QUESTION:
So Muons have a tiny mount of mass right? And we accelerate them to near the speed of light in particle accelerators right? And the faster something goes the more mass it has, until it hits infinite which is why we can't get it all the way to the seed of light right?
Assuming I am correct above, doesn't that mean if we wanted to say create artificial gravity on a space ship, we could accelerate a single Muon until it's moving so fast that it has the gravity equivalent of earth?
Could this also be used to warp space and create a warp drive essentially? Wouldn't it create like a spiral gravity field that moves around the acceleration track with the Muon? Wouldn't this create some kind of gyroscopic gravity effect?
If you could stop it and start the acceleration it seems like it would start causing near by objects to accelerate towards it and then past it.
Likewise if the Muon keep being accelerated closer to the speed of light maybe it would enter it's own gravity wake which would be warped, and thus allowing it to accelerate faster then the speed of light?
ANSWER:
Your idea will not work for many
reasons. First, look at an
answer
to an earlier question very much like yours; the muon is
so tiny, even at huge speeds it still only gets more
massive by like a few hundred thousand times, and ends up
still tiny. Also, the muon is not stable and although
time dilation would cause it to live much longer than if
at rest, it would still only live a few seconds. Also,
how are you going to accelerate it to such great speeds.
Also, if you could get it massive enough so that the
field would be equal to earth's at say 20 m away, the
field at 10 m would be 4 times bigger.
FOLLOWUP QUESTION:
So You mentioned the field would be bigger, does that mean that the particle would get larger as it moves faster? I'm not sure I understand what you mean by that. I guess I assumed the particle would stay the same size, but act gravitationally as if it was larger. I guess if it actually has to grow to the size of earth it would not be useful. Then the other challenge is energy, I agree that's a limiting factor, if energy wasn't a barrier could it still be used to produce 1G, or would the particle become earth sized?
ANSWER:
The field being larger means the force you would feel is larger, not the size of the particle is larger.
QUESTION:
Cool, thanks! So If I can get for all the energy needed I got artificial gravity figured out ??
ANSWER:
You have obviously not read my answers
carefully. The most important reason that your idea for
artificial gravity will not work is that the source is a point mass and the field falls of quadratically This means that force you will feel will only be what it would be on earth if you were one distance from the particle; move closer and it gets bigger fast, move farther and it gets smaller fast. What you need is to have a uniform gravitational field, which means that the force you feel
should be the same everywhere like it is on the earth. Furthermore, how are you going to keep this particle inside
your ship if it is moving, for all intents and purposes, at the speed of light? Just for the heck of it, let's figure out how big the mass
M of your particle would have to be for the force on a mass
m a distance 10 meters from
M is equal to mg where g=9.8 m/s^{2} is the
acceleration due to gravity on earth.
mg=MmG/r^{2},
so,
M=gr^{2}/G=9.8X10^{2}/6.67x10^{11}=1.5x10^{14}
kg
Here G
is Newton's universal gravitation constant. If you were
1 m away from this mass, the force on you would 100
times your weight. The very presence of a mass this
large would likely result in the collapse of the entire
ship. The energy you would have to give a proton would
be about 10^{31} Joules; it would take all the
power plants on earth about 100 billion years (much
longer than the current age of the universe) to deliver
this much energy. Please do not tell me that you have
"artificial gravity figured out"!
QUESTION:
Hello! I'm coming without a question about Mariana Trench. Since it's so deep and pressure is so strong, would it be possible for a £2 coin to fall to the very bottom of it? And would it be transformed in any way?
ANSWER:
A coin and water are both
essentially incompressible; in other words, the coin and
the water both will have the same density at the bottom
as at the surface. The only forces (ignoring drag if it
is moving) on the coin are its own weight and the buoyant
force which is equal the the weight of the water with the
same volume as the coin. So the motion of the coin does
not depend on the depth and it will fall to the bottom
because its weight is greater than the weight of the
displaced water.
QUESTION:
The weak force is stronger than gravity. Bosons reportedly mediate or generate the weak force.
Wiki says that "the high masses of bosons limit the range of the weak interaction". Perhaps the reason this seems odd to me is because I'm conceiving the weak force to function similarly the way gravity functions. Looking at it from that perspective, the greater the mass of an object, the greater its gravitational pull, right? So, if the weak force is stronger than gravity, why would the greater mass of bosons limit the range of the weak force? Or perhaps what I should be asking is: how does the function of the weak force differ from that of gravity?
ANSWER:
I should first warn you that it
is very difficult to compare gravity with any of the
other three fundamental forces because it is the "odd guy
out"; our best theory of gravity is general relativity
and the associated field has never been quantized while
the others have. Therefore, it is not really possible
talk about the gravity in any quantummechanical detail
because we do not have any gravitational boson mediator
to compare. However, it is generally agreed that we
should call this (unobserved, unpredicted) boson a
graviton. I will answer your question assuming gravitons
do exist, so you should be wary.
Let's first talk about the electromagnetic force. The
boson which mediates this force is the photon and it has
zero mass. The electrostatic force is proportional to 1/r^{2}
where r is the distance between a point electric
charge and where the force is felt. All indications are
that gravity is also a
1/r^{2} force and that
gravitational waves travel at the speed of light which
would mean that the graviton is also massless. It is also thought that the
gluon has approximately zero mass whereas, as you note,
the bosons associated with the weak interaction are massive. So, you see, the mass of the
boson has nothing to do with the relative strength of the
force.
The range of the force is not the same as the strength of
the force. The mechanism in field theory for force is
that "virtual bosons" act as mediators. But of course,
these bosons have energy so if one of them suddenly pops
into existence, doesn't this violate energy conservation?
Yes it does! The way this can happen is via the
uncertainty principle which permits a change in energy ΔE
as long as it only does so for a limited time
Δt where ΔEΔt
≈ℏ
where ℏ is the rationalized Planck constant. So a
massless boson has only its kinetic energy and travels
with the speed of light, c; a massive boson has
its rest mass energy as well as its kinetic energy and
must travel with speeds smaller than c so it
will exist for a shorter time and the range of the force
will be therefore be smaller.
Finally, the only reason gravity, the weakest of all, is
the only one we are normally aware of is that there just
happen to be huge chunks of mass, the source of gravity,
around us.
QUESTION:
I'm trying to get a handle on particle "spin". Most of my sources say that scientists long ago gave up the notion that particles actually spin on an axis the way the earth does. Wiki depicts it as a 720 degree rotation. https://en.wikipedia.org/wiki/Spin%C2%BD
Then other sources say that the spins of bosons, electrons, nuclei (and the quarks contained therein) are not the same spin. And then there's the 1/2 spin, etc. So what makes spin a spin?
ANSWER:
I suggest you look at three
earlier answers, 1,
2,
3.
QUESTION:
i am trying to invent something.
a tank of water, six inches deep,
20 feet in diameter. 157 cu ft.
Tank rotates 60 rpm, 62 fps.
What psi does outer wall have on it?
ANSWER:
There is no single answer because
the pressure in a fluid in a gravitational field depends
on the depth. Also, note that it will take some time for
the whole volume to come to the same angular velocity,
i.e. rotate as a rigid body. Now, there is an important
effect which will occur when you start the tank spinning: a
centrifugal force will push the water away from the axis
and the surface will become parabolic in shape.
Unless the tank has walls higher than the depth of the
water, the water will spill out over the top. Your proposed angular velocity is pretty large (1
revolution/second) and this will make the problem
extremely difficult to do. This is illustrated by using
an equation which I
found which expresses the the height
h above the fluid in the center and upper edge of the water
at the wall of the tank as a function of the angular
frequency ω and the radius R of
the cylinder, h=ω^{2}R^{2}/(2g)
where g is the acceleration due to gravity. I
prefer to work in SI units and will switch back to
imperial units in the end:
h=[((2π)^{2} s^{2})(3.05^{2} m^{2})/(2x9.8 m/s^{2})=18.7 m=61.4 ft!
What this means is that virtually all the water in the
tank would spin out to the walls and spill out. I must
admit that this surprised me. To make this a doable
problem we must make the surface of the water stay flat;
we can do this by putting a flat lid right at the surface
of the water. That is what I will now do to answer your
question; note that the pressure at the outer surface
will depend on the depth.
I will assume that you are interested in the gauge
pressure, the pressure in addition to the atmospheric
pressure; if you want absolute pressure, add about 14.7
psi. First, assuming the tank is not rotating at all,
write the pressure of the water (P_{w}) a depth
d below the
surface;
P_{w}=ρgd where
ρ=10^{3} kg/m^{3} is the density of water.
So, at the surface (d=0) the pressure is zero
and at d=6 in=0.152 m,
P_{w}=1490 N/m^{2}=0.216 psi.
The pressure at any depth d is P(d)=(0.216/6)d=0.036d
where d is in inches. That was gravity without rotation; next do rotation with
no gravity. (I am going to do this using calculus, so if
you don't know it you can just jump to the end.) All
infinitesimal
volumes of water (dV=rdφdzdr) will have a centrifugal force dF_{c}=(v^{2}/r)dm
where r is the distance from the axis of
rotation, v=rω is the speed of the tiny volume,
and dm=ρdV=ρrdφdzdr is its mass.
You will note that the pressure depends only on r
and if you do all the integration you will find that P_{c}(r)=½rω^{2}ρ.
I find that P_{c}(r)=6.02x10^{4} N/m^{2}=8.73
psi. Finally you can simply add P_{c}
and P_{w} to get P(d)=(8.73+0.036d)
psi.
QUESTION:
I am
in the middle writing what I believe will be a series of speculative fiction novels which take place in the far future. I've been working on them for over 20 years while I was working my normal job. Now that I'm disabled, I've become ENABLED to focus 100% of my energies on completing this first book.
I'm fairly decent at basic math but I was never taught how to utilize mathematical formulae  I'm forever sorry about it.
As such, I was wondering if it's possible to put a basic math series of steps together for me to be able to fill in the slots to figure out how fast a large rotating ring would need to spin to achieve the equivalent of 1G (9.81 m/s2) for an orbiting ring with a radius of 132,500,000 kilometers?
I have all the statistics of my habitat ready to go, I just need to know where to put which number and then a.s.m.d. which to which to get to the needed number.
ANSWER:
The acceleration a of an object
moving with speed v on a circle of radius R is a=v^{2}/R so
v=√(Ra)=√(1.325x10^{11}x9.81)=1.14x10^{6}
m/s; the acceleration is a vector directed at the center
of the circle. Suppose that object was a person of mass
m. The ring would have to exert a force F=ma on her to keep her
moving in the circle; she would interpret the force she
felt from the ring as a force force pushing her against
the ring (centrifugal force), and she would feel like she
felt on earth. With a speed of 1.14x10^{6 }m/s, the rotational
velocity would be ω=v/(2πR)=1.3x10^{5} revolutions/second=425
revolutions/year.
Have you ever read
Ringworld by Larry Niven? Your world sounds a
lot like his.
QUESTION:
I have a piece of metal that is designed to pass an electromagnet at very high speed. The set up is that the electromagnet is designed to manipulate this piece of metal as it passes it.
I need to know how quickly an elctormagnet can be turned on and off and on and off again in order to affect the peice of metal  the important fact is it is travelling very quickly indeed. For example, if this were to be posed as a circlar movement of the metal as the hand of a clock and the electromagent sitting at one side outside of the perimeter of the clockface, say at 8 oclock, this metal piece could easily be achieving speeds akin to a high powered motor. Yet at the same time the magnet musty be able to give full effect, turn off and given this thought example turn back on again at the next pass. Can an electromagnet react this quickly?
ANSWER:
I don't know the details of your
device, but I can tell you that very rapid changes of
magnetic fields from electromagnets are very difficult to
achieve. The physics reason is that electromagnets are
always coils of wire which have an inductance which,
because of Farraday's and Lenz's laws, resists changing
the current in the wire. Even a current in a simple
circuit cannot be instantaneously turned on or off
because of the inductance of the single loop of
currentcarrying wire. I would encourage you to study up
on inductance.
QUESTION:
When we throw light on opposite wall in an elevator by a torch, the light bends . Is is it because the photons leaving the torch move with constant velocity but the opposite wall of the elevator accelerates and we see it as bend?
ANSWER:
The light does not "bend" if the
elevator is moving with a constant speed, only if it is
accelerating. Also, you must imagine the elevator in the
middle of empty space, no gravitational field. This is
going to be one of my longwinded answers because it is a
good learning opportunity on several levels. I will
change your question slightly because it is more
conventional to consider this problem if the light source
is outside the elevator and enters through a small hole
in the side; the light moves perpendicular to the
velocity of the elevator as seen by an observer at rest
outside.
First
let's look at the case where you are in the elevator
moves with constant velocity v and and there is
also an observer at rest watching you go by. You, in the
elevator see the light follow a path shown by the solid
red line. The length of this line is d=√(L^{2}+(vt')^{2}) where
t' is the time your clock measures from when it entered until it hit the opposite wall.
The time it takes t'=d/c where c
is the speed of light. It is a straight line because v is constant and, for example, the elevator has moved
d/2 at t'/2. Meanwhile, the stationary
observer sees the light move straight across (dashed red
line) and strike the wall at the same place because the
wall has moved up in the time it takes to go across which
is t=L/c where t is the time
measured using his clock.
Therefore, the clocks of the two observers do not run at
the same rate. Now, the time the observer measures is t'=√(L^{2}+d^{2})/c where
d=vt'
is the distance the elevator has traveled.
Now, putting it all together and doing some alegbra, (t'/t)^{2}=(1+β^{2}(t'/t)^{2})
which leads to t'=t/√(1β^{2}) where
β=v/c. This is the famous
time dilation formula.
Finally,
let's consider the case of the accelerating elevator. I
will assume that you, in the elevator, have measured the
acceleration to be a and the the velocity at the
instant that the light pulse enters the elevator is zero;
therefore, v'=at'. The outside observer sees
exactly the same as he saw with constant velocity: the
light goes straight across and then strikes the wall
wherever you saw it strike the wall. You, however, see
the light follow a parabolic path (it would seem*) since
the position of the elevator is z'=½at'^{2}.
This was the thought experiment by Einstein where he
realized that there is no experiment you can perform in
an accelerating system in empty space which would not
have the same result in a uniform gravitational field,
the equivalance principle, a cornerstone of general
relativity.
*The previous part of my answer (accelerating) is
technically wrong but conceptually correct. The actual
mathematics and physics become quite technical. The main
reason it is not easy is that acceleration does not play
the important role it does in Newtonian mechanics because
it is not the same for observers. Also contributing to
the fact that the path is not exactly a parabola is that
the speed everywhere on the path is c whereas
the speed increases as it falls for an object with mass.
Also emphasizing the difficulty here is that, when
Einstein first calculated the deflection of light by a
star, he made an error of a factor of two.
ADDED THOUGHT:
If you want to learn more
about how acceleration and force are treated in
relativity, see the links on
this subject on the FAQ page.
QUESTION:
I have had this theory for about three years now and I was wondering if you could see if this is valid or not. My question is, if everyone person except the pilot on a passenger plane were to all stand up and jump at the same time, would the plane drop or be disturbed at all? I feel like it would be disturbed as average mass it at 6070kg each therefore surely it should move.
ANSWER:
This is not a "theory", it is
straightforward Newtonian phyics. For an isolated system
(which we can consider the plane plus passengers to be,
approximately)
the linear momentum must be conserved. Suppose that there
are 200 passengers each with mass 80 kg in a Boeing
737100 with mass 28,000 kg. If each jumps so that they
go up 40 cm, they need to have a speed of v=√(2gh)≈2.8 m/s. Their momentum is
p=80x200x2.8=4.48x10^{4}. The plane must recoil downward with the same magnitude of momentum, P=28,000V=4.48x10^{4} so
V=1.6 m/s=3.6 mph. The plane is flying with an
approximate speed of 600 mph, so this would not be very
violent, certainly not nearly as much as bumpiness due to
turbulance; and when everyone returned to the floor there
would be no vertical speed at all again.
QUESTION:
I understand that the speed at which time passes varies with altitude, or more accurately, with the strength of the surrounding gravitational field. I also understand that the speed at which time passes changes for an observer as their velocity changes (I.e. time dilation effect as you approach the speed of light). So... is the a difference in the speed time passes between a point at sea level at the equator versus at sea level near one of the poles?
ANSWER:
The first part of your question
alludes to gravitational time dilation. However, your
question asks only for the comparison of clocks where the
gravitational field is the same. So my answer will
discuss only the normal time dilation. Suppose that you
have a clock on the north pole. You observe a clock at
the equator which has a speed of about 464 m/s. When if
you observe your clock to tick off 1 s, the clock at the
equator will click off a time t=1/√(1(464/3x10^{8})^{2})≈1+1.2x10^{12}
s, a very tiny bit slower than your clock.
QUESTION:
I'm researching issues of physics and wanted just a clarification, specifically with regards to gravity. I've heard it said what actually keeps gravity from crushing us is the electromagnetic force  that the electrons in our bodies are pushing against the other atoms, which keeps us from being crushed. Is this true?
ANSWER:
Well, interactions between atoms
are primarily electromagnetic and interactions betweeen
atoms can bind them together to make a solid. If there
are to be living beings, that binding must be strong
enough so that it is not broken when the gravity tries to
push each atom in a solid down; any evolution will
certainly be driven by the environment and gravity is a
very important part of the environment. An instructive
example might be an ice cube. It sits on the table top
not being crushed. But as it melts the bonds break and
the water spreads out on the table as gravity "crushes"
it. If you did not have bones you would probably end up
pretty flattened on the ground.
QUESTION:
Can gravity be used to create energy, and if so, can dark matter be used to create energy?
ANSWER:
Drop a ball and it gains energy;
gravity is giving it that energy. Water from a reservoir
runs down and spins a turbine to generate electricity;
gravity is causing that energy to be generated. You have
to be careful with the word "create" though. The total
amount of the energy in the universe always remains the
same and our manipulations simply convert one kind of
energy to another. I wouldn't speculate about dark
matter, though, because nobody actually knows what it is
in any detail at all.
QUESTION:
I keep reading/hearing on videos that there have been experiments to prove that the speed of light is the same for all observers. A sample claim is "Since Maxwell's work, numerous experiments have been performed to test the prediction that electromagnetic radiation travels at the same speed for all observers  and none have failed." The articles or videos never go into any additional details on the experiments.
ANSWER:
Let's first talk about the
origins of the idea that light in a vacuum has the same
speed regardless of the motion of the source or the
observer. Maxwell's equations, the equations of
electricity and magnetism, unambiguously predict this
behavior. When Einstein developed his theory of special
relativity, perhaps the most important postulate was the
constancy of the speed of light, c. The
predictions made by special relativity were revolutionary
and would be false if c were not a constant for
all observers. So, certainly the most important and
accurate experiments verifying constant c were
experiments checking the predictions of special
relativity. No experiment was ever performed which
contradicted the predicitions. If you are not
familiar with special relativity, look up
time dilation and
length contraction. Incidentally, no prediction of
Einstein's theory of general relativity (our best theory
of gravity) has ever been found to not agree with
experiments.
QUESTION:
I only went through Physics 2 in college so I am a relative rookie, albeit a long in the tooth rookie.
My Question is related to Quantum Mechanics.
If a particle, electron or whatever is spinning one way; and we send it through a magnetic field of one direction or the other; and the object goes up or down or right or left and supposedly we know it's spin.
I understand that sometimes, however, we send it through again and the spin is different.
The question may be stupid but for a rookie we think of simple things like English on a pool ball, it can put the opposite spin on an object it collides with.
So the question is do physicists consider friction and it's effect on spinning objects when measuring these things?
ANSWER:
This is a really hard question to
answer. Electrons do indeed behave like they have an
angular momentum. We generally refer to this angular momentum as intrinsic spin or just spin. We are therefore led to believe that they are actually tiny spinning balls, just like the English on a billiard ball. But, as is almost always the case,
we generally run into trouble if we try to apply our classical physics to quantum physics. For example, consider a proton whose spin quantum number is, like the electron, ½; its actual angular momentum
S is S=ℏ√[½
(1+½)]=0.87ℏ where ℏ is the rationalized Planck constant. Suppose now you put in reasonable numbers for the mass and radius of a proton, assume that it is a uniform sphere, and calculate the rate of rotation. You find
a totally nonphysical answer; the surface of the sphere would have to move faster than the speed of light!
Certainly when particles collide with others their spins
will be affected depending on how they interact; to find
out the details you would have to do quantum mechanical
calculations. Friction is a wholly macroscopic concept
and has no existence in a microscopic world. Also, the
way we observe spin is to observe the requisite magnetic
moment of the particle; recall that a current loop has a
field just like a tiny bar magnet and a charged particle
with spin will be a tiny current loop. A bar magnet in a
uniform magnetic field will experience a torque but no
net force; you must make the field non uniform for the
particle to see a force. It seems that you are a little
hazy on how this observation is done and you should study
the
SternGehrlach experiment to sharpen up that
understanding.
QUESTION:
Plane 1 and Plane 2 have same mass and surface area.
If plane 1 starts its flight by applying 500N force (numbers are rough and just for illustration), Then it would accelerate. But when speed increases  air resistance increases, if the plane applies the same 500N force throughout its Flight then soon acceleration decreases and the plane will reach terminal velocity due to the opposing force of air resistance. ( I have applied Newtons second Law )
If plane 2 starts its Flight with 1000N force, then it would accelerate more than Plane 1. As the speed of plane 2 increases faster than plane 1  so will its air resistance. If the flight applies the same 1000N force throughout its flight  eventually its acceleration decreases and the plane will reach terminal velocity due to the opposing force of air resistance. ( I have applied Newtons second Law )
So how come two flights with different force reach their destination at the same time according to Newton's second Law???
ANSWER:
They do not reach the destination
at the same time because the terminal velocities of each
are not the same. Terminal velocity v is
achieved when F=Cv^{2} where F
is the applied force and Cv^{2} is the
air drag; so v=√(F/C). But you told me
that C is the same for both planes (same mass, area), so
v_{2}/v_{1}=√(1000/500)=1.4.
QUESTION:
Is refraction an actual physical phenomenon, or is it a strictly biological perception, like with touch?
To elaborate, does light actually bend from its path when its refractive medium is changed, or do we only see it a such?
ANSWER:
What a strange question! Of
course it is an actual thing. If refraction didn't really
result in bending the path of light, your eye would not
work; your eye captures light from some object and bends
that light (using a lens) to form an image on your retina
which is then interpreted by the brain. Likewise with
cameras and telescopes. I have attached a photograph of
light rays being bent to a focus by a convex lens; the
camera which took this photo would not be able to see a
"perception". Maybe you are thinking of whether a stick
stuck into water is really bent like it appears to be;
no, but it looks like it is because the refraction is
bending light rays.
QUESTION:
Newtons cradle: When you release 2 balls and they send their kinetic energy down the line,2 balls on opposite side move. But why doesn't it just send one ball much higher?
ANSWER:
Newton's cradle is usually
explained assuming all collisions are perfectly elastic.
Therefore, both kinetic energy and linear momentum must
be conserved. Suppose that two balls, each of mass m,
move together with speed v_{1} when they hit the other
balls. Then the initial linear momentum is p_{1}=2mv_{1} and the
initial kinetic energy is K_{1}=½(2m)v_{1}^{2}=mv_{1}^{2}.
Now, suppose a single ball exits the other side with
speed v_{2}; its linear momentum must be
equal to
p_{1}=2mv_{1}=p_{2}=mv_{2},
so v_{2}=2v_{1}. Now,
what is the final kinetic enregy?
K_{2}=½mv_{2}^{2}=2mv_{1}^{2},
twice the initial energy. Therefore energy is not
conserved so this cannot happen.
QUESTION:
If a singularity is a small point inside of a black hole where there's infinite density, does that mean if we take an iota of that singularity (presuming its tangible or transportable in its ways) would that be able to sustain gravity within a spacecraft or would that iota of singularity overpower everything in its way
ANSWER:
You are asking if you could use
some small black hole to provide a gravitational field in
a spaceship. Suppose that we put the black hole a
distance R=1000 m from the captain's chair. What
mass M would be required to make the
gravitational acceleration the same as on the surface of
earth (9.8 m/s^{2})?
So, a=F/m=GM/r^{2}=6.67x10^{11}M/10^{6}=9.8
or M=1.5x10^{17} kg. But the field
would decrease rapidly away from the captains chair and
increase rapidly toward the black hole; e.g., 10 m away
from the captains chair there would be an acceleration 9.8(100/110)^{2}=8.1
m/s^{2} and 10 m closer to the black hole 9.8(100/90)^{2}=12.1
m/s^{2}. Also, how would you hold it there?
Anything close to the black hole would would experience
huge gravitational forces tending to obliterate whatever
was holding it.
QUESTION:
For example I'm going to home with my car at 40mph and I open my car's lights.
These photons from my car won't go over the speed of light and my 40mph wont gonna add to this speed so what happend to this speed to this energy.
Where is it go?
ANSWER:
You are right, the photons
emitted forward from a source moving forward have more
energy than if the source were at rest. You are also
right that either photons will be going the same
speed. So, how could their energies differ? The energy of
a photon is E=hf=hc/λ where h is Planck's
constant, c is the speed of light, f is the frequency and
λ is the wavelength.
So the wavelength of the gets shorter if the energy gets
larger. This is just the Doppler effect for light and in
this case it is referred to as a blue shift. No energy
has been lost. (By the way, since 40 mph is so small
compared to c, it would be really difficult to observe
any effect at this speed.)
QUESTION:
I have a question that is driving me crazy. I'm a children's author; my stories have appeared in Highlights magazine.
I'm working on a middlegrade novel in which the protagonist plays lacrosse. The boy's father is an engineer who is fond of explaining how things work, and at one point he and his son have a discussion about what keeps a lacrosse ball in its pocket when a player is cradling (i.e. the rocking motion a player makes with a lacrosse stick to prevent the ball from falling out of the pocket).
I've researched this endlessly online and most explanations claim centrifugal force is what keeps the ball in the pocket. But a few claim that it's actually centripetal force. I tend to think that is the correct answer: the ball wants to fly off in the same path as the head of the lacrosse stick as the player swings it back and forth, but the pocket's mesh netting gets in the way and keeps the ball in the pocket. It would be the same force that stops water in a bucket from following the path of inertia and spilling out and when you swing it over your head in a big circle.
