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QUESTION:
I've heard that a bullet falls to the ground
at the same speed [in the same time] no matter the charge pushing it forward. More powder and it moves faster but vertical speed is the same. I imagine two equal weight gliders of 20/1 and 30/1 lift/drag ratios would also express the same vertical speed?
My question is, wouldn't this gravity effect also hold true underwater? An underwater glider expressing 30/1 L/D would travel 50% further, faster than a 20/1 L/D in a given vertical?
ANSWER:
I will assume that we are talking about horizontally fired bullets. All
bullets do not hit the ground with the same speed, rather they all take
(approximately) the same time to get to the ground; I have corrected your
question because I believe that is what you meant. Your comparison to
gliders with different lift/drag ratios is not appropriate, because the
thing which results in the times of vertical fall being the same is the
assumption that there is no lift at all. In other words, the vertical speed
of a bullet will always be much less than the terminal velocity so there is
no appreciable component of the drag in the vertical direction. (This is not
true if the gun is fired from a very high altitude since there will be
enough time for the verticle component of the velocity to approach the
terminal velocity.) Underwater (I am just talking about bullets, not
gliders) the bullet will not behave the same because the terminal velocity
is much smaller than in air, in other words the drag is much larger.
QUESTION:
Quantum mechanics has popularized the idea of very small particles existing in two or more points in space during the same instant in time. Is this to be taken at face value as a fundamental aspect of quantum particles, or is this simultaneous existence just a concept used to express our inability to observe these particles without inadvertently altering their speed or location?
ANSWER:
See an earlier answer.
QUESTION:
I read from my history book that when nukes were first created we didn't know that the radiation their explosions produce is harmful to one's health, even to one's life maybe. So my question is, how come neither physicists or medical personnel ever figured out that radiation can be dangerous until we had first victims of radiation?
ANSWER:
Ever hear the expression "hindsight
is 2020"? It is unrealistic to expect that when something new is
discovered that we should somehow know all the effects that might have on
anything else. At the time radioactivity was discovered, nobody even knew
what atoms were composed of or what their structure was. And it was found to
be very tiny bits of matter (e.g. what we know as electrons today)
and who would have thought that getting hit by something trillions of times
lighter than a speck of dust could be harmful? It took experience before it
was appreciated how dangerous it could be. One of the best known examples of
such experience is the case of
radium watch dials.
Radium, mixed with a phosphor, was painted by women onto watch dials and it
would glow in the dark. The workers were encouraged to point their brushes
with their lips to make the fine lines required. Subsequently many became
ill with cancer and other radiation sickness. It seems stupid now, but the
dangers were not known until they were discovered. Also many of the health
effects were longterm effects, the effects not appearing until years or
decades after exposure. One example I know about from personal experience is
the effect of radiation treatment for acne which was popular in the 1950s. I
very much wanted to have this done but my father forbade it, making me
furious with him. Years later people who had had this treatment started
coming down with cancer—thanks,
Dad! Yet another example is the shoefitting xray machine. When I was a kid
it was really fun to buy new shoes because you could look down at an xray
of all the bones in your foot and how well the shoe fitted you; again, when
more was learned about radiation, these machines were all sent to the scrap
heap. When it began to be appreciated that there were dangers, studies began
to be done to try to set allowable dosage levels. But, it is unethical to do
such experiments on people, so lab animals had to be used which always
presents problems with scaling and other variables. Much of what we know
today was learned afterthefact by doing longterm statistical analyses
over decades of medical records.
QUESTION:
Magnetic flux according to my book is total no. of magnetic field lines passing through a given area in magnetic field. ok but why there are not infinite no. of magnetic field lines, because magnetic field line are defined as the path that a magnetic north monopole would take if left in north part of magnet, so if i take a monopole and leave one atom away from the previous position then it should take a slightly different path and that path should be considered as magnetic field line,so in this way i can draw millions of line.
ANSWER:
Magnetic flux is well defined:
Φ_{M}≡∫∫B·dA,
the area integral of the magnetic field. If the integral is over a closed
surface, the flux is zero; this is the famous situation which tells you that
there are no point sources (called magnetic monopoles) of magnetic fields
like there are for electric fields. If you like, you may interpret this as a
number, but that is not really fundamental. If you talk about uniform
magnetic fields which are perpendicular to a plane surface, Φ_{M}=BA
, is the flux through an area A. Then, if you interpret
Φ_{M}
as a number, you would simply say, for example, there were 10 lines through
an area of 1 m^{2} if the magnetic field is 10 T. You could ask the
same question about electric flux,
Φ_{E}≡∫∫E·dA.
