QUESTION:
ANSWER:
This is simply a reference to classical electricity and magnetism (E&M).
The way students are generally introduced to electromagnetic radiation
(light, radio, x-ray, etc .) is to imagine an electric charge in
empty space oscillating, maybe like being attached to a spring which is
vibrating; hence Feynman's "shaking". The result of this shaking is
electromagnetic waves. What he was interested in was that if another charge
was in the vicinity, it would feel forces because of the radiation and it
would start shaking itself which meant that the first charge would feel the
radiation from the second. In other words the first would feel forces which
it originally caused. What he wanted was an explanation for another effect
and it didn't pan out.
QUESTION:
ANSWER:
Time dilation with respect to whom? For example, suppose you are on earth and Elon Musk is on Mars. The details of why you and he happen to be moving like they are is not important. Relative to you he will have some velocity;
you would observe his clock to be running more slowly than yours as determined by using that velocity to calculate the time dilation. For most applications this is purely academic because the speeds are all very small relative to the speed of light.
QUESTION:
ANSWER:
Here is a nice video of two black holes colliding. Each black hole has
an event horizon which defines the region, inside of which, nothing can escape. As the two spiral into each other, their event horizons
merge into one. Nothing inside can escape, but much of the huge energy
release will be outside.
QUESTION:
ANSWER:
I presume that when you say
"set to swing at 30 mph" the speed of the bat at the point
where it hits the ball is 30 mph relative to its own rest frame. The only
thing that matters is the relative speed of the ball and bat; so yes,
machine speed =20 mph, relative speed=40 mph; machine speed =30 mph,
relative speed=60 mph; machine speed =10 mph, relative speed=40 mph;
machine speed=5 mph, relative speed=35 mph; etc .
QUESTION:
ANSWER:
When you solve a linear algebraic equation you get an answer. When you
solve a quadratic algebraic equation you get two answers. The thing in
quantum mechanics which tells you the nature of state of some system is the
wave function which is the solution(s) of the equations of quantum
mechanics. These are partial differential equations which often will have
more than one solution. So you are faced by a problem: Which is the
appropriate solution for the situation? The answer is that the most general
solution is a linear superposition* of all possible solutions. You may, for
a given situation, be able to determine what the constants of the
superposition by applying constraints. The square of the total wave
function must be 1 because it is the probability of the system being in the
state being in a given configuration determined by the A s.
*Super position means: if ψ n is the wave function
for the nth state possible, the form of the total wave function
is Ψ=A 1 ψ 1 +A 2 ψ 2 +A 3 ψ 3 +….
QUESTION:
ANSWER:
Seriously? A 45°
slope sounds pretty steep to me! The only reason you would slow down would
be friction, and it is hard to do a calculation without knowing lots of
things you don't mention. I will later try to add in some reasonable
variables to estimate the effects of friction. For starters, let's do the
calculation assuming no friction. In the picture I have drawn, you are
going down a hill of angle θ . The weight of you plus your
board is denoted by W W x ), -W sinθ , is accelerating you down the
hill and the component perpendicular to the surface (y ), -W cosθ ,
is pressing you into the surface. Newton's third law says that there
is a force which you exert on the road, the road exerts an equal and
opposite force on you; this is denoted by the force N m .
Because there is no acceleration normal to the surface, N=W cosθ
and, for your situation, N=W cos(15°)=0.97W . But, in
the x -direction, there is an unbalenced force, -W sinθ=ma where a is your
accleration down the hill. Without going into further detail, a=- 9.8sin(15°)=-2.54
m/s2 . To make this easier to relate to, the acceleration is 5.7
mph per second—your speed down the hill increases at the rate of 5.7
mph as every second ticks by.
Except for air drag, just about all frictional forces are proportional to
the normal force. This means that f=μN where f is the
frictional force and μ is called the coefficient of friction.
Looking at some physics of skateboarding sites I found that μ is generally around 0.002-0.005
for skateboards. Taking the normal force to be about 200 lb
and using
the biggest μ , f =1 lb.
But W sinθ is about about 52 lb. So friction due to
everything but air drag is pretty negligible sort of at about the 2% level.
Finally let us consider air drag which depends basically on the square of
the speed v
and the area A presented to the onrushing wind. If you use SI
units a pretty good estimate is f =¼Av 2 .
Using A =1 m2 and v =20 mph I find f =11
lb. So this is not negligible! But in your case, starting the 15° slope
going 20 mph, the net force is still down the hill, about 52-11=41 lb.
Eventually, as your speed continues to increase, you will reach the speed
where the air drag is 52 lb and you will then
continue at that speed with no further acceleration. (This is called the
terminal velocity.)
QUESTION:
ANSWER:
This is an academic question! The sun, from its birth to death, will lose
0.034% of its original mass; this includes both the solar wind and the loss
due to fusion. The effect on planetary orbits of this change would be
negligible.
QUESTION:
ANSWER:
The problem here is that you do not know
relativistic kinematics. The
whole thing can be understood, but it is a
First, note that there is no definite result of an annihilation; the
results depend on the energies of the incident particles and could be
massive particles other than photons. And I believe that there could
be more than two photons. For illustration purposes I will assume that
there are two photons after the annihilation. In the little sketch I have
included, the red vectors are photons, the blue vectors are the electron
and positron. You specify suppose the electron, e.g. , has an
energy of 5 eV; but you must do the calculation relativistically because
massless photons are involved, so you need to know the total energy E of
each of the particles. For particles with mass E =√[(mc 2 )2 +(pc )2 ];
for photons E=hf ; here c is the speed of light, f
is the frequency of the photon, m is the rest mass of the
particle, p is the magnitude of the linear moment of the particles
with mass, and h is Planck's constant. Linear momentum for a particle with
mass is p=mv /√[1-(v /c )2 ]; for
massless particles (photons) it is p=E /c . Momentum must
also be conserved, total momentum before and after the annihilation must be
the same. But, lest we forget, momentum is a vector and, in my picture,
each momentum vector will have x and y components, so to conserve momentum
you can separately conserve their components. So now, supposing that we
know the energies and momenta of the electron and positron, there are three
equations (conservation of energy, conservation of x and y
momentum) but six unknowns (the energy of each photon, the x and
y momentum component of each photon); so you cannot know what the
photon energies are nor know the magnitudes and directions of the photons;
there are constraints, e.g ., the total photon energies are the sum
of the photon and electron energies. In this situation you would probably
measure the energy of one of the photons and, since p=E /c ,
you could infer the momentum components from where you have placed the
detector. Now you would have three unknowns (the other photon) so you could
predict where the second photon went and its energy.
I said I would say why you can't have a single photon after the
annihilation. The reason is that it is impossible to conserve both the
energy and the momentum. A simple case is the case of the two particles
both at rest when they annihilate: the photon would have to have an energy
equal to the sum of the rest mass energies of the two particles and no
momentum, not possible.
QUESTION:
ANSWER:
The simplest way to understand this is that the inertia (mass) m of an
object increases as the velocity v increases. The equation
describing this is m=m 0 /√(1-(v 2 /c 2 ))
where m 0 is the mass if v =0 and c is
the speed of light. Note that if v=c the mass is infinite meaning it would
take infinite energy to get there. I have always maintained that
accelerators are poorly named because they aren't doing any significant
acceleration; they would be better named energizers.
QUESTION:
Since this all occurs within the framework of General Relativity, and General Relativity notoriously does not conserve energy (photons in an expanding spacetime lose frequency and become less energetic and that energy goes nowhere, dark energy expands the universe and adds more dark energy, etc.), why do we insist on believing that the emission of radiation from the vicinity of the black hole must necessarily diminish the mass-energy of the black hole? From my understanding of Hawking radiation's close relative Uhru radiation the same thermal bath is expected to be experienced by an accelerated observer and yet (as far as I can discern) that energy isn't being subtracted from anything. It's just being created. Couldn't the same apply to Hawking radiation?
Regarding black hole evaporation, even if Hawking radiation is subtracting (mass-)energy, since at the horizon time is infinitely dilated to a standstill, how could a remote observer ever witness any evolution (evaporation) of the black hole? Isn't an infinitely time-dilated region essentially frozen in time and therefore unable to evolve from the standpoint of an observer at infinity?
ANSWER:
First, the site clearly states that I do not do
astronomy/astrophysics/cosmology. In addition, highly specialized,
technical questions are discouraged. I will speak to a couple of your
points, but not give anywhere close to a complete answer.
It is, by no means a settled issue that in gravitational redshift the
energy simply "goes nowhere".
The virtual-pair model is as deep as my knowledge of Halking radiation
goes. A summary of my understanding can be found in the two links in the
question following this one.
Although the question of energy conservation for the universe is not
definitively answered, energy in a local, relatively compact volume is
certainly conserved.
It is important to keep in mind, if you are studying physics, that what we
know about the laws of physics can be assumed to only be valid for about 5%
of the universe. There are no theories which describes the physics of dark
energy or dark matter.
QUESTION:
ANSWER:
Hawking radiation can be anything which results from a
particle-antiparticle pair; for example, if an electron-positron pair is
created just outside the event horizon, one will be captured and one not
captured, the not captured one being the radiation. The nature of the
vacuum outside is such that such pairs are constantly popping into and out
of existence; although you might think that would violate energy
conservation, the uncertainty principle allows it if the pair exists for a
short enough time before annihilating back to the vacuum. Everything I know
about Hawking radiation can be seen here and the
earlier answer linked
to there. (I remind you that according to site ground rules I usually do
not answer questions on astronomy/astrophysics/cosmology.
QUESTION:
ANSWER:
All the objects you talk about are spherically symmetric spheres,
which means that the density depends only on how far you are from the
center. The gravitational force from an object is proportional to its mass;
i.e. , if the mass of the earth were twice as large as it is, your
weight would be twice as big. But, the force is also inversely proportional
to the square of the distance from the center of the object; i.e .,
if the radius of the earth were twice as large but its mass the same as it
is, your weight would be one fourth as large. The sun's mass is hugely
bigger than the earth's; but your distance from the center of the sun
squared is hugely bigger than your distance from the center of the earth
squared. The effect from the distance falls off as distance increases much
faster than the effec from the mass increases as mass increases. The net
result is that the gravitational fields of very distant objects at points
on the earth are neglibly small for most purposes.
So, as an example I thought I should do a calculation of the force on you
here on earth due to the sun. To my surprise, I found that the acceleration
you would experience due to this force would be about 5.9 m/s2 !
This is about 60% what your acceleration would be due to the earth, 9.8 m/s2 ,
hardly neglegible?! But I quickly realized that you would be completely
unaware of this force; you, like the earth, are moving in an almost
circular orbit around the sun and this force, pointing toward the center of
the sun, is the force keeping you in orbit. This is just like the
"weightlessness" experienced by astronauts in orbits around earth where
they don't feel the gravity even though it is there. If you were placed at
rest somewhere in space on the earth's orbit but without the earth, you
would accelerate toward the sun with an acceleration 5.9 m/s2 .
QUESTION:
ANSWER:
It does! When moving with constant velocity an object is in equilibrium
just as if it were sitting still; i.e. , all forces on it must add to
zero. So whatever is keeping moving upward is exerting an upward force
(normal) equal to the object's weight. Not "rather interesting"!
QUESTION:
ANSWER:
Assuming that everything else but the pry
bars is identical, you get no
lifting force advantage with a larger area. However, the force (2000 lb) is
distributed over a bigger area. So if what you are lifting is breakable or
deformable you might be better with a larger area. (By the way, area is
measured in square inches, non inches.)
QUESTION:
ANSWER:
See a recent answer .
FOLLOWUP QUESTION:
Thanks, but I was directed to the answer below, and it doesn’t answer my
question. At least I don’t understand it. It seems to me that the
flashlight batteries energize the bulb, and then the bulb
instantaneously emits photons at the speed of light. How can two humble
flashlight batteries produce that much energy?
ANSWER:
Okay, let's try a different approach to answering your question. Suppose we
have a 10 Watt light bulb. That means that this light bulb, when on, is
consuming 10 Watts of energy per second, 10 Joules. This is the rate at
which energy is being taken from the batteries, a very modest amount. The
battery is a storage for energy and when all that stored energy is used
up the bulb will stop emitting light. The energy of a car depends on its
speed, but it can go at many different speeds so if you go faster you have
more energy; but a car also has mass and its energy depends on its mass
also. But light has no mass and only one speed, so it's a whole different
can of worms. What you can get from the earlier answer is that that the
light doesn't need to be "speeded up" to 186,000 mps, it is already there
when emitted.
