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 rR.
(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. Iï¿½m 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=6.4x10^{6}
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 m^{3}, 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 1618k 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 1.4x10^{18}
m^{3} 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 wellknown, though, to be about
5.15x10^{18} kg; so you can calculate the volume
if the whole atmosphere were at sea level pressure where
the density is 1.225 kg/m^{3}, 5.15x10^{18}/1.225=4.2x10^{18}
m^{3}. 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, 6.4x10^{6} 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} m^{3}kg^{1}s^{2}
is the universal gravitational constant and c=3x10^{8}
m/s is the speed of light. For example, let the location
be on the surface of the earth, M=6x10^{24}
kg, R=6.4x10^{6} m; then, t'=t√[17x10^{10}]≈t[13.5x10^{10}].
So, for example, the clock in space would click off 1
second while the clock on the earth would tick off
13.5x10^{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 R_{S}, the time
dilation equation becomes t'=t√[1R_{S}/R]
so time stops when R=R_{S}. If R=2R_{S},
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
online 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
incity 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 boldface, 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=ABcosθ_{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=Fï¿½d.
The scalar product is commutative, i.e.
A·B=Bï¿½A.
The vector product or cross product is defined
as
A×B=uABsinθ_{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=Bï¿½A.
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 wouldnï¿½t 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 isnï¿½t this just relativity? If you are moving that fast everything seems like itï¿½s 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 wouldnï¿½t we realize it instantly? So why canï¿½t you move faster than the speed of light? Wouldnï¿½t that just mean things relative to you would seem to be stopped for longer, meaning time doesnï¿½t 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√(10.99^{2})=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 earthbound 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 spacetime at
all, so I will then give a little more
background on how spacetime (four dimensions)
arises. When special relativity is mathematically
represented as suggesting a fourdimensional
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/√(1β^{2}).
So, two of the coordinates in our 4space get
mixed up together if the other frame is moving.
Does anything analogous happen in everyday
3space? 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 zaxis, the new x' and
y'axes
both depend on what θ is and they
depend on both x and y are:
x'=xcosθ+ysinθ
y'=xsinθ+ycosθ.
Hence the Lorenz transformation may be
interpreted as a rotation in the xt
plane. This is the impetus of mathematically
working in a 4dimensional space (spacetime).
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 ï¿½starting lines.ï¿½ 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 havenï¿½t learned about Lenzï¿½s 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 freefall 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
zintercept of our linear trendline is positive, not negative. When an object goes from rest to terminal velocity, I had shown them that the
zintercept 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/s^{2}. 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, mgbv=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 zintercept" 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 supercooled 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 arriveï¿½the 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
1/100^{2}=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 Iï¿½m not a physicist but a Biologyï¿½s 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=10^{4}/(3x10^{8})^{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 √(10.9^{2})=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 currentlyoff
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 W_{x} 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})=(2x10^{30}/6x10^{24})(3.8x10^{8}/1.5x10^{11})^{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}/10^{2}=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 lbï¿½mph 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=6.4x10^{6} m
(radius of the earth)

h=4x10^{5}
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/s^{2}=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:
v_{R}v_{R+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 onedimensional
wave equation are always of the form f(xvt)
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 exoplanets 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 crosssectional 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, classicalphysics 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 thirdgrader. 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 thirdgraders 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, ΔxΔp≤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 spacetime under general relativity be correct without the massenergy 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
spacetime" 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
wellknown 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
μmgcosθ. 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 (mgsinθ) is equal to the Normal force (mgcosθ). 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}mgsinθ=0
N_{1}+N_{2}mgcosθ=0
(d_{1}+d_{2})N_{2}+hmgsinθd_{2}mgcosθ=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 N2=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}=56.3^{0}. 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.