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QUESTION:
My brother died recently and I took great comfort in the Law of Conservation Mass/Energy. I'm guessing this isn't a typical question you'd get, but want to tattoo this equation on my arm. I've seen a few different variations of this equation, so I'd love to hear from a physicist on the best equation to represent this law.

ANSWER:
My condolences on your loss. Well, the total
energy of a closed system is indeed conserved
which would be expressed (kinda boring) as E _{i} =E _{f
} where i denotes initial and f denotes
final. But, there is no such law as conservation
of mass because Einstein showed that mass is
just a kind of energy. So the mass of a system
need not be the same before and after. Chemists
still use a "law of mass conservation" because
the mass changes in chemical reactions are too
tiny to measure; still, as we know in physics,
the mass does change in chemical reactions. Now,
you know the famous equation E=mc ^{2} ,
so how about
m _{i} c ^{2} +E _{i} =m _{f} c ^{2} +E _{f
}
where m _{i}
and m _{f} are the amount of
mass initially and finally and now E
means all energy
except mass energy.

QUESTION:
Hi, I'm confused about how linear-moment (LM) and angular-momentum (AM) can both be conserved simultaneously. For example, just before a bullet sticks in a stationary beam on a pivot, the system has only LM (that of the bullet) and the system has zero AM (because the beam is stationary). After the bullet sticks in the beam, the whole system will rotate, so now we have only AM and no LM, right? It seems to me that the LM was converted to AM and therefore not conserved. What's wrong with my thinking?

ANSWER: (I will assume that the beam is
like a uniform beam and hinged frictionlessly so
that the beam can swing in a horizontal plane.)
Your main mistake is that the
bullet does have angular momentum before the
collision. Consider the bullet as a point mass;
a point mass has a moment of inertia I _{bullet} =mr ^{2} where
m is the mass and r is the distance to the axis
about which you are computing the angular
momentum, in this case the hinge of the beam.
The angular momentum of the bullet at the
instant of impact is L=I _{bullet} ω =(mr ^{2} )(v /r )=mrv .
Because the hinge has no friction, there are no
external torques on it and therefore angular
momentum will be conserved. The beam has a
moment of inertia I _{beam} =ML ^{2} /3
where M is its mass and L its
length. After the collision the beam plus bullet
have an angular velocity ω' .
Conserving angular momentum, mrv= (ML ^{2} /3+mr ^{2} )ω'
which can be solved for ω' .

What about
linear momentum? It is not conserved because the
hinge exerts a force on the system, the absense
of which is the condition for linear momentum
conservation. If this were all happening in
empty space, both angular and linear momentum
would be conserved; afterwards the beam+bullet
would rotate about their center of mass and the
center of mass would move in the same direction
the bullet came in with a speed v'=mv /(m +M ).

The bottom line is that linear momentum is
conserved if there are no external forces on the
system and angular momentum is conserved if
there are no external torques on the system

QUESTION:
Sir, the average density of nucleus of an atom is very high in the range of 10^17kg/m^3.However the density of elements which are made of these atoms is low why

ANSWER: Because the volume occupied by the nuclei is extremely small.

FOLLOWUP QUESTION:
my question has been misunderstood.
Let me put in this way. There is a wooden empty box having some mass. Now if in this box we put a small volume of dense material say mercury. Then the mass of the box with mercury will be more even if mercury occupies small volume. The overall density will be greater than the density of empty wooden box. It will simply add the density of wooden box.
Similarly in case of nucleus of atom being denser up to the level of 10 raise to power 17 kg/m3 and the matter consisting of millions of atoms, how it does not reflect in the over all density of that matter. I mean there is a huge difference.
In 1 cm cube of water the mass comes to 1 gram
In that 1 gm of water there will be more than 6.023x10^23 no of atoms having these massive denser nucleus, each with density of 10^14 gm/cm^3, then how is it that water in that 1 cubic cm volume comes to be 1 gm only.

ANSWER:
I have not misunderstood your question. You
misunderstand how to calculate density. suppose
the box has a volume of 1 m^{3} and a
mass of 1 g=10^{-3 } kg. Its density is
10^{-3} kg/m^{3} . Now, put into
that box an object whose mass is 1000 kg and
whose volume is 10^{-3} m^{3} ;
the density of this object is 10^{6}
kg/m^{3} , but the density of the box
plus object is 1000.001/1 kg/m^{3} , 1000
times smaller. So, now let's do the same thing
for a single atom (which can be approximately
scaled to a macroscopic object). I will take
your number for the nuclear density, 10^{17}
kg/m^{3} ; I am sure that you know that
the mass of an electron is about 2000 times
smaller than that of a proton or neutron, so the
nucleus mass is roughly 4000 times greater than
the mass of all the electrons; therefore I will
approximate the mass of the electrons to be
zero. The ratio of the radius of the atom to the
radius of the nucleus is about 10^{5}
and therefore the ratio of their volumes is
about 10^{15} . Therefore the density of
the atom will be on the order of 100 kg/m^{3} ;
this number is comparable to typical macroscopic
densities. This is just a rough approximation
but the average
density of an atom is simply its mass divided by
its volume, and that is what matters when you
determine the density of some object made up of
a great many atoms, not what the density of some
little part of it is.

QUESTION:
State the condition at what height and what depth the acceleration due to gravity remains same?

ANSWER:
I have previously
answered this question.

QUESTION:
I have a bit of doubt relating to a scenario and I would really appreciate your assistance. If a person of known mass jumps off a wall of known height, but his body comes to rest a while after his feet strike the ground (perhaps 0.15 seconds), when determining the force, impulse,momentum and stopping distance, how do you take that 0.15 seconds into account?

ANSWER: You can calculate how fast he is
going when he hits the ground. Then you need to
assume that the force he feels from the ground
is a constant. Knowing the time to stop you can
deduce the force needed. Once you have that
force, you can calculate the acceleration,
impulse, and distance. Don't forget about his
weight.

QUESTION:
I am a layman, with a strong interest in natural Sciences. Lately light has caught my attention, and I am starting to understand the complexity of the topic and how far we are from a definitive answer on its real nature. Is EM waves energy dependent only on amplitude or amplitude and frequency?

ANSWER: When you say "…how far we are from
a definitive answer on its real nature…"
does "we" refer to you or to everyone? Just
asking because it is important for you to
understand that electromagnetism is perhaps the
best and most completely understood field theory
in all of physics. Regarding light
(electromagnetic radiation, in general), I would
recommend that you read something about the
photoelectric effect to understand where the
notion that the energy of an electromagnetic
wave was proportional to the square of its
amplitude was found to fail. The solution was
simply that the EM field is quantized which
means that the radiation has a minimum amount of
energy, called a photon, in which it can exist.
In essence this means that if you have a
particular wave and wish to reduce its energy,
you cannot do it smoothly but only in steps of
the energy of one photon at a time. This had
never been observed in a laboratory because the
energy of a single photon is so tiny compared to
the energy of a macroscopic wave. So, if the
wave interacts in a macroscopic way, for example
a radio wave interacting with a radio antenna,
you find that the energy is indeed dependent
only on its amplitude. On the other hand, when
you try to make that wave interact with an
individual electron, the electron sees and
interacts with individual photons of energy
hf as you note in your original question.
How can we reconcile these two facts? Simply
that if the radiation interacts as a wave,
amplitude matters, but if the radiation
interacts as a photon, frequency matters. If you
have a blue wave and a red wave with equal
amplitudes, they carry the same energy; but the
blue photons have a higher energy than the red
photons, so there are more red photons in the
red wave than there are blue photons in the blue
wave.

So, you asked whether the energy of light
is "…dependent only on amplitude or [on] amplitude
and
frequency…"; the answer is neither, it
depends on amplitude
or frequency!

QUESTION:
Why do the planets rotate is a disk pattern and not a spherical one?

ANSWER: Imagine a huge cloud of gas and
dust, trillions of miles across. It has formed
randomly, gradually growing because of the
gravitational attraction for material in its
neighborhood. It is almost impossible to think
that this cloud has no angular momentum, that
there is not some net spin. Even if the net flow
of particles around the center of mass of this
cloud were, say, 1 cm/s in its far reaches,
because it is so far away it has a large angular
momentum. Now the cloud begins to get smaller,
being pulled together by its own gravity.
However, if that cloud has angular momentum, the
result is that it will spin faster and faster as
it gets smaller and smaller. But, imagine a
squishy ball which is spun faster and faster—it
flattens out and eventually it is more like a
pancake than a ball. The center of this squished
ball is the most dense and eventually is dense
enough to ignite to become a star; the outer
reaches of the squished ball contain the matter
from which planets will eventually coalesce and
form.

QUESTION:
Regarding Kinematics, I understand that the outer edge of a car tyre is travelling between 0mph where it is contact with the road to 2x the speed of the car when that point is furthest from the road.
My first questions is:
If the car was driving at the same speed on two wheels, so right side was off the ground, would the drive wheel tyre still continue to have the velocity change if it was not in contact with the road?
The reason I ask is that someone mentioned that we, on the Earth should feel the velocity change due to the rotation and orbit of the Earth.
I'm sure it is due to the fact that the velocity has to be relative to a point, ie the road in the case of the car tyre, but I have been looking online for an answer to this problem and so far no luck!

ANSWER: Actually, your first question has
nothing to do with your second question. I am
assuming that the wheel off the ground is
spinning at the same rate as the wheel on the
ground. Then, relative to the road, the bottom
point of the off-ground wheel will be observed
to be at rest just as if it were touching the
ground. For your second question, you should, in
principle, be able to feel the effects of the
earth's motion; however, the effects are too
small for you to be able to "feel" them. But
those effects are due to the acceleration you
encounter, not the velocity. One simple example
would be the centrifugal force you feel due to
the earth's rotation. If you were standing on
the equator you would find that you weigh about
0.35% less than if the earth were not rotating;
if the earth spun fast enough, you might just
fly off into space.

ANSWER: Actually, your first question has
nothing to do with your second question. I am
assuming that the wheel off the ground is
spinning at the same rate as the wheel on the
ground. Then, relative to the road, the bottom
point of the off-ground wheel will be observed
to be at rest just as if it were touching the
ground. For your second question, you should, in
principle, be able to feel the effects of the
earth's motion; however, the effects are too
small for you to be able to "feel" them. But
those effects are due to the acceleration you
encounter, not the velocity. One simple example
would be the centrifugal force you feel due to
the earth's rotation. If you were standing on
the equator you would find that you weigh about
0.35% less than if the earth were not rotating;
if the earth spun fast enough, you might just
fly off into space.

FOLLOWUP QUESTION:
I realise now that I may not have fully explained my question.
I understand your answer as to why we don’t feel the rotation.
As the Earth’s translational motion of the orbit around the sun is an average of 30 kilometres per second, would there be a change from 0Km/s to 60Km/s?
And if so, why do we not feel that acceleration/deceleration?
I have attached a small diagram incase I have not explained it properly!

ANSWER:
Ah, now I understand! But, no, this is not the
same situation as the car wheel. If the earth
were rotating on its axis such that the equator
had a speed of 30 km/s, it would be so. However,
the actual speed is 0.46 km/s. Therefore the far
side is going about 30.46 km/s and the near side
about 29.54 km/s.

FOLLOWUP QUESTION:
One final part, could you explain why it is not the same situation as the car wheel?
I need to understand the difference between the physics of rolling without slipping and the orbit of a rotating planet.

ANSWER: In my original answer I said "I
am assuming that the wheel off the ground is
spinning at the same rate as the wheel on the
ground." That is the necessary condition. In
other words, if the earth were spinning such
that the equator moved with speed 30 m/s
relative to its axis, the sun-side of the planet
would be moving 30 m/s backward relative to the
30 m/s orbital speed forward.

QUESTION:
I have a hose that is 4 inches in diameter and a thousand feet long. I want to roll it up on a reel. It wieghs 4375 lbs. It is stretched out on a concrete road. I am going to reel it around a drum that it 6 feet in diameter after i attach it to the drum. How much torque will i need to reel this thing up?
I have looked at the coeffiecient of rubber/concrete and it is .65 That would mean it would take 2843 ft lbs of torque to reel it in. I don't think this is right because the whole 4" of rubber is not sitting on the ground. I would think that you have to factor in the area of hose that is in contact with the ground.

ANSWER: First of all, the way we often
calculate friction, F=μN , is not a
physical law, it is just an approximate method
for getting the frictional force if an object is
sliding and based on experimental measurements.
N is just how hard the
two (sliding and stationary) surfaces are pressed together
and μ depends only on what the two
surfaces are. Note that the area of contact is
nowhere in this equation, so that is not
relevent for your question. Secondly, the number
you calculate is not a torque, it is the force
you would have to exert to just pull the whole
hose. The torque is determined by how long the
handle you crank with is. I envision a crank of
length L attached to the axle of the
drum on which you exert a force F at
its end and perpendicular to its length; I
imagine a ratchet which will lock the crank to
the drum when turning counterclockwise (taking
in hose) and slipping when the crank is
repositioned by moving it clockwise. The
resulting torque about the axle of the drum is
FL. Meanwhile the hose is exerting a
force T , as shown, on the drum and its
torque is -TR . (This is negative
because it is trying to spin the drum clockwise
whereas the torque due to F is trying
to turn it counterclockwise and we have called
that torque positive.) If the hose is at rest or
moving with constant speed, the sum of the
torques is zero, so FL-TR =0 or F=T (R /L ).
In your case, T =2843 lb, so if, for
example, L =5R , F =569
lb. The resulting torque would be 17,058 ft lb.

Practical notes:

The
more hose you bring in, the smaller the
friction will be;

as hose
piles up on the drum, R gets
bigger; and

static
friction is bigger than kinetic friction, so
it will be harder to get it moving than to
keep it moving.

QUESTION:
The question in which I asked keeps me up for hours. There's a lot more to my question than I put which also doesn't add up. Could Einsteins theory of general relativity be flawed, at least when it comes to gravity. I'm ot saying I know how gravity works but I just have a niggly feeling that the whole warping of spacetime by a massive object just doesn't work in so many ways. I am not an argumentative person so don't worry. I'm just tired of this thought pestering me constantly for years. Just your thoughts would be much appreciated and maybe I could finally be rid of this thought wish bothers me so much.

ANSWER:
It is not at all clear what it is that is
bothering you, your "niggly feeling". I also
have been bothered by the notion that gravity is
apparently just geometry, not really a force,
yet one of the most important efforts in physics
today is the attempt to quantize it as if it
were a force. You must admit, however, that
considering it as "flawed" is overstatement
inasmuch as it has passed every experimental
test over the last century. I
recently answered a question along these
lines in which I found that "warping of
spacetime" is just one possible interpretation
of the nontrivial mathematics which is the
framework of general relativity; this
visualization is attractive because it allows
persons like yourself to comprehend the
implications of the theory. However, the
mathematics of the theory may also be
interpreted as a field theory, and fields can be
quantized. The
essay linked to is short and not too
technical; reading it may put you in a state let
you might be able to get a good night's sleep!

QUESTION:
On Reditt I made the staement "emissions from Earth are limited by temperature" in regards to it's continuous radiation. Given that the average temperature is 15C the shortest continuous wavelength of radiation would be IR at
about 10 μm.
I've been inundated by people who insist that Earth radiates continuously at all wavelengths right up x-rays. Every source says that's not possible. Would you please settle this and post the answer so that I can link to it?

ANSWER: The figure shows satellite
measurements (solid line) of the radiation
spectra over northern Africa at midday. The
temperature is near 320 K (about 115^{0} F).
If there were no atmosphere, this spectrum would
look like black body radiation, the dashed line
labeled 320 K. Note that its peak is near 15 μm.
But there is atmosphere and it absorbs a lot of
the radiation resulting in large discrepancies
between observed and predicted spectra. Most
notable is CO_{2} , the prime culprit in global
warning, which reradiates this infrared (heat)
back to the earth. Also ozone (O_{3} ), methane
(CH_{4} ), and water vapor absorb a lot of enery. I
do not understand what the argument is; I do not
know what is meant by

I do know that data clearly show that the earth
radiates at all wavelengths, some of which do
not fully escape the earth because of
interactions with the atmosphere; wavelengths
far from the maximum black body intensity for a
given temperature will be small enough intensity
to be negligible.

QUESTION:
My question may be too simpple for you and it is this: My dog, 22 Kg weeight, whle at play with another dog crashes against my ribs at almost 20 kilometers per hour breaking two of my ribs in the process. What force in pounds/feet was exerteed against my ribs? I am 75 years old so this is not a
'homework' question but mereely onee of intereest to me. My dog, Dylan, had done nothing wrong other than not look wheree he was going

ANSWER: There is no way to calculate this in any exact way. You need to know how long the collision lasted and how elastic it was. You could make some simplifying assumptions like the collision was totally inelastic (essentially that you stick together immediately after the collision) and maybe the collision lasted 0.2 seconds.
I will take your weight as 78 kg so you plus
Dylan weigh 100 kg. The dog hits you going a
speed of 20 km/hr=5.6 m/s. Conserving momentum,
5.6x22=100V where V is the
speed of the two of you right after the
collision, so V =1.23 m/s. Now, the rate
of change of your momentum was equal to the
force on you in Newtons, so (1.23x78-0)/0.2=480
N=108 lb. You see that the time of collision is
crucial, for example if you used 0.1 seconds the
average force during the collision would have
been 216 lb. For 0.01 seconds, the force would
be over a ton.

QUESTION:
How much energy (heat/cool) is lost through a 6’ X3’ piece of privacy glass 2” thick? I’m considering this for home construction.

ANSWER: It seems to me that what you
really need is not energy but the rate of energy
loss. The equation for thermal
conduction is R =kA ΔT /d
where R is the rate
of energy loss (typically Watts in SI units), k
is the thermal conductivity of the material, ΔT
is the temperature difference, A is the
area, and d is the thickness. For
glass, k is about 1 W/(K⋅m). For
your case, A =18 ft^{2} =1.67 m^{2
} and d =2 in=0.0508 m. Therefore
R =32.9ΔT. For example,
if ΔT =(20-0) ^{0} C=20
^{0} C=20 K (about (68-32) ^{0} F),
R =656 W=0.622 BTU/hr.

QUESTION:
I understand that the universe has no absolute frame of reference and that
the velocity of any object must relative to some other object.
I also understand the following thought experiment to be valid: If superman
were to fly to a distant star (say, 5 light years away) at very close to the
speed of light, and then fly back to Earth also at close to the speed of
light, then when he arrives back on Earth, all of us would all be much older
(around 10 years in this case) while superman would be still be almost the
same age as when he left, all due to time and space dilation.
So here is my question: If there is no absolute frame of reference, then
why does superman get to be the one who didn't age?

ANSWER: Go to the
faq page and look for
twin paradox .

FOLLOWUP QUESTION:
Well I don't see the answer to my question under
"twin paradox".

ANSWER: Your question is
precisely a simple variation of the twin paradox.
Below I have worked out the answer to your specific question (using 99% of light speed for
"close to the speed of light").
This follows exactly the method of my earlier
answer on the
twin paradox
but using your numbers, although I have not
redrawn the graph for your numbers because that
would be impossibly tedious.

My favorite way to understand the twin paradox is to suppose that Superman and someone on earth, each using his own clock, sends 100 light pulses per year to the other. Each receives all the pulses from the other, but Superman sends fewer since, because of length contraction, he has less far to travel in his frame than
the earthbound person observes; Superman travels to a star
and back which is 5 light years away with a speed of 99% the speed of light. Since Superman, because of length contraction, sees a distance to the star of only 5√(1-0.99^{2} )=0.71 light years, he sends out
71 light pulses on the way out and 71 on the way back, and all are received by the earthbound observer, so both agree that he has aged
1.42 years. The earthbound someone sends out
990 pulses and Superman receives them all, so both agree that he has aged
9.9 years.

I have done this in the frame in which both the
earth and the star are at rest. If you want, you
can do it in the frame where Superman is at
rest, but the result would be the same because
the distance between the earth and the star
would still be contracted to 0.71 light years
for Superman.

QUESTION:
I was playing a game called war thunder and it said that boats took longer to stop and start because they have more inertia than things on land. Is this correct? My take on a boat taking longer to start and stop is because of the lack of friction to slow the boat down and the lack of traction to propel the boat. Am I correct or is the game correct?

ANSWER:
When talking about acceleration, it is pointless
to just talk about inertia (mass) or about the
force applied. After all, Newton's second law is
a=F /m . If you compare two
objects with the same mass, the one with the
largest force applied wins. Similarly, if you
compare two objects encountering the same force,
the one with the smallest inertia applied wins.
Maybe your game is designed such that boats are
always much heavier than land vehicles.

