed during impact would be about 70.2 m/s. My
feeling is that this would have a negligible effect on
the speed of the ball.
FOLLOWUP QUESTION:
I did some more research and I have a follow up.
This article states that Force equals mass times acceleration.
It indicates that applying a force to an object over time creates momentum.
Therefore, it seems to me that a club head that is accelerating at the point of impact transmits Force and, therefore, momentum to the golf ball. And that this momentum would be in addition to the momentum of the club head.
A club head that is not accelerating will transmit no force to the golf ball.
It seems to me that a club head accelerating is going to hit the ball further, velocity at impact being equal.
ANSWER:
Sorry, but you have this all wrong. (This is what comes
from using formulas when you do not really understand
what they mean or when they are applicable!) F=ma
is Newton's second law and simply says that if you push
or pull on a mass it will accelerate. Indeed, if you
apply a force over a time you will accelerate it and
therefore change its momentum. But these do not
imply that the source of the force itself needs to be
accelerating. Here is what happens: when the club strikes
the ball, it exerts a force on the ball; Newton's third
law says that if the club exerts a force on the ball, the
ball exerts an equal and opposite force on the club. For
the short time that they are in contact (after which the
forces disappear), the ball will speed up and the club
will slow down if there are no other forces on it (which
may be the case*). If the average force they exert on
each other is F and the time of contact is t ,
the ball will acquire a linear momentum of p=Ft ;
the ball will experience a change in its momentum but by
how much is not so simple since other forces act on it.
Note that F=p /t=mv /t , so the
smaller t , the larger F , all else being
the same. In my answer to your original question, using
v =70 m/s and t =0.0005 and noting that
the mass of a ball is about m =0.046 kg,
F =0.046x70/0.0005=6,440 N=1448 lb.
*If the club is accelerating when it strikes the ball
there must be some force on it—ultimately that
comes from you.
QUESTION:
I am writing a video response to Dr. William Lane Craig, a well known apologist, and so I am answering every position he states from a debate between him and Christopher Hitchens at Biola university some years ago.
I write to you now because Dr. Craig says something that I don't understand. He states,
"First, when the laws of nature are expressed as mathematical equations, you find appearing in them certain constants, like the gravitational constant. These constants are not determined by the laws of nature. The laws of nature are consistent with a wide range of values for these constants."
I don't understand this at all.
ANSWER:
(This is not the whole question submitted, but gets to
the heart of what the questioner wants to know. Reminder
that site ground rules specify "…concise, well focused
questions…") I consider that the most important
"laws" of nature are not equations, they are
proportionalities. In order to keep this discussion
focused on my basic description of nature, I will focus
solely on classical physics to make my points. In physics
we begin with three fundamental concepts to lay the
foundation of describing nature: mass, length, and time.
In the SI unit scheme we choose kilograms (kg), meters
(m), and seconds (s). With the chosen units we can define
velocity as the rate of change of position with units of
m/s; then we can define acceleration as the rate of
change of velocity with units (m/s)/s=m/s2 . So
if an object changes its speed from 2 m/s to 6 m/s in a
time of 2 s, its acceleration is 2 m/s2 . Mass
is trickier to understand but it is a quantity which
measures how resistant an object is to being accelerated
if you push on it; it is harder to accelerate an object
of mass 1000 kg than it is one of 2 kg. Note that we have
not yet quantified that push (or pull) which we normally
call force. However we can certainly imagine figuring out
a way to push or pull on something twice as hard, or
three times as hard, etc . Whatever force is,
suppose we push on something with a constant force F
and vary the mass; we will find that if we double the
mass we halve the acceleration, if we halve the mass we
double the acceleration, if we make the mass 10 times
larger, the acceleration is 1/10 of its original value,
etc .
We have found that, F being constant,
acceleration a is inversely proportional to mass,
a ∝1/m .
Now suppose we vary F but keep m constant; we find that
if we double the force we double the acceleration, etc .,
i.e. a ∝F. We can now put
the two together and get a∝F /m or F∝ma .
This is what I consider to be Newton's second law; it
tells us an important connection of how quantities depend on
each other, it is a law of physics; and, you really did
not have to have defined the meter and second and
kilogram for it to be true, you just had to understand
conceptually what mass, length, and time are. If you know
even a little physics, you do not recognize this as
Newton's second law—any textbook will tell you that
it is F=ma . Now, how can one turn a
proportionality into an equation? It is simple, just
introduce a proportionality and voila! F=Cma
where C is an arbitrary constant since we
have not defined how we will measure F . Don't
think of it as a "fundamental constant" because if we had
already defined how to measure force, we would have to
measure C . For example, if we had defined a unit
of force to be a pound but measured m and a
in SI units (kg and m/s2 ), F=ma would not be a valid
statement of Newton's second law.
Now let's discuss an example of when a constant is a
fundamental constant. Newton's law of gravitation
expresses the magnitude F of the equal and
opposite forces exerted on each other for two point (or
spherical) objects of masses m 1 and
m 2 separated by a distance of r :
F ∝m 1 m 2 /r 2 .
Now, to make this an equation, we must introduce a
proportionality constant G : F=Gm 1 m 2 /r 2 . But can we choose it to be whatever we like? No, because
there is nothing in this proportionality which is
undefined; we can take two 1 kg masses, separate them by
1 m, and measure the force between them (a really hard
experiment!). In other words, we must measure the
constant G . This is a true
fundamental constant of nature, it measures the strength
of gravitation. Of course, its numerical value depends on
how we measure mass, length, and time, but it is still a
universal constant. In SI units it is 6.67x10-11 N·m2 /kg2 .
You might be interested in two earlier answers,
1 and
2 , along similar
lines as yours.
QUESTION:
I wondered what happens if one would make a magnet spin really fast. Could it be theoretically possible that it could create visible light.
ANSWER:
The frequency range of visible light is about 4-8x1014
Hz, so you can be sure that you cannot do this with any
macroscopic magnet. One way to get magnetic radiation is
to fabricate something called a
split-ring resonator (SRR) but technical problems
limit SRRs to about 2x1014 Hz, near infrared.
Another technique uses spherical silicon nanoparticles
and these can generate visible magnetic light. The method
is too technical to discuss here, but you can read about
it at
this link .
QUESTION:
i had a doubt, to detect an inertial frame we need to define 0 force, but to define 0 force we need an inertial frame,0 force is defined as the sum of real forces should be 0, and we can identify real forces if we believe that the real forces drop with distance,and this could be a possible solution to the above problem, but it based on an assumption, and even if we consider that we cannot detect an inertial frame, and all the physics upto special relativity is just based on if there is an inertial frame, then how come we could do experiments to confirm the theory and how did physics work uptill general relativity, if we cannot solve this basic conceptual problem.
ANSWER:
Technically, there is no such thing as an inertial frame.
No matter where you go in the universe there are always
electromagnetic and gravitational fields. That does not
mean we cannot define an inertial frame as one in which
Newton's laws are true at low velocities. In fact we can
do better by defining an inertial frame as one in which
the total linear momentum (relativistic
definition ) remains constant; then you are not
confined to low velocities. Then we will find that very
good approximations to inertial frames can be found in
nature. There is nothing wrong with basing a theory on an
idealized (fictional) concept as long as the ultimate
theory agrees with experimental results in the real
world.
QUESTION:
The idea of dark matter and dark energy is puzzling me. It is understood that dark matter acts like glue to hold galaxies together. On the other hand, dark energy is responsible for expansion of universe. Scientists say inside galaxies dark matter overcomes dark energy so gravity wins and keeps galaxies together but when it comes to intergalactic space, it is said that dark energy overcomes dark matter and wins the fight, so galaxies are moving away. How do you explain this contradiction? Why dark energy wins in intergalactic space but not inside the galaxies.
In other words, what makes dark matter overcomes dark energy in galaxies but loses in the space between galaxies.
ANSWER:
Even though I state on the site that I do not do
astronomy/astrophysics/cosmology, I will take a stab at
this one. Of course, keep in mind that nobody knows what
dark matter or dark energy are in any detail. Dark
matter, as the name implies, is thought to be some kind
of mass and therefore subject to gravitational forces. It
therefore would be not surprising to find it more
concentrated in the vicinity of large masses like
galaxies; so its density is likely to be smaller in
intergalactic space. Even less is known about dark energy
but, if it is anything like uniformly distributed, it is
more likely to lose a "tug-of-war" with dark matter
inside a galaxy than outside.
QUESTION:
What happens to the gravitational effect of mass when it transforms into energy, as in e=mc^2?
ANSWER:
In general relativity, our best theory of gravity, the
bending of spacetime, which is what gravity is, is not
caused only by mass. Any local energy density will cause
gravitation.
QUESTION:
I would like to ask about gravitational mass.
I know inertial mass is changing by motion (speed) according to m=mo/(1-v2/c2)^0.5 And also that is inertial mass which sits in E=mc2.
If the statements above is correct, now how about gravitational mass? Does it change with motion (speed)? And what mass should be used for general gravitational formula F=GmM/r2? should we use mo (rest mass) regardless of speed of the object? Or should we use m=mo/(1-v2/c2)^0.5 to substitute in F=GmM/r2?
In other words does mass equivalence principle (inertial mass=gravitational mass) hold in hight speeds?
ANSWER:
Read the answer to the previous question. The moving mass
has more energy than the stationary mass and therefore
has a stronger gravitational field as measured by a
stationary observer. Newton's gravitational equation is
amazingly accurate for speeds small compared to the speed
of light. However, if you are at very high speeds you
should not use it at all. The whole notion of force is,
itself, not useful in relativistic circumstances. We
might ask why force is a useful concept in classical
physics. Newton's second law, the central concept in
classical mechanics, is F=ma . Why? If someone is
moving past you and measures the acceleration a'
of something in your frame, she will measure exactly the
same value as you. Therefore you will both conclude the
same force is being applied to m, if a=a' then
F=F' . But if relative speeds are not small,
a≠a' so Newton's second law is no longer true.
Instead we rely more on conservation principles more in
relativity theory. In particular, we want to
redefine momentum so that it is conserved in a closed
system.
QUESTION:
I have recently learnt about the LIGO detector and watched a few videos on how it works, I understand how they detect the stretching and contracting of space time due to the gravitational wave only stretching and contracting a few hundred times per second and the new photons entering this space must travel farther or shorter deviating from the normal time travelled.
I would like to ask how these new photons enter this stretched space time since the gravitational wave itself is travelling at the speed of light does that not mean the stretched space will either move with a certain packet of photons, and since the wave travels at light speed, there wont be enough contractions to allow detection to occur.
ANSWER:
You just make this too hard to visualize thinking of photons.
Think instead as
electromagnetic waves . (Although
the speed of the gravity wave does not matter for the
interferometer to work, it is just as if somebody were
jiggling the mirror ends, I want to note that it is only
assumed that the speed of gravity is c ; the
speed of gravity has never been measured, we just know
that it is very fast. Since the gravitational field has
never been quantized, we can only assume that the
graviton has no mass. Note that for decades it was
assumed that neutrinos had no mass, something we now
understand is not true.)
QUESTION:
Who first derived the relativistic energy-momentum relation
E 2 +p 2 c 2 =m 2 c 4 and in what publication? Wikipedia says that it was Dirac in 1928, but I cannot find evidence of this in his famous 1928 paper, "The Quantum Theory of the Electron."
ANSWER:
As far as I can tell, Dirac's 1928 paper used the
energy-momentum relation as a jumping off point for his
derivation of the Dirac equation. Certainly this equation
was known long before 1928 since special relativity
itself was introduced in two papers by Einstein in 1905. The
first paper did not mention momentum at all. In the
second paper he shows that if a particle of mass m
radiates energy L that the mass changes its
kinetic energy by L /c 2 ,
hence E=mc 2 , but does not mention
momentum. Finally in 1907 he writes an
article in which
he shows that if relativistic momentum is p=mv /√[1-(v /c )2 ],
then it is a conserved quantity as in classical Newtonian
physics. From here, and knowing E =mc 2 /√[1-(v /c )2 ],
it is
straightforward to get the energy-momentum
relation.
QUESTION:
Why would lifting my feet off the ground make me heavier. Let’s say I were to do a push up while my friend is on my back for extra weight. If the friend was to plant
her foot on the ground, why would the push-up become easier.
ANSWER:
In the first figure I show all the forces on the man in
yellow. They are: N is the force which the
ground must exert up on the man, W is his
weight, and F is the force which the woman
exerts down on the man. (I have drawn N as one
vector, but it is really distributed among his two feet
and two hands. Nevertheless, the size of N is
proportional to the force he must exert to lift himself.)