BUT ... I'm simply not sure.
ANSWER:
It all depends on who is
observing the ball. Let us, to keep the situation simple
but still illustrate the basic principles, assume that
the ball is moving in a horizontal circle of radius R
with a speed v. We focus our attention on the
ball as we observe it watching the game. (I have drawn three Cartesian coordinates shown in green to help visualize.) First, find all the forces
on the ball. (Ignore
the dashed vector labeled C for
now.) Two things exert forces on the ball, the earth
(gravity) and the netting. The earth force is called the
weight as W in the diagram; the
netting force I have broken into its vertical (F_{V})
and horizontal (F_{H})
components. The vertical component is equal in magnitude
and opposite in direction to W
to ensure that the ball moves horizontally rather than
fall vertically as the weight would have it do; the
horizontal component (centripetal force) keeps the ball
moving in a circle and provides the acceleration of a=v^{2}/R
which points toward the
center of the circular path. You can now calculate the
force F_{H}
from Newton's second law, F_{H}=mv^{2}/R
where m is the mass of the ball. So, there is
your answer: the ball continues to be in contact with the
net because the force the net exerts on the ball causes
the ball to accelerate; as long as the player keeps
changing the direction and/or speed of the ball, it can
be kept in the net. This is the centripetal argument.
But, what about the centrifugal force which seems to be
the majority opinion? Is it wrong? Not if you understand
what a centrifugal force is. Imagine that you are the
ball and you wish to use your coordinate system to
understand the physics. In your coordinate system you are
at rest so all the forces on you, according to Newton's
laws, must add up to zero. What are all those forces?
Exactly the same as when you observed the ball from the
stands. There is nothing else but the earth and the net
which can exert forces on you. What this means is that
Newton's laws are false if the observer is accelerating
relative to some coordinate system in which they are
true. So what do we do? We invent and add in a force
which compels the system to obey Newton's laws. That
force is called a fictitious force and in this case it is
the socalled centrifugal force. So we add the vector
C=F_{H}
to make the sum of all vectors equal to zero even though
it does not exist! So this is the centrifugal argument
since clearly C is holding the
ball to the net in this picture. Is it "cheating" to just
add a nonexistent force? Not if you do it for a good
reason. Newton's laws are great and greatly help
understanding what is happening. As long as you know what
you are doing, if you can force Newton's laws to work,
you can often more easily solve the problem.
Finally, just a little linguistics!
Fugo is
Latin for I flee (the root of fugitive);
centrifugal is centerfleeing. Peto is Latin for I
seek; centripetal is centerseeking.
QUESTION:
I am 17 years old and I want to know why like charges repel and unlike charges attract. I know this is due to the exchange of virtual photons but I am interested to know the deep facts in simplified way. I have read
this,
but it is still complicated to me. I would be grateful if you simplified it.
ANSWER:
Gravity has only one kind of
force, attractive, because there is only one kind of
mass. However, electrostatics has two kinds of force,
attractive and repulsive, so there must be two different
kinds of electric charge. Measurements are able to
determine the forces quantitatively (Coulomb's law). One
can build up everything we knew before the early 20th
century about classical electromagnetism by taking the
attractive/repulsive nature of electric charge as
axiomatic and based on measurements; physics, after all,
is an experimental science at its heart. But just like
Einstein sought an answer to "why" masses attracted each
other, physicists (e.g. Feynman, Dyson,
Schwinger, Tomonaga, et al.) developed quantum
electrodynamics (QED) which quantized the electromagnetic
field and therefore provided a theory for the interaction
between electric charges. Unless you are an extraordinary
genius, at age 17 you will not have the mathematical
toolbox needed to understand the theory beyond the
qualitative understanding you have (exchange
of virtual photons); if you are that extraordinary
genius, you should not need me to explain it to you!
QUESTION:
It's a simple question, but I can't find an explanation. What is the total energy of a block which is at rest, at ground level on a flat, frictionless surface? Intuitively it should be zero, since it is not moving (no KE) and it's "height" above the ground is effectively zero (so V=mgh=0) However, I thought an object with mass must have energy.
ANSWER:
More specifically, we would say that (provided that we have chosen the ground to be zero potential energy) the total
mechanical energy of the block is
zero. Still, it has rest
mass energy mc^{2}.
QUESTION:
I have a question for you. In your book
"Atoms and
Photons and Quanta, Oh My!"
appendix B page B2 Energy in special relativity, you
derive E=mc^2 using an approach I've not seen before.
What I don't get is the 2nd line where F = m (d[v/(1v^2/c^2)]/dt), where m is the rest mass.
Can you show the steps you took that aren't showing (to save space)? I don't see how you arranged this expression to include the differential operator
"d". I've seen/used several derivations using differentials & integration by parts & so forth from old physics texts & online.
ANSWER:
Here is a screen shot of the page
the questioner is referring to:
So, keep in mind here that
m is rest mass, F
is force, p is linear momentum, K is kinetic energy,
W is work, v is
velocity, c is speed of light. In classical
mechanics, p=mv. But, if v is not very
small compared to c, p must be redefined or
else it will not be a quantity which is conserved if F=0. The
correct definition in relativity is p=mv/√(1(v^{2}/c^{2})).
So the second line (after F=dp/dt
) simply inserts the
expression for p. The third line multiplies and
divides by dx and rearranges so as to have v=dx/dt
appear as a factor. The fourth line rearranges and
replaces
dx/dt by v.
Finally, the fifth line is the result of evaluating
the derivative with respect to x:
m(d/dx)[v(1(v^{2}/c^{2}))^{1/2]}
=m(dv/dx)(1(v^{2}/c^{2})^{1/2}+(½)(v)(1(v^{2}/c^{2}))^{3/2}(2v/c^{2})(dv/dx)
=m(dv/dx)(1(v^{2}/c^{2})^{1/2}[1+(v^{2}/c^{2})/(1(v^{2}/c^{2})]
=m(dv/dx)(1(v^{2}/c^{2}))^{1/2}[(1(v^{2}/c^{2}))^{1}]
=m(dv/dx)(1(v^{2}/c^{2}))^{3/2}
To do this last step you need to know your elementary differential calculus, (d/dz)(f(z)g(z))=f(dg/dz)+g(df/dz)
[product rule] and (d/dz)f(g(y)=(dg/dy)(dy/dz)
[chain rule]; and, of course, a bit of algebra.
QUESTION:
If a raindrop is affected by the earth's gravity and it should accelerate at a rate of g(9.8m/s^2) downward. Why does when it hits the ground it doesn't result in a big impact?
QUESTION:
Can gravity slow down the atom ?
ANSWER:
You are asking, essentially, if
atoms feel gravitational forces. Anything which has mass
will feel the gravitational force and atoms have mass;
so, yes, atoms experience the gravitational force and
thus can be made to slow down. For example, if you shoot
an atom straight upwards, it will slow down just like a
baseball thrown upward with the same speed. The very fact
that we have an atmosphere is proof that gravity is
acting, keeping all those atoms from speeding off into
space.
QUESTION:
What is the smallest change in frequency within the visible light spectrum that the human eye/brain can discern? My new TV says it can show over a billion colors. Can humans even see 1 billion colors!
ANSWER:
The way a tv works is by mixing
the three primary colors, red, green, and blue, in
varying amounts. There are an infinite number of possible
ratios. All the colors you can see are not all the
possible wavelengths you can see (also an infinite
number) but mixtures of many many wavelengths. When the
manufacture refers to "billions" he is probably referring to how
accurately the electronics can mix the three. For
example, if the relative amounts of the three were 10, 7,
3, the electronics would have an uncertainity in how
accurately they could set these numbers; maybe the
numbers could
only be set to a 0.01% accuracy, the ten would be 10±0.0001.
Then you could estimate the number of possibilities you
could have over the whole visible spectrum.
QUESTION:
I have been thinking about this for countless hours and though not a physics student, and haven’t taken any classes of the sort I have wanted to know this question: we know black holes are brought down to a point of singularity. So could and atom being so small do the same? And if this is true would that be the last atom we see on the periodic table?
ANSWER:
The force which holds a star
together is gravity; a normal star does not contract to a
black hole because it has collapsed to the point where
burning hydrogen (fusion) in the star heats it up so that
the pressure balances the gravity trying to push it
smaller. As the fuel runs out, the star starts getting
smaller. At this stage, various things can happen,
depending on various conditions, mainly the total mass of
the star; one possible thing is eventually collapsing all
the way to a black hole. An atom is a completely
different kettle of fish. We should really look at the
nucleus since that is where nearly all the mass is.
Unlike a star, energy is not being produced in the
nucleus, its energy remains constant. An atomic nucleus
is held together not by gravity but by the strong nuclear
force. Unlike gravity which is a very longrange force,
the strong interaction is very short range, so each
proton and neutron see only their near neighbors, not all
the other particles in the nucleus. The result is that
the heavier a nucleus becomes, on average, the less bound
it comes so its tendency is to break apart, not collapse.
By the way, the nucleus does not collapse because the
nuclear force at extremely short distances becomes
repulsive, not attractive.
QUESTION:
How much does a thought weigh? And how much does one molecule of carbon dioxide weigh? I want to compare the two.
ANSWER:
A molecule of CO_{2} has
a mass of about 7.3x10^{26} kg (about 1.6x10^{25}
lb). A "thought" is a collection of billions of
electrical impulses and chemical reations in the brain.
Even if these could be assigned a "weight", which they
can't, all "thoughts" are not the same and would have
different weights.
QUESTION:
Franklin first coined the term negative and positive in 1750 and Volta invented the battery in 1800. Did Volta assign a positive and negative end to the battery? Or did he think the negative and positive applied to electrostatics and not "flowing" electricity. You see where I am going with this. I need to go up the chain to J.J. Thompson to see where the goof up was on adoped circuit diagrams. The explanations I have read are non sense. They do not actually describe why the convention had to be reversed. All the examples I read use circular reasoning and thus far no smoking gun of how and when the goof up happened. I thought if I could start with Volta I might be able to unravel it.
ANSWER:
I really don't understand your
perception of "goof up". There are two different "kinds"of
electric charge.
Coulomb's law shows that the magnitude
of the force F which a point
charge q feels due to the presence
of another point charge Q is proportional to the
product of the charges Qq and inversely to the
distance r between them, F∝Qq/r^{2}.
Also, if the charges are of the same "kind" they repel
and of different "kind" they attract; since attract/repel
result in exactly different directions for the vector
force if we randomly assign signs to the two "kinds", we
can write F∝1_{r}Qq/r^{2}
where 1_{r} is a unit
vector in the direction of the vector r
from Q to q, 1_{r}=r/r.
This proportionality does not depend on
which "kind" we call positive and which we call negative,
there is no right and wrong choice for the sign for an
electron. Franklin made a random choice which resulted in
electrons as having negative charge; he might have made a
different choice if he knew electrons existed and were
usually the charge carriers in currents (nobody knew
then), but it was not a wrong choice. Later, when
currents and magnetic fields were studied, all of
electromagnetism simply continued the original choice
with no problem whatever. All this means that positive
electric currents in wires actually are almost electrons
moving in the opposite direction which is endlessy
confusing to students. I really don't know anything about
Thompson correcting "a goof up" in circuit diagrams other
than pointing out that currents shown moving in one
direction (from + to  battery terminals) is actually
electrons moving in the opposite directions. No physics
was changed.
QUESTION:
What will be the motion of the earth if the sun disappears after t=0 sec?.
I was asked this question in an exam and the answer is not tangential to the motion
ANSWER:
It is assumed, although never
very accurately measured, that the speed of gravity is
the same as the speed of light. Therefore, the earth
would continue in its orbit until t=8.3 minutes,
the time it takes light from the sun to reach the earth.
At that time the earth would go dark and proceed in an
approximately straight trajectory.
QUESTION:
How do gravitons that exit a certain mass cause other objects to be attracted to that mass. I can imagine Einstein's idea of curved space being the cause for the seeming attraction, but what does that have to do with gravitons?
ANSWER:
There is no successful theory of
quantum gravity; therefore gravitons are hypothetical
particles, never observed, just an idea. Gravity is the
only one of the four fundamental forces of nature which
has not been quantized. See an
earlier answer and the links in that answer. You will
see that the theory of general relativity is not
unambigously described by the notion of warped spacetime
but is also a field theory. Any field theory should be
able to be quantized; in a quantized field theory the
"messenger" of the field would be called a graviton, just
as the messenger for the electromagnetic field is the
photon, and is the gluon for the strong nuclear field.
QUESTION:
Why does a rounded pencil I put on a flat ground make a small harmonic movement before stay steady?
ANSWER:
One side of the pencil is heavier
than the other. For example, if you had a uniform ball
and glued a weight to its surface, it would always rotate
so that weight would go to the ground. With some pencils
the lead in the middle is not exactly on the axis. The
"lead" is actually graphite (carbon) with a density of
2.26 g/cm^{3}; wood has a density of about 0.6
g/cm^{3}. So the lead will seek its lowest p
ossible point it can possibly find. The figure
illustrates this situation. The weight of the lead will
cause the pencil to roll to the right; but when it gets
there it will be moving and so the pencil will continuing
to roll until the lead stops and then begins rolling to
the right. It will oscillate forever if there is no
friction; but there is so it will quickly damp out with
the lead directly below the center axis of the pencil.
QUESTION:
I am a glass artist. When I fume gold (vaporize in the torch flame) and it collects onto the borosilicate glass it appears pink/purple. If I encase this gold with clear, it then looks green. I can throughly manipulate the glass and maintain the green color. I read that Faraday obtained green gold in a thin film with stress on the film. I also fumed gold onto a rod and stretched it at a fairly cold glass working temperature to stress the gold and avoid annealing the gold. The gold took on a green appearance. I have been doing this for over 25 years and have not been able to find out why gold turns green when encased (or stressed like with the stretch) I am asking a physicist because this has to do with light and gold, plasmon resonance etc are pretty heavy topics. If you can answer this question I’d love to gift you one of my art works! Why does a thin film of gold turn green when encased in glass?
ANSWER:
I am not sure why it is
pink/purple when you first deposit it. One video
suggested that has to do with the amount of oxygen in the
film; it is also possible that it is a thinfilm
interference effect where the thickness of the film is an
integral multiple of wavelengths of the color you are
seeing and the light of that wavelength which is
reflected from the front and back of the film add up to
become the dominant light you see (the same thing going
on in the pretty colors you see on an oil slick). There
such things which use thinfilm interference, for example
the magenta interference filter shown in the figure.
Now, to the green issue. Why is gold the color we see
with just a block of gold? The reason is that when white
light strikes the surface, red/yellow/orange are
preferentially reflected back making the gold look
golden. But some of the light will continue into the bulk
of the gold, and, coincidentally, red/yellow/orange are
preferentially absorbed. This means that once you get to
a certain depth in the block the only light surviving is
the blue/green part of the spectrum; eventually all the
light gets absorbed and you will see no light coming from
the back side which originated on the other side.
However, if you have a "block" which is thin enough,
smaller than the distance where everything is absorbed
but thick enough that all the red/yellow/orange has been
absorbed, the film will look blue/green. I am quite sure
that this is what is going on but, since I don't know the
geometry of your art, I do not have a detailed
explanation for your particular situation; e.g.,
does it look green when seen from any direction? Maybe if
you sent a photo or two I could go farther.
I have a question for you. I am a
stainedglass hobbiest and find that red glass is
often more expensive than other colors. I have been told
that the reason is that gold is required to get a rich,
pure red glass. Do you know if that is true and/or why?
QUESTION:
A thought occurred to me. From what I know about particle accelerators, scientists speed up accelerator particles of matter and crash them into slower moving or stationary particles of matter so that they can examine the breaking apart of all into other elementary particles. This then allows them to find more and more basic particles of matter.
I have read that the speeds these accelerator particles are hoped to reach is the speed of light, hence the effort to build more powerful particle accelerators.
So, why not simply have photons of light (that are travelling at the speed of light since they are the elementary particles of light) crash into those other particles of matter? Why use so much energy trying to push accelerator particles to reach the speed of light, when we have light in the first place and photons already moving at 186,000 miles per second?
ANSWER:
It is not the speed of the
incident particles which matters, it is their energy.
There are no good available sources of highenergy,
highintensity photons. Also, photons interact only via
the electromagnetic force and we often want projectiles
which interact via the strong nuclear interaction. So the
accelerators are actually poorly named because they do
almost no change in velocity when they increase their
energy by a large amount; I have maintained that they
would better named energizers. For example, a proton with
speed 99.999% the speed of light accelerated to 99.9999%
the speed of light increases its speed by 0.0009% but its
energy increases by 216%.
QUESTION:
A photon hits an atom perpendicularly to its speed v and it is absorbed as is known immediately. So it can not act after the time when it is not perpendicular (e.g. the force is 0 after that). Certainly all its energy goes to the atom when it is perpendicular to v.
But the impact (absorption) applies a force on the atom and it is postulated that a perpendicular force can not do work. So work has not been done on the atom. Consequently its energy can not change.
If all photon energy goes to increase internal energy of an orbit electron from Eo to E1 how then is the DIRECTION of velocity of the atom changed?
Isn’t this a contradiction?
ANSWER:
Your first paragraph really makes
no sense. Talking about "perpendicular" in a collision
really has no meaning; I think what you mean is a headon
collision, one where the recoil of the atom is in the
same direction as the incident photon. In a case like
this you do not want to think about force and work.
Rather you should ask if energy and momentum are
conserved; if so, you should solve your problem that way.
Suppose that the system we are interested in is the
photon (mass zero, energy E=hf, and momentum p=hf/c)
plus the atom (rest mass M, energy E=Mc^{2},
and at rest p=0). Because there are no external
forces on this system, its total energy and momentum must
remain constant. After the absorption, the mass of the
excited atom is M', its kinetic energy is K,
and its momentum is P'; there is no need to
think about what goes on microscopcally inside the atom.
So we have two equations:
E_{1}=E_{2} or
hf+Mc^{2}=M'c^{2}+K
p_{1}=p_{2} or
hf/c=P'.
The notions of force and work are seldom useful when
looking at atomic sizes. In this case, the photon carries
energy into the atom so the atom's energy has to
increase.
QUESTION:
If l video an energy source such as a manmade Ultraviolet Curing light, can watching the video hurt your eyes? And is there a way of capturing this light to make it even though it is invisible?
ANSWER:
The video camera is not sensitive
to UV light and therefore records none. The monitor is
designed to emit almost entirely visible, not UV light
and therefore radiates none. So no, the video will not
hurt your eyes. There are certainly detectors which do
detect UV radiation but it is not really 'capturing' it
in the sense that you could 'make' it.
QUESTION:
This question has resulted from considerable
backandforth communications between me and the
questioner. The essence of the question is this: Two
cylinders are rigidly attached to each other. The smaller
has a radius R and its center axis (B in the
figures below) is separated by a distance d from
the center axis (A) of the larger. A torque is applied
about the axis of the smaller cylinder but exerts no net
force. Find the resulting torque about the axis of the
larger cylinder.
ANSWER:
The
figure shows one possible way of exerting a torque about
point B. The four tangential forces all have magnitude
F/4 so that the torque is τ_{B}=RF.
There is a net force of zero. The torques about the
point A are:
τ_{1}=RF/4
τ_{2}=(d+R)F/4
τ_{3}=RF/4
τ_{4}=(dR)F/4
When all four
torques are added,
τ_{A}=RF=τ_{B}.
A
more general approach to the problem would be to have a
uniform tangential force all around the perimeter of the
smaller cylinder as shown in the second figure. If we
write the magnitude of the vector dF
as dF=[F/(2π)]dθ, the
integral ∫dF from 0 to 2π will be
F
and τ_{B}=RF. Now the
torque due to dF about A will
be dτ_{A}=r×dFdθ=[RF/(2π)][(dsinθ/R)+1]dθ.
Integrating this from 0 to 2π yields, again,
τ_{A}=RF because ∫sinθdθ=0
over one whole cycle.
There are two important provisos
here: there must be no other forces present which exert
torques about axis A and the forces exerting the torque
about axis B must sum to zero net force.
Sometimes I like to interject
examples of how science is done and sometimes progress
hits snags. Most often preconceptions are the biggest
stumbling blocks. When I first started to work on this
problem I assumed that the answer had to depend on d.
As you can see, this is not a particularly difficult
problem. I quickly tried the general approach and got the
answer as I did above; but I was so sure that d
should be in there somewhere that I assumed that I had
done something wrong. So I did the calculation for each
quadrant individually; these are tedious calculations,
not difficult but certainly prone to algebra and
arithmetic errors. I spent hours! Finally, when I was
sure that everything was right I found
τ_{A}=RF.
That was when I tried
the simplified torque with four forces (above) to
convince myself that it was correct.
QUESTION:
I am trying to clarify in my head an idea regarding electromagnetic force. It is usually explained in very superficial terms. One way to explain my question is this: I understand that two positively charged particles will repel each other, for example two protons. They both have a positive charge, but if they are at rest there will be no magnetic field. Do electrical charges repel like magnetic charges? Are they the same thing? We measure magnetic strength in one set of units such as Teslas but electrical potential in another set of units. I understand the strength of a permanent magnet to be somehow related to the arrangement of its electrical charges. Is that correct? Is a proton at rest a permanent magnet? Is there some simple relationship between a static electrical charge and a permanent magnet?
I'm sorry the question is so long but it is all part of one big question.
ANSWER:
Your question violates site
ground rules for "single, concise, wellfocused
questions" in spite of your saying it is "one big
question"; even so, it is neither concise nor
wellfocused. I am sorry, but I cannot give you an entire disquisition on electromagnetic theory in a single concise answer. You have a number of misconceptions. First, there is no such thing as a magnetic charge. I always like to like to refer to an answer to a question I answered long ago:
Answer: The laws of electromagnetism are perfectly symmetric: a changing
magnetic field causes an electric field and a changing electric field causes a
magnetic field. The first of these is called Faraday's law and the second is part of
Ampere's law. You seem to think that only a permanent magnet is magnetism. In
fact, any moving electric charge causes a magnetic field. The most common
source of magnetic fields is simply an electric current. Here are some facts about
electric and magnetic fields:
•electric charges cause electric fields,
•electric currents (moving charges) cause magnetic fields,
•changing magnetic fields cause electric fields, and
•changing electric fields cause magnetic fields.
Magnetic fields are not caused by a charge like electric fields are, they are caused either by electric currents (moving charges) or changing electric fields. So then, what is a permanent magnet? On the atomic level, think of electron orbits as little currents and electrons as tiny spinning charges, also currents. In a material where all these tiny magnets align with
each other results in a sum of all their magnetic fields. Regarding your writing
"but if they are at rest there will be no magnetic field", this has nothing to do with any force they exert on each other by virtual of their charges. Indeed protons do have a tiny magnetic moment because they can be roughly be thought of as spinning, but those moments are very weak. If you want to really understand all this, you need to seriously study electromagnetism.
QUESTION:
As solar energy is becoming a major, maybe eventually the main source of electricity, I wonder what the effects of capturing and retaining significantly more of the Sun's energy in the system of our planet could be. Would the overall effect be comparable to the warming effect of greenhouse gases, where less energy is radiated back to Space? What about other forms of renewable energy, that ultimately just harness the Sun's energy?
ANSWER:
In greenhouse warming, radiation
is absorbed by the earth and then radiated back into the
atmosphere; some of that energy goes back into space and
some bounces around in the atmosphere, heating it.
Factors, including the amount of CO_{2} in the
atmosphere, determine the fraction of reradiated energy
which is trapped. Now, if we capture this energy and
store it in batteries, it stores the energy in a form
which is not heat; then this energy is used to drive
cars, run our utilities, lift material to build
buildings, etc., all of which will convert a
fraction of the energy to heat, but a much smaller amount
than if all that energy were absorbed by the ground and
then was reradiated into the atmosphere. And, using this
energy instead of fossilfuel burning will further help
the situation by not adding to the CO_{2} in
the atmosphere.
QUESTION:
I am trying to calculate the energy consumed by an electric bike when it is charging at home. It has a 48V, 14Ah battery and takes about 6 hours to charge. Is the electricity in my power bill going to be 672 watt hours (48 x 14) or ~4,000 watt hours (48 x 14 x ~6)?
ANSWER:
I got exactly the same answer as
you but took a different path to get there. Since an
ampere is a Coulomb/second (C/s), 1 A·h=3600 C, so 14
A·h=5.04x10^{4} C. The energy to move that charge
across a potential difference of 48 V is E=QV=48x5.04x10^{4}=2.4192x10^{6}
J=2.4192x10^{6} W·s(1h/3600 s)=672 W·h.
QUESTION:
I learnt about the Rutherford experiment where they disproved the
plum pudding model for atoms where they used a gold leaf which was
apparently 100 atoms thick. I have since read about a gold leaf that
was 2 atoms thick. If atoms are mostly empty space why wouldn't you be
able to see right through said gold leaf.
ANSWER:
Gold looks golden because it
reflects yellow, orange, and red if illuminated with
white light. Interestingly, if an object reflects a
particular color, it is also a very good absorber of that
same color. What that means is that, for gold, yellow,
orange, and red will be exceedingly unlikely to get
through gold leaf; if any light at all gets through it
will be other colors. Indeed, light transmitted through a
gold foil will be bluish in color. Typical gold leaf is
more than ten atoms thick and two atoms is a very recent
achievement, I believe. I would think that the thinner
the foil, the more of the gold colors would also transmit
through.
QUESTION:
The Laws of physics say nothing can travel faster than the speed of
light. Given the universe is expanding, two stars on opposite sides of
the universe would be travelling away from each other at much faster
than the speed of light. From the perspective of planet a, planet b is
travelling faster than the speed of light; how is that possible.
ANSWER:
I presume that you are using
Galilean relativity to deduce the speeds of two objects
relative to each other if you know their speeds relative
to something which you will call "at rest". Let's do an
example: suppose that, relative to earth planet a is
moving away from you with a velocity V_{ay}=0.8c
and, on the other side of the universe, planet b is
moving away from you with a velocity V_{by}=0.8c
where c is the speed of light. You conclude
that the velocity of a relative to b is V_{ab}=1.6c
because the velocity addition formula for Galilean
relativity is V_{ab}=V_{ay}+V_{yb}
and V_{yb}=V_{by} (or,
you might say, just by "common sense"). However, for
objects moving with speeds comparable to the speed of
light, Galilean relativity, and indeed, all of Newtonian
mechanics, are not valid laws of physics. The correct
velocity addition formula is