QUESTION:
I have a question about Quantum reconstruction. But first, I'm a junior in high school and don't possess enough knowledge of the Quantum Theory or Quantum mechanics to base my theory on, if you can even call it that.
Theoretically speaking, what if we had a computer that could store data on a microscopically low scale, like down to an atom. What if you found a way of coping that same data to another position using the stored atoms as reference. It might sound like an unintelligent theory, but it's all I got. Am I saying that this could be a beginning into teleportation... NO. I believe that it's purely fiction. An example; Say we have a meatal cube. We take the cube and use some sort of device to break it down to its atoms, store that information and then transfer the atoms through an electromagnetic cable. We can use the cable to manipulate the electrons, protons, and neutrons in the said atom. Then use a device to reconstruct the atoms into the form that was deconstructed. Like I said this is not based on any test that I know about, just got curious.

ANSWER: "Beam me up, Scotty!" So, is
teleportation possible? There is a very
entertaining book, The Physics of Star Trek
by Lawrence Krauss, which has a lengthy
discussion of teleportation.

QUESTION:
If I am traveling at 99% the speed of light, and I have a special flashlight with a very high powered light on it, and I flashed a message in Morse code to an anticipating target, will the Morse code message arrive there faster than if I sent it out while stationary? I understand that the speed of light is constant, so the light will leave your flashlight at its usual speed relative to you. But since you're moving at 99% the speed of light, will this reduce the time it takes the light to reach its target?

ANSWER: Any given pulse you send will not
get there any faster than it would if you were
standing still at the position from which you
sent that pulse, so you are definitely not going
to get your message there faster. However, the
pulse you just sent was from a position closer
to your target than the previous pulse, so if
they were sent 1 s apart by you, the target
would receive them much closer together. So, if
your whole message were sent in 1 minute, the
target would receive it in a shorter time, but
not earlier. You might like to read the
answer to a question similar to yours but
with considerably more detail

QUESTION:
Does CERN effect earths gravity. i am talking about when the protons are at full speed but before they crash. I am not talking conspiracy . but the mass of the protons and the magnetic field used to control them. would it be possible to get the American gravity sensing satellites to record the earths gravitational field when its at full speed. I am only a cleaner who likes to learn and this question has been bugging me.

ANSWER: Any effect from the increased
mass of the high-speed protons would be
immeasurably small. See an
earlier answer .

QUESTION:
It is commonly taught that the B (magnetic) field and the E (electric) field, propagate together in a way that shows them as equivalent in magnitude.
If a B field is greater when I (current) is greater, and E is greater when the charge differential is greater, then I come to the logical conclusion that the B field should be 0 when the E field is max and vise versa.
Is my logic flawed here? I ask because I am trying to study and understand what is actually happening when an EM wave reaches an electron. The entire interaction between the two is usually summed up as “photon hits electron” but this situation implies that the photon strikes the electron in some perfect manner. I know that EM waves have amplitudes of significant size compared to the size of an electron and the entire interaction baffles me.

ANSWER: There is no way that any picture
can accurately represent the relative values of
the fields since the two have different units
which measure them. If you take the ratio of the
two fields in any electromagnetic wave at any
time and any point in space, you find that E /B=c
where c is the speed of light. Since
this is a constant, your conclusion that the
E and B fields are not in phase
must be wrong because two waves out of phase
have different ratios at different places. Where
you went wrong is thinking of B as just being
created by some current; in fact, another origin
for a B field is a time-varying
electric field. Similarly, a time-varying
magnetic field causes an electric field. In some
sense, the E and B fields are
keeping themselves going, sort of a perpetual
feed-back loop.

The second
part of your question involves the
photoelectric effect . What happens when you
send electromagnetic wave into matter cannot be
described by thinking about waves. Understanding
that it could only by acknowledging that
sometimes these waves fo not behave like waves
at all, but must be thought of as particles
called photons. Do not waste your time trying to
understand the photoelectric effect in terms of
electromagnetic waves.

QUESTION:
Me and my friends were arguing about a planes carry weight and someone asked what would happen if you had 15 pounds of feathers and circulated them through the air of the cargo hold would it still weigh down on the plane?

ANSWER:
Look at one feather. The feather will be moving
with a constant velocity because it has a very
low terminal velocity (that is, if you drop a
feather it will at first accelerate down but
rapidly stop accelerating and fall with a
constant speed called the terminal velocity). An
object with a constant velocity will have, by
Newton's first law, a net force of zero. What
are the forces on the feather? There is its own
weight W and an upward
force D due to the air
drag and they must be equal and opposite because
the net force, as noted, must be zero. Now,
Newton's third law says that if the air exerts a
force up on the feather, the feather must exert
an equal and opposite force down on the air.
Therefore the feather exerts a force of
magnitude D down on the air, but since
D has the same
magnitude as W , the
feather exerts a force equal to its own weight
on the air. Finally, since the air is also in
equilibrium, the force it exerts down on the
floor of the airplane will be the weight of the
air plus the weight of the feather. In other
words, the 15 lb of feathers floating around is
still felt by the plane just as if they were all
sitting on the floor. For a really exhaustive
examination of this kind of problem see an
earlier answer . For another question similar
to yours, see another
earlier answer .

QUESTION:
What would happen if a solid sphere of electrons ( a sphere of electrons so densely packed, that it's not only large enough to be visible, but also to weigh a considerable amount) was suddenly brought into existence?

ANSWER:
This question is so vague that I should have
just discarded it. What does "large enough to be
visible" mean? What is "a considerable amount"
of weight? What I can tell you is that whatever
happens will be, since electrons repel each
other, that this sphere will probably blow apart
more quickly than "suddenly". Also, "suddenly
brought into existence" could be thought to
violate site
ground rule forbidding "questions based on
unphysical assumptions". Anyhow, I can talk a
bit about uniformly charged spheres and
demonstrate how unphysical this question really
is. The total energy it takes to assemble a
uniformly charged sphere of radius R to
a total charge Q is E =3kQ ^{2} /(5R )
where k is Coulomb's conatant and is
about 9x10^{9} N m^{2} /C^{2} .
Now, the mass of an electron is about 9x10^{-31}
kg and its charge is about -1.6x10^{-19}
C. So I will call 1 gram=10^{-3} kg
"considerable" and will therefore need about 10^{27}
electrons; their total charge will be Q =-1.6x10^{8}
C and Q ^{2} =2.6x10^{16}
C^{2} . Suppose we jam them into a sphere
of radius 1mm=10^{-3} m, plenty big to
"be visible"; then the energy to assemble these
electrons would be E =1.4x10^{28}
J. If "suddenly" means 1 s, this is a power
requirement of 1.4x10^{28} W. The total
power consumption of the earth is 44x10^{12} ,
miniscule compared to that the create this
electron ball. Furthermore, there is no force to
bind this ball and it will blow itself apart
with the energy you put into it; you would not
want to be around it!

ADDED
THOUGHT: I am aware that the energy to
assemble a sphere full of discrete charges is
different from the energy to assemble a
uniformly charged sphere. Doing the problem this
way would essentially lead the same conclusions.

QUESTION:
I have a thought experiment that I've been puzzling about lately.
Say, that we make a narrow hole from the surface and vertically down to the centre of the earth, and fill it with air. If we put a human in this centre spot, he/she will experience weightlessness as the net graviation will be zero, i.e. equal in all directions.
But what will the air pressure be in the centre point given it is 1 atmosphere at the surface.
We may assume a homogeneously dense earth with equal mass per m^2 all the way through.

ANSWER: First of all, you should
acknowledge that the density of the earth is not
uniform, far from it; for more details about
this, see an
earlier answer . So what we are doing is an
academic exercise not really very close to
reality. Still, it is an interesting exercise.
Above the earth the
prediction of the pressure as a function of
altitude is fairly standard. I urge you to look
over that derivation since I will follow the
same procedure underground. The result there is
P=P _{A} exp(-mgh /(kT ))
where h is the altitude, m is the mass
of a molecule of air, g is the
acceleration due to gravity, k is
Boltzmann's constant, P _{A} is atmospheric
pressure, and T is the absolute
temperature. This derivation assumes that g
is constant over the range of h that contains
most of our atmosphere; that is not a bad
approximation since h is much smaller
than the radius of the earth. It is important to
note that there are really two unknowns here,
P and T . This model is usually
used assuming that T is constant; since
T actually decreases with h , if you measure
P and
then solve for T , you will find that
the measured pressure is larger than the
predicted pressure.

So, to predict the pressure in your hole I will
follow the derivation given in the
hyperphysics
site and start with ΔP=-ρg (r )Δr
where ρ is the density of the air, r
is the distance from the center of the earth,
and g (r )
is the acceleration due to gravity at r .
For a uniform sphere it is easy to show that
g (r )=gr /R
where R is the radius of the earth and
g =9.8 m/s^{2} . Now use the
ideal gas formula to write ρ=mP /(kT ) so
that the differential equation for P is
dP /dr =[mPg /(ktR )]r ;
integrating, I find that P =P _{A} exp[-(C (1-(r ^{2} /R ^{2} ))]
where C=mgR /(2kT ). Choosing
T =300 K, I find C≈ 350
so P /P _{A} ≈ exp[-350(1-(r ^{2} /R ^{2} ))].
I have plotted, below, the predicted
pressure both inside (red) and outside (blue for
constant g , black for g (r )=g (R /r )^{2} )
the earth; note that the
graph covering 0<r /R <2 is
logarithmic because the pressure is almost zero
very rapidly. Both inside and outside the
pressure falls very quickly away from the
surface, faster for inside because of the
quadratic rather than linear term in the
exponential function. The other graph shows the
curves 1% above and below the earth's surface;
inside the pressure is nearly zero, outside 80%
below atmospheric pressure.

The center of the earth has a temperature of
about 6000 K which would result in a much larger
pressure of about 10^{-8} P _{A} ,
but still very small. Just as the case of the
prediction of P above the earth's
surface has difficulties related to the
temperature profile, so does the prediction of
P below the surface.

FOLLOWUP QUESTION:
I thought a uniform-earth assumption would make the calculations easier, but still give a reasonable idea. I guess the earth density and temperature will follow a similar pattern as the air pressure; the physics are similar?

ANSWER:
You are right, the uniform-earth assumption does
make it easy to calculate and will give you a
very good estimate of what happens in the first
few kilometers. The second graph is the one
which gives you the most information; what you
can say about the pressure deeper than a few
tens of kilometers is that the pressure is
trivially small. The same is true for the
pressure profile above surface: after a few tens
of kilometers the atmospheric pressure is, for
most purposes, zero. For both cases, knowing the
temperature profile would improve the accuracy
of the model but there is no likelihood of
knowing that more than about 12 km below the
surface where the temperature is alread at about
500 K. It is harder to know what the temperature
profile actually is because we have not way to
get instruments down there. The model we have
been playing with would be correct if we could
perfectly insulate the walls of the tube. The
6000 K comes from the fact that we know that the
core is molten iron.

ADDITION:
There was an error in my original post in
the calculation for r>R which has been
corrected in the graphs. I also did the
analogous calculations for r>R where
the variation of g was included.

QUESTION:
What determines an object’s speed when falling with an open parachute?

ANSWER: What determines any objects
motion is the forces acting on it. A freely
falling object has two forces on it, its own
weight down mg and a
drag force up D . D
is proportional to the square of the speed v ^{2} .
So, if you drop something it starts with very
little drag and the weight is the main force
which causes the object to accelerate downward;
eventually v gets big enough so that
the drag is equal to and opposite the weight so
the object then falls with a constant speed,
called the terminal velocity, for the rest of
the fall. There is a well-known expression for
the
drag force which works well in most
every-day situations, D =½ρv ^{2} C _{D} A.
You can learn more about it at the link given
above, but for our purposes here, for an object
which presents an area A to the onrushing air at
sea level, a fairly good approximation is D =¼Av ^{2} .
(Warning: this equation only works if you work
in SI units.) So, we can estimate the terminal
velocity, v _{T} =√(2mg /A ).
Let's estimate the mass of a parachutist plus
unopened parachute as 100 kg and her area about
1 m^{2} ; then you could estimate v _{T}
for a sky diver plus unopened parachute as about
45 m/s=100 mph. Now, when the parachute is
opened maybe the area increases to 20 m^{2} ;
now v _{T } is about 10 m/s=22
mph. These are just rough approximations to
emphasize that the area is key. Actual numbers
will depend a lot on the design of the
parachute.

QUESTION:
Ok this is gonna sound weird but say your falling from an airplane and you have your phone in your hand. If you throw your phone upwards right before impact would it still break or at least suffer less damage? I can’t tell if this is like the jumping in a collapsing building situation or if it would actually help.

ANSWER: The terminal velocity of a human
body is about 120 mph. Do you think you could
throw your phone with that speed? Certainly not—a
big-league pitcher can throw a baseball only
about 100 mph and even he could not do that with
his feet not on the ground. If you could give
the phone that speed straight up it would have
zero velocity relative to the ground. I doubt
that you could possibly give the phone more than
about 20 mph vertical velocity meaning it would
hit the ground with speed around 100 mph.

QUESTION:
We have a large puddle outside our back door. Yesterday, the sun was casting a reflection from the puddle onto the ceiling of our kitchen through our sliding doors. It was pretty so we were watching it for awhile. What we noticed was that although the puddle was very still to our eyes, the reflection on the ceiling showed constant movement. We saw ripples and swirls and lines of movement, but the puddle appeared completely still. Little bits of dust/pollen/leaves which were floating in the puddle weren't moving at all. If it matters, the water was very clear, it was about an inch at it's deepest, and the puddle was sitting on a very white roofing liner which was just installed last week. (Our back door leads to a roof.) Is the reflection showing movement under the surface that we cannot otherwise see?

ANSWER: The rays from the sun come in
parallel. If the puddle were perfectly flat,
like a mirror, the reflected rays would also be
parallel. But if there are very small ripples in
the water, too small to notice, the reflected
rays would not be parallel. Even if the angle
between two rays was very small, their
separation would be greatly enhanced over the
large distance they have to travel to the
ceiling. This magnifies any "nonflatness" of the
puddle.

QUESTION:
I am writing scifi fanfic and I need help in physics.
In the fic planets are protected by a sort of a future version of coastal artillery when waships are not available. So the problem I have is the needed quantity of the guns.
Assuming a gun is placed at the height of a height of 25 meters on a turret tower, and the gun can aim straight up or straight ahead effectively having a 180 degree firing angle, how many guns would be required on minimum to cover the entore space around the Earth with no blind spots?
Would a thousand suffice?

ANSWER: This is not physics. I find
description of the way a gun can point to be not
clear; any direction above horizontal seems to
be what you can aim at or see. The 25 m height
seems to be superfluous unless the point is to
extend the horizon; but then horizon to horizon
would be greater than 180^{0} and the
change in horizon would be trivially small. I
assume that you are looking for something coming
in from far away, that is it is ok to have blind
spots near the earth as long as anything far
from the earth can be

seen.
Imagine one gun at each pole; they see anything
above a plane tangent to the earth at the pole.
Imagine one gun at the equator, another at the
equator on the opposite side of the planet; they
can see anything above their tangent planes.
Finally station guns 900 longitude from the
other two on the equator. You have made an
imaginary cube around the planet using only six
gunners into which no one can enter undetected.
You are screwed, though, if somebody gets
inside. It just occurred to me that you could
put the planet in a tetrahedron with only
four gunners.

QUESTION:
If you could make a radio with a frequency inside the visible spectrum, would you be able to see it?

ANSWER: Yes, that is what our eyes
evolved to do—see light in that wavelength
range. If the radio were designed to operate at
a given frequency, like AM radio is, that is the
color you would see. You would have trouble
creating an antenna, the dimensions of which are
usually the same size as the wavelength.

QUESTION:
Hello! I have a really quick question: what are the simplest forms of the formulas for each of Newton's 3 laws of motion?

ANSWER: It breaks the heart of a
physicist to think that students think that the
best way to learn physics is to memorize
"formulas". Newton's laws are monumental
concepts and should be not thought of as
formulas. I will try to specify these
conceptually, not mathematically.

An object which experiences zero net force
will remain at rest or moving with a
constant speed in a straight line.

The acceleration (change of its velocity) of
an object is proportional to the net force
applied to it and inversely proportional to
its mass.

If one object exerts a force on a second
object, the second object exerts an equal and
equal force and opposite on the first.

QUESTION:
So I work on boats and we entered a problem with barge drafts in the Mississippi river. And we have a 9 foot 6 inch draft on our barges but we generally have a 9 foot draft. The river was running at 4 mph. We stalled out and eas going 2mph backwards and the captains were blaming a 9 foot 6 in draft. Had me curious what the differnece in drag does 6 in make in water against a 4 mph current.

ANSWER: The operative equation is the
equation for drag force D moving with
speed v in a fluid with density ρ
and speed
v _{R} m/s, D =½ρ (v +v _{R} )^{2} C _{D} A ;
C _{D} is called the drag
coefficient and depends only on the shape of the
barge and A is the area presented to
the oncoming water. I have assumed the barge is
moving upstream and this force, of course, is
downstream. Now, what you want to know is what
is the increase in drag when you change the
draft. The only thing changed by the the draft
is A and A is proportional to the draft. So, if
we take a ratio D _{9.5} /D _{9} =9.5/9=1.06;
so the drag increases by about 6% if the two
draft situations have the same speed. If they
have different speeds, then you would have to
calculate that, D _{9.5} /D _{9} =1.06[(v _{9.5} +4)/(v _{9} +4)]^{2} .
For example if some force (tow boat) pulls the
barge and the speed is 5 mph for the 9.5 ft
draft and 7 mph for the 9 ft draft, D _{9.5} /D _{9} =1.06[(5+4)/(7+4)]^{2} =0.71;
there is about 29% less drag for the 9.5 ft
draft because it is going slower. I can't get
any more quantitative without more information
about the barge and the way it is being towed.

QUESTION:
Is there a way to find initial velocity with just the time and the angle of the projectile motion?

ANSWER: Suppose we choose our coordinate
system so that the projectile starts with speed
v _{0} at the origin. Then we have four equations
describing the subsequent motion:

v_{x} =v _{0} cosθ

v_{y} =v _{0} sinθ -gt

x =v _{0} t cosθ

y=v _{0} t sinθ- ½gt ^{2}

What do you know? t , θ , and g .
What do you not know?
v_{x} , v_{y} ,
x , y , and v _{0} .
Since you have more unknowns than equations, you
cannot solve the problem. The answer to your
question is no.

QUESTION:
If you drop a 10 pound cooking pot on your foot is it just 10 pounds?

ANSWER: See the figure. When the pot hits
your foot there are two forces acting on it, its
own weight W and the
force F which your
foot exerts on it. In order for the pot to slow
down and stop, there must be a net force upward.
Therefore, F is greater than W . Finally, we
invoke Newton's third law which says that if you
exert a force on the pot, the pot exerts an
equal and opposite force on your foot and so you
experience, for a brief time, a force larger
than 10 lb.

QUESTION:
Have you seen the movie 'Avengers:Endgame'
In this film Bruce Brenner says that if you go back in time and do something there it will only affect that particular timeline But in the back to future series its not as such? What will really happen if I go back in past and do something

ANSWER: The laws of physics, as we
currently understand them, forbid backward time
travel. Forward time travel is allowed.

QUESTION:
A gecko was clinging on my windshield and my friend who was driving the car decided to accelerate. 60 mph gecko was still on the windshield but at 64 mph poor little guy went airborne. I was upset of course and now wondering if the gecko would have a chance to make it ... being so light. Does mass matter? There was no wind and gecko most likely landed on the road.. yes the car is a moving object but the mass of the reptile is so light.

ANSWER: Geckos come in quite a range of
sizes. I am going to guess yours was some where
in the middle, say his mass is about m =2
oz=0.055 kg and his length is L =5
in=0.13 m. The initial speed is v _{0} =64 mph=29
m/s. Estimating his width as 1 cm=0.01 m, his
area is roughly A =0.01x0.13=1.3x10^{-3}
m^{2} . So you can estimate the force on him as
F =-¼Av ^{2} =ma =m (dv /dt )
or (dv /dt )=-v ^{2} x6x10^{-3} .
Integrating this differential equation, 6x10^{-3} t= (1/v )-(1/29)
or t =167(1/v -.035) seconds. Oh
dear, I started this answer expecting to
reassure you that the little fellow would be
fine. But, suppose that we wanted him to have a
speed of 2 m/s to be ok when he hit. Then the
time to slow to that speed would have been about
78 s and he would have hit long before that. So,
I calculated his terminal velocity, the speed he
would attain if he had simply been dropped from
some large height. That would be v _{terminal} =√(2mg /A )=28.8
m/s, just about the speed he already had when he
became airborne! If there had been no air drag
at all and you dropped him from about 3 m, he
would have hit with a speed of 7 m/s=15 mph and
it would only have taken him about 0.4 s. In
this time, there is no hope of his slowing down
appreciably. I am afraid that the gecko met an
unhappy end.