The man is in equilibrium, so the forces on him must add
to zero, N-W-F =0 or N=W+F . In white are
the forces on the woman: w is her weight and
f is the force which the man exerts up on the woman
which by Newton's first law equal and opposite to the
force which she exerts down on the man so F=f .
Since she is also in equilibrium, the sum of the forces
on her must add to zero, so f-w =0 or f=w=F .
So, finally, we see that the total weight the man must
cope with is the sum of his and the woman's weight,
N=W+w .
So now I put a blue brick under the woman's foot so
that it rests on the brick. This is a new force
n up on her, shown in red. Each
person's weight is the same as before, your
weight is just the force down due to gravity.
But now there are three forces on the woman
adding up to zero: n+f '-w=0 or f
'=w-n . The force f ' exerted
up on her by the man, which is the same
magnitude of the force F '
which she exerts down on the man is smaller than
before. So now,
N '-W-F ' =0
or N '=W+F '=W+w-n. He has a smaller
total force to contend with than before.
The mistake you made is that you assumed that the
force down on the man was the woman's weight.
But the woman's weight is not a force on the
man, it is a force on the woman. This is
accidentally correct in my first example because
if only the man is supporting her, it just
happens that the force she exerts down is equal
to her weight in magnitude. But in the second
example the brick is supporting part of her
weight so the man supports less. She does not
become lighter!
QUESTION:
I am writing a fantasy novel currently, and at the ending a giant Dragon, with scales made out of stone is falling into the sea infront of a city and I'm imagining a tsunami destroying the city. The dragon weighs around 2,000 tons, how high would the wave be and how far would it reach inland? (the city is basically a flat plain)
ANSWER:
Well, I have to make some wild approximations to address
this at all. I will perhaps get you within an order of
magnitude of how big the wave might be. Maybe I should sort of do a rough
general case first and then I can throw in some
reasonable numbers. If the dragon drops from a height
h and has a mass m , then the energy she
starts with is E 1 =mgh , where
g is the acceleration due to gravity. When
she hits the water she will have a smaller energy because
of air drag falling down and when she comes to rest she
will lose some energy to heat and sound and damage to her
body; I will say that a fraction f of the original energy
goes into the energy E of the wave created,
E=fE 1 . Now I will assume that all this
energy goes into a circular pulse which has a width of
w
and a length of 2πR (circumference of a
circle) where R is the radius of the circle at
the time of interest, namely when the wave hits the
shore; so R is the distance from shore where the
dragon hit the water. Now, I found an expression on
Wikepedia which relates the energy of a wave to its
height: E /A=ρgH 2 /16
where A =2πRw is the total
horizontal area of the wave, H is the height of the wave,
and ρ= 1000 kg/m3 is the density
of water. If you put it all together you will find
H =√[8fmh /(ρπRw )].
Suppose h =5000 m, m =2x106 kg (2000 metric
tons), f =½, and w =20 m; then I
find H ≈25 m. This is in the same ballpark as the
largest tsunami wave ever observed which was
about 100 ft. There is no way I could estimate
how far inland the wave would travel; it would
depend a lot on the terrain, obstacles, etc .
QUESTION:
This question is in reference to quantum entanglement.
When a Super positioned photon is measured to determine spin does it's spin stay in that orientation as long as they a measuring it or does it immediately go back to a Super positioned state?
In other words if you determined the spin of a quantum entangled particle at say 12:00 pm and constantly measured it, would it's spin continue to point in the same direction at say 12:02 pm. or is the measurement only good for the exact moment it was initially measured?
ANSWER:
When you observe it, you put it in a single state (and
simultaneously, put the other entangled photon in a
single state). Unless it interacts with something else
later, they remain in states you put them, they do not
reëntangle.
QUESTION:
What is the maximum distance between two electrons where they will notice each other? They would both be at rest to each other in a vacuum with no external stimuli or forms of energy. At what distance do the electrons "See" each other.
ANSWER:
This question has no meaning because "notice each other" is not quantified. In principle, the Coulomb
force extends all the way to infinity.
You could, for example, ask at what distance would the force
each electron feels would result in an acceleration of 1 mm/yr2 =[1x10-3
m]x[(1
yr)x(365 day/yr)x(24 hr/day)x(3600 s/hr)]-2 ≈9x10-22
m/s2 .
In general, the force is F=ke 2 /r 2 =ma ,
so r =√[ke 2 /(ma )]
where k =9x109 N·m2 /C2 ,
e= 1.6x10-9 C, and m =9x10-31 kg. I
calculate, approximately, that r ≈5x1011
m≈300,000,000 miles which is about 3 times the
distance to the sun. Of course, you could never do such
an experiment since you would need to be in a universe
which contained nothing but these two electrons.
QUESTION:
Aren’t "constants" simply a "fudge number" to make equations work? It seems to me, constants are values of a concept we have yet to identify or understand, but need to make equations with known concepts and values work out. Since these equations work correctly with a variety of known variables, they are generally accepted. It seems to me, non-electrical magnetism (i.e. that which is retained in magnetic material like iron) and gravitation have a variety of descriptions and equations which predict their effects – yet we still are devoid of actual cause.
ANSWER:
Constants are not just "fudge numbers" as you suggest.
They can be, but in general constants play a very
important role in physics. I will give you a few
examples.
The laws of physics are stated most elegantly as
proportionalities rather than equations. For example,
Newton's second law states that the acceleration (a )
of an object is directly proportional to the applied
force (F ) and inversely proportional to the mass
(m ), a∝F/m ; so you double the
acceleration if you push twice as hard and halve it if
you double the mass. Now, to calculate we prefer to have
equations rather than proportionalities, so we introduce
a proportionality constant, call it C , a=CF /m.
What does C mean? Since we know how to
measure a (m/s2 ) and m (kg),
your choice of C determines how we will measure
force. I would like the force to have the magnitude of 1
if a =1 m/s2 and m =1 kg, so I
choose C =1. The constant C is not a fudge
factor, rather its choice defines how we measure forces.
(We call the unit kg·m/s2 a Newton, N.)
Newton also discovered the universal law of gravity: if
two masses (point masses or spheres) m 1 and
m 2 are
separated by a distance d , they exert equal and opposite
attractive forces on each other the magnitude of which is
F ; then F ∝m 1 m 2 /d 2 .
So the equation for universal gravitation would be of the
form F=Gm 1 m 2 /d 2 .
where G is some constant. But we know how to measure mass
and distance and force, so we cannot choose G to
be any old thing we want as we did for Newton's second
law, we must measure it. It turns out to be G =6.67x10-11
N·m2 /kg2 . This is one of the
most fundamental constants in all of nature and most
certainly not a "fudge factor".
Constants are often used to quantify a particular
situation. For example, the frictional force f
of one object sliding on another is found, approximately,
to be proportional to the force which presses one object
to the other (often called the normal force, N ).
This is not a physical law, just an approximation of
experimental measurements, and it obviously involves only
the magnitudes of the forces since their directions are
perpendicular to each other. So, including a
proportionality constant μ , called the
coeficient of kinetic friction, f=μN . This
constant tells you how slippery the surfaces are; its
value for rubber sliding on ice will be much smaller than
for rubber sliding on asphalt.
Sometimes, as you suggest, constants are added to
fit data and their physical meaning is unknown,
you might label such constants as "fudge
factors".
You may be interested in learning about Planck
units discussed in an
earlier answer . where new units are
expressed by combinations of the five
fundamental constants of nature.
QUESTION:
In a example of a electrical generator. Central in the generator are rotating magnets. I understand that as the magnet moves the cores of the surrounding coils will react producing a magnetic field. Also the copper wire winding around the coil will react.
My question is does the copper wire winding react to magnetic field according to the field produced by the core or the moving magnet ? In a lenz law sense.
Plus would I be correct in thinking if there is no circuit on the copper winding there no organised field would form on the copper wire winding itself.
ANSWER:
The best way to understand the priciples is to look at
the simplest possible generator. The figure shows a
generator consisting of a single loop rotating in a
magnetic field which is approximately uniform. There
would be no difference in the results if it were the
magnets rotating around the coil as you describe. Since
the induced EMF in the loop is proportional to the time
rate of flux through the loop and the flux is
proportional to the cosine of the angle θ between the field
and a normal to the area of the loop, the induced EMF
will be sinesoidal. The angle may be written as θ=ωt where
ω is the angular velocity of the loop (or magnets)
and t is some arbitrary time, so the
voltage is proportional to cos(ωt ). If
there is a load across the terminals, and therefore a
current through the loop, there will be a field due to
the current which does not affect the output voltage if
the source of power maintains a constant angular
velocity. If you are drawing power from the generator, it
is harder to "turn the crank" than if there is no current
flowing. If you need detailed information on "real-life"
generators, it is more an electrical engineering question
than physics.
QUESTION:
How did Einstein derived general relativity math when he understood that the gravity is just the space bending around an object how did he applied this to the world of math ?
ANSWER:
You can learn about how Einstein struggled with the
mathematical methods necessary to fully describe general
relativity and with which mathemeticians he consulted in
his biography by Walter Isaacson, "Einstein: His Life and Universe ".
QUESTION:
I have been thinking about clocks lately and found myself in a sort of a pickle over the velocity of a dial. (I am not native so sorry if my description makes it very complicated). Suppose we have a disc with a dial starting at its center. The dial draws two circles on that disc as it turns. The first circle has its radius equal to r, whereas the second circle has a radius of 2r. Now suppose that 1 full turn takes 1s (which would be the same for both circles since they are being drawn by the same dial). Now because of the difference of radii, there is also a difference in their circumference
c. And so we've got t=1s, c=2πr and c'=4πr (2π2r). Now here's the question - if the time is the same, but the circumference or distances are different does that mean the velocity of these two points on the same dial is different? How is that possible since the dial turns with a set velocity which in my understanding would be the same for each point on that dial?
ANSWER:
It is simply because all points on a rotating rigid body
have the same angular velocity, (revolutions per second,
e.g .) but not the same translational velocity (miles per
hour, e.g .). In general, v=rω
where v is the translational speed, r is the distance
from the axis of rotation, and ω is the
angular velocity in radians per second.
QUESTION:
I was just wondering why does the centripetal acceleration formula work a=v^2/r?
ANSWER:
Why does it work? It "works" because it is correct. Maybe
you wanted to ask "where does it come from?" or "how is
it derived?" I will give you a brief standard derivation.
On the left of the figure I have drawn the particle
moving with constant speed v around a circle of radius R.
In some time interval Δt the particle moves through
an angle theta and has velocities v 1
and v 2 but both
speeds are equal to v , v 2 =v 2 =v .
On the right I have placed the velocities tail-to-tail
and drawn the difference vector v 2 -v 1 =Δv .
So now I see two isoscoles triangles, each with the same
angle between their equal sides; they are therefore
similar triangles and so we can write v /Δv =R/Δs.
If I now rearrange and divide each side by Δt ,
I find Δv /Δt =(v /R )Δs /Δt.
Now, Δv /Δt is the magnitude
of the average acceleration; to get the instantaneous
acceleration, we must take the limit as Δt approaches zero.
But, when Δt approaches zero, Δs /Δt approaches
v . So, finally, a=v 2 /R .
QUESTION:
Is Planck time a concept derived from the theory of relativity or its premises?
ANSWER: No, that is not where the concept
came from. The origin of
Planck units came from the
desire to have a system of units which are based only on
constants of nature, not on the specific units like the
meter, the second, etc . You take the five
universal constants:
the speed of
light in vacuum, c,
the universal
gravitational constant, G,
the rationalized
Planck's constant, ℏ,
the Boltzman
constant, k B , and
the Coulomb
constant, k e =1/(4πε 0 ).
You now form
combinations of these constants which have the correct
dimensions of the units you want:
length,
mass,
time,
temperature, and
electric charge
I have copied the table from
Wikipedia showing the values of these Planck units in SI
units:
It is hypothesized that, at
distances smaller than l P =1.6x10-35
m, the usual physics as we know it will not work. I think
you might say that the origin of the idea of Planck units
is quantum field theory.
QUESTION:
A toy top is a disk-shaped object with a sharp point and a thin stem projecting from its bottom and top, respectively. When you twist the stem hard, the top begins to spin rapidly. When you then set the top's point on the ground and let go of it, it continues to spin about a vertical axis for a very long time. What keeps the top spinning?
ANSWER: There is a property of spinning
objects called angular momentum. For example, the angular
momentum of your top is a vector approximately equal to ½MR 2 ω
where M is the mass of the top, R its radius, and ω
is the rate at which it is spinning; the direction of the
angular momentum vector is straight up if it is spinning
counterclockwise if viewed from above, straight down if
clockwise.