QUESTION:
ANSWER:
Since
your creatures are participating in your thought experiment, it would be
more illuminiating to let them be sentient creatures who also have the same
fantastic telescope model as yours. We have established that these
creatures existed 12 gigayears ago. So what you are seeing is the distant
past of this planet. I am bothered by your statement "…from
their perspective are they are being viewed by people from the future that
technically do not even exist yet?" They have no "perspective" because they
aren't viewing you at all, they would be viewing your world 12 billion
years ago in their world. If they thought about a bit they would say, "That
place we are looking at might be looking at us in 12,000,000,000 years in
their future if evolution allows them to be clever enough to make a
telescope as cool as ours is."
QUESTION:
ANSWER:
First, since a photon has no mass there is nothing to prevent it from
instantaneously going from zero to c . However, since a photon is
not a point particle, it will not be created instantaneously but rather
over the time of the decay of the atom to a lower state. This time depends
on the time it takes the excited atom to decay; the photon is "created" but
created already traveling c , (This time, the time for the
transition to occur, is not be confused with the half-life of the excited
state.)
QUESTION:
ANSWER:
As clearly stated on the site, I do not do
astronomy/astrophysics/cosmology. I think I can answer this question,
though. First, it is probably safe to guess that, over large volumes, the
photons move in random directions. Therefore, the net force on anything due
to photons would be expected to be near zero. Second, it is unimaginable to
me that any swarm of photons would exert sufficient force to move an entire
galaxy or massive black hole.
QUESTION:
ANSWER: There are three different constituents of our known universe:
Normal matter: This is the stuff we see and can
directly measure or observe. It consists of mass and force
fields. Modern physics describes, pretty well, the universe we observe.
Dark matter: In recent decades there have been
some observations which are not well-described by modern physics. To
account for these anamolies the presence of matter which which interacts
with the rest of the universe only via gravity has been hypothecized.
We have never directly observed dark matter and there is
virtually nothing known about it. It is estimated to be 27% of the
universe.
Dark energy: More recently it has been observed
that distant galaxies are accelerating away from us. The only way, at
least as described by modern physics, that this could be is that there
is an unknown repulsive force which is pushing the universe apart while
gravity alone would be pulling so as to eventually pull it all back
together. Like dark matter, we are mostly ingnorant of what this dark
energy is or where it comes from. It is estimated to be 68% of the
universe.
So the bottom line for modern physics is that it
describes, pretty well, 5% of our universe! It would therefore be pretty
foolish to say that the effectiveness of prayer is impossible; at least it
would not be described by physics. I personally do not believe in the
effectiveness of prayer, but I acknowledge that we do not know everything.
QUESTION:
ANSWER:
The meaning of translucent is obvious: light passes through. The meaning of
opaque, in physics at least, is no light passes through. Their meanings are
complete opposites.
QUESTION:
ANSWER:
It's quite easy: the force of gravity is holding it from being thrown
off. However, if it were spinning much faster the gravity would not be able
to hold the water down.
If you want to go into more detail, I can do an example of water at the
equator; Think of a small mass m of water on the surface of the
ocean. It has a weight of mg and a centrifugal force of mv 2 /R
where g =9.8
m/s2 is the accelation due to gravity, R= 6.4x106
m is the radius
of the earth, and v= 463 m/s is the speed of
m as it travels around the equator. Then the centrifugal force is
mv 2 /R= 0.034m kg·m/s2
whereas the weight is 9.8m kg·m/s2 . Comparing the two
forces, W /C =288. The force holding the water to the
surface is almost 300 times a great as the force trying to throw it off.
For water to be thrown off, the centrifugal force would have to be larger
than the weight; if you do the arithematic you will see that when the speed
v is about 17 times larger than it is today, W /C =1
and water would start coming off the surface. This would be when the length
of the day would be 17 times smaller, 1 day would be 24/17=1.4 hours long.
It is more complicated at latitudes other than the equator, but the same
ideas lead hold sway.
QUESTION:
ANSWER:
Heavy nuclei tend to be unstable the heavier they get because of the nature
of the forces which hold nuclei together. For example, a uranium nucleus
has many possible modes of decaying to lighter elements; one possible mode
of decay is spontaneous fission (SF) where an isolated uranium nucleus
spontaneously breaks into two much lighter nuclei and several neutrons. So
in this case, it takes zero energy input for fission to get energy out.
However, this is a rare decay mode; for example, one gram of 235 U
will have about 14 SF/day. So even though you get a lot of energy compared
to fuel "burned", the energy out is puny. You probably know that you can
induce fission by having a very slow neutron absorbed by the nucleus. So if
you flood a fissile material with slow neutrons you still put a relatively
small amount of energy in but get a comparatively large amount out.
Now consider nuclear fusion. You must push two nuclei close to each other
for the nuclear force to be felt. As long as the final nucleus is lighter
than iron, the result will be the release of much more energy per amount of
fuel "burned" than for fission. The key here is that in order to fuse two
positively-charged nuclei, you must bring them very close to each other.
But there is a repulsive electrical force between them which means that you
have to do a lot of work (energy input) just to get them close enough for
the nuclear force to kick in.
QUESTION: could emit one more particle, they
would reach the speed of light (after approx .98 years by my math, from
their perspective). However due to time dilation, that particle will never
actually be emitted... the observer could sit outside and watch the ship
for an infinite amount of time, and it will never emit that last
particle... implying that time, for those on board time has been
effectively paused due to intense time dilation, just as they've approached
the limit to their velocity, they've also approached the limit to the
amount of time they can experience.
ANSWER:
Here is the problem: acceleration is not at all a useful quantity when
special relativity is important. Why is it important in Newtonian
mechanics? Because no matter what your speed is, a force will cause the
same acceleration. Therefore, Newton's second law, F=ma , will be
true. That is not true if your speed is comparable to the speed of light. I
have answered several questions in the past with very long answers which
would help you understand this issue. Look at the faq page, in particular "
QUESTION:
ANSWER:
First, what are all the forces which act on the branch? There are only two,
its weight W W H 1 =H 2 and V 1 +V 2 =W .
I will just talk qualitatively now about rotational equilibrium (sum of all
torques about some axis=0) because I don't have any numbers given. Suppose
I choose my axis about which to calculate torques to be at the bottom where
green forces are drawn. About this axis W H 1
because all the other three have zero moment arms; this torque tends to
make the branch rotate clockwise and could therefore hold the branch in
rotational equilibrium. (Since H 1 has
a much smaller moment arm than W H 1 needs to be
quite a bit larger than W H 2 would exert a torque, still
opposing the torque due to W
When there is an ice storm, branches break off because the weight of the
branch gets bigger than the ability of the trunk to hold on. Once in
early summer a really big branch of a pecan tree in my back yard had nuts
growing on it; this gradually increased the weight of the branch resulting
in its breaking with a resounding bang in the middle of the night.
QUESTION:
ANSWER:
First, I want to clear up some nomenclature you used.
When you say that "the
gravitational pull of earth is 10 m/s/s", you should have said "the
acceleration due to gravity is approximately 10 m/s/s provided that the
accelerating mass is very small compared to the mass of the earth and that
the position of the mass is close to the surface of the earth. Similar
statments can be applied to Jupiter. When you say "gravitational pull" you
are suggesting that 10 m/s/s is a force (we think as a push or a pull as a
force); as I am sure you know, it is an acceleration.
Now , what causes acceleration? Force! So we must
find the force which each experiences. Newton's third law says that bothe
experience the same magnituce force but in opposite directions. Both forces
are pointed toward the other. Newton's universal law of gravity is F=GMm /r 2
where G is Newton's universal gravitational constant and
r is the distance between the centers of the two planets. Now Newton's
second law says that the acceleration is the force divided by the mass, so
a earth =GM /r 2 and a Jupiter =Gm /r 2 .
All this has nothing to do with relativity.
Incidentally, it is lucky that these are spheres
Also they must be spherically symmetric masses also which means the density
only can depend on the distance from the center; you can't have a lot more
mass in the northern hemisphere than the southern. Newton himself proved
that you may do this calculation (as I did) assuming all the mass of the
planet may be treated as if it is all in a point mass in the centers of the
planets.
QUESTION:
ANSWER:
I have shown just a part of the figure you attached. No, that orbit is
impossible unless you you burn
a "ridiculous amount of fuel"; the natural orbit of one object orbiting
another is (Kepler's first law) an ellipse with the object which is being
orbited (earth in your case) located at the focus of the ellipse as shown in my figure.
The figure shows the semimajor (a ) and semiminor (b ) axes
which would be necessary for the observatory to be close to earth at some
point in the orbit. Kepler's third law relates the period T (one
year for your case) of the orbit the a : T 2 =(4π 2 /(GM )a 3 ,
where M is the mass of the earth and G is Newton's
universal gravitational constant. It is important to note that every orbit
with semimajor axis a has the same period. It is instructive to
look at a circular orbit of radius a; I find that if T is one
year, a =2.5x1010 m. It is easier to visualize if we
express this distance in earth radii, RE =6.4x106
m and therefore a =3900 RE , almost four
thousand earth radii;
for comparison, the distance to the moon is about 60 RE .
Unfortunately, the velocity at the closest distance to earth is the fastest
anywhere on the orbit, the slowest is when the observatory is farthest away
(Kepler's second law), just the opposite of what you need to dock
conveniently. The animation below shows this. The red line is how the
radius in a circular orbit would move; note that both the red and blue
lines take the same time to go around once.
QUESTION:
ANSWER:
a
boat moving east at 5m/s…" means exactly what it says: in spite
of the north-moving current of 3 m/s, the boat is moving east at 5
m/s. So 5 east is not what the boat would have in still water as you have
assumed. In some sense this is a trick question because what is given is
V BG which is the boat with respect to
the ground which is given as 5 m/s east, so of course the boat will go 10 m
in 2 s. We might as well go on and find out what we might want to know, how
would the boat be moving in still water so that it would end up going east
with a speed of 5 m/s. We have to know the velocity addition theorem which
is V BG =V BW +V WG .
Here, V V BW =V BG -V WG .
The first figure shows what you are given, the velocities of the boat and
water as seen from the ground. The second figure shows the application of
the velocity addition theorem applied to find what the velocity of this
boat has relative to the water and this is how the boat would move if there
were no current.
QUESTION:
ANSWER:
First, site ground rules specify that questions should not contain
physically impossible conditions, viz . "indestructible bar". But the answer
to your question doesn't doesn't depend on the indestructibility of
the bar. Second you specify how fast something something is rotation by an
angular velocity, e.g ., RPM (revolutions per minute); in physics
we usually specify angular velocity as radians/second and denote it as ω.
This question has been
answered before .
QUESTION:
ANSWER:
I
don't understand what you mean by 'temperature differences". We don't think
of there being a temperature you can assign to a single particle, rather
temperature is a statistical property describing the average kinetic energy
of a very large number of particles. I can tell you that a positron and an
electron have very similar (practically identical tracks in a cloud chamber
if they have equal kinetic energies; they have equal masses and equal
magnitudes of charges. The only thing different is that they have opposite
charges. The way you can tell them apart is by making them pass through a
magnetic field. If you see one track curve left, the other will curve
right. You should, of course, have access to both β + β -
QUESTION:
ANSWER:
I found, when teaching, that students have considerable difficulty
understanding what potential is, they are more comfortable thinking about
potential energy and, even better, work. Before the concentric shell is
brought in, how much work does it take it take to move a charge q
from infinity to the surface of the sphere? If the total charge on the
sphere is Q , the force on q when it is a distance r
from the center is kqQ /r 2 ;
the total work is -kqQ /R . If q and Q
are of opposite sign, the work you do is positive since you have to push
the charge in the same direction it is moving. Now, put the spherical
shell, inner and outer radii R i and Ro . The
total charge on the sphere is Q and the total charge on the shell is zero.
But the field from the sphere at r =R i will
cause an induced surface charge on inside of the shell but also the
negative of that surface charge on the outer surface. It is important here
to note that all the charge on a conductor resides only on the surfaces,
there is zero field inside a conductor. So the field seen by q ,
which you are imagining pushing in, is exactly the same for ∞<r <R o
and R<r<R i as it was before the shell was
brought in. But when R i <r<R o , there
is no force and therefore no work is done. So the total work done is
different from that done before the shell was added by the amount
±|(1/R o )-(1/R i )|
You could say all this by just saying that because there is zero field
inside any conductor there is no work done on q in the conductor which
means the electric potential inside the shell will be different from what
it was without the shell.