QUESTION:
This question is driving me nuts. Please answer it if you can. Here it comes.
Why doesn't the direction of the vibration of a vibrating tuning fork affect the direction of the soundwave it produces?
As you probably know, sound obeys the inverse-square law, which means sound waves spread from the vibrating object like a growing sphere, radiating out in every direction with equal velocity and intensity.
As I'm sure you also know, sound traveling through air is believed to be a longitudinal wave. That is, as the sound wave travels up from the tuning fork, that air vibrates up and down. As the sound wave travels left from the tuning fork, the air vibrates left and right, etc. Significantly, it makes no difference in what direction the tuning fork pushes and pulls the air around it, the direction in which the air moves depends only on its orientation to the tuning fork. Air to its left will move left and right, the air above will move up and down, etc.
In other words, where sound is concerned, the direction of the force does not equal the direction of the acceleration. How can this be?

ANSWER: You state that "As you probably
know, sound obeys the inverse-square law". What
I definitely know is that this is true only if
the source is a point or a sphere; a tuning fork
is neither. You also state that "sound traveling
through air is believed to be a longitudinal
wave"; that implies that we are not sure—trust
me, sound is a longitudinal wave. A longitudinal
wave is one where the medium vibrates
parallel to the direction of propogation. The
animated figure shows a representation of a
vibrating tuning fork which, when vibrating,
compresses the air in front of it and that
becomes a high pressure (higher density) part of
the wave and propogates forward. The figure
represents the waves as plane waves which is
not, strictly speaking, accurate, because of the
finite extent of the vibrating tines; however,
it is certainly more accurate than representing
them as spherical waves which would be the case
if the inverse square law applied. The second
figure shows roughly what the distribution of
intensity of sound is in a horizontal plane
(viewed from directly above the tuning fork)
sounds like if your ear is very close. Contrary
to what you state, there are places where the
sound is quiet (off in diagonal directions) and
where it is loud (left and right where you would
expect it, as well as forward and backward where
you might not expect it). In fact, if you move
the tuning fork farther from your ear you find
that the forward/backward loudness dies off more
quickly than the left/right intensity. The
reason for this is too technical to go into
here, but not unexpected intuitively. The next
two figures show the measured intensities in the
near and far observation positions; the distance
from the center of the circle of the curves at
any angle represents the loudness of the sound.

I really have no idea what you are trying to get
at when you talk about the motion of the air. It
is important for you to understand that there is
no net motion of the air itself is any
direction. Focus on any air molecule and it
simply moves back and forth parallel to the
direction the wave is moving there. The wave
moves in some direction, the air does not.

QUESTION:
If I have two barbells that weigh exactly the same but one is 7' in length and the other is 9' in length would there be a difference in how they felt to the lifter?

ANSWER: If
the lifter has his hands the same distance apart
for both barbells, I can see no reason why you
would notice any difference. The longer barbell
would have a larger moment of inertia because
the major part of the weight is farther from the
center than for the shorter barbell. What that
means is that the shorter would be slightly
easier to get rotating in a vertical plane; so
if the lifter applied a net torque to the
barbell, for example by lifting with a slightly
larger force with one arm than the other, the
longer barbell would be more stable against
rotating. I doubt that this would be noticeable
to the lifter. That is the only difference I can
see. (I have assumed that negligible bending
occurs.)

QUESTION: Considerable water mass is stored in Earths polar ice-caps. If the ice-caps melted, and the stored water released, would this raise the sea-level more in equatorial regions? If so, would this in turn slow the earths rotation and thus reduce the total days in say a century?

ANSWER: Indeed when glaciers and ice
sheets on land melt the redistribution of all
this mass will result in an increased moment of
inertia which would slow down the the earth's
rotation rate. The most reliable and most quoted
estimate (by a Professor Steven Dutch of UW,
Green Bay) is that the length of the day would
increase by about 2/3 of a second. I also found,
interestingly, that removing the weight from
polar regions would result in the crust spring
back upwards by some amount, analogous to
removing a bowling ball from a trampoline;
similarly, the increased weight at lower
latitudes would tend to push the crust down
there and toward the poles. At least one site I
visited stated that the resulting decrease in
moment of inertia would approximately nullify
the increase in day length due to the melting.

QUESTION:
I've got a question that bothers me, please pardon my level of physics ignorance. So:
Let's assume that lightning strike has 100MV.
If it strikes me directly, 100MV will go through my body into the ground.
Let's say that I hold conductive rod with R_{1} = 0.01 Ohm that's let's say twice as tall as me (doesn't really matter in my question, it has to

strike the rod first) and lightning strikes the rod. Let's assume that human body in this situation has resistance of R_{2} = 1,000,000 Ohm. If the lightning strikes the rod it will go through parallel resistor (me + rod) hence the total R will be 0.00999 Ohm. If we calculate total current it will be 10,000,000,100 A. 10,000,000,000 A will go through the rod, 100 A will go through my body. I know I simplified it a lot, I'm not even a decent at that, but the question is - there's a saying that electricity takes path of the lowest resistance - but why would I prefer to hold super conductive rod than to hold anything less conductive, if the current through my body will remain unchanged? I total = I_{1} through R_{1} + I_{2} through R_{2} , right? So that would mean lower R_{1} means lower I_{2} , but it doesn't, because with lower R_{1} , R_{total} is lower hence I is higher, so I_{2} remains unchanged. I believe it has to do something with power, I'm just not sure.

ANSWER: As best as I can tell, you are
asking what happens if the rod has zero
resistance. The resistance of you and the rod is
R _{net} =R _{1} R _{2} /(R _{1} +R _{2} ).
If R _{1} =0 then R _{net} =0,
all the current will flow through R _{1} .
No rod will have zero resistance, but as R _{1} ⇒0,
I _{2} ⇒0.

QUESTION:
Imagine that you are trapped in a room in a space station that rotates so that you are subject to a centrifugal force exactly to one G. You don't know if you are on planet earth or in deep space. Is it possible to tell if you are subject to gravity or centrifugal force? If yes, how do you do that, what instruments do you need to have?

ANSWER: The answer is no; the equivalence
principle, an important cornerstone of the
theory of general relativity, states that there
is no experiment you can do which allows you to
determine whether you in empty space
accelerating with an acceleration a or
you are in a gravitational field where the
gravitational field causes an acceleration a
at the point where you are performing the
measurement. Usually, like your rotating
cylindrical room, we think of it as having an
angular velocity ω and a radius
R and if
we adjust these so that g=v ^{2} /R =(Rω )^{2} /R =Rω ^{2} ,
then it will be just like being on the earth.
But that is wrong because the field close to the
surface of the earth is very nearly uniform—g
is very nearly the same at both your head and
feet. That is not the case in the rotating room
because the acceleration depends on R
and therefore your head "weighs more" when you
are standing up than when you are lying down. If
you could simulate a gravitational field which
increased linearly with R as you moved
away from some straight line (analogous to the
rotation axis of your room), there would be no
way you could determine whether you were in that
field or in your rotating room. So for the
rotating space station to be a good simulator of
gravity on earth, it would need a very large
radius. If you want to read much longer
discussions of rotating gravity simulators,
there are several
earlier answers to questions similar to
yours.

QUESTION:
Imagine that you are trapped in a room in a space station that rotates so that you are subject to a centrifugal force exactly to one G. You don't know if you are on planet earth or in deep space. Is it possible to tell if you are subject to gravity or centrifugal force? If yes, how do you do that, what instruments do you need to have?

ANSWER: The answer is no; the equivalence
principle, an important cornerstone of the
theory of general relativity, states that there
is no experiment you can do which allows you to
determine whether you in empty space
accelerating with an acceleration a or
you are in a gravitational field where the
gravitational field causes an acceleration a
at the point where you are performing the
measurement. Usually, like your rotating
cylindrical room, we think of it as having an
angular velocity ω and a radius
R and if
we adjust these so that g=v ^{2} /R =(Rω )^{2} /R =Rω ^{2} ,
then it will be just like being on the earth.
But that is wrong because the field close to the
surface of the earth is very nearly uniform—g
is very nearly the same at both your head and
feet. That is not the case in the rotating room
because the acceleration depends on R
and therefore your head "weighs more" when you
are standing up than when you are lying down. If
you could simulate a gravitational field which
increased linearly with R as you moved
away from some straight line (analogous to the
rotation axis of your room), there would be no
way you could determine whether you were in that
field or in your rotating room. So for the
rotating space station to be a good simulator of
gravity on earth, it would need a very large
radius. If you want to read much longer
discussions of rotating gravity simulators,
there are several
earlier answers to questions similar to
yours.

QUESTION:
Imagine that you are trapped in a room in a space station that rotates so that you are subject to a centrifugal force exactly to one G. You don’t know if you are on planet earth or in deep space. Is it possible to tell if you are subject to gravity or centrifugal force? If yes, how do you do that, what instruments do you need to have?

ANSWER: The answer is no; the equivalence
principle, an important cornerstone of the
theory of general relativity, states that there
is no experiment you can do which allows you to
determine whether you in empty space
accelerating with an acceleration a or
you are in a gravitational field where the
gravitational field causes an acceleration a
at the point where you are performing the
measurement. Usually, like your rotating
cylindrical room, we think of it as having an
angular velocity ω and a radius
R and if
we adjust these so that g=v ^{2} /R =(Rω )^{2} /R =Rω ^{2} ,
then it will be just like being on the earth.
But that is wrong because the field close to the
surface of the earth is very nearly uniform—g
is very nearly the same at both your head and
feet. That is not the case in the rotating room
because the acceleration depends on R
and therefore your head "weighs more" when you
are standing up than when you are lying down. If
you could simulate a gravitational field which
increased linearly with R as you moved
away from some straight line (analogous to the
rotation axis of your room), there would be no
way you could determine whether you were in that
field or in your rotating room. So for the
rotating space station to be a good simulator of
gravity on earth, it would need a very large
radius. If you want to read much longer
discussions of rotating gravity simulators,
there are several
earlier answers to questions similar to
yours.

QUESTION:
I was wondering how do we observer galaxies complete, If they are 100 light years in diameter then the light would be scattered in a trialing effect not a complete shape, yet we see them as entire complete shapes, as through images we see on Hubble. To me that would be impossible as relativity would suggest. Would it not?

INTERPRETATION
I believe the questioner must not be a native
English speaker since I had difficulty
understanding what he was asking. I believe that
the question is essentially the following. If we
observe a galaxy, the plane of which is not
perpendicular to our line of sight, how can the
image be correct if the light we see from the
far side came many years before light from the
near side?

ANSWER:
As stated on the site, I normally do not answer
astronomy questions, but this one intrigued me.
The questioner has made an error in the size of
a galaxy. Since spiral galaxies are as large as
about 400,000 light years in diameter, the Milky
Way being 100,000 light years across, how can an
image of a galaxy be even approximately correct
unless it is normal to our line of sight? The
far side will be what the galaxy looked like
tens of thousands of years earlier than the
image of the near side, greatly distorting what
it must "really look like". The answer, I
believe, is simply that the galaxy does not
really change much over times on the order of
tens of thousands of years. For example,
although our galaxy is 100,000 light years
across, its rotational period is on the order of
250,000,000 years; so the angle which galaxy has
rotated in the time it takes light to traverse
its diameter is 360^{0} (100,000/250,000,000)=0.14^{0} .

QUESTION:
How would I put the following statement in an equation to solve?
if an event occurs every 40 seconds how long would it take to reach 300,000,000 events?
I am a technician at a casino and working out an issue.

ANSWER: The
the time T required is T =(300,000,000
events)x(40 seconds/event)=1.2x10^{10}
seconds. That is the answer, but maybe we should
put it into different units of time, years,
let's say: T =(1.2x10^{10}
seconds)x(1 hour/3600 seconds)x(1 day/24
hours)x(1 year/365.25 days)=380 years.

QUESTION:
Please understand I have very little understanding in physics. Because of best one third law if I jump and land in the ground, I would apply force to the ground and it
would go back up at me, so if I go to the ground jumping with more force than my bodyweight gravity would hit back with more than my bodyweight. I would even if for less than a second weigh less right.

ANSWER: First of all I want to get
something well understood: weight is the force
which the earth exerts down on you and no matter
what else might be happening to you, your weight
is always the same. Therefore your statement
that you might "weigh less" is wrong. So, I hope
that I can help you understand the physics of
jumping. Look at the picture. One woman is in
the air and there is one and only one force on
her, her own weight W .
Because of Newton's second law (N2), there is an
acceleration in the same direction as the
net force which slows her down if she is
going up and speeds her up if she is going down.
The other woman has her feet in contact with the
box and is either just landing of just about to
go up. There are now two forces on her, her
weight W and the force
which the box exerts up on her which I will
denote as N . If she is
just landing, she is slowing down and so the
acceleration needs to be upward because she is
moving downward; in this case the net force
must point up, so N>W as shown. If she
is just about to go up, she is speeding up and
so her acceleration also needs to be upward
because she is moving upward; in this case the
net force must also point up, so
N>W as shown. Newton's third law (N3) now
states that, if the box exerts an upward force
of magnitude N on you, you exert a
downward force of magnitude N on the
box.

Here is what you said correctly: you said that
the force with which you would have to push down
on the box would be bigger than your weight.
Here is what you said incorrectly: you said that
"gravity would hit back with more
than my bodyweight". You were apparently
thinking that W and
N would have to be
opposite (correct) and equal (incorrect) because
of Newton's third law. But W
and N are not a N3
pair because they are both on you. N3 pairs are
always on two different objects which exert
forces on each other: if woman exerts a force on
box, box exerts an equal and opposite force on
woman.

If the woman
is just standing on the box, N
and W are equal and
opposite but N3 has nothing to do with it;
rather, because she is in equilibrium (not
accelerating) all forces on her must add to
zero, Newton's first law (N1).

QUESTION:
Would there be any change in the theories of relativity or other related theories if the speed of light was slower, like say, 30000 m/s?

ANSWER: There would be no change in the
theory of relativity, but it would make a huge
difference for most of the technology we are
used to which often requires very exact
measurements of and comparison of times or
distances. The number you suggest is 10,000 time
smaller than the actual speed of light, so it
would take about 2 months for light from the sun
to get to us. Also, since c plays an
important role in quantum mechanics, atomic
physics would be different, perhaps making the
formation of atoms impossible given the other
physical constants. And, since E=mc ^{2}
would still be true, stars, if any existed at
all, would burn out approximately 10^{8}
times sooner; for example, the sun will have a
total life time of about 10 billion years but
would only have lasted about 100 years with your
value of c . I believe that for atoms,
nuclei, elementary particle physics, etc . to
work at all, you would have to adjust all the
other physical constants to also be different.

After writing the above I googled around a bit
and discovered that it is a very complex
situation. A thread on
Physics Forums is enlightening. This should
give you some food for thought!

QUESTION:
I was wondering if you could help me out here. what unit is used to measure radioactivity around you(like if you're in a radioactive neighborhood/city, etc), I've heard about Roentgens and Curies but those aren't used anymore, also there are Becquerels, Grays, Sieverts and RADS but i'm kinda confused which unit is actually used to measure radioactive levels in an area, both in SI Unit and Non-SI unit.

ANSWER: I agree it is confusing, but
there are many kinds of radiation of many
different energies and effects. The Curies and
Becquerels basically measure the rate at which
the sources are decaying, therefore counting
everything. For most purposes these are not
useful in determining how dangerous the
radiation is. Maybe the source is only emitting
low-energy alpha particles which would have a
very short range in air and not penetrate past
the surface of the skin. Maybe 1000 Curies would
be the level but nothing to worry about. There
is a very good
summary from the CDC of which units are
useful for which situations.

QUESTION:
When the train makes a turn, its outer wheel and inner wheels covering a different distance at the same time. How come the train does not come off the tracks?

ANSWER:
Basically, it is because the wheel shape is not
like a cylinder but a section of a cone. As the
train goes around the curve, centrifugal force
moves the train outward making the effective
radius of the outer wheel larger than that of
the inner wheel. There are lots of web sites
which explain this. Here are three of them:

https://www.popularmechanics.com/science/a25581/science-behind-train-tracks-wheels/

https://www.quora.com/How-do-trains-turn

https://www.quora.com/When-a-train-makes-a-turn-isnt-its-outer-wheel-covering-more-distance-than-the-inner-one-How-come-the-train-doesnt-come-off-the-tracks

QUESTION:
How come glass store front windows eventuality sag but bricks can be laid into a wall thousands of feet high if matter is mostly empty space ? strong nuclear force ?

ANSWER:
Glass does not flow if that is what you are
referring to; this is a myth and you can read
about it at this
link . It is incorrect to say that matter is
mostly empty space. Although true that most of
the mass occupies a very small volume, the
intervening space is occupied by a "cloud" of
negative charge which the electrons occupy.
Bricks are stronger than glass because they are
"proper" solids, i.e . the atoms arrange
in regular crystaline patterns which tend to be
strong, whereas in glass they do not.
Macroscopic properties of matter have nothing to
do with the nuclear force, they are determined
by atomic forces among atoms in the material.

QUESTION:
Our weight in more in day or night .I saw Vsauce video weight of shadow in which they said weight is more in day than weight in night but it is opposite in Google . Please explain to me?

ANSWER:
I generally define weight to be the
gravitational force which the earth exerts on
me, but let's define it for this question to be
the net gravitational force which the earth and
the sun exert on me. In that case, during the
day these forces are opposite and during night
they are in the same direction; therefore you
would have less weight during the day. However,
the moon also exerts a force on you so you
really should define weight to be the combined
gravitational forces of earth, moon, and sun.
Then it would also depend on the phase of the
moon. A little googling and I found that the
force on a 70 kg person would be 686 N due to
the earth, 2.4x10^{-3} N due to the
moon, and 5.9x10^{-3} due to the sun. So
I will estimate
the effect on the total weight for four
situations, new moon (moon between the earth and
the sun) midday and midnight, and full moon
(earth between the sun and the moon) midday and
midnight.

New moon day: 686-2.4x10^{-3} -5.9x10^{-3}

New moon night: 686+2.4x10^{-3} +5.9x10^{-3}

Full moon day: 686+2.4x10^{-3} -5.9x10^{-3}

Full moon night: 686-2.4x10^{-3} +5.9x10^{-3}

So
you are "lightest" with a new moon during the
day and "heaviest" with a new moon at night.

QUESTION:
I know this is more along the lines of nuclear chemistry and engineering, but I figured it was worth a shot. In Marvel comics, Iron Man's suit is powered by an "Arc Reactor," which is about the size of a fist and glows blue. Assuming that the technology could be perfected and miniaturized, is it plausible that such a reactor could be made by a self-sustaining hydrogen fusion reactor that gets its fuel from the moisture in the air by electrolysis? In more direct terms, a fuel cell harvests hydrogen from moisture in the air, feeds it to a fusion reactor, which, once kick-started with initial energy, can feed its own power needs plus extra, which is then sent to the suit.

ANSWER: There is an
earlier answer also addressing the
possibility of a very small fusion reactor.
Regarding getting the fuel from moisture from
the air, any practical fusion reactor cannot
operate on single-proton hydrogen, ^{1} H,
but requires deuterium, ^{2} H and/or
tritium,^{3} H. There is very little
deuterium in plain water and virtually no
tritium.

QUESTION:
How much time (s) would it take a 40 kg object to accelerate from 0 m/s to 40 m/s if 100 N of force is applied (assuming 0 friction). I can't figure out how to solve the formula F = ma for time (s).
[I just submitted a question to you about acceleration and time. I am an 8th grade science teacher and trying to remember how to solve for time. Please don't ignore the question. I would like to teach it to my students tomorrow but need to remember how to do it myself first haha!!!]

ANSWER: OK, Newton's second law gives you
the acceleration, a=F /m =100/40=2.5
m/s^{2} . There are two equations you
need to work with for uniform acceleration if you
start at x =0 from rest, x =½at ^{2}
and v=at . In your problem you know that
v =40 m/s so you can solve for t ,
t =40/2.5=16 s. You can also find the
distance traveled, x =2.5x16^{2} /2=320
m.

QUESTION:
Assuming:

That an open vessel is filled with water and fitted with a submersed pump;

That the water in the vessel is under normal atmospheric pressure, and at all times, remains so;

That the vessel is immersed in, and is level with, a surrounding pool of water;

That the surrounding pool of water is infinite in size;

That the surrounding pool of water has a pressure of 2 atmospheres;

That friction losses are ignored;

That the pump operates at 100% efficiency.

I need to know how many joules of energy would be required to pump 1 litre of water, from the vessel, horizontally, into the surrounding pool of water.