This is perhaps a little easier to understand because you can have point
charges. For example, if you have a 1 C point charge, the electric field 1 m
from it is E=Q/(4πε_{0}r^{2})=1/(4x3.14x8.85x10^{12}x1^{2})=9.27x10^{9}
N/C. The flux at a distance of 1 m from the charge would therefore be Φ_{E}=EA=9.27x10^{9}x4x3.14x1^{2}=1.17x10^{11}
Nm^{2}/C. Incidentally, this is the flux regardless of where you
measure it because the area of the sphere
(4πr^{2})
surrounding the point charge appears both in the denominator of the field
and in the numerator of the flux and cancels. Hence, you could say that
there were 1.17x10^{11} lines of
electric field emanating from a 1 C point charge. You can see this from
Gauss's law,
Φ_{E}=∫∫E·dA=Q/ε_{0}
if the integration is over a closed surface enclosing the charge Q.
QUESTION:
Why degree of monochromaticity is always non zero?
ANSWER:
The practical reason is that you cannot make a laser which is perfectly free
from electronic instabilities and noise. But even if you were able to make a
perfect laser, you could not know its frequency perfectly. The more
fundamental reason is that monochromatic light means light with a single
frequency. But, the linear momentum of the photons is proportional to the
frequency of the light. Because of the Heisenberg uncertainty principle, the
only way you can know momentum of the particle perfectly is to be completely
ignorant of its position; you need an infinitely long wave to know its
frequency perfectly.
QUESTION:
This question is about relativity. If person leaves earth and heads toward a star that is 10 light years away, and he travels at near the speed of light, time will slow down for him. The round trip to the star may take him only one year, but when he returns, maybe 100 years have passed on earth. When he started out to the star, he knew it was
50 [I think you meant 10,
right? The Physicist] light years away. It seems that for him, he will have traveled 20 light years in one year. That is faster than the speed of light. What don't I understand?
ANSWER:
First of all, time for the traveler will not slow down; rather, her clock
will run slower as observed by an observer on the earth. To her, time will
run perfectly normally. However, she will observe the distance to the
destination shortened because of length contraction. You have rounded things
to what they would be if she were going the speed of light, but let's do the
whole problem as if she were going with a speed v=0.999c. The
earth observer would see an elapsed time of 100x0.999=99.9 years (your 100
years). The traveler will see the distance shrunk to D'=10√(1.999^{2})=0.447
light years, so the time of travel is t=2x0.447/0.999=0.895 years
(your 1 year). You should read the earlier answer on the
twin
paradox.
QUESTION:
Supposing a weightless container is filled with water. I am sure the pressure at the bottom of liquid, P1 = atmospheric pressure + height of liquid x density of liquid x g = Patm + hdg, where Patm is atmospheric pressure and d is density of liquid. We can calculated this pressure as if liquid in region A and Region C does not exist.
But how about the the pressure at the base of the container, that is P2. Is P2 same as P1? For P2, do we need to consider the whole weight of the liquid, that is inclusive of the weight of water in region A and C?
ANSWER:
It depends on what the force on the bottom is. If the container is in
equilibrium, imagine it sitting on a table. The table would exert an upward
force equal to the weight W of all the water, so P_{2}
would be W/A_{bottom} where A_{bottom}
is the area of the bottom of the container. This assumes that atmospheric
pressure is the same everywhere in the vicinity of the container; in other
words, I have ignored the buoyant force due to the air on the whole
container because it will surely be much smaller than W.
FOLLOWUP
QUESTION:
Indeed the container is resting on the table. For P1, I use h x d x g. But for P2 you use W / Abottom, So can I say P1 not equal to P2 ?
ANSWER:
Yes, but I have to admit that my answer was misleading in that I gave you
the gauge pressure, the pressure above atmospheric. So I should have said
that P_{2}=P_{atm}+W/A_{bottom}.
There is no problem that P_{2}≠P_{1}
because the force which the container exerts on the table is not P_{1}A_{bottom}.
Think about it—the
sides of your container exert a downward force on the bottom of the
container.