QUESTION:
Is it practically possible to build such a detector? Would it sense the electric field of one single electron? If so it would need to be extremely sensitive. Also, it would be impossible to sense the electric field without affecting the electron's path, wouldn't it? Which would ruin the experiment.
From the electron's point of view, what is the difference between the detector and the slit arrangement? Doesn't the electron become entangled with the slit setup as well? Thus the interference pattern wouldn't appear even without the detector. Yet it does, why?
ANSWER:
I was going to begin with a statement about how this experiment could not
be done as described because the wavelengths of the electrons are too small
at reasonable speeds and the slits would have to be much narrower and
closely placed that we would not have the ability to fabricate them. But
Richard Feynman, around 1960, said it much better than I could: “We choose
to examine a phenomenon which is impossible, absolutely impossible,
to explain in any classical way, and which has in it the heart of quantum
mechanics. In reality, it contains the only mystery. We should say right
away that you should not try to set up this experiment. This experiment has
never been done in just this way. The trouble is that the apparatus would
have to be made on an impossibly small scale to show the effects we are
interested in. We are doing a 'thought experiment', which we have chosen
because it is easy to think about. We know the results that would be
obtained because there are many experiments that have been done, in
which the scale and the proportions have been chosen to show the effects we
shall describe”. But
Davisson
and Germer , around 1927, showed interference of electrons from
scattering from the surface of single crystal nickel where the atoms played
the role of the slits and were separated by a small enough distance to deal
with the small wavelengths of the electrons. But, the ability to actually
fabricate extremely narrow and close together slits had already been done
and Feynman was apparently unaware. It is done by milling gold plated on a
very thin substrate of silicon e.g ., with an atomic beam.
Now,
I have no idea what "the
detector and the electron’s wave function become entangled" means.
I believe that is much easier to say that if a detector detects it, it
is all there , i.e ., it is behaving like a particle,
not a wave. In order for the interference pattern to happen, the electron
must interfere with itself, i.e ., pass through both slits,
something only a wave can do. Regarding whether or not it is possible to
actually have such a detector, I would say that I am quite confident that
is would be straight forward to detect the presence of a single electron
QUESTION:
ANSWER:
Nothing will happen. There is no temperature difference so there will be no
heat flow. The ice cube will not melt, the water will not freeze.
QUESTION:
ANSWER:
This question has a number of possible answers, it is not unabiguous. If
you mean the best estimate of the observable mass of the universe, it is
about 1.5x1053 kg. Taking the volume to be about 5x10-4
m3 ,
the density would be about ρ =3x1058 kg/m3 .
There is estimated to be about 5 times more dark matter than normal matter
in the universe, so that would increase the density to 1.8x1059
kg/m3 . (Of course, dark matter has never been directly measured,
just inferred.)
You can ask
how much energy it takes to assemble a sphere of radius R and mass
M in classical mechanics. Because the force is gravitational, it
would assemble itself because gravitation is attractive! But we should
really ask how much work must be done such that each piece of mass we bring
in arrives with zero kinetic energy; in that case the energy you supply is
negative because the force you would have to exert is opposite the force of
the gravity. E =-(3/5)GM 2 /R where
G =6.67x10-11 m3 /(kg·s). You could go
ahead and calculate this energy but it would be a purely academic exercise
because matter can not exist at such high density. That's why black holes
exist: when you get huge amounts of mass, like the remnants of a star which
has burnt all its fuel, it will collapse into a neutron star or a black
hole. You may be sure that this much mass in this small a volume would have
collapsed into the most massive black hole anyone has ever seen. So
technically the answer to your question is when you try to make your
"softball" it will collapse long before it has all been assembled and the density will ever thereafter be
infinite; the energy to do this will be zero.
Finally I should remind you that my site ground rules state that I do not
do astronomy/astrophysics/cosmology and I usually discard such questions. I
thought this one would be fun to play with. You may choose to believe that
everything I have said is BS!
QUESTION:
ANSWER:
This question has previously been answered . No
information can be transmitted using entanglement.
QUESTION:
ANSWER:
Your problem can be understood from your sentence "exactly
what mass is being converted into energy?" You are thinking that mass is
something which gets converted into energy , but the fact is that mass is
energy . I sometimes find that these kind of issues can be
understood best from a simple gedanken (German for a thought
experiment). Let's consider a deuteron. If you measure its mass you will
find (as you seem to already know) it has a mass less than the sum of the
masses of a proton+neutron. Now, these are bound together by a force or
they wouldn't stay together. Now you wish to pull them apart; what do you
have to do? You have to pull them apart, and this is positive work because
you are pulling in the same direction the particles are moving. If you are
doing positive work on the system that means that the energy of the system
is increasing. Finally they are far enough apart that they no longer feel
the force which formerly bound them, which need not be very far apart
because the nuclear force is very short-ranged. (One micron might as well
be infinity!) So now you weigh them and find that their masses are now to
the masses these particles have when free, i.e ., more than when
they were bound. Your question: where did that mass come from? My answer:
you gave it to them by doing positive work. Now, suppose you want to put
them back together. So you push them together, but as soon as you get them
close enough to feel the force you will have to hold them back, that is,
doing negative work. The result is that the deuteron is going to have less
mass once it is bound because you took it away by doing negative work. So
bound particles in general do not have the same mass as free particles. And
in a heavy nucleus, you can't say "oh, that proton over there must have
less mass than it would outside"; what you can say is "that total
collection of protons and neutrons has less mass than they would if they
were all unbound."
QUESTION:
A 1kg sphere rolls at 1 m/sec along the X-axis. The center of mass of the
sphere will have a fixed y value, but will pass through an infinite number
of x values each second. However briefly, the ball will have each of these
(infinite) x-coordinates. If the ball is traveling at 2m/sec in the
X-direction, it will be 'located' at each of the infinite X-coordinates for
half as long as when it was traveling at 1 m/sec. How should I think of
this motion? Simple limits (as t > 0)? If so, I would apply the same limit
rational for both the 1m/sec ball and 2m/sec ball?
ANSWER:
You are messing around with infinity. The mathematics of this number is not
the same as usual numbers. You have stumbled on one quirky property when
the you conclude that the time per point is half what it is when you double
the speed. And think of this pecularity: there are an infinite number of
'points' in an inch but there are also an infinite number in a mile. The
simple way to answer your question is that a point spends zero time on each
of an infinite number; so if the path is one meter long the time spent will
be 1 s=(0 s)x(∞ points) which is a perfectly fine equation because 0x∞
can be anything. Students often think that it is meaningful to ask what the
probability of finding a point particle at some specific point in space at
some particular time. In fact, it is zero. To get what you want you need to
do a little calculus: consider an infinitesimal volume dV , apply
the physics, take the limit as dV → 0;
this will give you a function showing the general behavior of probability
as a function of location and time, but it will not give you what the
probability is of finding the particle at any single point. You get a
probability density as a function of time and you can find the probability
of the particle being in a volume which you can make arbitrarily small.
QUESTION:
I have a question related to time dilation. I have a good idea on how it
works and its principles, but I could not find any information regarding
the question. Here it is:Since the closer you go to the speed of light,
time passes more slowly for you, and quicker for the people you're
travelling away from. What is (if there is one) the fastest speed a
human could travel through space and come back without noticing any
change?
ANSWER: Actually, time passes perfectly normally for you in your frame of
reference. I suggest that you read my answer on the twin paradox.
Your question is qualitative and therefore unanswerable. The giveaway is
the phrase "noticing any change"; change in what? And the change will
depend on how far he goes. The satellites which support the GPS system have
speeds of 8000 m/s which is about 0.00003 of the speed of light. Is that
noticeable? It surely would be if you are trying to get somewhere and
the system did not correct for time dialation. Another consideration, even
if you defined what noticeable is, is that an 80 year old person could look
60 or 90 but, relative to more reliable clocks he carried along, he has
actually aged 80 years.
QUESTION:
ANSWER:
Momentum is not conserved because there was an external force (whatever
caused the bike to stop). The man will probably not have a speed of 10 m/s
because he will probably have touched the bike as he moved forward and the
friction between him and the bike will slow him down a bit. Still, ignoring
the man-bike detail, inertia is the approximately right answer.
It is always important to know the conditions under which conservation laws
are obeyed. Energy of a system is conserved if there are no external forces
doing work on that system; linear momentum of a system is conserved if
there are no external forces on that system. Angular momentum is conserved
if there are no external torques. Rotational energy is conserved if there
are no external torques doing work.
QUESTION:
ANSWER:
The figure shows the simplest block and tackle; once you understand this
you can just keep adding more pulleys. You do not need to know much physics
to understand this, just Newton's first law: If a body is at rest, the sum
of all forces on that body must be zero, called equilibrium. So I assume
that W W down from the earth (gravity) and upward forces from
strings 1 and 2; the left hand pulley has an upward force from the ceiling
and downward forces from strings 2 and 3; so string 3 must have an upward
force from the second pully and a downward force from your hand.
The right hand pulley plus W must have T 1 +T 2 -W =0*
which makes sense because the strings are the only things holding it up. If
1 and 2 were different there would be an unbalanced torque on the pulley
and it would start rotating. So T 1 =T 2 =T =½W .
Similarly, T 3 =T 2 =T =½W.
Since T 3 =½W , you must be pulling down with a
force of ½W , half the force you would have to exert if you lifted
it unaided.
You might ask "What is the price of this method lifting?" You have to pull
twice as far as you would just lifting it (in the example above, two
pulleys). (My dad would say "There is no free lunch". A physicist would say
"The energy used to lift a weight to some height is the same regardless of
how you do it.")
*Called the tension in a string. For a light, unstretchable string, this is
normally the same throughout.
QUESTION:
ANSWER:
Any number of particles may be entangled, each must share quantum states
with some or all of the other particles. If you made a measurement which
put one particle into one state, the other two would lose that state but
remain entangled sharing the other two possible states, but in a
complicated way.
QUESTION:
ANSWER:
Sorry to disappoint you, but there is no real answer to your question; this
is just the way nature is. The way our universe is constructed is that
objects like particles are 'aware' of other particles because of the forces
they feel and exert on each other. Without any forces the universe would be
very boring, just many objects none having any 'awareness' of the other.
Electric charge is a property that some particles have which lets the
interact with any other particle which has that same property. There are
two kinds of electric charge because there are two ways the forces can be
seen, either attractive or repulsive. Consider the sun and the earth.
Obviously the earth feels the presence of the sun or else it would not
revolve around it. The reason is that they each have mass and masses
attract each other but never repel; therefore there is only one kind mass.
But it is pointless to ask 'why' do some objects have mass? There are other
kinds of "charge" that quarks have which have been named "color", as well
as mass and electric charge.
QUESTION:
ANSWER:
Nothing happens because there is nothing moving the speed of light or
faster. See a recent answer for more explanation.
QUESTION:
ANSWER:
I think a brief discussion of angular momentum is appropriate here. Angular
momentum is the product of the angular velocity (the rate at which an
object is rotating about its rotation axis, e.g ., RPM) and the
moment of inertia (MOI) of the body. The MOI is a little trickier to
understand, but it is basically a measure of how hard it is to get the
object to start or stop rotating about the axis. If you think about it, the
closer to the axis the object's mass is, the easier it will be to change
its angular velocity. Now, it turns out that if there are no external
forces which exert a torque on the spinning object, its angular momentum
will never change; this is called conservation of angular momentum. So when
the skater pulls in her arms, her MOI gets smaller and, if you ignore
friction, her angular velocity must get bigger to keep the angular momentum
constant. If she doesn't pull in her arms she will slow down and eventually
stop because the friction between her skates and the ice will be exerting a
torque slowing her down.
The kids on the roundabout is a classic problem. The roundabout has a MOI
of I 1 ; all the kids have a collective MOI I 2 .
At the beginning the kids are all standing still on the roundabout, i.e .,
they all have the same angular velocity of the roundabout. Of course, you
can always choose the system you want to focus on as whatever you like.