ANSWER:
I shall assume that the volume of the
low-pressure fluid is very large compared to 1
litre so that the level of the fluid does not
change appreciably when the fluid is pumped out. I imagine moving water out
through a small hole of area a between
the two fluid containers which is at a depth
d below the surface of each; the volume of
moved water is La , imagined as the
volume shown in the diagram. The pressure at
depth d is P _{A} +ρgd
inside the hole and 2P _{A} +ρgd
outside, so the pressure difference is ΔP=P _{A}
and the force necessary to push water through
the hole is a ΔP=aP _{A} .
The work to expel the fluid is then W =aLP _{A} =VP _{A} .
Taking P _{A} =10^{5} N/m^{2}
and V =1 l=10^{-3} m^{3} ,
W =100 J. It becomes a more difficult
problem if the volume of the low-pressure water
is not very large; in that case, as the level
fell the pressure difference would become
larger.

QUESTION:
Suppose you are driving at speed v and find yourself heading straight for a brick wall that intersects the line of your path at 90°. Assuming that the coefficients of friction for stopping and for turning are the same, are your chances of avoiding a crash better if you continue straight ahead while braking or if you simply turn along a circular path at a constant speed?

ANSWER: Suppose that the coefficient of static friction is
μ
and the car has a mass m , weight mg , and the frictional
force is at its maximum, f=μN=μmg for both
crash-avoidance options.

The straight-ahead option starts
with kinetic energy K =½mv ^{2} and has work done on it
W=-fd=-μmgd before it stops with
no energy, where d is the distance to stop; therefore, ½mv ^{2} -μmgd= 0
or d=v ^{2} /(2μg ).

The turning option moves in a circle of
radius R and the centripetal force
is μmg=mv ^{2} /R .
Therefore, R=v ^{2} /(μg ).

Since d=R/ 2, straight ahead is a much
better option!

QUESTION:
I am a caregiver. I need to figure out how
many pounds a force it takes to push someone up
a ramp. I can't find anything online that will
give me a simple answer. May I please give you
the parameters and could you please figure it
out for me. The ramp length is 6 ft long ,the
height of the ramp is one foot tall, the objects
I am pushing up the ramp is a 40 lb wheelchair
and a 200 lb woman. What I need to know is how
many pounds a force am I pushing. Also when I
bring her down the ramp I am walking backwards
and pulling her down the ramp using the same
parameters can you figure out how much weight is
on me taking her down the ramp. I can't find
anyone else to ask. Can you help me. My
livelihood as a caregiver depends on it at this
point in my life.

ANSWER:
You are probably not interested in the
mathematical details, just the final answer, but
I always include the physics in my answers. If
you want, you can just skip to the end and see
what the final answer is.

Assuming you
go up with approximately constant speed, this is
a Newton's first law problem—all the forces on
the wheelchair plus patient must add up to zero.
There are four forces on the wheelchair plus
patient—the force F
which you exert up the ramp, the weight
W of the patient straight
down, the normal force N
the ramp exerts perpendicular to the ramp, and
the frictional force f
down the ramp. The angle of the ramp is
θ =tan-1(1/6)=9.46^{0} , so the
component of the weight down the plane is W _{x} =-240sin(9.46^{0} )=-39.4
lb; the component of the weight normal to the
plane is W _{y} =-240cos(9.46^{0} )=-237
lb. The most problematic force is the friction;
the most important thing is that f is
proportional to N , that is f=μN
where μ is called the coefficient of
friction. What this constant is depends on
things like the quality of the bearings in the
wheels, the nature of the surface you are
rolling on (a rug would have a bigger μ than a
hard surface, for example), how inflated the
tires are, etc . Wheelchairs are
presumably designed to minimize friction so I
will make a guess that μ ≈0.1. Now, the
physics:

y
forces add to zero, so N =237 lb;

therefore, f =-0.1x237=-23.7 lb;

x
forces add to zero so F =23.7+39.4=63.1
lb;

if
there were no friction, F =39.4 lb.

For going
down the ramp, the force diagram is just the
same except now f points up the ramp rather than
down—friction now works with you, not against
you. N and W are the same but
now F =39.4-23.7=15.7 lb. I noted that
you said you are "pulling her down the ramp" but
with my choice of coefficient of friction I find
you are actually "holding her back" from
accelerating down the ramp. If you need to pull
her down the ramp, then my estimate of μ
is too small which would mean you have to push
more than F =2x39.4=78.8 lb when going
up.

FOLLOWUP QUESTION:
I truly don't mean to look completely ignorant but.....from your answer,are you saying that I am applying 63.1 lbs or 78.8 lbs of force onto my spine when I am pushing my client in her wheelchair up her ramp. Also are you saying it's only 15 lbs of force when I am guiding her and her wheelchair down the ramp. Also how do I know this part of your answer. "find the force (call it G) necessary to move her with constant speed on level ground (with the same kind of surface as the ramp) "
I'm guess I'm going very slow and steady as to not allow her to fall off the side or myself to lose my balance as her weight pushes on my body. You can probably imagine that my legs are slanted far back to keep my balance,if that helps at all.

ANSWER:
Let me clarify what I mean by all those forces F
(e.g . 63.1, 15.7, 78.8 in
various situations). F is the
force which you must apply on the
wheel chair to move it up or down the incline with a
constant speed. Newton's third law says that if you exert a
force on the wheel chair, the wheel chair exerts an equal
and opposite force on you. This is a force on your hands.
There is no way I can deduce what this implies about your
spine, your shoulders, your legs, etc .; the body is
a very complicated machine and how you push might be
different from how somebody else might push, so there is no
simple answer. [For example, consider the simpler example of
lifting of a box. If you ask me for the force you need to
lift it, I could tell you that you must exert an upward
force equal to the weight of the box. You have probably
heard "lift with your legs, not with your back". So for two
diferent people doing the exact same thing but in different
ways, one might end up with an injured back and the other
not.] Regarding the 78.8 lb force I
derived, you can probably simply ignore it because it is
only for the special case of very high friction in the
wheelchair which I think unlikely. You can easily test
whether the friction is very high by carefully letting go of
the wheelchair at rest: if it begins begins moving down the
ramp, the friction is pretty low, but if it just sits there,
the friction is large. High friction would mean that you
have to exert a force up the ramp on the chair to move it up
the ramp of at least
78.8 lb (pointing up the ramp). The condition for high
friction would be μ ≥39.4/237=0.17. Keep in mind that I am only guessing about
the coefficient of friction μ ; if you do the
test above and find the chair will roll down if you do not
hold it, at least you will know that μ <0.17.
I have removed the part about a force
G in my original answer because it is too confusing.

ADDED
THOUGHT:
Look at the second picture. Is that the way you
bring her down the ramp, you being below the
chair? If not, I will need to redraw that
picture with you above the chair on the ramp.
The numerical answers will not change, though.

QUESTION:
Hello, my question is regarding radioactive
decay. How exactly does an individual atom decay
into byproducts? What is the actual cause of
this decay? Is it the strong nuclear force
becoming weaker over time and then at a certain
point the nucleus can no longer be kept
together? Perhaps like with the earth and the
moon, with the moons orbit slowly decaying.
There must be a cause and there must be an order
to why atom x decays at this time even if we
can’t know how.

ANSWER:
Radioactive decay is a sudden event, not one
that slowly occurs like the moon's orbit.
Suppose that there is an atomic nucleus which is
bound together. It is stable, technically only
if there is not some allowed process which would
result in energy release if it occurred. There
are lots of nuclei which are bound but not
stable. Let me give you just one type of decay
which can happen and discuss it. A proton in the
nucleus can spontaneously turn into a neutron,
an electron, and a neutrino. This is called beta
decay; it cannot happen to a free neutron since
you would have to add energy to a proton to make
it happen, but it can happen inside a nucleus.
The example I will use is the decay of ^{14} C
which is commonly used for carbon dating of
archeological items. Now, the decay is a
statistical process and since this nucleus is
bound you might have to wait for a while for the
decay to happen. But, somehow the nucleus knows
that if it can turn one of its protons into a
neutron, making ^{14} N, 0.156 million
electron volts of energy will be released;
depending on the detailed structured of this
nucleus, you may have to wait a long while. In
this case you will have to wait 5730 years for
half the ^{14} C nuclei to beta decay to
^{14} N. ^{14} N is stable against
beta decay which would be to ^{14} O,
which means that if you want it to happen you
would have to add energy. Incidentally, there is
another type of beta decay where a neutron turns
into a proton, a positron (a positively charged
electron), and a neutrino. So, if a nucleus has
too many protons it will beta minus decay, too
many neutrons it will beta positive decay. A
neutron, unlike a proton, will also beta decay
when outside a nucleus.

QUESTION:
If quantum theory states that shining red/blue lights on metals
knocks out electrons, can these be used as a shrinking ray?

ANSWER:
Actually, visible light does not knock electrons out of a
metal; you really need ultraviolet and beyond to see the
photoelectric effect. But why would you think it would
shrink something? Removing electrons just result in the
metal becoming positively charged and it would eventually
have a high enough voltage to cause a breakdown cascade in
the air around it, called a spark, to replentish the lost
electrons.

QUESTION:
I understand that a Rocket moves upwards, because it's an opposite reaction of the gases escaping, but why is that? Can we explain the Newton's 3rd law of motion using any other theories? Is it because electrons in the gas repel from each other, thus creating this repulsive force upward? Best i can visualize it is a rocket going up, by building a platform underneath it to repel from, but i would like to know how accurate that is!

ANSWER:
It is conservation of linear momentum. To simplify things,
just consider a 100 kg astronaut floating in empty space
holding a 20 kg ball. He tosses the ball and it flies away
at the speed of 5 m/s. The astronaut recoils with some
velocity -v in the opposite direction (the negative
sign denotes velocity opposite the ball). Linear momentum is
given as mass times velocity, so the ball has momentum
5x20=100 kg m/s and the astronaut has a momentum 100v .
But the ball plus astronaut had zero momentum before she
threw the ball and they must have zero momentum afterwards,
so 100-100v =0 or v =1 m/s. That's all a
rocket is: you throw stuff out and recoil in the opposite
direction. It has nothing to do with platforms repelling or
electrons in the ejected gasses. (Incidentally, conservation
of linear momentum is easy to derive just using Newton's
second and third laws. It is only true if there are no other
forces acting. But the recoil idea is true of a rocket going
up in gravity, it is just that a given amount of fuel
ejected in space will have a larger acceleration than when
leaving earth.)

QUESTION:
If I am at the lake on a boat, drop a line 50 ft straight down with
4oz weight and the boat is traveling at 4mph at what depth will the 4oz
weight be. And with an 8oz. If I want the weight to be at 25 ft in depth
while traveling at 4mph what weight should I be using?

ANSWER:
You cannot do this calculation, even approximately, without knowing the size
and shape of the weight.

FOLLOWUP QUESTION:
Please find pictures of the weight Used. This is a 2oz.

ANSWER:
This requires a calculation of the drag force on an object
moving through water. It should be appreciated that such
calculations are always approximations and generally not
reliable to better than perhaps 20% or more without doing sophisticated
aerodynamic calculations on a super computer. In that
spirit, I will push forward to get estimates; I enjoy
puzzles like this. In the end, however, I
will advise you to determine the answer by trial and error!
I will, as most scientists do, work in SI units but will return
to English units in the end. The drag force D on an object with a cross-sectional area A presented to the oncoming fluid
and moving with speed
v through a
fluid with density ρ , is given by

D =½ρv ^{2} C _{D} A

where C _{D} is the drag
coefficient, a constant which depends only on
the object's geometry and orientation relative
to the fluid flow. In our case, ρ =10^{3}
kg/m^{3} and v =4 mph=1.8 m/s. Now we come
to your weight; it is clear that it is made of
lead which has a density of 1.13x10^{4}
kg/m^{3} . The problem is that the drag
coefficient for this shape is not known and you
really do not know what angle it would assume
relative to the direction of the line, so you
would either have to do extensive measurements
or make simplifying approximations; since we
just want a rough approximation, so I am going
to assume that the weight is a sphere of radius
R , C _{D} =0.47 and A=πR ^{2} .
(You may be amused by
"consider a
spherical cow ".) R for a solid
sphere of mass m may be
calculated using the density of lead. Putting this all
together, I find that D =9.2m ^{2/3} .

So that takes care of the fluid dynamics part of
the problem, the part where approximations are
made. The rest is pretty simple first-semester
Newtonian physics. The weight W is in
equilibrium so W=T sinθ and
D=T cosθ . For your
desired situation, sinθ= 25/50=0.5,
so θ =30^{0} . Doing all
the arithmetic and algebra here I find that
m =0.067 kg=2.4 oz.

Finally, I emphasize again that this is a rough
calculation. My advice would be to go out in
your boat with a protractor, drop the weight,
and go 4 mph trying varous
weights. Since you want the sinθ= ½,
get the angle to near 30^{0} . Choose the
weight which is closest. I am guessing it will
be greater than 2.4 oz because I think the drag
coefficients for your geometry will be smaller
than for a sphere. If you do this, I would be
interested in what you find.

QUESTION:
I am a high school forensic science teacher, and could use your help.
My students do a lab in which they use a laser to measure the diameter of hair, using diffraction. The problem is that I’ve seen conflicting instructions for doing this and I am hoping that you can tell me which is correct.
We do this by shining a laser on the strand of hair, them taking measurements from the diffraction pattern. The set-up is laser—card holding hair—(1 m)—screen.
The diameter of the hair is calculated using the formula d=λL /x , where λ is the wavelength of the laser (around 632 nm) and
L is the distance between the hair and the “screen.” (This formula is derived from
d=λ /sinθ .) The confusion comes from measuring
x . According to some sources, we should measure the distance between the center of the central bright band and the 1^{st} order maximum (the center of the first bright area on either side). Other sources say to measure from the center to a minimum, either the center or an edge of one of the dark bands!
Here’s my question: Which is it (to a maximum or a minimum), and why? What measurement do we use?

ANSWER:
Here
is the confusion, I think. Some references say that mλ=d sinθ
locates maxima; those authors believe the pattern is the same
as a double slit with spacing d . Other references say that mλ=d sinθ
locates minima; those authors believe the pattern is the same
as a single slit with width d . The first of these
is flat-out wrong. There is a well-known rule, Babinet's principle,
which states that the diffraction pattern from some
obstruction is the same as the diffraction pattern from an
aperture of the same shape. The only difference is that the
central area of the pattern is much brighter since the part
of the incident beam which would have been stopped by
missing the slit carry right on through. Approximating sinθ∼x /L , d= mλL /x _{m}
where x _{m} is the location of the m^{th}
minimum relative to the central maximum. I have never done
this experiment, but it occurs to me that it would be
difficult to get a good estimate of the center spot because
it is so bright and broad. Rather, I would try to get the
distance x _{m,m+1} =x _{m+1} -x _{m} , the distance between two
adjacent
minima, and write d=λL /x _{m+1,m} .

QUESTION:
I'm trying to solve a little problem related to two vessels held under
pressure. Specifically, where one vessel (a fluid dispenser ... like a beer
keg) has a volume of 64 ounces and must be held at gauge pressure of 40 psi and the other
vessel is the compressed air source with a volume of 26 in^3. What is the
equation or procedure to determine the psi that the compressed air source
must start at to maintain the fluid dispenser at 40 psi through the
dispensing of a total volume of 448 ounces.

ANSWER:
How can the dispenser have a volume of 64 oz but dispenses 448 oz?

FOLLOWUP:
It’d get refilled after it emptied.
My crack at it uses Boyle’s Law and because you’d have to open the dispensing vessel after it empties, you’d effectively have to stuff 448 oz x 2 of the gas at STP into the compressed air tank and because the vessels must remain equalized after pushing the full volume of liquid I added 40 psi to my answer.
So my semi-educated guess is 954 psi. Am I right?

ANSWER:
(This turned out to be much more involved than I originally
thought. Get ready for one of my very long-winded answers!)
Boyle's law will not work here because the amount of gas in
either of the containers is changing. Where does Boyle's law
come from? It is

a special case of the ideal gas equation, PV /(NT )= constant, where N is some measure
of the amount of gas in a container of volume V , pressure
P ,
and temperature T ; N can be anything you like—number
of molecules, mass, number of moles, etc .—because,
as you will see, only ratios matter in my analysis. Boyle's
law assumes that N and T are constant
which is not the case here. In these calculations it is
important that T be absolute temperature (Kelvins)
and P be the absolute pressure which is the gauge
pressure plus atmospheric pressure which is 14.7 psi.
I am assuming that there is a valve or other
mechanism between the two tanks which keeps the
pressure at 40 psi in the beer keg as the beer
is dispensed.

I think it is instructive to first do the case
where only 64 oz of beer is dispensed and
both vessels are at 40 psi at the end. First, I
need to convert all volumes to in^{3} .
64 oz=116 in^{3} and 448 oz=809 in^{3} .
In the compressed air tank the initial absolute pressure
is P , the final absolute pressure is 40+14.7=54.7
psi, the initial amount of gas is N ,
and the final amount is N-n where n
is the amount of gas filling the dispenser after
all the beer has been dispensed. Therefore, 54.7/(N-n )=P /N .
Rewriting, 54.7/P =1-(n /N ).
Now the ratio of the volumes is 116/26=n /(N-n )
because when all the beer is out, this ratio is
the same as the ratio of the amounts of gas in
each volume; don't forget that the pressures are
now equal. Solving I find that 1-(n /N )=0.183
and so P≡P _{1} =299 psi (gauge pressure
is 284 psi).

Suppose now that we want to have enough pressure
so that two 64 oz tanks of beer can be
dispensed. The previous paragraph now represents
the second of the two dispensings, so the first
dispensing would finish with P _{1} =299
psi in the tank and start with an unknown
pressure P _{2} . The calculation
proceeds the same as for the calculation of
P _{1} except that the ratio n /(N-n )
and the ratio for the volumes are no longer
equal because the pressures are not the same.
Following the same procedure as for P _{1} ,
we now have 299/P _{2} =1-(n /N ).
I will now denote the pressure and volume in the
beer tank as p =54.7 psi and v =116
in^{3} ; similarly, P _{2
} and V =26 in^{3} for
the compressed air tank. Therefore from the
ideal gas equation, P _{2} V /(pv )=(N-n )/n.
So, 1-(n /N )=1-pv /(P _{2} V+pv )=P _{2} V /(P _{2} V+pv ).
So, finally, 299/P _{2} =26P _{2} /(26P _{2} + 6350)=P _{2} /(P _{2} + 244),
a quadratic equation for P _{2} :
P _{2} ^{2} -299P _{2} -73,000=0.
Solving, P _{2} =458 psi (gauge
pressure 444 psi).

The above procedure can be generalized as

P _{j-1} /P _{j} =1-(n /N )=P _{j} /(P_{j} + 244)

P _{j} ^{2} -P _{j-1} P _{j} -244P _{j-1} =0

P_{j} =[P_{j-1} +√(P_{j-1} ^{2} +976P _{j-1} )]/2

So,

P _{3} =[458+√(458^{2} +976x458)]/2=634
psi (620 psi gauge)

P _{4} =[634+√(634^{2} +976x634)]/2=822
psi (807 psi gauge)

P _{5} =[822+√(822^{2} +976x822)]/2=1019
psi (1004 psi gauge)

P _{6} =[1019+√(1019^{2} +976x1019)]/2=1220
psi (1205 psi gauge)

P _{7} =[1220+√(1220^{2} +976x1220)]/2=1428
psi (1414 psi gauge)

ADDED THOUGHT: When I first began to
think about this problem I thought that just
doing the problem by finding the pressure
required for a 448 oz=809 in^{3 } keg
ala the calculation for P _{1} above. I
was never able to convince myself that this
would be equivalent, so I trudged through the
step-by-step solution (which wasn't so bad after
I figured out P _{2} ). When I
had finished I did the one-step calculation:
809/26=n /(N-n ). Solving I find that 1-(n /N )=0.0321
and so P _{7} =1702 psi which is
not correct.

QUESTION:
I need to lower a 500lb barrel.straight down from a 2 story building and I have 1 minute to do it. Are there any mechanical tools that could do this?

ANSWER:
Block and tackle.

QUESTION:
How fast would Lady Valarian be traveling to go 17 light yrs in nine minutes?

ANSWER:
There are about L =9x10^{6} light minutes in 17 light
years. The speed v must be adequate to shrink, via
length contraction, L to L' =9 light minutes. So,
L'=L √[1-(v /c )^{2} ].
I find v =√(1-10^{-6} )c ≈(1-0.0000005)c =0.9999995c .

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QUESTION:
Why does value of small g maximum on the surface of the earth and decreases with height and depth ?

ANSWER:
Newton's Universal Law of Gravitation states that the force
F of attraction between two spheres (masses M
and m ) is inversely proportional to the distance
r between their centers, F=MmG /r ^{2}
where G is the universal gravitational constant. In
the case you are asking, M is the mass of the earth
and m is the mass you are measuring F with. If
g denotes the acceleration of m , then g (r )=F /m=MG /r ^{2} ;
g (r ) indicates that the value of g
depends on what r is. Note that if r=R
where R is the radius of the earth, g (R )=9.8
m/s^{2} . So this explains why g decreases
for r>R— as r gets bigger, g
gets smaller. But shouldn't that now imply that as
r gets smaller, g gets bigger? The answer is
no because of something called Gauss's law which says that
only the mass inside where you are contributes to the force,
not the whole mass of the earth. If you assume that the
density of the earth has a constant value of ρ=M /(4πR ^{3} /3),
it can be shown that
g (r )=4πρGr which decreases
linearly to zero at the center of the earth.