QUESTION:
Think of words (incapital letters) that can be read properly both with a mirror and without a mirror.
What are these words?
ANSWER:
Well, this is not physics and I usually do not answer such questions.
However, it is a cute puzzle. First, any letter in the word which is
symmetric upon reflection about a vertical axis will look the same in the
mirror: A, H, I, M, O, T, U, V, W, X, Y. Second, any such word must be a
palindrome composed of the allowed letters, e.g. AHA, OTTO, WOW,
YAY, MOM, TIT, TUT,….
But, wait, it is somewhat more difficult than that! Look at the pictures of
the Camel cigarette package and its reflection, in particular, the word
CHOICE on the side. So words composed of letters with symmetry upon
reflection about a horizontal axis will also reflect the same: B, C, D, E,
H, I, K, O, X will have a reflection which is the same as the original, but
only if turned upside down. There is no restriction for these words to be
palindromes; besides CHOICE, some others would be BED, HEX, BOX, BIKE, HIKE,
CHIDE,…
QUESTION:
What is meant by virtual image?
ANSWER:
It is an image formed by light which appears to be coming from
somewhere but is not actually coming from there. When you look into a
mirror, it appears that the light from the image of what you are looking at
is coming from behind the mirror, but it is actually coming from the mirror
itself.
QUESTION:
Supposing that initial
velocity is 0 Is a displacement vs. time (squared) equivalent to a Velocity
vs. time graph? I tried to find the slope of displacementtime^2: d/t^2 =
V/t which equals acceleration But what I read everywhere is that: d/t^2 =
a/2 according to the equation d= (initial V)(time)+ (1/2)(at^2) I don't what
is wrong with my calculations?
ANSWER:
If you plot d vs. t^{2}, the slope of the line
will be ½a.
You know that d=½at^{2}, so suppose that I call t^{2}=u.
Then clearly the slope of the line d=½au is ½a. Your
equation d/t^{2}=V/t is clearly
wrong: d/t^{2}=½at^{2}/t^{2}=½a
and ½a≠v/t.
QUESTION:
Two of us disagree on part of a sol'n given by two people with Physics background, and I want to know if I am correct, or if I am missing something in the analysis of the problem...in case I have to explain it to a student.
Question concerning Forces/impulse..... 50kg person falling @15m/s is caught by superhero , and final velocity up is 10m/s.
Find change in velocity.
Find change in momentum .
It takes 0.1 sec to catch them.....ave Force is ?
answers are: vel = 25 m/s change in mom.. 1250 kg*m/s, and ave Force = 12,500 N.
Here's where we disagree: Person B says that 12,500 N is equiv. to 25 g ?????
They try to explain that 250 m/s^2 accel. corresponds to 25g.... I said it makes No sense at all, [ I know the accel. is 250, but that doesn't in any way imply a 25 g "equivalence" to me ]. They then went further to "prove" their point........Here is their argument...
500 N/g = 12500N / ( )g ..... I agree the ( ) = 25, but say there is No justification for the 500 N / g in the first place...... any ideas where it comes from , or how to justify that value ?
BTW I teach physics on and off at the HS level.... person B is an Engineer , I think
ANSWER:
Person B is wrong but has the right idea. (As you and your friend have
apparently done, I will approximate g≈10
m/s^{2}.)
We can agree that the acceleration is a=250 m/s^{2} and that
is undoubtedly 25g. Now, we need to write Newton's second law for the
person, mg+F=ma=500+F=12,500, so F=13,000 N. This is
the average force by the superhero on the person as she is stopped, so the answer that the
average force is 12,500 N is wrong. When one expresses a force as "gs of
force", this is a comparison of the force F to the weight of the
object mg, F(in gs)=F(in N)/mg=13,000/500=26
gs; this simply means that the force on the object is 26 times the
object's weight. So neither of you is completely right, but if there is any
money riding on this, your friend should be the winner because the only
error he made was to forget about the contribution of the weight to the
calculation of the force. I am hoping that superman knows enough physics to
make the time be at least 0.3 s so that Lois does not get badly hurt!
QUESTION:
It is said gravitons are the
expected particles to exchange the gravitational force. Do these particles
exist in real?
ANSWER:
There is no successful theory of quantum gravity, so gravitons are "expected
particles" but I would not call them "real" at this stage; hypothetical
would be a better word.
