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QUESTION:
Around 1905 when Lorentz came up with his tranformation equation, it was assumed that c was constant, but was it also assumed that c is the speed limit of the universe?
Why didn't he assume that v could be greater than c in some cases? Why didn't he put a corrective constant next to c in the equation to allow for the case that time might stop BEFORE or After v reaches c?
The reason I'm asking this question is because the Lorentz transformation has time stop exactly when v=c. How did he know that should be the case?
ANSWER:
You have your history a little skewed, I think. Lorentz
was working with electromagnetic theory, Maxwell's equations in particular.
The transformation which bears his name was arrived at empirically with no
reference to the speed of light. He noticed that if you took Maxwell's
equations and transformed them into a moving frame of reference that they
took on a different form which was unacceptable because they had internal
inconsistencies which could not be borne out by experiments. He found that
if he used a somewhat different transformation, now called the Lorentz
transformation, the transformed equations were self consistent; he had
stumbled on a truth without understanding what he had. And he certainly did
not consider the speed of light constant because he was one of the foremost
proponents of the luminiferous æther, the special medium which
supposedly supports light waves. To my mind it is unfortunate that the
transformation is named for him since it was an accidental empirical
discovery, not based on any fundamental physics. Some books try to refer to
the LorentzEinstein transformation, but that hasn't really caught on. To
understand why c is a constant and why it has the value it does, go
to my FAQ page; there you can also see earlier questions about why c
is the speed limit. Your question about a "corrective constant" makes no
sense since universal constants by definition do not need correcting. Since
no clock can ever reach c, time never "stops". I doubt that Lorentz
ever thought about time stopping in any context.
QUESTION:
What do you think may be some possibilities if heat all of a sudden ceased to exist? Or, is it possible to completely neutralize heat within a contained space?
ANSWER:
First of all, read an
earlier answer regarding what heat is. Heat is the transfer of energy,
not the content of energy. Therefore the answer to the first question you ask is that if
there were no transfer of energy, everything would simply stop, I guess.
Regarding your second question, heat flow can be stopped or slowed down for
isolated systems; for example, a thermos bottle.
QUESTION:
Perhaps you can help solve a disagreement we have at work. The question being "Does a person's initial velocity during a jump equal their final velocity once the land?" My contention is "no" in that the jumper could theoretically produce any velocity on the way up, but downward would be limited to terminal velocity. Who's right?
ANSWER:
Technically, you are correct. If air drag is present,
energy is lost which results in the landing speed being less than the launch
speed. In practice, however, for a person jumping into the air the height
acquired is not high enough for this to be a measurable effect; that is,
this is an example where we can say, as we often do in an elementary physics
course, that air drag is negligible. A typical terminal velocity for a human
is about 120 mph≈54 m/s. If you jumped with this speed you go over 100 m
high, obviously not in the cards. I did a rough estimate assuming the
maximum height you could jump would be about 2 m; if the person drops from 2
m his speed at the ground would be about 6.32 m/s without drag, 6.30 m/s
with drag, a 0.3% difference. For comparison, dropping from 100 m the speeds
would be roughly 44.1 m/s and 37.1 m/s for no drag and drag,
respectively. It is good to be precisely correct as you are, but it is also
good to be able to make reasonable estimates in realworld situations.
QUESTION:
It is said that charges are quantised.Also if we bring two identical solid
conducting spheres in contact with each other,their charges are equally
distributed among them. Now suppose if we have a body A with 5e charge and
body B with 0 charge.Now what will be the charge distribution between the
bodies if we bring them in contact and then separate them?Since charges are
quantised,we cannot have 2.52.5 distribution.So will it be 32 or 23
distribution or what?
ANSWER:
The rules you learn like the one you quote apply only
when the charge can be thought of as a continuous fluid which can spread
itself out on any surface, no matter how large, and have zero thickness.
Because the electron charge is so small, these rules work very well to
describe electrostatics for normal circumstances. Obviously, a case like you
describe, with 5 excess electrons is in no way like a fluid and I would
guess that anything from 50 to 05 could be the distribution over a time.
In principle, the 5 electrons would arrange themselves so that each was as
far away from the other 4 as possible, but there would not be a unique such
distribution.
QUESTION:
I was looking at a few years old burn scar on my hand today and it got me to wonder what happened on a subatomic level to the electrons and protons of my hand when I received the burn. So could you tell me what happened to the electrons and protons of my hand when I accidentally touched the inside wall of my oven when takibg the food out?
ANSWER:
What happens when you cook meat? Chemistry. Adding
sufficient heat will cause some molecules in the meat to break apart and
make new molecules. That same thing happens in your hand when you burn
it—just a lot of chemistry going on. Then later biology takes over and takes
all that burnt flesh and sheds and absorbs the burnt flesh and creates new
molecules specialized to be scar tissue, again basically chemistry. So,
electrons and nuclei of all just move around as molecules are destroyed and
then rebuilt. It is all just a bunch of chemical reactions.
QUESTION:
If I consider a
tube both end open and
dip one end in water
(like pipette in chemistry lab)
and close the other by
thumb, water remain
hanged in the tube. If we say it is because
the atmosphere that
pushes up on the water
in the tube is same as
that of remaining air in
tube pushing down on
the water. Won't the
water fall out due to its
own weight as the
upward and downward forces are the same?
ANSWER:
As you lift the tube out of the water, the weight of the
water in the tube pulls it down and the volume of air between your thumb and
the top of the water in the tube increases. Because the volume increases,
the pressure decreases so that when the bottom end of the tube is pulled out
of the water the pressure at the top of the column of water is smaller than
atmospheric pressure at the bottom. Therefore, the water column has three
forces on it which are in equilibrium: its weight down, a force down due to
the pressure at the top, and a force up due to the pressure at the bottom.
Your error was in assuming that the pressure at the top of the column is
atmospheric.
QUESTION:
If an electron in an atom can only orbit in fixed orbitals at certain frequencies, how does a gas molecule increase its speed when heated. It implies that the gas pressure also would be allowed only at certain energy levels but the pressure seems to increase in a linear way and jump to fixed pressures when the molecule has attained enough energy.
ANSWER:
The question is, how can energy be added to a gas? Let's
think about a single molecule in the gas. It has, essentially, two kinds of
energy—internal energy which are allowed states of the molecule and are
quantized, that is restricted to certain discrete values, and the kinetic
energy associated with the molecule's motion as a whole as it hurtles across
the room. This kinetic energy is not quantized. So, when you heat up a gas
you are adding to its kinetic energy for normal temperatures. You have to
get to extremely high temperatures before you start excite atomic states.
You might be interested in why internal states are quantized and kinetic
energy is not. In quantum mechanics, systems which are bound (like electrons
in atoms) are quantized whereas systems which are not bound (like your
freely moving molecule) are not.
QUESTION:
When a ball is thrown vertically upwards ignoring air resistance, and another ball is also thrown upwards with air resistance, the time taken is less for the ball with air resistance to reach max height. Why is this "because average acceleration/force is greater"? Wouldn't there be less acceleration/force because the air resistance cancels some out?
ANSWER:
The reason is that the ball with air resistance does not
go as high. The force on the ball without resistance is the weight of the
ball pointing in the downward direction; but the downward force is greater
for the ball with air resistance because the drag force is also pointing
down. Therefore the ball with resistance slows down faster so it stops more
quickly. Think of an extreme example: if you throw the ball upwards in honey
which has very great resistance, it stops almost immediately.
QUESTION:
If I had a graph of an object's momentum against time, and at t_{1} momentum is p_{1} and at t_{2} momentum is p_{2}. p_{1} is a positive y value but p_{2} is negative. Wouldn't the average force be (p_{2}p_{1})/(t_{2}t_{1}) instead of (p_{1}p_{2})/(t_{2}t_{1})?
ANSWER:
The fact is, both force and momentum are vectors so the
signs of these quantities mean something in one dimension which is evidently
what you are asking about. A onedimensional vector with a negative value
points in the negative coordinate system direction. Let's take a particular
example of your supposition, p_{2}=5 kg∙m/s, p_{1}=1
kg∙m/s, and t_{2}t_{1}=1
s. This means that the particle started out moving in the positive direction
and ended up moving in the negative direction; therefore the average force
causing this must have been pointing in the negative direction, right? The
average force F is defined to be F=(p_{2}p_{1})/(t_{2}t_{1})=(51)/1=6
N. Your proposal would give
F=(p_{2}p_{1})/(t_{2}t_{1})=(51)/1=4
N, obviously pointing in the wrong direction. (I have no idea where your
F=(p_{1}p_{2})/(t_{2}t_{1})
came from; this is also wrong.)
QUESTION:
Why do physicists refer to the nonvisible portions of the electromagnetic spectrum as "light"? I thought only the visible portion was considered "light".
ANSWER:
This is really just semantics and of no real importance.
We often refer to "the speed of light" which is universally understood as
the speed of electromagnetic radiation. I never heard of anyone use the word
light to refer to gamma rays or AM radio waves, for example. If we want to
be very clear, we say "visible light" when referring specifically to the
visible spectrum.
QUESTION:
How is it possible to measure the the distance between an electron and a nucleus? I don't mean calculate, but measure. In other words, what is the operational definition of the Bohr radius?
ANSWER:
It is not possible to measure the distance because it has no fixed value. All you can do is predict the probability of finding the electron some distance from the nucleus.
There is no operational definition of the Bohr radius, it is just defined
guided by history (Bohr model); you cannot measure it.
QUESTION:
Whenever we roll a ball or spin a quarter it will slow down and eventually stop, since energy cannot just dissapear where does it go?
ANSWER:
The kinetic energy is being taken away from the ball or
coin by friction. That energy shows up as thermal energy, the
ball/cointableair all get a little bit warmer. Also, since you can hear
the ball rolling and the coin spinning, some of the energy must be lost to
sound.
QUESTION:
How can the speed of light be constant, when time is not?
In my understanding, the rate at what time passes changes, relative to speed of motion and gravitational forces.
So if the speed of light is 186,000 a second, but a second could flow at different rates, how does this affect the speed of light?
ANSWER:
This is a chicken & egg sort of thing. Time not being
"constant", moving clocks running slow, is a result of the fact that the
speed of light is a universal constant. See my earlier discussion of the
light clock which demonstrates this. Not only do clocks run slower, but
lengths get shorter by exactly the same factor. So, if it takes a pulse of
light 1 s to travel between two points A and B separated by 3x10^{8} m for one observer, another observer moving with
respect to the first might observe the same light pulse take 0.5 s to travel
between A and B which are now separated by 1.5x10^{8}
m, exactly the same velocity.
QUESTION:
If I raise a mass above the earth, you would say that I'm increasing the potential energy by mg x delta(h). So, on a macroscopic scale , if I raise a mass above the earth, doing work, Im actually adding mass to the earthmass system? and the increase in potential energy (height classically) is actually a small, virtually undetectable increase in mass of the earth/mass system?
ANSWER:
Let's look at several ways to understand this situation.
Start with the workenergy theorem which says that W_{ext}=ΔK=½mv_{2}^{2}½mv_{1}^{2},
work done by external agents equals the change in kinetic energy. So, when
you lift it to height h, gravity does W_{earth}=mgh
and you do
W_{earth}=mgh units of
work and so kinetic energy is unchanged. This ignores a couple of things,
one being that the earth is not flat and the field is not uniform and
another is that you do work on the earthmass
system (EMS) thereby increasing the mass of
the EMS. (Since you do work on the EMS, it does the same magnitude of
negative work on you, so your mass decreases by the same amount as the EMS
mass increases; energy is conserved if all forces are internal, that is if
we think of you as part of the earth.) The work done on the mass is
enormously larger than the work done on the earth because both experience
the same force but the earth hardly moves at all while the mass goes
(almost) h, so almost all of the mass increase will be in the mass,
not the earth.
To get an idea of the magnitudes of these
effects, I will do a specific example, m=1 kg, h=1 m, R_{earth}=6.4x10^{6}
m, M_{earth}=6x10^{24} kg.
 Work done on m is mgh≈10 J.
 Increase in mass of m is mgh/c^{2}=10/9x10^{16}≈10^{14}
kg.

Earth will move only about 10^{23} m, work you did on it will
be only about 10^{22} J, mass increase about 10^{37}
kg.
 I estimate that g will be
smaller by about 3x10^{5} m/s^{2} at 1 m above the
surface, a very small difference. But, it is huge compared to the mass
increase. The effect on g is a 3x10^{4 }% effect and
the mass increase is a 10^{12 }% effect.
Note that I have not said anything at all
about potential energy yet. I have always thought, at an elementary physics
level, of potential energy as just a clever bookkeeping device to
automatically keep track of work done by a force which is always there,
gravity in this case. To illustrate,
W_{ext}=ΔK=W_{gravity}+W_{you},
so ΔKW_{gravity}=W_{you }so we define
the potential energy difference to be ΔU≡W_{gravity} and,
voilà, ΔE≡ΔK+ΔU=W_{ext}
where external work now excludes work done by any force for which we
have introduced a potential energy function; we have simply internalized the
work done by a force always there. This has no effect on my discussion of
mass changes above.