Suppose we choose the roundabout+kids as the system such that the system
has a MOI of I=I 1 +I 2 . Then if the
kids move toward the axis I gets smaller so the whold system
speeds up just like the skater. Suppose that we choose the roundabout alone
as the system. We already know that when the kids go into the middle the
roundabout it speeds up, so its angular momentum changes. Why is angular
momentum not conserved? Evidently, the kids who are external to the system
have exerted torques which increase the angular momentum of the roundabout.
QUESTION:
ANSWER:
First of all, you realize that this is a very academic problem. It is
normally assumed that there is no friction (nobody is braking) and that all
collisions are perfectly inelastic (stuck together after the collisions).
In addition to these, I will assume that all cars have the same mass M ,
the oncoming car has a velocity V 1 , and that the
spacing between cars is D . The important physical principle here
is conservation of linear momentum (P , mass times velocity), the
momentum is the same after a collision. Before the first collision the
momentum is P 1 =MV 1 and after it is
P 2 =2MV 2 ; since P 1 =P 2 ,
V 2 =V 1 /2. Now note that, since there
is assumed to be no friction, the speed of the two cars will remain V 2 =V 1 /2
until they hit the next car, regardless of what D is. For the second
collision P 3 =P 2 =MV 1 =3MV 3 .
so V 3 =V 1 /3. You can probably see
here that the final collision will have V 4 =V 1 /4.
The conclusion is that, under these conditions , it makes
absolutely no difference what D is. It is instructive to look what
happened to the energy as a result of these collisions. The kinetic energy
K of an object with mass M and speed V is K= ½MV 2 .
The incoming energy was K 1 =½MV 1 2
and the final energy was K 4 =½(4M )(V 1 /4)2
=MV 1 2 /8, so the change of kinetic
energy is ΔK =K 4 -K 1 =-3K 1 /4,
3/4 of the incident energy was lost. Where did this energy go? It was the
energy used to do the damage to the cars.
Now suppose that there is friction in the problem. Now the cars lose speed
between collisions. The collisions themselves will still conserve momentum
but the lost energy will be larger because friction takes away energy away.
It would be fun to do this analysis, but tedious; but now we can understand
qualitatively. The energy lost in the second and third collisions will be
smaller than without friction because the friction slows the cars down
between collisions. Therefore the energy lost via collisions will
be less, hence less damage to the cars. If D is large enough, only
the front of car 1 and the back of car 2 will be damaged because friction
would stop them before they got to car 3. The larger D , the
smaller the damage. So as you can see, what you have always thought is just
opposite of what would happen.
[This reminds me of a good-driver rule which lots of folks don't know. If
you are waiting for oncoming traffic to clear so you can make a left turn,
never turn your wheels while you are waiting. If someone should hit you
from behind it would push you right into that oncoming traffic.]
QUESTION:
ANSWER:
Light, photons, are created by atoms, molecules, and atomic nuclei. This
creation decreases the energy of the source, giving it to the light. Once
light is created, it needs no force to keep it going, it will fly through
empty space forever at the constant speed of 300,000,000 meters per second
until it collides with something (like your eye).
QUESTION:
ANSWER:
Yes, it's the photon. It turns out that electricity and magnetisim are
different parts of a single force, electromagnetism, and the mediator of
the fields is the photon.
FOLLOWUP QUESTION:
ANSWER:
I guess that you could make that
kind of argument about any theory, that it is just some mathematics which
happens to describe everything it describes perfectly. It's hard to refute
something like that. All I can tell you is that electromagnetism is the
most fully developed theory in physics. Mostly in the 19th century a huge
number of experiments and theoretical proposals on electricity and
magnetism were performed. Classical E&M theories summarized by Maxwell
almost completely described this huge bank of data and hypotheses. A few
anomalies were described very well in terms the quantum physics being
developed in the early 20th century. This was not fully developed until the
middle of the 20th century, QED (quantum electrodynamics), by quantizing
the electromagnetic fields. QED is where the virtual photons come into
play; I admit that it sounds a little shaky, like a fairy tale or
something, but this kind of language is only a cartoon of the underlying
mathematics to appeal to make it more imaginable. Physicists believe that
we have a totally correct description of electromagnetism. Philosophers
often wonder about the connection between mathematical theories and
physical "reality"; that is a perfectly valid question to ask. Physicists
are less interested in such arguments. We are interested in if our theories
give "perfect" descriptions of nature, of all relevant data we have
observed.
It occurs to me, why don't you have similar questions about electric forces
or gravity? Do you think of these as "magical"? They are forces too, and
just compare the gravitational force the earth exerts on you, probably much
bigger than your puny magnetic force.
I do respect your scepticism, a sign of an active mind pondering deep
questions.
QUESTION:
ANSWER:
Well, all we know about particle physics says that there will not be one
single particle that builds everything. For example, we know that there is
a thing called electric charge which comes in two varities, plus and minus.
Your building block would have to be neutral so charge would not exist
unless you broke it in half, making it not the most basic particle. There
are two kinds of particles, fermions and bosons; again, how could a single
particle contain them both. Also the electron is believed to have no
internal structure, i.e ., it is already a fundamental particle but you
certainly build other known particles from just electrons.
QUESTION:
ANSWER:
I have personally baked scores of loaves of bread in my life. Often I would
bake four loaves at once. I had three black loaf pans and one cream colored
pan. I never noticed any differences among the pans and never burned the
crust in any of them. Follow the baking instructions in the recipe and you
can't go wrong. (Ovens vary, so you need to get to know whether you should
maybe use a slightly higher or lower temperature, slightly longer or
shorter times. )
FOLLOWUP QUESTION:
ANSWER: Okay, then, I have failed with my "anecdotal" answer! I shall dig a little
deeper. Let's start with what you call "thermal heat" by which I presume you mean
heat transfered between surfaces, C at temperature T C and
B
at temperature T B , a distance L apart in some
particular medium, during a time t. The energy Q transmitted
through an area A is proportional T B -T C =ΔT ,
A , and t , and inversely
proportion to L . Q =K ΔTAt /L
where A is the area through which the energy passes, and
t is the time elapsed; K is the proportionality constant and
is called the thermal conductivity and depends on what the medium is.
Radiation is much more complicated because the emission or absorption
depend on the wavelength of the radiation. Qualitatively, though you can
say that the heat (energy absorbed or emitteded) is proportional to the
area A of the transmitter and to the fourth power of its temperature, Q /t=εσAT 4
where σ is the Stefan-Boltzmann constant and ε(λ) is the emissivity of the radiator/absorber.
If you can see anything which would justfy "thermal heat behaves as radiant
heat does" in my descriptions of the two above, you're a better man than I!
It turns out that professional bakers do find differences between dark and
light bakeware. A nice
article from King Arthur Kitchens discusses this, although for cakes
rather than bread.
There, is that better?!
lQUESTION:
ANSWER:
What is a "insane laser"?! This is not a stupid question because we have
had it ingrained into our brains that nothing can have a velocity greater
than the speed of light. But that beam is not something, it is a sequence
of things (photons, say) at very short time intervals of hitting hitting
the ground; but there is nothing actually moving in the same direction as
the spot. You could not use such a phenomenon to send information from one
side of the earth to the other. (It would be easier if you took took your
insane laser outside and swept the beam across the moon!)
QUESTION:
ANSWER:
A photon is not a wave or a
particle, a photon is a wave and a
particle. This is called wave-particle duality. Suppose that some
measurement, such as having determined that it has just passed through a
slit, has "determined" it be a particle. If you now do another measurement
looking for a wave you will determine that it is a wave.
QUESTION:
ANSWER: So It looks like AI got you the right answer for the ground speed. But,
since I am neither a pilot nor familiar with some of the jargon, I have
started from scratch to see what the ground speed would be. As I understand
it, airspeed is the speed relative to the air. The first figure shows the
velocity vector, magnitude 80 knots, in still air; I have shifted to a
frame of reference where the wind is zero, i.e ., moving west
(negative x -direction) with speed 20 knots. Now, to see what is
seen from the ground I transform back to the ground frame. The effect is
that the wind has only an x-component in, w x =-20, as
shown in the second figure. As shown,
v is v x =v cosθ =80cosθ -20.
The y -component is
unaffected by the wind, so vy =- 80sinθ . So the
ground speed is v =√[(80cosθ -20)2 +(- 80sinθ )2 ].
A little algebra yields v =√(6800-3200cosθ ).
This is plotted in the graph. This is what you would expect: at θ =0,
v θ =180,
v θ =90 or 270,
v
So everything I have done so far has been something which you already knew
from your interaction with ChatGPT, but I had to be sure that I understood
for myself what the groundspeed is. The speed is constantly changing, so
the debate is simply: what is causing the acceleration ? Who's
right or partially right? The folks that say it isn't the wind because "wind
isn't a force" are mostly wrong although the wind isn't a force; but the
wind CAN EXERT A FORCE. This force, usually call air drag, is what is
solely responsible for the changes in velocity as seen from the ground.
Imagine that the pilot has some kind of autopilot which will cause the
plane to fly in a circle with constant speed in still air; if he has such a thing it is
set—he cannot change the throttle, flaps, rudder, anything—because
that would not be consistent with the constraints of this problem. That
drag force is continually being exerted on the plane in the negative x
direction. The winners of the dispute are those that said it is just the
wind causing speed
Those who mistrust AI to do physics problems must now "eat crow!"
After I, a human physicist, had worked through the problem, I asked AI to
do it and he came up with the same answer I did.
ADDED THOUGHT:
Here is a picture showing the path of the
airplane as seen from the ground for five revolutions; west is to the left.
By the way, I had AI integrate the velocity components for me to get the
parametric equations for x and y to plot.
QUESTION:
ANSWER:
Physics is, at its heart, an experimental science. So I suggest we
determine the speed of the air as a function of our distance from it.
First, stand right in front of the fan and see how much speed the air has
by the way it feels, hitting you hard or softly? Then go to the other side
of the room. The air is hitting you much more softly and is therefore
moving more slowly. As is almost always the case with aerodynamics, it is
complicated and best done with supercomputers. A rough understanding can be
achieved by noting that the air spreads out as it gets farther from the
fan. To illustrate this, I will imagine the air flowing through a narrow
pipe with some speed and then coupled to a larger pipe. Note that there
will be more molecules in a given length of the larger pipe. Because the
total number of molecules per second leaving each of the two pipes must be
the same, those leaving the larger pipe must be going more slowly.
QUESTION:
ANSWER: At
first I was going to send my usual reply to questions like yours: "As stated on the site, I do not do astronomy/astrophysics/cosmology."
But it interested me so I asked an astronomer friend of mine. It turns
out that this is a question to which the answer is not
known
certainly. One hypothesis is that, because the universe is expanding at an
accelerating rate, the energy of the universe is not conserved but
increasing; the lost energy of the gravitational redshift is just absorbed
by the whole universe. But the most accurate answer is simply that we don't
know.
QUESTION:
ANSWER:
The reason AI is giving you two different answers is that your question
makes no sense. The word 'matter' really is not a physical quantity, just
qualitative. In some sense energy is everything*. Any object which has mass
is energy, you know, E=mc 2 ! Your 'definition' of what
'matter' is is news to me and saying that "particles
do not in fact take up any space" is wrong (maybe not electrons or
photons). At any rate, whoever says that particles are energy is
correct.
*I am being a little glib here. Since we have no idea what dark matter and
dark energy are and how they might change our view of how the universe
works, nobody is really certain of some
thing involving energy
except
locally.
QUESTION: 2 came to mind. (I have a very vivid 3D visualization kind of mind, long story.)
Then, my thoughts showed me the following, in relation to E=MC2 2 3· C7
ANSWER: Short answer: meaningless gobbledeygook! Here is a longer
answer. Whenever one has an equation and you want to know if it is really
an equation you can check the dimensions which must be the same on both
sides. By dimensions I mean the three fundamental quantities of mechanics,
mass (M), length (L), and time (T). The left side of your equation is
energy squared.You know that E=Mc 2 so E 2 =M 2 c 4 ;
mass is just mass so its dimension is M2 , c is the
speed of light and a velocity is length/time, so c 4 has
the dimensions L4 /T4 . so the left side of your
equation has dimensions
M2 L4 /T4 . The dimensions of the other side
of your equation, by the same arguments, are M3 L7 /T7 ,
clearly different from the other side. Therefore, this is not an equation
at all because of what we mean by equal, the same in every regard.
QUESTION:
ANSWER: I have
previously answered this
question.