QUESTION:
Why does it take more time to go to
Dubai from Mumbai than to come from Dubai to Mumbai?
ANSWER:
There are winds at high altitudes called
jet streams. The
prevailing direction of these winds is easterly and their speeds are as
large as 100 mph. Therefore when you travel east (as from Dubai to Mumbai)
you have a tail wind, and when you travel west (as from Mumbai to Dubai) you
have a head wind. Since the greatest speed an airplane can fly, say 600 mph
for commercial jets, is airspeed, that is relative to the air, the ground
speed with a 100 mph wind would be 700 mph going east and 500 mph going
west.
QUESTION:
Why is the direction of angular displacement along the rotational
axes?
ANSWER:
There are two ways in which you can rotate something about some axis, either
clockwise or counterclockwise (as viewed from one side or the other). So, an
angular displacement has two possible directions only. It is like a
displacement in one dimension, like a bead on a long straight wire. You then
call one direction on the wire plus and the opposite direction minus. The
only way to get a similar directional assignment for a rotation is to note
that the rotation axis is just like a onedimensional axis, one way being
assigned to be positive, the other negative. Usually this is done with the
righthand rule such that counterclockwise is a positive vector on the
axis.
QUESTION:
Recently I came across such an article that Gravity isn’t actually a Force, It's the bending of spacetime that cause objects to move on curve paths.
If two objects of equal shape and masses are there they will bend the spacetime in same way by same amount.
Then according to that there shouldn't be any kind of force or something acting between them which isn't true i guess. Gravity forces will act b/w those two. Could you please
explain me this contradiction.
ANSWER:
I do not understand your logic that because they both have the same effect
on spacetime, they will feel no force. Although it is just a cartoon to help
you qualitatively understand warping of spacetime, in this case it is
illustrative. If you place a bowling ball on a tranpoline, what happens? The
surface of the trampoline is warped. Now, suppose that you place two bowling
balls a few inches apart from each other. What happens when you let go of
them? They will behave as if there were a force between them and roll toward
each other but what they are really doing is responding to the shape of the
trampoline.
QUESTION:
I have a question about gravity, or more specifically how to calculate it for the purpose of a scifi fanfiction I am making.
So Earth has the density of 5.5 grams per cubic centimeter, total mass of 5.97219×10^{24} kilograms and radius of 6371 kilometers. So how do I put these together in a formula to get the 9.81m/s gravity?
ANSWER:
The density is irrelevant. You need to use Newton's universal law of
gravitation for two point masses. The force F which each feels is
F=GMm/r^{2} where G is the universal gravitational
constant, G=6.67x10^{11} N·m^{2}/kg^{2},
M and m are the masses of the two masses (take M as the
mass of the earth), and r is the separation between M and m.
This also works if the masses are spherically symmetric. But, of course, you
also know in your case that F=mg. So, if you solve for g, g=MG/r^{2}≈6.7x10^{11}x6.0x10^{24}/(6.4x10^{6})^{2}=9.8
m/s^{2}.
QUESTION:
Why do electrons fall back to the ground state after jumping into a higher orbital?
ANSWER:
The simple answer is that a system will always seek a way to move to a lower
energy, like a ball rolling down a hill and not up it or sitting still. The
more complicated answer addresses whether there is, indeed, "a way" to
achieve this. For atoms, the radiation of photons in the transition is
usually electric dipole radiation and there is a corresponding
quantummechanical operator, let's call it O_{E1}. Then, you
have to look at what is called the transition matrix element for this
operator, <ΨO_{E1}Ψ'>=∫Ψ·O_{E1}·Ψ'dτ
which is an integral over all space and Ψ' and Ψ are the
excited and ground states, respectively. If this is nonzero, then the decay
will decay with some half life determined by the value of the matrix
element. If it is zero, the state will still probably decay to the ground
state but via a different kind of transition.
QUESTION:
We all know about second law of motion.
The example is about the coin which is placed above the cardboard and cardboard is placed above the beaker... if the cardboard is quickly pulled then the coin falls in the beaker but if the cardboard is slowly pulled then why does the coin come with the cardboard?
Why does this happen?