It is interesting to note, though, that it is a poor
approximation to assume the earth has uniform density. In
fact the core of the earth is much denser than the mantle
and crust. See an
earlier answer for more detail on this.

QUESTION:
When I was in 9th grade, I had completed an assignment early in science class and went up to the front desk of the teacher for grading as he would grade assignments immediately. Out of boredom I was messing around with the faucet on his desk and had a stream of water flowing out at a rate just enough for it to not be separate water drops, but a constant stream. Being a silly 9th grader, out of curiosity I pushed the tip of my pencil (the end with the led) into the faucet where the water stream was coming out of and to my surprise, instead of the water spraying out everywhere or evenly coming out around the pencil, it instead spiraled around it almost as if it were a spring. I asked my teacher about this and he said it was happening because the earth is spinning which I don't know is correct or not, and if it is, how? I'm 23 now and this still bothers me today because I've never been able to find an answer that can fully explain why the water reacted this way

ANSWER:
For starters, it certainly had nothing to due with the
earth's spinning. He was perhaps thinking of the Coriolis
force but it is far too weak in this situation to be the
reason for any phenomenon easily observed. Most likely the
explanation was that your pencil had a static electric
charge. See a
video
demonstrating this on YouTube.

QUESTION:
For hydrogen with one proton and one electron, what keeps them from collapsing onto each other due to the strong attractive electric force at that short distance? I would guess that the answer would NOT be the centrifugal force on the electron (like twirling a ball at the end of a string around you) spinning around the nucleus as this would assume the validity of the outdated Bohr atomic model. If that is indeed not the case that what is the answer?

ANSWER:
See an earlier answer . This also
discusses electromagnetic theory and quantum theory as
applied to the H atom. By the way, simple models like the
Bohr model can be useful in getting an understanding of the
crux of something; even if they are "outdated" it does not
necessarily mean that everything about them is wrong.

QUESTION:
If you have a large wheel and a small wheel can the small one be the force to spin the bigger wheel. Lets just say the small one is offset and a spring is pushing it for the force. Will it spin as long as the force is constant?

ANSWER:
This is the way gears work. In the little animation, a
constant torque (forget the "force" you allude to) is being
applied to the smaller gear which then drives the larger
gear. Note that the smaller gear is spinning with a larger
rpm than the larger gear. This would be like low gear in a
vehicle where the engine spins fast but the wheels spin
slower. It is just the same for wheels provided there is
enough friction to keep them from slipping on each other.

FOLLOWUP QUESTION:
This pertains to my previous question: If you have a large wheel and a small wheel can the small one be the force to spin the bigger wheel. Lets just say the small one is offset and a spring is pushing it for the force. Will it spin as long as the force is constant? So the force I speak of is the torque=radius x force. I am trying to make the torque.

ANSWER:
The only spring I know of which would work for you would be
something like a watch spring, a spiral spring like the one
shown in the figure. However, as it unwinds it does not
exert constant torque; there are
ways
to get around this but they are pretty tricky. If you want a
constant torque, why not just have a weight falling weight
as in grandfather clocks? Torque and force are not the same
thing. It would maybe help me answer your questions if you
gave me some idea of what you are trying to do.

QUESTION:
[Note: This is a condensation of several communications with the
questioner. It is an accident where a car
continued driving on a closed road under
construction.]
The picture with the orange piece of machinery is setting, just to the right of that is where the highway ended, she drove off the highway at that point, she flew 82 ft horizontally until she hit the bridge support where you see the circled black mark on the support.

ANSWER:
OK, the pictures helped. However, I still can only make a guess as to how far she fell before hitting the bridge support. Assuming that the point she left from was horizontal and level with the top of the bridge support she hit, I am guessing that she dropped 6 ft. In that case, the equations to be solved are 6=16t ^{2} and 82=vt where
v is the speed she left the road and t is the time to hit. I find that
t =0.61 s and v =134 ft/s=91 mph. Other guesses would be drops of 7 ft, 8 ft, and 9 ft giving speeds of 85 mph, 79 mph, and 74 mph,
respectively.
Hope this helps. There is no additional information which I can add.

QUESTION:
I understand that when something's in orbit around the earth it basically travels far enough in the x axis each second that the surface of the earth 'curves away' under it by the same amount it's fallen towards it. What I don't quite intuitively understand is why if objects dropped from say a cliff continue to accelerate and gain vertical velocity, those in orbit around the earth seem to stay at their initial 1/2 gt squared speed in the y axis and don't eventually spiral into the earth?

ANSWER:
When an object moves in a circle with constant speed it is
still accelerating. Acceleration is a vector and velocity is
a vector. Acceleration is defined as the time rate of change
of velocity. There are two ways velocity can change: by
changing speed (what you probably think of as acceleration)
and by changing its direction. In the case of a satellite in
a circular orbit of radius R and with a speed v ,
the acceleration points toward the center of the earth and
has a magnitude a=v ^{2} /R ; this is
called the centripetal acceleration. Of course there has to
be a force which provides the acceleration and that is
simply the weight of the satellite. If you are close to the
earth's surface (compared to the radius R _{E
} of the earth) the weight is mg and so
F=ma or mg =mv ^{2} /R _{E} ;
so the speed you must have to orbit close to the surface of
the earth is v =√(gR _{E} ).
The value of the acceleration g gets smaller as you
get farther away from the surface.

QUESTION:
If i am driving 55 mph and throw trash weighing 4 grams out of my passenger window, how far will the trash travel?

ANSWER:
Depends on the speed and direction with which you throw it. 4 grams is quite small so air drag would also be an issue; to calculate that you would also need to know its geometry. But why are you littering anyway? Wait to find a trash can.

QUESTION:
How large would a sphere made of lead need to be to atract a 1cm sphere of lead in deep space so as to capture it in it's orbit?

ANSWER:
There is no limit. Two 1 cm spheres of lead could be in
orbit around each other but since the gravitational force is
so small, their speeds would be very tiny. A space probe,
Rosetta , was sent to a comet and orbited it;
the escape velocity from the comet is only about 1 m/s so
the probe orbitted with a smaller speed than this.

QUESTION:
If you roll a ball down a gradient it will accelerate due to gravity, but
is there a theoretical shape of slope that will allow the ball to roll at
constant speed? (disregarding friction or drag.) The only thing I can think
of is a banked spiral track.

ANSWER:
There must be zero net force in the direction down the slope. With no
friction and drag, the only shape which seems to me to work is a horizontal
flat plane.

FOLLOWUP QUESTION:
Thank you for your reply.
I've been thinking about this problem some more and if the ramp was made high enough and long enough and parallel to the curvature of the earth, the ball would continue to roll/fall at a constant speed for ever. It would simply be in orbit around the earth.

ANSWER:
Very clever! Actually, our answers are the same but you were
thinking globally while I was thinking locally. A horizontal
plane is the same as your ramp because it would have to be
bent as it gets longer in order to remain "horizontal".
However, it does not need to be high and most folks would
not call this being "in orbit around the earth". Orbiting
usually means an object has only the gravitational force
(weight) acting on it. With your ring around the earth there
is also a normal force N
in addition to the weight W as
shown in the figure. Newton's second law says mv ^{2} /r=W-N
or N=W -mv ^{2} /r . So, when
v gets big enough N =0; so the ring could
be taken away and this would be the right speed for a "real"
orbit at this r . And the ring would not work
for speeds larger than the orbital speed, the object would
leave the ring; any smaller v
would result in the speed remaining constant as the mass
rolled around.

QUESTION:
We manufacture metal roofing panels and many orders have lots of different lengths of panels. For this reason the longest panels in a stack are rolled out first so the pieces get shorter as the stack grows. I need a formula to find the center-point of balance for any given stack. The panels weigh 2.08 pounds per linear foot and the length and quantity of each panel will feed into the formula. They are always aligned at one end so the stack is much heavier at one end than the other.

ANSWER:
Maybe you just want the final formula, but I will outline
how it is derived.
Each panel (numbered i) I will denote to have a length L _{i} ,
a weight W _{i} , and a center of gravity
r _{i} . Since the panels are uniform and have
weight density of 2.08 lb/ft, r _{i} =½L _{i}
and W _{i} =2.08L _{i} . I
will also specify the number of panels with the length L _{i}
to be N _{i} . The center of gravity of a
stack of panels of length L _{i } is r _{i} = ½N _{i} W _{i} L _{i} /(N _{i} W _{i} ).
Now, suppose there are two different lengths stacked. Then
the position of the center of gravity is

r _{CG} =(½N _{1} W _{1} L _{1} +½N _{2} W _{2} L _{2} )/(N _{1} W _{1} +N _{2} W _{2} )

=½x2.08x(N _{1} L _{1} ^{2} +N _{2} L _{2} ^{2} )/(2.08x(N _{1} L _{1} +N _{2} L _{2} ))

=½(N _{1} L _{1} ^{2} +N _{2} L _{2} ^{2} )/(N _{1} L _{1} +N _{2} L _{2} ).

So, if you have j different lengths, the final answer is

r _{CG} =½(N _{1} L _{1} ^{2} +N _{2} L _{2} ^{2} +...N _{j} L _{j} ^{2} )/(N _{1} L _{1} +N _{2} L _{2} +...N _{j} L _{j} ).

The more compact way to write this is

r _{CG} =½ΣN _{i} L _{i} ^{2} /ΣN _{i} L _{i} .

Here Σ denotes summing from i=1 to i=j.
Note that the denominator is the total weight of
the whole stack divided by 2.08 and that the
weight per foot is not needed to find the
location of the center of gravity.

QUESTION:
With increased bad weather, especially windy weather, I wondered. Does windmills, take any energy from storms, if only a tiny tiny tiny bit. If I recall, storms have more energy in them than nuclear bombs, so I know it would be insignificant. However, would you theoretically be able to have an infinite amount of windmills on earth to reduce storms?

ANSWER:
If anything increases in energy, including windmills, that
energy (in equal amount) must supplied by another source.
The source obviously must be the wind. Now, the site
groundrules forbid unphysical questions, and an infinite
number of windmills is obviously unphysical. However, let's
investigate whether a reasonable number of windmills could
have a significant impact on the weather. At the end of 2018
the total global wind power was about 600 GW, that's 0.6x10^{12}
W; the total world power production is about 44x10^{12}
W; an average hurricane has about 1.5x10^{12} W; the
Nagasaki atomic bomb was about 10^{13} W (10^{14}
J and I guessed the time of the explosion was 10 s). It has
been
estimated that if we put wind farms covering all land
and all off-shore water, 80x10^{12} W would be
generated. That same
article estimated that you would need to have enough
wind mills to generate more than 250x10^{12} W to
have a noticeable effect on weather, about four times the
number feasible.

FOLLOWUP QUESTION:
I guess I wasn't clear enough, in my question, though your answer is really cool.
Does windmills decrease the energy of a storm? Does windmills have any effect on storms?

ANSWER:
I think my answer does answer that question: "The source
obviously must be the wind." If the wind is the source of
the energy taken by the windmills, that means that the
windmills take energy away from the storm. It is, as you
speculate, an extremely small amount of energy. I have
earlier estimated the fraction of energy which is taken out
by turbines in the
Bay of Fundy during tidal changes; it is harder for the
windmills because it is not easy to estimate the energy of a
"typical" storm, so the following calculation is a very
approximate one. A typical windmill generates about 3x10^{6}
W and guess that a typical strong storm is 10 times smaller
than a hurricane, 1.5x10^{11} W; so the percentage
by which the energy of the storm is reduced is 0.02%, so
5000 windmills would use up most of the wind energy.

QUESTION:
We like to jump rope for exercise. Many times we use a weighted rope (the heaviest being 3 lbs). I don't think I have this right but if I swing the rope around my body (standard jump, nothing tricky) at say 10 mph and it's a 3 lb rope it has a force of 30 lbs if it hits something. However that is not the same as the rope having a weight of 30 lbs while swinging it. What would be the equivalent weight of a rope swinging at 10 mph? The faster you swing it, the heavier it feels (centrifugal force).

ANSWER:
You are right, you don't have this right! There is actually
no way you can tell the force the rope will hit without
knowing the details of the collision. To say the force is
the product of the weight in pounds and the speed in miles
per hour makes no sense; the units of force should be pounds
but your 30 would have the units of lb·mi/hr. What
you ask for is quite hard to calculate accurately. First,
what does "swinging at 10 mph" mean? In the figure I show
one piece of the rope in blue; this piece of the rope is
moving in a circle of radius r about the axis with
some frequency f (f times around every
second) and, if r is in ft, the speed it travels is
v =2πrf ft/s. For example if you are
doing 2 revolutions per second and r =2 ft, v =25
ft/s=17 mph. But, since the speed depends on r ,
every point moves with a different speed, the top being the
fastest. I have shown also the weight w and
centrifugal force c for
the piece we are examining; presumably, you want to know the
forces F with which the rope exerts on you; w
is the same for any piece of the rope the same size but
c depends on r , the largest being at the very
top in the figure and also at the very bottom as. When the
rope is straight up as shown, the forces F are
their smallest because c and w are in
opposite directions; when the rope is straight down, the
forces F are their largest because c and
w are in the same direction. But to actually
calculate the force is a formidable problem because we need
to add all the pieces which are all different (this adding
is integration in calculus if you have ever studied it).
Just to give you some idea of the answer you sought I will
assume that the rope has a negligible weight and the whole 3
lb is located in the middle of the rope and is a distance of
3 ft from the axis. You want v =10 mph=14.7 ft/s,
and c=mv ^{2} /r =6.75 lb. (Here
m is the mass which is 3/32 lb/(ft/s^{2} ), but
you don't need to worry about that.) So, at the top, the
force to hold it is 6.75-3=3.75 lb and at the bottom it is
6.75+3=9.75 lb. This is what you are calling the "equivalent
weight", I guess.

QUESTION:
How much power is needed to throw a 20 pound ball 14.5 ft straight up in the air?

ANSWER:
Power is not what you want. Power is the rate at which energy is delivered. You must deliver 20x14.5=290 ft lb of energy. The power depends on how long you are pushing; for example, if you deliver the energy over a half second,
P =290/0.5=580 ft lb/s=1.055 horsepower=786 W.

QUESTION:
If you had two incredibly large pieces of paper on the same plane that were just out of parallel to each other. One of these pieces is translating toward the other so they begin to overlap, and the point of overlap progresses from one end to the other. The question is if this physical point of overlap can travel faster than the speed of light? It does not seem to violate general relativity.

ANSWER:
Look at the figure. The two solid blue lines are the edges
of the paper at some time, and the dashed line is the moving
paper's edge a time Δt later. The moving
paper moves with velocity v perpendicular to the
not-moving edge and the intersection point moves with
velocity u along the not-moving edge. The distance
moved is v Δt and the distance
between the earlier and later edges of the moving paper is s =v Δt cosθ .
During this time interval, the intersection point has moved
a distance u Δt and s =u Δt sinθ.
Therefore,v Δt cosθ=u Δtsinθ
or u=v /tanθ. Suppose we want
u =2c and cause the paper to move with a
speed of v=c /2; then θ= tan^{-1} (v /u )=tan^{-1} (0.25)=14^{0} .
So, indeed, the point of intersection can move faster than
the speed of light. How can this be? Isn't c the
universal "speed limit"? It is the limit for transmitting
information or for any object with mass; this intersection
neither has mass nor transmits any information. There is a
similar example where something can have a speed greater
than the speed of light without violating special
relativity. Imagine pointing a laser at the moon and
sweeping the spot on the moon across the surface by rotating
the laser; because of the long arm, it would be pretty easy
to get the spot going faster than c but, just as in
your example, the spot has no mass and conveys no
information.

FOLLOWUP QUESTION:
You make the comment in your answer that this point of intersection does not contain any information/data, this is what I am questioning... What if instead of sheets of paper they were just lines (as you drew in your figure) and instead of just one line on the top there were a number of parallel lines. The distance between these parallel lines would not be equal to one another would represent some value (lets say a distance of 1 or 2). So it seems like you could transmit some data string i.e. 1212221. It seems like this difference in line spacing could be measured when the points of intersection reach the other end thus passing data faster than the speed of light?
Obviously could be digital or analog....

ANSWER:
I was a little puzzled by this one myself, so I opened a
thread on PhysicsForums.com. You can see the
discussion there.

QUESTION:
I don't understand the first thing about light. If there was a single photon emitted from a light source, would it be seen only by the person it was "aimed" at, or by any and everyone no matter where they were? If the latter, how could that possibly be?

ANSWER:
A photon is not the "first thing about light", more like the
last thing! I would say the first thing about light is its
wave nature. It was not until the 20th century that it was
discovered (by Einstein) that light really is both waves and
photons; this is called wave-particle duality. So, the
single photon in your question still is wave-like;
therefore, in spite of how you might aim your source, there
is no guarantee that it will end up where you aimed it. So
anybody in the room could see it, but once it has been
observed nobody else could see it too.

QUESTION:
To me(observer from earth), since your speed
is greater than c , the light from your flashlight illuminates the back of the spacecraft.
please reply sir this is from the book of sir arthur beiser book=modern physics (page no 8) .Thier is a diagram which confuse me?

ANSWER:
Reply to what? Nothing can have a speed "greater than c", so this is a nonsense statement. The author is likely trying to make a point
ad absurdum .

FOLLOWUP QUESTION:
Sir, I know that anything can’t travel faster than light. but Only I need to now why flashligh’s light going in the backward direction(sir see the figure which I told you).if you have read the book then ok but If you not I am sending the book so that sir you can understand I am asking a valid question or not.

ANSWER:
It is exactly as I said—the author is using an impossible example to demonstrate that you get ridiculous results if something has a speed faster than the speed of light.
Usually I will not answer questions which have something
moving faster than the speed of light, but now I see that
the point of the author is to demonstrate that it is
impossible. I can see now how you got confused; I was also
confused at first. If you apply Galilean relativity to the
example, you would say that the light was going forward with
a speed v+c , still illuminating the
front of the ship. But that is not what he is doing. Instead
he is saying that there will be Cerenkov radiation (from the
flashlight, I guess) which is like the shock wave of a
supersonic aircraft. This is in the form of a cone behind
and that is the light supposedly illuminating the back. I
find this a very far fetched example and would advise you to
simply ignore it!

QUESTION:
If you have an Entangled particle That falls into a black hole would it communicate With it’s entangled partner outside the black hole.

ANSWER:
First of all entanglement does not mean the particles
"communicate" with each other. But, once anything is inside
the Schwarzschild radius of a black hole, it cannot
communicate with anything. Furthermore, since you have no
access with the captured particle, there is no measurement
you can make on the uncaptured particle which would tell you
anything about its entanglement. A much more detailed
discussion may be found
here .

QUESTION:
Why earth has gravity?

ANSWER:
Every object with mass has gravity. The earth has a very
large mass so it has very noticeable gravity.

QUESTION:
How birds fly by overcoming the gravity of earth even with a large mass difference?

ANSWER:
The gravity exerts a force down on a bird which is called
the weight of the bird. For the bird to fly, there must be
an upward force greater or equal to its weight. When the
bird flaps its wings down they exert a downward force on the
air; because of Newton's third law, the air exerts an equal
force up on the bird. It is this upward force which allows
the bird to fly.

QUESTION:
Please help me settle a long-running argument: A motorcycle carrying two men is traveling, under engine power, in a straight line at a constant velocity on a flat road. The passenger falls off, and the driver continues applying the same constant amount of throttle to the engine. Will the motorcycle accelerate?
The f=ma equation has been tortured and abused by both sides of the argument to prove their point, with no resolution in sight.

ANSWER:
I have shown the most important forces on the motorcycle
plus men in the figure. These are W _{1} ,
the weight of the rider; W _{2} ,
the weight of the driver; W _{3} ,
the weight of the motorcycle; D ,
the air drag force on the system; f ,
all other friction forces retarding the motion, rolling
friction by the wheels, friction in wheel bearings, etc .;
N _{1} , the force up on the
rear wheel by the road; N 2, the
force up on the front wheel by the road; and F ,
the static friction force between the rear wheel and the
road which is the force driving this system forward. Since
the system is in equilibrium, N _{1} +N _{2} =W _{1} +W _{2} +W _{3}
and F-f-D =0. The most important thing to appreciate
is that f is proportional to the normal force, f=C (N _{1} +N _{2} )
where C is some constant. So when the rider falls
off, W _{1} disappears and so N _{1}
and N _{2} get smaller and so f gets
smaller. F, however, does not depend on N _{1}
because it is static friction; it is determined by how much
power is being delivered by the engine so it does not
change. I will denote the new frictional force as
f' and note that it is smaller than
f . The sum of the forces is no
longer zero, so F-f'-D =Ma where M
is the mass of the rider plus bike, M _{2} +M _{3} =(W _{2} +W _{3} )/g .
where g is the acceleration due to gravity.
Finally, the bike has an acceleration a =[(F-f'-D )/(W _{2} +W _{3} )]g .