FOLLOWUP QUESTION:
So, when the mass falls back down thru the distance, the very small change in mass we had on the way up gets converted into Kinetic energy.
now, lets do work in the horizontal plane, so gravity is not a factor. I have a mass, I perform some work on it (adding energy) now energy has mass, but at very very low speeds, compared to light, nature allows MOST of this work supplied to go into a velocity increase, with a very small amount (undetectable amount) to go into mass increase. As such, at low speeds, almost all of the energy supplied (work done) shows up as a velocity increase. Now, at very high speeds approaching light speed, nature in order to not exceed the speed of light, takes most of the supplied energy (work done) and puts it into a mass increase, with very little going into a velocity increase. Is this a correct picture I have of this?
ANSWER:
You have it about right.
There are some subtleties, though: if you are going to talk about mixtures
of kinetic and mass energies and introduce particles with large speeds, you
are going to have to start doing things relativistically. I have
emphasized before that you have to be careful about what you call mass.
My own preference is to work only with rest mass which I usually denote as
m. In your earlier question, all particles which we were looking at
were at rest, so when I talked about increase in mass, that was increase in
rest mass. If the object is moving, what do we mean by mass? In spite of
what many text books say, I do not say that mass increases with velocity but
rather that linear momentum has been
redefined in such a way that it is no longer simply mass times velocity.
Now, the energy E of a particle with rest mass m and speed
v is E=√(m^{2}c^{4}+p^{2}c^{2})
where p is the linear momentum defined as p≡mv/√[1(v^{2}/c^{2})].
Now, you will note, if the particle is at rest,
E=mc^{2 }and p=0,
and as long as v<<c,
p≈mv. Also note that kinetic energy is
no longer ½mv^{2},
rather it is that amount of energy which is not rest mass energy, K=Emc^{2};
you can easily show that if
v<<c, K≈½mv^{2}.
For example, v=0.8c=2.4x10^{8} m/s, m=1
ng=10^{12} kg:
 p=10^{12}∙2.4x10^{8}/0.6=4x10^{4}
kg∙m/s, wheras mv=2.4x10^{4} kg∙m/s;
 pc=1.2x10^{5} J;
 mc^{2}=9x10^{5} J;
 E=9.08x10^{5} J;
 K=8x10^{3} J, whereas ½mv^{2}=2.9x10^{4}
J.
YET
ANOTHER FOLLOWUP QUESTION:
I just wanted to add that in my previous question, concerning mass changes with increasing velocity, what I'm calling "mass increase"is not a " physical structure" change of an object. I view this increase of mass with velocity as a " change in how the object interacts with space time.
Therefore, I believe that the gamma correction at large velocities is how the object is changing how it interacts with space time.
I think that some people believe this "mass increase" with velocity is an actual accumulation of matter or a physical enlargement of the object. I hope my thinking is correct on this?
ANSWER:
You
are essentially correct. If you want to interpret the new definition of
momentum as meaning that the mass m' of a particle with rest mass
m and moving with speed v is
m'=m/√[1(v^{2}/c^{2})],
that is fine; with this interpretation you are saying that the momentum is
still mass times velocity but that the mass depends on velocity. But m'
should be thought of as the inertia of the particle and not some measure of
"the amount of stuff" as you suggest. The faster something goes, the harder
it becomes to make it go faster (which means that its inertia is increasing)
until eventually you run against the wall that v must always be less than c.
You may be interested in a
previous answer regarding the kinematics of a particle which experiences
a constant force.
QUESTION:
Generally, we are trying to determine how much electricity will be generated by falling water . With that in mind, an engineering group has proposed a project whereby they place one of their machines inside a tube. We have all of the electrical equations worked out from the movement of the blades inside the machine  the question for the Physicist is this: what is the speed of the water if it falls 5M or 10M or 20M? If the diameter of the pipe is an important variable, the answer is that we can make the mouth of the pipe as wide as we want: 5M, 10M, 20M etc. Bottom line, we want to achieve a water speed of at least 6 meters/second.
ANSWER:
You have not given me any details about the source of
this water. This sounds like a classic Bernoulli's equation problem in
elementary physics where you have a deep reservoir and there is a hole at a
depth h in the dam; what is the speed with which the water squirts
out? This hole could be your pipe and the cross section of the pipe does not
matter as long as its diameter is small compared to h. Bernoulli's
equation states that ½ρv^{2}+ρgy+P=constant
where
ρ is
the density of the fluid, P is the pressure, v is the speed of
the fluid, y is the vertical position, and g=9.8 m/s^{2}≈10
m/s^{2 }is the acceleration due to gravity. In your case, I would
choose y=0 at the bottom of the dam where your pipe is, so y=h
at the surface; the velocity at the surface of the lake is
approximately zero and the velocity in the pipe is v; the pressure at
the top and the bottom is the same, atmospheric pressure P_{A}.
So, Bernoulli's equation is
½ρv^{2}+ρg∙0+P_{A}=½ρ∙0^{2}+ρgh+P_{A
}or, ½ρv^{2}=ρgh, or v=√(2gh).
Interestingly, this is exactly the speed the water would have if you just
dropped it off the top of the dam.
For example, at a depth of 5 m the speed
should be about 10 m/s before you put your machine in it.
Bernoulli's equation is, essentially, just
conservation of energy for an ideal fluid; water is not an ideal
fluid, but close enough for this to be a pretty good approximation. However,
you will be asking this moving fluid to do work on your generator which will
take energy away and slow the water down. So, maybe it is useful to
calculate the energy which this moving water has. The quantities in the
equation are energy per unit volume of the water, so E=
ρghV where E is the energy
contained by a volume V of the water. The power P (not to be
confused with pressure), is the rate at which the moving water is delivering
energy, P=dE/dt=ρgh(dV/dt).
If the cross section of your pipe is A, then
dV/dt=Av
so P=ρghAv.
Here is an example: taking h=5 m with
v=10 m/s as above, using
ρ=1000 kg/m^{3}, and assuming a
pipe with a diameter of A=1 m^{2}, P=5x10^{5}
W=500 kW. You could not get any more power than that from this water.
QUESTION:
In Galiean Relativity, there is no such thing as absolute velocity, as all velocities are relative. However, there is such a thing as absolute acceleration in Galiean Relativity.
In Special Relativity, there is no such thing as absolute velocity, however there is also no such thing as absolute acceleration (unlike Galiean Relativity). Isn't the connection/correction to acceleration the Lorentz Transforms Gamma correction, G=1/square root 1v/c^2?
ANSWER:
The Lorentz transformation for acceleration is very
complicated and I will not write it because it really has very little
interest or applicability in the theory of special relativity. As you note,
all Galilean observers will measure exactly the same acceleration. The
reason that this is so important is that Newton's second law, if written as
F=ma, on which so much of classical mechanics hinges,
is true only if all observers measure the same acceleration because surely
all observers will agree on the force applied. Interestingly, Newton did not
write his second law in this form but rather in the form F=dp/dt
where p is linear momentum. You can learn more about
relativistic momentum
here and here. Also
of interest here is the motion of a particle which experiences a
constant force; the acceleration cannot be constant because the particle
cannot go faster than c.
QUESTION:
Does Newton's third law apply only to contact forces or to noncontact forces (e.g. gravity, electrostatic force)? Our teacher says that it applies when the bodies come in contact with each other, but I think it should be applicable to both. Which is the case?
ANSWER:
Well, it depends on how you state the third law. If you
state it as "if particle 1 exerts a force on particle 2, particle 2 exerts
an equal and opposite force on particle 1", then there are certain cases
where this law does not apply: see an example for electromagnetism in an
earlier answer. However, when
fields are present, Newton's laws are much more subtly stated because the
fields themselves contain energy and momentum. You can read a more detailed
discussion, concerning linear momentum conservation which crucially depends
on Newton's third law, in another
earlier answer.
QUESTION::
My question is concerning the energy required to pry apart (binding energy) of molecules ( not nuclear). Such as the burning of hydrocarbon fuels. When you do work against gravity, the work (energy) you put in goes into potential energy above the earth. When you pry apart the molecules, you are doing work against the attractive forces holding the molecule together( binding energy)Does that work appear as potential energy (distance) between the molecules ( if the forces remain attractive) OR does the attractive force between the molecules change from attractive to repulsive or vanish with distance? If the forces remain attractive then the work supplied should become potential energy between the molecules. If the forces vanish, or become repulsive, the work (energy) supplied, prying them apart, should manifest as a small increase in mass of the molecules forced apart. I'm not sure how the forces between bound molecules behave with distance as you separate them.
ANSWER:
Molecules are held together by electromagnetic forces, so
it is useful to get an orderofmagnitude idea of the binding energy
compared to mass energies. Consider a hydrogen atom with an electron and a
proton bound together by their electric attraction. The energy necessary to
move the electron very far away is 13.6 eV. The restmass energy (mc^{2})
of a hydrogen atom is about 1 GeV. Therefore, since you have added energy by
ionizing the atom, the atom is lighter by about 100x13.6/10^{9} %≈10^{6
}%. Any molecular binding energy will be of the same
orderofmagnitude. The energy we get from chemistry comes from mass and it
is extremely inefficient. So, although E=mc^{2} is at the
heart of things, you usually do not have to worry about mass changes in
molecular chemistry because they are so tiny. To do detailed calculations of
chemical reactions usually requires that you do things quantum mechanically
which requires a potential energy function. These calculations can be very
complex and approximate models are used to simulate the potential energies
of the molecular systems. Once you get beyond the simplest atoms and
molecules, the calculations can only be done numerically and approximately
on computers. An example of a potential energy function, the Morse
potential, for a diatomic molecule is shown in the figure to the left. The
form of this potential is V_{Morse}=D_{e}[1exp((rr_{e}))]^{2};
note that, for one of the atoms in the molecule, the force (slope of the
potential energy function) is repulsive for r<r_{e} and
attractive for r>r_{e}. This is expected since the molecule
has a nonzero size because of repulsion but is bound because of attraction.
A first approximation often used for bound molecules is a harmonic
oscillator potential (masses attached to a spring).
QUESTION:
Suppose an ice cube is suspended in a gravity free region (having room temperature). Now as the ice melts,what shape will the whole system(including melted water) will attend and why? What shape will the solid ice have(i.e will it maintain its cubical shape as it melts)?
ANSWER:
First, think about an ice cube sitting on a table. Why
does it melt? Because heat flows through the surface to the small volume of
ice right below the surface causing that small volume to melt when
sufficient heat has been added. But, a small volume near the edge of the
cube has a larger area and a small volume near a corner has a larger area
yet. This is illustrated in the figure to the left: each little cube on the
surface of the big cube has a volume V and faces with areas A.
As you can see, the cube on the face of the big cube has an area A
through which heat can flow, one on the edge has an area 2A, and one
in the corner has an area 3A. Therefore, the cube will melt fastest
on the corners, next fastest on the edge away from the corner, and slowest
elsewhere. That is why a cube gets more rounded as it melts. I cannot think
why this would be any different in a zerogravity situation except that the
water would not flow away. So, you would have a layer of water around the
still unmelted ice but, eventually, when there was enough water, you would
have the unmelted ice inside a sphere of water.
QUESTION:
If I observe an object which is in relative motion it will have increased mass and due to length contraction an increased mass density. So can it become a potential "relativistic" black hole?
ANSWER:
No. In the frame of reference of the object, everything
is perfectly normal. Besides, a black hole is a singularity, infinite
density and zero size, so the object would have to be going the speed of
light which is not possible.
QUESTION:
The speed of a mass thrown from a moving object is effected + or  by the speed of the thrower. Not so with the speed of light. If light speed was limited by some unknown phenomena, that would cause questions in one area  but, with a source moving opposite to the direction of the light, the speed is still the same. Light is not affected + or  by the movement of the source. It cannot travel faster, nor can it travel slower based on external forces, though it can be vector shifted by gravity and the medium thru which it passes. Why? Unless we can find a reason, say in quantum mechanics, like the metaphysical identity of an electron, these known truths appear to contradict the laws of cause and effect.
ANSWER:
The speed of light in vacuum, as you note, is constant
regardless of the motion of the source or of the observer. Strange, I know,
but you seem to think that the reason for this is unknown. Indeed, it is
fully understood why the speed of light in a vacuum is a universal constant;
see earlier answers.
QUESTION:
I was watching Babylon 5 and in there there was a description of an Earth Alliance space ship weapon. The weapon was a gun that has two very conductive parts on both sides and a conductive armature in the middle and electricity somehow launches the projectile. The gun is 60 meters long, has two barrels with each capable of firing two shots per second simultaneously. The projectile is 930 kg in mass (1.7 m long 20 cm in diameter) to a velocity of 41.5 km/s. The barrel of this electric gun is 60 meters long. Is this kind of gun physically possible to build?