ANSWER:
This is usually used to demonstrate inertia, Newton's first law; since you
are interested in why it does not work if you move the cardboard slowly, the
second law also must be used. To move the cardboard from rest it must
experience some acceleration. If the coin is to experience an acceleration,
that is move with the cardboard, some force must be exerted on it. The only
force on the coin horizontally is the static friction and the greatest it
can be is μmg
(where μ is the coefficient of static friction) so the greatest
acceleration the coin could have is μg; for example, if μ=0.3,
and you caused the acceleration of the cardboard to be any greater than
about 3 m/s^{2}, the cardboard would slip under the coin. But, if
you gave the cardboard an acceleration of 0.2 m/s^{2} the coin would
move with it.
QUESTION:
How can it be said that the universe is only 13.7 billion years old? This seems ridiculous to me that anyone can make this assumption. Taking in the fact that there are black holes almost as old as the universe itself how is this possible? Stars live billions of years so the first group of black holes would be billions of years after the "big bang". Also due to the fact that the universe is expanding way faster and not slowing down like predicted it would seem to me the current model is wildly inaccurate. Can you please explain this to me? My understanding is that stars form and die out all the time so to be able to accurately determine the age of the universe another method would need to be devised. It is my understanding that galaxies form with the help of super massive black holes.(Which aren't even holes at all but super compacted spheres of matter so dense light cannot escape created by countless other bodies of mass.) Which attracts matter into tight clusters of stars we call galaxies. These galaxies would have to take billions of years to form after the creation of the first stars ever created and after the first black holes were created from some of those stars. Does the current model explain this?
ANSWER:
As clearly stated on the site, I do not do astronomy/astrophysics/cosmology.
I can at least say a little about your question, though. The age of the universe is not an "assumption", it is based on careful analyses of many measurements. The stars in the
early universe were much more massive than the sun; this
resulted from the fact that the early universe was essentially all hydrogen and a bit of helium.
Stars began forming only 100200 million years after the big bang. Very large stars burn much hotter than smaller stars and therefore have much shorter lifetimes, a few million years rather than
a few billion years for less massive stars (e.g. the sun with an
expected lifetime of about 10 billion years) which would explain why many
black holes are very old. The rate of expansion of the universe is not
inconsistent with the model of a big bang about 14 billion years ago (in
fact, the rate is one of the measurements used to estimate the age of the
universe) even though we do not understand all the details (e.g. dark
energy and accelerating expansion).
QUESTION:
I would like to ask this question because I do not agree to this but it is written in my school handout. It says that work needs only two things to be calculated which is weight and height of displacement.
Now , there is this question that says that two persons who have the same mass went to the top floor of eiffel tower. One used the stairs and the other one used an elevator. The thing is that our handouts states that they both exerted the same amount of work done because they have the same weight and they both have the same height distance covered.
I do not agree simply because the man using the stairs carried his own weight up the distance while the the other one was being carried by an elevator and I immediately thought that the elevator was the one doing the work and not the person which is why they cannot have the same work done.
If I am wrong then I would humbly accept it. However , if our physics teacher is wrong, I would need a legit source stating about the problem and saying that it was wrong or similar problem like the one given with an answer so I can correct my grade. I would really appreciate anyone's help right now.
ANSWER:
The point is that the net work done on each person was the same.
Where the energy came from is different. On the stairs, the energy is
supplied by what the guy ate for lunch. In the elevator, the energy is
supplied by the motor lifting the elevator. Regardless where it came from,
mgh of energy must be somehow supplied.
QUESTION:
How surface tension overcome insect weight and help it to stay on liquid surface, as surface tension is along liquid surface?
ANSWER:
The weight of the insect causes the water to be depressed where the legs
touch the surface, so the surface tension has a vertical component.
QUESTION:
Can we make something like antigravity on Earth, or something similar to that? And i was wondering if we could make something like antigravity with magnets, to actually reduce the weight of some object. I was thinking to make a chamber with a magnet at the bottom, and take magnetized objects and put them on the top, so it will levitate.
ANSWER:
Your device will not work. The magnet on the bottom can be made to exert a
repulsive force on a magnet at the top which tends to lift your chamber.
However, Newton's third law requires that the top magnet exerts an equal and
opposite force on the bottom magnet, so the net force on the chamber is
zero.
QUESTION:
Is rotation a reason for earth being round?
ANSWER:
No. The reason that all large astronomical objects are nearly spherical is
because the force which holds them together is spherically symmetric. If you
have a point mass, the gravitational force another mass feels depends only
on how far apart they are, not on the direction. If you want a lot more
detail, see an earlier
answer about
gravity of a cylinder which shows how the tendency is toward a spherical
shape.