V_{ab}=(V_{ay}+V_{yb}
)/[1+V_{ay}V_{yb}/c^{2}]=1.6c/1.64=0.976c.
QUESTION:
I am emailing you with a specific question my 5 year daughter asked my husband and I about rainbows. I’m hoping you might be able to help us, or point us in a direction that would help us with an explanation. We found you online, and we need a professional.
We showed our daughter a prism refracting the light just like a raindrop would. She asked, if we hung 100 prisms from the ceiling that should make 100 little rainbows, right? I said correct. BUT!!!! she asked "If each raindrop can refract the sunlight into the colors of a rainbow. Then why aren't a million rainbows in the sky? Why do we only see one big rainbow? How do the reds stay together, the orange together etc...Please help. I haven't been able to find any information on the web.
ANSWER:
The prism demonstrates dispersion, the fact that different colors of light
travel with different speeds in any medium. The prism is
made of glass and the rain drop is water; both
demonstrate dispersion but the details are different
because the rain drop is a sphere. All the light rays
from the sun come in essentially parallel. In my figure,
several rays are shown and one is followed as it refracts
when it enters the water, disperses into the colors, and
strikes the back of the drop; when it reaches the back,
part of it goes back into the air (not drawn here), and
part of it is reflected back through the water as shown;
then it reaches the surface again and part of it refracts
back into the air as shown, part of it reflects into the
water again (not shown). For this particular ray, and all
others from millions and millions of other raindrops for
which light enters at the same place on the surface,
the
emerging light constitutes the rainbow you see. But, note
that the angle between the direction of this incoming ray
and the exiting "rainbow rays" is 42°. But, what if I had chosen one of the other rays? I would get a different path for the ray bouncing around inside the drop and
the rainbow would emerge in a different direction. So, the whole sky would be covered in overlapping rainbows! And, it actually is. But not all of these infinite rainbows have the same brightness; it turns out that at 42° the emerging light is the brightest. You can actually derive this angle of maximum brightness if you know calculus and trigonometry; see the Wikepedia article on
Rainbow.
So, if you are standing in just the right place to see
the colors from this drop, then all other drops which sent their rainbow in your direction will also be seen but as coming from different points in the sky. The locus of those points will be a perfect circle. If you are in an airplane it is possible to see the full circle.
One last issue: it turns out that we often see a second,
dimmer rainbow above the bright one. The reason for this is that some of the incident rays will reflect twice instead of once inside the drop before they come out; these form the second rainbow. I hope this is not too complicated for a fiveyear old, but
hopefully you can digest it and explain it in terms she would understand.
QUESTION:
It says in a lot of places that wavelength is inversely proportional to frequency. Would it not be possible to increase the speed of the wave, (and therefore the frequency) so more waves pass through a point per second, without increasing the wavelength?
ANSWER:
In general, v=fλ and you can hold any one of the three constant and ask
what happens to the other two. Here is a simple example:
A guitar string of length L vibrating with its
fundamental frequency is a standing wave with λ=2L.
You can change the speed of the wave by changing the
tension in the string (which is what the tuning pegs do).
But the wavelength cannot change so the frequency must.
QUESTION:
Does the gravity of earth comes from molten core.and if so can we create artifital gravity by making molten core simillar to earth by rotating it at high speed.
ANSWER:
Gravity is caused by mass. The
earth's core constitutes approximately 1/3 of its total
mass and rotational speed has nothing to do with it.
Neither does the "moltenness" have anything to do with
gravity. So if you had a molten core alone, rapidly
rotating, its gravitational field would be just the same
as if you had a solid sphere of the same mass; there
would be nothing "artificial" about it.
QUESTION:
what force are acting on a ping pong ball at the top of its bounce is there just gravity or is the kinetic energy from the ball still at play.
ANSWER:
The first thing to note is that
kinetic energy is not a force. And I don't know if when
you say "top of its bounce" whether if is in purely
vertical motion or is in a trajectory during which it
will still have a speed at the top. If it is the former,
it is at rest and the only force on it is its weight
(gravity). If it is the latter there will also be an air
drag force opposite its direction of motion. (Air drag is
quite important for a ping pong ball.)
QUESTION:
My question is about the actual energy consumption that is required to produce a certain amount of electrical output:
Framing the setup for the question:
If I have an electrical generator that can produce 1kw of electrical energy and I spin the armature with NO load, the amount of work I am producing to spin the generator is fairly nominal. It's simply the frictional losses of the bearings, the amount of energy to accelerate the mass of the armature to a particular rpm (lets say 1000 rpm), and perhaps some wind resistance. This will have some numerical value, I am assuming in joules or horsepower. (please correct me if I am wrong).
Now, if I throw a switch that connects 10, 100 watt lightbulbs. The back EMF is tremendous. Anyone who has tried to spin a generator by hand can attest to how difficult it is to try and light just 1 100 watt light bulb.
Assuming the generator is 98% efficient; how much work (or energy) is required to produce the 1kw of electricity to light the bulbs?
I would like to compare the "NO LOAD" power requirement to spin the armature to the "FULL LOAD" requirement to spin the armature.
ANSWER:
I don't think there is much of a
mystery here. Let's forget about your first paragraph
because when you say 98% efficient, that presumably
includes all the losses you enumerate. This generater can
generate 1 kW of power at 1000 rpm. Now you are asking it
to give you 1 kW if you attach 10 100 W bulbs. You ask
how much energy you have to put in but that is not the
right thing to ask; you should ask how much energy per
second (Watts) you need to supply, power. Since the
efficiency is 98%, you need to put in 1 kW/0.98=1.02 kW
of power.
QUESTION:
Hello, i was curious why science hasn't utilized the physics of "slipping" into generating energy...
I understand there are road blocks, but is the potential of slipping to generate energy not extreme enough to warrant overcoming them?
Can you not generate a lot of energy with slipping? or is it just that in comparison to other forms of generating energy, it just doesn't compare and it's not as exponentially capable of generating energy as im assuming. Like when i imagine moving a large object at high speeds once you've propelled it to the speed you want it wouldn't the physics of slip, give it the continuous push it needs to maintain or even exceed the level you had put it too..and would not the physics of slipping then take over the propelling of the object...and if we could create or find..a suitable force that creates slip and resists wear and tear long enough to warrant using would that not enable us to drastically improve our capacity to generate energy??
ANSWER:
I presume you mean friction?
Friction generally takes energy away from a system. For
example, a box with kinetic energy sliding across a floor
stops because friction takes the energy away from it.
That energy which the box had does not really disappear
but shows up as heat (the floor and box will get a little
warmer) and sound (while sliding you can hear it). Are
you suggesting that we could use the heat and sound
energy somehow? Why not just use the energy the box
already had? Usually we try to eliminate friction as much
as we can to maximize other sources. One example is the
brakes of a convential car which transform all the
kinetic energy of the car into heat; electric and hybrid
cars use magnetic induction to put that kinetic energy
into the batteries rather than into heat in the drums or
disks of brakes.
QUESTION:
Hi, today I read an article on Armstrong Limit and I have a question about it, if saliva/tears boil in 36 degrees Celsius in the Armstrong Limit why is it a problem, does it cause any health issues?
ANSWER:
Your body is about 60% water. How
could you ask if it would be any health issues if all
that water started boiling?
QUESTION:
If you could dissassemble a human atom by atom, and then store the human in his or her dissassembled atom state, could you reassemble said human in the future without them aging essentially creating some form of atomical time travel? Also if it were achievable, would they retain their conciousness, memories and all that?
ANSWER:
The answer is no. To read a
discussion of the physics of the impossibility of the
"beam me up Scotty" problem, I suggest The Physics of
Star Trek by Lawrence Krauss, HarperCollins
Publishers, 1996.
QUESTION:
I am wondering the following;
1. Will the electron of atoms always apear when looked for in its cloud, or can it sometimes be "missed"?
2. Will it actually only be in two places at once or could it be where ever it is looked for simultaneously?
3. Is the "quality" or "strength" of an electron compromised when seen simultaneously in its different locations?
ANSWER:
The problem with your questions
is that you have the idea that you can even talk about
the electron being at any particular position. All you
can know, until you make a measurment, is the probablity
of the electron being in any particular arbitrarily small
volume. A measurement means that you look at a particular
small volume and if the probability of finding it there
is 1% you will only see it once every 100 times you look
there. It is never in two places at the same time, it is
really smeared over all places until you observe it and
then it is where you observe it to be. It can never be
observed at the same time in two places; otherwise there
would mysteriously be two electrons even if you know
there can only be one.
QUESTION:
I have a potentially unusual question to which I am currently unable to find an answer and I thought that you may be able to help. I was riding my motorcycle today and turned a corner and the front tyre came into contact with a large patch of gravel on the surface of the road. I had touched the front brake immediately prior to this and, as the front tyre hit the gravel the bike began to slide and to pitch sideways. I put both feet on the ground and steadied the bike, effectively holding it upright and avoiding a spill. Now for the question: My body is very sore after around 5 hours after the event and this has made me wonder what kind of energy did I absorb in my attempt to avoid the accident? The bike weighs 220kg, I weigh 125kg, I was travelling at 20mph (after I had released the brakes )on relatively flat ground and stopped the momentum of the bike within 2.5 yards (after I had released the brake). I am completely unable to fathom a calculation for this, I’m afraid . Can you help me at all?
ANSWER:
So, we have m=345 kg moving with a
speed v=20 mph=8.94 m/s and stopping in d=2.5 yd=2.3 m. The
initial kinetic energy was E_{1}=½mv^{2}=2.73x10^{4}
J and the initial linear momentum was p_{1}=mv=3.08x10^{3}
kg·m/s; the final energy and momentum are zero. Change in
energy is equal to work done, so W=E_{1}=Fd=2.3F
where F is the average force exerted by the
ground on you (frictional drag); so F=1.19x10^{4}
N. You could also say that the change in momentum equals
the Ft where t is the time to stop, so
t=1.19x10^{4}/3.08x10^{3}=3.86
s. The average acceleration over the stop was F/m=34.5
m/s^{2}, about 3.5 times the acceleration due to
gravity, 9.8 m/s^{2}. So why were you so sore?
Because Newton's third law says if the ground exerts a
force on you, then you exerted an equal but opposite
force on the ground. So you were exerting a force on the
order of 10,000 N, about 2000 pounds, for about 4
seconds. That sounds like an awful lot to me. You said
the brakes were not engated as you traveled 2.5 yards but
was the bike skidding sideways? That would have been like
having your brakes on and have lessened the amount of
force you needed to apply to stop the ride.
QUESTION:
My question is about Newton's 3rd law. I understand it pretty good, so I'm making myself questions trying to understand the limits. Here is my question.
Let's suppose I'm in a target range, shooting at the target. Obviously the bullet go through it. In the very precise instant that the bullet hit the card, how can that be explained by the 3rd law. Is there an opposite force, equal in magnitude, opposite in direction, even though the bullet go through the target?
ANSWER:
During the time the bullet and
target are in contact with each other, the bullet exerts
a force on the target; this force is evidently stronger
than the target can withstand and that is why it tears
and the bullet passes through. Now, when that is
happening, Newton's third law states that the target
exerts an equal but opposite force on the bullet. It is
not very hard to punch a hole through a sheet of paper,
so the force is not very big in this situation;
nevertheless, the bullet feels that force and as a result
emerges on the other side with a slightly smaller speed.
If you had 100 sheets of paper to make your target, the
bullet would be going with a considerably smaller speed
when it came out the back; with a thick or strong enough
target the bullet would just stop in the target.
QUESTION:
In this
video, starting at 2:00, a pair of 2liter bottles suspended over 2 loudspeakers are made to orbit about a fulcrum on which they are balanced when excited at their resonant frequency purely by sound waves. But such wave motion is back and forth, so how can it impart momentum in just the forward direction? I must know how this works!
ANSWER:
Yes, the pressure at any point in
the bottle, including the open end, will fluxuate with
the frequency of the air in the bottle. When the pressure
is low, air from the outside is sucked into the bottle;
when the pressure is high, the air is pushed out. But the
two air motions are not analogous. As shown in the
figures below, when air flows in it comes from many
directions so this gives the bottle only a small thrust
to the right. But when it flows out, the neck tends to
align the air motion causing a larger thrust to the left.
A cavity with a narrow, open neck at one end is called a
Helmholz resonator.
QUESTION:
Einstein's theory claims that a large object in space will create a gravitational field that will bend SpaceTime. My question is  if indeed a large body like the Earth does create a bend in SpaceTime do we exist in another aspect of time than that of the surrounding space which is empty and thus does not have a gravitational effect on that around it?
ANSWER:
The rate at which clocks run
depends on the intensity of the gravitational field. A
clock 100 m above you runs at a different rate as yours.
So I guess you do "exist in another aspect of time"
(whatever that means!) from any region where the
magnitude of the gravitational field is different than
yours. This effect is often referred to as the
gravitational red shift. It turns out that gps systems
must correct for this effects since timing between you
and satellites at high altitudes must be extraordinarily
accurate. I should note that the differences in time are
extremely small, even for something as large as the sun
which is incredibly massive.
QUESTION:
If at the start of the universe all the matter was within a cubic meter how didn't it create a black hole
ANSWER:
I state clearly on the site that
I do not normally do astronomy/astrophysics/cosmology, so
take with a grain of salt that I am no expert. I would
guess that there is so much energy in this cubic meter
that whatever forces might be acting are simply
inadequate to reverse the expansion. Also, I used the
word "other" because we do not know what the laws of
physics were in the very young universe.
QUESTION:
got a discusion with a friend, about the riding wind versus side wind.
when driving, the air thats in front of the care needs to move, and puts presure on the car. does that increase when you drive harder, and will a side wind have more or less effect when you drive harder?
ANSWER:
If you are driving into a
headwind there is a force back on you due to the speed of
the wind. You would therefore have to press harder on the
accelerator to keep going with the same speed you would
go in still air. But that does not have a handling the
car, merely holds you back causing you to use more
gasoline. For example, if your car has a speed of 60
km/hr into a 20 km/hr headwind, it would be the same as
driving 80 km/hr in still air. In a crosswind, however,
handling is more of an issue. There is a tendency to push
the car in the same direction as the wind is blowing; if
the road is wet or icy and the wind is strong enough, the
friction of the tires could be inadequate to keep you
from sliding across the road. Even if the road is dry,
the tendency for the car to turn with the wind and you
need to slightly steer into the wind. Also keep in mind
that a strong headwind will not necessarily keep blowing
exactly opposite your direction or else you may need to
do a maneuver where the wind now is partly blowing across
your path. So, always be extra careful on a windy day.
QUESTION:
I respect that Heisenberg's Uncertainty Principle can't be violated. . But a daily lab event seems to say "no you can". Here's the situation. Take the cathode ray tube in an oscilloscope. An electron is ejected from the cathode, deflected by coils around the tube's neck and then impacted at a precise screen location. The UP says I can never accurately know the electron's position and momentum at T. But I do accurately know P, it's relativistic mass is constant and so is it's velocity. In order to place that electron at the precise point on the tube face, the circuit must know its exact location to know when to increase voltages to the neck coils for proper deflection. Why is this scenario not a violation of UP?
ANSWER:
Well, how accurately can you
actually know P? Certainly not better than maybe 0.1%.
And position? Maybe to a micron or so. But, the P
you are talking about and the position you are talking
about are not the proper quantities to be talking about
in terms of the UP. Suppose we call the line between the
electron gun to the center of the screen the zdirection.
Then the UP is ΔzΔp_{z}≥ℏ
but the position on the screen would be perpendicular to
z, could be either x or y,and ΔxΔp_{z}
has no uncertainty principle associated with it.
QUESTION:
So I'm in high school and I'm in physics. I'm also a
musician (piano, singing, etc.). I was wondering about
pendulums, but specifically metronomes. If I wanted to
use a metronome as a model in an energy conservation
project, would I simply apply the same rules as normal
pendulum situations? I guess I'm just saying that I'm
having trouble with the difference between metronomes and
other pendulums. I tried to find the answer online, but
whenever a metronome is used in a physics example is
always is talking about simple harmonic motion or
synchronizing metronomes, which isn't really what I'm
asking. I'm not advanced enough yet to figure out the
physics behind a metronome by myself, either. Sorry about
the long paragraph for what's probably a pretty simple
question! I'm basically just asking if, when discussing
energy transfer/conservation, work, and force in physics
does a metronome abide by the same rules as a normal
pendulum, and if not, how does it differ?
ANSWER:
A metronome is certainly not a
simple pendulum which is basically a weightless stick of
length L frictionlessly pivoted at one end and
attached to a point mass m at the other end; the point
mass oscillates below the pivot. As you have probably
learned, this seemingly simple problem cannot be solved
analytically but can be approximately solved is the angle
is small. The result is that the frequency f (cycles per
second) is approximately f≈[√(g/L)]/(2π)
where g is the acceleration due to gravity.
Note that the mass does not matter, a bit of a surprise
perhaps.
Now the metronome is clearly not a simple pendulum
because the mass is above the pivot point. But, how can
that be? If you just stuck a mass on the end of a stick
and rotated the mass to a point above the pivot, would it
oscillate about the very top? Of course not, it would
still oscillate about the very bottom but with an
amplitude which was very big. So, how does the metronome
do this? Well, I have earlier worked
out how a pendulum works. You may want to have a look
at this but, being in high school you probably are not
ready for the math there. And maybe your physics class
has not even gotten to rotational motion yet so you would
not know about moment of inertia, angular acceleration,
etc. But you find out the nature of the pendulum in a
metronome: there is a bigger mass below and out of site.
So this is really a "double pendulum" and as long as the
hidden mass is bigger, it (the bigger) will oscillate
about the bottom while the smaller oscillates about the
top. The picture here of the metronome made of plexiglas
shows that bottom mass. You can also get the final answer if the "stick" has
negligible mass compared to the two masses, f=ω/(2π)=[√(g(ML_{M}mL_{m})/(ML_{M}^{2}+mL_{m}^{2})]/(2π)
where M (m) and L_{M}
(L_{m}) are the mass and distance from
the pivot of the lower (upper) weight.
Now to your question, whether energy conservation applies
to a metronome. No realworld pendulum as its mechanical
energy conserved because friction of some sort is always
present. However, if you put it in a box from which no
energy can come in or go out, the total energy will be
conserved, even if the pendulum stops, the air in the box
will heat up a little bit and if you were to measure all
the energy contained in thermal energy you would find all
the energy which the pendulum originally had. So the
answer to the "basically
just asking" is yes, the two pendula "abide" by the same
physics rules. Also note that there is a little
springdriven motor you can see inside; when you wind it
up you do work to give it potential energy and it then
gives this energy to the pendulum to replace energy lost
to friction.
QUESTION:
Let's consider crushing some item with scissors blade. It is easier to do it when the item is positioned closer to the pivot(assuming the force of squeezing the scissors is constant). I am asking to check the following explanation:
Is it so because to avoid being crashed the item must generate reaction force impacting the blades of such magnitude that the blades are not moving(theirs torques have to equal zero).Let's assume that force momentum coming from squeezing the scissors is constant(we don't change force of our fingers nor the distance from the pivot) and potential of generating reaction force by the item also. Then the only thing we can do to decrease this torque coming from reaction force(and therefore making the item less "resilient") is to shorten the distance between item and pivot.
My wonder is if it all comes down to considering movement of scissors blades around axis, meaning: blades are moving the item gets crushed, blades are still(torque equals 0) the item is in one piece.
ANSWER:
You can understand the principle
by just considering one half or the scissors; then the
other does just the same but with opposite forces. All
your talk about "reaction force" is wrong; look at my
diagram—the force labelled F_{1}
is the force your thumb exerts on the scissor and the
reaction force is the force which the scissor exerts on
your thumb which is the same magnitude but oppisite
direction. I will assume that there is some small object
you want to "crush" located somewhere along the blade.
This gets a little confusing, so read the next sentence
carefully. If you exert the force F_{1
}on the scissor and the scissor is not closing, the
sum of the torques exerted on the scissor must
equal zero; if the object is located near the pivot, like
where F_{2} is, the
object must exert a force opposite to F_{2}
but of the same magnitude since F_{2
}is the force the scissor exerts on the object which
is what you are interested in. (Ignore F_{3}
for now.) To calculate the magnitudes of the torques you
need to multiply the force times the moment arm (yellow
lines). The moment arm for F_{1} is much
longer than for F_{2} and therefore F_{2} should be much larger than
F_{1}.
Now, consider placing the object out near the end of the
blade where F_{3} is
drawn; its moment arm is larger than for F_{1}
so its magnitude is smaller. I hope it is now clear why
placing the object closer to the pivot results in a
larger force on it than farther away.
QUESTION:
I have a question about the weak nuclear force.
Why do we call weak force a force?
All what's said about the weak force is that it cause decays by changed one particle into another (or at a fundamental level, by changing the flavour of quarks).
But , "what makes it a force?"
Do we call it a force only because it involves W/Z bosons or there's something else.
ANSWER:
It is better named the weak
interaction. The concept of a force is generally not
useful in quantum physics, but, since we all have a gut
feeling what a force is in macroscopic classical physics,
it is natural to think of interactions between objects as
resulting in forces. But, to treat interactions between
objects quantum mechanically we use the concept of fields
to describe those interactions; fields are also used in
classical physics as well—think electric, magnetic,
and gravitational fields. At the quantum level fields
alone are useful, not forces. And when you have a field
it can be quantized and the quanta are often viewed as
the "messengers" of the field; the field quanta are
photons for the electromagnetic field, gluons for the
strong interaction, and W^{±} and Z bosons
(as you note) for the weak interaction.
It is interesting that the fourth "force" in nature is
problematical: no one has successfully quantized the
gravitational field and achieving that is one of the holy
grails of physics.
Also interesting is that the notion of force is not
useful in the theory of special relativity because the
usefulness of force depends on the usefulness of
acceleration. At high speeds where special relativity is
important, different observers will see different
accelerations for the same object; if we think of force
in terms of Newton's second law (F=ma),
different observers would deduce different forces. If,
instead, we thought of Newton's second law as force
equals time rate of change of linear momentum, force
could be invisioned but only if linear momentum (mass
times velocity) were differently defined. But I
digress...!
QUESTION:
I am a 180 lb man competing 180 lb weight class five time world champion powerlift. I bench 500 lb and I would like to know how much force am I executing to lift 500 lbs?
ANSWER:
To hold 500 lb at rest or to move
500 lb straight up with constant speed requires a force
of 500 lb. Since
the weight you lift is initially at rest and momentarily
at rest when your arms are fully extended, you must exert a
force larger than 500 lb to get it started
moving upward and smaller than 500 lb for it to
stop at the top. I have watched videos of the bench press and
it appears that during most of the time the bar is moving
with a constant speed; so most of the time the force you
exert is about 500 lb. To do a rough calculation of the force
necessary to get it moving, I will use a total lift distance of
about 1 m in a time of about 1 s so the speed is roughly
v=1 m/s; 500 lb has a mass of about m=227
kg. (I will use SI units, like scientists prefer, and
convert back to pounds in the end.) Suppose that the time it
takes to accelerate the bar to 1 m/s is t=0.1 s;
so the acceleration is a=(1 m/s)/(0.1 s)=10 m/s^{2}.
So the force F you must exert is calculated
using Newton's second law, ma=(FW)
where W=mg is the weight of the bar and g≈10
m/s^{2} is the acceleration due to gravity; F=227(10+10)=4540
N=1000 lb, twice the weight you are lifting.
Similarly, if it takes 0.1 s to stop the bar, the force
you would apply would be F=227(10+10)=0, the
weight will stop itself because of gravity. Keep in mind
that these are estimates using reasonable numbers; the
actual numbers depend on the way the individual does the
lift.
MY ANSWER IS WRONG, OR AT LEAST INCOMPLETE. SEE
BELOW
QUESTION:
Recently when cleaning, I opened the cover of our ionisation smoke detector, as I was opening the smoke detector cover, dust fell from the inside of the smoke detector into my eye. The whole inside of the cover was covered in this dusta brown colour dust, not grey colour. I was concerned that this dust has been given off by the Americium due to the slats on the ionisation chamber which allow particles to escape from the radiation. Could this be so? I was concerned as it said online that it was dangerous to ingest or inhale the americium from smoke detectors and this dust entered my eyes?
ANSWER:
I have recently answered a
question about smoke detectors; you should read
that. The americium is in a
sealed container and cannot get out. Also, the color of
dust depends on where it comes from. There is probably
some source of brown dust in your house and you wouldn't
notice its color except when something like your smoke
detector accumulates a relatively large amount over a
long time.
CORRECTION:
I received an
email with additional
information about which I was not aware. Essentially it
said that the recoiling americium nuclei could collide
with and eventually damage the casing around the source.
Since your smoke detector is 27 years old and the rule of
thumb is that the useful lifetime of americium smoke
detectors is about 10 years, you should dispose of this
one and get a new one. (Use the recommendations of the
answer you included in
your question to dispose of it safely.) I still believe
that you need not obsess over the small exposure you
received when you were cleaning, given the very small
numbers I quoted in the earlier answer about smoke
detectors; the unit quoted there,
rem (Roentgen equivalent man), is specifically meant
to indicate exposure to ionizing radiation of human
tissue and is therefore more useful than Curies used by
the author of the
answer which just measures the amount of radiation
emitted without consideration of their health effects.
QUESTION:
A piece of cork is thrown vertically downwards from a sky scraper 300 meters high.
Its initial velocity is 2 m/s. Air resistance produces a uniform acceleration of 4 m/s^{2} until the cork reaches a terminal velocity of 5 m/s.
Why does the cork reach terminal velocity?
ANSWER:
There is something terribly wrong
about this question. I see only two ways I can interpret
this question.