There is a physical law which states that if there are no
torques on a spinning object, its angular momentum never
changes. There are no torques on the top if its spin axis
is vertical and so it does never stops.
QUESTION:
The third dimension is 3D (spheres, etc.). The first dimension is a dot. Everything that has a front and back, has an edge. So isn't a dot 3D?
ANSWER: A dot has zero dimensions. The
formal definition of the dimensionality of a space is an
answer to the question "how many numbers does it take to
specify the location of a point in that space?" You may
heard of Cartesian coordinates, (x ,y ,z );
they were named in honor of the 17th century philosopher
and mathematician
René Descartes . Legend has it that he
was once watching a fly buzzing around his room and asked
"what is the minimum number of numbers I must specify to
describe the position of the fly at any time?" He figured
it was three: the distance up from the floor, the
distance from the back of the room, and the distance from
one of the side walls; his room is said to have three
dimensions. One dimension is a line; here you simply
measure the distance from one point on the line to
wherever the particle is. The line need not be a straight
line. A circle is one-dimensional and the location of a
point on the circle is usually specified by an angle
relative to some point we call zero. Two dimensions is a
surface. The surface of the earth is two-dimensional, the
position being angles, latitude and longitude. One
surface of a flat sheet of paper is two-dimensional and
the two numbers are usually Cartesian coordinates (x ,y ).
The spatial universe in which we live is
three-dimensional, just like Decartes' room. A solid
sphere has three dimensions usually measured by the
distance from the center and two angles (longitude and
latitude).
QUESTION:
Why there is only time dilatation and no time shrinkage? For me I think there should be time dilatation regarding objects traveling away from each other and shrinkage regarding those closer to each other. As what makes one gets older and the other still young for both of them are moving with same speed relative to each other.
ANSWER:
Thinking how you think something should be does not make
it true! You should read my answer about the
twin paradox . How fast a clock runs and how fast it
appears to actually runs are two different things; see
earlier answer .
QUESTION:
While we use the combination of rear and front brakes to stop a motorcycle, I mean we can chose between the the front and rear without any effort, why do cars primarily use the front brakes to stop? Why does the brake pedal engage only the front brake? The rear brake is only seem to be used when the car is parked
ANSWER:
All modern cars have four-wheel braking. So where did you
get the idea that the brake pedal engages only the
front-wheel brakes? Parking brakes engage only rear wheel
brakes. Your confusion may be that the braking causes the
front-wheel tires to wear more than the rear-wheel tires;
this is due to the car slightly "rocking forward" when
stopping and also because in a front engine car more of
the weight is likely to be supported by the front tires.
QUESTION:
This is a kinematics question about
the motion of wheels; specifically, about how friction between a wheel
and a surface causes forward motion of the entire wheel. There's
something unsatisfactory to me about how this forward motion is
typically explained.
Consider the rear wheel of a bicycle; it is driven by torque applied to the wheel (via chain drive connected to pedals, etc.). In the classical treatment of this scenario, it is claimed that the friction between the wheel and the road surface, which is a
forward-pointing force, causes forward motion of the wheel. But this frictional force is tangential! So wouldn't this tend to be a rotational force instead of one of displacement? It would seem to me that somehow the tangential rightward force of friction at the point of contact needs to be transformed into a rightward force applied at the *center of mass* (i.e., at the axle) of the wheel in order to explain an induced forward/rightward motion, but no source that I have found has attempted to explain this gap.
ANSWER: The figure shows all forces on
the rear wheel of a bicycle:
C ,
the force exerted by the chain on the sprocket
(radius r ),
W ,
the weight of the wheel (radius R ),
N ,
the vertical force exerted by the ground on the
wheel,
f ,
the frictional force exerted by the ground on the
wheel, and
F ,
the force exerted by the rest of the bike on the
wheel.
Note that I have
resolved F into its horizontal,
H , and vertical, V ,
components. For any body which cannot be approximated as
a point, Newton's second law (N2) has two parts, translational
and rotational:
The sum of all
forces on a body is equal to the product of the mass
(m ) and the acceleration (a ) of the
center of mass of the body;
The sum of all
the torques on the body about any axis is equal to
the product of the moment of inertia (I ) of and the
angular acceleration (α ) about that axis.
You seem to
think that a force which exerts a torque about the
axis you have chosen (I will choose the axle for
the bike wheel) only can cause angular
acceleration; in fact, every force on the object
contributes to the translational acceleration.
Suppose now that you are standing on the pedal but
also applying the front brake such that the bike
remains at rest; both translational and angular
acceleration are zero. So the translational N2
yields two equations, N-W-V =0 and
C+f-H =0; the rotational N2, choosing the axle
as the axis, yields Cr-fR =0. (F ,
W , and N exert no torque about
the axle.) If you now release the brake,
H will get much smaller and
the wheel will start moving forward and rotating.
If you are
interested in the entire bike, not just the rear
wheel, the force C has no contribution to the
motion because it is an internal force; the
chain exerts a force C
on the rear sprocket but a force -C
on the front sprocket netting zero net force.
FOLLOWUP QUESTION:
You say, "in fact, every force on the object contributes to the translational acceleration" but the force, f, in your answer, is not directed at the center of mass of the wheel.
I guess my question is simply, "why is f treated as a force directed at the center of mass and not as a torque on the wheel"? I don't understand where the forward acceleration of the *center of mass* of the wheel is coming from.
ANSWER: You did not read my answer
carefully. The translational N2 is "The sum of all
forces on a body is equal to the product of the mass
(m ) and the acceleration (a ) of the
center of mass of the body". Note that it does
not say
…the sum of all forces which are directed through the
center of mass … Let me give you a simpler
example. Suppose that you are in the middle of empty
space and you have a wheel which has a string wrapped
around its circumference. You pull on the string so that
the tension in that string is the only force on that
wheel and it is tangential . Surely you do not
think that your pulling will cause the wheel to only spin
and not accelerate toward you too. Here is another
example, a yoyo. There are two forces on it, its own
weight which passes through the center of mass and points
down and the tension of the string which you are holding
which points up and does not pass through the center of
mass. If the string does not affect the translational
acceleration, the only force on the yoyo would be its
weight so it would fall with acceleration due to gravity,
g , which it clearly does not do.
SECOND FOLLOWUP QUESTION:
In classical billiards problems, for example, when a force is applied to a ball, the force has to be decomposed into its normal component directed at the center of mass (call it f_n) and the tangential component (call it f_t). Only f_n will cause translation, right? It's my understanding that f_t will only contribute a torque, but not to translation. So why is the bike case any different? Contact friction between the wheel and the surface is a tangential force, so, according to my understanding, it can not cause translation, because it has no component directed at the center of mass of the wheel...
ANSWER: Look at the figure*. It is much
better to resolve F into its
horizontal and vertical components rather than normal and
tangential components. It is wrong to insist that a force must be
directed toward the center of mass for that force to
contribute to tranalational motion; my examples above
should have convinced you of that. Now, a billiard ball
is confined to move on the table, so there will be no
translational motion in the vertical direction. So the
ball is in equilibrium in the vertical direction, so
N-V-W =0. But in the horizontal direction there will
be an acceleration, ma=H-f . Only f
and F will exert torques and it
will be a little tricky to calculate the torque due to
F but just geometry. I have
drawn the line of force to be below the center of mass;
if it is through the center of mass, it will exert no
torque and only f will cause
angular acceleration.
Here is the take-away for you: just use N2 as I have
given it to you, always look at all forces on the object,
and do not continue insisting that a force which exerts a
torque does not affect translation.
QUESTION:
A hammer hitting an anvil creates energy (sound) which moves in all directions at about 700 mph in atmosphere. Sound energy is also created by a large fire. In space, vaccuum, that same energy must also be present if a hammer hits an anvil in space, same as a star (a large fire ball) should also create sound energy, along with all the other energies it creates, so if sound energy in space, a hammer hitting an anvil, does not have atmosphere to slow it down, or to make noise, how would one measure that energy, its speed and pressure. Could this be what is called "Dark Energy".
ANSWER:
Sound is a wave that exists in a solid, liquid, or gas.
If you are in a vacuum, there is no sound. The energy
which the sound carried away in air is instead retained
as sound in the anvil (and hammer). But this sound
eventually damps down and where that energy goes is an
increase in the temperature of the anvil (and hammer).
This energy eventually radiates away, not as sound but as
electromagnetic waves (mainly infrared) until thermal
equilibrium is achieved. It certainly has nothing to do
with dark energy.
QUESTION:
So my question is a bit out of reality question as its not possible to do this in real life. But if you were able to turn a cannon ball into a musket ball and fire it from a flintlock, having the ball turn back into a cannonball apon exiting the gun, would the cannonball retain its velocity after changing or would it lose all energy and drop to the ground?
ANSWER: Essentially what you are asking
is if an object with a certain mass m and speed
v suddenly loses or gains mass, what happens?
The mass either disappears without trace or appears
without a source. Suppose your musket ball has a mass
m and speed v and your cannon ball has a
mass M . Since there are no external forces on
the ball, linear momentum must be conserved, that is
mv=MV where V is the speed of the cannon
ball. So V=mv/M , much smaller than v .
However, the energy is not conserved: E 1 =½mv 2 and
E 2 =½MV 2 =½mv 2 (m /M )=(m /M )E 1 ;
a large amount of energy is lost.
QUESTION:
How many pounds per inch does a lacrosse ball exert when it hits a surface? For info: Ball is traveling 90mph. A lacrosse ball weighs 8 ounces. The ball is 5.25 inches in circumference. As a sidebar, does it matter if the surface the ball hits is stiff like a wall or springy like a bounce-back? If so, could you expand on the difference in pressures against the two different surfaces?
ANSWER: If it is pressure you want, you
should ask for pounds per square inch. But I do not think
that is what you really want; it is more straightforward
to estimate the average net force the ball exerts on
whatever it hits. Then you could think about pressure,
how that force is spread over the area which the two
surfaces have touching; pretty hard to do that, though.
If something with mass m with a velocity v 1
collides with a much more massive object at rest and
rebounds with velocity v 2 , the
average force F can be estimated as F≈m (v 2 -v 1 )/t
where t is the time the collision lasts. ("Much more
massive" means the struck object does not recoil
significantly.) You seem to want to get answers in
imperial units (lb, ft, s); without going into details,
m =0.5 lb/(32 ft/s2 )=0.016 lb·s2 /ft=0.016
slug
and v 1 =90 mph=132 ft/s. Suppose that
v 2 =0, called a perfectly inelastic
collision; then F =-132x0.016/t =(-2.1
lb·s)/t . So, the smaller the time, the
larger the force. (The fact that the force is negative
means that this is the force the struck object exerts on
the ball because I have chosen the incoming velocity to
be positive. The force the ball exerts on the object is
positive and equal in magnitude.) That addresses your
second question: the ball will certainly stop in a much
shorter time when hitting a hard surface than a soft one.
Of course that should not be too surprising to you
because you know that if you fall onto a mattress it
hurts much less than if you fall on a concrete floor. So,
if t =1 s, F =2.2 lb whereas if t =0.01
s, F =220 lb. Now, if the ball hits a hard
surface it does not penetrate very far and therefore the
force is spread over a small area and pressure is large.
If the ball hits a soft surface, it penetrates deeper and
therefore the force is spread over a larger area and
pressure is smaller.
QUESTION:
Since this isn't technically a homework question and was a challenge put out to the class to see if anyone can do it, I hope you will consider it.
I'm attempting to do some fun physics maths as a challenge from our professor. It may be a trick question though since she's done that in the past. The problem assumes that c is only 30000 kmph. The problem was to determine the time to reach 29999 kmph from 25000 kmph with a total of 6.48MN of force in a vehicle that weighs 168000kg (168t) in the vacuum of space, too far away from any celestial body to have gravity noticeably affect it.
So just to clarify, there is no air resistance or gravity or time dilation. Relative mass increases when reaching said 30000kmph, and the goal is to find the time it takes to accelerate to 29999kmph from 25000kmph.
ANSWER:
I don't see the point of this odd choice of c ;
it makes the problem neither easier nor harder, just
unphysical! You will still have to do some
work! I am going to point you in the direction of solving
it yourself. First, go to an
earlier answer ; you will see that I have solved your
problem but starting at rest. If you start at some
initial velocity v 0 , the result is
that Ft=p-p 0 . Here, p is the
relativistic linear momentum, p=mv /√[1-(v /c )2 ]
where m is the rest mass, 1.68x105 kg
in your case; similarly, p 0 =mv 0 /√[1-(v 0 /c )2 ].