QUESTION:
Using a spring balance, weigh (separately) a piece of wood and a container with water in it. Then weigh them again, but with the wood floating in the water. Does the reading on the balance change? There will be an upthrust on the wood (Archimedes' Principle), causing it to weigh less. I guess there will be an equal downforce exerted by the wood, resulting in the same weight being shown in both cases. This seems a fairly easy question, but I can't find the answer clearly stated anywhere.
ANSWER:
The measured weight will be the sum of the individual weights. The buoyant
force is the force which the water exerts on the wood; Newton's third law
requires that the wood exert an equal and opposite force on the water. As
you correctly "guess", these two forces cancel out.
QUESTION:
Good afternoon. I started learning quantum mechanics and I have a question about Heinsenberg's principle. I was thinking about a photon created by an electronpozitron annihilation. As I understand (and I hope I am not wrong here) all observers, no matter their state, can specify the point (and all will specify the same point) where the photon appears, and I think in a Wilson chamber we can do this thing. Anyway, we can specify its position with an error less than infinity. But at the same time we know it's speed with absolute precision, as, assuming we measure it in vacuum, it's speed is invariably c. So the product between (delta)x and (delta)p will be 0, as (delta)x is not infinity and (delta)p is 0. In other words we can predict it's position with a certain accuracy knowing at the same time it's speed (I am not sure, but i think that the (delta)p is not a vector but a scalar quantity). What is wrong about this as it seems the Heinsenbrg's principle is violated?
ANSWER:
The mass of a photon is zero but it has momentum. Therefore your notion that
linear momentum p is mass m times velocity v must be
wrong; the relativistically
correct expression for momentum is p=mv/√[1(v/c)^{2}]
where c is the speed of light. For a photon, this is a little tricky
because p=0/0, but you can also write that E^{2}=p^{2}c^{2}+(mc^{2})^{2}
where E is the energy of the a particle of mass m, so p=E/c
if m=0; since E=hf for a photon, where h is Planck's
constant and f is the frequency, the linear momentum of a photon is
p=hf/c . Therefore the frequency of the photon must be
uncertain according to the uncertainty principle. Since the energy of a photon is
hf, there will be an uncertainty in the energy of the photon.
QUESTION:
The units of the Hall coefficient is m^3/C that's, cubic meter per coulomb. How so we get this?
ANSWER:
You just need to know the
definition of the
Hall coefficient R_{H}, R_{H}≡V_{H}t/(IB)
where V_{H} is the Hall voltage, t is the thickness of
the sample, I is the current, and B is the magnetic field. The
Hall voltage may be
shown to be V_{H}=IB/(nte) where n
is the charge carrier density (units [m^{3}]) and e is the
charge (units [C]]) of a charge carrier. Therefore R_{H}=1/(ne)
so the units of R_{H} are m^{3}/C.
QUESTION:
What is the true shape of the Sun?
ANSWER:
I was surprised to learn that the sun is almost a
perfect sphere.
QUESTION:
Why do the planets revolve only around the sun?
There are stars which are bigger than the sun, and apply more gravitational force, but other planets do not attracts towards the stars.
Please tell why?
ANSWER:
I guess you have not been following news articles over the past dozen years!
In fact hundreds of
exoplanets (planets revolving around stars other than the sun) have been
observed.
QUESTION:
I have something I have been wondering about.maybe you can answer. I recently drank a glass bottle of rootbeer and then set the empty bottle on the hood of my car that was slanted slightly (1980's model). Within seconds the empty bottle started to vibrate slightly and then"walked" down the hood of my car. The engine was turned off. I was trying to tell my girlfriend that it was from heat convection, but she disagreed. How can this happen? I did it a second time and the empty bottle did it again. Not my imagination.
ANSWER:
Here is what I think: The bottle probably had condensation (water) which was
fairly cool and the hood of your car was warm or even hot. The viscosity of
water depends quite sensitively on its temperature. If the water is heated
from 25^{0}C to 50^{0}C, for example, the
viscosity
decreases from about 890 to 547
μPa·s which will result in a
much slipperier contact. (Think of a very heavy oil and a very light oil to
get an intuitive feeling for how viscosity would affect the frictional
force. Or, think of the old expression "…as slow as molasses in January…".)
