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QUESTION:
what happens to a satellite when it reduces its mass, or to any rotating object in an orbit? Is angular momentum conserved? velocity of the satellite does not depend on its mass (gravitational force = centripedal force) so if the velocity doesn't change when mass is reduced, how is angular momentum conserved, does the radiues change? if angular momentum is not conserved, what causes a net torque?

ANSWER:
Mass does not just disappear. If the satellite jettisons
some mass, that mass will carry away both linear momentum
and anglular momentum. Depending on how the mass was
ejected, the orbits of both parts will now likely be
different from the original orbit.

QUESTION:
If I'm holding a flashlight and I'm running with it turned on, does the light from it move at my speed + the speed of light?

ANSWER:
No, the speed of light is a universal constant. Its value is
completely independent of the motion of the source or the
observer. On my FAQ
page you will find many links to earlier answers which
relate to this.

QUESTION:
Why does the smoke of a burning incense stick first follow a straight path and then disperse?

ANSWER:
The answer is sort of conceptually simple but quantitatively
very complicated. Just about anything involving fluid
dynamics in the real world is pretty difficult to calculate
quantitatively and accurately. The simple concept is that a
fluid, under the right circumstances, is laminar
which means that if you focus your attention on a small
volume of the fluid it will move along a smooth line and be
followed and preceded by many other small volumes following
the same smooth line. Under other circumstances, the fluid
will become turbulent , moving chaotically. So
something happens which causes the smoke from your incense
stick to change from laminar to turbulent at some point.
This is usually quantified by a dimensionless variable
called the Reynolds' number, Re=ρuL /μ
where ρ is the density of the fluid, u
is its speed, L is a length which describes the
geometry, and μ is the dynamic viscosity of the
fluid. So, Re changes in such a way that the
original laminar flow becomes turbulent after rising a
certain distance from the source of the smoke.

QUESTION:
How is buoyancy affected by zero gravity? Specifically, if I put a steel ball-bearing in a jar of water in zero g, where will the ball bearing rest in the jar?

ANSWER:
Buoyancy is a purely gravitational effect, does not exist in
zero gravity. Your ball bearing will stay exactly where you
put it in the water. (One proviso is the equivalence
principle which says that there is no experiment you can
perform to determine whether you are in a uniform
gravitational field with a gravitional acceleration of a
or in zero field but having a uniform acceleration a ;
buoyancy would be observed in an accelerating rocket ship in
empty space.)

QUESTION:
This is not a homework question, I'm just trying to sharpen my knowledge for my ap test on Tuesday. The question puts two identical blocks, block A and B. These blocks have the same mass and velocity, except Block A's velocity is directed straight down, where Block B's velocity is directed right or parallel with the ground in the positive x direction. The answer, according to Khan Academy, is that both blocks will hit the ground with the same speed, but I just can't wrap my head around this since Block A will have the advantage of an initial velocity when using the kinematic vf^2=vi^2 + 2ay, whereas block B's velocity wouldn't count for Vi since it is perperndicular to the accerleration. Please help me to understand.

ANSWER:
The easiest way to do this is with energy conservation.
E _{i} ^{A} =½mv _{i} ^{2} +mgh ,
E _{f} ^{A} =½mv _{f} ^{2
} and so v _{f} =√(v _{i} ^{2} +2gh ).
Exact same calculation for E ^{B} .

QUESTION:
Let's assume a rocket (or flying saucer) could hover above an atmosphere-free planet - (that avoids all the added concerns that air would add for a helicopter). I'm asking whether one might consider the work done by thrust and gravity to be offsetting work that results in no Net Work.
Also, I believe thrust adds KE at the same rate gravity converts KE to PE when traveling up at Constant Velocity (or when at hover). Am I correct in assuming that once you reach constant velocity, you do no NET work as long as you maintain CV (again, noting that the work done by thrust offsets the work done by gravity)?
Lastly, I often see the comment, "Work adds energy to the object", yet if we drop an apple, gravity does work on it (W = f x d and there is a displacement in the direction of gravity). But gravity adds no energy of course - it simply converts KE to PE (when traveling UP) and PE to KE (when the object is traveling down).
Should the definition of work be expanded to include energy conversions?

ANSWER:
For physics at this level there is no need to introduce
potential energy at all; I always tell my students that PE
is a clever bookkeeping mechanism to keep track of the work
done by some force which is always present (like gravity in
your examples). (It should be noted that you cannot
introduce a PE function for some forces, but gravity is not one
of them.) In your flying saucer example, no work is being
done by either gravity or thrust because neither are moving
and so d =0 in your (simplistic) W=f x
d definition of work. If the saucer is rising with a
constant speed v, then when it has gone some distance d, the
thrust T has done W _{saucer} =Td
and gravity has done work W _{gravity} =-mgd
because T=mg for it to be rising with constant speed. The
net work W on the saucer is still zero, W =W _{gravity} +W _{saucer} =0.
Your last question "…simply converts KE to PE…"
is again using PE unnecessarily. Gravity does work on an
object as it falls because there is a force mg
acting down.

QUESTION:
Which trip takes longer when you throw a ball upwards,the trip down or the trip up?

ANSWER:
On the trip upwards, the air drag and the weight are both
down, whereas on the trip down the weight is down but the
air drag is up. So the speed decreases more rapidly going up
than the speed increases on the way going down. It follows
that the trip up will be shorter than the trip down.

This is the most popular way to explain the answer without
resorting to very opaque mathematics, but I find it just a bit
unsatisfying even though it is certainly correct. I found a more
convincing argument put forward by Eltjo
Paselhoff on the website
quora.com . Suppose we plot the velocity v as a
function of time t for the upward trip. At t =0
the magnitude of the slope must be larger than g
because the air drag is slowing the ball down as well. As the
ball gets higher, its speed decreases so the air drag gets
smaller so that finally at the top of its trajectory it is at
rest and the magnitude of the slope is equal to g . Now,
the ball falls back down and speeds up because air drag is
smaller than weight; eventually, though, the two forces become
close to equal and the speed becomes constant at the terminal
velocity v _{T} . One more thing you must
understand now is that the area under the velocity curve is the
distance traveled by the ball, so since up and down go the same
distance the green and red areas need to be the same. The only
way this can be is if the time to fall is larger than the time
to rise.

QUESTION:
In a tug of war, if a team exerts a force on the second team, then the second team exerts an equal and opposite force one the on the first.
Then how is it possible that one team wins?

ANSWER:
To simplify things, each team is one person. How does the
red-shirt guy exert a force on the blue-shirt guy? Via the
rope. So, the forces which the two exert on each other are
equal and opposite because of Newton's third law (F =- F
in the figure). Are there any other forces on the guys? Yes,
each is pushing against the ground using the friction
between their feet and the ground. I have drawn it so that
the guy in the red shirt is able to exert a bigger
frictional force on the ground than the guy in the blue
shirt, so the net force on everything is to the right—the
guy in the red shirt wins. You could think of this happening
because the red-shirt guy has stronger legs or because the
ground under the blue-shirt guy is slipperier.
(Incidentally, the lower case f s
are the forces which the ground exerts on each guy; not
shown are the equal and opposite forces which the guys exert
on the ground, Newton's third law again. Similarly, the
uppercase F s are the forces which
the rope exerts on each guy, not the forces which the guys
are exerting on the ropes. In other words, the only forces
shown are the forces on the guys, both of whom will
accelerate to the right.)

QUESTION:
I'm wondering why there is no drag in frequency response in an electromagnetic loudspeaker. Given a fixed input frequency, say 1KHz audio sine wave, given the friction of the air, the inherent mechanical resistance in the paper of a speaker cone and the constantly changing momentum of the vibrating cone, why isn't the frequency of the sound coming out lowered some amount by all those factors? This is something that has puzzled me for a long time. BTW, I have a B.S. in math and have taken college and univ. physics, so you can go into a little depth if you must.

ANSWER:
This is a classic intermediate-level classical mechanics
problem, the driven damped oscillator. Your having been a
math major should make it pretty straightforward to follow
the details. I found a very good
web site which gives the details of the
solution of the physics problem but is still quite readable.
For the casual reader, I will include here an overview.
Newton's second law is an inhomogeneous second-order
differential equation. What makes the equation inhomogeneous
is the presence of the driving force, the signal your
amplifier sends to the speaker; with no driving force, the
solution of the problem would be x _{t} (t )=A _{t} e^{-γt} sin(ωt+φ _{t} )
where ω is the natural frequency the speaker
would vibrate with on its own, γ is a parameter
which describes the damping of the speaker, and A _{t}
and φ _{t} are constants
determined by the initial conditions. If the
driving force is present, F=F _{0} cos(ω _{0} t +ψ ),
the solution is a linear combination of the what is called
the steady state solution , x _{s} (t )=A _{s} sin(ω _{0} t+φ _{s} ),
and the solution for the undriven oscillator x _{t} (t )
which is referred to as the transient solution: x _{t} (t )=A _{t} e^{-γt} sin(ωt+φ _{t} )+A _{s} sin(ω _{0} t+φ _{s} ).
The constants A _{t} and φ _{t}
are determined by the initial conditions (speed and position
of the speaker) and the constants A _{s} and
φ _{s } are determined by the driving
force (F _{0} and φ _{t} ).
Because of the damping term e^{-γt} ,
the transient term dies out in a very short time (assuming
the speaker has been sensibly designed) and you are left
only with the term which has the same frequency as the
output from the amplifier.

Below are shown a couple of examples: The
figure below shows the transient (red), steady-state (blue), and
total (black) solutions separately.

These show the total displacement for different initial
conditions. In both cases the speaker started at rest. The first
shows the starting position at zero, the second shows the
starting position at 0.5. The important thing is that they both
end up, after the transient has died, as the same.

QUESTION:
I have been searching for help/answers with no success, so here goes:
My dog competes in AKC dog diving distance jumping. Yesterday’s event my dog was jumping a distance of 21’, (basis being where the base of the dogs tail hits the water), with the height of the dock being 16” over the water.
However AKC rules state the dock height is supposed to be 24”. Seems logical to me that the increase in dock height would increase the distance to impact. If not, ok, and sorry for wasting your time. If so, can you tell me how you would calculate that, and what the increase would be?

ANSWER:
There are three important factors which determine how far
the dog will go: the height of the dock, the speed the dog
leaves the dock, and the angle relative to the horizontal of
his initial velocity. Of course, I have no idea what these
are but if the dog goes horizonatlly 21 ft I have to assume
that he has a running start, not from a standstill. Also,
you probably know that if the dock were level with the
water, the maximum distance for a given initial speed would
be maximized if the angle were 45^{0} . I found a
little
calculator you can play with; a screenshot is shown
below. All you would need to vary are the velocity, the
angle, and the height of the dock. I found that if I chose
the initial speed to be 26 ft/s (about 18 mph) the distance
the dog would go for a 45^{0} jump would be about 21
ft from a zero-height dock. Now, when I do the calculations
for the 24 in=2 ft and 16 in=1.33 ft, I find the best angles
and distances (for 26 ft/s) are (43.5^{0} , 22.9 ft)
and (42.5^{0} , 22.3 ft), respectively. All in all,
there is just about no difference.

Note that I have not included any air drag. I believe the speed
of the dog is low enough to make drag negligible.

QUESTION:
We know that looking at the stars is like looking at the past because we are seeing the light that left the star many years ago. However, the faster something travels, the slower time passes for that object. So, while for that light 1.000 years have passed since it left the star, for us, here on earth, many more years have gone by. My question is: when we see the light from a star 1.000 light years away, are we seeing 1.000 years in the past or actually, that number is a lot bigger because of relativity?

ANSWER:
No, that is wrong. Light emitted from a star 1 light year
away will arrive, using earth's clocks, in 1 year.

QUESTION:
Why does distance increase torque? I am aware of the equations, and how a lever behaves, but why does it behave that way? Why does me putting my force on a point farther away from the fulcrum multiply my force?

ANSWER:
I think it is easiest to think of a static situation where
the system is not experiencing any motion whatever. Consider
the little setup shown in the figure. A weightless stick of
length D is hinged to a wall. A force F
is applied down on the end of the stick; clearly this exerts
a force which, if there were no other forces acting (except
the hinge) would cause the stick to experience a clockwise
angular acceleration, in other words a torque around hinge.
Now suppose that we exerted a force T
a distance d from the hinge; we would find
experimentally that if T =F (D /d )
the stick would not rotate. We would conclude that the net
torque about the hinge is now zero and that this torque
changes linearly with the distance. Of course, the torque
exerted by F would have to be opposite that exerted by
T . If
we choose clockwise torques as positive, counterclockwise
torques would be negative. Newton's first law for rotational
equilibrium would be Στ=FD-Td= 0.

This is not the most general case but I feel you wanted to
be convinced that torque depends on where the force is
applied relative to an axis about which you are computing
torques, which was easy to see.

QUESTION:
Can someone please really answer the question about a 747 on a massive treadmill that is set up in a manner to match the speed of the wheels? This seems to be different then the Mythbuster's version and no one really just outright answers it from a physics position, and maybe because it is just simply it will take off.
However, with the weight of the plane, the friction of tires and wheel bearings and the treadmill's speed changing I can only assume two scenarios, the plane remains in place because the engines cannot produce enough thrust to overcome the friction generated by the wheels spinning at a ridiculous speed (and the reality is tires would blow and bearings would seize) or the wheels and treadmill spool up to infinity.
It just kinda sucks that no accredited organization has stepped up an answered this.

ANSWER:
I cannot, for the life of me, understand how this is
apparently such a big mystery. I have answered this twice
before,
first and
second times. In a nutshell, if there is not air moving
over the wings with sufficient speed, there is not
sufficient lift on the airplane for it to leave the ground.
I agree that Mythbusters'
version of this is not the same as I understand your
question and previous questions where whatever the conveyer
is doing, the plane is not moving forward through the air;
the Mythbusters have the conveyer at rest relative to the
plane but moving forward with the same speed as the plane,
so the wheels are not spinning.

I guess it is possible
that a propeller plane could take off from rest since the props
send a wash of air over the wings, but probably not unless there
were a pretty hefty headwind; you do see some planes, designed
for flying into and out of tight places, which take off in a
very short
distance.

FOLLOWUP QUESTION:
Regarding the 747 Jet on a rapidly moving treadmill, I can only find your answer unfathomable. The four General Electric CF6-80s 44,700 lbs of thrust each or 178800 lbs forward thrust total would push the plane forward off the treadmill almost instantly. It's difficult to believe that the friction of the wheel bearings would/could be enough to hold it back. If that were the case, the 747 would be forever grounded and could never take off, which we know full well to be false. Furthermore, lift is not a factor because lift does nothing to negate the 178,800 lbs of thrust pushing the plane forward. Wheel bearing friction isn't going to stop this monster from going ahead. You'll need well over 100,000 lbs of counter-thrust to do that. I seriously doubt that you can spool up a treadmill fast enough to lock up the wheel bearings before the CFC-80s were brought to bear and force the plane off the treadmill, especially given the monstrous size and weight of the treadmill and mechanism that would be required to support a 747, loaded or empty. You seem like one of those physicists who hasn't had a lot of real world experience with tremendous forces and loads.