So now you know everything except t , so just
solve for it; this is just algebra/arithmetic. I would
advise you to convert everything to SI units (kg, m, s)
before doing your calculation. Then t will work
out to be seconds.
I do not understand what you mean by "…there
is no…time dilation…" since time dilation
is simply not of any interest here but would come into
play if the whole thing were seen by an observer in a
different frame. Also, I almost never apply the idea of
relativistic mass because I view the linear momentum as
being the quantity redefined in relativity, not mass.
Read my discussion of this in
one of the links in the earlier answer.
QUESTION:
Atoms are mostly empty space. How can they form solid objects?
ANSWER:
First of all, atoms are not mostly empty space. See an
earlier answer . Although almost all the atom's mass
is in a volume much smaller than the volume of the whole
atom, the electrons fill the rest of the space in a
"cloud". Bohr's model of the atom, tiny electrons in tiny
orbits, is wrong. If you need a simpler explanation
without reference to the "electron cloud" think of all
the atoms connected to their nearest neighbors by tiny
springs. Each atom connects to its neighbors by some
interaction which can be approximated as a spring. The
atoms close enough to interact with each other will stick
together and thus form a solid object.
QUESTION:
2 balloons, one filled with 1 cubic meter of air, and the other with 1 cubic meter of Helium, are lowered to a depth of 50 feet under water and simultaneously released.
Will they reach the surface at the same time?
ANSWER: With the information you gave me,
I will have to put stipulations on the problem. The
baloons are rigid so they do not get compressed under the
water, they have identical shapes, their masses (without
being filled) are the same, and both are filled to
atmospheric pressure. There are three forces on each
balloon: its own weight -mg , the buoyant force
of the water B , and the drag force opposite the
velocity v
which has the form -Cv 2 where C is
some constant which depends only on the shape of the
balloon. So the net force on each balloon is -mg+B-Cv 2 =ma
where I have put this equal to the mass times the
acceleration, Newton's second law. When you release each
balloon, each will accelerate up and, as v increases,
eventually a =0 and it will have a constant speed
v t ,
called the terminal velocity, thereafter. It is easy to
show that v t =√[(B-mg )/C ]
The only difference between the two is m which
is smaller for the helium which therefore has the larger
terminal velocity. The helium balloon reaches the surface
first.
QUESTION:
does a bouncing ball possess simple harmonic motion?
ANSWER: A bouncing ball is not an example
of simple harmonic motion (SHM). By definition, the
ball's motion must be describable by simple sinesoidal
functions of time t , i.e. like y (t )=A sin(ωt )+B cos(ωt )
where A , B , and ω are
constants and y is a variable describing the motion, for
example the height above the ground in your case. If the
ball were perfectly elastic, the motion would be
harmonic, but not SHM, because y (t ) would be a train of
identical downward-opening parabolas. If it were a real
ball, it would not even be harmonic because each
subsequent bounce would be a little smaller that the
previous bounce.
QUESTION:
What exactly is the 4th dimension? How do we know it’s real? How are space and time related?
ANSWER: I have
recently answered this question in some detail.
QUESTION:
As I understand it, electricity moves at the speed of light. Conductivity is where I'm getting stuck on. Are we able to measure speed from it? I tried searching "siemens/meter to kph" with no results.
My questions here are these: If silver has a conductivity of 63x10^6 siemens/meter, and copper has a conductivity has a conductivity of 59x10^6 siemens/meter, does that mean energy in silver is travelling faster than copper, or something else? But, silver can't go, say 101% the speed of light. So, there's obviously something else I'm not understanding.
I guess the point I'm getting at is this: What does conductivity mean, and can you convert siemens/meter into a speed calculating how many particles move through 1 meter of wire in x amount of time, thus making copper's speed x time/meter, while silver is y time/meter?
I'm sorry if this is a bit confusing, or not even your field. But, I figured electricity is, in some part, in the field of physics.
ANSWER: I am going to give you a
qualitative overview since you seem to not have any
understanding of what goes on in a conducting wire. If
you want a more detailed discussion of the simple model
of electric currents, you can find it in any introductory
physics text book. A conductor has electrons which are
moving around at random in the material, not really bound
to any of the atoms; these are called conduction
electrons and they can be thought of as a gas of
electrons. But there is no flow of current because the
electrons move randomly so there are just as many going
in one direction as there are going in the opposite
direction. When this wire is connected to a battery an
electric field is established inside the wire which
points from the positive terminal to the negative
terminal. When you say "electricity moves at the speed of
light", what that means is that the establishment of the
field moves with speed c ; the electrons do not
move at that speed. So all electrons see the field turn
on almost immediately. All electrons, having
negative charge, begin accelerating in the direction
opposite the field. If the only force seen by the
electrons was due to the field, they would accelerate
until they reached the positive terminal and left the
wire. But, even though they are not bound to atoms, they
certainly see the atoms and when they hit one they are
momentarily stopped or scattered. So each electron is
constantly being accelerated, then stopped, then
accelerated and the net result is that, on average, all
the electrons participating in the current move with a
very small velocity, on the order of one millimeter per
minute; this is called the drift velocity. All the
electrons are continually gaining and then losing kinetic
energy and that lost energy shows up in the heating up of
the wire. What determines the conductivity is the number
of conduction electrons per unit volume of the material
as well as its crystaline structure and how individual
atoms scatter the electrons.
QUESTION:
The classic escape velocity formula takes in consideration an uniform acceleration of gravity, g. What would be a mathematical model for escape velocity where the force of gravity varies as the body distances from the planet?
ANSWER: "The classic escape velocity formula"
that I know does not assume a uniform gravitational
field, it assumes the correct field. Let's see where it
comes from. The potential energy for a particle of mass
m a distance r from the center of a large
spherical mass M is, choosing zero potential
at r =∞ is V (r )=-mMG /r
where G is the universal gravitational constant.
If m has a speed v at r=R , where
R is the radius of the
earth, the total energy is ½mv 2 -mMG /R=E .
Now, suppose we want v to be big enough that m will go
all the way to r =∞ before it finally
comes to rest; then E must be zero. Therefore
v e (R )=√(2MG /R );
but, MG /R=gR , so v e (R )=√(2gR ).
That is the escape velocity at the surface of the planet
(ignoring any other forces, notably air drag) where g
is the acceleration due to gravity. Now, you want it
everywhere else. That's easy, just replace R by
r , v e (r )=√(2MG /r ).
You could use g=MGr if g now means the
acceleration due to gravity at an altitude of r-R .
(Incidentally, this does not work anywhere inside the
planet.)
QUESTION:
I have a bit of a debate with colleagues and looking for a professionals take. I�m a plasterer and when mixing up the plaster introducing too much air to the mix makes it less workable. Some people I work with think mixing counter clockwise introduces less air to the process than clockwise, I think it would make no difference.
ANSWER: Well,
I just so happen to have a plaster/paint stirring tool
which is powered by a drill. Examine it carefully and you
will see that clockwise and counterclockwise are not the
same in their effect; if the direction is clockwise (as
viewed from above), the blades push down on the mixture
and if the direction is counterclockwise, the blades push
up. I do not know for sure which, if either, of these
would introduce more air. My best guess would be that for
the clockwise rotation there would be a whirlpool into
the surface, whereas for the counterclockwise rotation
there would be more of a hump on the surface; I would
think the whirlpool would pull in more air. I must also
note that not all stirring tools are of the same design
as mine.
QUESTION:
Why does this toy move in a straight line? Shouldn't the toy rotate due to gyroscopic precession?
https://youtu.be/hs4iv7IHvGg
ANSWER:
Well, it does not move
in a straight line. One thing I noticed is that the toy
starts moving immediately when released; this indicates
to me that the surface it is on is not level. Just
because it is a gyroscope does not mean it will precess.
There must be an external torque acting on it; imagine an
axis running between the two legs about which you would
calculate torque. In the figure that I have clipped from
the video you can see the motor and battery and the cds
which are rotating; the toy is leaning such that I would
certainly expect the center of gravity to be beyond the
being above the axis and, were the discs not spinning the
toy would certainly fall. If you could put the center of
gravity directly above the axis, it would not precess.
You can see several instances in the video where the toy
is moving in a curved path, and in one case the path is a
pretty tight circle; these are due to precession because
of the torque due to the weight.
QUESTION:
Would you be marginally taller when on top of a tall building/mountain or at the bottom of the ocean. Does the pressure/gravity change and if so which one makes you taller?
ANSWER: If you are in the the vicinity of
the earth there are two forces which are different
between your head and feet, tidal force and pressure
force due to the fluid you are in. The tidal force is due
to the fact that the gravitational field at your feet is
slightly larger than at your head, both pointing down;
therefore the net force on you tends to stretch you
taller. We can estimate the tidal force because the
gravitational field falls off like 1/r 2 : F feet /F head =(r+h )2 /r 2 =[1+(h /r )]2 ≈1+(h /r )2
where r is the distance to your feet from the
center of the earth and h is your height; I have
used the binomial expansion because h /r <<1. So, F feet -F head ≈F head (h /r )2 .
Clearly, either force is going to be approximately equal
to your weight, so the tidal force is about mg (h /r )2 ;
if I take h =2 m, mg =1000 N, and r =6.4x106
m, this is about 10-10 N (2.2x10-11
lb).
The pressure force is
down on your head and slightly larger and up on your
feet; so the pressure force tends to compress you
shorter. The net force on you is simply the buoyant force
which equals the weight of the displaced fluid. If I take
your volume to be about 0.02 m3 , the buoyant
force force is about 1000 N in water and 1 N in air.
So, as you can see, the stretching due to the tidal force
is negligible to forces due to fluid pressure. So, you
will be shorter than your "real" height, but more so in
water than air. Therefore you are taller up high in the
air than down low in the water.
QUESTION:
I know that at sea level, atmospheric pressure is 14.7 pounds per square inch. I know that the higher in elevation you go, the thinner the air gets and a corresponding decrease in the amount of oxygen such that trying to survive above 16-18k feet is not very easy. My question is this: If you where to drain all of the oceans so that the existing atmosphere before draining was now occupying the void left by the water, would you be able to survive at what was sea level? My thought is that the atmosphere would become so diluted with the oceans gone that you would actually have to travel way below what was sea level to able to survive the atmospheric pressure change and the oxygen dilution. Wondering about this has always bugged me lol.
ANSWER:
The volume of all the oceans is about 1.4x1018
m3 and their average depth is about 3700 m
(about 12,000 ft). Now, dealing with the volume of the
atmosphere is tricky because the density of the
atmosphere gets smaller with altitude so you have to be
careful how you describe volume. The mass of the whole
atmosphere is pretty well-known, though, to be about
5.15x1018 kg; so you can calculate the volume
if the whole atmosphere were at sea level pressure where
the density is 1.225 kg/m3 , 5.15x1018 /1.225=4.2x1018
m3 . So, if the atmosphere were uniform, about
1/3 of it would flow into the emptied oceans. But, it is
not uniform so less than 1/3 would come; there would be
some base pressure 3700 m below the sea level (since
there is no sea now, when I say sea level I mean what it
was) which I would estimate to be about the same as the
pressure at sea level was because 3700 m is small
compared to the radius of the earth, 6.4x106 m
(0.06% smaller). Since the oceans occupy about 70% of
earth's surface, sea level pressure would be about what
the current pressure is at about 12,000 ft, the height of
a modestly high mountain and certainly habitable. The
highest habitable place on earth is about 16,000 ft,
so lots of places (like Denver) habitable now would not
be. And certainly nobody would be climbing Mt. Everest in
this new world!
QUESTION:
What are the signs that show scientists that some particles have a quark structure and others do not?
ANSWER:
Quarks are something which were hypothesized after most
of the elementary particles and their properties were
known. There are three kinds of particles, hadrons,
leptons, and field quanta. Leptons are few, electrons,
muons, and neutrinos. Field quanta are also few, photons,
gluons, Z bosons, W bosons, and the Higgs boson. Hadrons
are many, the proton and neutron being the best known,
but what has been referred to as a "zoo" of other
hadrons. To try to systemize and explain this array of
particles, first group theory and then, building on that,
introducing hypothetical particles called quarks was
extremely successful. There is no way I could answer
"what are the signs" because the "signs" are all the
hadrons and their properties.
QUESTION:
What is the difference between a force and weight
ANSWER:
A force is a push or a pull. Weight is the name for a
specific kind of force, one which is caused by gravity.
For example, your weight is the force which the earth's
gravity pulls you down; the gravity on the surface of the
moon is smaller and therefore your weight is smaller
there. All weights are forces but all forces are not
weights.
ADDED
COMMENT: You should be aware that a few textbooks
define weight as being what a scale measures. So if you
are in an upward accelerating elevator you have a larger
weight. I find this to be total nonsense.