QUESTION:
ANSWER: I do
not understand your first sentence. Conservation of energy says that the
total amount of energy in a closed system is a constant. Taking the
universe as a closed system, that means the total amount of energy of the
universe is constant. When your body is no longer alive, it will undergo
changes. Most of those changes will be due to chemical reactions which
occur in decay and are exothermic (the reaction heats its environment). But
any energy which that heat carries results in the same amount of energy
decreasing in your remains. Far and away, most of the energy your body your
body contains is mass energy, E=mc 2 , but that remains.
However if you carefully measured the mass before and after your total
decay (not forgetting the mass of mainly soft tissue which ends up as
different molecules which have escaped into the environment) you would find
that it weighs just a tiny bit less than before you decayed; this tiny bit
of mass energy would be precisely equal to all the heat energy which
escaped. A short answer to your question: yes. One proviso: you say
"forever"; nobody knows the untimate fate of the universe, so I only know
about how things are going right now.
QUESTION:
ANSWER: If you
have any two quarks of different colors, the force is both attractive and
repulsive. There is an attractive component which is necessary for the
quarks to bind and there is a repulsive component to keep them from
collapsing into each other. Two quarks with the same color repel each
other.
The repulsive force is very large and very short-range
and
is evident in the nucleon-nucleon force as the "hard core" part of the
force. For quarks at large distances, the farther apart you push them the
stronger the attraction becomes. That is why you can never observe a single
quark.
QUESTION:
ANSWER: The easiest way to think about entangled particles is that they are both parts of the same system. The particle states are correlated such that the system all changes together. That, of course, is Einstein's
"spooky action at a distance". So rather than thinking of the second particle
changing its state when the other is measured because it somehow detects
this change, it is simply doing what it is required to do because the
two particles are really one system. Because the changes are apparently
instantaneously happening, there is no need to worry about the clocks of
either of them. This is why they are called entangled—they are
really a single quantum system. Indeed, two entangled particles are
entangled regardless of their positions and frames of reference. An
observer in any frame of reference will see the change of quantum states to
be instantaneous and if that is true, clocks are irrelevant—instaneous
is instantaneous.
QUESTION:
ANSWER: The
equation to calculate the angle of deflection θ is θ =4GM /(rc 2 ). I get θ =(1.7x10-9 )°.
Something tells me that you will have no chance of observing a nanodegree!
Don't even bother messing with the percentage.
QUESTION:
ANSWER: What
matters in quantum mechanics is what the 'system' is. Once you have
specified the system, you need to solve Schrödinger's equation for the
system. The wave function for an oxygen with 5 electrons stripped off is
different than the wave function for an un-ionized oxygen atom; therefore,
the answer to your question is no, changing what the 'system' is will
change what the wave function is. Along these same lines, the energy
spectrum of the O5+ will be different than un-ionized oxygen.
QUESTION:
ANSWER: The
time you are interested in is the time in your frame of reference. In your
example of the "universe moving at a million mph", it has no relevance to
whether the laws of physics are true in your frame. (Actually making that
statement is not meaningful because what is the whole universe moving
relative to?) You seem to think that in your moving frame something would
change if you accelerated to a higher speed. Once you got to that speed and
turned off your engines, there would be no experiment which you could
perform to detect that change. There is no such thing as "your absolute
velocity". All times in frames relative to yours would depend on their
velocities relative to you; your time would, for you just tick away like it
always has. I guess you have not studied special relativity which reveals
to us that there is no such thing as "absolute time", there is no god
somewhere with the master clock which everyone compares to.
QUESTION:
ANSWER: I
don't know about your 2% number but I do know that an accelerating
charge radiates photons; this is called
bremsstrahlung
QUESTION:
ANSWER:: I have
redrawn the picture you sent, replacing your thread by a fulcrum which does
not matter. My notation::
being measured.
g is the acceleration due to gravity.
D is the length or the beam.
f is the fraction of the length of the beam on the left of the fulcrum.
μ is the mass density of the beam, mass per unit length, so
μD is the mass of the beam.
θ is the angle
which the beam makes with the horizontal.
Let's sum all torques
about the fulcrum,
MgfD cosθ +(μgfD 2 /2)cosθ -(μg (1-f )D 2 /2)cosθ =0.
This is a scale, so we
want M as a function of θ. Solving the above
equation for M I find M=μD [1-(1/(2f ))]. Note
that M is not a function of the angle, so your scale will not work.
Once you have found the right f for the given M , it will
be in equilibrium regardless of the angle. If you want to learn more about
balances and their designs, there is an earlier answer to a question
which you might like to
link to.
QUESTION:
ANSWER: I
edited your question to take out a lot of extraneous rambling. You clearly
do not understand how the pickup works, but you are in good company—when
I was googling I found that easily half the explanations about the physics
were wrong or so vague as to provide no explanation. To understand what is
going on, you need to understand magnetic induction of electric current,
first discovered by Michael Faraday. In a nutshell, if you have a changing
magnetic field you will induce electric currents in nearby conductors;
also, because of relativity, if the nearby conductors are moving and the
magnetic field is constant, you will still find that currents are induced
in the conductor (as what is happening with your quarter in the MRI). Of
course, if you have currents they will have their own magnetic fields; this
plays an important role in the guitar pickup too. So let's look at what is
going on:
There
is a magnet for each string and each magnet is a small bar magnet;
there is a coil of wire (hundreds/span > or
thousands of turns) and the two ends of the wire will be accesible to be
the output of the coil to amplify.
of course, there is a
string, which is which is fabricated from a conducting metal;
if the string is not
vibrating there are no currents anywhere because the magnetic field is
static, not changing.
now the string is plucked
so the string vibrates so it is moving through the unmoving field of its
magnet;
because the string is
a conductor there will be an electric current induced in it and it will
be varying with the same frequency as the vibration;
now we know that
electric currents give rise to magnetic fields so the vibrating string
will have its own magnetic field;
this induced field will be seen by the coil;
since the current will
be constantly changing, its magnetic field will be changing;
this changing field
will now be passing through the coil and there will therefore be an
induced current, same frequency as the string;
the current will be passed through a load
causing a voltage which can be amplified
electronically and sent to the speakers.
QUESTION:
ANSWER: The
atmosphere moves, for the most part with the rotating earth. The airplane
flies relative to the air, so it's like paddling a canoe downstream in a
river; if you don't paddle, you move along with the river but if you do
paddle you move faster than the water. Also, the air does move at different
speeds, it's not just a static blanket of air; winds blow depending on the
conditions of the air (weather). The most permanent high-speed (more than
100 mph) winds are called the jet stream. Flying across North America or
to/from Europe to North America, because jet planes fly at the altitudes
(30,000-40,000 feet) where the jet stream is, the flying time is almost
always shorter going W to E than E to W because the
jet stream mainly goes to the east.
QUESTION:
ANSWER: Most
people, like you, have no concept about how huge the energy of macroscopic
objects going close to light speed are. I have answered questions like this
before which might interest you (See
faq.html#faq game scifi guns ). Apart from the vacuum inside the LHC,
there is no reason for it to be in this question. Let's just say that two
bearings in a vacuum, each of mass m =10-2 kg and moving
with a speed of 90% the speed of light. Each has energy E =[10-2 /√(1-0.92 )](3x108 )2 =2x1015
J. The Nagasaki bomb had a total energy of about 1014 J, so each
of these bearings would have about 20 WWII era atomic bombs. If you took
2x1015 J and delivered it over the period of one day, it would
be equivalent to a power of P=E /t =2x1015 /(3600x24)=23x109
W=23 GigaWatts; that's about 6 times larger than the biggest power station
in the US. You may assume that the "structural
integrity of the ball bearings would remain stable for the duration of the
experiment" but there is no way that could possibly happen if there is this
much energy. We are talking about 4x1015 Joules of energy which
is the energy of a typical H bomb.
QUESTION:
ANSWER: So, by best resolution you probably mean the narrowest
width possible. And, it seems that you are having trouble even seeing the light on the
screen because the light line is so narrow that it has insufficient light to be seen well. You are barking up the wrong tree, I believe. There are actually two things you need to understand:
T
he
width for geometric optics. If the slit is much bigger than the
wavelength of the light, the image is simply identical to the slit as
shown in the figure. The angle can be written as θ1 =s /R =d / R
Eventually the slit will become comparable to the
wavelength. Significant broadening of the light passing through the slit
will occur due to diffraction. The single slit diffraction pattern looks
like the next figure:2 =λ /d .
where λ is the wavelength of the light. Finally I show a graph of θ2 (red), θ1 (black), and their sum (blue). The narrowest possible
image can be shown, by finding
the minimum of the blue curve, to be θbest =√(λR ).
If you are interested you can read a beautiful
example of an application of all this, the eye of a bee, in Section 36-4 of
the Feynman
Lectures . This will also give you more detail filling in my "it can be
shown"s.
I suspect that you will not be able, with your
apparatus, to get out of the geometrical optics region. But from your
question I suspect that you will need to make a compromise because you
probably don't have enough intensity to do your measurements.
QUESTION:
ANSWER: You
don't define what an "instant" is. It usually means "in zero time" as in
instantly. The answer, then, is infinite energy. It is physically
impossible for anything to move through some distance instantly.
QUESTION:
ANSWER: The
speed of light in a vacuum has the same regardless of what its wavelength
is. Most experiments would measure the speed in air, not a vacuum, but air
has an index of refraction extremely close to 1; therefore the answer to
your first two questions is, for all intents and purposes, no. Your third
question has no meaning ("…completely red light environment…")
QUESTION:
ANSWER: "…consider
bar magnets…"? Bar magnets are not electric charges* . You need to
consider two electric charges, q and Q. Imagine Q
to be held fixed. The work done to move q from r 0
to r is W=kqQ [(1/r 0 )-(1/r )];
if r 0 =∞, W=-kqQ /r. The
potential energy U is defined as U (r )=-W =kqQ /r.
Note that I have chosen U ( r 0 =∞)=0.
So, if q and Q have the same sign, U (r )>0 for all r and
if q and Q have the different signs, U (r )<0
for all r . I have left off all the steps leading to W =…
and U=-W ; you can find these in any physics text or on the web, e.g.
this
link .
*Because magnetic forces are velocity
-dependent, they are not
conservative forces so you cannot introduce a potential energy for them.
QUESTION:
I'm currently in the process of Worldbuilding my own character's story. [I
have no idea what this means!]
If the universe, instead of a vacuum, were an infinite "block" of matter, expanding forever in every direction, how would gravity behave inside a spherical hole (at the theoretical center of this infinite matter) within this infinite block of matter?
So far, I have only ever found mathematical formulas to determine gravity
inside an empty sphere but I am unsure if that behaves different if there
is a vacuum outside the sphere (which, in my case, there is not). [This
question was originally way too long and I have edited it because the
answer is simple.]
ANSWER: I
assume that the mass of your block is uniform and that it has spherical
symmetry; to assure
spherical
symmetry, the block should be a sphere. Physicists like to avoid
infinities so just make it extremely large (like billions of light years). The gravitational
field inside a thin uniform spherical shell is zero everywhere. This was proven by Newton, in
fact it is supposed that this question was the main reason why he invented
calculus. There are several consequences of what is called the shell
problem. First, the field inside a thin spherical shell is zero. Second, a
thin spherical shell will give rise to a field outside which is the same as
if all the mass of the shell outside were concentrated at the center of the
hole. Third, the shell need not be thin since you can imagine adding many
thin shells like an onion making a thicker and thicker shell, each having
the shell theorem applied to it. The gravitational force in the matter will
linearly increase the farther you get from the center (another reason for
not being infinite). If it is expanding, is mass being created somehow? Or,
if mass is conserved, it must be getting less dense. And if it is getting
less dense, it must be compressable which means that it is getting more
compressed at larger distances than closer to the hole; this would still
result in zero field inside the hole but the field in the matter would not
increase linearly anymore. If you added any more holes, all bets are off
because that would destroy spherical symmetry; however, if the holes were
much farther apart than their sizes it would be an excellent approximation
that the fields would be zero.
QUESTION:
When I multiply the volts x Amps,
I get 0.71 Amps 115Vx0.71Amps=81.7 W=0.109
hp.
T
he
hp rating of the pump is 0.04 hp=29.8 W.
Why the discrepancy?
Do I have to use the RMS value of the AC line voltage to get the right value?