ANSWER: My goodness, you do not need to get so upset
that you have to be
insulting! The problem here is that you and I are clearly
looking at different aspects of this problem. You are more
interested in the engineering aspects of whether it would be
possible and/or practical to build such a treadmill and
could the bearings in the wheels withstand the punishment
they would have to take as the wheels spun faster and
faster. I agree that it would be folly to try to build such
a contraption. The issue which interests me is if this
airplane does not move forward would it be possible for it
to become airborne? As I suggested in all my answers, the
treadmill is not necessary at all to discuss that question—just
lock the brakes, give the engines full power and ask what
happens. First, ask if you can get enough friction to hold
the plane. The weight of a 747 is about 800,000 lb and the
coefficient of static friction between rubber and dry
asphalt is about 0.8, so the maximum
frictional force you could get before the plane started
skidding forward would be 0.8x800,000=640,000 lb, far
greater than 178,800 lb of maximum thrust. So you could run
at full thrust and remain at rest by just locking the
brakes. So, now, your statement that "lift is not a factor"
is nonsense because if the plane is to fly there must be a
lift greater than its weight and I sure don't know where
that lift will come from if there is not air passing over
the wings. When you say that lift is not a factor, you are
assuming that the plane has broken away and is roaring down
the runway which then violates the premise of the whole
situation that the treadmill gizmo will work and ensure that
the plane does not move forward. One important thing which
you have gotten wrong is that it is the friction in the wheel
bearings which will hold the plane back. The friction
which would hold the plane back would again be static
friction between the tires and the treadmill; even though
the tires are spinning and the treadmill is moving, they are
still not slipping on each other and so it is still static
friction which would keep the plane from moving forward if
this thing were practicable to build. Also, if you were to
try to do this crazy thing, you would not just open your
throttle at the outset but gradually increase thrust while
the treadmill adjusts to match the wheel rotation; could you
imagine a treadmill which would keep the plane at rest if
the thrust were just right that the plane would move forward
at 20 mph if not on the treadmill? Although the wheels on
this (impractical to build) treadmill would probably end up
spinning faster than on normal takeoff, it would probably
not be destructive to them until you stayed there too long.

Let's just call a truce on this whole thing. A 747 at rest for
any reason is not going to take off. And practical problems with
the fabrication and utilization of this stupid treadmill will
ensure that it is never constructed or tested.

QUESTION:
In Star Wars there is the Executor class star dreadnought that is 19 kilometers long from bow to stern. And a few kilometers wide.
So I was wondering how large would it appear in our sky if it was at the low orbit of 100km altitude? As big as the moon does?
How about the Death Stars which were 120km and 160km in diameter?

ANSWER:
The principle you need here is that of subtended angle of an
object with length L a distance d from where you are viewing
it. You can see that the angle subtended by the moon is 31
arcminutes≈0.52^{0} . The angle is expressed in
radians as θ=L /d. To convert that to degrees,
multiply by 180/π , so θ= (L /d )x57.3
in degrees. For example, your 19 km dreadnought 100 km away
would have θ =10.9^{0} , about 21 times
larger than the moon.

QUESTION:
Why don't electrons get pulled into the nucleus? What forces prevent electrons from being pulled to the protons?

ANSWER:
See an earlier answer which answers
the same question except for the earth-moon and sun-earth
pairs. For your question it is the electrical force which
holds the electron in its orbit. There is an important
difference, though. The electron-proton system, if you look
at in classical electromagnetic theory, should radiate
energy (it is essentially a tiny antenna) and fall into the
proton. This does not happen; the reason is that for such
small systems classical physics does not work and this was
one of the milestones of physics when Niels Bohr proposed
his model
of the hydrogen atom where certain orbits simply did not
radiate. To understand the details of quantum mechanics,
which eventually evolved from the Bohr model, would require
a much longer explanation.

QUESTION:
Why don't the gigantic cruise ships at sea today simply flip over? It appears there is a lot more above the water than below.

ANSWER:
Because the heaviest things are at the bottom—engines,
fuel tanks, drinking water tanks, sewage tanks, ballast
tanks into which sea water may be pumped, etc . For
a more detailed explanation, see
this link .

QUESTION:
Why is angular momentum conserved for a charge in an electric field?
I don't know, I just can't see how the relation between distance and velocity could justify that.
It made sense in the gravitational field, since when a satellite gets closer it also gets faster. Now, if I have a stationary positive charge and a smaller positive charge in its field, the first charge will accelerate the other to repel it, so with the increasing distance of the second charge there's also an increase in velocity. positive charge and a smaller positive charge in its field, the first charge will accelerate the other to repel it, so with the increasing distance of the second charge there's also an increase in velocity.

ANSWER:
Angular momentum is not necessarily conserved in an electric
field, but the problem you are describing (which is
essentially
Rutherford scattering ) does result in the
angular momentum of the moving charge q about an axis passing
through the fixed charge is conserved. It is not hard to
understand. What is the condition for which angular momentum
is conserved? There must be no torques on q . The
expression for torque is τ =r xF =qr xE
where r is the position vector of
the moving charge and E is the
electric field due to the stationary charge. But,
r and E are
parallel so r xE =0
and there is therefore no torque on q and its
angular momentum will remain constant.

QUESTION:
If a mass is accelerated it will create gravitational wave.The earth is always accelerating as it moves around the sun. Does always creating gravitational wave? If yes, then where is this energy coming from?

ANSWER:
This exact problem has been worked out on
Wikepedia : "Gravitational waves carry energy away from their sources and, in the case of orbiting bodies, this is associated with an in-spiral or decrease in orbit. Imagine for example a simple system
of two masses—such as the Earth–Sun system—moving slowly compared to the speed of light in circular orbits. Assume that these two masses orbit each other in a circular orbit in the
x–y plane. To a good approximation, the masses follow simple Keplerian orbits. However, such an orbit represents a changing quadrupole moment. That is, the system will give off gravitational waves.
In theory, the loss of energy through gravitational radiation could eventually drop the Earth into the Sun. However, the total energy of the Earth orbiting the Sun (kinetic energy + gravitational potential energy) is about 1.14×10^{36} joules of which only 200 watts (joules per second) is lost through gravitational radiation, leading to a decay in the orbit by about 1×10^{−15} meters per day or roughly the diameter of a proton. At this rate, it would take the Earth approximately 1×10^{13} times more than the current age of the Universe to spiral onto the Sun. This estimate overlooks the decrease in
r over time, but the majority of the time the bodies are far apart and only radiating slowly, so the difference is unimportant in this example."
So the answer is that the energy comes from the orbiting
earth which causes its orbit to get smaller as it loses
energy; however, as the calculation shows, the loss of
radius of the orbit is trivially small.

QUESTION:
Assume a stationary satelite in outer space at a distance of 299,792,458*3,600 meters from the earth.
So basically the distance is the speed of light times 3600 seconds (or 1,079,252,848,800 meters) This satelite has a big lantern atached to itself that is pointing to the earth.
The lantern is initially turned off, but can be turned on by sending a "turn on" signal from the earth, through radio waves.
So, a "turn on" signal is sent from the earth.
We assume that the "turn on" signal will reach the satelite in 1 hour and will turn on the flashlight instantly.
So, when the signal reaches the satelite (after 1 hour) we send a rocket from earth that points to the direction of the light beam that is emitted from the flashlight.
The speed of the rocket relative to the earth is 10,000 m/s.
The question is: How much time will it take the rocket to see the light beam? Is it 1 hour? Or is it less? If it is less, does it mean that the light beam traveled faster than the speed of light relative to the rocket?

ANSWER:
So, using a clock on the earth, we will receive a light
pulse exactly two hours after we sent the first pulse toward
your satellite. You are thinking that the rocket will see
that pulse a little bit earlier because it will have met the
light before it gets to earth. That is not surprising, but
that is from the perspective of a clock on the earth. But
you have chosen a really tiny speed for the rocket—10,000
m/s=0.000033c where c is the speed of
light. In order to see the effect which I think interests
you, you need to choose a much larger much larger speed for
the rocket; I will choose v =0.8c , 80% the
speed of light. I will choose my unit of length to be one
light hour and my unit of speed to be light hours per hour,
so c =1, v =0.8, and L =1. First,
look at what an earth-based observer will measure with his
clock. If t is the time when the light
arrives at the rocket, then vt =0.8t =x and
ct=t =(L-x )=(1-x ); solving, x =0.8/1.8=0.444
and t=x /0.8=0.556. Also, note L-x =(1-x )=0.556.

But, what you asked for is how long a clock on the rocket
takes to see the light. The important thing is that the
distance between the earth and the satellite is not 1 light
hour any more, it is length contracted such that the
distance seen for L by the rocket is L'=L √[1-(v /c )^{2} ]=0.6
rather than 1. Of course the same factor occurs for x' =0.6x =0.267
and the time this happens is t'=x' /0.8=0.333
because even though the distance to the satellite is
shorter, it still zips past the rocket with a speed of 0.8c .
Finally, c' =(L'-x' )/t' =(0.6-0.267)/0.333=1!
So, you see, c really is the speed seen by all
observers.

QUESTION:
If I throw a bottle forward in a car as we are going 50 miles an hour what speed is the bottle going after it leaves my hand? Over 50 miles an hour or as fast as my hand can throw as if I wasn’t moving?

ANSWER:
So many times have I answered this question or its
equivalent! You cannot ask what a velocity is unless you
specify velocity relative to what . As a concrete
example, suppose that you throw the bottle with a velocity
of 20 mph relative to you (the speed you would throw it if
you were at rest on the ground). But you are are at rest
relative to the car, so you and anybody else in the car
would see it moving forward with a speed of 20 mph. But
someone by the roadside would see it moving forward with a
speed of 70 mph. Somebody in a car going in the same
direction as you with a speed of 70 mph would see the bottle drop
straight down to the ground (no horizontal velocity). If you
threw it backwards (relative to the direction of your car)
with a speed of 50 mph, someone by the roadside would see it
drop straight to the ground (see a
Mythbusters episode).

QUESTION:
How would one calculate the buoyancy force/pressure of a static fluid acting on an object entering said fluid from the side?

ANSWER:
I see no reason why Archimedes' principle would not work,
the buoyant force is equal to the weight of the displaced
fluid.

QUESTION:
If u had a grain of sand with the density of the universe, next to ur average gravel driveway rock would the grain of sand have a greater mass?

ANSWER:
The density of the universe is about 10^{-29} g/cm^{3} ,
so it isn't even close.

QUESTION:
Would the black hole in the image released today have appeared to be a circle, rather than an ellipse, regardless of the position from which it was viewed? And if so, why?

ANSWER:
Read the following question/answer to get a detailed
description of the experiment. The resolution achieved was,
in fact, much better than what was expected and there is no
way you could call this dark region anything other than a
circle or close to it. Certainly, if you viewed this object
edge on you would not have seen the dark area at all; if
your viewpoint were somewhere between edge on and straight
on, it would have appeared to be elliptical.

QUESTION:
About this new descovery I have several questions
https://www.bbc.com/news/science-environment-47873592
Why this cannot be done by one single telescope and how the group of telescope functions in order to work (I cannot think why it needed so many to be focused in a single point). Also the earth is rotating so most of them are not aligned with the point of focus.
Second question is why we only have an image? if we can take a photo, we can surely take more photos and create a video. In that way we can see the moving plasma or whatever is there.

ANSWER:
The first thing you need to understand to appreciate this
problem is that if you were much closer to the black hole
where you did not need a telescope; instead what you would
see would be a very bright and large source of light which
would hot have that black hole in its middle. So you cannot
use optical telescopes at all. The actual telescopes in the
experiment are all detecting photons in the millimeter
range, on the order of 10^{-3} m wavelength; this
wavelength is expected to originate only from the accretion
disk which forms around a black hole. Now a large optical
telescope has a size on the order of 10 m in diameter and detects
photons around 500 nm=5x10^{-7} m. Increasing the
size of a telescope gives it better resolution, so the size
S for a millimeter telescope comparable to the best
optical telescopes would be S =10^{-3} (10/5x10^{-7} )=20,000
m=10 km; but, looking at an object so far away (55 million
light years) would be impossible for visible light, but
since the earth has a diameter of about 13,000 km, we can
get very much better resolution with a large millimeter
telescope array by having the individual telescopes have
separations comparable to the size of the earth. Although the size is very
huge, the total area of all the radio telescopes is
extremely
small, so trying to make an image would be like blacking out
99.999% of the mirror in an optical telescope—you
would still get an image but you would have to wait a very
long time. So the data from all the telescopes in the array
were gathered over many hours during a 10 day observation
period. Then, all the data had to be put together using a
computer to reconstruct the image; the data analysis took
two years to complete, a very difficult problem.

Finally, you should ask what the dark circular region in the
center is—it is not the black hole itself which has
zero size. The outer edge is what is called the photon
radius, the distance from the black hole where a photon will
orbit. Somewhere inside of the photon radius is the event
horizon, the sphere inside of which nothing, including light
can escape.

A warning: as I clearly state on the site, I do not usually
answer complex questions in
astronomy/astrophysics/cosmology, so take what I say with a
grain of salt!

QUESTION:
How has the Bristlecone Pine been able to survive the Earth's pole shifts if they occur every 3900 Years?

ANSWER:
The oldest bristlecone pines have lived more than 5000
years. However, the current polarity has been unchanged for
about 780,000 years. The frequency of polarity changes are
notoriously and unpredictably variable, typically from
80,000 to 780,000 years over the last 5 million years. I can
still answer your question, I think, supposing that the
period were within the lifetime of the plant. Most life is
relatively insensitive to magnetic fields at all and would
not be significantly affected by changes in magnitude or
direction of ambient fields. I believe some birds navigate
using the earth's magnetic field and they would get messed
up and maybe fly north in wintertime were the field to flip.
There has been some speculation about a lack of magnetic
field resulting in higher radiation levels from cosmic rays,
but no link has been found between polarity changes and
extinctions.

QUESTION:
I am confused about the actual model of the atom. Some say that electrons move in circular paths about the nucleus.Other times its a wave and then
I read about orbitals with different shapes. What is the actual motion of the electron about the nucleus. I am a high school student.

ANSWER:
You need to understand a little bit of history of atomic
physics to fully appreciate the places of various models in
the overall scheme of things. First came Niels Bohr who
proposed a model which had electrons moving in circular
orbits; but not just any orbit but only those with
particular values of their angular momentum—the angular
momentum L was quantized, L=n ℏ where ℏ
is the rationalized Planck constant. This was actually
wrong, but it gave the right answer for the spectrum of
light observed from a hydrogen atom and that was a great
leap forward; but purely empirical. Improvements

and
extensions of this model were made, but they were all
equally empirical. Still, this first empirical model set
everyone on the right track to achieve a more fundamental
model. During this time, it was beginning to be understood
that trying to imagine a model based on classical ideas,
ideas like an electron orbiting in a circular orbit, were
invalid at such tiny scales—that realization ultimately gave
rise to the development of quantum mechanics. You can no
longer even think of an electron as a little point charge
when it is bound in an atom. Rather, you solve an equation,
known as the Schrodinger equation, which gives you a
mathematical function, called the wave function, which tells
you the likelihood of finding the electron at any particular
place in space. So the best way to think of an electron in
an atom is akin to picturing a cloud over which the electron
is smeared. The shape of this cloud is dependent on four
quantum numbers which quantize the energy, the angular
momentum, the projections of the angular momentum and the
spin angular momentum on some axis. Some examples are shown
in the figure. To visualize the "clouds", imagine rotating
each around a horizontal axis through the center; so, e.g. ,
the upper right cloud is donut-shaped.