QUESTION:
If time slows down near massive objects, how fast is time in the space between Galaxies ?
ANSWER: Suppose that your clock ticks off
a time t if you are in a region of space where there is
essentially zero gravitational field. Now imagine bringing
your clock a distance R from a spherical,
nonrotating object of mass M . The same time interval will
now be given by t'=t √[1-(MG /(Rc )]
where G =6.67x10-11 m3 kg-1 s-2
is the universal gravitational constant and c =3x108
m/s is the speed of light. For example, let the location
be on the surface of the earth, M =6x1024
kg, R =6.4x106 m; then, t'=t √[1-7x10-10 ]≈t [1-3.5x10-10 ].
So, for example, the clock in space would click off 1
second while the clock on the earth would tick off
1-3.5x10-10 seconds. To see significant
gravitational time dilation requires a larger M ,
for example a black hole; for a black hole with a
Schwartzschild radius RS , the time
dilation equation becomes t'=t √[1-RS /R ]
so time stops when R=RS . If R= 2RS ,
t' =0.7t .
FOLLOWUP QUESTION:
Previous answer did not address this question :
If time slows down near massive objects, HOW FAST IS TIME IN THE SPACE BETWEEN GALAXIES?
Could you do the math at a distance of 10,000 light years from any mass, please.
ANSWER: There is no answer to that
question because time depends on the observer, it is
relative. Any observer in any frame of reference and in
any gravitational field will say that it takes any clock
in his frame 1 s to advance 1 s. Let's take the case of
the black hole example above. The observer in empty space
sees his clock advance 1 s but observes the clock of the
girl near the black hole to advance in 0.7 s in that
time. The girl near the black hole sees her clock advance
1 s but observes the clock in empty space to advance 1.4
s in that time.
QUESTION:
In the theory (?) of quantum entanglement, maybe you could clear up some basic things for me: a) in order for two particles to be entangled - no matter how far apart in space they are - must those same identical two particles have been physically 'partnered' up at some point in the past?
b) If a particle in a spin up/spin down position is OBSERVED, it chooses (?) to become spin up or spin down. WHY does the act of observation force it to make that choice? Or is that even the right question? I understand that this idea of observation causes change goes to the heart of quantum physics, but I'm still trying to get my head around the WHY.
ANSWER: The
answer to your first question is yes, the particles
become entangled by interacting with each other. So, for
example, if you had two electrons, each with a wave
function which is half up spin, half down, unless they
have been put into those states by a mutual interaction,
they are not entangled. For the second question, the
particle does not "choose" anything. Rather, the
measurement puts the particle into the state
which you observe. The result of a measurement is to
observe a single state, not an admixture. So if you have
an electron whose wave function is half up, half down,
your measurement will either give one or the other, not
both; if you observed a large number of electrons, all
with the half and half wave functions, you would observe
spin up for half your measurements and spin down for the
other half.
QUESTION:
The longer the garden hose the less the pressure at the end - Is this a true statement, if so, how much less?
It's not a homework problem, it's a heated discussion with spouse, we're 76
ANSWER:
There is a pressure drop for a fluid flowing through a
pipe. The main reason is that there is frictional loss
due to viscosity of the water and the rubbing of the
water on the walls of the hose. The loss depends on the
size and length of the hose, the density of water, the
flow rate, the viscosity, a constant characterizing the
friction for the material, and the change in height if
the hose is not horizontal.
The physics is very complicated, but there is a handy
on-line calculator where you can get at least a
qualitative estimate of how big an effect it is. I did a
calculation for water in a 1 inch diameter, 100 foot
rubber hose, and for a flow rate of 3 gallons per minute;
I assumed that the hose is horizontal and straight. The
results are shown in the figure. The calculated pressure
drop is 0.38 psi which you can compare with a typical
in-city water pressure of about 40 psi. So the drop is
only on the order of 0.1%. If the water flow were much
faster, say 20 gallons per minute, the drop would be
about 10 psi, a 25% drop.
QUESTION:
I have a question, if the pulling force of mass or acceleration or electromagnetism can cause time dilation... would a pushing force of the repelling force of electromagnetism or the expansion rate of the universe cause reverse time dilation?
ANSWER:
Only speed (magnitude of velocity) matters for
time dilation. It is irrelevant how that speed
was acquired, attractive or repulsive force.
QUESTION:
What is the meaning behind multiplication in physics? Is multiplication in physics purely mathematical or there is a physical explanation to it? How do we explain the product for example, s=v.t? Is there any meaning behind this? For example, I can say that "Distance is defined as the product of velocity 'times' time"? But what does this even mean?
ANSWER: Suppose you are in 6th grade.
Multiplication is explained as simplified
multiple addition, the product of 5 and 3 is 5
plus 5 plus 5. Then we learn all our
multiplication tables to not have to figure out
all the simple products we might need. This is
arithmetic and, indeed, we are taught that it
often means something, like in the example
above, if we have three baskets, each containing
5 apples, we have 15 apples. And then it starts
touching on physics. The area of a rectangle is
the height times the width; or, like your
example (but not exactly) the distance traveled
is the product of the speed times time. But the second
example is sort of ambiguous if you think about
it. What does "distance traveled" mean? It might
mean the distance between the starting and
ending points; but if you get there on a winding
path, that is not the really the distance you
traveled. The problem is that before you get to
physics, there is only one kind of number,
called a scalar quantity, which you can specify
using only a number. Examples of scalars are
time, speed, length, area, etc . But, in
the physical world, some quantities require more
than one number to adequately describe them. The
simplest are vectors which require a
specification of both magnitude and direction.
Examples are force (e.g. 10 lb straight
up), velocity (e.g. 25 mph due north),
displacement (e.g. 5 ft straight down).
Vector quantities are denoted as bold-face, for
example F =10 lb
straight up. Now multiplication takes on new and
different meanings, and the ways we multiply all
have a meaning. The easiest to grasp is how to
multiply a scalar times a vector: as in
arithmetic, multiplication is just repeated
addition, so 3F =F +F + F .
But, what is the multiplication of two vectors?
There are two ways we could imagine multiplying
vectors, namely the product is also a vector or
the product is a scalar. The scalar product or
dot product is defined as
A·B =AB cosθ AB
where θ AB
is the angle between the vectors
A and
B and A and B
are the magnitudes of the vectors. An example of
a scalar product is the work done by a force
F which acts over a displacement
d , W =F�d .
The scalar product is commutative, i.e .
A·B =B�A .
The vector product or cross product is defined
as
A×B =u AB sinθ AB
where u is a
dimensionless vector of magnitude 1 which is
perpendicular to the plane defined by
A and B .
You can do additional research on your own if
you want to understand whether u is "up" or
"down" relative to the plane, but vector
products are not commutative,
A·B =-B�A .
An example of a vector product is torque
τ caused by a
force F a displacement
r from an axis,
τ =r×F .
QUESTION:
Something that confuses me is that scientists say it takes approx. 8min for light to travel from the sun to earth. So if the sun were to explode we wouldn�t know till 8min after. However Einstein gathered that the faster an object travels the more time slows down so as you approach the speed of light time stops. But isn�t this just relativity? If you are moving that fast everything seems like it�s not moving cause your moving so fast. If time were to actually stop how could it take 8min for us to realize the sun exploded wouldn�t we realize it instantly? So why can�t you move faster than the speed of light? Wouldn�t that just mean things relative to you would seem to be stopped for longer, meaning time doesn�t stop when you reach the speed of light?
ANSWER: See the
faq page
for an explanation why you can't move faster
than the speed of light. Suppose that someone
was moving with speed 99% of the speed of light
from the sun to the earth, a distance of about 8
light minutes. Because of length contraction the
distance would, for her, be shortened 8x√(1-0.992 )=1.13
light minutes and the time for her to get to
earth would be 1.1/0.99=1.14 minutes; so she
would observe the sun's demise to hit earth in a
time of 1.13 minutes, and she would arrive 0.01
minutes later. But, just because she observed
the time to be 1.13 minutes does not mean
someone on earth does; the earth-bound do not
see the distance to the sun to be shorter but
see it as 8 light minutes.
QUESTION:
Suppose you have two persons watching universe expanding, one of them is on earth, the other is moving fast towards the end of universe. Both of them are watching it all the time. Does this mean, that the "end" of universe exists twice and each is in a different "position"?
ANSWER: You miss the whole point of
special relativity, that the time and position
of some event do not have absolute values. That
the traveler will observe the end of the
universe to be in a different location and at an
earlier time than you do does not imply that the
universe ends twice at two different locations;
rather, it implies that time and position are
different for him than for you.
QUESTION:
I was telling my kids that time is space and explaining the block universe and they asked me, if time is space, then how many minutes are in a mile? It sounds stupid, like it does not have an answer, but the more I think about it, the more it seems like if space is time, it should have a specific answer, and if not, why not?
ANSWER: I will first give the simple
answer without much detail since you are dealing
with kids. But your kids must be fairly
sophisticated to be thinking about space-time at
all, so I will then give a little more
background on how space-time (four dimensions)
arises. When special relativity is mathematically
represented as suggesting a four-dimensional
space, the fourth dimension is not represented
by time (t ) but by x 0 =ct
where c is the speed of light. So the
fourth dimension has units of length just like
the original three, x 1 =x ,
x 2 =y , x 3 =z .
So, if x 0 =1 mi=t ·186,000 mi/s,
then t =1/186,000 s=5.38x10-6
s=9x10-8 minutes, approximately one
nanominute.
Before Albert Einstein developed special
relativity, it was generally assumed that time
and length are universal and independent. One
second for me is the same for anyone else in the
universe; similarly, one meter is the same
regardless of the position or motion of the
meter stick I am measuring. If someone is moving
with a speed v along my x axis and I measure the
length of his meter stick, I find it is shorter;
similarly, if I measure the rate his clock runs,
I find that it is slower than mine. We are
talking about meter sticks and clocks which, if
located in my reference frame, are identical to
mine. Mathematically this is expressed in what
we call a Lorenz transformation,
x' 0 =γ (x 0 -βx 1 )
x' 1 =γ (x 1 -βx 0 )
where the
primed coordinates are in the moving frame as
measured by the stationary frame, β=v /c ,
and γ =1/√(1-β 2 ).
So, two of the coordinates in our 4-space get
mixed up together if the other frame is moving.
Does anything analogous happen in everyday
3-space? The answer is yes; if you take a
coordinate system with three axes, (x,y,z )
and rotate it through an angle θ
about the z -axis, the new x'- and
y' -axes
both depend on what θ is and they
depend on both x and y are:
x'=x cosθ+y sinθ
y'=-x sinθ+y cosθ.
Hence the Lorenz transformation may be
interpreted as a rotation in the x-t
plane. This is the impetus of mathematically
working in a 4-dimensional space (space-time).
QUESTION:
I teach high school physics, and I recently performed a lab to investigate Lenz's law. Given the current state of things, this was done in my own home, with students on video chat timing my experiment with their phones.
My experiment involved dropping a cylinder of neodymium magnets (about 3 cm long and 1 cm in diameter) down a 1.5 m long copper pipe about 2 cm in diameter. I marked out different lengths on the outside of the cylinder, and used a steel nail to hold the magnet in place at these various �starting lines.� So I could directly drop the magnet 153 cm from the top, and 130, 110, 90, 70 and 50 cm using the nail. For each distance, we did three trials. For each of those trials, seven students recorded through the Google Meet. I expected a bit of a hectic disaster, but the precision of the results was amazing.
My idea was to use the graph of distance vs. time to discuss what was happening (they haven�t learned about Lenz�s law yet) in terms of forces. I talked about different types of graph shapes for different situations and how they relate to what is happening with forces: uniform acceleration, uniform velocity and free-fall with air resistance (terminal velocity). My fingers were crossed that our data would approach linearity like the latter, and it did! A valuable learning experience for all. The one hitch is that the
z-intercept of our linear trendline is positive, not negative. When an object goes from rest to terminal velocity, I had shown them that the
z-intercept of the linear portion of the graph is negative, because the speed only increases to this section of the graph. With our results, I am forced to conclude that the magnet slows down before achieving linearity, because the average slope of the graph is steeper prior to the domain of our results.