ANSWER:
I always like to look first at the energetics when
answering questions like this. Assuming that the acceleration of the
projectile is uniform, I find that the time it would take to traverse the
barrel is 0.029 s and the average acceleration is 1.4x10^{6} m/s^{2}.
Thus, the average force on the projectile would be F=ma=1.4x10^{6}x930=1.3x10^{9}
N=280,000,000 lb. I do not think you could have a projectile which would not
be destroyed by such a force. But, suppose the projectile could withstand
this force; the energy which you would have to give it would be E=½mv^{2}=8x10^{11}
J. Delivering this energy in 0.029 s would require a power input of P=8x10^{11}/0.029=2.8x10^{13}
W=28 TW; for comparison, the current total power output for the entire earth
is about 15 TW. Or, if you think of the energy being stored between shots,
and there are four shots per second, P=8x10^{11}/0.25=3.2x10^{12}
=3200 GW; the largest power plant currently on earth is about 6 GW. And,
this power source needs to be on a ship? I do not think this gun is very
practicable!
You might be
interested in similar earlier
questions I have answered.
QUESTION:
Something
ridiculous I thought of, if the Moon suddenly stopped moving and began to
fall toward the Earth, how long would it take to impact? I'm stumped as to
how to calculate this, as the force on the Moon gradually increases as it
falls, and the Moon also pulls the Earth toward it, and the radius of each
object would have to be included.
ANSWER:
I guess I am going to have to put
questions like this one on the
FAQ page. You should read the details of these
earlier questions since I do
not want to go over all the details again. It is tedious and uninstructive
to try to do this kind of problem precisely. I, being a great advocate of
"back of the envelope" estimates, use Kepler's laws to solve this kind of
problem; I have found that a very excellent approximation to fall time can
be found this way. I note that the mass of the moon is only about 1% of the
mass of the earth, the period of the moon is about 28 days, and the moon's
orbit is very nearly circular. The trick here is to use Kepler's third law
and recognize that a vertical fall is equivalent to the very special orbit
of a straight line which is an ellipse of semimajor axis half the length of
the line. Kepler's third law tells us that (T_{2}/T_{1})^{2}=(R_{2}/R_{1})^{3}
where T_{i} is the period of orbit i and R_{i}
is the semimajor axis of orbit i. Now, T_{1}=28 and R_{2}=R_{1}/2
and so T_{2}=T_{1}/√8=9.9 days. But this is
the time for this very eccentric orbit to complete a complete orbit, go back
out to where it was dropped from; so, the time we want is half that time,
4.9 days.But this is not what you
really wanted since I have treated the earth and the moon as point masses.
What you really want is when the two point masses are separated by a
distance of the sum of the earth and moon radii, 6.4x10^{6}+1.7x10^{4}≈6.4x10^{6
}m. To see how much error this causes, I can use the equation for the
velocity v at the position r=6.4x10^{6} m if dropped
from r=R_{moonorbit}=3.85x10^{8} m which I derived
in one
of the earlier answers: v=√[2GM(1/6.4x10^{6}1/3.85x10^{8})]=1.1x10^{4}
m/s. It would continue speeding up if the collision did not happen, but even
if it went with constant speed the time required would be about t=R/v=6.4x10^{6}/1.1x10^{4}=580
s=9.6 min. This is extremely small compared to the 4.9 day total time, so,
to at least two significant figures, 4.9 days is the answer to your
question.
An important part of doing physics, or any
science, is knowing when to eliminate things which are of negligible
importance!
QUESTION:
If our solar system was formed from a cloud of dust, created by numerous super nova explosions, where we get various elements, and the sun is comprised of the greater part of this cloud, how is it that the sun is comprised of hydrogen burning to make helium? How did the sun separate all the other elements?
ANSWER:
Again, I am not an astronomer/astrophysicist. However, I
was just reading in the April 2014 Scientific American an article
about stars early in the universe. When the universe was young, it was
composed almost entirely of hydrogen with only a tiny amount of helium and
lithium. When stars form, they contract gravitationally and heat up as they
do so. But, heating up, the pressure increases and keeps the star from
collapsing further until it can shed some of the heat so that it can
continue contracting. Eventually, a core dense and hot enough forms where
the fusion can ignite. It turns out that hydrogen is not very good at
getting rid of the heat and so more and more hydrogen accumulates;
eventually ignition occurs but the typical early star, because of the
inabilty of hydrogen to cool, is hugely bigger than the sun, anywhere from
100 to 1,000,000 solar masses. These giant stars now eventually die after
burning much hydrogen and creating lots of heavier elements up to iron; they
explode in supernovae and fall back into black holes, scattering the heavy
elements (including heavier than iron made in the supernovae) into space.
Now, to answer your question, the sun has lots of the heavier elements in
it, just a smaller fraction than the planets. It turns out that these
heavier elements are much more effective in cooling the protostar as it is
forming which allows much smaller stars like the sun to form.
QUESTION:
Would it be possible to suspend an electron via electrostatic levitation in
a uniform magnetic field? And if the voltage was decreased enough so that
the electrostatic force on the electron was lower than the force due to the
electron's weight would the electron then experience a 'fall' due to
gravity?
I'm asking this because I'm wondering if placing a positron in a uniform
magnetic field (in a vacuum) and lowering the voltage to a very small value
so that the electrostatic force is lower than the weight would cause a
positron to also experience a 'fall'. And if it did experience a fall it
could be ascertained whether it would fall upwards or downwards.
I'm 17 years old and in the UK about to study physics at university in september and was just curious about this since one of our topics this year was electrical phenomena and we talked about millikan's oil drop experiment which featured a similar sort of suspension when the electrostatic force and force due to gravity were balanced.
It's a thought I had when wondering if antimatter
was affected by gravity the same way matter is.
ANSWER:
It is always nice to see young folks asking interesting
questions. First, I need to correct one thing: everywhere you refer to a
"magnetic field" you should say "electric field"; a magnetic field exerts no
force on a charge at rest which is what you want to observe, à la
Millikan. Now, it is known to extraordinary precision that the inertial mass
of a positron is equal to the inertial mass of an electron. By inertial mass
I mean the ratio of the force you apply to it divided by its acceleration,
in other words its resistance to being accelerated. (A more correct way,
relativistically, to say this would be that they have equal momenta for
equal speeds.) I believe it is true that nobody has ever "weighed" a
positron by measuring the force it experiences in a gravitational field.
But, if the gravitational mass were different from the inertial mass, this
would fly in the face of the theory of general relativity. But, let's talk
about the feasibility of your experiment. The mass of an electron is about
10^{30} kg so its weight would be about 10^{29} N (taking
g≈10 m/s^{2}). The electron charge is about 1.6x10^{19}
C and so the electric field required to levitate an electron would be 10^{29}/1.6x10^{19}≈6x10^{11}
V/m. Suppose we use a parallel plate capacitor to create this field. The
charge density σ on a plate with field E is about σ=ε_{0}E≈10^{11}x6x10^{11}=6x10^{22}
C/m^{2 }which would correspond to an electron density on the plates
of about 6x10^{22}/1.6x10^{19}≈0.004
electrons/m^{2}! This would correspond to about one electron for
every 250 square meters! That would not give a very uniform field would it?
There is no such thing as a uniform surface charge density because charges
in nature do not comprise a continuous fluid; so really tiny uniform fields
are not possible. I did all that just for the fun of it, but there is an
even more serious consideration—the earth itself has an electric field near
the surface of typically 100 V/m pointing down, so an electron would be
repelled upward. To do your experiment you would have to get rid of that
field and I do not believe that it would be possible to be assured that you
could make the residual field much less than your
6x10^{11} V/m. Back to the drawing board! Keep asking those hard
and interesting questions, though, and good luck with your university
studies.
ADDED
NOTE:
A
recent article discusses a new proposal to compare matter and antimatter
weights.
QUESTION:
If there is a planet on the opposite side of the sun, directly across from the earth,
rotating at the exact speed as the earth, could it go unnoticed?
ANSWER:
Certainly not in this day and age when we have space
probes all over the solar system. It seems that in earlier times such a
planet would be detectable only by its gravitational effects on other
bodies, for example comets or asteroids, which have orbits which cross
earth's orbit.
QUESTION:
I'm a Science Olympiad coach trying to optimize the performance of our "Scrambler", a car which must be accelerated by only a falling mass.
Most competitors simply tie a weight to a string and route that string over a set of pulleys (using no mechanical advantage to convert the vertical falling acceleration horizontal.
…Read a whole lot more!
ANSWER:
Sorry, but if you read site groundrules you will see that "concise, wellfocused questions" are required.
FOLLOWUP QUESTION:
I was hoping you'd like the challenge of a motion/force problem that must span across several formula  PE, KE, PEspring, velocity solved by
acceleration and distance only, etc. Something to sink your teeth into...
ANSWER:
It is really not that interesting to work the whole thing
out, but on second thought it is interesting to talk qualitatively and
generally about the questioner's proposal; so I will do that. I will
summarize the situation since I am sure none of you loyal readers will want
to read the whole original question. By using a falling mass M
attached to a car of mass m, it is wished to maximize the speed v
of the car for M having fallen through some some distance H.
The car moves only horizontally. The simplest thing to do is to have the two
simply attached by a string over a pulley. Then, using energy conservation,
0=½(M+m)v^{2}MgH or v=√(2MgH/(M+m)).
What the questioner proposes is to hold the car at rest and insert a rubber
band in the string so that the falling weight stretches the rubber band
which has been carefully chosen to be just right that, when M has
fallen H, it has just come to rest and is held there. Now,
presumably, the rubber band has a potential energy of MgH. If the car
is now released, the rubber band will presumably contract back to its
original length giving its potential energy to the car, MgH=½mv^{2
}or v=√(2MgH/m), a considerable improvement. My
suggestion would be to use a spring rather than a rubber band since a rubber
band has much more damping (energy loss due to internal friction) and
hysteresis (will not return to its original length). Since the rules fix
M and H, one obviously wants m to be as small as possible.
QUESTION:
What is the moon's orbit around the earth? I was wondering if you could send me pictures and diagrams or whatever you could about the orbit.
ANSWER:
I am not sure what you want. The moon has a nearly circular
orbit of radius about 385,000 km. It takes about 28 days to go around earth
once (which is how it came to be that a month was a standard time
measurement). The same side of the moon always faces earth which means it
also takes about 28 days to rotate once on its axis. The picture above is
drawn to scale. For more detailed information, see the
Wikepedia entry.
QUESTION:
Why is charge a scalor quantity if different charges are given positive and negative signs and also if the flow of current which is mainly due to the flow of electrons but has direction,by convention,same as that of protons??
ANSWER:
What makes you think that a scalar quantity cannot have a
sign? How about Celsius temperature, 40^{0}C or +20^{0}C?
How about time, 5 s is 5 seconds before t=0. The fact that electrons
have negative charge is just an accident of history. All important aspects
of electromagnetism would be just the same if you called the electron charge
positive. The reason that current is defined to flow in the direction of
positive charges has to do with the definition of
current density
which is a vector quantity.
QUESTION:
What is the size of an image as a function of its constant velocity ? i.e. what is the percentage increase of the image of a square travelling at 1 m/s towards a fixed camera? Is there some equation to calculate this?
ANSWER:
Since you do not give any details about the camera, I assume
it has a fixed focal length f. I will call the size of the object
L, the size of the image h, and the distance from the camera to
the object R. You wish to relate the rate of change of h, u=dh/dt,
to the rate of change of R, dR/dt=v where v is
the speed the object is approaching the camera; note that if the
object is approaching the camera, R is decreasing, i.e. dR/dt<0.
(If you do not know calculus, you will not understand my work here but you
will end up with a formula for the rate at which h changes.) Because
of the geometry of the situation (shown to the left) we can write L/R=h/f
or h=Lf/R. Differentiating, dh/dt=Lf(dR/dt)/R^{2}
or u=Lfv/R^{2}. This is the rate at which h
is changing. Suppose that f=5 cm=0.05 m, v=1 m/s, L=10
cm=0.1 m, and R=2 m; then h=0.05x0.1/2=0.0025 m=2.5 mm and
the image is growing at the rate of u=1.25 mm/s. After 1 s, the
object has moved 1 m and so now h=0.05x0.1/1=0.005 m=5 mm and is now
increasing at the rate of 5 mm/s. So, you see, the rate the image increases
in size depends not only on v but also on how far away the object is
(R) and how far the image is from the lens (f). If you need to
know what h is as a function of t, you need to also know where
the object was at some earlier time which I will call t=0; then R_{0}=R(t=0)
and so R(t)=R_{0}vt. Finally, we can
write h(t)=Lf/R(t)=Lf/(R_{0}vt).
For the example above and choosing R_{0}=2 m, the graph on
the right shows the image size as the object approaches the camera; note
that both the size and the rate of growth approach infinity as R
approaches zero (t approaches 2 s).