ANSWER: I
actually had no idea because I did not know how water pumps are rated. They
are not rated by power but rather by head (H) times flow (F); on a
site
I found the statement is made that "Unlike
motors, pumps are rated by head and flow, not by power. There's no such
thing as a 50-horsepower pump "H, F,
and P is (P=H xF )/e where e
is the efficiency of the pump. I believe that where you went wrong is to
compare the input power to the output power which will not be equal unless
the pump is 100% efficient which it never will be. If I understand all your
numbers, the pump efficiency will be e =1-(29.8/81.7)=63.5%. The
site I found most helpful is
here .
QUESTION:
ANSWER: This
can be confusing! The word moon means any natural satellite
orbiting a planet; here natural means it is not man-made. The space station
is a satellite but not a moon. "The Moon" refers only to the moon which
orbits earth. Note the capitalization of the Moon which sets it apart from
all other moons; the Moon is a moon! Astronomers often refer to the Moon as
Luna, the Latin word for Moon.
Io ,
Europa ,
Ganymede , and
Callisto are moons of Jupiter and were the first moons of another
planet to be discovered (by Galileo in 1610). Hundreds of exoplanets
(planets outside our solar system) have been observed but no undisputed
evidence for exomoons has been found.
QUESTION: The more I read about quantum mechanics, the more I think that electromagnetic waves (as described by Maxwell) really don’t exist and is purely a mathematical contrivance (although Maxwell’s equations do great job predicting what happens at a macro level). I am aware that photons contain discrete amounts of energy.
If we look at an antenna and why the electrons start moving back and forth and produce a small voltage when radio waves hit it is because the waves have oscillating magnetic and electrical fields. Because electrons have electric and magnetic moments, these fields “push” the electrons around (pretty classical description of a force).
The quantum mechanics description of what is happening seems very different. My crude interpretation seems to be that a quantum particle called a photon binds to an electron and gives information to the electron (hey, Mr. Electron, that transmitter says you move that way).
Some of the literature seems to imply that energy is just information and the electrons move because that’s what they were told to do and that’s what they do.
ANSWER:
Classical Electricity and Magnetism (E&M) is what mathematicians call
a classical field theory. It is probably the most fully understood theory in
all of physics. It is beautifully expressed and, in its most complete form,
fits in perfectly with special relativity (which most physicists consider to
be a classical theory replacing Newtonian mechanics). But in the early years
of the 20th century a couple of experiments seemed to not be in agreement
with E&M. First was the photoelectric effect and the other is Compton
scattering. I will leave it to you to learn about these results in detail if
you already haven't. In a nutshell, they provided evidence for photons.
The photoelectric effect: Shining ultraviolet light on the surface of a metal results in ejection of
electrons which have energies consistent with the existence of photons which
have given all their energy to an electron in the metal. Compton
effect: Photons can also be observed to interact with with an electron by simply scattering from it,
thereby losing a some of its energy,
and that is also observed. I don't know where you got "…a photon binds to an
electron and gives information to the electron…",
but it is not correct on any level. Photons interact with electrons. Finding
the constituents of a field which are the smallest possible amount of energy
that can still be identified as being in that field is indicative that the
classical field can be "quantized", expanding it to a quantum field. This
was not achieved mathematically until the late 40s, early 50s. In a
quantized field the quanta (photons in your case) of the field may be
thought of as the "messengers" of the field, the way that the field operates
mechanically by communicating forces between electric charges. With
quantization of the field, E&M is physics' most well-understood and complete
theory. QUESTION:
Since the the 1970’s air conditioning has become standard on all new vehicles and in all buildings. In my car the AC is set on auto and my windows are closed so anytime the temp is warm the AC cools the interior.
I know for every action there is an equal opposite reaction. Cool air inside equals hot air expelled outside. So are all these air conditioners all over the world contributing to global warming by emitting heat into the atmosphere?
ANSWER:
What an air conditioner does is move heat from inside to outside.
There has been no net change in the overall energy of the atmosphere. (Heat
is the flow of a certain amount of energy from one place to another.
However, this doesn't just happen, it must be made to happen. You use
electrical energy to cause this zero net change in energy, to move that
energy. That energy is likely to come from an electrical power plant pouring
tons of carbon into the atmosphere. QUESTION:
how many H2O molecules does it take to make water ... with the characteristics of water, i.e., solvent, liquid, etc.?
ANSWER:
One of the main concerns of nanotechnology is to answer this question. Of course, there is no clear answer since there will be a continuous transition to the bulk value of each property being measured
as more molecules are added. A very important consideration is the the ratio of the surface area of the cluster
to the volume; this is because the molecules on the surface are interacting with fewer neighbors than those inside are.
QUESTION:
I am NOT a scientist, just a curious layman. I'm 74 years old; this isn't homework. On the question of singularities—doesn't quantum theory pretty much disprove them? I'm
guessing that if energy (and pretty much everything) can be considered a
wave function, then doesn't any wave have to have some amount of volume in order to exist? Or is this question just highlighting the well known problem with the existence of singularities.
ANSWER: e.g. freezing of water) in certain observable
properties or their slopes may instaneously change, an artifact of the
theory.
QUESTION:
I am a freshman in high school and in science class we were learning about planets, moons, and stars, and their sizes compared to the size of earth. I was wondering how we can determine the sizes of different planets, moons, and stars with them all being at varying distances away and all looking relatively the same size in the sky?
ANSWER:
see. So,
suppose that you have a telescope which allows you see Jupiter, for
example. Now, that telescope has nice crosshairs in it so you can first aim
it at one side and then rotate it to the opposite side. You measure the
angle (A in the figure). Now, the angle is related to the diameter
D and the distance to Jupiter R by A=D /R . If you are a
freshman you may not be familiar with the units A has in this
equation. The units are called radians and there are 2π radians
(π =3.1416 as you probably know) in a whole circle just as there are 360 degrees in a whole circle. So
let's see what A is for Jupiter: D =1.428x108 meters=88,732 miles and
R =7.14x1011 meters=444 million miles. So now we can calculate
the angle, A =1.428x108 /7.14x1011 =0.0002
radians=0.0115 degrees. This is a really tiny angle and you can understand
how you would see it as only a dot with your naked eye. We say that Jupiter
subtends and angle from earth of about 0.0115 degrees.
Of course, this gets really hard to do for smaller objects like moons or
very distant objects. Today, these are determined by getting up close to
those objects with probes which we have sent to space for the past 70 or so
years. But once you get closer you still use the subtended angle method.
Stars are harder still to measure. If the star is big enough and/or close
enough, you can still do the subtended angle direct measurement method;
this is done with a really powerful telescope where you can take a photo
and then just measure the diameter of the image. There are other methods.
Like if you find a binary star pair (two stars rotating about each other)
you can measure the time it takes for an eclipse to occur and calculate the
size of the star. If you can see a planet in orbit around a star you can
infer the time it takes the planet to cross in front of the star.
I could go on, but maybe this is a good place to stop. Say hello to Ms.
Welch from me!
QUESTION:
In Einstein's book The Theory of Relativity, I read that gravity is not a force but rather an acceleration or curvature of spacetime that pulls objects toward regions with high mass or energy density. If gravity is not a force but a field curvature, then why are physicists trying to find a theory that unites gravity with the other three forces? After all, if gravity were a force, black holes wouldn’t be able to attract massless photons. Why are we trying to unify something that is not a force with forces?
ANSWER:
earlier answer is the first time I answered it (just the paragraph starting
with "…While
researching this question…" is relevant for you). Unfortunately the
link to the essay is no longer on line, but you can get the gist of it from
my summary. The bottom line is simply that a mathematical theory, general
relativity, can be interpreted in more than one way; in this case, GR could
be interpreted as geometry or as a quantum field theory.
QUESTION:
I've never realised it before but the solar system is constantly moving. I saw an animation that almost scared me. It's the sun travelling through space and the planets following slightly behind spiraling around it. I found it amazing. My question is if we're constantly moving, are there areas in space that could be warmer, cooler or have more radiation in it. I'm not sure how to explain it. Almost like an atlantic fish showing up in the pacific and we wouldn't be able to toleratee it basically.
ANSWER: animation you
saw, I think. [There seem to be several of them.] The video I watched was
very "pseudoscience". For example, saying that the usual picture of the
solar system is incorrect is, well, just incorrect. The usual picture shows
how the solar system looks when viewed from the framework of the sun. If
you now view from the framework of, say, the center of the galaxy, and look
at the paths of the planets, you see those very pretty helixes; now the
narrator waxes eloquently about the profound meaning of his "vortexes",
calling what he sees the helical model of our solar system, as if this were
some meaningful new law of physics. In fact if you just didn't draw the
paths you would simply see the usual picture of the solar system moving
past you. Also, since the solar system is rotating in an orbit relative to
the galaxy center, most of the stars and other nearby stuff is also in
about that orbit so that our neighborhood moves with us keeping our region
of space more or less constant. And space in our region of the galaxy is
more or less empty on scales much bigger than the solar system, so don't
spend your time worrying about suddenly being a "fish out of water" (to use
your metaphor). If you want to see a very good and scientifically accurate
video about the path of the solar system throught space, I would recommend
this PBS video.
QUESTION:
What is the optimal size (circumference) and weight of a
ball such that it could be thrown faster than any ball currently thrown by
a human.
I'm assuming that using physics there is a way to determine the amount of force imparted on the baseball thrown by Aroldis Chapman, and then extrapolate that to optimize a ball size/weight such that his throw could exceed 105.8 mph with perhaps this optimized ball.
(I have edited this question to make it more concise and well-focused, but
it doesn't matter since there is no possible answer.)
ANSWER: this link
to see this discussed in some detail; you will see that the mass plays an
important role in calculating the speed reached but the size is irrelevant.
One thing which would make a larger ball matter would be how awkward it
would be to throw it. Also, the human arm doesn't apply a constant force
over the distance the ball moves while the throw is in progress and this
would likely vary.
QUESTION:
How would I calculate for the μ of kinetic friction with the force of kinetic friction and the angle on which a wood slider starts sliding down an angled wood board.
ANSWER:
fs ≤μs N,
fk =μk N,
and μk <μs .
In the figure there are three forces on the block if it is not slipping:
its weight W
f s , and the normal force
N N=Wy =W cosθ and
f s =Wx =W sinθ. Now
keep slowly increasing the angle until at some angle θ 0
the block starts sliding, accelerating down the incline. At this angle,
fs =μs N=μs Wcos θ=W sinθ ,
so μs =tanθ . So you can measure the coeficient of static friction this way, but not the
coeficient of kinetic friction.
QUESTION:
I think you can make a microscope using a regular 3,000X microscope and two drops of water in a tube of fluorinated oil. Fluorinated can be used as a lens. They can adjusted the density, to be very near the density of water. You can search the internet: Can you make a lens with a drop of water in fluorinated oil? It says it is potentially possible. So I think you could put two sphere drops of water in a tube of fluorinated oil. Then you could look into the tube with a 3,000X microscope. And, you could put a layer of water in the tube, below the two drop lenses. Each drop could be 100X. So all together, you could magnify, 30 million. You could see viruses while they are alive, and through them. The big scopes, don't do both of those.
ANSWER:
QUESTION:
It is said that acceleration is equal to meters per seconds squared. a = m/s^2.
I agree that the explanation is that it is the rate of change of a velocity.
But why is the rate of change based on the time parameter - seconds.
Why can I not say that the rate of change is per meter travelled ?
What if I say: a = m^2/s. Meters squared per second, so the travelled distance increases per second. (?)
So to use a counter-intuition argument: this is what we observe, time ticks the same amount of ticks (not 1/squared), and speed/distance increases
ANSWER: displacement rate of change of velocity , but it isn't acceleration which is universally accepted to be
time rate of change of velocity.
QUESTION:
My brother was taught in a driving class that if two identical cars collide head-on at 50 mph, that the collision would be equivalent to a 100 mph crash. Now, I assume the instructor meant that the occupant would receive the same damage/injuries as if they had a 100 mph collision. I disagree, I believe the damage would be the same as if either car had hit an immovable wall at 50 mph.