ANSWER: The questioner said in his
initial message that the graph of the data would
be included but it was not. I have not been able
to get him to reply to any messages, but I spent
some time working this all out, so I thought I
would go ahead and post it. In doing my
calculations I approximated some of the
variables with reasonable guesses. I had three
neodymium approximately 1 cm magnets which when
stacked were about 1 cm; the mass of three was
10 grams so three cm would be about m =30
grams=0.03 kg. I took the acceleration to be
about g =10 m/s2 . I made a
wild guess at the terminal velocity by watching
videos on the internet and chose v =20 cm/s=0.2
m/s. I also note that many videos
seem to indicate that terminal velocity is
achieved very quickly. The resistive force is
given by F=-bv (not proportional to
v 2 as most air drag problems
are) where b is a constant to be
determined. I choose the coordinate z to
increase in the positive vertical direction and
the initial position to be at z =0 and
the initial velocity to be v =0. This is
a classic problem and I will just state the
results which result from solving the Newton's
second law equation, -mg-bv=m (dv /dt ):
v =(mg /b )[e-bt/m -1]
=0.2(e-50t -1)
z =(mg /b )[(m /b )(e-bt/m -1)-t ]
=0.2[0.02(e-50t -1)-t ]
Note that, for large bt /m , v≈-mg /b=- 0.2
which is the terminal velocity. This can be
solved for b , b =1.5 N/(m/s).
The plot above for v shows that it
takes only about 0.1 s to approximately reach
terminal velocity. The plot for z shows z (t )
in black and z (t ) if the
magnet moved with the terminal velocity the
whole time; the linear portion of the curve is
shifted up by 4 mm, possibly responsible for the
"positive z-intercept" observed by the
questioner. Of course these plots are not what
is being plotted by the questioner because each
new datum starts at rest; but every datum
plotted has the same behavior as the one I have
shown, so all the individual runs are shifted by
the same 4 mm. The shift is so small compared
with the total fall distances, the non linear
time is so small compared with the total fall
times that the experiment should be able to
extract a reasonably good measurement of the
constant b . Unfortunately, I do not
have the times of the various data points, so I
cannot be more specific; the time of fall from
the top if the terminal velocity is 20 cm/s
would be 7.5 s.
QUESTION:
Is following the line of steepest topographic descent (as stream would do) an example of conservation of energy?
ANSWER:
Let's take a stream as the specific example to
discuss the question. The reason the water flows
downhill is that there is a force acting on it,
the downhill component of the force of its own
weight. Generally a stream of water will tend to
follow the path where it finds the greatest
force impelling it down, that is the steepest
terrain. Now, if that were the only force on the
water, it would accelerate and the water would
be going faster at the bottom. But, in order to
do this, the stream would have to get narrower
as you went down because water is basically
incompressible and if you have, say, 100 gallons
per minute flowing at the top of the hill you
need the same flow rate at the bottom where it
is flowing faster. (Think of a hose nozzle where
the water flows much faster out the nozzle than
in the hose.) But this is not what happens
because there are drag forces which impede the
flow. What you generally find is that, if the
stream is about the same depth and width as it
moves down the hill, the speed is pretty
constant. Therefore the kinetic energy at the
bottom is the pretty much the same as at the top
but the potential energy has decreased;
mechanical energy has decreased, not been
conserved. However, if you consider the whole
system of the stream, the stream bed, the banks,
etc ., the energy is conserved because
the lost potential energy of the of the water
will show up as heat in the environment and the
water. However, this would be true regardless of
whether or not the water flowed down the
steepest path. So, I would not use the choice of
steepest path to be an example of energy
conservation.
QUESTION:
I tried posting this question on a
different website and didn't get much of a response.
For some reason there doesn't seem to be a lot of
information available about this subject, or maybe I just
can't find it. Anyway, I am hoping you can help shed some
light on the subject. Please forgive the formatting -
I've posted a couple of different questions, but if you
need a single, concise question to answer, the main one I
have is this: What's going on here?
I have a bottle of water that I put in a freezer for a
period of time, until the water becomes super-cooled and
is still liquid. Now we know that when we shake or
disturb the bottle, some of it crystallizes. But it
doesn't turn into solid ice. Instead it turns into slush
- a mixture of tiny ice crystals and liquid water. Here
are my questions: What determines what parts of the
water will become ice, and which will stay liquid?
What is the temperature of the slush mixture? Is the
water part slightly warmer than the ice part? or is
it all pretty much the same exact temperature? If I
poured out the liquid water into a different bottle, and
left the slush / ice in the original bottle, would there
be any chemical differences between the two? Is the
explanation that the liquid water has impurities such as
minerals, and the frozen water is pure H20? If that
is the case, then if we performed this experiment with
pure water, then it should turn directly into solid ice,
and there'd be no slush, right? I have read about
Fractional Freezing which seems relevant. But this
doesn't seem to fully describe what is happening. I am
hoping for more of a general explanation rather than a
list of individual answers.
ANSWER:
I recommend that you google
supercooled water videos and watch a few. It seems
pretty clear to me that your problem with slush is that
your water is not pure enough. Possibly not cold enough
either so that generated heat warms the water just above
zero before it freezes. This is a very touchy sort of
experiment, one of the videos has the teacher admitting
it took him a number of tries to get a good enough take
to post. By the way, I thought the answer on Reddit was
pretty good.
QUESTION:
Hi, I have a bee in my bonnet about a real world home improvement matter concerning wall tiles and the surfaces they are stuck to.
There is much ado about the supposed weight of tiles per square metre that the sub surface can support and arguements are made for using concrete backer boards which can take greater weights than plasterboard (or drywall if you prefer). So plasterboard can take 30kg/m2 and backer boards 50kg/m2 or more.
The base of wall tiling always finishes at a horizontal surface, such as a floor etc. So given there are no gaps between tiles and that the wall is truly vertical, what is the relevance of the supporting weight figures for the subsurface board?
I can't see that there would be any force acting that would cause any board to fail assuming the tiles end on a horizontal surface that can take the weight. I follow that there will be compressive force down through the tiles and thus through the backing board by virtue of the adhesive to some degree. But surely this compressive force can't affect the backing board as the weight is carried down through the tiles to the horizontal surface that they connect with.
ANSWER:
Well, this isn't really physics, but having had
a recent experience (a backsplash behind our
range), maybe I can make a few comments. If the
wall is sheetrock and sealed or painted, it is
best; this is because the surface is just paper
and if it is necessary to provide vertical
force, it could be too weak. But, as you state,
if the tiles extend all the way to the floor (or
other support like a cabinet as in my
backsplash), there should be no reason to apply
the concrete boards. On the other hand, I would
not put tiles on a bathroom sheetrock wall
because if moisture gets behind the tiles you
would have a mess. The same thing goes for bare
wood. I once had laid large terra cotta tiles on
my kitchen floor on top of an old pine floor; it
turns out I had a slow leak in the ice maker
which I did not discover until the tiles started
coming up from the saturated spongy wood floor.
I had to rip up half the floor and leave it to
dry for several weeks before reinstalling (too
late for a subfloor because it would have raised
the floor too far). When I tiled our laundry
room I did use the backing board and it is still
in pristine condition after 40 years.
QUESTION:
This is really not a question about astrophysics but something simple I am missing. I know that as two massive objects orbit each other they lose energy by emitting gravitational waves and that eventually the orbit decays and they will collide. However as the Earth and Moon rotate around each other and the system loses energy by tidal forces the Moon moves AWAY from the Earth. Why the difference?
ANSWER:
It is understandable that you would think of
this because it would appear that the moon is
gaining energy rather than losing energy. In
fact, it doesn't just appear to gain energy, it
is gaining energy. The reason is rather subtle.
The moon causes a bulge in the oceans which is
responsible for tides. The bulge is actually two
bulges, one toward and one away from the moon as
shown in the figure (greatly exaggerated); this
is because of the
tidal force . Because there is friction
between the oceans and the ocean floors, the
rotation of the earth drags the bulge forward,
so it is slightly ahead of the moon as shown in
the figure (again greatly exaggerated). Now the
bulge exerts a force on the moon which causes it
to accelerate a tiny bit; that's where it gets
the added energy to increase its orbital radius.
Similarly, the moon pulls on the bulge which
causes the earth to slow down. If this were the
whole story, the energy would be conserved, the
earth losing energy and the moon gaining the
same amount. But, as noted above, there is
friction between the earth and the oceans which
takes energy away from the system, giving a net
loss of energy but the moon ends up with more
than it started with.
QUESTION:
If subject A is traveling through space at a high rate of speed and subject B is traveling at a much slower rate of speed would you expect to see a difference in the rate at which either subject can process information. Example: Would it be easier for subject A to process information than subject B or vice versa?
ANSWER: This is too vague. What does
"process information" mean? What do you mean by
"easier". Now, whatever the answers to these
questions are, if each subject is presented with
the same information stream in his frame of
reference and has the same equipment to process
it, they will be equally easy to process. But if
you mean that A sends an information stream
which he processes to B to process (or vice
versa ), they will not process the
information at the same rate. For simplicity's
sake, I will choose the "slow" frame B to be at
rest; in relativity, only the relative velocity
is important. Let's choose a simple concrete
example of information stream: the information
is a stream of ten pulses separated by 1 second
as seen in the reference frame B; B will process
them in 10 seconds. Now, how does A process that
same information which B processed in 10
seconds? Suppose A is moving away from B; then
the time between pulses will be longer than 1
second because he moves some distance away while
waiting for the next pulse to catch up with him—the
information is processed more slowly. You should
be able to see that if you interchanged A and B
everything else would be the same so B would
process the information more slowly. Suppose A
is moving toward B; then the time between pulses
will be shorter than 1 second because he moves
some distance toward B while waiting for the
next pulse to arrive�the information is
processed more quickly. You should be able to
see that if you interchanged A and B everything
else would be the same so B would process the
information more quickly. An
earlier answer would be of interest to you.
QUESTION:
Matter wave of a particle depends on its momentum,doesn't that mean different observers with different speed will measure different wavelength?
ANSWER:
Yes. This is called the Doppler shift.
QUESTION:
If I am driving in a vehicle in a straight line at a constant speed and activate a drone in the vehicle so the drone is hovering inside the vehicle, open the back of the vehicle and increase my speed so the drone drifts out the back of the vehicle, what then happens to the drone once it is not longer inside the vehicle?
ANSWER: There will be a period when the
drone is leaving the car when there will be lots
of turbulence and other currents of air; I am
answering neglecting this transition period and
also assuming the air outside the car is still.
You have adjusted the drone so that, even though
it is moving forward with the car at speed v ,
it hovers in still air. When it suddenly finds
itself outside it sees a headwind of v ,
not still air. This headwind will exert a force
which will tend to slow the drone down so that
eventually the drone will be hovering at rest
relative to the ground.
QUESTION:
Once a spacecraft breaks free from earth's gravity, how much propulsion does it actually need to reach a destination, say Mars? Would the craft have to continually accelerate through the vacuum of space, or once it reaches its maximum velocity, would its propulsion be turned off?
ANSWER: Technically, you never "break
free from earth's gravity". If you acquire a
speed greater than the escape velocity, then you
will never fall back to earth, but you will slow
down continually forever (assuming you have no
other objects exerting forces on you).
Nevertheless, the force gets very small pretty
quickly. If you are 100 earth radii away from
earth, the weight (force of earth gravity) is
1/1002 =10-4 times smaller
than on earth. And 100 earth radii is not all
that far in the whole scheme of things; the
distance to the moon is 60. Anyhow, a spacecraft
always coasts nearly all the time on a mission
somewhere in the solar system. Propulsion is
used mainly for getting up to speed and then
slowing down to orbit or land on your
destination; a much smaller use is to steer,
changing direction.
QUESTION:
I have had this question in my mind for over 10y, and I would be very grateful if you could help me to get an answer.
The theoretical experiment goes as follows:
1) Bring to outer space 1 round glass chamber, 1 pure iron ball, and a source of heat.
2) Place the iron ball in the center chamber, and heat it until it gets bright red.
3)Go out of the chamber, close it, and generate a total vacuum.
As far as I have been able to go: Q1: Can we see the bright red iron bar?
A1: Yes, and considering that the ball is in an absolute vacuum the light must be behavior as a particle. *Quantum mechanics dictate that given the duality of light a media is not necessary for light to travel, thus radiation.
Q2: If the light behavior as a particle it must have a mass. Where does this mass come from?
A2. Given that the iron bar is in an absolute vacuum, my logic dictates that the only source of mass for the photons must be the mass of the iron bar itself... Is this correct? Q3. After an X amount of time when the iron bar has finally cooled down. Will the iron bar be lighter than at the beginning of the experiment?
** Consider that the energy in the iron bar can only be released in the form of light.
A3: I have been stuck in here for several years.....
Thank you very much for your time, please notice I�m not a physicist but a Biology�s with a deep interest in physics.
ANSWER: I do not see why all the stuff
about a glass chamber, heat source, etc .
is necessary to ask your question which is,
simply: if an object is hot and cools soley
by radiation, does the object have less mass
after cooling?