QUESTION:
I don't know if you get this question alot, but a man standing on the ground is subjected to 2 forces. Gravity and normal reaction force. Do they form an actionreaction pair of forces?
ANSWER:
This is one of the most misunderstood aspects of Newton's laws. Let me state
Newton's third law for your man in a couple of ways:

The
earth exerts a downward force on the man (his weight) and
therefore the man exerts an upward force on the earth of the same
magnitude as his weight.

The
ground exerts an upward force on the man (the normal force) and
therefore the man exerts a downward force on the ground of the same
magnitude as his weight.
Both of these are correct
statements of Newton's third law. Notice that the pairs of forces
("actionreaction") are never on the same object. The two forces you give
are both on the same object (the man) and can therefore not be an
"actionreaction" pair. So why are they equal and opposite? Because Newton's
first law states that if the man is in equilibrium the sum of all forces on
him must be zero; therefore the ground must exert an upward force to balance
the weight. Your example has nothing to do with Newton's third law.
QUESTION:
Why in the michelson and morley experiment that air does not have any effect on the experiment. To me air would be the current medium that the light is using to propagate and would yield a null result. The main key to this line of thought is that light travels slower through air then in a vacuum. The same as the speed of sound through different medium.
ANSWER:
The MichelsonMorley experiment does not attempt to measure the absolute
speed of light. Rather, the idea is to find the difference in speeds between
two different directions of travel. That measurement would not depend on the
presence of air.
QUESTION:
Does a submarine have to work harder to travel at the same speed in deeper water?
ANSWER:
The drag
force in water is approximately proportional to the speed; it also
depends on the shape and on the viscosity of the water. As you go deeper
into the water, the pressure increases and the temperature decreases.
Viscosity only very weakly
depends on
pressure but increases significantly as the temperature decreases. After remaining
constant around 20^{0}C until a depth of about 200 m, temperature
rapidly decreases to about 4^{0}C at a depth of about 1000 m; the
viscosity, and therefore the drag,
nearly doubles at
that depth.
QUESTION:
Do permanent magnets have an electric current surrounding them?
Permanent magnets seems to have different properties to electromagnets, such that electromagnets can be used for induction and energy transfer if a conductor is placed within their changing magnetic field.
So I understand that an electromagnet will have a changing magnetic field, which in turn generates a changing electrical current in a conductor placed within this magnetic field.
My question is, do permamant magnets generate this same electric current if a conductor is placed within their magnetic field?
I.e. if I had a strong permanent magnet, would I be able to generate a current in a coil if it was brought into the permananet magnet's magnetic field?
ANSWER:
A permanent magnet may be thought of as having "bound" currents. These are
not currents you could actually connect an ammeter to and measure. If you
are interested, there are, in general, two types of bound current, bound
volume current density J_{b} and bound surface current
density K_{b}. If the magnetization of the material is
M, you can write J_{b}=curlM
and K_{b}=Mxn where
n is a unit vector normal to the surface and pointing out of the
volume. I do not understand any of your rambling about induction. I will
simply say that any changing magnetic field can induce currents and a
permanent magnet is a source of a magnetic field; moving that magnet,
through a wire ring, for example, will induce a current in the ring.
QUESTION:
Can the equations for Compton Scattering be applied to a photon colliding with a neutron? If no, what equations can be used?
ANSWER:
Yes, you just have to replace the electron mass with the neutron mass.
However, the effect for this much larger mass would be very difficult to see
because the change in wavelength will be much smaller than for electrons. To
have any hope of observing the effect you would need to use extremely
highenergy gamma rays.
QUESTION:
If matter above absolute zero emits thermal radiation—and thermal radiation is actually electromagnetic radiation—what temperature is required for matter to emit visible light?
ANSWER:
Technically, visible light is emitted by all objects because the spectrum of
emitted radiation is continuous and includes all wavelengths. However, you
can say at what temperature the visible part of the spectrum is more intense
than any other emitted radiation, about 5000 K. Interestingly, that is about
the temperature of the surface of the sun.
QUESTION:
if an object ( like a bullet ) is fired vertically at constant speed. Where does it land?