QUESTION:
If I could travel away from a light source at the speed of light looking in the direction from which I came would the photons that hit my eye continue to be visible or bounce off and away from my eye at the speed of light in the opposite direction leaving nothing to see as the photons behind it would not reach me?

ANSWER:
If you had bothered to read the site
ground rules , you would have seen that "I no longer answer questions which are based on premises like:
'...if we could travel faster than the speed of light...' or
'...ignoring the fact that nothing can go faster than the speed of light...',
'...if I were traveling at the speed of light...', etc."
Suppose that you were traveling at 99.99% the speed of
light. The light which you would see if at rest would be
visible, but in your moving frame the wavelength would be
Doppler-shifted to a longer wavelength, into the region of
far infrared which your eye cannot detect.

QUESTION:
I am currently working on a school science project for what is the equivalent of High School in Norway. It is about trying to measure different materials ability to repel radioactive materials. We do this by pointing a radioactive source toward a geiger counter,
and sticking a materials to block between. All distances and thicknesses are fixed. Our results after some 100 different test, pose some problems. We were expecting to see that most Alpha radiation were completly blocked because our materials were 13mm. But the results show that alot of the materials, like cotton and cardboard, does not stop the radiation. In addition to this, the results form water showed that water, on all radiaion sources, were the worst at blocking radiation. We are having some trouble explaining these phenomena because they are very stable and clear. Do you have any answers to why the results are the way they are?

ANSWER:
There is no way the alpha radiation will get through 13 mm
of either cotton or cardboard. I am quite sure that the
reason the Geiger counter detects radiation in the anomalous
cases in your experiments is that any alpha source will also
have gamma radiation associated with it. For example, above
I show the decay scheme of ^{241} Am. In addition to
the four energies of alpha particles there are gamma rays of
energies up to about 100 keV. I cannot explain why water is
the least absorptive than any other material you tested.

QUESTION:
We say that a electron has +1/2 or -1/2 spin.What is this spin actually and how do we get the value?

ANSWER:
There are actually two answers on this page which answer
this question. The first is here , the
second is linked to from that answer.

QUESTION:
As a hot rod building enthusiast and garage mechanic I have a torque question that has never been adequately explained to me. Why does the definition of torque differ between a physicist and engineers\mathematicians?
Engineers and mathematicians say a physicist's definition of torque is what to them is the definition of a torque moment. To the engineer and mathematician torque is defined as a measurement of "the amount of change" between torque moments and when there is no change measured, torque is at equilibrium and equals zero. Wikipedia acknowledges this difference by disciple on their torque page but doesn't discuss why.
I've always accepted the reason the definitions differ has to do with the solutions needed for problems each discipline is trying to solve. If possible. I'd like to understand why the different definitions are needed with more substance than my assumption. What's never been adequately explained to me is what is it about the problems being solved by disciple that requires the different definitions of torque?

ANSWER:
It is mainly simply semantics. In physics, rotational
Newtonian physics uses definitions which end up with
physical laws which are wholly analogous to the laws in
translational Newtonian physics. Newton's second law is,
perhaps, the best example. In translational physics, force
equals mass (inertia) times acceleration, F =m a .
In rotational physics, moment of inertia I plays
the role of mass and angular acceleration α plays the roll of
a, and if torque (due to a force F applied a distance
r from
the axis about which the object rotates (or does not)) is
defined as τ =r x F ,
then τ =Iα .
I do not know much about engineering notation semantics but
if what you say is correct, that torque means that which is
zero if the object is not rotating at all (or with a
constant angular velocity), then that is what would be
called net torque in physics which seems to make
more sense to me. According to Wikepedia, mechanical
engineers (and also British physicists) refer to (physics)
torque as moment of force , not torque moment as you
state. Similarly, the definition by mathematicians is that
torque is equal to the rate of change of angular momentum
which is exactly the same as τ =Iα
and so torque, again contrary to your
statement, is the same as in physics. So, I read the
Wikepedia article differently from you and, to me, there is
no difference in anybody's definition of what I would call
torque other than semantics. And I read nothing there which
implies that the word torque means the sum of all moments of
force to an engineer. So my final answer is that everybody's
definitions of the physical quantities are the same, only
the words used to label them differ.

QUESTION: Why doesn't the moon fall to the earth?

QUESTION:
A six year old has just walked up to you and asked you why the Earth keeps spinning around the Sun but never goes out of orbit. After all, there's nothing that holds them together, right? What explanation do you give?

ANSWER:
There is something which holds them together—it's
called gravity. I think the best way to understand why the
earth does not fall into the sun, nor the moon fall into the
earth, nor any artificial satellite fall into the earth is
Newton's Mountain thought experiment. If you are on a very
high mountain and shoot a cannon horizontally, it will
eventually fall to the ground; as you give it more and more
speed, it will go farther and farther; eventually, with
enough speed, it will go all the way around, always falling
but never hitting the ground. Click on the picture to see
how this works.

QUESTION:
Two drops of water of the same size are falling through air with velocities of 10 cm/s.
If the drops combine to form a single drop, how do you find the terminal velocity?

ANSWER:
The things you need to know:

the volume of a sphere of radius R is 4πR ^{3} /3;

the cross sectional area A of a sphere of
radius R is πR ^{2} ;

the air drag force on an object having speed v
is proportional to Av ^{2} ; and

the terminal velocity U of an object of mass
m is
proportional to √(m /A ).

I
will use subscripts 1 and 2 to denote the original and final
drops, respectively; so m _{2} =2m _{1} and
R _{2} /R _{1} =2^{1/3} .
Therefore, A _{2} /A _{1} =2^{2/3} .
Finally, U _{2} /U _{1} =[(m _{2} /m _{1} )(A _{1} /A _{2} )]^{1/2} =[2⋅ 2^{-2/3} ]^{1/2} =2^{1/6} .
So, finally, U _{2} =10⋅ 2^{1/6} =11.22
cm/s.

QUESTION:
Am interested in calculating the B field a certain perpendicular distance from a large stream of moving like charged particles.

ANSWER:
The questioner had considerable confusion regarding how to
do this, so I have not included most of his question. After
a couple of communications with the questioner I believe
that the question asks for the magnetic field for a long straight
circular cross section wire of radius R with the current
I distributed uniformly throughout the wire. The
figure shows the field due to the wire, the vector
B having a constant magnitude at a
distance r from the axis of the cylinder defined by
the wire and always tangential to the circle of radius r
around the wire. Hence field lines, as shown are circles,
the direction defined by the right-hand rule as shown. The
magnitude of the field is determined by Ampere's law,

∫B ⋅ ds =μ _{0} I _{enc}

where the integration
is all the way around the path defined by the circle and
I _{enc} is the amount of current which passes
through the area defined by that closed path. Given the
cylindrical symmetry of this special case, the integral is
simple because B is constant and B
and ds are, everywhere on the
integration path, parallel,
∫B ⋅ ds =2πrB=μ _{0} I
so B_{r>R} =μ _{0} I /(2πr )
for r>R . This is only correct outside the wire
itself because the integration path to determine B
inside the wire encloses less current. Inside, I _{enc} =I (r ^{2} /R ^{2} )
so B_{r<R} =μ _{0} Ir /(2πR ^{2} )
for r<R. Note that when r=R both results
give the same value of B , B =μ _{0} I /(2πR ).
So, the field increases linearly with r inside the
wire and decreases like 1/r outside.

The questioner also
wanted to express this in terms of the velocity
v of electrons with charge Q=-e and having a density of
n
electrons/m^{3} . The fact that the electrons have
negative charge means that, in the figure above, the
direction of v is opposite of the
direction of current flow. In a time Δt the
electrons go a distance ΔL and so sweeps out a
volume ΔV which contains a total charge ΔQ =-ne ΔV =-ne ΔL (πR ^{2} )
and therefore I =ΔQ /Δt=-ne (ΔL /Δt )(πR ^{2} )=-nevπR ^{2} .

FOLLOWUP QUESTION:
Thanks for the answer. Am not sure but am thinking you missed the fundamental part of my question, I am looking for the B field adjacent a large stream of like charges in free space, not in a wire. That I know. I also however think I figured out how to do it this morning using the formulas I provided earlier.

ANSWER:
My answer was exactly the same as if there were many uniformly distributed point charges moving with speed
v in a cylindrical beam of radius R . The second part of my answer showed you how to relate
I in the first part of the answer to the charge, speed, density of charges, and
R . So the quantity you seek results from setting
I=nQvπR ^{2} into
B=μ _{0} I /(2πR ) (assuming that "adjacent" means right on the edge of the moving charge distribution).

SECOND FOLLOWUP QUESTION:
I am trying to compare the "B" fields associated with two moving electrons in free space, both travelling in the same direction and at the same velocity.

In the first case the two electrons are very close together and are moving to the right at say 50 m/s, what would be the
"B" field be 50 mm directly above, i.e. in the positive y axis above these electrons?

In the second case, same two electrons traveling to the right at 50 m/s but this time they are separated. Call the first electron,
"e1"and the second "e2".
The first electron e1 is located directly 1 mm below the point of interest where I want to calculate the combined
"B" field. The second electron e2 is located directly 98mm below e1 or 99 mm below the point of interest where I want to determine the combined
"B" field. In effect from the first case I have moved e1 closer to
"B" by 49 mm and e2 farther away by 49 mm.

I was using the formula B=μ _{0} NqV /(4πR ^{2} ) where
μ _{0} is permeability of free space, N is the number of charges, q is charge on an electron and V is the velocity of the electrons, R is the vertical distance between the charges and the point of interest for
"B" the strength of the magnetic field.

ANSWER:
The equation you are using for the field, a special case of
the Biot-Savart law, will work for the specific case you
state. It is important that you understand that this gives
only the magnitude of B and not
its direction. Also the position vector R
of the point where you calculate B
must be perpendicular to V . The
correct form of the Biot-Savart law is B=μ _{0} Nq V xR /(4πR ^{3} );
note that the magnitude of V xR is
VR sinθ where θ is the angle between
V and R . So, for
θ =90^{0} , your equation is correct. Be
careful, though, because the direction matters when you are
adding vectors from two sources; the case you are looking at
has the point the same side of both electrons so the
magnitude is the sum of the magnitudes, but if you look at a
point between the electrons, the magnitude will be the
magnitude of the difference. The form of Ampere's law which
I used in the earlier answer is simply not applicable in
this situation because making reference to a current I only
makes sense in magnetostatics. Finally, I should note that
all this is only approximately correct when you are in a
nonrelativistic regime which you certainly are for V =50 m/s.

QUESTION:
Regarding the twin paradox. I read an example where the traveling twin aged 6 years while the earth bound twin aged 10 years. Since a year is 365 earth rotations, 10 years would represent 3650 rotations, while 6 years much fewer. But if they are both counting the days by watching the earth rotate, their counts must be the same when they meet, I would think. Will they both count 3650 or the 2190 days for 6 years? As a corollary, whichever count they get, they would also count the same number of birthdays during the trip.

ANSWER:
You have fallen for the trap of assuming that there exists
some "real clock" which tells you the "real time", in your case
the rotation of the earth. I hope you have already read my
explanation of the
twin
paradox (I have attached here the figure from that post for
your convenience) which shows how the traveling twin "sees" how the
earth clock runs by having the earth-bound twin send him one
message per year; this is different from your situation where
essentially a daily rather than an annual message is sent, but
the ideas are identical. When he arrives home, the traveling
twin has "seen" the same number of years (days) on earth pass as
the earth-bound twin has but, as you can see, a different number
of years (days) have passed on his on-board clocks. "Watching"
some clock in another frame of reference is not in any way
relevant to time in your frame; the traveling twin would count
his birthdays using his clock, not earth's clock. Note also that
the traveling twin "sees" the earth spinning very slowly on the
trip out but spinning very rapidly on the trip home.

QUESTION:
My 9 yo son come up with this idea that there is a limit at how hot can something get becase heat are particles in movement and nothing can move faster than the speed of light hence the limit to how hot can something get.
Either if he is correct or not could you tell us why?

ANSWER:
This is a good question. In fact, even though there is a
limit on the speed that a particle can have, there is no
limit on the energy it can have. The temperature of
something depends on the average kinetic energy of the
particles, not their average speed. The kinetic energy in
Newtonian physics is ½mv ^{2} ,
so it is natural to think that the maximum possible kinetic
energy would be ½mc ^{2} . But
Newtonian physics is not valid at speeds close to c ,
and in relativity the kinetic energy is ½mv ^{2} /√[1-(v ^{2} /c ^{2} )].
To learn more, see an
earlier answer to the same question; you might also look
at the answer following that one.

QUESTION:
How is quantum spin defined? and how did scientists get the idea of this?

ANSWER:
See an earlier answer . Spin is nothing more than a
specification of the angular momentum of a system; e.g. ,
the spin quantum number of an electron, proton, neutron, and
other so-called fermions is s =½. The actual
angular momentum is given by S =ℏ√(s (s +1))
where ℏ is the rationalized Planck's constant. Usually, we think of spin
as the intrinsic angular momentum of an elementary particle
or entire quantum system like an atom; particles may also
have orbital angular momentum, for example an electron
orbiting around a nucleus in an atom. But each angular
momentum will be quantized (have a specific quantum number)
and the total angular momentum J of any angular
momentum number j will be J =ℏ√(j (j +1)).
Orbital angular momentum quantum numbers are integers rather
than half integers like electrons; so, an electron in a
"p-orbit" in a Bohr atom has an orbital quantum number 1 and
a spin quantum number ½. As to where the idea came from,
there were several things which suggested it. The experiment
which showed that spin of an electron was the
Stern-Gerlach experiment , but there were also hints from
spectroscopy and, notably, that there were twice as many
states in quantum systems (like atoms) as expected from the
Pauli exclusion principle .

QUESTION:
What direction was the earth rotating during the ice age? And is the north pole now on a different axis?

ANSWER:
The earth's axis is pretty-well fixed relative to the earth
itself. However this axis, in absolute space, precesses much
as a spinning top does. The earth's axis is inclined by
about 23.4^{0} relative to the normal of the plane
of its orbit but that axis precesses around the
perpendicular axis about once every 26,000 years. This
precession is caused by torques exerted on the earth by the
sun and the moon. Currently the axis is pointing almost
exactly to the the north star, Polaris; however, in a few
thousand years there will be some new star taking that role.
The last ice age was about 2.4 million years ago and lasted
much longer than 26,000 years, so the answer to your
question is that it pointed many directions during the ice
age. You are asking about the geographic north pole which is
essentially constant in its location (although the land
masses move over the surface because of plate techtonics).
The gravitational north pole, though, is forever wandering
around and has actually jumped to the southern hemisphere
around Antactica and back many times.