I can interpret
everything completely literally as written. Of
course, this cannot be physical because the air
resistance F_{A} must depend on the speed
v (usually F_{A}∝v^{2})
of the cork and it will certainly not, in the real
world, result in an acceleration which is constant.
If the mass of the cork is m, then F_{A}=4m;
there is also the force of gravity F_{G}=mg
where g=9.8 m/s^{2}. So the net
force is F_{net}=ma_{net}=m(49.8)=5.8m,
so a_{net}=5.8 m/s^{2}.
So the cork will fall with a downward acceleration of
magnitude 5.8 m/s^{2} and never have a
terminal velocity since it just keeps speeding up.

Maybe you didn't
really mean to say it has a "uniform acceleration"
but that F_{A}=kv^{2}
at the instant you release it with speed down of 2
m/s and the acceleration due to that force is 4 m/s^{2};
so k(2)^{2}=4m or k=m.
Now we have F_{net}=ma_{net}=(mv^{2}mg);
when v=√g=3.13 m/s, a_{net}=0,
so if this is the interpretation of the problem, the
terminal must be 3.13 m/s, not 5 m/s.
I cannot think of
any interpretation of this problem which would be
selfconsistent.
QUESTION:
How much energy in joules would it take to propel an object with a mass of 14 grams to a speed of 1.25 kilometers per second over the course of 0.15 seconds? Further, how much power would it take to create this amount of energy?
I know this sounds like a homework question, but it isn't. this is a question from someone who barely understands any of these physics terms and who has hardly any idea how to calculate the simplest of formulas, yet is trying to get a concrete idea of what it would physically require to complete this hypothetical task.
ANSWER:
I think it is a homework question
and you are trying to pull the wool over my eyes; nobody
just wonders how much energy a 15 gm object going 1.25
km/s has. I will
outline what you need to solve, but just this once and
not in complete detail.
The kinetic energy E of an object with mass
m and speed
v
is E=½mv^{2}. If you
want the energy to be expressed in Joules, m
must be in kilograms and v must be in meters per
second. If the object takes a time of t to
achieve an energy E, the average power expended
to do so is P=E/t. If you want the
power to be in Watts, E must be expressesed in
Joules and t must be expressed in seconds.
QUESTION:
If a full cart is pushed at the same time as a empty cart with the same force witch one will stop in motion first? And why?
ANSWER:
I assume that the carts are
identical except for the loads; and that the force "does
the same thing" to both of them. The heavy cart has mass
M, the light cart m. There are two ways you can
apply the force F:

Push over the
same distance d for each. In that case you
give the same kinetic energy to each, ½MV^{2}=½mv^{2}, so
V=v√(m/M) where V(v)is the starting speed of
M(m).
Note that, as you would expect, v>V.

Push for the same time for each. In that case you
give the same linear momentum to each, MV=mv,
so V=v(m/M).
Again, v>M but by a different factor.
Now, what stops the
carts? Friction f which is proportional to the
weight of each car. For the loaded cart, f_{M}=μMg
and for the empty cart, f_{m}=μmg
; here the negative sign indicates that the force is
slowing the cart down, μ is the coefficient of kinetic friction, and
g is the acceleration due to gravity. Because of Newton's second law,
the accelerations of the two cars are A=f_{M}/M=μg
and a=f_{m}/m=μg.
Because the accelerations of the two cars are the same,
the one which started fasted (empty) will go farther and
take a longer time to stop.
QUESTION:
My question is hypothetically ( or exactly ) does or would an "alien" be able to negotiate a 90 degree turn and not be splattered on the inside of the vessel.??
ANSWER:
That would depend on two things.
First how long does it take the vessel to turn which
would determine the average acceleration? Second, what is
the physiology of the alien, in particular how much
acceleration can her body endure?
QUESTION:
Can you please explain why Magnetic force is NonCentral when the Electromagnetic forces are Central forces?
On the same note, Why is the friction non conservative when the EM forces are considered conservative in nature?
ANSWER:
You have some misstatements here.
For "central" I think you mean conservative because, for
example, the electric field for a uniformly charged wire
does not all come from a single point (which is the
definition of central fields). Also, when you say
"electromagnetic" I think you mean electric. Now, the
force due to a static electric field is conservative, but
not all electric fields are conservative. For example,
the electric fields induced by changing magnetic fields
are not conservative. You ask why the magnetic force is
like it is—because that is the way nature is. Be
aware that electric and magnetic fields are
manifestations of one field, the electromagnetic field,
and not separate fields but intimately linked. Regarding
why friction, admitedly due to electromagnetic forces, is
not conservative, I refer to the first part of this
answer where I emphasized that there is no reason to
assume that electric or magnetic forces are conservative.
QUESTION:
If black is the absence of light, and thus of color, how is it one can mix primary colors together to get black paint?
ANSWER:
Because mixing paint is not the
same thing as mixing light. It makes sense
simplistically: red paint absorbs everything but red,
blue paint absorbs everything but blue, so red+blue paint
absorbs everything—black. Like I said, this is
simplistic because they are not purely red nor purely
blue, but it gives you an idea of why mixing paints would
look black.
QUESTION:
Can something rotate if it is not symmetrical?
ANSWER:
Yes. It is easiest to visualize
an object of any shape you like which is in empty space
at rest. You now give it a kick somewhere on its surface.
Unless the direction of the force is directed directly at
the center of mass, the kick will cause the object to
move away from you and be also rotating as it moves. The
rotation will be about an axis which passes through the
center of mass. You could make the object rotate about
any axis you wanted but you would have to hang on to the
"axle".
QUESTION:
Do we know how a nucleus of an atom is structured with its protons and neutrons? Is it merely a discombobulated mess, or is there actual structure to it? Are they in movement within that space between itself and the electrons (spinning, rotating, doing a disco dance, etc)?
This question has been on my mind for a while and I cannot find any resources on it, and graphical 'representations' just show a mass of random protons/neutrons.
ANSWER:
I will note at the outset that
this question is in violation of site ground rules: "…single,
concise, wellfocused questions…" Nuclear
structure is an entire subsection of physics and it would
take a whole book to give you even an overview. I spent
my entire research career, more than 40 years, studying
nuclear structure. The answer is that it is not "merely a
discombobulated mess" but pretty well understood. I will
give you a few examples.

The force
between nucleons (neutrons and protons) is very
strong and results in the nucleus being extremely
small compared to the the atom (nucleons and
electrons). The size on an atom is on the order of
Angstroms (10^{10 }m) where the nucleus is
on the order of femtometers (10^{15} m). If
the atom were about the size of a football field, the
nucleus would be about the size of a golf ball.

One of the first
successful models of the nucleus was the shell model.
Because of the average force due to all the other
nucleons, each nucleon moves in orbitals like
electrons do in atoms under the influence of the
Coulomb (electric) force.