ANSWER: i.e. , all cars and
walls are stuck together after the collision. Linear momentum of an object is
MV , mass times velocity; keep in mind that velocity is a vector, so if
a car is going south with velocity 50 mph, a car going north with the same
speed will have a velocity -50 mph. Also, if a system has zero net force
acting on it, its momentum remains constant (momentum conservation). If you
look at one of the colliding cars, its momentum is not conserved because
during the collision whatever it is colliding with exerts a force on it
which slows it down. But if you look at both the cars momentum is conserved
because they experience equal and opposite forces.
The initial case the two identical cars going 50 mph in opposite
directions. The situation:
Their total linear momentum is zero, the
magnitude of each car's momentum is 50M
Since they are identical, they will crush at the
same amount in exactly the same time.
Average force is the average time rate of change
of linear momentum. Since magnitude of the time rate of change of
momentum for both cars is 50M , the forces are all equal in
magnitude comparing the two cars and their occupants.
Next, compare the case of one car going 100 mph and
the other at rest:
The total linear momentum is 100M
After the collision the momentum is 2MV
where V iw the mass of both cars, stuck together now.
V may be determined using momentum
conservation, 100M= 2MV , so V =50 mph. So the
at-rest car and the 100 mph car both have a magnitude of momentum change
of 50M , the same as the
50 mph in opposite direction cars, so same forces. So you and your
brother misunderstood the instructor's lesson.
Finally, what if a car hits an immovable wall?
QUESTION:
I have a question regarding quantum tunnelling and trying to understand how a particle can cross a barrier and appear on the other side. Let' take a helium nucleus for example that is fired at an energy barrier. From my understanding, the particle's wave function provides for a non-zero probability of it appearing on the other side if it is fired at a barrier instead of just bouncing back as is most probable. I am having trouble conceptually understanding the mechanics of this. Does the particle's ability to pass through the barrier have anything to do with the frequencies,
'spin' coherence or otherwise of the subatomic constituents being in phase to allow it to traverse the barrier? If not please explain how it can pop out on the other side.
ANSWER: x direction. to
remind you of what's happening, I have drawn a cartoon of the particle
below but it doesn't belong on this graph because this is an energy vs .
position graph. So the particle moves in from the left with a constant
kinetic energy ½mv 2 until it gets to x =0
where it sees a linearly increasing potential energy which corresponds to a
force in the negative x direction; therefore it begins to slow down
and comes to rest momentarily. Now the force starts pushing the particle
back, speeding it up until it is now moving in the negative direction with
the same speed it came in with. For obvious reason the point where the
total energy is equal to the potential energy is called the turning point;
it is classically impossible for the party to penetrate beyond there.
Now, let's examine same problem but using quantum mechanics instead of
classical mechanics. Again we imagine an incident beam moving toward a
triangular wedge potential. Now we cannot just think about a particle
because all particles are also waves and to understand how tunneling works
we must calculate the particle's wave function using the Schrödinger
equation. The particle can exist only if there is a solution to to the
equation; in the classical case the particle could not exist below the
potential energy because it would imply that the particle had an imaginary
velocity. However, the Schrödinger does have have a solution in the
classically forbidon region, it just doesn't look like a wave; it can
therefore re-emerge as a wave beyond the barrier, hence tunneling. Because
of nuances of quantum mechanics, you cannot actually directly observe the
particle when it is in the "forbidden zone". When you are solving these
equations you must pay particular attention to the boundaries where the
wave function and its slope must be continuous.
FOLLOWUP QUESTION: I'm looking over the diagrams you provided in your explanation, which are very helpful. Can you tell me if I'm getting it right:
Regarding the quantum mechanics diagram it tells us that:
the particle's kinetic energy has to be greater than the potential energy of the barrier/force, and when the particle traverses the barrier it loses some of its energy in the process and appears on the other side with a reduced kinetic energy.
The shroedinger equation solutions would show us that that the reason not all the particles cross the barrier is that the odds of crossing it are so small
Would it also follow then that the greater the difference in the energy of the particle and the barrier, the more likely the odds of the particle(s) appearing on the other side and/or in greater numbers?
If a particle does actually appear on the other side of the barrier - will it have left a hole in the barrier (if it were a physical object) where it passed through?
ANSWER:
Note that the vertical axis is the energy axis, just as it was for the classical case.
The horizontal line represents the total energy of the particle. The
wave function, the solution to Schrodinger equation in each of the three
regions, is drawn on this graph even though its amplitude is not an
energy; rather you may think of the amplitude squared as the probability
of finding the particle left, right, or inside the barrier. The particle
has the same energy everywhere. The wave function of a particle is
smeared all over this one-dimensional space, it is both untunnelled and
tunneled but it is liklier to be untunneled. It may be more usefull for
you to imagine ithe wave function as being a flux of many particles.
then the wave function gives you the probability that an incident
particle tunneled out.
The width of the barrier is what determines how
likely tunneling is. So, for my triangular barrier, tunneling is less
likely near the bottom than the top of the barrier. I would rather not
get into the energy difference dependence since it gets too technical.
Think of it as a flux of particles. Now it gets a
littll tricky because we have only been thinking of the particles coming
in from the left; but particles which do not tunnel will be reflected
and move to the left. So if you add the tunneled flux to the reflected
flux you should get the incoming flux.
I’m looking over the diagrams you provided in your explanation,
which are very helpful. Can you tell me if I’m getting it right:
QUESTION:
I am a curious 49 year old who has become rather obsessed
with the nature of reality. My question, I think, concerns what counts as
an observer. For example, if we consider a single calcium atom in the bone
of my arm. It is bonded with other atoms.Are the 20 electrons in a
superposition? Or, because they are bonded or close to or in a relationship
with another atom(aka "observed), they are in a definitive
position/velocity/etc...? I think where I am eventually going is an attempt
to picture vibrations in fields moving through the universe. As I wave my
arm in front of my face, should I imagine vibrations in superpositions
moving through a field? Or should I imagine wavicles that have
relationships to other wavicles moving through a field?
ANSWER: superposition .) Now if you
add more atoms to your arm bone, they bond to make the bone and it is the
electrons which are responsible for holding the atoms together—all
the electrons' interactions form a solid bone. In principle, the bone is
one system which must be antisymmetrized and every electron has a "piece"
of each other electron in your bone. Of course that would be impossible to
compute and normally only "nearest neighbors" are considered which surely
have the most significant contributions. And, indeed, the electrons in both
of two atoms have pieces of all the electrons in the pair. I do not
understand what your statement about "observed" means and they are
certainly not "…in a definitive position/velocity/etc…" I am
afraid everything after this makes no sense to me.
*An antisymmetric wave function is one in which you interchange the
position coordinates of any pair of particles, the wave function remains
unchanged.
QUESTION:
ANSWER: Poiseuille's law describes the flow of a viscous fluid through a hose
of length L , radius R , viscosity μ , and volumetric flow rate
F : P =8μLF /(πR 4 ) where P is the pressure drop over the length.
Everything else being equal, P∝R -4 or P 12 /P 10 =(R 10 /R 12 )4 =0.48,
So P 12 =0.48P 10 =96 PSI. This
demonstrates that it should be less. Since I do not know how MAWP is
defined, I don't know if "…Everything else being
equal…" is a correct assumption . If, perhaps it is
assumed that the velocity v of the fluid is the same in each, then
the flow rate would be F=vπR 2 and
Poiseuille's law would be written P =8μLv /R 2 .
In this case you would calculate that P 12 =0.69P 10 =139
PSI which is essentially what your numbers are.
QUESTION:
ANSWER: V =ρ· dρ· dz ·dφ .
Volume elements having different distances from the z -axis, ρ , have different sizes.
QUESTION:
ANSWER:
FOLLOWUP QUESTION: Well that was an enormous waste of 5 minutes. It is not really a cosmological question. In the absence of particles, can there still
be heat?
ANSWER:
The main problem is that you do not understand what heat is. Heat is not energy, heat is the flow of thermal energy
usually from high temperature to low temperature. If you look at your question you will see that you talk about heat as being thermal energy. This is an important distinction. The flow of thermal energy is accomplished three ways.
Solids can conduct energy because thermal energy (in the simplest sense) is the kinetic energy of the atoms/molecules. So at the hot end the particles have oscillations larger than their neighbors which the transfer to their neighbors causing thermal energy flow (heat).
Usually gases and fluids move macroscopically, currents, winds, etc. This is called convection. And, yes, without particles there is no convention, hence no thermal energy flow (heat).
Finally, all objects are constantly emitting and absorbing electromagnetic radiation. Classically, EM radiation is waves, not particles. In that sense thermal energy flow (heat) can occur without the presence of particles—EM waves carry energy. Of course you could split hairs (and you seem like the kind of guy who would!), and say that EM waves are composed of photons which we usually call particles, but they have no mass.
QUESTION:
We are going to build a mobile chicken coop so they can graze over different parts around the farm. This is what we are looking at:
We plan on having solar panels on the roof near the back(above the wheels). We also plan on having a battery bank to help heat the water in the coop or run fans depending on season. The question I have is would the weight of the batteries be better situated higher or lower in the back of the coop to make it easier to pick up by the doors?
ANSWER: w H above the ground and acting straight down. The center of gravity ⊗
is where all the weight W w C from the axle of
the wheel. The ground exerts upward forces of N F D from the wheel. We can find what the unknown forces
F
and N are:
Sum of forces is zero, F+N-W-w =0.
Sum of torques is zero (I choose wheel axle as
axis of rotation), FD-WC =0. (Note that w and N
exert no torque.)
Solve: F=W (C /D ) and
N=w+W [1-(C /D )].
Note that N and F do not
depend on H .
Next you lift at the door end until the floor makes
an angle of θ relative th the horizontal. All forces remain
vertical (assuming you lift vertically). F now is the force you
exert to lift the coop rather than the force the ground exerts at the
doors. Now proceed in the same manner as before:
F+N-W-w =0.
Sum of torques is zero (I choose wheel axle as
axis of rotation), θ-WC cosθ+wH sinθ =0=F-W (C /D )+w (H /D )tanθ
Solve: F =W (C /D )-w (H /D )tanθ.
Here is your answer: as H gets larger, F gets
smaller. Therefore the higher up you put the battery, the easier it
will be to lift the coop. It is easy to see this intuitively because
note that you and the battery are both tending to rotate the coop
counterclockwise, i.e. , the torque from the battery is tending
to "help you out".
I have one proviso here though: the angle you
will need to lift it is likely small and the weight of the battery is
very small compared to the whole coop, so in the real world the
difference between low and high mounting is likely to be negligibly
small.
QUESTION:
I have more of an abstract mind as opposed to a more mathematical sort, so I'm sure you can help me here.
Regarding Hawking radiation, virtual pairs of particles and anti-particles manifest and annihilate. I wonder if that's due to the anomalous nature of 'zero'. The formula is there 1-1=0, particle plus anti particle equals nothing. However, by definition, nothing can't exist. Space is constantly trying to resolve an equation that can't be balanced, thus it forever 'boils' no matter how much or how little energy there may be.
As a consequence to this, I imagine going well beyond the universe and Into the vast empty space. A place so cold that only these virtual parings can exist. In this space no force is greater than the other, and the 'tug' on the fabric of space is uniform. This makes me think about a sort of 'tug of war' where both sides are of equal strength. In that scenario it isn't either side that will give in, instead, the rope will eventually break. This 'cosmic 'tug' would eventually result in a massive reaction of chaotic expansion we think of as 'the big bang'.
So, I believe there are multiple expansions allover the infinite void. This might explain why the cosmic background radiation image was so asymmetrical.
I don't have any of the math to really back this concept up, but I'd be deeply grateful if you consider it.
ANSWER: Your question makes no sense to me. But it is not too hard to understand Hawking energy without any math at all. You just need to understand what the Heisenber uncertainty
princlple (HUP) is. Secondly, if some process can happen (that is, it is not forbidden by any physical law), it will happen although perhaps with an extremely low probability. Consider a particle-antiparticle pair
suddenly pop into existence; are any physical laws broken? Yes, perhaps the most inviolable of all,
energy conservation. The pair will have energy which was not there before they appeared, so they have violated energy
conservation. But the HUP says that you cannot simultaneously know the energy of the pair and the time it exists; i.e ., you can violate energy conservation as long as it exists only for the short time
which the HUP says it can exist. When you understand these two, go to
this link to see how Hawking radiation works.
QUESTION:
I have a question in the realm of mechanical physics (I think?).
I'm a quad skater, old school style with four wheels aligned in square pattern rather than inline.
Let's say I'm going down a straight flat road, build up speed and then coast on that speed without propelling myself forward.