QUESTION:
If you have a glass tube with an object inside and the tube contains a vacum, why can you see the object in the tube and why doesnt the inside of the tube go dark like outer space. What are the photons interacting with?
ANSWER:
The air has
almost no effect on light—light entering your tube behaves almost as if the
air
were not there. In space, the vacuum itself is not dark and transmits light
just fine. What is dark is the "sky" because although the air on earth has
almost no effect on light, a very tiny fraction of light does scatter
from the air and, because the sun is so intense, the whole sky is brightened
by our seeing this scattered light.
QUESTION:
If an infinite number of bullets were fired from a gun at a solid target, would one of those bullets "pass through" without making a hole/dent ?
ANSWER:
No. See earlier answers.
QUESTION:
If the speed of light is constant for all observers, then why does
the Doppler effect for light take place?
ANSWER:
Because light frequency is a clock subject to time dilation
and light wavelength is a distance subject to length contraction. Because of
the constancy of the speed of light, however, the Doppler effect for light
is independent of whether it is the source or the observer who is moving,
unlike the situation for sound.
QUESTION:
How far apart must electrons be from one and other before they stop exchanging photons or does that only happen when atoms are sharing electrons in the same valence shell (assuming my question has a basis  I have no formal physics background, just what I read in places like this great site and others!)
ANSWER:
Photon exchange should not be taken too literally. It is a
cartoon attempting to demonstrate how the field quanta (photons) convey the
force (electromagnetism). And, it has nothing to do with whether or not the
electrons are sharaing some shell. Any time charges are in an electric field
there are virtural photons. Regarding the distance, there is theoretically
no cutoff distance where "they stop exchanging photons" because the electric
field from a single electron extends all the way to infinity.
QUESTION:
Without having wings, how does a helicopter turn in air?
ANSWER:
First of all, a helicopter does have wings. The rotor is
shaped like a wing and moves through the air by spinning thereby creating
lift. Imagine the path of the overhead rotor as a disk. The rotor is
connected to a plate called the swashplate (the thing between the rotor and
the fuselage in the picture) which can be tilted relative to the main shaft
to the motor and which causes the disk of the rotor to tilt relative to the
horizontal plane. Tilt it forward and the helicopter goes forward, to the
right and it goes right, etc. The swashplate is controlled by a
joystick in the cockpit called the cyclic. The tail rotor is also used in
turns, controlling yaw (rotation about a vertical axis); yaw is controlled
with foot pedals.
QUESTION:
Suppose there are 2 mirrors facing each other and I light a laser perpendicularly to one of the mirror and instantly remove my hand . Will the light beam continue striking from one mirror to the other?
ANSWER:
See an earlier
answer.
QUESTION:
How do we convert electron volts to wavelength or vice versa, example: the wavelength of 50 KeV xrays?
ANSWER:
The energy of a photon is E=hν=hc/λ where h=4.14x10^{15}
eV∙s is Planck's constant, c=3x10^{8} m/s is the speed of
light, ν is the frequency, and λ is the wavelength. So, E=1.24x10^{6}/λ
eV or λ=1.24x10^{6}/E
m.
QUESTION:
Does acceleration have momentum?
In other words if you fire a rifle, does the highest velocity of the bullet occur as it exits the barrel or does the acceleration increase after it leaves the barrel?
ANSWER:
In terms of physics nomenclature, you first question has no
meaning. But your second question seems to clarify what you mean: if
something has an acceleration does it keep accelerating even if there are no
forces on it? The answer is an unequivocal no. The only thing which causes
acceleration is force and when the bullet exits the barrel of the gun the
force which was accelerating it disappears. If there were no new forces on
it, it would continue with the same velocity it had when it exited. There
are, however, two important forces on the bullet when it is outside the
gun—gravity and air drag. Gravity causes it to accelerate toward the ground
and air drag causes it to slow down.
QUESTION:
I would like to know if amplitude is a scalar or vector quantity. The definition of amplitude of a wave is written as the maximum displacement of a point from the rest position but why doesn't the calculated amplitude have a plus or minus sign before the magnitude? If we have a type of sound waves that have a maximum negative displacement of 5 cm and a positive displacement of 3 cm, what is then the amplitude of the wave?
ANSWER:
Amplitude is a positivedefinite scalar and is defined as "the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position".
In the example you cite, amplitude is 5 cm. For a traveling wave, there is
one amplitude of the wave; for a standing wave, each point in the wave has a
different amplitude ranging from zero (nodes) to some maximum value
(antinodes). A closely related quantity is the root mean square (RMS)
displacement where displacement is squared, averaged over one cycle, and
then the square root taken.
QUESTION:
We have a metal ruler 1 yard in length. If held in exact balance center then "pinged" it causes vibrations. nothing new there BUT at exact equal distances to either side there is a point at which it appears that the vibrations stop (a calm spot) for about 1 inch then the vibrations start again and continue to the end. We've tested and the vibrations don't stop, they are just a much smaller wave length. What is is this "calm spot" phenomenon? What causes it? Does it happen with earthquakes too?
Really geeked out about this! Way cool!
ANSWER:
You are exciting standing waves when you "ping" the stick. These are waves
which bounce back and forth from the ends of the stick and, for special
wavelengths, are just right to to resonate like a guitar string or an organ
pipe. The various wavelengths for which resonance occurs are called the
modes of oscillation. For a stick clamped at the middle, the lowest mode,
called the fundamental, has approximately 1/4 of a wavelength on either side
of the center as shown by the upper part of the figure above. A point with
zero amplitude, the center for the fundamental, is called a node and points
with maximum amplitude, the ends for the fundamental, are called antinodes.
What you are seeing is the next mode, called the first overtone, which has
approximately 3/4 of a wavelength on either side; this mode has three nodes
and four antinodes. To the right are animations for a stick clamped
at the end but they are exactly what your stick is doing on one half. Here
the nodes are near the darkest blue. Earthquakes are traveling waves and
therefore do not have nodes.
QUESTION:
I have points A, B, and C which are moving in 3 dimensional space. If I use point A as a frame of reference, I know the direction and velocity of points B and C. If my points are moving arbitrarily close to the speed of light, what equations would I use to calculate the relative velocity of C from B's frame of reference, and the difference in clock speeds between those two points?
I understand how I would do this if the speeds were nowhere near the speed of light, but things get a tad complicated after accounting for relativity.
ANSWER:
I have previously dealt with relativistic
velocity addition if the velocities of B and C are colinear: in your
notation, V_{CB}=(V_{CA}+V_{AB})/(1+(V_{CA}V_{AB}/c^{2}));
read V_{XY} as velocity of X relative to Y. This may be
generalized
to noncolinear motion as V_{CB}=(V_{CA}+V^{‖}_{AB}+γ_{CA}^{1}V┴_{AB})/(1+(V_{CA}∙V_{AB}/c^{2}))
where V^{‖}_{AB}
and V┴_{AB
}are the components of V_{AB}
parallel and perpendicular, respectively, to V_{CA }
and γ_{CA}^{1}=√[1V_{CA}^{2}/c^{2}].
Note that if the speeds are small compared to c, V_{CA}∙V_{AB}/c^{2}≈0
and γ_{CA}^{1}≈1
so V_{CB}≈(V_{CA}+V^{‖}_{AB}+V┴_{AB})=(V_{CA}+V_{AB}).^{ }The reason that the transformations are different for parallel and
perpendicular motions is that there is length contraction for the parallel
motion but not for the perpendicular motion. (Note: V_{XY}=V_{YX}.)
QUESTION:
I've been reading about rotational space habitats for a while now and haven't found an answer to this question by googling. So how would a space habitat be rotated? By cogs? By propulsion systems?
ANSWER:
If you want to read more about the details of such habitats,
see my earlier answers (1 and
2). In answer to your question,
you would need thrusters to get the ring spinning and then to occasionally
correct minor changes, but once it was spinning, it would continue to spin
just the same forever if there were no external torques on it as would be
the case in empty space.
QUESTION:
The moon's gravity is one sixth that of the earth. Thus if you kicked a box with a force of 60 N across a frictionless floor on earth, the box would travel the same distance in 1 second as when the same box was placed on a frictionless floor on the moon and kicked with a force of 10 N. Am I wrong?
ANSWER:
You are wrong on both your conclusion and on your
"…kicked…with a force of…" premise. First the premise: you need to go the
FAQ and read the link
from the question about how much force does it take to make something move
with some speed. Just knowing the force you cannot know the resulting speed;
you need either how long the force was applied or over what distance it was
applied. Now, your question implies that you think it will be easier to get
the box on the moon moving with the same speed as a box on earth with that
speed. But, in fact, the box has the same mass on both the earth and moon
and you are not lifting it against gravity, so it is equally easy to move a
box horizontally on earth or the moon. It is six time harder to lift a box
on earth as on the moon. If there were friction, however, it would be harder
to move the box on earth than on the moon because the frictional force on
the moon would be six times smaller.
QUESTION:
I know that neutrons is placed in a wax container to contain the neutrons, but why is this an effective way to contain the neutrons. I want to take in account the linear momentum and the energy, but I'm not sure how to begin or where to do the research?
ANSWER:
The wax does not "contain" the neutrons. Most neutron sources
yield fast neutrons and most uses for neutrons need slow (called "thermal")
neutrons. Thus the source is encased in a block of paraffin which serves as
moderator, i.e. the paraffin slows the neutrons down. The reason that
paraffin is good is that it is a hydrocarbon and has lots of hydrogen in it.
Hydrogen is good to slow down neutrons because the best way to slow down a
fast moving object is to collide it with an object at rest which has about
the same mass (think of a headon collision of two billiard balls); hydrogen
has about the same mass as a neutron.
QUESTION:
How do CDs create rainbows when we shine white light on them?
ANSWER:
Because they have many closely spaced lines which act like a
diffraction
grating.
QUESTION:
What is the difference in time dependent and time independent Schrödinger equations?
ANSWER:
The solution of the timedependent equation is the wave
function as a function of space and time; the solution to the
timeindependent equation is the wave function as a function of only space.
QUESTION:
why does a piece of chalk produce hideous squeal if you hold it incorrectly?
ANSWER:
It is caused by friction. If the conditions are right
(holding it incorrectly), the chalk will stick, slip, stick, slip, stick,
slip… very rapidly, hundreds or thousands of times per second. This results
in a resonant response of the chalk itself, vibrating loudly. The same
slipstick friction is responsible for how a violin bow works.
QUESTION:
How can the speed of light be the fastest thing in the universe if it is a relative measurement? I mean that if you had two light particles going in opposite directions, wouldn't one go twice the speed of light relative to the other?
ANSWER:
See my FAQ page. It is
a relative measurement, but your idea of the relative velocities is
incorrect at large speeds. You want to use
v'=(u+v),
but the correct expression is
v'=(u+v)/[1+(uv/c^{2})]
where
c
is the speed of light; notice that if
u
and
v
are both small compared to
c,
your expectation is very nearly correct. For your particular example,
u=v=c
and so
v'=(c+c)/[1+(c∙c/c^{2})]=c.
You can also find other answers about light speed, the maximum velocity
possible, etc. on the FAQ page.
QUESTION:
I read a question on your site about using magnets to propell a spacecraft. I want to expand on that question. How would it affect the movement of the ship if we'd replace the bar magnets with electric magnets and then make one magnet stronger than another? Shouldn't the direction with weaker magnet start to move since the the weaker magnet can not completely nullify the effects of the stronger magnet?
ANSWER:
Newton's third law always applies: if one magnet exerts a
force on the other, the other experiences an equal and opposite force from
the one. All forces interior to the spaceship must add up to zero.
QUESTION:
I am confused about 1/2mv^2 and mgh.
If an object is lifted vertically up at a constant speed, the formula is mgh. But wouldn't the object also gain k.e. too due to the constant velocity? So wouldn't the total energy be mgh+1/2mv^2?
ANSWER:
If it is still moving when you get to h, then its
energy is indeed mgh+½mv^{2}. In a textbook, we
usually envision the mass being lifted from rest at one location to being at
rest at some higher location. During this process, the total amount of work
you do is mgh and the total amount of work the weight (gravity) did
is mgh so the total work done is zero and the change in kinetic
energy is thus zero.
QUESTION:
If you have two permanent cylindrical magnets (the kind with a hole in the center) and you stack them with poles opposite on a pencil, the top magnet will "float" above the bottom magnet. Energy is being expended to keep the magnet "up" the pencil. Where is the energy coming from? The bottom magnet will be pushing down with an equal but opposite force, but that does not cancel the energy needed to float the top magnet as far as I can see.
ANSWER:
I am afraid you do not understand energy. The lower magnet
exerts a force on the upper magnet. The force holds it there in equilibrium,
it does not require energy to hold it there. It is no different from saying
that if one of the magnets were hanging from a string, where does the energy
to hold it there come from? Or, if one of the magnets were sitting on a
table top, where does the energy to hold it there come from?