QUESTION:
I’ve been having a discussion with a friend of mine, a professor of philosophy at the Autonomous University in Mexico, for the past two months about aesthetics. I had just finished reading Phillip Ball’s new book about quantum physics as well as Sabine Hossenfelder’s critique of
"beauty" as a criterion for judging theories in physics. I told my friend that I thought
"quantum mechanics is an art form". He asked me - as good philosophers do - why I thought so. And that has led me on a happy quest into philosophy as well as what physicists have to say about what they do. But I have not read a physicist saying that they thought they were creating
"art". So I thought I’d ask a real physicist. If what you do can be both
"beautiful" and "elegant", then do you believe you are creating "art"?

ANSWER:

My third
book , PHYSICS IS … The
Physicist Explores Attributes of Physics , has a chapter
titled Physics Is Beautiful , devoted to beauty and
elegance. However, I have never thought of it as art. A
sunrise over the Sangre de Christo mountains is beautiful
but you would certainly not call it art. I found that it is
impossible to find an unambiguous or not vague definition of
art; art is qualitative which sets it apart from science
which is quantitative. And, consider this: There are many
paintings of the of Chartre Cathedral, four of which I have
copied here, and all are indisputedly art; there is no
"correct" painting of this subject. The theory of
electromagnetism, on the other hand, is beautiful and the
only one of its kind; you might think up another
electromagnetic theory, but it would be wrong even though it
might be considered very imaginative, creative, even
beautiful. There is no well-defined test which specifies
which painting is the best; there is a test as to which is
the theory of electromagnetism —it must
quantitatively describe what really happens in nature and if
it fails to do so, it is false. I have always thought of
string theory as not a theory (and not beautiful) because it
makes no attempt to quantitatively interface with the real
world and is therefore not falsifiable.
QUESTION:
What is the distance to stop a truck weighing 7200 lbs. at 45 mph ?

ANSWER:
That depends on what the composition and condition of
surface on which the truck is moving and on the composition
of its wheels. I will choose a truck with rubber tires on
level dry asphalt and that the truck has antilock brakes.
The maximum force which can be exerted by the friction on
the tires is approximately F=- μ_{s} N
(negative because the force is opposite the velocity. where
μ_{s} is the coefficient of static
friction between rubber and dry asphalt which is about
0.5-0.8, and N is the normal force of the truck on
the road which, on level ground, is the weight; I will
choose μ_{s} = 0.65, halfway between the
extremes. At this point I will switch to SI units which
scientests prefer: weight N =7,200 lb=32,027 N, mass
M =32,027/9.8=3,268 kg, and speed v =45 mph=20.1
m/s. Now, using Newton's second law, F=Ma =- μ_{s} N ;
so acceleration a =-6.37 m/s^{2} . Now, you
can compute the time to stop: v =0=v _{0} +at =20.1-6.37t,
so t=3.16 s. Finally, the distance traveled is d =v _{0} t +½at ^{2} =20.1x3.16-½x6.37x3.16^{2} =31.7
m=104 ft. It is interesting that this answer is independent
of how heavy the truck is. The reason is that the
acceleration is proportional to the weight/mass which is
just 9.8 m/s^{2} .

QUESTION:
How a rubber ball and an iron ball with same mass hits on the same ground with same force but the rubber ball gets more bounce without violating Newton's third law of motion?

ANSWER:
What makes you think that they experience "the same force"?
The force which the ground exerts on the ball is determined
by what the mass of the ball is, how long the collision
lasts, what the nature of the gound and ball are, and how
fast the ball was moving when it first encountered the
ground. But all that is beside the point if what your
question is concerned with Newton's third law. Whatever
upward force from the ground which a ball experiences during
the collision, the ground is experiencing an equal and
opposite force down from the ball. The iron ball may not
bounce at all, but these two forces are always equal and
opposite. The rubber ball might rebound with a speed half
the speed it came in with, but, during the time the ball and
the ground are in contact, Newton's third law is still
obeyed.

QUESTION:
Would an object that weighs 4.5 ounces going at 2,500 feet per second kill a person? I'm asking because my friend asked how fast you have to throw a beanie baby at someone in order to kill them, and I'm slightly concerned for them but I want to know

ANSWER:
Scientists prefer to use SI units, so I will switch and then
convert back when I have finished" 4.5 oz=0.128 kg, 2500
ft/s=762 m/s. What matters is, when it hits you how much
force do your experience over the time that the collision
was occuring. Suppose that the beanie baby flattened by
about ½ inch or 1.27 cm=0.0127 m before it came to
rest. If you assume that the object comes to rest by having
a uniform deceleration over a half inch, you can show that
time to stop is about 3x10^{-5} s and the
acceleration is about 2.3x10^{7} m/s^{2} .
The corresponding force to stop the beanie baby is
F=ma=.0128x2.3x10^{7} ≈3x10^{5} N≈67,000
lb. I think that if the beany baby hit you anywhere except
possibly your limbs, you would be killed. Keep in mind that
this is a very rough calculation, but certainly in the right
ballpark.

QUESTION:
We know if V= 4πr ^{3} /3 is
the volume of a sphere and if it is increasing with a constant rate dV /dt=C ,
then C =4πr ^{2} dr /dt .
Solving, dr /dt=C /(4πr ^{2} ) which is velocity at which the
surface is expanding.
So the acceleration of the surface is (d^{2} r /dt ^{2} )=-[2C /(4πr ^{2} )]dr /dt=- 2C ^{2} / [(4π )^{2} r ^{5} )] or, combining all constants,
kr ^{-5} . I just don't see if the radius of the
sphere is expanding at a rate of Cr ^{-2} , how it is accelerating at kr ^{-5} ?

ANSWER:
This is your question which I have edited to make it more
clear what you have done. I see no errors in your
differentiations or integrations. So, your constant k is k =- 2C ^{2} / (4π )^{2} .
I think that the problem you are having is that you have
carried the minus sign into your constant. But that minus
sign is important because it tells you (since everything
else in k is constant) that the radius of the surface is
increasing, but the rate at which it is increasing is
decreasing as r changes. So, if the volume of
a sphere is increasing with a constant rate, its radius is
increasing like 1/r ^{2} , but the rate at
which the speed of the surface is decreasing goes
like 1/r ^{5} .

QUESTION:
Can you tell me why/how this snow is holding together and
curling backwards after sliding off a metal roof? Facebook
pic mystery.

ANSWER:
This is not really so rare, at least the sliding off the
roof part. You can see, in the other picture, my son at his
home in Vermont with a similar (not curled) situation. The
roof is warmed from inside and the snow slides down. If
conditions are right, the wet snow on the bottom will
refreeze and hold it from breaking off and also refreezing
to halt the slide. I would guess that the curling would
result from repeated cycles of sliding, pausing, sagging
repeatedly. I cannot really see why the end of the slab
would be being pushed back toward the house.

QUESTION:
When you let a lead ball and an aluminium ball of the same radii to free fall from a height of 100 metres which will ground first?

ANSWER:
Assuming that the balls are not hollow, the lead ball will
have more mass. At a given speed, both balls will experience
the same drag force. But that force will have a bigger effect on
the ball with the smaller mass (Newton's second law). So the
lead ball will reach the ground first. If you are interested
in more detail, a recent answer
is very similar to yours.

QUESTION:
I am a man of some girth, 490 lbs to be exact. I also enjoy "competing" in 5K race events.
As I struggle to maintain a 3 mph pace, Cutler, skinnier people zip past me unphased and undaunted by hills, heat, etc. I have often pondered but do not have the requisite knowledge to figure the answers.
My question is how much force is required to propel a body, weighing 390 lbs, at 3.1 mph. Compare that to someone at 130 and 180...or share the formula.
Also, how can I calculate KCAL consumed at different external temps and humidity.

ANSWER:
To figure out the force is very difficult because running is
a series of accelerations (when you push off with one foot)
and decelerations (when the other foot hits the ground). So
your speed is constantly speeding up and slowing down as you
run and the forces which the ground exerts on you is
constantly changing in both magnitude and direction. It is
daunting to try to think about such a calculation. But, I
believe what you really want is energy expended over a
certain distance in a certain time. I found a handy little
calculator which you can look at
here . For example, for your current weight, a 5K race at
3.1 mph (which is about equal to 5 km/hr, so it takes one
hour for you to complete), I find "You burned 1151 calories
on your 5 kilometer run. At that pace, your calorie burn
rate is 230 calories per mile and 1151 calories per hour."
Sorry, I could not find any information on temperature and
humidity.

QUESTION:
The total amount of energy is always same in the universe. But if there were two observer who were travelling at different velocity,would their measurement of total energy differ?

ANSWER:
Kinetic energy is not invariant, different observers see
different values. Imagine a universe which is entirely at
rest and you are at rest in that universe; the universe has
no kinetic energy. Now imagine that you move through that
universe with some velocity v ; you
now observe that universe to have a kinetic energy of ½Mv ^{2} .
where M is the mass of the universe except for your
mass. And just a minute ago that energy was not there.
Obviously, the total energy of that universe cannot be the
same in all frames. But, if you stay in one frame and keep
track of the total energy of the universe in your frame, it
will not change.

QUESTION:
if a person dove off the back of a boat going forward at 35mph would that person actually be pulled backwards? I say no he would just de accelerate

ANSWER:
The answer would be easy if you dived from helicopter moving
35 mph —you would have a forward velocity of 35
mph and the water would be at rest, so you would experience
a drag force opposite your motion, backwards, which would
cause you to slow down (decelerate) in the horizontal
direction. But, hydrodynamics can be very complicated, and
it is quite possible that the water behind the boat is
moving with the boat; I cannot imagine, though, that it
would be moving faster than the boat (and you), so it would
not pull you toward the boat.

QUESTION:
Is the pull of gravity the same for every part of Earth? (32 feet per second?)

ANSWER:
The number you are stating is not the "pull of gravity" but
rather the acceleration due to gravity. And, the units are
ft/s/s, not ft/s. But it is an indication of how hard
gravity is pulling, so I can go ahead and answer your
question. We (physicists) usually prefer to work in SI
units, meters instead of feet; in SI units, g ≈9.8
m/s^{2} . The acceleration due to gravitation does
vary at various places on the earth. The animation shows you
how it varies. The variation is small, no larger than 50
milligals; a gal is 1 cm/s^{2} =0.01 m/s^{2} ,
so 50 milligal=0.005 m/s^{2} . So, compared to 9.8
m/s^{2} , the variations are only on the order of 0.05%. There
are several reasons for gravitational variations, notably:

altitude (the farther you are from the
center of the earth, the weaker the gravity);

local geology (large volumes of more
dense earth will increase the gravity);

the earth is
not a perfect sphere.

If you want a lot
more detail, see the
Wikepedia article on the earth's gravity.

QUESTION:
when a stone is thrown up with a velocity and it collides with a particle in mid air(the particle being at rest during collision) , in this situation , is vertical momentum conserved?

ANSWER:
Linear momentum is conserved only if there are no external
forces. But there is an external force acting, the force of
gravity, so the vertical component will not be conserved for
a "real" collision by which I mean one which lasts for some
nonzero time. But, you have stated that the struck particle
remains at rest during the collision; the only way this can
be true is if the collision is instantaneous. So the
momentum will be conserved because the collision lasts for
zero time. The reason this is not a physically possible
collision is that it requires the struck particle to
instantaneously acquire its velocity which requires an
infinite force. Nevertheless, these kind of problems with an
external force present are often approximated as having
momentum conserved if the collision happens very quickly.
Another example is shooting a bullet into a block of wood
sitting on a horizontal surface which has friction; only if
the collision happens very quickly will momentum be
approximately conserved.

QUESTION:
Newton's third law of motion says that for every action, there is an equal and opposite reaction. If i throw a ball towards the wall with some Force it will not Bounce back to me with the same force. Why is that? I mean like if i have a bowling ball and I threw it Towards the ground it will not bounce back. This is driving me crazy

ANSWER:
You are thinking of Newton's third law incorrectly. It
applies only to forces during the collisions, it does not
tell you how fast the rebound will be. When the ball hits
the wall it exerts a force on the wall, and the wall exerts
an equal and opposite force on the ball. But knowing the
force on the ball does not tell you how fast the ball will
be going after the collision. What happens is that some
energy is lost by the ball during its collision so it comes
off the wall more slowly than it went into it. The bowling
ball is an example of the same thing except it loses all its
energy when it collides with the ground. These are called
inelastic collisions, the bowling ball is a perfectly
inelastic collision. If the ball were to leave the wall with
exactly the same speed it came in with, it would be called
an elastic collision.

QUESTION:
What the heck is spin? Quantum objects don’t literally have angular momentum, do they? Does the concept of spin have any relationship to angular momentum? Or is it just a name like “color” that isn’t descriptive of what’s happening?

ANSWER:
They do "literally have angular momentum". So, how can we
observe experimentally whether an object has an angular
momentum? One way is that if you can find a way to exert a
torque on an object which has angular momentum, the torque
will cause the angular momentum to change. For example, a
spinning top has angular momentum and if the top does not
have its symmetry axis vertical, there will be a torque
because its own weight exerts a force which makes it want to
fall down. But, because it has angular momentum, it does not
fall but instead precesses about the vertical. Now, we know
that an electron has an intrinsic angular momentum, a spin.
Presumably because of this spin the electron also has a
magnetic dipole moment —it looks like a tiny bar
magnet. So when we put a magnetic dipole in a magnetic
field, there is a torque exerted on it which tries to line
it up with the magnetic field. This torque, because the
electron has angular momentum, causes the dipole to precess
just like the top did. Without going into detail, it is
possible to infer this precession from measurements. What is
not possible, though, is to understand this intrinsic
angular momentum classically. For example, if you try to
make some classical model imagining a little sphere with
some reasonable radius and mass of an electron spinning on
an axis, you will get nonsensical results like the surface
of the sphere having a speed greater than the speed of
light.

QUESTION:
If a space vehicle left earth at 18000 mph in a direct line to the sun, how fast would vehicle accelerate to taking gravitational pull into consideration?

ANSWER:
The escape velocity from the earth's surface is about 25,000
mph, so your vehicle would fall back to earth.

QUESTION:
If you built a perfect sphere in space that was 187,000 miles wide, the inside was a perfectly polished mirror, and you fired a Laser inside for 1 second, how long would that beam of light bounce around?
Provided the sphere was dust free and a vacuum inside.
This question has bothered me for 35 years.

ANSWER:
You have chosen the distance to be about the distance which
the light would move in 1 second. I cannot imagine the
reflectance of the sphere would be greater than 99%, so if
you solve the equation 0.99^{N} =0.001, N would be the number of
seconds it would take for the intensity of the light to drop
to 0.1% of its original value. Using
Wolfram Alpha , I
find N ≈687, a little more than 11 minutes.

QUESTION:
If two objects that weigh the same are attached by a string will they sink at the rate of combined mass or will they sink equally?