A later, also
successful, model is the liquiddrop model where the
particles all move collectively. Imagine a liquid
drop which, when bumped will oscillate in some way.
Or imagine a nucleus which is deformed like an
American football. It could rotate about its center
of mass. Many heavier nuclei have rotational band
structures. In these cases the nucleons all move in
choreographed unison, maybe more like a line dance
than a disco dance.
Hope this gives you
the idea that structure of nuclei is fairly well
understood. The examples I give are over simplified
because, among other things, when inside a nucleus,
nucleons lose their individuality.
QUESTION:
I was recently doing an OCD Exposure homework and now I am unsure of what I did. I opened the smoke alarm cover and touched around all the inside parts of it. I touched all the sides of the outside of the ionisation chamber with the radiation symbol on it for around 56 minutes. I did not open the ionisation chamber. I want to find out
1. Did this expose me to a lot of radiation?
2. Will this increase my risk of taking cancer in the future.
Can you please tell me honestly, even if it is not what I want to hear, I just need the facts.
ANSWER:
The radiation from smoke
detectors is trivially small. Even if you had removed the
source and kept it around it would have given you
negligible radiation exposure. A study by the Nuclear
Regulation Commission showed that "…a teacher who removed
the source from a smoke detector could receive a dose of
0.009 millirems per year from storing it in the
classroom. The teacher would get another 0.001 mrem from
handling it for 10 hours each year for classroom
demonstrations, and 600 mrem if he or she were to swallow
it…" To put this in perspective, you receive
approximately 2.2 millirems per year from natural sources (from space above, ground below)
if you are at sea level, even more if you are above sea
level. Don't swallow it, though!
QUESTION:
I have a question about the operation in drift chambers. Charged particles enter the chamber medium to ionize the gas atoms. The electric field is applied to drift resulting electrons and ions. Also, there is a magnetic field applied to measure the particle momentum. However, all the papers use the Lorentz force on the charged particle to calculate its momentum in this form F= qvb and don't include the force due to the electric field.
In other words, why the electric field doesn't have an effect on charged particles entering the drift chamber?
ANSWER:
The figure above shows a
schematic sketch of one layer of a wire chamber. The
particle ionizes atoms near wires it passes close to. The
wires carry a charge which creates an electric field near
them which then collects charged particles and sends a
pulse to a computer. The electric field is fairly strong
near the wires but quickly becomes nearly zero in a very
short distance distance from the wires. Shown in the
second figure are equipotential lines due to the voltage
on the wires; note that at distances about equal to the
wire spacing the equipotentials are constantly spaced
indicating nearly zero fields. A second consideration is
that the particle being detected has an extremely large
energy which makes electric fields of the magnitude near
the wires practically invisible to the detected particle.
QUESTION:
We know that our universe has fine tuned gravity constant so my question is can a universe born with a different gravity constants other than fine tuning or there is no possibility other than fine tuning constant
QUERY:
This is a very deep question, and
you greatly underestimate the situation. In fact, the
properties of a universe are extremely sensitive to the
values of many fundamental constants, not just G.
So you cannot just think of how things would change if
G were changed. Usually the question is not
about whether a universe could be "born" if the
fundamental constants were different (partly because we
really do not know the mechanism of the big bang);
rather, the question is usually "how much could we change
the fundamental constants and still have a universe where
life is possible?" Here is an interesting example I read
about of the consequences of changing fundamental
constants: If the strong nuclear force were increased by
2%, the diproton (two protons bound together) would be
possible. The consequence would be that stars could not
exist because the protonproton cycle which produces the energy in
stars would not occur if two protons could just stick
together.
QUESTION:
We know from Maxwell equations that c = 1/(e0m0)1/2 which indicates that
electromagnetic fields propagate in vacuum with the speed of light. Thus,
light is an electromagnetic wave as proved later by Hertz. Consequently, Can
induced electromagnetic fields change the energy carried by light waves if
they were exposed in a certain way to them?
QUERY:
It is not clear what you are asking. "...in a certain way..."?
REPLY:
Would light interact with electromagnetic fields or not?
ANSWER:
Yes, light will interact with any
electric or magnetic field. First, there is the
superposition principle; at any time and place the net
electric or magnitic field is the sum of all fields from
all sources. Here is another example how "light" can
interact with an electric field: a photon with sufficient
energy, if it passes close to a nucleus where the
electric field is particularly intense, may convert into
an electron/positron pair.
QUESTION:
I ride motorcycles and do have BS degree in Biology and am familiar with simple physics.
There is a new airbag technology that has been recently released to consumers. This technology was initially developed for motorcycle racers close to 15 years ago.
The early tech used a cord attached to bike and rider, when the rider and bike were separated the airbag deployed.
Due to advancements in electronics new system use a small computer with appropriate algorithms sensors to determine when a crash is happening and activate the airbag.
This technology is now available to consumers like myself at reasonable cost $7001,100 + for a system.
The European Union has developed standards for impact protection pads for motorcyclist's. EN 16212:2014 defines the exact spinal pad impact reduction capabilities, standards and testing requirement's.
I have found it difficult to find technical data on the new airbag system with regard to exact impact reduction capabilities.
What I have found is that rider report being in a crash with the airbag system and it registered a IMPACT force of 18 Gforces. The rider was not injured.
Assuming that the rider suffered a TRANSIMTED force of the no more than 12kN (as per the EN 16212:2014).
How much energy in Kn did the airbag system absorb ?
What is the energy absorption capability difference between the airbag and a EN 16212:2014 pad with 12 kN transmitted energy ?
Problems
1) I have been unable to determine the EN 16212:2014 IMPACT test force, I only know the TRANSMITTED test force is no greater than 12kN
2) Rider crash data uses Gforce to measure IMPACT force, and provided no transmitted force data, due to there being no injury, I feel it safe to use a transmitted force of 12 kN .
3) Some forces are given in kN and some Gforces, I do not understand the difference of these force measurement systems and I do not know how to correctly convert from one measurement system to another to solve my problem
ANSWER:
I am sorry, but your question violates one of the rules of the cite, "...single, concise, wellfocused questions..."
I can, however, help you with the single question of confusion between kN and
gforce. The gforce is technically not a force at all, it is an acceleration. As you likely know, the acceleration due to gravity is
g=9.8 m/s^{2}. The gforce is usually expressed in
gs; for example, because of some force you experience an acceleration of 10
gs, so that means that if your mass is 100 kg, the force you experience is
F=ma=100x10x9.8=9800 N=9.8 kN.
QUESTION:
I was wondering at sea level, given variable air densities. What is the rate at which a human would slow in relation to the earths rotation if suspended in the air for an extended length of time. If you were able to nullify the air density acting on your body could you theoretically travel the globe at 1000 mph opposite the earths rotation. Am I forgetting variables other than loss of contact with the earth and air density? I know that inertia would keep you moving with the earth however you would begin to shed that inertia quickly correct?
ANSWER:
Any way you could be "…suspended in the air…" would rely on
something attached to the earth. Even a
hovering helicopter is being held up by the
atmosphere which rotates with the earth. No, you cannot
travel around the earth by staying still.
QUESTION:
Why does a volume of any randommotion particles begin spinning when it is compressed, either by outside force, or by gravity? Why they heck do things go into a spin? Is this some form of conservation of momentum?
ANSWER:
The
things do not "go into a spin", they were already
spinning but more slowly, perhaps too slow to notice. It
is, indeed, because of a conservation principle, the
conservation of angular momentum. The principle states
that if there are no external torques acting on a system
of particles, the angular momentum will remain constant.
Let us take a very simple case, a single point mass
moving with speed v in a circle of radius R. It is
attached to string which passes through a hole in the
table; you are holding it in its orbit by pulling down
with just the right force F (which happens to be
F=mv^{2}/R,
but we don't need to know that now). Its angular momentum
is equal to L_{1}=mvR. Now you
pull down with a bigger force so that the new radius is
R/2; as you did this, no forces on m exerted any
torque so the angular momentum was conserved. L_{2}=mv'R/2=L_{1}=mvR,
so v'=2v., spinning faster than before
your force pulled it in. But, you argue, if there are
many particles with random velocities there is, surely,
for each particle a second particle which is rotating in
the opposite direction so the net rotation is zero. But,
this argument will only work exactly if there are an
infinite number of particles. In a very large assembly of
particles even a very small difference between a pair of
particles can cause a large angular momentum if they are
far away from the center of mass of the assembly.
QUESTION:
What is the terminal velocity of a 5 inch hail stone, and how many psi would be necessary to launch that roughly 1.1 pound ball of ice down a barrel style tube? I'm looking to do some stress testing on some roofing material simulating the most severe weather.
ANSWER:
I have a good way to estimate the
air drag force at atmospheric pressure: F=¼Av^{2}
where F is the force in Newtons, A is
the cross sectional area in m^{2} of the
object, and v is the velocity in m/s; you must
work in SI units because the factor ¼ contains
constants like drag coefficient, density of air, etc.
Now, F=ma=¼mAv^{2} where
a is the acceleration. The hailstone has a
downward force on it, its weight mg where g
=9.8 m/s^{2} is the
acceleration due to gravity; when v gets big
enough that the drag force (upwards) equals the the
weight down, the acceleration becomes zero. This leads to
the terminal velocity v=2√(mg/A).
I calculated the mass to be about 0.74 kg=1.6 lb and the
area about 0.0167 m^{2}, so v=48 m/s=107
mph.
Regarding the cannon you want to make, you cannot
just know the psi but the distance over which the force
from that pressure will act. If you want to push it with
a force F over a distance d with a pressure
P, you will
(ignoring any friction) give it a speed v=√(2Fd/m)=√[2PAd/m]
where m is the mass of the projectile and A
is the cross sectional area of the tube. So, you can
either push it with a big pressure over a short distance
or a small pressure over a long distance. Let's use the
number I used above, m=0.74 kg, A=0.0167 m^{2},
and use a d=2 m long tube; then v=48 m/s, 48=√(2xPx2x0.0167/0.76).
Solving, I find P=2.83x10^{5} N/m^{2}=41
psi. Don't forget that there is atmospheric pressure,
14.8 psi, on the front of the ball so a rough estimate
would be 56 psi. The actual required pressure would be
quite a bit larger than this because friction (including
air drag) would not be negligible.
QUESTION:
When actors are shot off of roof tops, why do they fall forward when the bullet should propel them backwards?
ANSWER:
I have already
discussed this question in some detail. The bottom
line is that when a bullet hits a man the recoil is
negligibly small.
QUESTION:
i have come up empty in trying to figure out a certain issue ..
I am doing a project with air cylinders and need a calculation based on how much compressed air would be in a cylinder thats 2 x 6 inches and yes i searched google and YouTube and there is mention on these but it appears different folks have different formulas and different conclusions... i am startled because of the lack of information on this, i wouldn't assume this would be THAT elusive.. most answers I got was  P1 x v1 = p2 x v2 formula and that seems something i will learn at one point in time after i learn the basics but here is what i got so far if i may
Pressure times volume divided by atmospheric or : P x V / 14.7 .. seem ok at first when i calculator a pressure of 40 psi times the volume of my cylinder being 18.85 Cu inch then divided by 14.7 atm
which comes to 51.3 Cu inch...but if i change 40 psi to 10 psi things get hairy... 10 (psi) x 18.85 (volume) / 14.7 (atm) = 12.82 Cu inch.. but this is less air than regular atmospheric pressure .. so the formula must be faulty
ANSWER:
The relation P_{1}V_{1}=P_{2}V_{2}
(called Boyle's law) which you state is correct provided
that the temperature and the amount of gas inside the
cylinder do not change. But then you do not apply it
correctly. I take it that you are not adding
air to the cylinder but are compressing the air in the
cylinder. I am also assuming that the cylinder has a
height of 6 in and a radius of 1 in and that so the volume is
V_{1}=6x3.14x1^{2}=18.8 in^{3}.
I also assume that when the cylinder is at 18.8 in^{3}
there is 1 atmosphere of pressure, P_{1}=1 atm=14.7 psi, so
P_{1}V_{1}=276
in·lb. We now have that V_{2}=276/P_{2};
if P_{2}=40 psi, V_{2}=6.90
in^{3}; if P_{2}=10 psi, V_{2}=27.6
in^{3}. Notice that for the 10 psi case the
volume is bigger than the the original volume because the
pressure is smaller than atmospheric pressure. If you
can't make the volume any smaller than 18.8 in^{3},
the only way to reduce the pressure is to remove some gas
or cool it. If you are interested, the most general
expression for an ideal gas is PV/(NT)=constant
where T is the absolute temperature and N
is some measure of how much gas you have.
ADDED THOUGHT:
Rereading your question I
am thinking that it is not a cylinder with a piston but
maybe of constant volume 18.8 in^{3}. In that
case, keeping V and T constant, the appropriate relation
would be P_{1}/N_{1}=P_{2}/N_{2}
or N_{2}=P_{2}N_{1}/P_{1}.
Suppose we call the amount of gas you can put in the
cylinder at atmospheric pressure 1 gas unit. Then if you
fill the tank with 10 times atmospheric pressure, you
will store 10 gas units.
QUESTION:
Imagine a 10
m^{2} plate of aluminum (thermal conductivity 225.94 W/mK)that is
0.1 m thick. In the center of the aluminum plate protrudes a long skinny aluminum bar that is 20
m high, 0.1 m wide, 0.1 m long. This would look like a dirt tamper. The temperature of the large plate is at 100°C and the top of the long skinny pole is 10°C. How can I calculate heat flow in this system?
I certainly appreciate your help.
QUERY:
Is the plate maintained at 100 and the top of the bar maintained at 10 and you want the rate of heat flowing through the bar? Or do you want to know the final temperature is allowed to come to equilibrium insulated from the environment? And if you actually want an analytic expression for any point in the system as it is coming to equilibrium, it would require that I have the shape of the plate; this problem probably cannot solved analytically and would require a numerical solution which I am not able to do.
REPLY:
Plate is maintained at 100; top heats up but I understand that is hard to model because the rate would decrease as the top increases in temperature (eventually becoming zero when the top reaches 100). I guess I'm really interested in the heat flow if the top is at 10 (so we could imagine it is held at 10)
ANSWER:
I am
going to assume that there is no heat leakage from the
sides of the bar. Initially I will assume that the top
end of the bar is kept at a constant 10°C, so heat flows
through the bar at a constant rate. We
can figure out the heat rate as a function of the
temperature difference, ΔT=100T.
I will neglect any edge or geometry effects in the bar,
in other words I will treat it as a onedimensional
problem where the heat flow vector in the bar is in the
direction of the bar and uniformly distributed across the
crosssectional area. In that case the equation for the
rate R is
R=ΔQ/Δt=kAΔT/L
where A=0.01 m^{2} is the cross sectional area,
k=226
W/(m·K) is the thermal conductivity, and L=20 m
is the length. So R=0.113ΔT W.
For ΔT=90, R=10.2 W. Now,
if the top remains at 10°C, that means that energy at
the top is being taken away at the rate of 10.2 W; since ΔT∝L,
the temperature must increase linearly along the bar
when equilibrium has been achieved.
TIME DEPENDENCE:
The questioner indicated
interest in the more difficult problem of no heat
flowing out the top end of the bar which started out
10°C. What I did was to just give the steadystate
solution like you would learn in any elementary physics
course. So I dug into the transient case; I learned a
lot! Below I find the general solution to the heatflow
problem and apply it to the case in question, the bar
starting out at 10° and ending up at 100°. I
have left out a lot of details but have provided links
to derivation of the 1d heat equation and its general
solution. Algebraic steps I have omitted in applying
boundary conditions could be filled in by anybody
interested in these details.
The onedimensional heat equation is
∂T/∂t=c^{2}∂^{2}T/∂x^{2}
where c^{2}=k/(C_{p}ρ)
where k is thermal conductivity, C_{p} is
specific heat
at constant pressure, and ρ is the mass
density. (You may find a derivation
here.
) The general solution of this equation is
T(x,t)=exp(c^{2}s^{2}t)[Asin(sx)+Bcos(sx)]+Cx+D
where A, B,
C,
D, and s are constants to be determined for
the specific problem. (For a derivation, go
here. Go to Section 2.2)
Suppose the boundary conditions are that
(3) the temperature of the bar is at some constant value T_{1},
(1) one of the bar is held at a temperature T_{2},
and that (2) the other end of the bar (and the sides) are
insulated:

T(0,t)=T_{0}

∂T(x,t)/∂x_{x=}_{L}=0

T(x,0)=T_{1}
These boundary conditions lead to the
following:

Bexp(c^{2}s^{2}t)+D=T_{0
}B=0, D=T_{0}

exp(c^{2}s^{2}t)[sAcos(sL)]+C=0
C=s(exp(c^{2}s^{2}t)[Acos(sL)])
s≠0 so C=0
A_{n}cos(s_{n}L)=0⇒s_{n}=½nπ/L,
n odd

∫(T_{0}T_{1})sin(s_{m}x)dx=∫Σ{A_{n}sin(s_{n}x)sin(s_{m}x)}dx=(L/2)A_{n}δ_{mn}
A_{n}=(2/L)(T_{0}T_{1})∫sin(½nx/L)dx
=8(T_{0}T_{1})sin^{2}(nπ/4)/(nπ)
So, finally,
T(x,t)=T_{0}Σ{exp(c^{2}s_{n}^{2}t)A_{n}sin(s_{n}x)}
c^{2}=k/(C_{p}ρ),
A_{n}=8(T_{0}T_{1})sin^{2}(nπ/4)/(nπ)=4(T_{0}T_{1})/(nπ),
s_{n}=½nπ/L,
n odd
We should tabulate the constants to be used:
T(x,t)=100Σ{(115/n)exp(0.00206n^{2}t)sin(nπx/40)}
(t is in hours here.)
The plot of this function including just
the first three terms of the series (1,3,5) is shown
in the first figure. Note that because the bar is so long it
takes the system several hundred hours to fully equilibrate. The
second figure shows the calculation for the first hour; this is
clearly wrong since the whole bar is at 10° at the
beginning. The reason for this is that so few terms have been
included in the infinite series. However, to understand the
longterm behavior, n>1 plays almost no role because the n^{2}
behavior in the exponential damps out higher n contributions,
particularly at large t.
One more calculation, for the
initial temperature profile linearly decreasing over the
length of the bar may be seen
here.
QUESTION:
I'm writing a novel and want to make sure I'm describing a scene correctly. A spaceship built in a "tower" formation (rocket at the bottom, levels stacked on top of each other to the bridge at the top/front) is forced to land on Earth in an emergency. It initially enters at a steep angle nose first, causing rapid deceleration in the atmosphere, and then does a flip and burn, the engines faced to the ground and firing to slow them down. My question is around the gravitational forces an occupant would endure, sitting in a crashcouch/seat that cushions them and rotates to ensure they're always being pressed into it. Which way would these forces push/pull on them from start to finish, particularly around the flip and firing of the engine? I can't seem to wrap my head around it.
ANSWER:
Here is the trick you need to understand: Newton's laws,
by which we usually do classical physics problems like
what you are describing, are not valid in systems which
are accelerating. When your astronaut is in empty space
and turns on her engines, she thinks she feels a force
pushing her into her chair; but she cannot understand
this because she is at rest in her frame and Newton's
first law says that if she is at rest all the forces on
her should be zero. If we look at it from the outside we
see the chair pushing her, that being the force which is
accelerating her in accordance with Newton's second law.
However, she can do Newtonian mechanics in her frame if
she invents a force which is equal in magnitude to her
mass times the acceleration of the system (rocket) but in
the opposite direction as the acceleration; this is the
"force" pushing her into the chair and in physics we call
it a fictitious force. You have probably heard of a
centrifugal force,
the force which tries to throw you from a merrygoround;
that is a fictitious force because rotation is a kind of
acceleration.
In the following examples, the green dotted line vector
indicates the orientation of the chair desired for most
comfort for the astronaut. Note that in each instance
this vector points exactly opposite the net acceleration
vector.
First the initial reentry, entering the atmosphere at a
steep angle. The rocket has some velocity v.
When it encounters the atmosphere the result is a
retarding force opposite to the the direction of the
velocity which causes an acceleration a of the rocket
(and astronaut). There will also be an acceleration
g due to the gravitational force
(weight); if there were no air, this would be the only
acceleration. The net acceleration is the sum of the two
and is labeled a_{net}.
The chair will orient with the green aligned with the net
acceleration as shown.
Next
we have the situation where the rocket is rotating to be
tail down for landing. It is still falling with some
velocity v so there is still an upward acceleration due
to the air drag, labeled a_{v}
here. There is still the acceleration due to gravity
g. But now there is also a rotation
about the center of mass of the rocket so there is a
centripetal acceleration a_{c} experienced by the astronaut
which points toward the axis of rotation. Add all three
to
get the net acceleration a_{net}.
Now the chair will be aligned as shown.
Finally, the landing position which is the easiest. There
is still the gravitational acceleration g.
There is now an upward acceleration due to both the
engines and the air drag but the engines will be the main
contributer as the speed decreases. The net acceleration
now points straight up and the chair orients as shown.
Keep in mind that I have not made any attempt to have any
particular relative values of the various accelerations
but they are not unreasonable. They will vary
considerably depending on the specific conditions at
any particular time.
QUESTION:
So from the movie space cowboys. Two pilots ejected from a plane and deployed their parachutes at different times. Assume they deployed their parachute at terminal velocity. They started at 100,000 feet. Assume same size parachute. If pilot A weighs 230lbs and deployed their chute 5 seconds later than pilot B who weighs 250lbs. How possible is it that pilot B gets to the ground first?
ANSWER:
Well, this is kinda tricky for a
couple of reasons. First, "terminal velocity" is not some
constant number; it depends on geometry of the object
(which you try to keep the same by having "same size
parachute"), the density of the air, and the mass of the
object. So, falling from 100,000 ft they will experience
very significant change in the density of the air. The
second reason is that, because of the first reason, your
phrase "deployed…at terminal velocity" doesn't
really tell me much. Also, terminal velocity is not
attained at a particular time so you would have to
specify something like 'at 99% of terminal velocity'.
All that aside, let's do the calculation assuming that
the density of the air is independent of altitude,
the same as at sea level. The air drag is proportional to
v^{2} where v is the speed and
I will call the proportionality constant k; k is where both the density of the air and the
geometry of the falling object are 'hiding'. Then
Newton's second law may be written as ma=mgkv^{2
}where
m is the mass, g is the acceleration of gravity,
and a is the acceleration. Terminal velocity v_{t
}is when a=0
or v_{t}=√(mg/k).
Since we have stipulated that k is the same for
both A and B and is altitude independent, only their
masses would affect v_{t}. If they both
jumped and never opened their parachutes, B would clearly
be the winner (loser?!) since his terminal velocity is
larger than A's as would be his speed at all times.
Similarly, if each deployed his parachute immediately, B
would get to the ground first. And, certainly, if A
deploys first he will reach the ground after B since he
was already behind and will end up falling more slowly
than B. So, finally we come to your scenario where A
deploys after B. What will happen depends on the
altitudes where each deploy and how far ahead B is when A
deploys. If A does not pass B during the 5 second free
fall, B will get to ground first. If A does pass B he
will end up below B but going more slowly; if they fall
for a very long time B will pass A after some time. If
that time is less than the fall time left for A to hit
the ground, A will win.
Now, falling from 100,000 ft will be quite different
because terminal velocities at very high altitude will be
much greater than at atmospheric pressures. That means
that the distanced between the two will be much greater
when they have both deployed. If they wait until they are
fairly close to the surface, it seems to me quite likely
B will hit first.
QUESTION:
So, when earth moves faster around its axis (=one full rotation=1 day), does that mean we have a
"faster time"? If time is correlated to biological age, does that mean we
"age faster"?
ANSWER:
Time is not measured relative to
any astronomical motions. If the earth were to spin twice
as fast, that does not mean that time is running at twice
the rate; rather it means that a day would now be 12
hours long. Any clock would run at the speed it did
before the earth sped up. That includes biological
clocks. (I have neglected relativistic effects which are
exceedingly tiny at the speeds which the earth rotates.)
QUESTION:
In beta plus decay proton gets converted to neutron— where does the extra mass come from?
ANSWER:
A free proton does not (as far as
we know) decay for the reason you have inferred from the
masses; it must be inside a nucleus to βdecay. The nucleus begins with some mass
M. It emits a β^{+}, mass
m_{β} and kinetic energy K_{β};
and a neutrino, mass
m_{ν} and kinetic energy K_{ν}.
The nucleus now has mass M' and kinetic energy
K'. The energy of this isolated system must be
conserved,
Mc^{2}=(M'+m_{β}+m_{ν})c^{2}+K'+K_{β}+K_{ν}.
Essentially, the energy came from the mass of the
nucleus; you would find the new nucleus more tightly
bound than the original nucleus which you could interpret
as being why it decayed in the first place. Therefore,
β^{+}decay does not occur if the
new nucleus is less tightly bound than the original. (By
the way, the neutrino mass and kinetic energy of the
nucleus are negligible in most cases.)
QUESTION:
Just listened to a Mark Parker Youtube where is was stated (obviously) that at the top of a sphere (the earth), say near the north pole, you are travelling slower, i guess in some reference frame, than someone at the equator. ie. you travel a shorter 'distance' in a a 24 hour period than someone at the equator (larger circumference, same 24 hr period).
This is obvious with hindsight but not something I've ever tried to think about.
My question is where does the additional energy come from if I travel south of the north pole (or north from the south pole) that enables me to travel faster (along the direction of rotation).
I expect I'm misunderstanding the problem in some way. At the equator I travel 40000km in 24hrs or 1666.7km/hr (I'm assuming a spinning sphere and can ignore anything external to the earth).
I've randomly picked a latitude of 60degrees where the circumference is 20000km, which in 24 hrs = 833.3 km/hr The degrees don't matter except to show a smaller circumference travelled in the same time (slower).
So, as I travel north of the equator my rotational speed decreases. where does that energy go ? is the question valid.
If I travel south of some arbitrary northern latitude towards the equator my rotational speed increases.
Again, I'm grossly missing some key aspects of physics here. Am I actually (by travelling towards the equator) benefiting from the rotational energy of the earth, and when travelling away from the equator......actually is this a conservation of energy (or angular momentum) thing similar to pulling my arms in while spinning on a chair.
ANSWER:
You hit the nail on the head when
you mentioned angular momentum conservation. You plus the
earth are an "isolated system" with no external torques
acting on you (forget the moon, etc.) and the
angular momentum L of this system must remain
constant regardless of how you move around.
Suppose that you are at the north pole; then the angular
momentum of the system is L_{1}=Iω_{1} where I
is the
moment of the earth and ω_{1} is
its angular velocity. Now you walk down to the equator.
The angular momentum is now L_{2}=(I+mR^{2)}ω_{2}
where m is your mass and R is the
radius of the earth. But, L_{2}= L_{1} so
if you do the algebra you will find that ω_{2}
=ω_{1}/(1+(mR^{2}/I)).
So the earth slows down by the tiniest amount; I estimate
(ω_{1}ω_{2})/ω_{1}≈10^{29}=10^{27}%!
Regarding the energy, the initial energy is
E_{1}=½Iω_{1}^{2
}
and the final energy is
E_{2}=½(I+mR^{2})ω_{2}^{2}=½Iω_{1}^{2}/(1+(mR^{2}/I))
and so
E_{2}/E_{1}=(1+(mR^{2}/I))^{1}.
So, the energy has not been conserved but is a tiny bit
smaller even though your energy has increased by
½mR^{2}ω_{2}^{2}=½mv^{2}.
If you work it out you will find that the energy of the
earth alone is
½Iω_{1}^{2}/(1+(mR^{2}/I))^{2}.
So
we conclude that energy is lost by the system although you have gained
energy, so the earth lost more energy than you gained.
Energy is changed by forces doing work on the system, so
what force is doing negative work here? Since we are in
a rotating coordinate system, we will introduce the
appropriate fictional forces so that Newton's laws can be
used. If you are at some latitude θ, one
force you will experience is the centrifugal force shown
in the figure, C. It has two
components, a normal component N
which would reduce your apparent weight and a tangential
component T
which is trying to drag you toward the equator. In order
to keep T from accelerating
you, something must exert an equal but opposite force on
you; this is simply the frictional force f
between your feet and the ground. As you move toward the
equator f does negative work
thereby decreasing the energy of the system. Finally, the
reason you gain energy is that there is another component
to the frictional force which points also tangentially
but perpendicular to f such
that as you move south it speeds you up to keep the earth
from sliding away from you.
QUESTION:
I'm trying to explain to my gf that if the car in front of us is moving 60mph we would have to be moving 60mph also in order to maintain a 25 ft distance behind it. I know it
sounds dumb but I can't seem to make her believe it
ANSWER:
Ask her to imagine the car in
front of you is going 60 mph and is towing you with a 25
ft rope. Ask her to imagine looking at your speedometer.
QUESTION:
This may be super simple but I keep wondering about it. If a vehicle in the void of outer space fires its boosters, why does it move anywhere? Here's why I ask: movement on Earth requires friction and a medium through which to move. If I'm in a swimming pool and kick off from the edge with my legs, then I move far. But if I'm in the middle of the pool and make the same kicking motion without contacting the edge, I probably won't move. Well, not unless I move a lot of the medium around me (the water). If there's no medium in outer space, why does the rocket booster have any effect?
Here's my theory is it correct? There's no resistance behind my spaceship, but there's no resistance in front of it either. After, say, the first second of the ship firing its boosters, the next second's worth of focused, exploding fuel is contacting the exhaust and energy of the first second. This gives the later seconds worth of exploding fuel something to push against and, with no resistance in front of the ship, it moves forward easily. Is this right?
ANSWER:
It is a common misconception that
a rocket needs a "medium" against which to push, but it
is wrong. Your attempt to think of a way to "create" an
atmosphere to push against is therefore also incorrect.
In order for something to accelerate, i.e.
change its speed, all that is needed is for the sum of
all forces on an object to be not zero. Even if you were
on a surface with no friction and you were in a vacuum,
you could start moving if someone behind you pushed you.
The best way to understand a rocket is to consider the
following example using the above situation of you in a
vacuum with no friction on the floor.