QUESTION:
I read of "gravity assist" swingybys of Jupiter to speed a spacecraft up to reach the outer planets. As the spacecraft approaches Jupiter, it speeds up. But it retreats from Jupiter on a symmetric path (a hyperbola I think) and Jupiter will therefore slow the spacecraft down by the same amount on the outbound path. It appears there should be no net increase in speed, just a bending of the spacecraft path. But bending a spacecraft path also takes energy. So Jupiter is providing the energy in some manner though it is unclear to me how.
ANSWER:
The trick is that the planet, with much greater mass than the
spacecraft, is moving in its orbit and the boost comes from using the speed
of the planet to speed up the spacecraft. The figure shows the idealized
onedimensional interaction with planet; because the mass of the spacecraft
is much less than the mass of the planet, the spacecraft picks up twice the
speed of the planet. For those who have studied elementary physics, this
should look vaguely familiar: a perfectly elastic collision between a BB at
rest and a bowling ball with speed U results in the BB going with
speed 2U and the bowling ball still going U. (Of course, this
is only approximately true if the mass of the BB is much smaller than
that of the bowling ball; the bowling ball will actually lose a very tiny
amount of its original speed.)
QUESTION:
The number of known Mars Meteorites on Earth at last count I know of was greater than the number of Lunar meteorites. This is the opposite of what you would expect: The moon is 140 times closer than Mars with a weaker gravitational field to recapture the meteorites and no atmosphere to slow material blasted off the planet. The number of Lunar meteorites on Earth should be hundreds of times greater than Mars Meteorites.
ANSWER:
As always, a disclaimer that I am not an astronomer. The
origins of either lunar or Martian meteorites are major asteroid or comet
impacts; after impact, some of the debris has enough velocity to escape the
gravitational field. First of all, the cross sectional area of the moon is
about 4 times smaller than Mars, so all things being equal, the probability
of a major impact is 4 times greater for Mars. Second, Mars is closer to the
asteroid belt and would therefore, I presume, be more likely to suffer an
asteroid impact. Most major impacts occurred early in the history of the
solar system when embryonic planets were "sweeping up" material in their
orbits and there was a much greater potential of major impacts. These
early meteorites from the moon which landed on earth would have been eroded
away or taken under the crust by tectonic action; those from Mars would have
gone into orbits around the sun and land on the earth over a much longer
period, probably still landing today.
QUESTION:
My son and I are trying to build a Foucault pendulum. We have 11' ceilings so we need to dampen ellipsoidal motion and provide a drive mechanism to keep the pendulum moving. I've seen pendulum driver circuits that operate via magnetic induction with relays, transistors, etc. but do not prefer these; it's not clear that the pendulum and Earth are doing the magic, as opposed to the electronics.
I'm thinking of the following design: bend a 20' steel 1" conduit into a circle. Then wrap say 16 gauge bare copper wire evenly and entirely around the doughnut creating a precision ring shaped electromagnet. Then power the ring shaped magnet with a variable DC supply to create the proper pull on, say, a 6lb steel pendulum that swings inside the ring. The idea is to adjust the voltage to give the electromagnet enough pull to keep the pendulum swinging, but not so much that it overcomes the force of gravity pulling the pendulum back to the center, thereby sticking the bob to the electromagnet. That solves the problem of keeping the pendulum moving.
ANSWER:
After thinking about this a bit, I believe there are serious
flaws here and your idea will not work. If you put a ferromagnetic material
(e.g., your pendulum bob) in a uniform external magnetic field, that
field will polarize the material and essentially make it look like a bar
magnet with north and south poles. The south pole will feel a force opposite the
direction of the field and the north pole will experience the same magnitude
force in the direction of the field—a bar magnet in a uniform field
experiences zero net force (see left figure above). Admittedly, the field
due to your ring (see figure at the right above) is not uniform but, over
the size of your bob, it is approximately uniform. Forces due to the small
nonuniformity would be mainly in the vertical, not horizontal, direction.
QUESTION:
the bond between carbon and oxygen atoms in a carbon dioxide molecule behaves like an elastic spring and vibrates the molecule.why?
ANSWER:
It is easier to just talk about a diatomic molecule like O_{2}.
When you understand that you can generalize to any molecule. Think about the
forces between two atoms. If you try to pull it apart, this force tries to
pull it back together or else it would not be a molecule; and, since the
molecule has a size larger than the size on one atom, if you try to push the
atoms together, the force tries to push them back apart. This is exactly the
way a spring works and so a reasonably good model for a molecule is that all
the atoms are attached by tiny springs. Of course, they are not but this
still is a model which can help you understand many molecular properties.
So, one of the ways you can excite (add energy to) a molecule is to get it
vibrating.
QUESTION:
why does the ground frost go deeper when the surface begins to thaw?
ANSWER:
The way you frame this question is that frost has gone to
certain depth before the thaw and then goes deeper because of the
thaw. In fact, there is what is called a
frost front, the line between frozen and unfrozen soil, which moves
slowly downward if the air temperature is below freezing. When the air
temperature goes above freezing, the top thaws but the frost front
continues moving down for a while because there is still a frozen region
above it which prevents it from getting the information that the air has
warmed. So, the thaw does not cause the frost to deeper, it is just
continuing what it was already doing. Eventually the thawing will reach the
frost front and all the soil will be thawed; in some arctic and antarctic
regions, the thaw will never reach all the way to the frost line and
permafrost results.
QUESTION:
If an elementary particle has a mean lifetime of say 1 second and there are 1 trillion particles in the system then about how many particles will be left after say 100 seconds?
ANSWER:
See a recent question and then do the
calculation: N=N_{0}exp(t/τ)=10^{12}exp(100)=3.7x10^{32}.
Although there is a nonzero probability (3.7x10^{30 }%) that there
will be a particle left, for all intents and purposes there are none left.
(Incidentally, there are two meanings of the word "trillion". In the US it
means 10^{12} but in some countries it means
10^{18}. Look
here for
more detail.)
QUESTION:
During a discussion with my 6th grade class about the Law of Inertia and space travel, a student asked: "If a spacecraft leaves a solar system, for example Voyager 1, will its velocity increase due to the lack of gravity from the sun." Fairly certain the answer is no; however, it did spark a rather interesting debate. Can you explain?
ANSWER:
Perhaps the key is to note that "leaves the solar system"
does not mean that there is no longer any gravity from the sun.
Rather there is a boundary where the solar wind, particles like protons and
electrons, stops; this is called the heliopause and has been definitely
observed by the Voyager 1 spacecraft as shown in the graph to the right.
This shows the amount of solar wind the Voyager detected as it moved during
the months of 20112. In September 2012 it dropped to near zero. This is
where we define the edge of the solar system to be and it is about 50
Astronomical Units (AU) from the sun; the earth is one AU from the sun. But
the gravity from the whole solar system is still present and the craft will
continue slowing down, but ever so slightly since as you get farther away
from a mass the gravity gets weaker. This craft has enough speed that, if it
never encountered any other mass it would keep going forever. I did a rough
calculation and found that the acceleration of Voyager is about a=0.01
mi/hr^{2} which means that it loses about 1/100 mph per hour; but
the speed is about 40,000 mph, so I think we could agree that it is moving
with an almost constant speed. As it gets farther away, the acceleration
will get even smaller (physicists call slowing down negative acceleration,
not deceleration). Until it gets close to something else, like some other
star, it will keep going with an almost constant speed. When it does
approach another star it will begin speeding up. Your students should
appreciate that the only thing which can change the speed of something is a
force, a push or a pull.
QUESTION:
I am building a cold frame to keep veggies alive in the winter. It will be 3' x 6' with two "doors" (called lights) each 3' x 3' that will be hinged to the frame. The frame probably weighs about 10  15 lbs. The doors will be quite lightweight, possibly only 3lbs each. I wanted to use magnets to keep the doors closed at night or when I am not venting the cold frame. We often get very windy days with 40mph wind speeds and gusts to 60 mph. From what I've read, magnets have different "pull force" properties. I'd like a way to figure out what pull force the magnet for each door needs to have to withstand the winds we get. Please don't tell me to just hook the doors closed  shockingly, it appears magnets are a more cost effective solution.
ANSWER:
I must say that I cannot believe that you could not buy a
couple of simple hooks/latches for under $5, but I will do a rough
calculation for you to estimate the force you would need to apply at the
edge of the doors to hold it down in a 60 mph=26.8 m/s wind. When a fluid
moves with some speed v across a surface, the pressure is lower than
if it were not moving; this is how an airplane wing works and why cigarette
smoke is drawn out the cracked window of a car. To estimate the effect,
Bernoulli's equation is used: ½ρv^{2}+ρgh+P=constant,
where ρ is the density of the
fluid (ρ_{air}≈1 kg/m^{3}), P is the pressure,
g is the acceleration due to gravity, and h is the height
relative to some chosen h=0. For your situation both surfaces are at
essentially the same height so P_{A}=P+½ρv^{2
}where the pressure inside your frame is atmospheric pressure (P_{A})
and the velocity inside is zero. So, P_{A}P=ΔP=½ρv^{2}=½∙1∙26.8^{2}=359
N/m^{2}=7.5 lb/ft^{2}. This would be the pressure trying to
lift the door. So the total force on each door would be F=AΔP=9∙7.5=67.5
lb where A=3∙3 ft^{2}=9 ft^{2} is the area of the
door. But, this is not the answer since we want to keep it from swinging
about the hinges, not lifting into the air. So, assuming that the force is
distributed uniformly over the whole area, you may take the whole force to
act in the middle, 1.5 ft from the hinges, so the torque which is exerted is
1.5∙67.5≈100 ft∙lb. But, the weight of the door also exerts a torque, but
opposite the torque due to the wind (the weight tries to hold it down)
3x1.5=4.5 ft∙lb. So, the net torque on the door about the hinges is about
95 ft∙lb. To hold the door closed, one needs to exert a torque equal and
opposite to this. To do this, it would be wisest to apply the force at the
edge opposite the hinges to get the maximum torque for the force. The
required force from your magnets would then be F=95/3=32 lb. Note
that this is just an estimate. Fluid dynamics in the real world can be very
complex. Also note that, if my calculations are anywhere close to correct,
you should probably be sure the whole thing is attached to the ground or the
side of your house since the total force on the whole thing would be
67.6+67.53315=114 lb, enough to blow the whole thing away in a 60 mph
wind! Also, once the door just barely opens, the wind will get under it and
simply blow it up, Bernoulli no longer makes any difference.
QUESTION:
If a particle has a lifetime of say 100 years then what are the chances that that particle will decay in 1 year. How will one go about calculating particle decay probability?
ANSWER:
Lifetime has no meaning for a single particle. It is a
statistical concept as I will show. The rate at which a large ensemble of
particles decays is proportional to how many particles there are, dN(t)/dt=t/τ
where N(t) is the number of particles at some time t
and τ is the mean lifetime. The solution of this differential
equation is N(t)=N_{0}exp(t/τ)
where N_{0}=N(t=0). So, when t=τ,
N=0.37N_{0}, there are 37% of the original particles
left. For your question, you might want to ask what is the situation when
t=1 if N_{0}=1 and τ=100:
N=exp(0.01)=0.99. If you want to interpret that as a 1% probablility
that the particle has decayed, I guess that that would be ok. But, of
course, there is no such thing as 99% of a particle. Keep in mind, though,
that this is not linear; for example, if t=50=τ/2,
N=exp(0.5)=0.61, not 0.5.
QUESTION:
where does electron go that created by beta emission?
ANSWER:
The electron is not "created" in a vacuum, a proton and a
neutrino are also "created" and a neutron "destroyed." The net effect is
that there is still the same amount of charge, zero, after the decay as
there was before. One possibility is that the electron never leaves the
source and, since the atom it left is now positively charged, the electrons
in the source all move around until all the atoms have the correct number of
electrons. Or, if it leaves the source, it will encounter other matter with
which it will interact, lose energy and either find a positive ion somewhere
to join or stick to a neutral atom which becomes a negative ion. Charge can
build up to some limit on anything.
QUESTION:
Galileo was punished by the Church for teaching that the sun is stationary and the earth moves around it.His opponents held the view that the earth is stationary and the sun moves around it.If the absolute motion has no meaning, are the two viewpoints not equally correct or equally wrong? i.e, By the concept of relative motion can't we say both?if so, then why do we usually say that earth goes round the sun,and the other way round?
ANSWER:
If the sun and earth were both just moving with constant
velocity and not interacting in any way, you would have a point. However,
because of their gravitational interaction and the fact that the mass of the
sun is hugely bigger than the mass of the earth, there is no way you can
sensibly argue that the sun orbits the earth. Each exerts a gravitational
force on the other (equal and opposite) but because the sun is so massive,
the force has almost no effect on it. If the earth were not orbiting but
simply released from rest, it would fall into the sun. There would be no
question which object had the greater acceleration—it would be obvious to
any observer that it was the smaller mass, the larger mass practically
unaffected. So, if any frames are accelerating, they are not equivalent to
those not accelerating. What you call "the
concept of relative motion" applies only to unaccelerated frames.