ANSWER:
Since you refer to "sink" I will assume that they are
falling in water. There is no simple answer to this question
because the drag force on a sinking object depends on its
geometry. For example, suppose one object is a big flat
sheet and the other is a long thin needle suspended below
the sheet which behaves like a parachute; clearly, this will
sink with a much smaller speed than if both objects were
long thin needles. The most appropriate approximation of the
magnitude of the drag force f in water is a
quadratic
function of the velocity v , f= ½ρ_{w} C_{D} Av ^{2} , where
ρ_{w} = 10^{3}
kg/m^{3} is the density of water, A
is the area presented to the onrushing water as the
object falls, and C_{D} is a constant which depends on the
geometry of the object, called the drag coefficient. For a sphere of radius R
the drag coefficient is given by C_{D} =0.47 . The upward force f is
only one force on the object as it sinks; it also has a
weight down W=mg and a buoyant force up B=Vρ_{w} g
where V =4πR ^{3} /3
is the volume of the sphere. If the sphere is
solid and has a density ρ=m /V , we can
write B=mg (ρ_{w} /ρ ).
Also note that we can write R as R =^{3} √[3m /(4πρ )].
Finally, we can write the total force on a sphere: F =[mg (ρ_{w} -ρ )/ρ ]+½ρ_{w} C_{D} Av ^{2} .
Note that when a speed is reached where F =0, the
sphere will fall with a constant velocity called the
terminal velocity, v_{t} =√{[mg (ρ-ρ_{w} )/ρ ]/[½ρ_{w} C_{D} A ]} .

So, let's do a specific example to be able to answer your
question for a particular situation. Suppose we have two 1
kg balls of solid iron for which ρ= 7.87x10^{3}
kg/m^{3} . Then the radius of each ball will be R =0.0312
m. Doing the arithmetic, F =-8.56+0.72v ^{2} .
So each of two balls of mass 1 kg, tied together will
accelerate until their speeds would be v_{t} = √(8.56/0.72)=3.45
m/s. On the other hand, if we have one sphere of mass 2 kg,
its radius will be 0.0393 m. So, F =-17.1+1.14v ^{2} .
So one ball of mass 2 kg will accelerate until
v_{t} =√(17.1/1.14)=3.87 m/s.

You could do more examples to come to the conclusion that
the answer to your question is that, in general, two spheres
tied together will not sink at the same rate as one sphere
with their combined mass. I am sure, by carefully tailoring
the shapes of the objects, you could force the rates be the
same, but that is not you would find in general.

QUESTION:
If we go inside a mine and drop a 10 kg iron ball and 1kg aluminium ball from the top of a high platform. what will happen?

ANSWER:
I do not understand the point of going inside a mine. I will
assume that the atmospheric pressure and acceleration due to
gravity are the same as at sea level. In that case the air
drag f can be approximated as f ≈ ¼Av ^{2}
where A is the area presented to the onrushing air
and v is the speed. (You will note that Av ^{2}
is not the same dimensionally as a force; this approximation
only works if you work in SI units, A in m^{2} ,
v in m/s and f in N.) So the net force
F on each ball (which I assume to be solid) is F =¼Av ^{2} -mg
where m is its mass; and, by Newtons second
law, the acceleration a is a =(¼Av ^{2} /m )-g ;
and there is a velocity U , called the terminal
velocity, at which the drag equals the weight so the
acceleration is zero and U= 2√(mg /A ).

Iron and aluminum have different densities, but it is pretty
straightforward to compute the radii of each

ball and from that compute A for each. I find A _{Fe} =0.014
m^{2} and A _{Al} =0.0064 m^{2} .
The accelerations are a _{Fe} =-(9.8-3.5x10^{-4} v ^{2} )
and a _{Al} =-(9.8-1.6x10^{-3} v ^{2} ).
At any given speed the magnitude of the acceleration for the
iron ball is larger than for the aluminum ball. The terminal
velocities would be U _{Fe} =170 m/s and
U _{Al} =78 m/s. The iron ball always has a
larger speed than the aluminum ball and will hit the ground
first.

It is also enlightening to find the distance traveled as a
function of time, s= (U ^{2} /g )ln(cosh(gt /U )).
This may beyond your math ability (ln is a natural logarithm
and cosh is a hyperbolic cosine), but the graph shows the
behavior of each ball and clearly the iron ball is falling
faster. Note that for the first few seconds (6 or 7 s) the
balls pretty much keep pace with each other and after about
15 s both have nearly reached their respecitve terminal
velocities because the curves are nearly linear.

QUESTION:
Would the weight of any object become less as it approached the center of the earth? Neglecting the obvious environmental difficulties of survival could i fall thru a tunnel bored completely thru the axis of the earth and slowly decelerate to the center of the earth. Possibly oscillating back and forth near the center and finally come to rest in the weightless middle?

ANSWER:
Do you want the Physics 101 answer or the answer based on
our best understanding of the density distribution of the
earth? I will give you both.

When this kind of problem is
presented as an exercise in elementary physics, it is
usually assumed that the density of the earth is uniform
throughout—a cubic meter of earth from the surface has
the same mass as a cubic meter at the center. In this case,
it is easy to show that the acceleration due to gravity,
g (r ), is linear: g (r )=9.8(r /R )
m/s^{2} where r is the distance from the
center and R is the radius of the earth. Since the
acceleration is always toward the center, you speed up until
you reach the center (not decelerate as you suggest) and
then slow down until you reach the other end, neglecting any
frictional forces. More realistically there will be at least
air drag which will get bigger as you speed up until you
eventually reach a constant speed; then at the center, you
start slowing down until you stop somewhere before the other
end. Then you speed up back to the center getting there with
a smaller speed than the last time you were at the center,
etc. eventually stopping at the center as
you
suggest. One thing you might note is that, for the
frictionless case, your motion is exactly as if you were
attached to a spring with spring constant k =9.8/R
N/m.

But, the earth is not a uniform
sphere. The core is much denser than the mantle and
therefore the acceleration is not a nice linear function.
The figure shows that the acceleration is almost constant
until about one third of the way to the center. The above
qualitative description of your motion remains the same
except for the similarity to the simple spring.

ADDED
THOUGHT:
I never answered your first question, regarding the weight
of the object. In the uniform density earth the weight (mg )
decreases linearly to zero at the center. In the realistic
model, the weight remains constant for a while, then increases
slightly, then falls almost linearly to zero at the center.
QUESTION:
I am a professional tennis coach and would like to explain the relationship between mass and spin of a tennis ball. So in my words, I experience hitting the ball with allot of spin makes the ball heavier. When a fast flat ball is hit, I feel it is faster but on contact lighter than the spin ball. Does this mean the increase in rpm, will increase the mass exerted on the inside of the ball, thus making it heavier? And is there a formula for this?

ANSWER:
I am surprised, if you are a professional coach, that you do
not tell me what kind of spin the ball has. The two most
common spins are top spin and back spin (the reversed
direction of top spin); there is also apparently something
called slice spin.

First, though, let me
address your reference to the ball being "heavier" or
"lighter" depending on the spin on the ball. This is totally
incorrect as the ball always has the same weight, a force
which points vertically downward. The way that the ball
moves results from the forces which the ball experiences
after it leaves your racquet; there is the weight straight
down, the force the ground exerts on it when it hits, and
the forces which the air exerts on it. The figure above
shows a ball with top spin; the top of the ball is spinning
in the same direction as the ball is moving to the right.
(If you are confused, the diagram is shown in the frame of
the ball, so the air is shown like a wind blowing in the
opposite direction as the ball is moving.) Here the ball is
experiencing the force of its weight straight down (not
shown), a drag force to the left slowing it down (not
shown), and an aerodynamic force called the magnus force
F straight down. If the ball were
not spinning, there would be no magnus force; if the ball
had back spin, the magnus force would be straight up; if the
ball had slice spin, the magnus force would act
horizontally. (These are all also important for a pitched
baseball.) I am guessing that the ball you judged as
"heavier" is the top spin case because, due to F ,
the ball will drop faster than a nonspinning ball. I am
guessing you call the nonspinning ball "flat" and, indeed,
it will not fall as fast as the ball with top spin. A ball
with back spin would fall even less fast than the
nonspinning ball, so you would call it even lighter. (In
fact, I believe that the magnus force on a ping pong
ball with sufficient backspin, can actually be larger than
the weight of the ball.) But in all cases the ball has
exactly the same weight. (I also note that an actually
heavier ball, in the total absense of air, would move
identically to a lighter ball because the acceleration due
to gravity is the same for all masses.)

For
completeness I would like to talk about what happens when
the ball hits the ground because these are strategically
important in tennis. Shown in the second figure are the
forces (in red) which the ground exerts on the ball in top
and back spin cases. For clarity, I have not shown the
forces due to the weight or the air. In each case there is a
normal force N
upwards which is what lifts the ball back in the air; but
there is also a frictional force
f which slows down the spin
in both cases. In the the top spin case, the friction both
slows down the spin and speeds up the ball so it comes up
faster and lower than a nonspinning ball would; in the the
back spin case, the friction both slows down the spin and
slows down the ball so it comes up slower and higher than a
nonspinning ball would. If the spin were large enough, the
back spin ball could actually bounce straight up or even
backwards!

QUESTION:
When an electron absorbs photon does its mass increase?

ANSWER:
I guess you are thinking about the photoelectric effect. A
photon strikes an atom and disappears. An electron is
ejected from the atom. You are thinking that the photon is
totally absorbed somehow by the electron. But what really
happens is that the energy, hf , which the photon
had is given to the electron; the electron ends up with a
kinetic energy equal to hf minus the
binding energy it had in the atom. The rest mass of the
electron remains unchanged.

QUESTION:
"If a ladder you are standing on begins to fall...hold on... it will reach the ground slower than a falling body."
Is the above advise true?

ANSWER:
One could work out an analytic solution to this problem, but
that is not necessary if you just want to be convinced that
the advice is good. All the forces on you are your own
weight, straight down, and the force the ladder exerts on
you. The vertical force which the ladder exerts is upward,
so the net force in the vertical direction is smaller than
if you were just falling.

QUESTION:
Is there anything that can block the effects or decrease the effect of gravity? Any experiments that tested it? Magnetism can be blocked and why not gravity?

ANSWER:
Both magnetic and electric fields can be manipulated to be
zero, but that is much more difficult for gravitational
fields. The main reason is that there are two kinds of
electric charge but only one kind of mass. That is not to
say it is impossible. For example, there is a point between
the earth and the moon where the field is zero, where the
attraction to the earth is equal and opposite the attraction
to the moon. But, in order to have any effect on the field
near the earth, you need to manipulate giangantic masses. As
far as we know, there is no such thing as "antigravity".

QUESTION:
When I carry my cell phone (iPhone 7) in my left pocket and it vibrates, I feel the vibe on my right hand, while my left hand feels nothing. It happens quite often and I even check to see inside my right pocket (or in my tote bag on my right shoulder) I fear of having misplaced my cell there. What does the trick?

ANSWER:
This is not really so surprising. The brain is divided into
two halves and there is some crossing of nerve impulses
sometimes. I used to hear and feel a thumping in my right
ear when I tapped my left big toe. It seems to not do that
anymore.

QUESTION:
If you take out the spaces between all the atoms of the nuclei in your body, would you end up half the size of a flea but still weigh the same?

ANSWER:
Technically, there are no spaces between atoms in your body. I think what you are asking is that if we compressed all the matter in your body (as happens in a neutron star)
to nuclear density, how big would it be? Suppose your body
has a mass of 100 kg. The density of nuclear matter is about
2x10^{17} kg/m^{3} , so your volume would be
about 10^{2} /2x10^{17} =5x10^{-16} m^{3} .
So, if you are a cube, your dimensions would be ^{3} √5x10^{-16} =8x10^{-6}
m ≈10 microns. A flea has a size of about 1
mm=1000 microns, one hundred times bigger than the
compressed you. And, yes, you would still have a mass of 100
kg.

QUESTION:
If we are looking back in time through telescopes now at stars and galaxies, would it be possible to calculate the distance required away from earth in order to look back and see a moment in history here on earth, theoretically ?

ANSWER:
Of course, but you could never get there in time to see it.
An alien who was 5000 light years from earth would see, if
he had a good enough telescope, the earth as it was 5000
years before he was looking.

QUESTION:
Since even light cannot escape the gravitational pull of a black hole, does this mean that the theoretical particle, known as the graviton, have the potential to move faster than the photon?

ANSWER:
Sorry this took so long, but I had to consult with an
astrophysicist. The best answer he could come up with is at
this
link .

QUESTION:
Attached is a picture culled from the internet on an object rolling down a plane. I understand everything except how to explain and draw the forces to my AP physics students. Should this look like a box sliding down the plane? with mgsintheta down the plane and mgcostheta as the normal force and mg down and friction up? Or are some of these properly represented as torques?
Could you please draw how this should be properly represented?

ANSWER:
I have not included your picture because there is too much
stuff on it (the velocity, the R , the R ω ,
etc . all distract from the crux of the problem, in
my opinion). I have redrawn the problem showing some axially
symmetric object of radius R , mass m rolling down
an incline of angle θ without
slipping. There are only three forces acting on the object,
the frictional force f and the
normal force N , both of which act
at the point or line of contact, and the weight
mg which acts at the center of mass of the
object. There are three unknowns, N ,
f , and a , the
acceleration of the center of mass. To solve for these
unknowns, we must apply Newton's laws, both translational
and rotational (ΣF =ma
and Στ =Iα );
here I is the moment of inertial about the axis
about which torque is calculated and α
is the angular acceleration about that axis.
You are a teacher, so I will skip a lot of the details which
you can fill in easily. From ΣF =ma
I find N=mg cosθ and ma =mg sinθ
-f. The first equation nails N , one of our
unknowns, the second is one equation for the two remaining
unknowns, so we are not done. So the only place to get a
second equation is the rotational form of Newton's second
law; I think this is the trickiest part of this problem. I find many students
have an inclination to choose the symmetry axis to sum
torques; they have been told, when they learned statics of
non point objects, that you can choose any axis you like,
but this is not a statics problem and there is a very
important constraint—Newton's laws are not valid
in an accelerating system , and the center of the
rolling object is accelerating. (See correction below!) So Στ =Iα
yields Iα=Ia /R=mgR sinθ ,
or a= (mR ^{2} /I )g sinθ ;
knowing a , then f =m (g sinθ -a )
But, what is I ? It isn't what you look up in a book
because usually what is listed is the moment of inertia
about an axis passing through the center of mass (I_{cm} ).
Here the axis is a distance R from the center of
mass, so using the parallel axis theorem, I=I_{cm} +mR ^{2} .

I will do one example, a solid sphere where I_{cm} =2mR ^{2} /5,
so I =7mR ^{2} /5, so a =5g sinθ /7
and f =2mg sinθ /7. If you sum torques
about the center of mass, you get the wrong answer.

CORRECTION:
OK, I slipped up here, albeit in a fairly minor way. While
it is true that Newton's laws are not correct in an
accelerating frame of reference, there is one exception: the
sum of the torques about an axis passing through the center
of mass is indeed I_{cm} α .
A little detail I had forgotten about classical mechanics.
Let me work that out for my example: I_{cm} α= 2maR /5=fR ,
so, using the result above, f =2ma /5=m (g sinθ-a )
or a =5g sinθ /7. Then it
easily follows that f =2mg sinθ /7.
But you should try to sum torques about an axis halfway
between the point of contact and the center of mass—you
will not get the right answer.

Nevertheless, it is important that your students be careful
in choosing the axis. I have edited the original answer
somewhat.

QUESTION:
How massive is the largest black hole ever discovered?