First, your A2 is wrong. Photons do not have
mass even though they do have momentum and
energy. Therefore, your conclusion that the
object will be lighter because the photons carry
away mass is wrong. However, because mass is
just a form of energy, we can say that the
photons carry away energy (even though they have
no mass). But, where did that energy come from?
The object had to provide it. But there are just
as many atoms after the cooling as before, so
the object evidently changed mass into energy.
This change in mass is very difficult to observe
in the case of a red hot iron object cooling down.
Inasmuch as E=mc 2 , the change in mass would be
Δm =ΔE /c 2
where ΔE is the energy carried
off by the radiation. Suppose that ΔE
is a real big number, say a 10,000 Joules; then
Δm =104 /(3x108 )2 =10-13
kg=0.1 nanograms! This is approximately the mass
of a single yeast cell.
QUESTION:
For a space traveler that leaves Earth headed for a distant star @ 0.9c, wouldn't his onboard clock (that's @ rest relative to him) read the time passes normally......i.e. as if he were still on Earth? He would see no time dilation nor length contraction!
ANSWER: Wrong. the distance to the star
is measured from the earth; think of it as a
long stick. The stick is moving with speed 0.9c
as seen by the traveler, so he sees the distance
shorter by a factor √(1-0.92 )=0.44.
Since, as you correctly stated, her clock runs
normally, it only takes her 4.4 years to get to
a star 10 light years away. It would be worth
reading my explanation of the twin paradox which
is linked to from the
faq
page.
QUESTION:
Which is the correct term to refer to the speed of light (scalar) or the velocity (vector) of light. They are both used in books & literature. I think its should be the "speed of light", but just asking.
ANSWER: It depends on the context. One
context is "the speed of light is a constant in
all frames of reference and independent of the
velocity of the source or observer". Another
would be "when a ray of light passes close to a
massive galaxy, it is bent and so its velocity
changes although its speed does not."
QUESTION:
How do I calculate the kinetic energy of an object moving submerged under water? Specifically if water pressure is involved. Like if a cannonball was fired at 110 m/s under 9,000 feet underwater. Or If a submarine is moving 10 m/s at a depth of 800 feet. I know kinetic energy is factored in and pressure at that depth and possibly water drag. I'm just not sure if there is a specific equation for this. Can you please help me with this?
ANSWER: The kinetic energy of something
depends only on its mass m and its speed
v , K =½mv 2 .
It has nothing to do with its pressure or its
drag force. If there are any forces on the
object, those forces do work which might change
the kinetic energy. Because of the near
incompressibility of water, the depth will have
almost no effect in how something moves. The
effect of the pressure is simply the buoyant
force which is a force equal to the weight of
the displaced water which is nearly the same 100
ft down as 10,000 ft down. Similarly, the drag
depends only on the density, speed, and
geometry, not pressure, the drag will be about
the same at any depth.
QUESTION:
What's the physics behind
"walking ladders" that move down slopes by themselves?
I'm thinking maybe it's the same for slinkys that 'walk' down steps. Gravity and the weight of the ladder being moved around the different legs.
I tried Googling for an answer but can't find any scientific answers.
ANSWER: This was a fun problem to figure
out. First, let's think about all the forces on the
ladder when it is slightly tipped so that two legs are
off the ground. There is the weight W ,
normal forces (not shown) N front
on the front leg and N rear
on the rear leg, and frictional forces (not shown)
f front on the front
leg and f rear on the
rear leg. The weight acts at the center of mass of the
ladder which is centered relative to the two sides, but
is closer to the front than the rear because the front is
heavier. That means that the normal force on the front
leg is greater than on the rear leg; this is important to
remember as we go further. Now, I want to examine the
torque which the weight causes. I have resolved
W into components down the incline
(W x ) and normal to the incline (W y ).
W y causes a torque about the
axis I have labelled DROP AXIS causing the currently-off
legs to drop to the ground. The inertia causes the ladder
to keep going lifting the other legs off the incline; if
none of the legs ever slipped (lots of friction), the
ladder would rock back and forth without going forward.
Now, the other component W x exerts a
torque about the axis I have labelled TWIST AXIS; if no
legs slip, this torque has no effect on the motion.
However, for a couple of reasons, the rear leg may slip:
Because the normal force is smaller for the rear
leg, the maximum static frictional force will be
smaller than for the front leg.
The material contacting the ground may be
different for the rear legs than for the front,
e.g. the rear legs may have a smaller
coefficient of static friction and could slip
even if the normal forces were the same.
So, for the video, the rear legs must slip and
the ladder, with each "step" twists about a
normal axis. There are frictional losses which
would cause the ladder to lose energy and
therefore eventually stop, but since it is going
down an incline, this energy is replenished by
the work done by Wx as it
moves down the incline. It seems to me that it
would be pretty tricky to get this set up since
everything must be just right for it to work. ADDED
THOUGHT: A different possibility occurred
to me. I have treated the ladder as if it is
rigid. However, if it is kind of rickety the
lifted legs will tend, if they can, to move
forward because of gravity, even if the two down
legs do not slip. So when they come back down
they will have moved forward a bit.
QUESTION:
My Dad loves physics and always referred to the speed of light as 186,000 miles per second, per second or 186,000 miles per second sq. that is how he always spoke of it. Is that speed of light an excelleration? (even though nothing goes faster than light.)
ANSWER:
The dimensions for speed are length/time, not
length/time/time. Therefore, 186,000 mi/s/s is
wrong.
QUESTION:
E=m(c)2
What is the reason for twice the speed of light? Einstein tried everything and found it?
ANSWER: Oh dear, it is not twice
c it is
c squared. E=m·c·c=mc 2 .
And Einstein did not find it by trial and error;
how could he, since it was not really
appreciated at the time that mass was, in fact,
just another kind of energy? It was a natural
result of his theory of special relativity.
QUESTION:
According to Enistine space time is like a fabric and all masses round around the sun so why Moon revolve around Earth not sun?
ANSWER: Not just the sun, but every
object with mass, warps spacetime. It is much
easier to think about this in terms of Newtonian
gravity for which the equivalent statement is
that all objects with mass have a gravitational
field. The force which a mass M exerts
on a mass m a distance r away
is F=MmG /r 2 where
G is the universal gravitational
constant. You can calculate the ratio of the
force the sun exerts on the moon to the force
the earth exerts on the moon:
F sun /F earth =(M sun /M earth )(R earth /R sun )=(2x1030 /6x1024 )(3.8x108 /1.5x1011 )2 =2.1
So the sun exerts a force more than twice the
force the earth exerts on the moon. But if you
think about it, the moon does revolve around the
sun since it moves right along with the earth;
but because of the force the earth exerts, it
also revolves around the earth.
QUESTION:
according to law of gravitation less mass object attract toward heavy mass object so why we don't attract toward wall because wall have greater mass then a human mass ?
ANSWER: Actually, the law of gravitation
states that two masses attract each other with
equal and opposite forces. So the wall does
exert a force on you and you exert an equal and
opposite force on the wall. However, since
gravity is such a weak force, you do not feel it
at all. For example, if you and have a mass of
100 kg and the wall has a mass of 10,000 kg, the
force you experience, if you are 10 m away, is
about 10x10,000x6.67x10-11 /102 =6.67x10-8
N=1.5x10-8 lb.
QUESTION:
I'm trying to calculate what effect hitting a 1,500 lb water barrel at 30 mph in a 15,000 vehicle would have upon the truck's momentum?
Also, if I increased the truck's weight to 65,000 and drove it at 50 mph, what difference would this make on the momentum too?
I'm working on a road safety prototype in my shed and to be honest this kind of math is way beyond me, so any help would be much appreciated.
ANSWER: I will assume that the barrel and
vehicle move off together after the collision;
this is called a perfectly inelastic collision.
In a collision the total linear momentum, the
product of the velocity times the mass, remains
constant. Although lb is not a unit of mass, it
is a measure of mass. So, before the first
situation you state, the momentum before the
collision is 15,000x30=450,000 lb�mph and after
it is 16,500v where v is the
speed after the collision; equating the two and
solving for v , I find v =27.3
mph. For the second case, 65,000x50/66,500=v =48.9
mph.
QUESTION:
I have watched
Human Universe with Brian Cox. If you haven't seen it, in
part one he goes to Russian to meet with the Soyuz spacecraft. He tells his
viewers that only knowing two of Newton's equations (F=ma and the law of gravitation
F=GmM /r 2 ) he can calculate that the spacecraft only
needs to reduce velocity by 128 m/s to safely reenter orbit. I have looked
and tried to figure out how the calculations work but can't. Do you have
any clue how he would have done the math?
ANSWER: I will assume that the Soyuz is
returning from the International Space Station. First I will give some
information needed later:
R =6.4x106 m
(radius of the earth)
h =4x105
m (altitude of the space station)
G is
the universal gravitational constant (will not need
it)
M is the mass of
the earth (will not need it)
m is the mass of
Soyuz (will not need it)
g =9.8
m/s2 =MG /R 2
(acceleration due to gravity at r=R )
√(MG /R )=√(gR )
(this will be useful later and is why I do not need
to know G or M )
MmG /r 2
is, as you note, the gravitational force on m a
distance r from the center of the earth.
If orbits are
circular, a=v 2 /r and you can
show that v =√(MG /r ). I
am thinking that we need to find the difference of speeds
between orbits with r=R+h and r=R :
vR -vR+h =Δv
=√(MG )[√(1/R )-√(1/(R+h ))]
=√(MG /R )[1-(1+(h /R ))-1/2 ]
≈√(MG /R )[½(h /R )]
=½h √(g /R )
=247 m/s
Note that, because
h /R= 0.0625 is small compared to 1, I have used the binomial
expansion (1+x )n ≈1+nx .
My answer is about twice what Green
got. Of course, in the real world, air drag when entering
the atmosphere would significantly reduce the required
speed change due to the additional braking it would
cause; but you cannot estimate that using only those two
equations. So, I am at a loss to understand the
difference.
ADDED
THOUGHT: Maybe the Soyuz dropped to a lower orbit
before beginning its reentry, h ~200 km, after
leaving the ISS, but I could find no reference to its
altitude in the video.
QUESTION:
i am currently designing a fully adjustable boom arm for a microphone my prototype is fine and it is functional but i am interested on getting more weight suspension from my joints the joints im using are friction based and use clamping force to lock the joints in the desired position what material should i to use as washers to create more friction in my joints
ANSWER:
it is hard for me to be specific since i do not
know what the joints are made of what you need
to do is find a material with a larger
coefficient of static friction with the material
currently in the joints i think the very best
way would simply apply trial and error try
different materials to see how they work i would
start with obvious choices like rubber why dont
you use punctuation or capitalization it is very
annoying reading your question
QUESTION:
In a fire where does the photons come from? Are photons in a superposition of wave and quanta particles?
ANSWER: Fire is many chemical reactions
going on. The reactions are, for the most part,
exothermic and most of the energy comes out as
heat and light. Photons are particles which have
wave properties.
QUESTION:
I am confused in the formula of orbital angular momentum of electron. According to quantum physics the formula of orbital angular momentum is L=l√(l+1)h/2π B.M. while according to Bhor's postulate the formula of angular momentum is L= nh/2π. Both are giving different result. For example for hydrogen putting n=1 in Bohr formula and l=0 in quantum mechanics formula gives h/2π and 0 respectively.
ANSWER: I believe there is an error in
your question, L =√(l (l +1))h /2π.
The answer to your question is that the
Bohr model is simply incorrect. As luck would
have it, it describes the energy levels of the
hydrogen atom well, but it is incorrect
regarding angular momentum of the levels.
QUESTION:
Why would a heavy object like a bowling ball not float on top of water, but a wooden telephone pole, which is much heavier, would float? I'm sure the shape and the material the object is made from plays a role but I would just like to know the science behind it.
ANSWER:
The science behind this is over 2000 years old.
Archimedes was a Greek mathemetician who
discovered what we call today Archimedes'
Principle. It says that there is an upward force
(called the buoyant force) on an object in a
fluid which is equal to the weight of the fluid
which it displaces. It turns out that, applying
this principle, anything with a density smaller
than water floats and anything with a density
greater than water sinks. Wood is less dense
than whatever bowling balls are made of. You
might then ask why a steel battleship can float.
It is because it is hollow and therefore the
total density is much lower than the density of
steel; a hollowed out bowling ball would float.
QUESTION:
I read that metals conduct heat well because the electrons of metallic elements have free electrons in its outer shell that move quickly when heated. These quickly moving electrons transfer their kinetic energy to other electrons which vibrate quickly and collide with other electrons and transfer the heat energy and so and so forth.