QUESTION:
It occurs to me that a
'station-keeping'scenario to maintain a fixed altitude above a fixed point, (say several hundred miles and above the atmospheric envelope), above the Earth should still allow for normal gravity on the vessel.
My question: How much thrust in terms of g-factors would be required to maintain that fixed position?

ANSWER:
If you want something to hover a few feet above sea level,
the upward force on it must equal its weight. By weight I
mean the force which the earth exerts on something. But,
this force (weight) gets smaller as you get farther away
from the surface. So, the weight of a mass m at sea level is
given by F _{0} =mMG /R ^{2} =W
where R is the radius of the earth, M is
the mass of the earth, and G is the universal
gravitational constant; if you go up to an altitude h ,
the weight gets smaller, F_{h} =mMG /(R+h )^{2} .
Therefore, if you want to hover at an altitude h you must
exert an upward force of F_{h} = [R ^{2} /(R+h )^{2} ]W
where W is the weight at sea level. Even several
hundred miles is small compared to R which is about 4000
miles, so we can make an estimate by doing a binomial
expansion: R ^{2} /(R+h )^{2} =(1+(h /R ))^{-2} ≈1-(2h /R )+…
So suppose that h=R /20; then F_{h} ≈0.9W .
(The exact calculation would be 0.907W .) You asked
for the force "in terms of g-factors"; that would be the
factor of W , [R ^{2} /(R+h )^{2} ].

QUESTION:
I've just been reading up on LightSail2, a craft that unraveled a big "sail" made of aluminized Mylar to test the viability of light propulsion. The theory goes, though light doesn't have mass, it does have momentum, and can therefor be used to propel the sail in space. They say the experiment produced about the same force as a paperclip over an entire orbit of the earth. My question is. How can they be sure the propulsion force is being generated by the light striking the sail, and not by the solar winds that are striking the sail?

ANSWER:
The solar wind carries a much smaller momentum flux than
that of the solar radiation. The force which it exerts on
the sail is about
1% of that exerted by the radiation. By the way, light
having momentum is not a "theory", it is a well-measured
fact.

QUESTION:
I am an elementary physical education teacher. I would like to have my third through fifth graders record how many jump squats they do each semester. Then at the end of the semester I would like to ask the question: How long have you spent suspended in air this semester? (Assuming that there is a way to figure out how much time an object traveling against gravity spends transferring their direction from up to down???) Is there in easy math equation using the amount of jumps in a semester & maybe their weight to figure this out.

ANSWER:
Because the acceleration due to gravity is constant no
matter what the weight is, the weight is only peripherally
important; a light student is likely to be able to jump
higher than a portly student and will therefore spend more
time per jump in the air. So each student needs to do
several squat-jumps for some adult, maybe a parent; that
person would then estimate the typical height the student's
feet go above the ground. So let's call the average height
reached by a particular child H and the number of times she
jumped during the semester N . The total time, in
seconds, that
child spent in the air would be T =0.072N √H
if H is measured in inches and T =0.045N √H
if H is measured in centimeters. For example, if
the child typically jumped 8 inches and jumped 200 times,
the total air time is T =0.072x200x√8=40.7
seconds. I would only worry that an elementary school
student might not be familiar with a square root. (You only
asked for an equation; I can also post the derivation of the
equations I posted here if you are interested.

QUESTION:
If I am on a moving platform travelling at 20 mph in one
direction and a platform is moving towards me at 20 mph, I
understand that the relative speed is 40 mph but what would
happen if I stepped from one platform to another as they
crossed? Would the speed of the platform I moved to cancel
my speed in the opposite direction and therefore simply move
me at 20 mph in the new direction or would there be a
process of braking where I would fall off the back of the
platform unless it was long enough to sustain the braking
time?

ANSWER:
Imagine you are a passenger in a car going 40 mph. You
decide to step out onto the road. That is what it would be
like to step onto your crossing platform.

QUESTION:
I have a photo. It is a nature shot in which the sun is going down behind a hill and yet the full (sphere of the) sun is visible in the image. My question is how is this possible? Some sort of light refraction?

ANSWER:
I believe that this is a problem with focus. If you examine
the photo, it appears that the objects in the foreground are
in focus. Looking at a blowup of the area near the sun,
there is no evidence of an image of the sun above the
horizon because the whole region is saturated, too bright to
image. Nevertheless, the sun itself is brighter than the
region around it even though you cannot see it because of
the saturation. But there is an out of focus image of the
sun and it is this larger image which is visible below the
horizon. The sun itself is probably totally above the
horizon and with a smaller raduis than the radius of the
"bite" out of the hill.

QUESTION:
You are the captain of a pirate ship and you're shooting a cannon at land. If all parameters(air density, drag coefficient, etc.)are constant, and you could shoot the cannon ball from the exact same spot on the water, one shot while the ship is stationary and the other while the ship is in motion would the cannon ball land in a different spot or the same?

ANSWER:
See a recent answer . Think of the
passer as your pirate ship and the receiver as the target on
the ground. If the passer (ship) in the first figure is not
running (moving), the ball will go to the location of the
receiver (target) who will receive (which will be hit.) it
if not running. If the passer (ship) is running (moving), he
(it) will give his (its) velocity to the ball (cannonball),
in addition to the velocity he (it) imparts by throwing
(shooting); if the receiver (ship) is not running (moving),
he (it) will not catch it (be hit) because it will go to the
position he (the target) would have been if he (it) were
running (moving) too.

QUESTION:
There is a great debate in mountain biking that when going into a turn at a high speed whether to have the outside pedal down, so that if I lean my bike down into a right hand turn my left pedal will be down and my right pedal up. OR to go into the turn with level pedals as the bike leans.
Some things to consider here is that weight must be centered over the bike left to right and front to back to avoid tire washout. The leaning of the bike allows the outside knobs of the tire to grip the terrain. While centered weight allows for proper downward force maximizing grip. So I guess the question stated another way is where is your weight more centered when the bike is leaning, pedals flat or outside pedal (pedal opposite of the direction of the turn) down. (Also please try to respond in layman’s terms. I’m far from a physicist). :).

ANSWER:
I am assuming that "tire washout" means wheel slipping
because of loss of adequate friction with the ground. This
would occur if one or the other wheel was holding up
significantly less weight than the other. The front-to-back
location of the center of gravity of the bike
plus rider needs to be halfway between the two axles
for the wheels to have equal friction. The left-to-right
location would have nothing to do with having equal
friction, so that can be ignored for your purposes; however,
since the the legs are moving up and down and forwards and
backwards, not sideways, the left-to-right centering never
changes anyway. The first figure shows one pedal up and one
down; the second figure shows level pedals. In the first
case, the right knee is quite forward and the left leg is
straight down; in the second case the right knee is less
forward than in the first case and the left knee is slightly
less forward than the right but substantially forward from
the first case. To my mind it is a tossup as to which has
the center of gravity more forward; and besides, where is
the center of gravity of the bike rider system? Furthermore,
in my opinion, any difference between the two cases will
make a trivial difference regarding whether one wheel or the
other is more likely to lose traction. In addition, if you
have an extreme lean or you are on terrain which is sloping
upward in the direction you are turning, you are less likely
to catch your pedal on the ground if it is all the way up.
Do whatever you are most comfortable with; this issue does
not merit "a great debate"!

ADDED
INFORMATION:
Since you asked for layman's terms, perhaps I should be
clear what center of gravity is. It is the point where, if
you suspend the object from a rope, it will remain balanced;
like the fulcrum of a teeter totter with equal-weight kids.
In the figure below, the biker on the right is hanging from
a rope attached to her center of gravity (the cross). The
biker on the left is hanging from a rope attached to her
front axle; note that the center of gravity is directly
below the point of suspension. Only if the center of gravity
is halfway between axles will the total weight be equally
supported by each wheel.

QUESTION:
I know the solar system is traveling thru the galaxy at a certain speed. My analogy is the solar system is in a car, the car is moving down the highway. Say you throw an object out the window, that being a satellite going outside into interstellar space. My question is, will that satellite fall behind us? Kind of like the object being thrown out of a moving car on a highway.

ANSWER:
First of all, the only reason that the object you throw
straight out of the car only "fall[s] behind us" is the air
drag; if there were no air drag the thrown object would move
away from the car but keep pace with it as you will
understand below. The first diagram shows the situation in
the frame of the solar system: we have "thrown" the
satellite so that it is moving with some velocity to the
left in the figure. Now, if there is some outside observer,
she will see the solar system moving in some direction
indicated by the red vector attached to the sun in the
second figure. (The sun (car) is moving perpendicular to the
velocity of the satellite (object).) But the satellite will
have its same velocity relative to the sun but also the
velocity of the sun in the second figure; these velocities
(shown in blue) have to be added to get the satellite
velocity which will be seen from outside the solar system.
(Since you chose throwing something through the car window
as an analogy, I have chosen the satellite velocity to be
perpendicular to the motion of the sun as seen by the
outside observer. Other angles will give different results.)

QUESTION:
I was sitting behind a thick double glass door that coused light comming from other side of the bulb to be light coming from multiple bulbs and you can even distinguish each and every bulb and even count the number of bulbs but when the double glass was made to be single it seemed crystal clear that it came from single bulb . Why this happened , what will be the ray diagram of the cenerio . This happed at night.

ANSWER:
When light strikes the glass, a small fraction of it is
reflected but most goes through; the amount of reflection
depends on the angle the light strikes the glass. If there
is a single glass, the reflected light just is not seen on
the other side. However, if there is a second sheet of
glass, the light is reflected back toward the inside as I
show in the figure. You should also have noticed that the
images you see get dimmer as you go farther from the main
first image.

QUESTION:
Acceleration
is rate of change of velocity, then why a body is said to have a constant
acceleration due to gravity when a body is at rest at a plane horizontal surface?

ANSWER:
A body at rest does not have any acceleration. Nevertheless,
the value of acceleration due to gravity g plays an
important role for the body which, if its mass is m ,
has a weight W given by W=mg .

QUESTION:
If I can throw a tennis ball at 92 mph, how fast would a baseball travel with the same force acting on it?

ANSWER:
If
you exert the same force over the same distance for each
ball, you impart the same energy E to each, E =½mv ^{2} ,
where m is the mass and v is the speed.
Therefore, ½m _{t} v _{t} ^{2} =½m _{b} v _{b} ^{2}
or v _{b} =v _{t} √(m _{t} /m _{b} ).
Putting in the masses of the tennis ball and the baseball,
m _{t} =0.058 kg and m _{b} =0.145
kg, I find v _{b} =58 mph.

QUESTION:
If I have 1.5 ft H x 25 ft L x 8.6 ft D mass that is suspended by 4 cables, 6 ft long and the mass weight 3000 lbs. How hard does the wind has to blow before the vertical cables reach to 15 degrees?

ANSWER:
It is a lot easier for me to do this in SI units, so W =3000
lb=13,345 N, 1.5 ft=0.467 m, 8.6 ft=2.62 m, 25 ft=7.62 m, and
6 ft=1.83 m. The answer depends on which way the wind blows.
So, I will make the calculation for the wind blowing
directly on the largest surface whose area will be A =7.62x0.467=3.56
m^{2} . There is a pretty good approximation for
calculating the drag force F on an object of cross
section A in a wind with speed v , F =¼Av ^{2} ,
so v =2√(F /3.56). (N.B. ,
this works only in SI units.) The forces on your mass are
shown in the figure. When I do the analysis I find F =3576
N=804 lb, and v =63.4 m/s=142 mph. Please keep in
mind that this is an approximation as are all calculations
of air drag.

QUESTION:
Air conducts heat and goose down is an effective insulator because it prohibits airflow. It seems to me that it may also be effective at keeping cold things cold in a warm environment. Would high-fill down be an effective insulator to keep cold things cold? If not, why not? Do water evaporation effects come into play?

ANSWER:
Any material which keeps something hot in a cool environment
would be equally effective in keeping something cold in a
warm environment. There is a caveat, though. Imagine a
hollow container which has surface area A ,
thickness t , and temperatures T _{in}
inside and T _{out} outside. The rate of
heat flowing through the material depends on four things:
what the material is, what A is, what t is,
and the temperature difference ΔT =|T _{out} -T _{in} |.
(This assumes that conduction through the container is the
most important means of having heat flow, not convection or
thermal radiation.) So take two examples: inside cold water
at 1^{0} C, outside room temperature at 20^{0} C,
ΔT =19^{0} C; and inside hot soup at 90^{0} C,
outside room temperature at 20^{0} C, ΔT =70^{0} C.
So the soup would cool down faster than the cold water would
heat up. Why would evaporation have anything to do with it?

QUESTION:
My questions is, how do we rationalize the fact that the speed of sound has to do with modulus and density. As we know that solids have greater density than liquid and air, does the space between each particles effect how fast sound propagates?

ANSWER:
To illustrate your primary question, "modulus and density",
let's just consider solids. A simple model of a solid is
that the molecules are connected by by tiny springs. Sound
passes through this solid by one molecule moving and
pushing, via one of the springs attached to it, on
its neighbor which then propogates the push to the next
neighbor, etc . Clearly, the speed of this sound,
displacements of molecules, will depend on the spring
constant of the springs; but the spring constants are an
approximate measure of the appropriate modulus. If we think
of the spacing of the molecules as being roughly constant
regardless of the composition of the solid, the mass of a
molecule is going to be proportional to the density. But,
the mass of a particle on a spring determines the speed of
the motion of the particle—the greater the mass, the more
slowly the particle moves when attached to a spring. By this
reasoning, the speed of sound depends on both modulus and
density.

QUESTION:
I have a question about space travel. Since there is little matter in space, and it would have very little slowing effect on a spacecraft. So, can a space craft increase in speed incrementally? What I mean is, we launch a spaceship from Earth. It takes a straight line trajectory once leaving Earth's gravity field. It reaches a certain speed, and cuts it's engines to coast. It wouldn't slow down any real amount while coasting. So, if it fires it's engines again for an extended period of time, would it not go faster than it was while coasting? And if this process was repeating dozens of times, wouldn't that spacecraft have basically a limitless top speed until it ran out of fuel for the burns? I would think that it would.

ANSWER:
Of course, burning more fuel would give you more speed.
There is a limit, however, to how fast you can go, a limit
set by special relativity—the speed of light.

QUESTION:
if I take a photo-voltaic cell/panel and connect both of its terminals to a load (say an LED or bulb), without any battery in the circuit, and irradiate with light, will the photo-voltaic cell be able to drive the load? Or do I need a battery in the circuit for the photo-voltaic cell to deliver power?

ANSWER:
A photo-voltaic cell is a battery. As long as it
can supply the current needed at the right voltage (power),
no other battery need be added.

QUESTION:
Assume that I stood at the boundary line of a football / baseball / cricket ground(without any turf/pitch/grass in it) - just a plain ground - like play ground. I have a ball that can bounce when thrown on this ground.
My question is - When I throw the ball at an angle to ground/horizontal plane, the ball takes a trajectory- which is very clear. However, after hitting the ground (that is at first bounce), it takes another trajectory and so on and so forth. Are all these trajectories are same in their shape, except for the amplitude/ maximum height it can reach?

ANSWER:
If you consider the effect of air drag to be negligible,
each segment is a parabola with the equation y=x tanθ -(½g /(v _{0} cosθ ))x ^{2}
where the acceleration due to gravity is g =9.8 m/s^{2} ,
θ is the launch angle, and v _{0}
is the launch speed. In fact, since speeds are relatively
modest, air drag will probably be negligible. If you include
air drag, each bounce segment will be a different shape
because the drag depends on the speed of the ball.

QUESTION:
According to coloumbs law every body attracts other body but how come the electrons arranged in a magnet in such a way that they have such a great attractive force that we notice the attraction of other bodies towards it ?

ANSWER:
You misunderstand Coulomb's law. This law applies to
electrically charged bodies, not "every body". And the force
between two charged bodies can be either attractive (if the
charges are of opposite sign) or replulsive (if the charges
are of the same sign). Inasamuch as a magnet has no net
electric charge, you can conclude that Coulomb's law is
irrelevant. The reason the magnet can exert forces on other
bodies is because the way the electrons are "arranged" in
such a way that a different force (magnetism) is coming into
play because of the magnetic field caused by the electrons.

QUESTION:
In rugby, you cannot pass the ball forward, only backward. We often witness passes that appeared to travel forward across the ground, but are deemed not to be forward passes because the ball was not thrown forward by the passer, but rather it drifted forward due to the momentum it carried by virtue of having been thrown by someone running forward. My question is, how much forward speed/momentum would a runner need to have to affect the trajectory of the thrown ball enough to cause this to happen?

ANSWER:
In the diagram I show the velocity of the passed ball in
blue as seen in the frame of the passer. The red vector
shows the velocity of the passer in the frame of the ground.
The diagram in the prasser's frame shows a legal pass since
the receiver is behind the passer and the ball is thrown
directly at him. However, this pass, as observed by a fan
watching from the the ground frame, will have the velocity
of the passer added to the velocity of the ball in the
passer's frame; the velocity of the ball in the ground frame
is shown in green. The answer to your question is that the
runner needs to be running forward with a velocity greater
than the backward component of the ball's velocity in the
passer's frame.

QUESTION:
Hello. I was watching a show about the moon landing last night where they said that the moon's gravity is 1/6 that of the Earth's. I mentioned that to a coworker today, and he said "So if a 180 pound man can bounce like that on the moon, then I would bounce on Earth if I was 1/6 that weight at 30 pounds." I mentioned the fact that you don't see all the 30-pound children floating around in their front yards, but he countered with "I mean, if I had my current muscle mass and weighed 30 pounds then it would work" and I really can't see how that's possible - but I also couldn't come up with the right argument to defend my position. We both know there's a difference between weight and mass, but this is something different.
Which one of us is correct, and what's the best way to explain this concept? If no one could bounce on Earth like they do on the moon regardless of how much mass they have, then how do you phrase the reason why not?

ANSWER:
It is pretty absurd to suggest that a 30 lb (14 kg) tot
could have the same muscle strength as a 180 lb (82 kg) man.
But, let's talk about how you jump straight up. You squat
down, a distance maybe half the length of your legs (I would
estimate about 0.5 m for the adult, 0.25 m for the child)
and spring up. What springing up means is that you exert
some force F down on the floor in
addition to your own weight W so
that the floor exerts an upward force on you which is equal
to F'= F -W
and it is F' which accelerates you
upward so that when your legs are straight again, you leave
the floor with some speed v . Your coworker is right
that the force which provides the acceleration,
F' , is greater if F
is the same for both adult and child because W
is 6 times smaller for the child than for the adult. The
calculations are straightforward but a bit tedious, so I
will just give you the results for the situation where
F has a magnitude twice the weight of
the adult, approximately 1640 N (360 lb), so F' =820
N. If this force acts over 0.5 m for the adult, he launches
with a speed of about 3.2 m/s, while the force acting over
0.25 m launches super child with a speed of about 7.3 m/s.
With these initial speeds, the adult goes up about 0.5 m while
the child goes up about 2.7 m. Now, we must calculate the
height the adult will rise on the moon exerting the same
force over the same 0.5 m; but now the launching force is
F'=F-mg /6=1640-137=1503 N. I find the launching
speed to be about 4.3 m/s and the height to be about 5.5 m.
So the adult on the moon goes about 5.5/2.7=2 times as
high as child with equal leg strength does on earth, and
about 11 time higher than he could jump on earth; the reason
for this large factor of 11 is twofold—first, the fact
that the weight is much smaller means that the given leg
strength will be much more effective in accelerating during
the launch; second, once launched the weaker gravity cause
him to slow down more slowly. So your coworker's
guess about how the compared jumps would be is pretty far
off, a factor of two in terms of height achieved. However,
he got it qualitatively correct that a smaller person with
equally strong legs could jump higher on earth; it just
doesn't work out quantitatively when comparing to the jump
on the moon.

But, what is the real difference? If you look at the videos
they look like they are taken in slow motion, but they are
not. Because the gravity is much smaller, it will take much
more time to get up to and back down from a given height;
even if the height achieved by the super child on earth were
exactly the same as for the adult on the moon, the two would
appear very different because the child would be back to
ground level much faster than the adult would be.

QUESTION:
A cone is standing inverted on a friction less surface at equilibrium. (tip of the cone touching the ground.) Now a perfectly horizontal force acts "at a point on the surface of the cone" such that line of force passes through the center of mass of the cone. Will it topple or just move in horizontal direction?