ANSWER:
This is a standard intermediatelevel classical mechanics problem for
accelerated reference frames. All the mathematical background is much too
complicated to include here but the answer is that it lands a distance
d=4ωv_{0}^{3}cosλ/(3g^{2})
west of where it was shot; here ω=7.27x10^{5} s^{1}
is the angular velocity of the earth, v_{0} is the initial
velocity of the bullet, λ is the latitude where the gun is located,
and g=9.8 m/s^{2} is the acceleration due to gravity. This,
of course, neglects any air drag or wind corrections. For example, for λ=45^{0}
and v_{0}=390 m/s (typical 9 mm muzzle speed), d≈60 m.
At the poles (λ=90^{0}), it goes up and comes back down
perfectly vertically.
QUESTION:
what is the cause of buoyancy.?
we faced many confusions about then we worked on it and did some experiments
and found an another reason which is satisfying all the conditions and
aspects...but we wanted to consult about it......
ANSWER:
The reason for buoyancy is simply that the force due to the fluid up on the
bottom of something is greater than the force of the fluid down on the top
because of the pressure difference.
QUESTION:
Why does stirring a liquid makes it cool faster? Shouldn't the reverse happen, as the kinetic energy imparted to liquid in stirring should also change eventually to heat?
ANSWER:
The liquid in the cup loses energy to its environment (cools) by conduction
and radiation from its surfaces. The energy inside the surfaces is
transmitted to the surfaces (to continue cooling) by convection. Convection
is a relatively slow process for transmitting heat. Stirring more
effectively moves hot liquid to the surface. Energy you add by stirring is
trivially small compared to the increased cooling rate.
QUESTION:
Suppose light is travelling in a straight line parallel to Yaxis takes time t to reach from y1 to y2 in a frame of reference S. Let there be another frame S' which is travelling parallel to Xaxis with velocity v relative to frame S. Then a seen from S' the light ( which was travelling parallel to Yaxis in frame S ) now takes a longer path which is greater than y2y1 (as now it will be travelling obliquely in frame S' ). Now due to time dilation time taken in frame S' is lower than t as seen by frame S due to time dilation. Then speed of light in frame S' as seen from S will be greater than c as length is higher and time is lower. then how to explain the constancy of speed of light in this case ?
ANSWER:
You are making this too hard! The fact that c is a universal constant
is where you should start, not end. When you move with your S' frame with
speed v in the positive x direction, the light beam acquires
an xcomponent of c_{x}=v. If the speed of the
light is to remain constant, this must mean that the ycomponent must
therefore decrease such that c^{2}=c_{x}^{2}+c_{y}^{2}.
So,
c_{y}=c√[1(v^{2}/c^{2})]
which means that the angle θ which the light makes with the yaxis
is θ=tan^{1}[β/√(1β^{2})] where
β=v/c. See an earlier question for
an example of this phenomenon.
QUESTION:
If I have two hoses a varying diameters such as a garden hose and a firehose that are vertical and approximately 20 feet long that are filled with the same amount of water and have the same size opening at the bottom, using just gravity, will water flow through each hose in the same amount of time?
This question came up during a recent visit with a customer. I sell feeding supplies for neonatal patients. The current product being used is a large bore tube compared to the smaller bore that my company sells. However the opening at the distal tip of the feeding tube is the same size for both. Again, using just gravity, shouldn't the speed at which formula flows through the distal end be the same since it bottlenecks there?
ANSWER:
Your first question ("…will
water flow through each hose in the same amount of time?") is ambiguous, so
let me answer the question by finding how the speed of the delivered formula
(labelled V_{bottom} in the figure) depends on the geometry.
The operative physics principal is Bernoulli's equation, P+½ρV^{2}+ρgy=constant
where P is pressure, ρ is fluid density, y is the
height above some chosen reference level, and g=9.8 m/s^{2}
is the acceleration due to gravity. In your case, P is atmospheric
pressure both at the top and at the bottom, I will choose y=0 at the
bottom so y=h at the top. Therefore Bernoulli's equation becomes V_{top}^{2}+2gh=V_{bottom}^{2}
or V_{bottom}=√(2gh+V_{top}^{2}).
There are two ways that V_{bottom}
will be independent of the geometry: (1) if h is held constant by
replenishing the formula at the top or (2) if the area of the bore is much
larger than the area of the distal tip A>>a. Both of these result in
V_{top}≈0 so
V_{bottom}≈√(2gh). If
neither of these is true, it is a much more complicated problem.
To see questions and answers from longer ago,
link here.