ANSWER:
See the Wikepedia article List of
Most Massive Black Holes . This is the most
up-to-date information I could find.

QUESTION:
Does the core of the earth spin? If is does how does it spin faster than the mantel with the viscosity of magma and the 4+ billions years old the earth is? I just don't understand what I am being told.

ANSWER:
The core of the earth is composed mainly of iron and is
comprised of an inner core which is solid and an outer core
which liquid. Apparently, from what I understand from an
article I found, recent research indicates that the
outer core rotates with the same speed as the mantle (once a
day) but the inner core's rotational speed is slightly
faster by about 0.3-0.5 degrees per year.

QUESTION:
If I dropped a circular, triangular, and square parachute from the same height which will land first assuming the parachuted object's weight is the same and the surface area is the same for all three parachutes?

ANSWER:
I am assuming that your parachutes are flat, horizontal
plates and that drag forces due to the load and lines
attaching the load to the parachute are negligible or at
least the same. The drag coefficients C_{D}
for the three shapes, determined experimentally, are shown
in the figure. The drag force is given by F_{D} = ½ ρv ^{2} C_{D} A
where ρ is the density of the air, v
is the speed of descent in still air, and A is the
area. Therefore the drag force is least on the circular
parachute so it reaches the ground first. (A proviso is that
the values of the drag coefficient depend on a quantity
called the Reynolds number which depends on the velocity.
However, the values given are for low Reynolds numbers which
are suitable for parachutes.)

QUESTION:
How did Einstein get the idea that the speed of light must be constant for all observer?

ANSWER:
Einstein's original
paper on special relativity, On the
Electrodynamics of Moving Bodies , states two main
problems in electromagnetic theory at the end of the 19th
century:

The first
paragraph states, essentially, that Maxwell's equations
will not have the same form in a frame of reference
moving with constant velocity relative to a frame where
they are the laws governing electromagnetism if you
transform them using Galilean transformations.*

The
beginning of the second paragraph acknowledges "…the unsuccessful attempts to discover
any motion of the earth relatively to the 'light medium' [luminiferous æther,
e.g. the
Michaelson-Morley experiment]…"

Einstein had a
strongly-held philosophical belief that, for something to be a
true law of physics, it must be the same in all reference frames—regardless
of where you are or how you are moving, the laws of physics for
you are the same as for anybody else in the universe; this is
the principle of relativity . Since he believed
Maxwell's equations to be laws of physics, he concluded that
Galilean relativity was not correct and this led to his
postulate that the speed of light was independent of the motion
of the source or the observer. To my own mind, I believe that
the constancy of the speed of light is not a necessary
postulate, rather a
consequence of the principle of relativity applied to
Maxwell's equations.

*A Galilean transformation is, for
example, observing a car moving 60 mph east relative to the
ground to be moving 100 mph east if you are in a car moving 40
mph west relative to the ground.

QUESTION:
What type of impact will cause the metal in the car to "splash" (eliminating all hope of a simple conservation of momentum calculation) should the New Horizons interplanetary space probe hit your car?

ANSWER:
The only thing which prevents linear momentum from being
conserved is an external force. If there are no forces other
than those exerted by the car and the probe (or pieces
thereof), linear momentum is conserved. For example, if the
collision happens in empty space with the probe of mass
M moving with velocity V in a
frame where the car is at rest, the total momentum of all
the pieces will be MV after the
collision or any time during the collision. What will not be
conserved is the kinetic energy ½MV ^{2} .

QUESTION:
In a recent answer , you mention the Cosmic Microwave Background and say that it "provides a frame of reference for the whole universe".
It was my understanding that one of the underpinnings of the basic theory of relativity is that there CANNOT be such a universal frame of reference (UFOR). I assume you're not claiming Relativity has been disproven (well, shown to be a special case of something else), so can you explain the difference between the frame of reference meant for relativity and the CMB as you described it?

ANSWER:
Good question! Special relativity rests on the assumption
that the laws of physics must be the same in all inertial
frames (the principle of relativity); no frames are ruled
out as long as they are inertial. In other words, no one
frame is better than all the rest. But suppose that we find
a frame of reference which contains a "gas of photons" which
are all moving in random directions (average velocity of all
these photons is zero) and this "gas" pervades the whole
universe. Then, if our frame happens to be moving with some
velocity v through this gas, we
will see that gas having an average velocity -v
which would be determined by looking at the doppler shift of
the photons; this is analogous to feeling a wind on your
face when you run in still air. Having found such a frame
allows us to specify an "absolute velocity" which, before
now was not possible. What makes this frame special is the
fact that, because it is a remnant of the big bang, it does
pervade the whole universe.

QUESTION:
I was curiouse at which height would a 6kg turkey have to fall for it to be perfectly cooked when it hits the ground?

ANSWER:
It is very difficult to do a quantitative calculation
because the air drag (which would cause heating) changes
with altitude as the air gets less and less dense. I doubt
that there is any height where this could work because,
giving it a high enough speed to compensate for the short
time (keep in mind that it takes several hours to roast a
turkey at 350^{0} F) of fall would result in the
outside being burnt and the inside being raw. It might be
fun to do a couple of very rough calculations just to get an
order of magnitude feeling for what you are interested in.

I will approximate the turkey to have the properties of
water, in particular that its specific heat is C =4.186
J/gram⋅^{0} C; so the energy which will raise
the temperature of a M =6000 gm turkey is E=MC ΔT =6000x4.186x56=1.41x10^{6}
J. (70^{0} F→170^{0} F is 21^{0} C→77^{0} C,
hence ΔT =56^{0} C.)
The change in potential energy in falling from a height
h is V=Mgh =6x9.8h= 59h , so if the turkey falls
with constant velocity the potential energy lost goes into
heat. If the heat is 1.41x10^{6} J, the height is
h =1.41x10^{6} /59=24,000 m. The atmospheric
pressure at this altitude is only about 3% what it is at sea
level; so it is clear that, if your turkey is given any
velocity at this altitude, it will not continue moving with
constant velocity. (Whereas, if the pressure were
atmospheric and you gave the turkey a speed equal to the
terminal velocity it would continue moving with that speed
as it fell.) If you dropped it from very far away
(essentially infinitely far), it would arrive at the upper
atmosphere going nearly the escape velocity, 11,200 m/s. At
this speed, which is faster than the reentry speed of the
shuttle which needed special ceramic tiles to shield against
burning up, the turkey would be burned to a crisp, probably
would not reach ground before totally burnt. Although there
would be some speed of entry to the atmosphere where the
turkey would absorb 1.41x10^{6}
J, it would almost certainly not be "perfectly cooked".

QUESTION:
In a nuclear fusion reaction hydrogen atoms combine to create a helium atom. But in a hydrogen atom there are no neutrons,then where does the neutrons in helium nucleus come from ?

ANSWER:
There is more than one answer to this question. We can look
first about how hydrogen (with no neutrons, just one proton)
becomes helium (two protons, two neutrons) in a star like
our sun. As shown by the figure, it begins with pairs of
protons which interact and fuse in a reaction which results
in one deuteron (heavy hydrogen, one proton and one
neutron), one positron (a positively charged electron), and
one neutrino. The deuterons thus created now fuse with
another proton to create a

light isotope of helium, He (one neutron, two protons), and also a gamma
ray. Two of these ^{3} He now fuse, throwing off two
protons in the process and one alpha particle (^{4} He,
two protons and two neutrons). The two thrown-off protons
are now free to continue the process, so this chain, called
the proton-proton cycle, is self-sustaining until the star
runs out of hydrogen which ends the life of the star.

For more earth-bound hydrogen fusion, like fusion reactors
or hydrogen bombs, single-step reactions are sought. The
most frequently used is the deuterium-tritium reaction,
involving the fusion of one deuteron and one triton (^{3} H,
one proton and two neutrons) shown in the other figure.

QUESTION:
If time ticks at 1 second per second in a 1g gravity at the earths surface. Then my question is would a clock tick twice as fast 2 second per second if the earth surface was 0.5g.

ANSWER:

Most certainly not! The expression for the time t
elapsed in the gravitational field of a nonrotating
spherically symmetric mass M is t=t _{0} / √(1- δ )
where δ= 2GM /(Rc ^{2} )=2aR /c ^{2} ;
here, R is the distance from the center, t _{0}
is the time elapsed when R =∞, G
is the universal gravitational constant, M is the
mass of the sphere, and c is the speed of light,
and a is the local acceleration due to gravity. If a=g
and R =6.4x10^{6} m (earth radius), it
is easy to calculate that δ= 1.39x10^{-9} .
Now, t _{g} /t _{½g} =[√(1- ½δ )]/[√(1- δ )];
using the binomial expansion
(1 +δ )^{x} ≈1+xδ+ …
if δ<< 1, we finally get t _{½g} ≈t_{g} (1+¼δ )=t_{g} (1+3.5x10^{-10} ),
quite a bit different from t _{½g} =2t_{g} !

QUESTION:
Why does steam rise from the cooking pot when I turn off the heat after the water starts boiling?

ANSWER:
Water evaporates from the surface and the higher the
temperature, the faster it evaporates. The higher the rate
of evaporation, the more likely it is that individual
evaporated molecules will coalesce with others to form
droplets of water large enough to be seen. That is what
steam is and what clouds are. But steam can be observed even
at low temperatures under the right conditions, for example
the morning mist on the surface of a lake or fog. The water
in the cooking pot, if you observe it carefully, will begin
to steam before it starts boiling. So it should be no
surprise to find that it still steams after you turn off the
heat.

QUESTION:
what makes my radiometer occasionally accelerate wildly when bumped? I can't duplicate the action but have seen it a few times. It has happened around a desk lamp with an LED bulb. Could it be part of an electrical emission around the radiometer? I was allays taught that the photons are what push the vanes

ANSWER:
Well, as it turns out, photon pressure is not what makes the
radiometer turn. You can learn about radiation pressure by
reading an
old
answer and the links it contains. I do not know what
"accelerate wildly" means in this context but, since you
mentioned "bumped", mechanical bumping probably caused
unexpected behavior of the radiometer.

QUESTION:
I wanted to ask a question regarding airfoils. When looking at videos online about them, the instructors say that airfoils work because the top of the wing turns air down due to the coanda effect and the downturned air also turns the air under the wing down to create lift. I think I understand the law applied here to be with every action there is an equal and opposite reaction however, with the explanation they gave, it seemed like they were saying that the air on the top that was turned down was the only thing that is turning the air under the wing down. I that true? I would assume that, even though that does happen, the lift generated from the bottom of the actual wing deflecting the air down was more and more important than the coanda effect on the top. The video also said that disrupting the flow on top of the wing through too high of angle of attack would cause it to stall, but wouldn't the actual reason being that the bottom of the wing converts too much lift into drag by changing the opposing force that is lifting it from mostly up to mostly back? There was another source that said that there was a vortex at the back of the wing that pushed the air at the bottom of the wing so that the net force would cancel out a bit, slowing the air down and making a pressure difference to lift the wing. What is that about?

ANSWER:
This question violates the site
groundrule for "single, concise, well-focused questions". I can tell you that what is learned in elementary physics courses that airplanes get lift from the Bernouli effect of lower pressure due to higher velocity of air on the top of a wing is a relatively small part of how an airplane is able to fly. As you seem to refer to in your question, the main thing is that air leaves the wing with a downward component
of its velocity; this means that the wing exerted a force down on that air which means, via Newton's third law, that the air exerted a force up on the airplane. To quote Wolfgang Langewiesche, author of
Stick and Rudder ,
the private pilot's bible on the art of flying, "The main fact of all heavier-than-air flight is this:
the wing keeps the airplane up by pushing the air down ."

QUESTION:
How many light-years or parsecs away from 2018 is 1961? The solar system, the earth and the galaxy would have been in a different part of the universe. Do we know the direction of where our galaxy, solar system used to be before we arrived here in the present moment? How many miles has the Earth traveled in space in fives minutes? Including the movement of the solar system and galaxy moving through the vacuum? Millions of parsecs? Billions?

ANSWER:
It never makes sense to talk about velocity of something
without specifying the frame of reference relative to which
the velocity is measured. And when you are talking about the
earth, the motion is very complicated because it rotates
about its center (about 0.5 km/s), revolves around the sun
(about 30 km/s), and the solar system revolves around the
center of the Milky Way galaxy (about 220 km/s). What about
the motion of the center of the galaxy? Until recently, you
could only measure that speed relative to another object
outside our galaxy, e.g. , another galaxy or the
center of our local cluster of galaxies. However, the
discovery of the cosmic microwave background (CMB) has now
given us a reference frame which seems to define the
reference frame of the whole universe. Very careful
measurements and data analyses from the satellite COBE have
shown that our galaxy moves with a speed of about 580 km/s
relative to the CMB.

So, what can we do to answer your question which is,
essentially, how far does the earth move in 57 yr=1.8x10^{9}
s? (Your 5 minute question is essentially impossible to
answer since it depends on the time of year. Over years the
orbital motion averages out to zero.) Relative to the center
of the galaxy, the distance is D =220x1.8x10^{9}
km= 4x10^{11} km=0.013 pc. The center of the galaxy
in that time will move 580x1.8x10^{9} km=10^{12}
km=0.032 pc. But I do not know the relative directions of
these two velocities, the sum of which would be the average
velocity of the solar system over these 57 years; this speed
would be anything between 360 and 800 km/s. However, in my
researching this question I found that the speed of sun
relative to the CMB is about 370 km/s, so the final answer
is that the earth moved about 370x1.8x10^{11} =6.7x10^{11}
km=0.022 pc. It is interesting to me that, at this time, the
sun and the center of the galaxy are moving in nearly
opposite directions (370 km/s compared to 360 km/s); since
the time for the sun to go all the way around the galaxy is
about 225 million years, in about 110 million years our
speed relative to the CMB will be about 800 km/s.

Keep in mind that I am not an
astronomer/astrophysicist/cosmologist! I have gathered these
data as best I can.

QUESTION:
I have a potentially stupid question about pressure in a vacuum. So I just found out that in a vacuum chamber, microns can be used for measuring vacuum pressure because it's representative of a DISTANCE- the distance being the displacement of columns of Mercury. In understanding Leak Up Rate tests in a vacuum furnace, the Leak Up Rate is a function of Difference of Vacuum Levels/Elapsed time. My question is, does the displacement of Mercury happen faster with more time? IE, does the displacement of Mercury gain momentum with extended periods of time under vacuum? Or, IE, is it exponential instead of linear?

ANSWER:
Very likely there is not a column of mercury being used to
measure pressure in your chamber. There is probably some
mechanical device using strain gauges which is calibrated in
mm or μm of Hg. So, you do not have to worry about what a
column of mercury is going to do as the pressure changes. 1 μm
(micron of Hg)=0.133322 Pa (N/m^{2} )=1.93368x10^{-5}
PSI=1.31579x10^{-6} atm.

QUESTION:
After a math class yesterday (homeschool) , my son and I were discussing an article I read a few days earlier regarding a exploded star that went super nova about 150 million years ago. The event bathed the earth in a stream of high energy cosmic rays for a couple of months and possibly caused an extinction level event that effected ocean life of the day.