QUESTION:
Two spherical bobs, one metallic and other of glass, of same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why?
ANSWER:
Two objects with identical geometries (same size and shape)
experience identical air drag force while moving with a speed v.
While falling, the force has a greater effect on the less massive causing it
to slow down more. Therefore the mass, not whether glass or metal,
determines which gets to the ground first; the winner is whichever has more
mass. You can find links to many old answers about air drag on the
faq page.
QUESTION:
Suppose you have radiation detectors fixed on the ground on Earth. Will they detect radiation coming from a charged particle in free fall near them?
The first answer that comes to mind is: Yes, they will detect radiation because the particle is accelerated, and electrodynamics predicts that accelerated charges must radiate in this situation.
According to the Equivalence Principle, this situation is equivalent to detectors fixed on an accelerated rocket with acceleration g moving in the outer space and far away from the influence of other bodies. If the answer to the previous question is yes, then the detectors on the rocket should also detect radiation coming from a charge in free fall as observed by the reference frame of the rocket. But a charge in free fall in this reference frame is at rest in the inertial reference frame fixed with respect to the distant stars, and a charge at rest in an inertial frame should not radiate.
Is it possible that detectors fixed on the rocket detect radiation but detectors at rest in the inertial frame do not? Is radiation something not absolute, but relative to the reference frame?
ANSWER:
This is a fascinating question and points to an experiment
which would seemingly violate the equivalence principle. The answer to your
first question is an unequivocal yes, an electric charge accelerating in
free fall in a gravitational field radiates electromagnetic waves, an
electric charge not accelerating does not radiate. But, suppose that you are
falling along with the charge; relative to you the charge is not
accelerating and therefore not radiating. Or, equivalently, suppose that you
are in a spaceship in empty space with your rockets turned on. If you
release an electric charge inside, it will "accelerate" toward the rear of
the ship and therefore radiate because the equivalence principle states that
there is no experiment you can perform which can distinguish between the
accelerating frame and a static gravitational field. However, the charge
will move with constant speed relative to an inertial observer nearby and
therefore not radiate. In both cases we have an electric charge both
radiating and not radiating, a seeming paradox. Although I had not heard of this
paradox before, apparently it has been a topic of many articles. The most recent of
these, by
Almeida and Saa,
has evidently laid the paradox to rest. They demonstrate in this article
that observers for whom the charge is not accelerating "…will not detect any radiation because the radiation ﬁeld is conﬁned to a spacetime region
beyond a horizon that they cannot access…" and "…the electromagnetic ﬁeld generated by a uniformly
accelerated charge is observed by a comoving observer as a purely electrostatic ﬁeld."
Like all "paradoxes" in relativity, there is not really a paradox; rather a
radiation field in one frame may be a static field in another. Basically,
you nailed it when you said "radiation
[is] something not absolute, but relative to the reference frame."
QUESTION:
What would happen if a needle that weighed as much as the Earth and everything on it was placed on the surface of the planet? Where would the pin end up? What would be the impact on the Earth? Would the outcome be different if it was set on land or water?
ANSWER:
This is a wacky question but I guess I can do a wacky
question now and then. I will take "placed on" as being synonymous with
"suddenly appears at the surface". It will also be easier to think about if
the needle appears at the equator. I will also assume that no matter what
happens, the needle retains its size and shape. If the needle just sat
there, the effect would be that the earth would now rotate not about its
axis but about a parallel axis which is halfway between the center of the
earth and the needle; the length of the day would increase to about 5 times
the current 24 hours. But, it would certainly not stay on the surface of it.
The force which the needle would exert on the surface of the earth would be
about 6x10^{25} N. But this force is over an area of a needle, very
tiny, so the pressure would be astronomical. This would cut into the earth
and the needle would end up, probably after oscillating back and forth for a
while, at the center of the earth. The length of the day would return to
about 24 hours and everything on the surface would weight twice as much. If
the needle were to appear at a pole, there would be no effect on the length
of the day at all.
QUESTION:
If a dogs bark measures approximately 6080 decibel, how much will it reduce traveling in air (average 20 Celsius) over 500 meter.
Could you please brake down the formula for me so I can calculate other distances.
ANSWER:
You should first read an
earlier answer for a detailed
explanation of what a decibel (dB) is. The main reason for loss of sound
intensity I (measured in watts per square meter, W/m^{2}) is
that the sound waves spread as they get farther away so the energy per
second (power) striking your ear gets smaller. The intensity falls off like
1/R^{2} where R is the distance from the source. You
do not specify the distance from the dog that the 6080 dB level is
measured, so I will arbitrarily put it at 1 m. Therefore the ratio of
intensities at 500 and 1 m would be I_{500}/I_{1}=1/500^{2}
or I_{500}=4x10^{6}I_{1}. But,
the catch is that dB is a measure of the level L (a logarithmic
scale), not the intensity. So we need the equations to convert between L
and I. These are L=10∙log_{10}(I/10^{12})
and I=10^{12}∙10^{(L/10) }where I is
in W/m^{2}. So, as an example I will choose L=70 dB, so I_{1}=10^{12}∙10^{(70/10)}=10^{5}
W/m^{2 }and I_{500}=4x10^{6}∙10^{5}=4x10^{11}
W/m^{2}. Finally, we find what the dB level of I_{500
}is: L_{500}=10∙log_{10}(4x10^{11}/10^{12})=10∙log_{10}(40)=16
dB. I believe that this will the main source of quieting with distance, not
any absorption of the sound by the air. Wind can also have an effect on the
intensity if it is across the direction the sound travels to reach you. By
the way, when the intensity reaches 10^{12} W/m^{2},
it will be below the "threshold of hearing" and you will no longer hear it;
that corresponds to 0 dB.
QUESTION:
My question is that whether gravitational field an be reversed
in Earth`s atmosphere? it would bring revolutionary changes in Aerospace
engineering.
ANSWER:
No.
FOLLOWUP QUESTION:
but space organisations train their astronauts on how to deal in zero gravity in camps. then they have to create artificial zero gravitational field. how is it possible??
ANSWER:
You can simulate zero gravity but you certainly do not do it
by reversing the gravitational field. We are stuck with the field we have
and to simulate zero field there are two ways. The first is to put the
astronaut trainee in a freefall situation. If you are in a freely falling
elevator, for example, it will seem like there is no gravity. The more
practical way to do this is to ride in an airplane which is moving over a
parabolic path it would follow if it were simply a projectile; inside that
airplane it will seem like there is no gravity. The plane which does this is
affectionlately known as the "Vomit Comet". A second way you can simulate
weightlessness is to build a giant tank of water and then the buoyancy will
provide a force opposite to your weight making the net force on you equal to
zero.
Finally, it should be noted that when you are in
orbit you are also not truly "weightless". In orbit you are constantly in
free fall so, although you still have your weight, you are in free fall just
as you were on the Vomit Comet.
QUESTION:
If we are given a graph showing pressure versus distance for a sound wave, does a higher pressure amplitude indicates a louder sound produced?
ANSWER:
Sound waves are longitudinal waves of pressure as shown in
the animation at the right. You are right that the greater the pressure
difference between the lowest and highest pressures, the louder the sound.
You should not think of the regions of high pressure being the regions of
loudest sound, though. Because the elapsed time between low and high
pressure hitting your ear is so short, your ear/brain averages the square of
deviation of the pressure in the wave from normal atmospheric pressure to
determine loudness. The information from the time difference is also
processed by the brain and interpreted as the frequency or pitch of the
sound.
QUESTION:
Let's say that we started mining for metals and such somewhere other than earth. What if any would the effects of large scale increase (and I mean really really big) in planet mass be? Also what would happen to the other planets we were robbing of minerals as their mass decreased?
ANSWER:
As long as the mass of the orbiting body is much less than
the mass of the orbited body, the orbital motion is independent of the mass
of the orbiting body. So, the length of the year would be unchanged.
However, increasing the mass and radius of the planet would increase its
moment of inertia, so its rotational speed would decrease and the day would
get longer. Keep in mind, though, that a huge amount of stuff would have to
be added for any real difference to occur.
QUESTION:
In case of string, when only a pulse is sent then why do the particles vibrate just above the mean position and not below the mean position?
Please answer in a way that i could understand. If your answer include some mathematics esp the calculus please simplify so that i could understand
ANSWER:
There is an equation, called the wave equation, which to a
fair approximation applies to waves on a string. One of the properties of
this equation is that the wave retains its shape as it propogates. Hence, if
you start with a pulse above the string, it will remain that way.
QUESTION:
I am trying to determine the buoyancy of a 10 kg polyethylene tray (density = 0.95 g/cm^{3}=950
kg/m^{3}) that is used in lobster pounds to hold lobsters. The tray holds 60 kg of lobster, and fills up with water up to the lid (which floats at the top of the water). The volume of the tray is 0.1223 m^{3}. The lobsters are held in seawater (density = 1026 kg/m^{3}). So, my question is how much air is required to be placed in cavities in the lid to keep the tray floating at the top of the water?
ANSWER:
I will take the "volume of the tray" to be the inside volume,
that is the volume occupied by lobsters and water. The density of a lobster
is sure to vary from animal to animal, but it certainly must be larger than
the water since the creatures dwell on the bottom. I actually found a
measurment of a lobster's density to be 1187 kg/m^{3}, so I will use
that as an approximation for all lobsters. So, the volume occupied by 60 kg
of lobsters is V_{lobsters}=60/1187=0.0505 m^{3}. So,
the volume of water needed to fill up the tray is V_{water}=0.12230.0505=0.0718
m^{3}; so the mass of the water is M_{water}=1026x0.0718=73.7
kg. The volume of the polyethylene is V_{poly}=10/950=0.0105 m^{3}.
So, the total mass is 60+10+73.7=143.7 kg and the total volume is
0.1223+0.0105=0.1328 m^{3}. The density of the full tray is then
ρ=143.7/0.1328=1082 kg/m^{3}; since this is greater than the
density of the water (1026 kg/m^{3}), it will sink, hence presumably
the point of this question. In order for it to "just" float, volume must be
increased until the density is equal to 1026 kg/m^{3}, i.e.
1026=143.7/V or V=0.1401 m^{3}; so, the volume of the
air should be V_{air}=0.14010.1328=0.0073 m^{3}=7300
cm^{3}. This would be a volume of a cube about 20 cm on a side. I
guess I would increase that by at least 50% as a safety factor. (Note that I
have neglected the density of the air which is about 1 kg/m^{3}.)
QUESTION:
What happens to gamma rays when they are blocked by materials like lead, and how come they're not blocked by some other materials?
ANSWER:
For all intents and purposes, gamma rays interact only with
electrons in the material. The two most important ways they interact are the
photoelectric effect where the photon gives all its energy to an electron
and Compton scattering where the photon scatters from an electron, giving
some of its energy to the electron. Which of these is more important depends
on photon energy and other factors. But, as you would think, the density of
electrons in the material is very important. To a very rough approximation,
atoms are all about the same size. That means that there are about the same
number of lead atoms in a cubic centimenter of lead as there are aluminum
atoms in a cubic centimeter of aluminum. But every lead atom has 82
electrons whereas every aluminum atom has 13 electrons. There are therefore
roughly 6.3 times more electrons in a cubic centimeter of lead than in a
cubic centimeter of aluminum; therefore lead will be much more effective in
shielding against gamma rays. Regarding "what happens" to the gamma rays,
they disappear completely if photoelectric effect occurs or become gradually
less energetic and changed in direction if Compton scattered.
QUESTION:
To tighten the loose head of a hammer, the base of the handle is sometimes struck on a hard surface. Explain the physics behind this maneuver.
ANSWER:
This sort of sounds like homework. The physics is inertia.
The head of the hammer is pretty massive and so it has a lot of inertia, it
wants to keep going. When the handle suddenly stops, the inertia of the head
keeps it moving so that it moves down a bit on the handle before stopping,
thereby becoming more tightly bound to the handle.
QUESTION:
What would happen to a beam of proton particles if the pass through a coil (perpendicularly) imagine a closed circle shaped loop. Currwnt is passing through this coil hencr generating a magnetic field. Will this B field make the particles vibrate? Accelerate them?
ANSWER:
I assume that you have the beam of protons very narrow and
moving along the axis of the loop. As you can see from the figure to the
right, the onaxis field is along the axis. Therefore, a beam of protons on
the axis would pass through unaffected because the force on a charged
particle is zero if the velocity and field are parallel. If the proton were
off axis, there would be a small force which, if the proton were in the
plane of the page here, would point either into or out of the page.