I also read that metals conduct electricity well because the electrons of the outer shell are free to move (as is the definition of electricity - the movement of charged particles).
So it would reason to stand that heating a metal should generate electricity or at the least facilitate the production of electricity as both heat conductance and electricity involves the movement of electrons in terms of metals. To my surprise after a google search, I learned that heating a metal does not generate electricity, and furthermore hampers (not facilitate) electric conductance.
Please dear physicist help me to understand why hot metals are poor conductors of electrons.
ANSWER: In a conductor, there are "free
electrons" which essentially behave as an
electron gas. There is normally no net current
because, just as there is no wind in still air,
the electrons move in random directions and the
net flow is zero. If you apply a voltage across
a conductor, these moving electrons, on average,
flow from the negative to the positive
terminals. Since they are now experiencing a
force due to the electric field, they
accelerate; but, obviously, they do not really
speed up the whole time that they move from one
end to the other of the conductor but, on
average, move with some constant speed called
the drift velocity. The reason is that they are
not really free; rather, they accelerate for
some short distance and then collide with one of
the atoms in the solid which scatters them, then
accelerate again, then collide again, etc .
Now, all the atoms in the solid are constantly
vibrating as if attached to tiny springs, and
the hotter the metal gets, the larger the
amplitudes of their vibrations becomes. This
means that they present bigger targets for the
electrons to hit and so collisions happen more
often which reduces the drift velocity and
therefore the conductivity.
QUESTION:
I was confused about the strict physics definition of a wave. Considering it is a traveling disturbance that carries energy, would wind on a field, an audience created "wave", and dominoes falling be waves? I thought the audience would not be a wave, and I was unsure about the dominoes because there is no wavelength or period.
ANSWER:
The "strict physics definition of a wave" is
that it is something which satisfies the wave
equation. The solutions of a one-dimensional
wave equation are always of the form f (x-vt )
where f is any function; for
example where x is the position, t
is the time, and v is the speed at
which the disturbance travels through the medium
which supports it. For your examples: dominos
falling and people in a stadium sitting and
standing are both waves, but wind is not. Wind
is an example of the entire medium moving but
that would not be described by a wave equation.
The stadium wave would not be a wave if the fans
all all just ran around together, like a
hurricane of people.
QUESTION:
In my understanding of time dilation if one has a clock in constant motion relative to me I will measure the time ticking off on that clock as slower. However from the other clocks view it is just as valid to say that I AM in motion so an observer traveling along with that clock would measure my clock as moving slower. How is that reconciled? Especially if at some time in the future both clocks are brought together at rest with each other to compare clocks? Am I missing the part that acceleration plays here?
ANSWER: There is no need to reconcile
that each observer measures the other's clock to
run slower unless you bring the two of them back
together into the same reference frame.
Acceleration has nothing directly to do with it.
So this is the "twin paradox" which we know is
not really paradoxical at all. Go to the faq
page and find "twin paradox ".
QUESTION:
We have been taught that the big bang created all heavenly bodies as we understand them. I'll disagree with that. Were the exo-planets we have discovered thus far created by the same creation event that brought forth us, Mars and Venus or was it something else that lead to the Gliese's and the Kepler's?
ANSWER:
Good grief, nobody claims that the big bang
created everything as it is right now. The big
bang created a huge amount of energy (from
where, nobody knows) and physical laws. From
that point the universe evolved over time
eventually atoms, stars, galaxies, planets,
etc . appeared. To see a timeline of the
universe, click
here .
QUESTION:
I'm designing a special suction tube for dental practitioners (I'm an orthodontist) and have a basic question I feel I should know. I want to maintain as MUCH air flow through the tip of the suction tube(end of the tube by the patient's mouth) as possible. The lumen of the suction at the dental unit has a cross sectional area of 75mm2. This tube sucks the air from the patient's mouth into the dental unit. If I use a tube that has a much smaller lumen EXCEPT at the patient's mouth where the cross sectional area would be 75mm2, will I lose air flow? Another way of asking this is: am I required to keep the cross sectional area of the ENTIRE tube at 75mm2 in order to maintain the same flow rate at the tip of the suction (patient's mouth) that I have at the dental unit?
ANSWER: I learned something new: in
physics, lumen is a measure of luminous flux.
Biolgists, though, use it as a measurement of
inner space of a tubular structure; apparently
it is just the cross-sectional area of the tube
since the questioner specifies the area of the
opening. If you were dealing with an ideal
fluid, incompressible, no wall drag, no
viscosity, you could argue that the flow would
be constant, the speed in the narrower part of
the tube would be larger, but the volume per
unit time flow would be the same. But air is not
incompressible and energy loss effects are not
negligible. In order for you to maintain the
same volume flow rate you would need a more
powerful pump which is pulling the air. Think of
trying to drink through a pinched straw–you
have to suck harder to get the same amount of
drink.
QUESTION:
I have a question for you. So from what I've learnt, Einstein's Theory of Relativity says that an object with mass will bend light towards it by way of its gravity and the source and direction of the light will therefore appear differently to an observer. Besides that, Einstein also says that E=mc^2 and thus m=E/C^2 so something like light, with measurable energy, would have a measurable mass. My question is, just like that object with mass pulls light towards it with its gravity, can light also pull that object towards it with its gravity? Similarly to how an object falling to Earth pulls on the Earth itself? Or does light not have gravity in the same sense as objects with actual mass on account that its mass is relativistic and not actual, classical-physics mass?
ANSWER: For the first part of your
question, see an
earlier answer . (You could also have found
this on my
faq page.) The answer to your second
question is that the theory of general
relativity finds that any energy density warps
spacetime, that is causes a gravitational field;
a photon has energy and therefore creates
gravity; keep in mind how small this is when
compared to any macroscopic mass.
QUESTION:
This is not a homework question, I am actually a teacher and saw this question in a textbook and would like to know if this is a valid question to ask a third-grader. I may be wrong but I do not believe that the pound and ounce are units of measurement for mass. I have tried looking it up online and did not find anything useful.
Which two units of measurement are used to measure mass?
Gram
Pounds
Ounce
Kilogram
ANSWER:
Kilograms (kg) and grams (g) (1 kg=1000 g) are
units for mass. Pounds (lb) and ounces (oz) (1
lb=16 oz) are units for force. This is very
confusing because of the ways weight is measured
in different countries. In countries where
Imperial units are used, weight (the force which
the earth pulls down on something) is measured
in pounds (which is correct) and we think that
this must also be a measure of mass (the amount
of stuff there is). In countries where SI units
are used weight is measured in kilograms (which
is incorrect) and we think that this must also
be a measure of force. I will leave it up to you
to decide whether third-graders can understand
this subtle difference!
QUESTION:
I know that in a solid the particles are still in motion, but I was wandering if their is any point in the state of matter where the particles are completely motionless? (something like before solid)
ANSWER: The Heisenberg uncertainty
principle stipulates that you cannot know to
absolute precision both the position x and
momentum p (mass times velocity). Stated
mathematically, Δx Δp ≤ℏ
where ℏ is the rationalized Planck constant
(a very tiny number and Δ denotes the uncertainy of the quantity.
So if the velocity (momentum) uncertainty is
zero (particle perfectly at rest), the
uncertainty of the position of the particle is
infinite—the particle could be anywhere in
the universe.
QUESTION:
how can the calculations for the curvature of space-time under general relativity be correct without the mass-energy associated with dark matter taken into account? I am just a retired attorney trying to educate my self about our physical world.
ANSWER:
Keep in mind that I specify on the site that I
do not do astronomy/astrophysics/cosmology, so
you may wish to take my answer with a grain of
salt. When you refer to the "curvature of
space-time" you are actually talking about the
gravitational field. And, you are certainly
right, dark matter, if it actually exists, would
affect the field. I believe that it is actually
the other way around: by studying the
gravitational field you can infer something
about the distribution of dark matter. The most
well-known such inference has to do with
rotating galaxies and how the angular velocities
of the stars varies with distance from the
center. If the only gravity in our Milky Way
galaxy were from stars and gas which we know are
there, the galaxy would not be able to hold
together, stars at larger radii would have
speeds far too large. The Wikepedia article on
dark matter halos would be of interest to
you.
QUESTION:
If a toy car hits a wooden block and pushes it along the track, do the wooden block and the car have the same level of energy or does the energy of the car decrease and the energy of the block increase?
ANSWER: You have described a perfectly
inelastic collision, one in which the two
colliding bodies stick together after the
collision. I am assuming that the block is at
rest before the collision. So linear momentum is
conserved and energy is not conserved. I will
denote the velocity of the car before the
collision to be V , the velocity of the
car/block after the collision to be U ,
the mass of the car to be M , and the
mass of the block to be m . Conservation
of linear momentum gives MV =(M+m )U
or U=V [M /(M+m )]. Note
that U<V , so the car has lost energy
and the block has gained energy.
If you are interested you can calculate the
change in energy (ΔE=E after -E before )
for each:
ΔE car = ½M (U 2 -V 2 )=½MV 2 [
(M 2 /(M+m )2 )-1]
ΔE block = ½MU 2 =½MV 2 [M 2 /(M+m )2 ]
For example, if M=m ,
ΔE car =- (3/4)[½MV 2 ]
and ΔE block =(1/4)[½MV 2 ].
The car lost three fourths of its initial energy
and the blocked gained one fourth of the initial
energy. In this case , half of the total
energy was lost. In this case , half the
speed was also lost.
QUESTION:
The question is about Maximum gradient a vehicle can climb.
The maximum friction force that can be transferred to the road cannot be more than
μmg cosθ . As far as I know, μ cannot be greater than 1 (irrespective of contact patch size) for a Rubber tyre on concrete road. At 45 degrees, the drag due to gradient (mg sinθ ) is equal to the Normal force (mg cosθ ). Any more than 45 degrees, and the drag force increases and Normal force decreases, thus making the vehicle's tyres slip due to loss of traction.
Now, I see videos of RC cars on
youtube
climbing gradients well over 54 degrees.
Could you explain this? Is the μ value greater than 1?
ANSWER: Your analysis is fine if the
vehicle is a point mass. But, the video you
refer to has the failure to climb a steep
incline due to the vehicle tipping over
backwards rather than slipping. I will assume the
truck shown above is at rest because it is
easier to visualize and exactly the same as if
it were moving up with constant speed. The whole
weight mg may be considered to act at the center
of mass of the truck (white cross) which is
located a distance h above the ground,
a distance d 1 behind the
front axle, and a distance d 2
in front of the rear axel, as shown. There are
three equations, sum of the forces along the
ramp equals zero, sum of the forces
perpendicular to the ramp equals zero, and sum
of the torques about the point where the rear
wheel touches the road equals zero:
f 1 +f 2 -mg sinθ =0
N 1 +N 2 -mg cosθ =0
(d 1 +d 2 )N 2 +hmg sinθ-d 2 mg cosθ= 0
Now, we are interested in when
the truck will start rotating backward about the rear
axle; when it is just about to do that, the normal
force N 2 =0. So, the third equation
tells us that tanθ=d 2 /h ;
therefore the maximum angle before tipping is θ tip =arctan(d 2 /h ).
For example, if d 2 =1.5 m and h =1
m, θ tip =56.30 . What
determines tipping is just where the center of mass is;
you want it as far forward and as close to the ground as
you can.
But, if you lose traction, slip,
before you rotate, you do not need to worry about
rotating. As you correctly surmised above, the truck will
start slipping when θ slip =arctan(μ )
where μ is the coefficient of static. I agree
with you that μ= 1 is a reasonable estimate
for rubber on dry concrete. However, the video you sent
is not done on a concrete ramp, rather the surface
appeared to be carpet. Since the tire tread can press
into the carpet pile, it is not surprising that μ >1
for those videos.
QUESTION:
How does a single speaker unit play an audio clip consisting of sound frequency of different magnitude? It's the same speaker and has only one 'vibrating membrane' to induce the sound in the air to propagate. How does this single membrane generate sound waves of different frequency and different amplitude simultaneously?
ANSWER:
Sound waves are linear, i.e. if several are
acting on a point in space, the net action is
simply the sum of all the soundwaves. As an
example, I have plotted three waves representing
the individual sounds, each with different
frequencies and different amplitudes, at some
point in space as a function of time. The net
sound at this point in space is simply the sum
of the three, shown by the black curve. Your
speaker need only be driven to vibrate like
this; this is pretty easy to do, usually with a
magnet attached to the speaker cone and inside a
coil, the current in which varies like the black
curve.
If you cannot find what you are looking for, it was
probably archived. You may find the most recent archived
file
here .