ANSWER:
Be sure to understand that this is a gedanken , a
thought experiment; the cone is at the point of unstable
equilibrium and you would not be able to actually balance a
cone on its tip in the laboratory. Any force which has no
vertical component and has a line of action through the
center of mass will cause horizontal translational motion
but no rotational motion. In fact, any force passing through
the center of mass will cause no rotational motion. The more
general case is shown in the diagram. The effect of the
horizontal component (F _{x} )
on the cone is to provide a horizontal acceleration. The
effect of the vertical component (F _{y} )
is to reduce the normal force (N )
which would otherwise be equal to the weight (mg ).
If F_{y} >mg , the cone would have an
acceleration in the y
direction as well, fly off the surface. The components of
the acceleration would be a_{x} =F_{x} /m and a_{y} = (F_{y} /m )-g.
In no case would the cone have any rotational motion.

QUESTION:
I have been trying to understand why an inflated tire does not lose energy and deflate. I am not trying to understand how air isn't lost through holes or permeation of the tire material. What I am trying to understand is the air molecules in a tire are energized they are deflecting outward causing pressure on the tire, and the car is providing downward pressure via gravity. So why don't the molecules "run out of energy" if energy is being used to keep the tire inflated?

ANSWER:
The air in the tire starts out at some temperature, volume,
and pressure. The temperature is an indicator of how much
energy the air has. I will assume the volume stays about
constant. Now, if the environment around the tire is at a
lower temperature than the air in the tire, the collisions
of the air molecules with the inside of the tire will, on
average, slow down the molecules' speed thereby reducing the
energy of the gas. When the temperature goes down, the
pressure in the tire will go down too. But when the
temperature of the air in the tire is the same as the
outside temperature, the collisions will on average leave
the speed of the molecules the same. So, you are right,
sometimes the air does lose some energy but that loss ends
when the everything has the same temperature. Conversely, if
the air inside the tire is cooler than the outside
temperature, the air inside will gain energy until the two
temperatures are the same.

QUESTION:
On 9th May 2019 Jeff Bezos made a presentation at the Air & Space Museum in which he said," it takes 24 times less energy to lift a pound off the Moon than it does on the Earth". Since the escape velocity of the Earth is 11.186 km/s and that of the Moon is 2.38, the ratio of their squares is 22.09. Apparently the Earth's escape velocity only includes gravitation, and not air or magnetic drag. But air drag is too small to account for the discrepancy between 24 and 22.09. One reference claims that magnetic torque is 10% of the gravitational torque on a satellite - which would take care of the discrepancy. But when I try to calculate the magnetic drag for a spinning rocket I get micro-newtons, which is too small.

ANSWER:
This discrepancy seems too small to worry about. Bezos was
probably just giving an approximate number some researcher
gave him. I find it difficult to believe that the earth's
magnetic field has any significant effect on escape
velocity. However, the air drag will not be small. The
projectile will be moving with a terrifically large velocity
at the beginning of its flight where the atmosphere is most
dense and encounter a very large drag force; this could very
easily account for the discrepancy.

QUESTION:
I was wondering if someone were to jump off with hoover dam inside a 100m wide ball if they would live or not. I believe that they would live personally because the hoover dam in meters is roughly 225m. You might consider this off the wall but I'd personally would like to know and to prove my coworker wrong.

ANSWER:
What does being inside a ball have to do with it? I envision you standing on the inside surface of the ball positioned on top of the dam. Now, you each start falling with the same acceleration. You hit the bottom with the same speed as you would have without the ball. Maybe you are thinking the ball would act as a parachute? Because it has so much surface area, if it was not very heavy it
might set you down gently.

QUESTION:
My understanding of physics and science in general is limited so please bear with me. Is it possible to create a controlled interaction between electrons and direct the force in a specific direction to create a type of propulsion? If so, could the force created between this interaction be significant enough to move large objects through the vacuum of space?

ANSWER:
Your question is confusing but I think you are asking if
electric charge can be used somehow for propulsion. The
answer is yes, and it is called an ion propulsion. You can
read a nice little article about
ion propulsion in this link by NASA.

QUESTION:
When a full bus takes a sharp turn, could it ever tip over? Are
there conditions under which this could happen, everyone sitting on one
side, wind, etc.? I am aware of the
Mythbusters episode debunking the scene from that Sandra Bullock movie, but I'm interested in the theoretical point at which it could happen.

ANSWER:
The problem in examining this question is: what constitutes
a "sharp turn". The best way to characterize the turn is to
specify its radius. That is the way I will proceed,
considering a 50 mph turn and calculating the radii for
which the bus will skid, or begin to tip, or actually tip
over. I will use 50 mph for the speed and make educated
guesses for the distance between left and right wheels, the
location of the center of mass (which I will assume is in
the center across the bus and some distance above the road),
and the coefficient of static friction.

The diagram above shows the forces on the bus if it is turning
right (as seen by the driver) around a circle of radius R
(center to your left), traveling with speed v ; the
coefficient of static friction between the tires and the
road is μ . The bus is just about to skid and so
the frictional forces are at their limits, μN . The driver's side tires have a normal
force N _{1} and a frictional force f _{1} =μN _{1} ;
similarly, the tires on the other side have a normal force
N _{2} and a frictional force f _{2} =μN _{2} .
The weight of the bus, W=mg , acts at the center of mass of
the bus; I will assume that the center of mass is halfway
between the left and right wheels. The bus is in equilibrium
in the vertical direction, so N _{1} +N _{2} =mg .
The bus has a centripetal acceleration to your left a=v ^{2} /R
so f _{1} +f _{2} =mv ^{2} /R.
The greatest that f=f _{1} +f _{2} =can
be is f=μ (N _{1} +N _{2} )=μmg=mv^{2} /R.
Therefore the smallest radius that the bus can have if
its speed is v is R _{min,slide} =v ^{2} /(μg ).
In your case, v =50 mph=22 m/s, g =9.8 m/s^{2} ,
and μ≈ 0.8 for rubber on dry asphalt, R _{min,slide} ≈ 62
m=203 ft.

But you are interested
in the radius when the bus will start to tip over; this
happens when N _{2} =f _{2} =0,
N _{1} =mg , f _{1} =mhv ^{2} /R .
Now we need to calculate
the torque about some axis and set it equal to zero, but
that axis must pass through the center of mass because the
bus is accelerating. Now, you need to know the height above
ground h of the center of mass and the distance 2d ,
between the left and right wheels. Doing that, the torque equation is (mhv ^{2} /R )-mgd =0
or R _{max,start} =hv ^{2} /gd.
This is the radius at which the bus will just start to
tip. For smaller radii the inner wheels will
come
off the ground but the bus will not tip over until you
reach some critical smaller radius. So let's just guess that
approximate values of h and d are

equal at
about 2 m each: R _{max,start} =v ^{2} /g =49
m=161 ft. This choice of h and d result in
the bus never tipping because it would slide before it
started to tip; to reach this point, f =μmg=mv ^{2} /R ,
μ= 22^{2} /(9.8x49)=1.01.

Consider the bus when it has tipped through some angle
θ relative to the horizontal as shown in the second
figure. Now the moment arms to compute torques are D=d cosθ-h sinθ
for N=mg and H =h cosθ +d sinθ
for f=mv ^{2} /R . (I
must admit that it took me a while of floundering with my
trig to figure these out!) So now the torque equation is
mgD =Hmv ^{2} /R or, R=Hv ^{2} /(gD ).
In the graph below you can see that, for d=h , the radius gets
smaller and smaller as the tip angle gets larger; all these
calculations assume that the bus does not skid, but unless there is
something else besides static friction to provide the
centripetal acceleration it will likely skid before it tips. Note that if the bus tips until the center of mass is directly above
the outer wheels, where D =0 and θ= 45^{0}
for d=h , it is on the verge of
fully tipping over. The bus
will roll over when the angle it makes with the ground is greater
than tan^{-1} (d /h ).
If you manage to get the bus on two wheels and at the angle
just before it tips over you can go essentially straight
(making small steering corrections), not turning as the
video
linked to above. The bottom line is that a bus is extremely
unlikely to roll over in a turn because it will likely skid
before it rolls. I think my choices of d and h are very
reasonable and probably, because so much of the weight is in
the frame, suspension, axles, transmission, and engine of the bus, the center of gravity is
likely to be lower than 2 m making a roll even less likely.
It would be a different story for a train on tracks where
the "friction" is almost limitless because the train is
constrained by the track from moving sideways ("skidding")
at all.

ADDED
NOTE:
You could probably, by suddenly steering violently to one
direction, cause the inner wheels to come off the ground
before it skidded much, but I doubt that you could flip it
over this way at 50 mph unless the skidding wheels "snagged"
on something like a curb or a hole. If you watch a movie of
a high-speed car chase you will note that the turns are
usually executed by skidding around the curve.

QUESTION:
What percentage of the time in motion is a pendulum or piston motionless?

ANSWER:
Zero.

FOLLOWUP QUESTION:
Thank you, but, they change direction without stopping? That sounds impossible. What am I missing? If it is just that the real number is so small that it might as well be zero, ok, but I was hoping to know what the real number might be.

ANSWER:
Well, dealing
with an engineer, I guess I have to consider the "real world"! A physicist will consider an idealized zero friction system for which the pendulum
does stop but only for zero time. But an engineer will have to take into account that a real pendulum might stick for a brief time, maybe a microsecond, at the extrema. But I cannot give you an answer to your question because the amount of stick time will depend entirely on the details of the construction of the pendulum.
(I believe that the main reason for sticking would be the
difference between static and kinetic coefficients of
friction.)

QUESTION:
I’m not a physicist, but am curious about “Planck's” constant. Everything seems to have a
"smallest" inadvisable value. Given that doesn’t it follow that numbers like
"pi" would have an absolute value and not continue toward infinity? I’m sure this must be pretty basic, but seems the
"small" has a limit so "infinite series" numbers or values would have a limit also?

ANSWER:
Planck's constant, h , plays more roles than just
setting the smallest value of something. Most famous of this
role is that, if you have electromagnetic waves with a
frequency f , the smallest amount of energy, called
a photon, which
such electromagnetic waves can have is hf . But,
this idea of minimum amount possible does not extend to
everything, certainly not to something like the number of
digits in any irrational number like π . The idea
that something has a minimum value usually has to do with
what we call "fields"; when the electromagnetic field is
quantized, the photon is the "messenger" which conveys the
field.

More interesting, I believe, is the role played by Planck's
constant in the structure of systems of particles like atoms
or atomic nuclei. The great hypothesis by Niels Bohr that the
hydrogen atom can only exist in particular energies owes its
inception to the idea of quantization. He assumed that the
electron in a hydrogen atom moved in circular orbits but not
in any old orbit but only in those for which the angular
momentum was quantized, L=nh /(2π ) where
n =1, 2, 3, 4…; the angular momentum is the
product of the electron mass, the orbit radius, and the
speed of the electron. This led to quantized energy levels;
this model excellently explained everything known about the
hydrogen atom at that time. See the Wikepedia article on the
Bohr
model. (The Bohr model, while easy to visualize, is a
very simplistic and incorrect compared to the correct modern
quantization of the atom.)

If you think about it,
energy quantization is an astounding surprise. It is like
saying that a pendulum may have an amplitude of 5^{0}
or 7^{0} but not 6^{0} . But nature does hand
us surprises sometimes.

But energy quantization does not apply to everything. If a
system of particles is not bound together it can have any
amount of energy.

What
is not known but speculated about is whether space and time are
quantized. Is there a shortest possible distance? Is there a
shortest possible time? Nobody knows but speculations called the
Planck
length and the
Planck time
are extraordinarily small, far smaller than current measuring
capabilities.

Finally, look
at the following Q&A for another example of the role of Planck's
constant.

QUESTION:
in quantum world, i know that particle doesn't have a definite position before measurement as per copenhagen interpretation, but what about momentum and energy, does a particle have definite momentum(or energy) before measururement?

ANSWER:
In quantum mechanics there are what are called conjugate
variables; these are pairs of observable quantities which
are coupled together such that you can never know the exact
value of either one. This leads to the famed Heisenberg
uncertainty principle. On such pair is position x
and linear momentum p , and the product of their
uncertainties cannot be smaller than the rationalized Planck
constant, h /(2π ). So, the more
accurately you measure either, the less accurately you can
know the other. Another pair are energy E and time
t . So, for example, if a radioactive system has a
half life of Δt , the most accurately you can
measure the energy is ΔE >[h /(2π )]/Δt .

QUESTION:
if action and reaction are always equal in magnitude and opposite in direction, why don't they always cancel each other out and leave no net force for accelerating a body?

ANSWER:
Because they act on different objects. If object A exerts a
force F _{AB} on object B,
object B exerts an equal and opposite force F _{BA} =-F _{AB}
on object A. F _{AB} tends
to accelerate object B and F _{BA}
tends to accelerate object A.

QUESTION:
if action and reaction are always equal in magnitude and opposite in direction, why don't they always cancel each other out and leave no net force for accelerating a body?

ANSWER:
Because they act on different objects. If object A exerts a
force F _{AB} on object B,
object B exerts an equal and opposite force F _{BA} =-F _{AB}
on object A. F _{AB} tends
to accelerate object B and F _{BA}
tends to accelerate object A.

QUESTION:
I have a 4ft long ramp, 1ft tall at the take off. My 3 year old hits this on his bike and clears 4ft when he lands.
How fast is he going at take off?

ANSWER:
You may not be interested in the details, but I will include
them anyway for the interested reader. The equations of
motion for horizontal and vertical position are:

These are two equations for two unknowns, v and
t , and can be solved using simple algebra. The
solutions are t =0.563 s and v =7.33 ft/s=5
mph.

QUESTION:
Consider 2 cases:
Case1: a moving train with constant velocity of v hits a stationary person. Case2: a person running with constant velocity of v and hitting a stationary train. In which case is the damage to the person more severe? And by how much?

ANSWER:
What determines "damage"? The laws of physics, in particular
Newton's laws of motion. Newton's laws are Galilean
invariant. What this means is that the laws are exactly the
same to two observers who are moving with constant velocity
relative to each other. In your question, Case2 is the same
as Case1 if viewed by an observer running along the train
with speed v . There is no difference in the
"damage".

QUESTION:
I read that during the 70 mega ton atomic
detonation, the blast wave was felt 300 miles away in L.A.
How is this possible since the Earth is curved, wouldn't it pass over them?

ANSWER:
OK, the earth has a radius of about 5,000 miles. The
angle θ subtended by 300 miles is θ= 300/5000=0.06
radians=3.44^{0} . Looking at the diagram, cosθ=R /(R+d )=[1+(d /R )]^{-1} ≈1-(d /R );
therefore d≈R (1-cosθ )=9 miles.
Now, you are probably thinking that all the sound waves move
in straight lines, that the sound waves coming from the
detonation are radiating out as shown in my second figure
thereby missing the earth's surface except very close to the
source. But, since sound is composed of waves, it is subject
to diffraction. Huygen's principle says that the wave a
short time later is determined by treating each point on the
wave front as a new point source. Finally, I have drawn one
of the wave fronts which is off the earth in the earlier
picture and how Huygen's principle predicts the wave a short
time later. Note that the wave where it originally ended is
now bending and lengthening; this is diffraction and allows
the sound to follow the contour of the earth. This is the
same principle where when you speak to someone on the other
side of a sound-absorbing wall with an opening in it (but
she is not visible to you), you can be heard just fine;
sound diffracts around corners.

QUESTION:
Is water pushed down a river or sucked down by water in front of it?

ANSWER:
To understand what is happening, you need to focus on one
tiny volume of the water and find all the forces on it. I
will focus on the center cube of water in the figure; these
cubes are, of course, imaginary but if I can understand the
motion of one tiny piece of the river I understand the whole
thing. There are three forces on this cube
which have components pointing in the direction of the flow:
the weight mg of the cube, the force
f _{down} which the upper
cube exerts on it in the direction of the flow, and the
force f _{up} which the lower cube exerts on it opposite the
direction of the flow. Notice that all cubes exert pushing
forces on their neighbors, not pulling, so "sucked" is never
what is going on. Now, the cubes are flowing downstream with
constant velocity, so the sum of all forces in that
direction must add to zero (note that only the downstream
component of the weight is mg sinθ
pushes the water downstream): f _{up} -f _{down} -mg sinθ =0.
So, f _{up} =f _{down} +mg sinθ.
So, you see, gravity and the water upstream push the
water downstream while the water downstream tries to hold
the water back. There is a greater force upstream from the
water itself than there is a force downstream. Without
gravity or with no slope to the riverbed the water would not
flow at all.

QUESTION:
What force of magnet do I need to suspend a 36 to 40 inch plastic ball 2mm
thick and weighing less than 15 pounds?
I am creating a home art item to be suspended about 4 feet off the ground.

ANSWER:
You need a magnet which can exert a force of 15 lb; I would
suggest building in a safety factor of 5 lb or so. If you go
to a supplier, the catalog should tell you the force rating
for each magnet.
Here is a nice explanation of how magnets are rated.

QUESTION:
I'm in the middle of watching a show that deals with a massive hole in the ground that the people in the show call "The Abyss". They don't know how deep it is. I was wondering if you were to travel into a hole like this would the Atmospheric pressure increase or decrease?

ANSWER:
What is atmospheric pressure? It is the force (per unit
area) which the air pushes on you. This force is the weight
of all the air which is above you. When you go up a high
mountain, the pressure decreases because there is less air
above you than when you are at sea level. If you are deep in
the earth there is more pressure on you because there is
more air above you than when you are at sea level.
Here is a real-world example.

QUESTION:
If you have a basketball that is filled with air, and a basketball filled with cement and you are in a weightless environment, and you throw each one as hard as you can; would they each travel the same distance?

ANSWER:
There is no gravity, but is there anything else? Like air?
Let's assume you are in empty space when you do the
experiment. "…as hard as you can…" implies that you
will exert the same force (your best) on each over the same
distance (the length of your arm). But the cement ball has
much more mass (inertia) so the effect of the force will
give it a velocity smaller than the air ball would get, and
both would move with constant speed in a straight line. The
air ball would go faster, not necessarily farther.

QUESTION:
This is a practical but I think possibly a complex physics question. I like to climb mountains. On the descent when the grades have reduced to walking speeds but still require retarding input, perhaps a 10 percent downhill grade, my experience indicates to me that it requires less energy to descend at for example 4 mph than 2 mph. Am I just imagining this?

ANSWER:
The human body is very complex and I doubt there is a simple physics answer to your question.

FOLLOWUP QUESTION:
Thank you for your answer, which I am sure is right on the money. One follow up question if I may. Would a ball rolling down a frictionless ramp, 10 degree grade, require less retarding force to obtain zero acceleration at 4 mph as opposed to 2 mph? Yes? No? Maybe?

ANSWER:
First of all, a ball will not roll down a frictionless ramp,
it will slide. But that does not matter regarding your
question. The unbalanced force which accelerates the ball
down the ramp is the component of its weight along the ramp; if the weight is
mg , that component is mg sinθ
where θ is the angle of the ramp, 10^{0
} in your case. The force pointing up the ramp which
must be applied is therefore mg sinθ.
Note that the speed is irrelevant.

QUESTION:
Since Dark Matter/Energy makes up most of the known universe, how can we know if the light reaching us from distant parts of space is an "unaltered version"?
More specifically, what are the chances that light reaching us has been gravitationally lensed (along the way), giving inaccurate trajectories to the stars and to their actual positions in the sky?
(And if this error cannot be corrected, does that mean that much of todays deep-space astronomy could very well be flawed)?

ANSWER:
First, unless the mass is very large and the light passes
very close to it, deflections will be very small. There are
numerous examples of gravitational lensing. See an
earlier answer . I am sure that astronomers are perfectly
aware that masses will alter the position of the images they
observe relative to the objects they are observing are not
perfect. However, the position recorded is simply the
position in our sky at this time .
Positions of objects can change simply because they are
moving.

QUESTION:
Physics in schools teaches two contradictory and mutually exclusive things: (1) That the upward lift force on an airplane in flight equal its weight (Lift = Weight = mass x gravity). This is based on applying Newtons 2nd law of motion (F = ma) to the airplane in flight. (2) However, modern commercial airplane like the Boeing 747–400 can fly with thrust-to-weight ratio as low as 0.3. Here the engine thrust is only 0.3x the weight of the airplane, but this thrust is sufficient to push the airplane forward and generate enough lift to fly. Therefore the upward force required for lift and flight must be a lot less than 0.3x the weight of the Boeing 747-400. (Lift < Thrust < Weight). Both statements cannot be true. Which is correct?

ANSWER:
You are assuming that all the lift comes from the thrust.
That is true for a rocket but not for an airplane. The
thrust generates very little lift; its main function is to
provide horizontal force to overcome drag and accelerate.
Lift for an airplane comes from the way that air passes over
the wings; see this
link .