QUESTION:
Isnt the gravitational pull of an object determined SOLELY on the mass of that object and NOT its size or density? obviously distance from said object plays a role just as size and density affect mass.... however, if what i believe is true, there can be no such thing as black holes, which even today their existence is only theoretical, not proven... anyway my point is this... i believe that if our sun was the size of a basketball but still had the same mass it has today, (giving it almost infinite density)..(what many scientists today consider a 'black hole') but the Earth was still the same 93,000,000 miles from the actual surface as it is today, the orbit of the Earth and the planets would remain, and the sun would not be considered a 'black hole', nor would it behave like one, (sucking in all its orbiting masses, and even light itself).. it would be much smaller, but given the exact same mass and distances surface to surface, little would change as far as orbits go... could this be correct theoretically?
ANSWER:
If the object is spherically symmetric and you are totally
outside the mass distribution, then you are right—only the total mass
matters. But, this does not mean that black holes do not exist. It is
dramatic to say that nothing can escape a black hole, not even light, if you
are inside a critical distance called the Schwartzchild radius, but that
does not mean that objects could not orbit the black hole, either inside or
outside that radius. If the sun were a black hole, its Schwartzchild radius
would be
R=2GM/c^{2}=2x6.7x10^{11}x2x10^{30}/(3x10^{8})^{2}=3,000
m; so all the planets would be outside and would orbit just fine; they could
even be dragged away if you wanted. Only objects inside 3000 m would be
"trapped" but even they could still orbit the black hole. The figure to the
left shows some orbits of stars around the supermassive black hole at the
center of the galaxy which have been observed by astronomers.
QUESTION:
I don't understand the hydrogen spectrum..the theory given in my book says that Balmer series will be observed when electron falls from n=3,4,5......(up to infinity) to n=2 but isn't n=2 for an electron of the hydrogen also an excited state,so it must fall back to its ground state?? Or am i getting something wrong here??
ANSWER:
Yes, it must get down to the ground state again once it
reaches n=2. But that is part of a different series, called the Lyman
Series. The radiation from that transtion is called the Lyman alpha line and
is in the ultraviolet region of the spectrum. The rest of the series is
n=3,4,5… going to n=1; all the lines in this series are in the ultraviolet.
Every level has a
series
terminating on it.
QUESTION:
if i create vaccum on earth ,like in a spherical shell,if pierced it will suck air around it, its not the case with atmosphere of earth,why it is not that space vaccum suck it?gravity is indeed much stronger near earth and atmosphere much more denser ?
ANSWER:
The reason that air does not leak into space is that gravity
holds it to the earth. Think of the air as a collection of molecules all
moving around with different velocities. The velocities are distributed
according to the MaxwellBoltzmann distribution.
At normal temperatures, most molecules (e.g., O_{2}, N_{2},
H_{2}O, CO_{2},…) have almost no molecules have velocities
large enough to escape the
earth's gravitational pull. Lighter molecules, in particular H_{2}
and He, have much higher velocities and do escape into space; you will find
almost no hydrogen or helium in our atmosphere.
QUESTION:
With respect to special theory of relativity, light speed c is an invariant, but wouldn't rest mass also be an invariant? If true, why is it not one of the postulates?
Velocity and acceleration are not invariants, but are there other properties that are invariants?
ANSWER:
Light speed is not usually referred to as an invariant; it is
simply a universal constant. Furthermore, as I have
stressed before, I see no need to call
the constancy of c a postulate; it is simply a
result of the principle of relativity
which states that the laws of physics be the same in all inertial frames.
Rest mass is the inertia something has in its own rest frame, so of course
it must be a constant. The word invariance usually refers to any
quantity which remains constant when a Lorentz transformation is performed.
For example, the total energy of an isolated particle E=√(m_{0}^{2}c^{4}+p^{2}c^{2})
is an invariant.
QUESTION:
I've been trying to calculate how much impact force an object weighing 2,000,000 lbs traveling at 120,000mph will generate. I've run calculation conversions that tells me that the Newtonian Force is a whapping 48665704243.2 but what does that mean? I read about how an 80 pound 1 foot long object traveling at 52,000mph hit the surface of the moon with a force equivalent to 5 Tons of TNT exploding, creating a 65 foot wide crater!
What would my 1,000 Ton 157.64 Mach object do?
ANSWER:
For the umpteenth time, you cannot calculate a force by
knowing the mass and velocity of something; see my
FAQ page. What you can do, as your
example does, is calculate the energy which the object carries in. I will
not do what I usually do and convert to SI units since you seem to like
English units. The kinetic energy something has is K=½MV^{2}
where M is the mass and V is the speed. Using your example,
K_{TNT}=5 if K=½∙80∙52,000^{2}=1.1x10^{11}
lb∙mph^{2}, so to convert energy in lb∙mph^{2}, to energy in
tons of TNT you need to multiply by a factor of 5/1.1x10^{11} lb∙mph^{2}=4.6x10^{11}.
So,
½∙2x10^{6}∙120,000^{2}∙4.6x10^{11}=6.6x10^{5}
tons=0.66 Megatons of TNT. For comparison, the bomb dropped on Nagasaki in
WWII was about 0.02 Megatons.
QUESTION:
Why are loops provided for transporting oil/water for longer distances?
ANSWER:
When the temperature of the pipe changes it changes length.
In the figure to the left, the pipe will expand if you heat it up and
contract if you cool it down. If it were just a straight pipe, the resulting
forces on the pipe along its length could be large enough to cause it to
buckle and fail. Inserting loops allows the length changes to be taken up by
the size of the loop as shown.
QUESTION:
Can u get pure energy? Can you make it visible with power modulation?
ANSWER:
Since mass is just another form of energy, everything you can
see or feel is pure energy. Maybe you mean energy without mass? In that
case, the answer would be electromagnetic radiation like light or radio
waves. I have no idea what your second question means.
QUESTION:
My question relates to time dilation and the speed of light. When we consider the passage of time, and the speed of an observer or traveler on Earth, how does the speed of our planet around the sun, our arm of the galaxy and the speed of the galaxy itself play into the speed of time for someone on Earth? We talk about a spaceship travelling at a certain speed, but isn't the Earth itself like a spaceship for its inhabitants? And isn't the Earth rotating around the sun at a certain speed, and the spiral arm of our galaxy also rotating at a certain speed? So isn't the Earth also?
ANSWER:
Time dilation is meaningful only as a relative thing. One
observer can measure the rate at which the clock of another observer runs.
Relative to a clock on the sun, an earth clock would run slowly. Relative to
the center of the galaxy, an earth clock would run slowly differently.
Relative to a clock on Andromeda galaxy, an earth clock would run slowly
differently. This is the thing about relativity—it is only relative
measurements that mean anything; that's why they call it relativity.
Underlying your question is the misconception that there exists some
absolute rest relative to which all clocks in the universe will run. There
is no such thing as "absolute rest"; any inertial frame may be thought of as
being at rest. Incidentally, for all but the most accurate measurements, the
time dilations I have alluded to are negligibly small because the relative
speeds are small compared to the speed of light.
QUESTION:
I don't understand angular momentum but as a first step here is my question. If you hold out weights and turn on a pivot and bring the weights in; are the weights moving at the same velocity and you are spinning faster so that the weights move at the same velocity or are the weights actually moving faster?
ANSWER:
The angular momentum L of a mass M a distance
R from the axis around which it is rotating is of L=MVR where V is its speed. It requires no torque to change
R, so L will stay the same if R is changed because
angular momentum of a system on which there are no torques is constant
(conserved). So, for example, if you change R to ½R, V will double. If there were no
person spinning with the weights and no external torques (like friction),
this would be your answer. It is more complicated if you have anything other
than the weights. You probably want to ignore the rest of the answer below,
but I want to do it for completeness.
If you take into account the presence of the
person you need to introduce the moment of inertia I. The angular
momentum now is written as L=Iω where ω is the angular
velocity (in radians per second) of the system. If a point in the system
is a distance R
from the axis of rotation and moving with a speed V, the angular
velocity is ω=V/R. So, comparing with the idealized situation above
you can see that the moment of inertial of a point mass must be I_{pointmass}=MR^{2}.
Now, the man holding the weights has some moment of inertia which I
shall call I_{man}; I will ignore the contributions from
his arms, so his moment of inertia does not change when he pulls in the
weights. The total angular momentum at the beginning is now L_{1}=(I_{man}+2MR^{2})ω_{1}=(I_{man}+2MR^{2})V_{1}/R.
So let's again reduce the distance to ½R. The new angular
momentum is L_{2}=(I_{man}+2M(½R)^{2})ω_{2}=(I_{man}+½R^{2})V_{2}/(½R);
this needs to be the same as the original angular momentum, so (I_{man}+½MR^{2})V_{2}/(½R)=(I_{man}+2MR^{2})V_{1}/R.
Solving, V_{2}=½V_{1}(I_{man}+2MR^{2})/(I_{man}+½MR^{2}).
Note that if I_{man}=0, you get the same as above, V_{2}=2V_{1}.
If I_{man}>>2MR^{2}, there will
be negligible change in angular velocity and so V_{2}=½V_{1}
because V=Rω and ω did not change. Finally, there
must be a case where I_{man} is just right that V_{2}=V_{1};
that would be if I_{man}=MR^{2 }and would
correspond to ω_{2}=2ω_{1}.
QUESTION:
Where can I find the assumptions for conservation of linear momentum? I'm specifically looking for a source which says that both bodies in a linear momentum problem have to attain a common velocity. The assumption or condition I'm looking for is so basic I can't find it.
ANSWER:
You can read my
earlier answer on
momentum conservation. I am afraid I have no idea what you mean by "both
bodies…have to attain a common velocity". In fact, that does not even sound
right to me for a general case.
QUESTION:
The question, "Why does water freeze at 32 deg. F?" has been proposed all over the internet. There is not an adequate answer to be found. The answers will describe how it freezes, and what takes place, but never why it freezes. The freezing point of hydrogen is 241.746 degrees celcius and the freezing point of oxygen is 218.79 °C. So why does water freeze at 0 deg C?
ANSWER:
Iron freezes around 1200^{0}C, mercury at 39^{0}C,
etc. The temperature of the phase transition between solid and liquid
depends on the material; it is not a simple thing to predict exactly where
that transition will occur, but it can be done in some cases. It is also
dependent on the pressure. But I am curious: Why do you particularly ask
about water? Why didn't you ask me "why does molybdenum freeze at 2620^{0}C?"
This question is "all over the internet"? Guess i missed that.
QUESTION:
Hello,when we look at objects with triangular prism instead of of eye glasses we see spectrum or rainbow on objects edges ,why this spectrum is always formed on the edges.
I mean the edges of objects for example chair,wall etc. I mean things i look at through prism instead of eyeglasses.
ANSWER:
The ideal situation for seeing a spectrum from a prism is if
you have white light, all coming from the same direction and preferably
through a narrow slit. If you just hold the prism up and look through it,
light of many colors is coming from all directions. When you have an
obstruction, like the edge of a table leg, for example, it lis like "half a
slit" and light coming from the edge is restricted to come mostly from one
direction. If that light has a fairly large fraction of white light, you are
likely to see a rainbow spectrum.
QUESTION:
Hi I'm a high school student and having hard time understanding something about electric potential energy. Do electrons moving in a circuit have potential energy because battery has done some work moving them against electric field (from low potential to high potential)? If yes, suppose that I just put the wires around the battery and think about the moment when it took 1 electron from low potential to high potential. That electron now has a potential energy equal to the energy spent when moving the charge from low to high potential. But since a battery doesn't generate electrons (and just providing force to move them) and all other electrons were already there in the copper wire, how did other electrons get their potential energy? The battery hasn't done any work for them (or did it?).
ANSWER:
The electric current in a conducting wire is not the best way
to learn about potential energy of charges because it is a rather complicated
process. What happens is that the potential difference across the ends of
the wire cause there to be an electric field inside the wire. Electrons see
this field and therefore each believes that it is in an approximately
uniform field and therefore it accelerates and gains kinetic energy as it
loses potential energy it has by virtue of the field. But what next happens
is that little electron almost immediately encounters an atom in the wire,
collides with it, and loses some or all the kinetic energy it has just
acquired and has to start all over again. So each electron bounces slowly
along the wire, repeatedly gaining and then losing kinetic energy. On
average, there is a net drift of electrons down the wire but it is really
quite slow and we consider the average electron to move with a constant
drift velocity. So any electron, moving from the negative terminal of the
battery to the positive terminal of the battery of potential difference V
moved, on average, with no change of kinetic energy but it has lost eV
Joules of potential energy. Where did that energy go? Put your hand on the
wire and you will see that it has warmed up. If you want a more lucid
example of electrons and potential energy, imagine a uniform electric field
of strength E (like in the gap between plates of a parallel plate
capacitor) with an electron released at some point. Then as the electron
moves, accelerating along the direction opposite the field (because it is a
negative charge), it loses potential energy eEz where z is the
distance it has traveled. After it has gone a distance z, it will have
acquired a kinetic energy ½mv^{2}=eEz.