As the conversation progressed, we considered the implications and began to wonder if it was possible to shield against cosmic radiation? To the best of my knowledge, I explained that no substance I know of can block cosmic radiation and that the Earth's magnetic field would have little effect either. The only effects I know of are when they impact atoms in the atmosphere or the earth. If I understand this, if it doesn't hit something fairly dense, it just keeps going, understandable given the space between atoms.

I have read about the gravitational lensing effects on light by massive objects like neutron stars and galaxies that bend light around them. I have also read about the possibility of warping space as a method space travel by changing the shape of space in front of and behind a craft so that it "falls" in a particular direction.

While we were discussing these effects, he looked at me and said something to the effect, "Maybe we could bend space and make them go around?". To me, this was a pretty profound leap in my book. So I decided to ask physicist if he could be right.

Please forgive the length of this post before asking our question. I wanted to provide sufficient context before asking. So, here it is. Is it possible to make cosmic rays go around something, either by gravitational or microwave warping of space?

B.T.W. I really really enjoy your web site. It is awesome!

ANSWER:
I am not a paleontologist, so I had to do a little research
on mass extinctions. I discovered that there is no known
mass extinction 150 million years ago. In addition, no known
extinction is associated with a nearby supernova. (The
timeline here is clickable for an enlarged picture.) I would
conclude that the article which you read is a "fringe
theory" of mass extinctions, not a mainstream fact.

But, I can speak to your main question and comment on some
of your statements.
Cosmic
radiation refers to a broad range of radiations but
mostly protons and light nuclei; but also it could refer to
neutrinos, electrons, positrons, antiprotons, gamma rays,
x-rays, and others. What is not the case, though, is that "no
substance…block cosmic radiation"; just about any
cosmic radiation can be blocked by a dense solid or liquid.
Neutrinos are the exception and most pass all the way
through the earth without any interactions. You may know
many neutrino experiments are performed in deep abandoned
mines to block out all other radiation. The atmosphere also
is very effective since a single energetic proton will
strike an atom and many other particles, all with less
energy than the original proton, will form a "shower" of
particles. And magnetic fields are not very effective
deflecting the very highest energy charged particles but are
known to be an effective shield for lower energy particles.

Finally, let's address
the question of warping of space as a way to deflect the
rays. The only way to warp space that I know of is to have
an enormous mass nearby. The earth already warps the space
around it which would cause a particle passing close by, but
not on a collision course with earth, to be deflected a tiny
bit (viz. gravity). But putting a star sized object near
enough the earth to "shield" us would obviously be a
catastrophic thing to do! The whole solar system would be
disrupted. I have no idea what "microwave warping of space"
means.

I commend your son for his hypothesizing and you for your
tutoring and encouragement of his curiosity.

QUESTION:
I understand that LED light won't refract a full spectrum of colors through a prism. Is this true? I want to combine a grouping of different wave length diodes to deliver a white light that will separate from indigo to red. Can this be done?

ANSWER:
Although a simple LED will not be monochromatic like a
laser, it will tend to have a fairly narrow range of
wavelengths emitted, not a full spectrum. The most common to
make a white LED is to take a blue or near-UV LED and coat
it with a material called a phosphor. The phosphor will
absorb some of the light from the LED and reemit that energy
in a broader spectrum creating white light. The details of
the spectrum depend on the phosphor used; varying the
phosphor used, for example, can create the whites described
as cool white or warm white or bright white, two of these
are shown in the figure. I do not understand what "separate
from indigo to red" means.

QUESTION:
As I understand it, electricity and magnetism are the same force; this is where "electromagnetism" comes from. I also understand that radio waves, light, and non-particle radiation like x-rays are all the same thing; waves in the magnetic field. However, they're carried by different particles: electrons for electricity, photons for magnetism (radio, light, etc). Furthermore, there is a significant difference between electrons and photons; electrons have mass, photons do not.
Wouldn't this mean that electricity and magnetism are actually separate forces which are so tightly coupled (have a very strong influence on eachother), that it's just easier or simpler to think of them as the same force most of the time? One example of this coupling in my hypothesis is how electrons gain energy when a photon collides with and is absorbed by an electron, and the reverse, when an electron emits a photon and looses energy.

ANSWER:
You have some serious misconceptions here. First, the waves
you refer to are not "waves in the magnetic field", they are
waves which are composed of magnetic
and electric fields —they
are electromagnetic waves. Second, both electric and
magnetic forces are "carried by" photons—electrons
never play that role. Indeed, there is only one field, the
electromagnetic field; for example, if you have a charged
particle at rest you will only see an electric field, but if
you give the charge some velocity you will now also see a
magnetic field and you will find that the electric field is
somewhat smaller which you could interpret as having
"morphed" into the magnetic field.

QUESTION:
I want to reach 90%
c in 30 days. How many Gs would I have to accelerate at to reach that velocity in the given time? (in a vacuum)

ANSWER:
You should read an
earlier answer to understand the answer
here. The graph here is from that earlier answer and shows
(in red) v /c as a function of a _{0} t /c
where a _{0 } is the acceleration measured in
the frame of the object which is accelerating and t
is the time since the object was at rest. Reading off the
graph, you can see that v /c =0.9 when
a _{0} t /c= 2. Now, t =30
days=2.6x10^{6} s, so a _{0} t /c=a_{0} (2.6x10^{6} /3x10^{8} )=8.64x10^{-3} a _{0} =2;
solving, a _{0} =232 m/s^{2} =23.6 Gs. I would not
suggest having anybody aboard this object!

QUESTION:
Everything being equal,including speed and mass, Like a man walking up a slope, does the amount of energy expended on say a 10% inclined slope double with a 20% inclined slope?, if traveling at the same speed and everything else the same?
In other words, is energy expended related arithmetically and constant according to slope of incline. and resistance, or is it proportional to the slope by some other factor, geometric increase.?

ANSWER:
In the figure the man of mass m is walking up a
hill with incline θ with a constant
speed of v . His potential energy is E=mgy
where y is his height above the bottom of the hill
and g= 9.8 m/s^{2} is the acceleration due
to gravity. The rate at which his energy is changing is dE /dt=P =mg (dy /dt )=mgv_{y} =mgv sinθ ;
so this rate (called power) depends only on the vertical,
not the horizontal component of his velocity. Now, you have
made things difficult for me because you have not specified
the angle, but rather the inclination which is actually the
tangent of the angle (rise/run) times 100%. I will call this
inclination the grade G =100tanθ. For
example, let's take a man with mass 100 kg walking with a
speed of 2 m/s up the 10% and 20% inclines you specify. The
angles would be θ _{10%} = tan^{-1} (10/100)=5.71^{0}
and θ _{20%} = tan^{-1} (20/100)=11.31^{0} .
So the power values are P _{10%} = 100x9.8x2xsin(5.71)=195.0
Watts and P _{20%} = 100x9.8x2xsin(11.31)= 384.4
Watts. The ratio of these two is 1.97, close to, but not
equal to, 2.0; for small angles the sine and tangent are
approximately linear functions but as the angle (hence
inclination or grade) increases this is no longer a good
approximation. The graph shows the power for grades up to
200% (about 63.4^{0} ); as you can see, the power
increases approximately linearly only up to about a grade of
about 30% which is about 17^{0} .

QUESTION:
A vehicle , unknown speed of travel, strikes a 14lb object and displaces it
29ft. Could you please tell me what the approximate speed of travel is and
how you come to that conclusion. I ask because a vehicle deliberately veered
to the opposite side of the street in order to run over my cat. The driver
was successful in killing my sweet lad and I am trying to collect as many
facts as I can before I request help in prosecuting the individual under
Arizona's Animal cruelty laws.

ANSWER:
I am afraid that you cannot deduce the speed of the vehicle from the
information you gave me. You need to know how elastic the collision was and
what the trajectory of the cat was.

FOLLOWUP QUESTION:
Thank you for responding to my question but I am at a loss as to what is meant by "elastic". As far as trajectory my cat was launched a distance of 29 ft from point of impact with the next point where he died on the street...… no evidence, ie; of blood in between with there being blood evidence only at the point of impact and final resting location. If it would help I can shoot a video of the street and the location of where Liam was struck and where he landed in relation to the surroundings.

ANSWER:
Elastic refers to the energy lost in the collision which
gives you information on how fast the cat was moving
immediately after the collision. The extremes are a
perfectly inelastic collision where the cat would be stuck
to the vehicle after the collision and perfectly elastic
which would have the cat moving with approximately twice the
speed of the car after the collision. Clearly, neither of
these would describe a car hitting a cat and I cannot
imagine how you could approximate it.

When I say "trajectory"
I refer to the path from collision point to the ground. The
horizontal distance gone from the point of collision to the
point of landing depends on the angle at which the cat
started; if launched at 45^{0} relative to the
horizontal he will go much farther than if launched at 20^{0} .
Again, unless you witnessed the accident, there is no way to
know the angle.

QUESTION:
If you are traveling on a spaceship close to the speed of light (pick a number e.g. 0.95c), With or without constant acceleration whichever works for the question.
Time will slow down relative to earth and you will be able to travel a greater distance than you would be able to without relativistic effects. You would also be able to return to earth with significant time passed relative to your experience.
My question is, what would it feel like on the spaceship...I know the standard answer is you wouldn't feel anything, but logically, if you were traveling through galaxy after galaxy in a relativistic time of human life then intuition says everything would look like it whizzing past.
Basically, what is the human experience of being able to travel interstellar distances?

ANSWER:
Of course
everything is whizzing past compared with what it would look
like if you were at rest relative to the earth. What you
would also see is that the distance to an object you were
moving toward is shorter by a factor of √(1-0.95^{2} )=0.31;
so if you went to a star 1 light year from earth, it would
only take you 0.31/0.95=0.33 years to get there but the
earth-bound clock would say it took you 1/0.95=1.05 years.
Also where the stars appear to be relative to where they
would be if you were not moving would be different; see an
earlier
answer . In the case where v /c =0.95, the graph shows
how the angle at which you see something depends on where it
would be if you were at rest; for example, a star actually
at θ =120^{0} (30^{0}
behind you) would be seen as being about θ' ≈30^{0}
in front of you!

QUESTION:
how much times earth to be compressed to become black hole?

ANSWER:
A black hole is a singularity, of zero size, so compression
is infinite. If the earth were a black hole, its
Schwartzchild radius, R =2GM /c ^{2} ,
would be about 1 cm. R is the radius within which
nothing can escape. So, if you could get the earth down to
that size, it would then collapse the rest of the way to a
black hole, I presume.

QUESTION:
If momentum is applied not on the center of mass on an object, what would its momentum be at the center of mass and is there an angular momentum at the point of impact?

ANSWER:
Linear momentum is not something which is applied, force is
what is applied. And, what happens also depends on the
details of how the force is applied. I will work out one
simple example which will illustrate the principles involved
when the force is applied in a simple collision. A mass
m approaches the end of a thin uniform rod of mass
M and length 2L with velocity v
perpendicular to the rod; when it collides with the end of
the rod it sticks. The center of mass of the rod is at its
geometrical center a distance L from where m
exerts the force. Linear momentum and angular momentum are
both conserved but, as we shall see, energy is not
conserved. After the collision the center of mass has a
velocity u ; conserving linear momentum, (M+m )u=mv
or u=v [m /(M+m )]. Before the
collision m brings in an angular momentum mvL .
After the collision, the rod plus mass has angular momentum
(I +mL ^{2} )ω
where I=M (2L )^{2} /12=ML ^{2} /3
is the moment of inertia of a thin rod about its center of
mass. Conserving angular momentum I find ω [m +(M /3)]L ^{2} =mvL
or
ω=mv /{[m+ (M /3)]L }.

Just for fun I put in some numbers to illustrate what would
happen in a specific situation: m =1 kg, M =3
kg, L =1 m, v =4 m/s. With these I find:
u =1 m/s, ω =2 radians/s=19.1 rpm. You can
also calculate the energy before and after the collision:
E _{before} = ½mv ^{2} =8
J, E _{after} =½(M+m )u ^{2} +½([m +(M /3)]L ^{2} ω ^{2} =6
J. So, energy was not conserved in this collision, 2 Joules
were lost.

CORRECTION:
I realized that an error had been made in my analysis. After
the collision the center of mass is no longer at the center
of the rod but displaced a distance x=mL /(M+m )
toward where m is stuck. It is this center
of mass which moves forward with speed u , and u
is still v [m /(M+m )]. The rotation
is about this center of mass but now the moment of
inertia is different; the net (rod plus mass) moment of
inertia is ML ^{2} /3+Mx ^{2} +m (L-x )^{2} ,
so the corrected value of the angular velocity is
ω=mvL /[ML ^{2} /3+Mx ^{2} +m (L-x )^{2} ].
In the example I did, x =1/4 m, so ω= 16/7 radians/s=21.8 rpm and E _{after} =46/7=6.57
J (1.43 J lost).

QUESTION:
I have been trying to figure this out, but having some trouble doing it: In general, let's say I had a vehicle that could accelerate me (according to my perception) at 1G for 1 year (again according to my perception),
then turn around and decelerate (by my perception again) for another year to return home. How far would I have travelled and how old would my twin at home be when I got back?

ANSWER:
The reason you are having trouble is that relativistic
kinematics involving acceleration, even if the acceleration
in one frame is uniform, is not easy; the main reason is
that, unlike Newtonian mechanics, acceleration in one frame
is not the same as acceleration in a different frame. As a
prelude to reading my answer, you should read an
earlier answer and the earlier answers referred to in
that answer. The graph here shows the results derived or
cited in that earlier answer. The three graphs all show what
is observed of your motion as seen by the twin on earth. The
red curve is your velocity v relative to c
as seen by the earth-bound twin, v /c =(gt /c )/√[1+(gt /c )^{2} ];
the blue curve is your acceleration a
relative to g , a /g =[1+(gt /c )^{2} ]^{-3/2} ;
the black curve is your position expressed (approximately)
in light years, gx /c ^{2} =√[1+(gt /c )^{2} ]-1.
Now, since I have set this up as seen from earth, let me
specify the time you travel as the time of your trip out as
observed from earth: earth-bound twin observes you to arrive
at you destination in two years. Then, as you can read off
the graph, gx /c ^{2} =√[1+(2)^{2} ]-1=1.24
ly; the return trip will be the same length of time, so the
earth-bound twin's clock will have advanced 4 years. Light
would take 2.48 yr to travel that distance, so your clock
will have advanced a time somewhere between 2.48 and 4
years. Your maximum speed would have been about v /c =(2)/√[1+(2)^{2} ]=0.89.
We can get a rough estimate of your clock reading by
approximating your average relative speed as 0.89/2=0.445;
so you would have seen, on average, the distance contracted
by √[1-(0.445)^{2} ]=0.9 and your time would
have been 0.9x4=3.6 years.

QUESTION:
What happend to accelaration due to gravity if the earth stops rotating?

ANSWER:
Because of the centrifugal force, g appears to be
somewhat smaller than it really is, but the amount is very
tiny. This effect depends on your latitude and is zero at
the poles and greatest at the equator. Also because of this
effect, the acceleration is not directly toward the center
of the earth except at the poles and equator.

QUESTION:
Hello, is the mass of a person on a spaceship (to provide a suitable upward force) important in designing a spaceship? My teacher says otherwise but upon researching I found out that additional fuel is required for every kg of payload that will be sent out into space.