FOLLOWUP QUESTION:
Can I follow up your response? If so: The principle of conservation of momentum means that your response is impossible. Upward lift force canot appear out of nowhere, for example by air passing over a wing. In this case, lift will arise due to the wing converting horizontal wing airflow (thrust) into vertical lift. I also note that you did not provide a scientific / reliable source as evidence for your response, but an uncredited / verified popular science video on youtube.

Kind regards

ANSWER:
Kind regards? After such a rude and ignorant response?
Really? Conservation of momentum is only true for an
"isolated system". If a cannonball is fired horizontally from a cannon
on a nearly frictionless track, the momentum of the ball is
clearly not conserved; but the ball is not an isolated
system, there was the cannon also. The ball acquired
momentum but the cannon recoiled backwards and acquired an
equal amount of momentum in the opposite direction; momentum
was zero before and after the cannon was fired. If air
passing over a wing starts horizontally and ends up with a
downward component, it does not mean that vertical momentum
has materialized out of nowhere, it means that something
exerted a downward force on it—the wing. The air is
not an isolated system because the wing is exerting forces
on it. But if we take the wing and air together as an
isolated system (approximately), then Newton's third law
says that if the wing exerts a force on the air, the air
exerts an equal and opposite force on the wing and that
force is called lift. Lift is, as you would probably figure
out, the force which lifts the airplane. Regarding
"scientific/reliable" and "uncredited/verified", I am not
writing an encyclopedia here where footnotes and references
are required. I am presenting myself as someone who has a
degree of expertise in physics who can answer questions
about the subject; it is your responsibility, since there
can be lots of BS on the internet, to convince yourself that
I do indeed have some physics credibility. To that end I
post information about myself, on the page
"The
Physicist" where you can get the information you need to
judge my qualifications. Incidentally, the reason I chose
the video which I did was because the Bernoulli aspect of
wing lift is presented as the whole story in a great many
introductory physics textbooks and only when I was taking
flying lessons years ago did I learn that "angle of attack"
is much more important in actually creating lift; I like the
way this video presented both of these important concepts.

QUESTION:
I am writing a sci-fi book based in the future. I would like to know what would happen to the other planets, such as Venus and Mars, if Earth was to be destroyed by a massive implosion.

ANSWER:
The effect would be minimal.

QUESTION:
Would supporting a barbell on your back weighing equivalent to your body weight be the same as feeling 2G forces downwards?

ANSWER:
No, it would not be the same. Look at your head, for
example. In a 2G environment it would be twice as heavy and
it is not if the weight is on your back. The only part of
your body which would be the same are your feet which have
to support twice your normal weight in either case.

QUESTION:
What force does a 1 pound rainbow trout swimming at 23 mph exert on a fishing line. I'm thinking about the shock exerted on a fishing line as quoted breaking strains are irrelevant when a trout strikes, or bolts at speed. I've read that a trout only takes one second to reach maximum velocity.

ANSWER:
(I will work in SI units, preferred by scientists, but will
convert to Imperial units, lb, when forces are calculated.)
What determines the average force is the time Δt over which the
fish is accelerating and the change in linear momentum Δp =m Δv ; F =Δp /Δt .
In your case, the mass is m =1 lb=0.45 kg and the speed is
v =23
mph=10 m/s so the change in momentum is 4.5 kg⋅m/s if
the fish starts at rest and accelerates to 23 mph or
vice versa . I
will assume that the line does not break and then compare
forces with the pound-test of the line; then I will compare
with a four pound test leader. If your
rod had zero flex, was perfectly rigid, the fish would be
unable to move at all unless the line broke, so assuming
that he could exert a force of four pounds or more which he
probably could, you would lose that fish. If he hit the fly
going 23 mph and stopped in 0.1 seconds, the average force
would be 4.1/0.1=41 N=9.2 lb—bye bye fish! But, that is unrealistic because
what really happens is that the rod bends and what that does
is lengthen the time the strike lasts; that means that the
force on the line will be smaller. Suppose that it takes 2 s for
the fish to get up to his maximum speed while hooked; during
that time the average force exerted will be 4.5/2=2.25
N=0.51 lb, and you have caught that fish. Now you understand
why fishing rods are designed to bend.

QUESTION:
Can a bridge made with two I-beams hold more weight than it's designed for if the object moving over the bridge goes at a high rate of speed where the weight load is just seconds?

ANSWER:
I cannot make any recommendations; if I was wrong, I might
be held responsible. Engineers always make the structures
they design to be stronger than the specified maximum load;
for example, if the maximum specified load is 2000 lb,
it might be 2500 lb before the structure started to fail.
And the design, I suspect, always is made assuming static
equilibrium, so a very brief load above the maximum might
very well not cause the bridge to fail; nevertheless, I
would be wary.

QUESTION:
While metal detecting I found a strip 1/4” thick 2” wide by 4” long I thought to be lead as it is heavy!
Upon doing the marking test it does not write or mark anything!
It is pliable and weighs 156grams
It is also non magnetic
Could it be silver or other valuable metal?

ANSWER:
I calculate the density to be about 4.76 g/cm^{3} .
The closest metal to this is germanium with a density of
5.32 g/cm^{3} ; but germanium is brittle and would
not be pliable. Lead has a density of 11.34 g/cm^{3}
and silver 10.49 g/cm^{3} , both about twice as dense
as your sample. So it must be something else. To sort it out
would require a chemical analysis, I guess.

QUESTION:
Why does an astraunaut experience weightlessness in outer space?

ANSWER:
Because he is in free fall all the time when he is in orbit.
If he is just in space, far from any sources of gravity, it
is because weight is the force which earth exerts on you and
there is no such force in space.

QUESTION:
Why are there so many types of baryons but the only nuclides that form atoms ( and I suppose molecules too for that matter ) are protons and neutrons?

ANSWER:
The simple answer is that the proton is stable and the
neutron is almost stable, so they are the only baryons which
hang around long enough to contribute to the formation of
nuclei. The more complicated answer is that, once bound in a
nucleus, they tend to somewhat lose their identity and form
a "quark-gluon plasma" from which many different types of
baryons can pop into and out of existence.

QUESTION:
How fast does a 3000 pound vehicle have to go to launch the vehicle in the air and travel 130 ft. I know the angle of the ramp you would have to know and I don't know that so use the optimum.
My grandson was driving a 2014 Mazada 6 and lost control and went in the ditch and totaled the car The police officer said he went in the grass on a rural road hit a driveway and launched him a 43 yards

ANSWER:
Assuming he lands on the same level as he launched and that
air drag is negligible, the
range is R=v ^{2} sin(2θ )/g
where R =130 ft, v is the speed, θ is the
launch angle, and g=32 ft/s2 is the gravitational
acceleration; the "optimum" launch angle is 45^{0}
for maximum range. So v =√(130x32)=64.5
ft/s=44 mph. For a smaller launch angle the speed would be
larger; e.g . at 30^{0} , v =50 mph.

QUESTION:
Could a spaceship traveling very,very fast suddenly stop and make a reverse without harming the occupent inside if somehow the occupent was suspended in a liquid,or a jelly substance?

ANSWER:
If the space ship experienced a very large acceleration,
greater than on the order of 50 Gs, you would have to cause
the acceleration experienced by an occupant to be much
smaller. This could be achieved by making the change in
velocity occur in a much longer time than for the ship. I
suppose that being suspended in a viscous fluid could
achieve this provided that the tank is large enough; if the
tank were not long enough, you would smash into the end
before you had lost enough speed.

QUESTION:
Hello, my question involves interstellar navigation. Assuming that such travel becomes a reality, how would one navigate accurately to a target star? We see a past instance of a star (or other distant object) based upon the light that can be observed from our distance to it. That object may no longer exist. If you were targeting the object by line of sight, it may not exist once you get to its location. I am assuming that light alone would not be enough for accurate and reliable navigation. Also, there is distortion to consider, so that, the object might not necessarily be in the observed location. What mechanisms would compensate for this? Also, since we see an older representation of an object due to the distance the light has to travel, how close must one be to an object before the image becomes "current", or, real-time?

ANSWER:
For a lengthy discussion of the problems of space
navigation, see an
earlier
answer . The issues you raise could mostly be handled by
making a judicious choice of target. Stellar evolution is
reasonably well understood and the life time and age of a
star could be decuced from its properties. For example, if
the star you have in mind is within a million years of
running out of fuel and is a million light years away, it
would foolish to plan a journey there. Regarding distortion,
the earlier answer emphasizes that problem but the most
important distortion is that caused by the high speed of the
vehicle itself; "distortion" due to motion of the target
star itself as well as gravitational lensing could be
adjusted for in flight if you could actually see and lock
onto the star at high speeds. Your last question has no
answer because you need to specify how small a lag you would
call "current"; we are quite close to the sun but light from
it takes about 8 minutes 20 seconds to get to us.

QUESTION:
How long does it take one ounce of water to hit the ground from twelve feet high, in Denver vs Los Angeles, in June?

ANSWER:
A very peculiar question! The only thing which will make the
answer different is a significant effect from air drag which
will be greater in LA because the density of the air is
higher there. I did a rough calculation assuming the water
fell as a sphere. The terminal velocities were 29.54 m/s in
LA and 31.90 m/s in Denver. If there were no air drag the
speed the water would have when it hit the ground would be
about 8.5 m/s; since the water is never anywhere close to
the terminal velocity, air drag can be neglected.
Extraordinarily accurate measurements would find that the
time is shorter in Denver but it is a pretty involved
computation to get those times and the theortical
differences are only approximations anyway.

QUESTION:
When a ball is thrown into the air, how long does it spend at its maximum height, not moving up or down? Some friends have told me the change from upward motion to downward motion is instantaneous. If that is true, how would it be possible to take a picture of the ball (or catch it) at its maximum height considering that zero time would be spent there?

(To give a background on why I asked the question. I have a habit of clicking my pens on my desk so they shoot a few inches into the air. I always try to catch them right at the height of their motion. I started wondering how long I actually had to do that. When I asked some engineering friends they said the pen instantaneously goes from traveling up to falling down. I’m not sure if that’s true but if so it introduces a bit of a crisis. If the pen spends zero time at its maximum height, is it ever actually there? And would I ever be able to actually catch it at is maximum height? Can something be somewhere but spend zero time there?)

ANSWER:
According to
classical physics, it spends zero time at any particular
velocity, including zero. So, is it ever somewhere if it is
only there for a zero amount of time? Of course it is, and I
see no "crisis"; the fact that it spends zero time at one
location simply means that it is "passing through". Only the
center of gravity of your pens will necessarily be at rest.
You can take a picture of your pen over the time interval of
your shutter and, because the speed is very small during the
times just before and after the maximum height, you can get
a pretty clear picture with a pretty slow shutter speed.
When the pen is moving faster, you need to use a faster
shutter speed.

To get more technical, what is the probability of finding
the object at a particular point? It is zero. That is
probably what is bothering you. So you could ask for the
probability of catching the pen during the interval 1/1000th
of a second before and after the zero speed time. I will not
get into quantum physics here, but you cannot know precisely
both the speed and location of your pen at any time; but
this truth implies an incredibly small uncertainty,
certainly not important for a macroscopic object like your
pen. One final thought, we do not know whether or not there
exists a shortest possible time or length, sometimes
referred to as the Planck length and the Planck time; if it
exists, the Planck time is expected to be about 5x10^{-44}
s, so perhaps your pen stands still for that time!

QUESTION:
If a small nail from little distance is thrown towards a man, it hurts. But why don't we feel hurt if a rain drop falls on our head with more speed and from long distance

ANSWER:
The terminal velocity of a raindrop is about 20 mph. You
could probably throw a nail with a speed greater than 50
mph. Perhaps even more importantly, the force which you feel
when something strikes you is inversely proportional to the
time the collision lasts—the longer the collision
lasts, the smaller the force. The nail, being solid, stops
very quickly and all the force is on a very small area, so
there is a large pressure over that area. The raindrop
spreads over a larger area and takes longer to stop, so the
smaller force is spread over a larger area. (Read the next
question and answer to see what terminal velocity is. You
also can find terminal velocity on the
faq page .)

QUESTION:
In uga physics class we learned that you can go up in the sky and drop an object. It will accelerate up to about 120 mph, and then it doesn't get any faster.
If you drill a hole from the surface of the earth down to the center of the earth, and then drop the object down the hole, how fast of a speed will it get up to?

ANSWER:
Actually, your instructor was getting you to see that,
because of air drag, objects dropped do not continue
accelerating at 9.8 m/s^{2} until they get all the
way to the ground. The air drag is a force on the falling
object upward (opposite the velocity) which gets bigger as
the object falls faster; eventually the drag force becomes
equal in magnitude to the weight which is a force down, so
the object falls with constant velocity after that. That
speed is called the terminal velocity and that depends on
both the mass and the geometry of the object. The speed of
120 mph is the approximate terminal velocity of a falling
person. A bowling ball would have a much larger terminal
velocity and a feather a much smaller one. If the earth had
no atmosphere, anything would continue accelerating until it
hit the ground. If you want to learn more about air drag,
you can find links to many of my older answers on the
faq
page . If there were no air in the hole you want to drill
and the earth were a uniform sphere (which it isn't), the
speed would be about 18,000 mph at the center.

QUESTION:
Are there different types of plasma? What I mean is, do chemical differences (i.e., water and mercury) result in different plasmas just as they make different liquids? The largest variation I have yet to see in plasma is between the blue electrical stuff in desktop toys and the pink haze in the core of that fusion reactor.

ANSWER:
Plasma is a state of matter, a vapor with a significant
fraction of the molecules ionized. Of course a plasma
depends on what it is composed of as well as other
distinguishing properties such as the average fraction of
singly ionized molecules, doubly ionized molecules, etc .
Is a block of iron a "different type" of solid from a block
of tin? Yes. Similarly, a plasma composed of iron ions is a
"different type" of plasma than one composed of tin irons.

QUESTION:
So, I am a bit of a martial artist and I fancy my self a bit of a scientist. However I am studied in the fields of anatomy and biology, not physics. So my question is this. If I were to punch or kick a person, would more force be transferred through my blow into their face if my whole body was in mid air or would there be just as much force or more if I leave my feet grounded.
Please feel free to leave a lengthy answer.

ANSWER:
Applying simple physics to situations involving the very
complex human body can lead to incorrect conclusions, so my
response should be taken with a grain of salt. The simplest
thing to try (and I mean really simple) is to compare
collisions between two objects which has one at rest and the
other moving with speed v . I will assume that the time which the
collision lasts, Δt , is the same for any situation
I consider. There will be two extremes for a collision—perfectly
elastic where energy and linear momentum are conserved, and perfectly inelastic
where the objects are stuck together after the collision and
energy is not conserved but linear momentum is. The impulse
received by the object originally at rest (your opponent) is
equal to the linear momentum he has after the collision and
is equal to F Δt ; since I specified that Δt
is the same for all collisions, the force experienced is
F=mv' /Δt , where v' is the speed of the struck
object after the collision. First consider a collision
between two equal masses m . For the elastic
collision the incoming mass is at rest after the collision
and the outgoing mass has a speed v ; for the inelastic
collision both masses (stuck together) have a speed ½v .
Therefore the force experienced by your opponent is given by
½mv /Δt ≤F ≤mv /Δt .
Now suppose that your mass M is much larger than your
opponent's mass m , M>>m . In this
case, the inelastic collision will have both objects moving
with approximately a speed of v ; the elastic
collision will have M moving
approximately with the speed v and m
moving approximately with the speed 2v . Therefore
the range of possible values for F experienced by
m is mv /Δt ≤F ≤2mv /Δt ,
which is twice the force for equal masses. So my simplistic
view is that being anchored to the floor is like increasing
your mass allowing you to deliver more impulse to your
opponent as compared with if you are flying toward him which is
more like the collision of equal masses. As I said, it is
much more complicated than this but this might give a little
insight. I would welcome any comment on this treatment.
(Lengthy enough answer for you?)

QUESTION:
If a gyroscope is undisturbed, possible mounted in a 3-axle gimbal the direction of the angular velocity vector remains constant? That makes them useful for navigation. But, the angular velocity vector is constant relative to what? Not a point on the earth, or in the solar system, or the galaxy? What point does it remain fixed upon? Does it define some absolute location within the ponderable mass of the universe?

ANSWER:
You can point it in any direction you like. Now, as long as
the gyroscope encounters no torques, it will remain pointing
in that direction.

QUESTION:
DEAR SIR WE HAVE A PRECAST CONCRETE MANUFACTURING UNIT IN COIMBATORE. NOW WE ARE IN A NECESSITY TO CAST A 7 METER BY 4 METER CONCRETE SLAB. THICKNESS 150 MM . SO THE TOTAL WEIGHT COMES AROUND 10.5 TON. MY QUESTION HERE WHAT WILL BE THE FORCE ACT ON THE BOTTOM SURFACE IF WE LIFT THAT SLAB. WHY I ASK THIS IS , WE USE RUBBER MATERIAL BETWEEN THE MOULD AND CASTING FOR THE SMOOTH FINISH . SO WHEN WE LIFT THE RUBBER MATERIAL COMES WITH THE SLAB INSTEAD OF STICKING TO THE MOULD. SO IF I KNOW THE PRESSURE THEN I CAN DECIDE HOW TO FIX THE PROBLEM.

ANSWER:
The pressure is the total weight divided by the area. The
total weight is (10,5000 kg)x(9.8 m/s^{2} )=102,900 N
and the total area is (4 m)x(7 m)=28 m^{2} . So the
pressure is 3675 N/m^{2} . Other units used for
pressure which you might prefer are 375 kg/m^{2} or
76.75 lb/ft^{2} .

QUESTION:
Why is gravity considered a Fundamental Interaction?
I would think that gravity is a feature of the geometries of space-time rather than a fundamental force of nature like the Weak, Strong and Electromagnetic Interactions.

ANSWER:
I believe that the main reason is that most theorists feel
that any field should be quantized and that general
relativity is a field theory. Although attempts to find a
theory of quantum gravity have been unsuccessful, it remains
one of the holy grails of physics. The best explanation to
reconcile these two viewpoints may be seen in an
earlier answer (skip down to the
last paragraph ).

QUESTION:
What is the reason of collapsing of an empty plastic bottle when air is sucked out of it?

ANSWER:
Before evacuating the bottle, atmospheric pressure, about 15
psi, acts both inside and outside the bottle, so the net
force on the bottle is zero. When you take the air out,
there is now a pressure of 15 psi on the outer surface of
the bottle which crushes it.

QUESTION:
When a charge radiates energy,does it radiate in all directions at the same time? (like when a pebble is dropped in water and it creates wave in all directions)

ANSWER:
A charge will radiate when it accelerates. Electrons
oscillating back and forth in a wire radiate; this is what
an antenna is. What the radiation pattern is depends on how
the charges are accelerating. One of the simpler systems is when you have
an electric dipole p =q d , equal positive and negative
charges ±q separated by some distance
d ; imagine these as vibrating
sinesoidally along the direction of the vector d
with some frequency. They will radiate most strongly in the
direction perpendicular to p and
radiate not at all along the direction of p .
A nice animation of this can be seen
here .

QUESTION:
On page 114, in Hawking's book, (Brief Answers to Big Questions ) it is stated that a chemical rocket with a thrust velocity of 3 km/sec, that jettisons 30% of its fuel, attains a velocity of 'about half a km/sec'. Using
Tsiolkovsky's rocket equation I get a value of 1.07 km/s.
Have I made a mistake?

ANSWER:
I have used this equation in an
earlier answer , v=u ln[(M+m )/M ]
where M+m is the initial mass, m is the
fuel burned, u is the exhaust speed, and v
is the speed acquired by the rocket. So I find v=3ln(0.7^{-1} )=1.07
km/s. I do not know how Hawking came up with his answer, but
it might hinge on what he means by "30% of its fuel"; you
and I have interpreted that to mean the exhausted fuel has a
mass equal to 30% of the mass of the rocket before the fuel
was burned, hence M /(M+m )=0.7. Also, maybe
he was thinking about a launch from earth which would
require some of the energy from the rocket to overcome the
gravitational force and the energy loss from air drag. Or
maybe he was guesstimating the effects of efficiency of the
engines. In any case, you did not make a mistake.

QUESTION:
What would happen if a plastic bottle was taken up in the air 10 000 meters?

ANSWER:
At sea level the pressure both inside and outside the bottle
is about 15 psi, so the net pressure is zero. At 10 km the
pressure outside drops to about 4 psi, so the net pressure
is about 11 psi. One site told me that, for the plastic in a
2 liter bottle to fail, the net pressure would be about 150
psi. So, nothing much would "happen". Actually, since the
temperature at 10 km is about -50^{0} C, if the air
in the bottle chilled to that temperature, the pressure in
the bottle would be smaller than 15 psi. Using the
Gay-Lussac Law, P /T =const., I find the
pressure would be about 11 psi, a net of about 7 psi. (If
you want to check this, you must use absolute temperature,
not ^{0} C.)

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.