You have a ball
in your hand;

you throw the
ball with some speed;

in order to give
it that speed, you need to exert some force on it;

Newton's third
law says that if you exert a force on the ball, the
ball exerts an equal and opposite force on you;

therefore you
end up moving in the opposite direction as the ball;

it turns out
that the speed you acquire is smaller than the the
ball's speed by the ratio of the ball's mass to your
mass;

for example, if
the ball has a mass of 1 kg and your mass is 100 kg,
your speed will be 1/100 of the ball's speed.
The rocket engine is
essentially throwing countless tiny "balls" (atoms,
molecules, ions) with very high speeds to propel the
rocket ship.
QUESTION:
If a system of two separate modules connected together with a linear passage(A spaceship carrying space traveler) is suspended in space with no gravitational influence and is rotating to create artificial gravity, at which point in the system will there be no force experienced by the accommodates while transitioning from one module to the other through the linear passage? It is the center I suppose, but I am unable to figure out the math.
ANSWER:
It depends on what the masses of
the modules are. It also depends if the masses M_{1}
and M_{2} of the modules themselves are
much larger than the mass m of the person moving
from one to the other. The pair will rotate about their
center of mass which, if M_{2}=M_{1},
is at the halfway point. If the person is very light
relative to the rest of the system, she will experience
no centrifugal force at the center as you guessed; the
reason is that the force is proportional the square of
the speed she is traveling and her speed will be zero at
the axis of rotation. But if m is not relatively
small the location of the center of mass will vary when
she moves from one module to the other.
The following is probably more than you want, but I like
to do it anyway. I will denote the masses of modules as
M_{2}
and M_{1},
including anybody who is in them, and will treat them as
point masses separated by a distance R; the
traveling astronaut has mass m and is a distance
d from M_{1} and will also be
treated as a point mass and I will assume
the passage has negligible mass. The center of mass is
located a distance r_{cm} from M_{1}.
Calculating the center of mass location relative to M_{1} is
straightforward and results in
r_{cm}=Σ(m_{i}r_{i})/Σ(m_{i})=[(M_{2}m)R+md]/(M_{2}+M_{1})
For example, if M_{2}=M_{1}=M,
r_{cm}=½R(m/M)(Rd);
and if m<<M,
r_{cm}≈½R.
Now, you are interested in when
d=r_{cm}:
d=R(M_{2}m)/(M_{2}+M_{1}m).
And again, for a check,
if M_{2}=M_{1}=M,
d=R(Mm)/(2Mm),
which is, if
m<<M,
d≈½R.
QUESTION:
Okay so lets say your free falling from some odd feet in the sky in a car. Dont worry about the reason why but that its happening. If you put yourself in a position to jump out of the car before it hits the ground and explodes, will you be able to change your momentum enough to not take so much damage from hitting the ground so fast?
ANSWER:
It depends on how far the car has
fallen. If it falls from say 10 ft, you could do it fine
but wouldn't even need to jump. But if the car fell from a
high enough altitude to reach its terminal velocity,
which I estimate to be about 200 mph, could you survive
it by jumping upwards at the last minute? How high can
you jump from the ground? The highest humans can jump is
about 6 feet and to do that requires that you can jump up
with a speed of about 13 mph. If you jumped that fast
relative to the falling car, your speed relative to the
gound would be about 187 mph; you would be just as dead
as you would have been if you hadn't bothered to jump.
QUESTION:
Hello. If a 200 pound man runs at 5 miles per hour, hits another man going same speed same weight, opposite. What is the force of impact?
Fyi. I'm 56. Past homework!
ANSWER:
There is no way to answer this
question because it depends on the details of the
collision. What matters most is the time the collision
lasts (or, equivalently, the distance over which each
moves during the collision) and whether the two
essentially stop or each rebounds backward. I can give
you a couple of suggestions. I will work in SI units (v=5
mph=2.25 m/s, m=200 lb=90.7 kg) which is usual for
physicists and will convert back to imperial units
(pounds) at the end. I will assume that the collision is
perfectly inelastic, that is both runners are at rest
following the collision; in that case all the kinetic
energy the two had (each has ½mv^{2}=230
J) is lost in the
collision.
Suppose, first that the two have lowered their
heads much like bighorn sheep do when fighting. Since
there is not much flesh on the forehead, they will stop
in a very short distance, let's say d=5 mm=0.2 inches.
The work done by the average force each man feels must
equal the energy he lost, Fd/2=230 J so F=460/0.005=92,000
N=20,700 lb; of course, this force will be spread out
over an area of several square centemeters. Still it is
pretty darned big.
Suppose the two men have big beer bellies and that is
where they collide. Then the distance over which the
collision occurs will be more like 5 cm and the resulting
force more like 2000 lb. And, the force will be spread
over a much larger area.
In the real world, such a collision would occur over a
large amount of the body so the total energy, converted
as a force, would be spread out over maybe a square
meter. places which are hard (like your head) would be
hurt more badly than places which are softer (like your
torso). Also, if the forces look unbelievably big, keep
in mind that they last for a very short time.
QUESTION:
If energy and mass are interchangeable or equivalent can E=mc^{2} be written as M=ec^{2} and mean the same thing? Thanks! I'm totally ignorant maybe there's is an obvious reason why not here.
ANSWER:
Mass and energy are not
"interchangeable or equivalent". That would be like
saying that electricity and wind, both forms of energy,
are the same thing. Mass is a form of energy and the
total amount of energy which a mass potentially has if it
could be totally converted into energy is given by E=mc^{2}. If you wanted to know how much
mass could be realized by converting an amount of energy
into mass, m=E/c^{2}. Another
way to see why your M=ec^{2 }is an
impossible equation is by doing dimensional analysis: The
units of mc^{2} must be kg·m^{2}/s^{2}
so energy must have units of kg·m^{2}/s^{2};
but your equation would have the units of energy being
kg·m^{4}/s^{4} which is not what
the units of energy are.
QUESTION:
I hope there really is no such thing as a stupid question, if there is you are really in for a treat. Okay, so if something is being propelled from the rear it is being pushed, and if something is being propelled from the front it is being pulled, correct? Is there a term for something being propelled from the front AND the rear at the same time?
ANSWER:
The words 'push' and 'pull' are
qualitative words, not physics words. Pushes and pulls
are both denoted a forces in physics. For example, if you
have a horse pulling on a cart with a force of 100 lb and
a man pushing on the back of the cart with a force of 10
lb, the net force on the cart due to these two forces is
110 lb forward. If the man is instead pulling on the back
of the cart, the net force is 90 lb forward.
QUESTION:
Light cannot escape a black hole. What is the nature of light inside a black hole, theoretically? Static photons? Can light be "static" and exist in only one point in space time? Perhaps it ricochets around inside the event horizon rather than static? Perhaps gets squeezed into a different subatomic particle? Am an old guy who has to much time and thinks of such things.
ANSWER:
Note that I usually do not answer
questions on astronomy/astrophysics/cosmology as I state
on my site. When light is captured by a black hole, it
ceases to exist as photons and basically becomes mass,
increasing the overall mass of the black hole. When you
when you ask what is going on "inside a black hole" it
can mean two things: either inside the Schwartzchild
radius (from inside of which no light can escape) or
inside the black hole itself. There is no answer for the
latter because at nearly infinite density we have no idea
at all as to what the laws of physics are. For the
former, you can say that a photon loses energy as it
falls from the Schwartzchild radius to the surface of the
black hole, the lost energy being converted to mass; the
photon continues moving at the speed of light but its
frequency decreases until the photon disappears at the
surface of the black hole, totally converted to mass.
QUESTION:
Say you have a rocket ship in a vacuum away from all gravitational fields firing its engines to maintain uniform circular motion. Since the engines are burning fuel, it makes sense to me to calculate the power output in watts. But the Force in the direction of the Velocity is zero so the power is zero. I'm confused. And since the Kinetic energy is not increasing, where does the energy expended by the engine go?
ANSWER:
You are going to have a rocket
engine which can fire perpendicular to your velocity
vector. None of the energy generated by the engine will
be given to the rocket ship because the force the engine
applies is perpendicular to the displacement, hence no
work is done. The energy generated goes into the kinetic
energy of the ejected gases. It will be a little tricky
to maintain uniform circular motion because mass is being
ejected, so as the ship loses mass you need to reduce the
force to keep velocity and radius constant: F=mv^{2}/R.
QUESTION:
How can universal gravitation and conservation of energy both exist together? If gravitation is universal there should be no where in the universe where a system is not acted on by an outside force (gravity).
ANSWER:
Conservation of energy is
applicable to systems with no external forces doing
work on them. You have to be very exact when
applying conservation principles. Suppose you choose the
solar system as the system; if the solar system were in
the middle of empty space its energy would be conserved,
but it isn't an isolated system in real life and the rest
of the galaxy exerts forces on it which might change its
energy. Suppose you choose the milky way galaxy as the
system; if the galaxy were in the middle of empty space
its energy would be conserved, but it isn't an isolated
system in real life and the Andromeda galaxy, the
nearest major thing which exerts forces on it, would
change its energy. You can see where I am going with
this. Suppose the entire universe were the system; then
there are only internal forces acting on this system so
its energy will not change.
QUESTION:
I have some understanding issues with Current and Voltage regarding Induction. When there is a change in magnetic flux there must be either a change in the area of the e.g. wire or the magnetic field. But how is there a flow of current then induced by it without any poles where I can measure the voltage. I guess, it's just me thinking wrong of voltage, but I really didn't understand that one if you ask me. Connecting to that, how must I think of a coil that is getting an induced voltage by changing magnetic flux. How are the electrons flowing through the coil, because right now i always try to think, that in every nth part of that coil there is some sort of electric field that is created because of the electrons getting pushed. I really do think it's a question of understanding, but the mathematic definition and explainition is not really enough for me, i want it visualized for me.
ANSWER:
When electricity and magnetism is
taught, there is usually a sequence:

First we talk about
electrostatics where all electric fields are constant and
conservative. A conservative
field is one for which, if you move an electric
charge from one point in the field to another, the
amount of work you do is independent of the path you
choose. This means that potential difference
(voltage) between two points is a meaningful number.
You are worried because you have been told that a wire loop with
no battery in it might have a current cannot flowing in
it.

Next we talk
about magnetic fields which are caused by electric
currents which are constant and presumably caused by
potential differences. This is called magnetostatics.

Finally we come
to the full theory of electromagnetism where the
fields are no longer constrained to be constant.
All of
electromagnetism is described by Maxwell's four
equations. It gets mathematically dense so I have
previously
qualitatively described these equations without any math.

Electric charges cause electric fields.

Electric currents cause
magnetic fields.

Changing electric fields cause magnetic
fields.