ANSWER:
If the ship and all the fuel are much bigger than the mass
of the person, then you could neglect the mass of the person
in any calculations you know; in other words, the rocket
launch would be, as close as you could hope to be measure,
identical whether or not an astronaut was on board. However
you are also right that the amount of fuel you need depends
on the payload you want to deliver, and that would include
everything which is not fuel. You might find the an
earlier answer
illuminating.

QUESTION:
How exactly do electrons in electric discharges generate radio waves?
Whenever someone uses an AM radio while there are electric sparks like lightning strikes nearby the radio waves they emit can be detected.

ANSWER:
Any time an electric charge is accelerated it emits
electromagnetic radiation. There are many electrons in the
spark which accelerate under the influence of the electric
fields they experience.

QUESTION:
I just read about the kilogram being defined by measuring the current of an elector-magnet to balance a scale. How can various electromagnets have exactly the same force to current ratio? Would variations in the purity of the conductor, variations in the exact coil shape, etc. cause variations in this ratio? What am I missing about this new designation?

ANSWER:
You cannot use any old electromagnet. There is an
exquisitely accurate balance called the
Kibble balance which is used. Also see
Wikipedia for more detail

QUESTION:
Is the following statement correct?
Net radiated energy is made up of emitted and absorbed energy.

I stated that it was incorrect based on the fact that
"made up of" usually means putting two or more things together (aka addition). Net radiated energy is the difference between emitted energy and absorbed energy. I have yet to see my exam paper but I believe that the mark I lost was due to this question on basis that “made up of” means that it is a result of a mathematical relation between the two values and not necessarily addition (in this case subtraction). What is the correct answer in this case and why?

ANSWER:
I do not believe that there is a hard and fast definition of
what "net radiated energy" is. It is a matter of semantics.
However, radiated energy usually means energy emitted and
absorbed energy means energy absorbed; therefore, I would
say that this is an incorrect statement because net radiated
energy would mean the sum of all radiated energy, absorbed
energy having nothing to do with it. Your reason is quite
shaky because "made up of" is even more semantically
ambiguous than "net radiated"; it would simply imply that
"net" means total energy flux.

QUESTION:
How is momentum conserved for single or double slit diffraction? in other words:I have a photon gun that I can dial the intensity down to fire one photon at a time. I fire my photon gun, the gun recoils in the -X direction, the photon flies off in the +X direction. The photon then passes thru two closely spaced slits (double slit experiment)instead of appearing on the screen directly behind the slits, the photon ends up to the side (after many, many photons the interference pattern appears) however, for the single event of a single photon, how is momentum conserved? dos the single event photon exchange momentum with the walls of the slit? Since the photon ended up to the side, it was no longer travelling in the purely X direction, therefore shouldn't the photon have exchange d momentum with the slit somehow? If the photon interacts with the slit, that counts as “which way” information, and the interference pattern vanishes. If the interference pattern remains, then how do we account for the photons change in momentum? This is for a single event, one photon, not expectation value of many, many photons.

ANSWER:
You are forgetting that the photon goes through both slits,
is being forced (if you like) to assume its wave-like
identity. It behaves like a wave until something makes it
act like a particle, so it is a wave spreading out in both
directions, so its net momentum in the direction parallel to
the screen is zero. When the screen is encountered it is a
"measurement" which collapses the wave function at some
point.

QUESTION:
I can imagine 2 similar objects rotating around a sphere at the same velocity but at different orbits. One object would rotate around the sphere faster than the other, right? What is puzzling to me is that if I think of both objects going now in a straight line one could not say that one object was going faster than the other. Can someone help me through this? Is there a subtlety here?

ANSWER:
You want to say same speed, not same velocity, but I get the
idea. Say one orbit is twice the radius as the other. Then
the smaller orbit has a circumference half that of the
larger. So each time the object in the larger orbit goes
around once, the one in the smaller orbit goes around twice.
But they still have the same speed. But the angular
speed of the object in the smaller orbit is twice that
of the object in the larger orbit even though their linear
speeds are the same. You cannot do this with planets or
moons, though, because the speed of a satellite in a
circular orbit depends on the radius.

QUESTION:
If space is curved, doesn't it need another dimension to curve into? If a 2d object is curved it curves into the 3rd dimension. So if 3d space is also curved it must curve into 4d.

ANSWER:
This
is really mathematics, not physics; I am not a
mathematician. But the definition of an N-dimensional space
is, I believe, that N quantities are required to specify the
location of a point in that space. Curvature of a space is
defined as a space in which the geometry is not Euclidean.
The physical existence of an N-dimensional space does not
imply the physical existence of an N+1 dimensional space.
You cannot make a generalization that curvature is
necessarily "curving into" a higher dimension. For
example, a planar sheet can be distorted without leaving the
2D. You also might "curve into" a higher dimension than N+1;
for example, a coil spring is a one dimensional space which
exists in a 3D space. You are trying to make generalizations
based on the only spatial dimensions which we know to exist;
but higher dimensions are defined by certain rules which in
no way prove that they exist physically.

While
researching this question I came upon a very interesting
little
essay about general relativity, curvature of
spacetime, gravity. Many nonscientists (and scientists as
well) are familiar with and accept as gospel that general
relativity demonstrates that the distortion of spacetime by
the presence of mass is what gravity is; the "cartoon" which
illustrates this is the "bowling ball on a trampoline
example in which a mass causes the distortion of the
two-dimensional space defined by the trampoline surface. The
implication is that gravity is just geometry. On the other
hand, many theorists are struggling to find a successful
theory of quantum gravity and to do that one must treat
gravity as a force. The take-away from this essay, though,
is that curved spacetime is not a unique representation of
what the mathematics which are the framework of the very
successful theory of general relativity mean. The author,
Rodney Brooks, notes that "…you can view gravity as a force field that, like the other force fields in QFT
[quantum field theory], exists in three-dimensional space and evolves in time according to the field equations."
He also goes so far as to say "…shame on those who try to foist and force the four-dimensional concept onto the public as essential to the understanding of relativity theory."

QUESTION:
A rear wheel drive car accelerates quickly from rest. The driver observes that the car noses up. Why does it do that? Would a front wheel drive car behave differently?

ANSWER:
The car with rear-wheel drive has an acceleration
a and a mass M . The forces on
the car are its own weight W (W=Mg )
acting at the center of mass a distance d from the
real axle; the normal and frictional forces on the rear
wheels, N _{1} and
f _{1} ; and the normal and
frictional forces on the front wheels, N _{2}
and f _{2} ; the radius of
the wheels is R and the distance between front and
rear axels is D . I will ignore the rotational
motion of the wheels, that is, I will approximate their
masses as zero. If the front wheels are not lifting up from
the ground, the sum of torques about the rear axel is (f _{1} -f _{2} )R +N _{2} D -Mgd= 0;
Newton's first law for the horizontal and vertical
directions are (f _{1} -f _{2} )=Ma
and N _{1} +N _{2} -Mg =0.
From these equations it is easy to show that N _{2} =M (gd-aR )/D .
This is interesting because it shows that N _{2} =0
if a=gd /R ; for larger accelerations, the
front wheels will lift off the ground. (Note that as N _{2}
approaches zero, f _{2} also approaches
zero.) I have ignored the suspension of a real car, so this
lift will not be abrupt; it will happen gradually as the
front springs decompress and the rear springs compress; so
overall the nose goes up but the rear end goes down.

If
it is a front-wheel drive, the directions of the frictional
forces reverse. The corresponding torque equation (this time
about the front axle) is
(f _{2} -f _{1} )R-N _{1} D+Mg (D-d )= 0;
Newton's first law for the horizontal and vertical
directions are (f _{2} -f _{1} )=Ma
and N _{1} +N _{2} -Mg =0.
From these equations it is easy to show that N _{1} =M (aR+g (D-d ))/D .
As a increases in magnitude, N _{1}
gets bigger and bigger until N _{1} =Mg
which necessarily means that N _{2} becomes
zero. It is easy to show that this happens when a=gd /R ,,
exactly the same result as for the rear wheel drive.

Finally,
it might appear that there is no difference between front-
and rear-wheel drive. That is not the case in the real world
at all. Keep in mind that the frictional forces are static
friction and the larger the corresponding normal force is,
the more friction you can get. Since in the rear-wheel
situation the normal force gets bigger, the amount of
friction you can get without the wheels slipping can keep
increasing so that eventually (with a powerful enough
engine) you could lift the whole front end far off the
ground. However, when you have front-wheel drive, the normal
force on the front wheels gets smaller as you increase a ,
so most likely the wheels will slip well before N _{2} =0
when f _{2} must be zero. For very large
accelerations, like required for drag racers, rear-wheel
drive is imperative.

QUESTION:
Gravity is what makes objects orbit around other objects,and gravity is a reflection of an object’s mass.So why doesn’ the mass of the objects appear in kepler’s third law??

ANSWER:
For the same reason why all objects have the same
acceleration in the gravitational field, i.e. all
objects near the surface of the earth have an acceleration
of 9.8 m/s^{2} . The acceleration (which is what
describes the orbit) is F /m but the force
is MmG /r ^{2} , so the acceleration
is MG /r ^{2} .

QUESTION:
If kinematics can give you data on a dropped object, would it be valid to use kinematics to calculate the acceleration experienced by the object upon impact with the ground? (the time from when it hits the ground to when it actually stops)

ANSWER:
When the object hits the ground until its velocity reaches
zero is some time, call it Δt . Its
momentum drops from mv (where m is the
mass and v is the speed when it hits) to zero
because an impulse I was delivered to it by the
ground. The impulse may be written as I=F _{avg} Δt .
where F _{avg} is the average
force. Therefore you can find the average acceleration a _{avg}
during the collision, but not the acceleration as a function
of time: a _{avg} =F _{avg} /m =I /(m Δt )=v /Δt.
So you cannot get detailed kinematics about the
collision without measuring how the speed decreases with
time.

QUESTION:
We know that air density decreases as you ascend into space, but what about looking out to the horizon? Is there a atmospheric lensing that occurs? Is there a formula for its calculation?

ANSWER:
Certainly there is refraction due to this density profile of
the air. There is no simple formula because the actual
density at any point is dependent on the pressure and
temperature of the air as well as altitude, and these vary
with time. One could calculate based on average density
profiles, but I would suppose it is a numerical calculation,
not an analytic formula. There is a nice description of this
phenomenon
here .

QUESTION:
I've been interested in physics for years but have never managed to get my head around the concept of space-time. I understand time and i understand space but somehow i could never get this. However today i think it finally clicked when i wasn't even trying. Can you tell me if my understanding of it is correct or not?
I thought of space-time as like if i were on a roller-coaster or train tracks. When i am on the long flats, i travel slower, and when i am on the declines, i travel faster. Is space-time like this analogy? In that, when you are on Earth for example, you experience time moving faster than if you were floating in space millions of miles away. This is because the presence of Earth "warps" space time, much like a bowling ball on a trampoline. This "dent" in space-time causes a gravitational pull, pulling time into it, much like if i were rolling down a hill.
I know that might sound very convoluted but it's the idea i have in my head and it has bothered me for YEARS, but i think i am on to an understanding here.

ANSWER:
Sometimes we make things more difficult than they need to
be. The main takeaway from the theory of relativity is that
time and space are not separate and universal things. In
Newtonian times, it was assumed that any clock in the
universe would run at exactly the same rate; now we know
that if a clock moves with respect to us that it runs more
slowly than ours —doesn't appear to run slower,
it runs slower. Similarly, space depends on the observer; a
meter stick moving by you is actually shorter than 1 m. What
these facts about space and time tell us is that time and
they are not different things, they are really entangled
with each other. You cannot talk about one without talking
about the other. They are two pieces of the same thing. So
the term space-time is simply a way of acknowledging that. I
do not find your roller coaster analogy very helpful. The
bowling ball on a trampoline is a classic but keep in mind
that it is really a cartoon and not rigorous. Generally
trying to picture something in four dimensions futile!

QUESTION:
What does a wind speed (a gust is okay)in miles per hour, need to be to launch a baseball at a speed of 80 miles per hour into an object like a window.
Given:
Ball is 60 feet 6 inches from the window that it hits traveling at 80 mph Baseball is 9.25 inches in circumference and 2.94 inches in diameter and weighs 5.25 ounces
My name is Karin and I'm testing to see if I use a baseball pitching machine from 60 feet 6 inches from a window protected with a protective panel, and set the machine to pitch at 80 mph, what would the hurricane wind speed need to be to get that baseball to fly into the window at 80 mph. I hope this is clear enough.

ANSWER:
If I understand your question, you are essentially asking
what wind speed would be required for a baseball to reach a
speed, starting from rest, of 80 mph in a distance of 60.5
ft due to the force of the wind on the ball. This is a
problem which is very tedious to do in full analytic detail
and not very illuminating. Also, all problems involving air
drag are approximate, so it would be appropriate for me to
find an approximate way to solve this problem. I am
going to work in SI units so I will rephrase your question:
what wind speed would be required for a baseball to reach a
speed, starting from rest, of 35.8 m/s in a distance of 18.4
m due to the force of the wind on the ball? In SI units you
may approximate the force at sea level which the wind exerts
on the ball as F= ¼A (u-v )^{2 }
where u is the speed of the wind, v is the
speed of the ball, and A is the cross sectional
area of the ball. Applying Newton's second law, ¼A (u-v )^{2} =ma =m dv /dt
. This differential equation may be shown to have the
solution t=cv /[u (u-v )] where
c =4m /A and t is the time it takes the
ball to reach the speed v if the wind speed is
u . I find that c =137 s^{2} /m. Note
that after a very long time the speed will be constant at
the wind speed; we are only interested in the time it takes
to get to 35.8 m/s (80 mph). I have plotted t for
three values of u , 70, 80, and 90 m/s.

Now comes the hard part. We are interested in finding the
value of u for which v will be 35.8 m/s when the ball has
gone a distance x =18.4 m. This means that we need
to write the equation for t as
t=c {dx /dt }/[u (u- {dx /dt })]
and solve that differential equation for t as a
function of x . That is actually doable, but the
solution is so messy that I looked for a simpler way to do it.
This took some trial and error, but I got a pretty good
estimate, I believe. Look at the black line in the graph; it
is not a straight line, but it is pretty close, so I am going
to approximate the acceleration as being constant over this
time interval of 1 s, a ≈(35.8 m/s)/(1 s)=35.8
m/s^{2} . Now, an object with constant acceleration
starting from rest goes a distance d = ½at ^{2}
in a time t , so if u =90 m/s=201 mph the ball will have a
speed of 35.8 m/s when d =35.8/2=17.9 m=59 ft. As
far as I am concerned, I have adequately solved this problem—it
takes a wind of approximately 200 mph to accelerate a
baseball from rest to the speed of 80 mph in a distance of
about 60 ft.

QUESTION:
Why is light speed the fastest speed and why would people see a train going backwards if it is traveling at a velocity greater than light speed?

ANSWER:
I will not answer the second question because the site
groundrules state
that I will not answer questions of this sort. Your first
question is equivalent to the question "why can nothing with
mass travel as fast or faster than light?" which is included
in the FAQ page.