QUESTION:
I am reading the ABC of Relativity. This is what I don't understand. If I get on a train in London and travel to Edinborough why is it equally true that Edinborough is travelling towards me? It am the one who initiates the motion. If Edinborough is really moving towards me, how is it also moving towards someone approaching it from the North, at the same time?
ANSWER:
The important thing to understand is that there is no test
you can do which determines who is "really" at rest—there is no such thing
as one frame of reference which is absolutely at rest. The way this is
expressed in the theory of special relativity is the laws of physics are
the same in all inertial frames of reference. (An inertial frame is one
which moves with a constant velocity in a straight line, no acceleration
allowed.) This means that there is no experiment you can do inside your
train which will have different result if you were in Edinborough or any other
train. All velocities are relative and it does not matter whom you consider
to be at rest to do physics. You, before you left London, were in the same
frame as Edinborough, and nobody would argue that it was you which caused
the change to a different frame since you were the one who accelerated and
consumed diesel fuel; that does not change the fact that the two frames are,
in all ways, equivalent once you are done accelerating.
By the way, it turns out that the laws of physics
have to be the same in all frames, including accelerating frames. This
is the basis for the theory of general relativity. But, you need not
worry about this is you are just studying special relativity. If you are
interested, see my FAQ page
QUESTION:
I was playing a game known as ''Fallout 3'' and in the game there are laser weapons. The laser weapons are powered by a marshmallow sized microfusion cells that are basically miniature nuclear reactors that fuses hydrogen atoms. In the game they produce enough power to turn a 500 kilogram bear into ash in one second. So could a reactor that small produce enough power for the gun and how energy much would a marshmallow sized blob of fused hydrogen produce? A normal microfusion cell in the game has enough energy to fire 24 of these shots. So would it be possible in any way for these laser weapons to be able to be this powerful with an energy source like the microfusion cell?
ANSWER:
I have no way to estimate the "power
to turn a 500 kilogram bear into ash in one second". I am sure you realize
that, with today's technology, the possibility of there being such a power
supply is zilch. Let's just do a few estimates to show how hard this is. One
gram of hydrogen fuel (deuterium + tritium), if fully fused into
helium+neutrons, releases something on the order of 300 GJ of energy; so, if
released in 24 one second pulses, each pulse would be about 10 GW. That is
probably way more than your bear burning would need, so let's say 100 MW
would do it; so, we would need about 10^{2} g of fuel. I calculate
that to confine that amount of gas in a volume of 10^{5} m^{3}
(about 1 in^{3}) would require a pressure of about 5,000,000
atmospheres! That, in itself, should be enough to convince you that this
machine could probably never be possible. If you need more convincing,
consider shielding: 80% of the energy produced is in the kinetic energy of
neutrons. How are you going to harvest that energy in such a small volume
and how are you going to protect yourself from the huge neutron flux? And
surely there needs to be some sort of mechanism to control the process and
convert the energy into usable electrical energy to power the laser; all
that is supposed to fit into 1 in^{3}? This truly is a fantasy game
with no connection to reality!
QUESTION:
I read that when no external forces act on a system, the internal forces are paired to balance the momentum.
So if the Universe is considered as such a system where momentum is indeed conserved, every action will have equal and opposite reaction. Is this a sufficient explanation to consider Newton's third law to be a special case of Newton's second law?
ANSWER:
Suppose that Newton's third law were not true. For any single
particle in a system of particles Newton's second law would be true but you
would not be able to apply it to the system as a whole. Without Newton's
third law, the total linear momentum of an isolated system of particles
would have no physical significance, it would not be a conserved
quantity. Newton's second law refers only to a single particle and is useful
for systems of particles only because Newton's third law is also true. I
fail to see how that makes the third law a "special case" of the second.
QUESTION:
If you were to take a 2 liter bottle of air, down to 10 meters and were to let a slight amount of water into the bottle and then sealed the bottle, what would happen to the bottle when you brought the bottle back to the surface?
ANSWER:
It is always hard for a scientist to deal with amounts like
"a slight amount"! If you did not let any air out of the bottle, the
pressure inside would be "a slight amount" higher. If you let in as much
water as would go in without any air escaping, the gauge pressure of the
contained air would be about 1 atmosphere, that is, twice atmospheric
pressure.
QUESTION:
If two deuterium atoms collide at an ideal velocity and at an ideal angle, are there conditions where fusion will not occur and some sort of scattering will take place? If so, why or how would scattering occur?
ANSWER:
I do not know what you mean by "ideal velocity" or "ideal angle". I can tell you that whenever the two deuterons come close enough together to interact, there are a great many things which can happen, all with specific probabilities ("cross sections" in the usual parlance). Essentially any possible reaction allowed by conservation laws and selection rules can happen. Nearly always, the most likely thing to happen is elastic scattering where two deuterons exit
after the interaction.
QUESTION:
Assuming I had a piece of string that was a light year long and I was holding it on 1 end and my friend was 1 light year away holding the other end, with no slack in the string. If I pulled it would they feel the immediately, or no? If not, why?
ANSWER:
You would not believe how many times I have gotten variations
of this question. See my faq page for a
similar question. In a nutshell, the information would travel at the speed
of sound in your string so it would take far longer than one year before
your friend felt your tug.
QUESTION:
What would happen to a fly while an airplane takes off assuming that the fly was already flying forward while the plane was at rest?
ANSWER:
The effect on the air due to the acceleration of the airplane
is quite small; the air pressure at the rear of the plane would
increase very slightly.
In other words, the air stays almost at rest relative to the airplane. Since
the fly flies relative to the air, he is mostly unaffected. He is, however,
accelerating along with the air and the plane, so he would have to fly a
little harder in the forward direction of the plane to continue moving the
same as he was; since a typical acceleration of a commercial jet is about 3
m/s^{2}, this additional forward force would be about 1/3 of his
weight.
QUESTION:
why are soap bubbles colorful?what colors are observed when a soap bubble is illuminated by monochromatic light?
ANSWER:
The reason is that a soap film has two surfaces from which
reflection of light can occur and they are very close to each other.
Therefore, light of certain colors will interfere constructively while light
of other colors will interfere destructively. A pretty clear explanation as
well as a calculator may be found on the
hyperphysics site. If you illuminate the bubble with monochromatic
light, you will only see that color or no light reflected at all.
QUESTION:
I just watched a nature document about the largest snake that ever lived on Earth that was the Titanoboa that could exert a pressure of 400 pounds per square inch and that lived in the water mostly. So my question is that could a Titanoboa destroy a submarine by coiling around it with the pressure of 400 pounds per square inch assuming that the submarine is small enough for the 48 foot snake to coil around it?
ANSWER:
The pressure under water increases by about 0.44 psi/ft. To
get to 400 psi, therefore, a submarine must go to a depth of 400/0.44=909
ft. WWII Uboats had collapse depths of 660920 ft. Modern submarines have
collapse depths of around 2400 ft. So, it depends on the submarine, but most
modern ones could withstand 400 psi. The only proviso is that they are
designed to have the pressure uniformly distributed over the whole surface,
not a narrow band where a snake would squeeze; a very localized pressure
could cause a structural failure at a lower pressure.
QUESTION:
I came across
this just now.
It implies a balloon full of air weighs more than the same balloon empty. That doesn't feel right to me as an inflated balloon is surely buoyant by the amount of air it contains. i.e. The extra weight of the air is cancelled out by the buoyancy. The site above is a respected organisation and I would be surprised if is recommending an experiment based on a false assumption. However, I cannot see how the instructions they give could possibly be used to determine the density of air.
ANSWER:
You are right, the experiment ignores the buoyant force on
the balloon; if the density of the air in the balloon were identical to the
density of the air outside the balloon, the experiment would fail to find
any weight of air. However, the pressure inside the balloon is greater than
the pressure outside and therefore the density of the air inside is larger
than normal atmospheric density. The experiment then measures the difference
between the mass inside the balloon and the mass of an equal volume of air
at atmospheric pressure. This would still be a reasonable orderofmagnitude
measurement of the density of air. A more accurate experiment would be to
measure the pressure to which the balloon was inflated; with that
information you could do a better measurement of the air density.
QUESTION:
I'm interested in why we loose our signal when two walkie talkies get too far apart even there is no obstacle between us. Range of reach of EM waves is infinite, so what's the reason?
ANSWER:
You are right, in a vacuum electromagnetic waves last
forever. You are not in a vacuum, but absorption by air is probably very
small for your walkie talkies. The problem is intensity of the waves. If you
are far away compared to the size of the unit, you can say that the signal
spreads out from the sending unit like a sphere. But, there is only a
certain amount of energy in that wave and so the amount of energy
intercepted by the receiving unit gets smaller and smaller the farther away
you go. Eventually, the receiving unit is not sensitive enough to detect the
signal.
QUESTION:
What should be the temperature of a gas molecule if it needs to escape out of earth's gravitational pull suppose if we take the case of oxygen at what temperature its average velocity will be enough to escape earth's gravitational pull?
ANSWER:
Temperature of a gas is
a statistical quantity, no single gas molecule has a temperature. A gas of a
particular temperature has a distribution of speeds called the
MaxwellBoltzmann distribution which contains all possible speeds. The
figure to the left shows this distribution for N_{2} for several
temperatures. The escape velocity from the surface of the earth is about
11,000 m/s, so you can see that a heavy molecule like nitrogen has almost no
molecules going that fast at normal temperatures (300 K), or even if T=1000
K. The picture to the right compares the distribution of speeds for N_{2}
with that for H_{2}, about 14 times lighter. You see that the most
likely speed of H_{2 }is about 4 times faster than N_{2}.
Comparing with the figure to the left, 300 K hydrogen gas would have a most
probable speed of more than 1200 m/s. Although there are still very few with
speeds higher than 11,000 m/s, there are still a few which escape.
Eventually, as the slower molecules speed up to fill in the distribution,
they would essentially all leak out of the atmosphere. That is the reason
why there is almost no hydrogen or helium in the atmosphere. Most hydrogen
on earth is locked up in water and other molecules, but since helium is
inert, the only source of it is from underground, usually as a byproduct of
natural gas wells. The form of the MaxwellBoltzmann distribution is given
by 4π[m/(2πkT)]^{3/2}v^{2}exp[mv^{2}/(2kT)].
QUESTION:
Assume we're in a very large hollow sphere. (let's say r = 1 light year).
Then we take a very powerful laser with extremely low diffraction and we fire it while effectuating a full rotation with it.
Wouldn't the laser point travel faster than light on the inside surface of the sphere?
Or does it fall as into the immaterial category and is exempted from the rule?
ANSWER:
You do not need such an extreme condition to do what you
want. The distance to the moon is about R=3.8x10^{8} m, so if
you swept your laser beam across the surface of the moon the speed of the
spot would be equal to the speed of light c=3x10^{8} m/s if c=Rω
where ω is the angular velocity you are rotating the laser. So,
ω=3x10^{8}/3.8x10^{8}=0.79 radians/second=45^{0}
per second—really easy to do. But that spot is not anything, really, because
the spot a second from now will not be "made of" the same photons that it is
made of right now. The most important thing is that there is no way that
this spot could be made to carry information from one point on its path to
another. And you are right, it has no mass, but even massless light cannot
travel faster than the speed of light.
QUESTION:
As a chemistry teacher, I often get questions from students that are best asked of a physicist. Is there a "short" answer to explain the nature of charge? Why are there only two charges? We understand that the assignment of the negative to the electron could be completely arbitrary, but what exactly is a/the "negative charge"?
ANSWER:
For every force field in nature there is a corresponding
source of that field: gravitational mass is the source of gravitational
fields, electric charge is the source of electric fields, electric currents
are the source of magnetic fields, quarks are the source of nuclear fields,
etc. Further, if there are two such sources in proximity of each
other, each experiences a force due to the field of the other. We first are
aware of gravitational fields and conclude after some experimentation that
the property something must have to be a source of that field is mass.
Inspired by our success in understanding the gravitational field, we start
looking around for other forces. Combing our hair one morning, we notice a
new kind of force which is obviously not gravity. After some experimentation
we discover that some objects in nature cause electric fields if they have a
property which we call electric charge. So, putting two electric charges
near each other, we find, not unexpectedly, that both experience a force due
to the other. But now there is something different. In gravity, the force
between two masses is always attractive, never repulsive. So mass is
a relatively simple thing because there is only one kind of it. But with
electric charges, sometimes the force is attractive and sometimes repulsive,
so we conclude that there must be two kinds of electric charge which are
most easily distinguished from each other by assigning a sign, + or . Of
course, further experimentation reveals that opposite charges attract and
like charges repel. Why are there two? Well, just because that is the way
nature is. You might just as well ask why there is only one kind of
gravitational mass. The way that quarks interact with each other is more
complex than just attractive or repulsive and three kinds of the quark
property called "color charge" are required. So the answer to "what exactly
is charge?" is that it is that property which something has which allows it
to both cause and feel an electric field.
To see questions and answers from longer ago,
link here.