Changing magnetic fields cause electric
fields.
So the answer to your
question is that you do not need charges to create the
electric field which will drive a current; a loop of wire
will have a current running in it if you cause the
magnetic field passing through its area to change. This
is called Faraday's Law.
QUESTION:
Okay, so i may dissagree with a master here, aka. Stephen Hawkings himself, but i have some questions that need to be answered, so here goes.
Hawking Radiation. It's basically a theory that says particles and antiparticles spontaneously materialize in space, seperating, joining, and annihilating each other. But then a black hole comes and sucks the a particle, leaving the other without a partner to annihilate with. This particle appears to be in the form of black hole radiation. Sooo, blackholes are not eternal. But what if all of the particles are sucked into the black hole? This is particle and antiparticle that we're speaking of, that has POSITIVE mass. It's not exotic particles that have NEGATIVE mass right? So does Mr. Hawkings law apply? Please answer this, i'ts kinda been bugging me for a while...
ANSWER:
First I will note that I do not
normally do astronomy/astrophysics/cosmology, as clearly
stated on the site. However, since I have answered this
and related questions many times before, I will accept
your question. I would say that you stop thinking about
mass and simply think of energy. A particle can have
negative energy without having negative mass. I think the
best and most detailed answer I have given is
here.
QUESTION:
why light cannot curve around object
ANSWER:
In fact, light does curve around
an object with mass. Originally it was known that light
had no mass and therefore would not be affected by
gravity. But Einstein's theory of general relativity
predicts that light passing near a massive object like a
star or a galaxy will be deflected. This has been
observed to be correct. There is a nice
article on gravitational lensing on Wikepedia.
However, for everyday life the bending is hardly
noticeable because it is small. Even the entire earth has
too little mass for this bending to be noticable (see a
recent answer).
QUESTION:
Assuming no prior information is given, except the formula for the time period of a pendulum (T=2π√(L/g)), would the time period of a swing be changed at
all if someone went from sitting down on the swing to standing up on the swing? Eg. would the mass distribution affect the time period?
ANSWER:
The formula you state is true
only for a point mass M attached to a massless
string of length L. Therefore you cannot solve
this problem using that formula. However, you could guess
that the period would be proportional to the square root
of the distance D from the suspension point to the center
of gravity of all the mass (the person, the swing, the
ropes). In that case, the period would get smaller when
the person stood up because the center of gravity would
be closer to the suspension point. The correct equation for the period
is T=2π√[I/(MgD)]
where I is the moment of inertia about the
suspension point.
QUESTION:
Under "miscellaneous" here on your site, you answer the following
question: "If the earth is curved how is it you can get a laser to hit a target at same height at sea level more then 8 km away? How is it that it's bent around the earth?"
Part of your answer states that the laser is perfectly straight.
But spacetime is bent (curves) within the gravity well of a massive object like the earth. Astronomers have shown that it's possible to see stars that are actually behind such massive objects because the light from the star is bent around the massive object as it necessarily follows (somewhat) the curvature of spacetime around the object.
So how is the laser beam in the question you answered perfectly straight?
ANSWER:
It was clear to me when I was
answering this question that the questioner was
interested in a classical physics question, not one
taking general relativity into account. You are right,
any mass (or other energy density) will cause a ray of
light to bend. But in this case the amount of
bending is very, very tiny. If you apply the angle
of deflection equation θ=4GM/(rc^{2}
to a beam of light tangent to the surface of the earth at
the surface of the earth, you will find that θ≈0.0006"=1.67x10^{7}°.
I think you will agree that this is negligible in the
context of the question I answered!
QUESTION:
It takes 375 joules of energy to break a human bone. How high must a 60kg person fall to break a bone?
ANSWER:
There is no answer to this
question. And it really does not make any sense because
it is force, not energy, which breaks a bone. If you
delivered 0.01 J/s over 37,500 s (approximately 10 hours)
would you break the bone? So what matters is the time the
stopping collision takes to deliver the energy impulse.
QUESTION:
Why doesn't your brain explode in a PET scan? Doesn't antimatter meeting matter create a lot of energy?
ANSWER:
Yes, matter/antimatter
annihilation releases the maximum amount of energy—when
a positron meets an electron, 100% of their mass is
converted into energy. But how much mass do they have and
how much energy is that? Each particle has a mass of
about m=10^{30} kg, so their total mass
is about 2x10^{30} kg. The energy is then E=mc^{2}=2x10^{30}x(3x10^{8})^{2}=1.8x10^{13}
J where c=3x10^{8} m/s is the speed of
light. To put this into perspective, this would be the
energy of a particle of dust which has a speed of about 2
inches per second. Furthermore, nearly all of the energy
(which is in the form of two xrays) escapes without
leaving any energy in the brain.
QUESTION:
Hi, hoping you can help settle a debate I'm having with a colleague. We know that when a solid object spins, radial and circumferential tension exists within it due to centripetal/centrifugal force. However, for a massive object such as a planet, whose gravitational acceleration far exceeds centrifugal force, does that tension still exist within the object? I say that the product of gravitational and centrifugal forces results in a net force towards the object's centre of mass leading to a net compression, and an object under compression cannot also be under tension. He states that the tension that would have existed due to the centrifugal force only would still exist and that the gravitational force makes no difference to this. As you can probably tell we're not physicists! Can you help answer whether tension (such as a hoop stress force) would still exist in such a massive object, or would the gravitational force
'overwhelming' centrifugal force prevent tension from forming in the first place?
ANSWER:
Whenever you want to understand
something, you need to include all the forces acting on
it. Centrifugal force is what we call a fictitious force,
it doesn't really exist. When you are experiencing an
acceleration, Newton's laws do not work. However, if you
add fictitious forces cleverly, you can do Newtonian
physics. If we are in a rotating system like the earth we
are accelerating because in physics acceleration does not
just mean speeding up or slowing down but also includes
changes in your direction which is constantly happening
to you as the earth spins (unless you are at a pole). To
see how this works, look at an
earlier
answer. So, suppose that you are standing on the
equator; there are three forces acting on you, your own
weight (gravity) which points toward the center of the
earth, the centrifugal force which points away from the
center of the earth, and the force (which points up)
which whatever you are standing on exerts to keep you at
rest. Suppose that your mass is 100 kg and the
acceleration due to gravity is approximately 10 m/s^{2};
then your weight is approximately 1000 N. The centrifugal
force is your mass times your speed (464 m/s) squared
divided by the radius of the earth (6.4x10^{6}
m), C=100x(464^{2})/(6.4x10^{6})=3.4
N. Suppose you are standing on a scale; since you are at
rest, the force which the scale exerts up on you is
10003.4=996.6 N, about 0.3% smaller than your weight.
This is a longwinded answer to your question: yes,
forces due to rotation apply to anything regardless of
its mass or size. (If you are not comfortable with metric
units, 1000 N=225 lb and 3.4 N=0.76 lb. If you weighed
yourself at the poles the scale would read 1000 N if the
earth were a perfect sphere.
Here is another example: the earth is not a perfect
sphere, it bulges at the equator. The reason is that the
earth, when it was just forming billions of years ago,
was very hot, almost molten, and therefore more
"plastic"; so the centrifugal force caused it to flatten
as it rotated. That would mean that your weight at the
poles would really be greater than what it is at the
equator because it is closer to the center of the earth.
QUESTION:
Hey, i wanted to discuss about the MPEMBA EFFECT. Mostly i have read that it has no valid and accepted explanation but i think it should be taken as common sense like if we draw an analogy with electrodynamics we can say that a body with higher temperature should be at higher potential and a cold body should be at lower temperature. So as the potential difference gets bigger the rate of flow of charge(current) gets higher, similarly the rate of flow of heat charge should also get higher. Hence, it should be taken as obvious that a hot body will cool down much faster than a relatively cold body.
ANSWER:
I first note that I have already
given a quite lengthy
answer to the "hotwaterfreezesfaster" hypothesis.
As you will see, what happens depends a lot on the
conditions of any measurement or experiment you might try
to do. Your attempt to bring in "potential" is pretty
muddled, so let's review rate of flow in electrodynamics
and in thermodynamics. Materials have a property called
electrical conductivity; this property tells you how much
electric current you will get if there is a voltage
(potential difference) between two points in the material—the
larger the conductivity, the larger the current will be.
Similarly materials have a property called thermal
conductivity; this property tells you the rate at which
heat travels throught the material for a given
temperature difference—the larger the temperature
difference, the larger the energy flow will be. But, this
will have very little influence on how quickly the water
will freeze. First, the change in thermal conductivity of water
changes by only about 10% between say 20°C and 70°C;
second, water is not a very good thermal conductor; and
third conduction is not the primary way water cools
because the density of water, unlike most materials in
the molten state, gets larger as it cools, the cooled
water at the surface sinks so most of the transfer of
heat inside the water is by convection. Certainly the hot
water loses energy at the surface faster than the cold
water, but it will eventually catch up with the cold
water and then the two will be indistinguishable unless
the cold water freezes before the hot water catches up
with it. As I explained in the earlier answer,
evaporation cools the hot water more and if a significant
amount of the hot water evaporates before it catches up
to the cold water, it will win the race because there is
less of it; but that is really cheating, isn't it. Sorry,
I do not know anything about the mpemba effect.
QUESTION:
I just want to ask if, can we solve the force of attraction between two identical pendulum bobs with only mass(0.35kg) as given? Is it solvable? I am just curious about this, my teacher discussed about this topic but with enough components to solve it. But with this problem, I really can't think of something to solve it with only one given component.
ANSWER:
You cannot calculate the
gravitational force without knowing the distance between
them. And you could not easily calculate the force unless
the bobs were spheres.
QUESTION:
I was wondering, in particle annihilation between say a electron and a positron, how long does it take to occur? I know it's refered to as being instantaneous, or happening in a immeasurable amount of time in clearer terms.
ANSWER:
I believe there is no good answer
to this question because when would you start and stop
the clock? And even if you could specify some particular
times, they would inevitably depend sensitively on the
initial conditions like how fast each was moving
initially. You can, however, measure the lifetime of a
positronium atom (one electron bound to one positron). In
vacuum the singlet state (spin zero) atom has a lifetime
of about 0.125 ns and the triplet state (spin one) has a
lifetime greater than 0.5 ns. The times for the atom in
various materials would be longer.
QUESTION:
If a carousel is out of control, spinning at high speed, and suddenly it is stopped, there will be chaos and horses and riders flying out of the carousel structure. How do you explain this in terms of physics? Are they ejected due to the centrifugal force? Loss of centripetal force? And what happens to the kinetic energy? Is it transformed to what kind of energy?
ANSWER:
If the carousel is moving in a
circle with constant angular velocity, the only forces
horizontally are forces necessary to provide the
centripetal acceleration. In the photo, the girl holds
the pole and presses on the side of the horse with her
leg to provide those forces; the horse is held by the
force of the pole to which it is attached. All horizontal
forces are toward the center of the carousel because the
speed of everything on the carousel is constant.
What happens if the carousel suddenly stop. All those
radial forces quickly drop to zero. The tendency is for
everything on the carousel to continue moving with the
velocity they had before, but now in a straight line.
But now, although the horizontal radial forces are gone,
each object experiences horizontal forces opposite the
direction of their velocities. The girl would initially
move forward until she smashed into the pole; the horse
would probably be held in place by the pole although the
force required to stop the horse could very possibly bend
the pole; someone standing on the spinning floor would
only have the friction of the floor to stop her and would
likely keep moving in a tangential direction. Nothing is
"ejected", it just keeps going in the direction it was
going when the carousel stops, unless something stops it.
QUESTION:
I push on a wall and am accelerated backward. Per Newton's third law, how does the acceleration of the wall manifest itself.
ANSWER:
You push on the wall with some
force F and, as you note,
Newton's third law says that the wall pushes with a force
F. So the magnitude of your
acceleration a is a=F/m where
m is your mass. The magnitude of the wall's
acceleration A is A=F/M where
M is the mass of the wall. For all intents and
purposes, the mass of the wall is infinite so its
acceleration is zero.
QUESTION:
Are electrons made of quark/antiquark pairs? If yes, which pair?
ANSWER:
No. To the best of our knowledge
electrons are elementary, i.e. they have no
components.
QUESTION:
What force pulls the train? I am doing a science project about this simple electromagnetic train. This project looks so simple yet so complicated, here is the
video
of the project that I am working on.
ANSWER:
The construction details and a
brief description of the physics are shown in another
youtube
video. Essentially, when the magnets touch the copper
wire the battery causes a current to flow in the coil
between the two batteries which causes a magnetic field
in that section of the coil; each magnet, now being in
that field, experiences a force moving it forward.
Important things:
The same pole, north or south, must point away from the
battery.
Be sure to use bare copper wire. Often copper wire has an
insulating layer on its surface and this would not allow
current to flow
Good luck on your project. It is really not as
complicated as you thought.
QUESTION:
I can't grasp how two waves can pass each other on the same piece of string. For a wave to travel on a string each piece of string is, let's say, pulled up wards by the preceding piece of the string, and the wave propagates forwards. If a wave moving in the +x direction meets a reflected inverted wave in the  x direction, a node will be formed as one wave pulls that piece upwards and the other wave pulls it downwards. Therefore the piece of string doesn't move, so how can either wave travel past this point?
ANSWER:
What you need to understand to
see why the two waves, moving in opposite directions are
able to do so is a little bit of the physics of waves on
a string. Any wave which moves through a medium is a
solution of a very famous equation, the wave equation:
d^{2}f(x,t)/dt^{2}=v^{2}d^{2}f(x,t)/dx^{2}.
Here, x the position on the string,
t
is time, v is the velocity of the wave, and f(x,t)
is the solution to the equation which will describe the
shape of the wave at a time t. You may not know
calculus and this equation is goobledygook to you, but
all you need to know is that, mathematically, any
function f is a solution as long as it of the form
f(xvt). The most commonly used example of
waves is sinesoidal, for example, f=Asin(kxωt)
where k is called the wave number and ω is
the angular frequency of the wave. Note that ω/k=v.
Also, to touch base with quantities you might be more familiar with,
k=2π/λ and ω=2πf where
f is the frequency of the wave and λ is the wavelength.
Now comes the the important part: because any f will be a solution to the wave equation, if you have a
wave traveling to the right, f_{right}=Asin(kxωt),
and an identical shaped wave traveling to the left, f_{left}=Asin(kx+ωt),
their sum will also be a solution to the wave equation.
As you can see from the figure, waves traveling
simultaneously right (red) and left (green) add up to a
"standing wave" (brown) which does not appear to move but
is still oscillating*. You are correct, there are nodes
which do not move but both component waves go right on by
them nevertheless, it just isn't apparent when you look
at there sum. The fact that the net motion of the medium
(string) is just the sum of all individual waves is
called the superposition principle.
*If you're handy with trigonometry, you can calculate
f_{right}+f_{left} which is
apparently not moving even though you know that before
you did the calculation you could certainly see that it
was two waves.
QUESTION:
Hi, I have been toying with the
concept of using gravity as a perpetual energy source,
and have had a lot of pushback in the process of trying
to ask questions. The conversation usually ends before I
can get an answer. The skepticism surrounding perpetual
motion is understandable. If I understand correctly the
issue is that closed systems lose energy to various
forces such as friction. Firstly, does it count as
perpetual motion if the energy is supplied constantly
from outside forces?Secondly, if not, is there a proper
term for what I've described?
ANSWER:
I am afraid that you have it
exactly backwards. In a closed system, defined as one
which has no external forces acting on it, energy is
conserved. In a closed system where friction is present,
when kinetic energy is lost (e.g. a spinning
wheel slowing down or a box sliding to a stop on a
surface), the lost kinetic energy shows up mainly as
heat. And, if you do work on a closed system to keep it
moving forever, that is not what we mean by perpetual
motion. Proper term for what? If you mean motion that you
keep going by pushing on it, it has no particular name.
QUESTION:
First, I am way out of my field of understanding here so please keep it simple. I watched some videos on E=MC2 which led to how light reacts differently than matter at high speeds causing time to slow down when moving fast. My question is, if I was to shine a flashlight perpendicular (90 degrees) to the direction traveled am I correct to say if I was moving at half the speed of light the beam would actually be at a 45 degree angle and when travelling at the speed of light the beam would be horizontal (0 degrees). This would also be true whether the beam was inside or outside of the spacecraft, correct?
ANSWER:
No, you have it wrong. The most
important thing to remember here is that the speed of
light, c, is the same for all observers. If you
are in the rocket and shoot a beam of light straight
across the ship, it will go straight across the ship in
some time t_{s} and the distance it goes
will be ct_{s}; if the width of the ship
is w, t_{s}=w/c.
Now, someone on the ground will see the rocket moving by
with some speed v, so the point on the inside of the ship
where the light beam will have moved a distance vt_{g}
when the light strikes it; but since this is light, the
distance it will travel is longer because the light has
to go farther. Now, to find the angle the light relative
to the perpendicular to the velocity v,
we note that sinθ=v/c (t_{g}
cancels out). In your question, since v=c/2, so
θ=sin^{1}(0.5)=30°. This
example also demonstrates time dilation because the two
observers see different times of transit of the light. If
you work it out, t_{g}=t_{s}/√[1(v^{2}/c^{2})].
QUESTION:
The radius of earth is 6440 km . Suppose, the angle between Canada and USA inside the earth is 10 degrees , then what is the distance between them? I have been stuck with this problem . Can't i answer this problem using The Cosines Laws? If i do so....does it would go wrong? please answer my question...? It is not a homework question . i am thinking to solve this one with a different way....
ANSWER:
This is a very strange question
because the USA and Canada are adjacent and therefore the
angle would be 0°. However, you can easily find the
distance between two points which subtend 10° just by
knowing the definition of radian measure of angles. The
angle θ subtends a distance s for a circle of radius
r. This angle, in radians, is defined as θ=s/r.
Since the circumference of a circle is 2πr and there are 360° in a circle, there are 2π radians in 360°.
So 10°=10x(2π/360)=0.175=s/6440,
so s=1124 km.
QUESTION:
Hi! I would like to settle a bet with my father.
My questions is that if I was sitting in a chair that was tied to a rope, and the rope was divided by a pulley on the ceiling, would it be possible to lift yourself up just by pulling down on the other side of the rope.
ANSWER:
The figure on the left shows all
the forces on you: T is the
force which the rope exerts up on you*, N is the force
which the seat of the chair exerts up on you, and
Mg is
the force of gravity (your weight) down on you. Choosing
+y up as indicated, Newton's second law in the
y direction is, for you, N+TMg=Ma
where a is your acceleration and M is
your mass. The figure on the right shows the forces on
the chair: N is the force you
exert on the chair, the same magnitude but opposite
direction as the force the chair exerts up on you because of Newton's third law;
T
is the force which the rope exerts up on the chair, the
same magnitude as the tension on you because the rope
and pulley are assumed to have negligible mass;
mg is the force of gravity (weight)
down on the chair. Newton's second law in the y
direction is, for the chair, N+Tmg=ma where
a is the acceleration of the chair (the same as
yours) and m is the mass of the chair.
Let's first assume you are exerting just the right force
T on the rope so that you are at rest, so a=0.
In that case, if you solve the two equations you will
find that T=(M+m)g/2. In other
words, you need to be able to exert a force down on the
rope which is equal to half the weight of you and the
chair combined to hold the chair from falling. All you
need to do is be able to pull just a little harder to
move upwards. That answers your question but it interests
me to look at a case where a is not zero.
If you solve the two equations for
a and N
you get
a=[(2T/(M+m))g] and
N=T(Mm)/(M+m).
A few observations about what these results tell us:

If you let go of
the rope, T=0, a=g, N=0;
you and the chair are in free fall.

If m=M,
N=0, a=(T/m)g;
the motion of you and the chair are identical even
though there is no interaction between you and the
chair.

You will
accelerate upwards if T>½(M+m)g,
half the total weight. Your strength is the only
thing limiting how fast you can accelerate upwards.

If 0<T<½(M+m)g
you will accelerate downward with an
acceleration smaller in magnitude than g. Of
course T<0 is not possible for a rope.
*It is important to
note that, because of Newton's third law, if the rope
exerts a force up on you, you exert an equal and opposite
down on the rope. T is a measure of how hard you
are pulling. How large T can be depends only on
how strong you are and how strong the rope is.
QUESTION:
We define image as the intersection point of reflected rays and the focal point is the point at which light will meet after reflection.
So, why is it that we have different positions of image when object is placed at different distances from concave mirror and not always at focal point except when object is placed at infinity?
ANSWER:
Because only rays which come in
parallel to the optic axis are reflected through the
focal point. This occurs only (approximately) for objects
very far from the mirror.
QUESTION:
Does radiation from your phone stay on your hands after you use it?
ANSWER:
Absolutely not. Any radiation
from the phone is electromagnetic, essentially radio
waves. It is transient and does not "stay" anywhere. Even
when you hold the phone the radiation is not "on your
hands" but harmlessly passing through them just like the
waves from some radio station is passing through your
whole body.
QUESTION:
When a uranium or plutonium atom is fissioned, energy is released visa E=MC2. I believe this energy is in the form of photons. What I do not understand is how these photons (light) create such high temperatures in an uncontrolled nuclear explosion. Can you please explain?
ANSWER:
Actually, emitted photons account
for only about 2.5% of the total fission energy. Emitted
neutrons account for 3.5%. Most of the energy is
contained in the kinetic enerngy of the fission
fragments, 85%. The other approximately 9% of the total
energy shows up later when the fission fragments, which
are not stable, decay radioactively, mainly from βdecay.
The high temperatures are due to the kinetic energy of
the fission products which have speeds typically 3% the
speed of light.
QUESTION:
I'm embarking on a horological project. I need to buy a very particular type of spiral spring. I guess it's a type of torsion spring. The only supplier I can find who sells these springs classifies them with two numbers. The overall diameter in mm
and the torque in gm/cm/100°.
(As noted below, I assume that this means the torque on the spring when
the angle is 100° and it should be gm·cm, not gm/cm.) I know the spring I need. But I only know in terms of the CGS System. I'm following an old book and I've ascertained that the spring I need has a
"CGS number" of 0.76.
So, my challenge is to find a way to convert between these two unfamiliar units of torque measurement. A
"CGS number" and gm/cm/angle.
There are plenty of calculators for converting between different units of torque. But I don't think it's as simple as that. I have two questions.
I know a bit about the CGS system and there is loads of info online. But I'm not clear what it is in the context of classifying a torsion spring. The book only says
"the CGS number indicates the restoring couple of the spring when its diameter is 1cm." So what is the CGS number exactly? Is it dynecentimetres? If so, then that's easy to work with since that's a measure of torque I'm familiar with. So I'm half way there. If it's something else, then perhaps you can help explain.
Then there is the question of gm/cm/angle. The problem is that none of the calculators I've seen have a notion of angle. This is specific to torsion springs. The torque, in this case, is the working load when the spring bends 100°.
Will I ever find the correct spring? Sadly, I have spoken to the manufacturer, and they were unable to help me.