If I want to try and maximize my forward force/distance and reduce the effect of friction slowing me down, would adjusting the amount of wheels in contact with the ground make a difference?
Part of me feels that if I can balance in two of the 8 wheels, there's less contact with the surface of my wheels and the ground, meaning less friction. But, the distribution of my weight across that surface of contact is far less, meaning there would be greater friction anyway. Which feels like a conservation of energy sort of thing but it's been a while since I've been in a physics class.
So, my question is ultimately, is there a technique where I could reduce friction and maximize distance by adjusting the amount of wheel on the ground, or is physics laughing at me for trying to outsmart it?
ANSWER: It is a fact that the frictional force is
proportional only to how hard two surfaces are pressed together and depends
on the nature of the surfaces (e.g ., there is a lot less friction
from steel sliding on teflon than steel sliding on rubber.) The force
pressing your wheels to the ground is equal to your weight and it makes no
difference the area over which you cause that force to be distributed. I
would guess that the air drag would more important in slowing you down than
the rolling friction of your wheels; the drag is proportional to the square
of your speed and the area you present to the onrushing air, so you could
minimize the area by squatting down, thereby maximizing the distance.
QUESTION:
Say you have 2 masses separated by a distance of 1 lightsecond.
Mass A begins moving away from B (relative to their common fixed inertial frame). Will B feel a weakening gravitational force from A immediately or starting 1 second later? A from B?
ANSWER: Although the speed of gravity has not been
measured precisely, everybody assumes it is the same as the speed of light
and many measurements indicate that it is so. Therefore, B will not observe
any change in the field for a second, the same time he would have to wait
to see A depart. A will observe the same sequence of events.
QUESTION:
If a rocket were to take off vertically from earth and travel in a straight line away from earth at 10000 mph continuously for an indefinite amount of time with unlimited fuel what would happen? Escape velocity is 24000ish mph so it shouldn't get into outer space, but it is always maintaining a tremendous forward motion. Would it hit an imaginary wall and just hover at the same point in the sky as the earth rotates under it, would it maintain the same height but orbit with the earth or would something else happen all together? This is assuming that the nose of the ship is pointed into outer space and its trajectory never changes. For directional sake say it's taking off from the very center of the North Pole or whatever the highest point on earth longitudinally is and it keeps flying away from the earth where would it go?
ANSWER: You have evidently never had a physics course. I
am going to assume there is nothing we need to focus on in the whole
universe other than your rocket and the earth. (I ignore any forces from
the sun, any planets you might encounter, any stars you might encounter,
etc .) The escape velocity is the speed which the rocket must start
with at the surface of the earth if your engines never fire again; this
rocket would immediately start slowing down because of the gravitational
pull of the earth on it and continue slowing down until at an infinite
distance where it would stop. Now, your rocket starts at some constant
value (it would not have to be large, maybe just 100 mph) because the force
the engines give it are always adjusted so that it is exactly the same as
the gravitational force trying to slow it down. Eventually you would get so
far from the earth that the force trying to slow it down is maybe one
ounce. Except for an occasional tiny pulse from your engine you could coast
almost forever and use almost no fuel.
QUESTION:
Many experiments have been done bombarding protons with energetic
electrons. Most results appear to be focused on how the electrons bounce or
recoil off the protons or their imputed constituents, perchance to
illuminate what's in the proton. Have any experiments been done focusing on
how many (what %) of bombarding electrons pass directly through the proton
without bounce or recoil? Or is it impossible to determine whether any do?
I would appreciate any light you can shed on this. Perhaps how narrowly the
beam is focused and the density of protons in the target mean one can't
distinguish between electrons that missed all the protons in the target
entirely, and those that might have passed through the protons.
ANSWER: This is not a collision of classical billiard
balls, it is an electrically charged electron interacting with a charge
distribution. "Bounce or recoil" really has no meaning in this context
because the electron has no option other than to interact with the field.
What can happen is that the electron might exit without changing the the
state of the nucleus; this is called elastic scattering and the electron
could be scattered at any angle including 0°. In fact this is, by far,
the likeliest thing to happen.
QUESTION:
I am a newbie to electricity and magnetism. I am confused about 1 electron charge is 1.64x10^-19 Coulombs and 1Coulmbs = 6.24x10^18 Elementary charge Unit. First of all what does it mean by elementary charge unit and what is the difference between two values 1.64x10^-19 and 6.24x10^18.
ANSWER: First of all, one of your numbers is wrong,
1.64x10-19 should be 1.602x10-19 . All electrons have
an electric charge or e =1.602x10-19 C/electron where C
is the unit of electric charge, the Coulomb. Until recently the Coulomb was
defined in terms of an Ampere (A) which is the unit of electric current; 1
C=1 A/s (s is second). An Ampere was defined in terms of the force which
two long parallel current carying wires exerted on each other. This was
always difficult because it was just too convoluted. Recently the SI unit
of electric unit was redefined as the magnitude of the charge of something
which contained as N electrons of charge e such that Ne would be
equal to a Coulomb: 1 C/electron=N x1.602x10-19
C/electron. Solving, N =1/1.602x10-19 =6.242x1018
electrons/Coulomb. (Coulombs per electron is the reciprocal of electrons
per Coulomb.) This is now how the Coulomb is defined, the magnitude of the
charge of 6.242x1018 electrons.
QUESTION:
Years ago, I used to listen to the NPR Saturday radio program "Car Talk" They always had a puzzler question that they asked (new puzzle) ant then answered the prior week's puzzel later in the program. A puzzler was asked that I never heard the answer to. I'll try to repeat as much of the details as I remember.
A guy is floating in outter space and is "buck naked". He is within reach of two iron bars. One is magnitized and the other is not. How can he tell which bar is magnitized?
They didn't say it, but I presume that the non-magnitized bar would be attracted to the magnitized bar, and vice versa.
That's as much as I remember. I have wondered for years what the answer was but searches of NPR "Car Talk" website only turn up re-plays of the radio programs and I don't know the dates of broadcast where I might be able to listen to the next-week's program to hear the answer.
ANSWER: Wow, I loved Car Talk . I really miss those two
guys. I have answered a few other questions involving that show*. Now, the
iron bars puzzler, two identical iron bars but one is magnetized, the other
is not. How to find which is which. I must admit that this took a little
thought to come up with the answer. Look at the figure below. If you grab
the magnetized bar it exerts the same large force, always attractive,
wherever you bring it up to the other bar. The pictures all look the same
if the south pole is up. Now, if you pick up the unmagnitized bar, the same
force will be found at the ends; but at the center there will be only a
very small force.
*Other mentions of car talk are at
carstuck ,
pickup ,
saltwater .
QUESTION:
I would like know what affect solar flares have on the earth's orbit. Since we talk about solar sails as a means of propulsion, wouldn't that mean the solar wind pushes against the earth as well?
ANSWER: In principle, yes. In the real world it is
negligible because the earth is so massive; there would be no observable
effect on the earth's orbit. A more interesting question is why aren't we
getting mortally fried by all the radiation we should be exposed to?
Actually, the earth's magnetic field serves like a shield to protect us
from the solar wind. So most of the solar wind which would have hit us is
deflected by the magnetic field and ends by just flying by. Still, some
gets through and is responsible for the northern and southern lights. The
intensity of the wind fluctuates a lot. I have read that, because the
orbits of satellites which GPS uses must be known to extraordinary
accuracy, the orbital calculations are crucial; the biggest errors in GPS
results come from when the solar wind is big and it does affect these
orbits and it is not as not possible to know when the satellites will be
affected.
QUESTION:
I recently thought of this question that I could not confidently answer myself, and I have googled as well to no avail. I apologise if this question seems trivial, but I am hoping to be blessed with some of your knowledge.
The question is: How does visual and tactile nerve input match in our perception?
For example; if I tap the back of my hand, I see it happening instantly, and feel it happening at the same time. The pathway of course as you know is light hitting the back of my retina, then having that nerve impulse travel through the optic nerve towards the brain, while the nerve input from the back of my hand does the same as well.
And apparently, nerve impulse conduction velocity does not even come close to the speed of light.
Is it because of the differing lengths of the nerves involved allowing them to somehow match each impulse upon arrival in the brain? I thought of this, but it feels so simple and incomplete.
ANSWER: This is really neuroscience, but I can give you
the physics slant on the question. I looked up what the speed of nerve
impulses: the speeds range from about 0.5 to 120 m/s. There are two kinds
of nerves, myelinated and unmyelinated. The myelinated are in the
peripheral nervous system and have generally much bigger speeds than the
unmyelinated. So, let's just suppose the speed of the impulse carrying the
information of feeling the tap is 120 m/s. The distance from your hand to
your brain is about one meter, so the time is about 0.0083 s. I am sure
that the time until your brain sees the tap is about the same. But I don't
think that your brain can comprehend a time interval that small.
QUESTION:
I was watching a film starring Arnold Schwarznegger and when explaining the futuristic weapons (90s film), an actress said the bad guys were using railguns as rifles. When I looked it up on google, it said that the concept was not viable. If it can be done and built as a rifle; what challenges do we need to solve and if not what are the unsolvable problems ?
ANSWER: I have already answered a question almost
identical to yours. You can see that answer
here . There are also links
to several more questions "about sci-fi and video games guns" on the faq
page.
QUESTION:
The fabric of spacetime itself is expanding exponentially. This would imply that eventually there will be an amount of spacetime with absolutely nothing within its volume. Eventually this area would be a complete void being too distant from any interaction with anything whatsoever. Not a neutron, not a photon or anything. Absolutely nothing.
What kind of events would occur in such an environment?
ANSWER: Actually, most of the universe already satisfies
your criterion: intergalactic space contains approximately one particle
per cubic meter. So you would have no trouble finding some cubic meter
where there is nothing inside its volume, a perfect vacuum. Since there is
no mass and no field inside your volume, the total energy in that volume is
zero. In classical physics you would say that the total energy remains zero
forever, energy conservation (no external forces doing work). But there is
something which says that energy conservation can be violated as long as
that violation exists for a short-enough time—the Heisenberg
uncertainty principle which can be applied to energy and time, ΔE Δt ≤ℏ
for example, if an M =5 kg bowling ball popped into existence that would be ok
as long as it disappeared before a time Δt =ℏ/ΔE
where ΔE =Mc 2 =5x(3x108 )=4.5x1017
J and ℏ=1.05x10-34 J·s, so Δt =4.7x10-51 s.
So a big mass would exist an impossibly short time. What normally happens
is that a particle-antiparticle pair, say an electron-positron pair, pops
into and promptly out of existence. This is what is called virtual pair
production and you cannot observe it directly. However, if you do some
electrostatic experiment in a perfect vacuum you will not get the correct
answer unless you account for the virtual pair production processes. This
is why this is sometimes called vacuum polariztion .
QUESTION:
Someone originally posited the question: "Is it possible for two different balls of identical mass and velocity to have different kinetic energies?"
At first I agreed with most others that it was impossible and used the KE equation as the basis for my opinion.
After some thought however, I came up with a scenario that makes me question this as follows: Imagine two spheres of identical masses, one with an internal void space of some amount. Impart to them both a spin and a velocity, both of which being equal. Now if you were to measure the KE of the homogenous spere as it moves, it would be a constant as one would expect. With the spere with an internal void spinning as it moves along its path, would this not create a kinetic energy level whose vale plotted on a graph would look like a sin wave whose frequency would match the spheres rotation and the amplitude would be a relationship between the total volume of the sphere and the amount in the void space. I think I belive that thte kinetic energy problems I have done all assumed the center of the item I was calculating for as an easement of solution. I think that if the ball is psinning under these circumstanes you would have t consider the sphee having a 'moent' that you have to calcullate its position for insead of the center and thus the sin wave effect.
Am I close to the solution or is it truly just the KE equation regardless of what is going on iwthin the mass?
ANSWER: You have the answer, but it is not necessary to
go through all the stuff about voids, sine waves, etc . You simply
say that one sphere is spinning as it also moves with speed v .
Even a ball not having a translational speed at all but spinning has
kinetic energy. The total kinetic energy of rotation is K rotation =½Iω 2 where
I is the moment of inertia of the rotating object and ω is the angular velocity. So if one
of the balls is spinning it has more kinetic energy than the other. If both
balls are spinning with the same angular velocity, the kinetic energies
could still be different because of the moment of inertia. For example, the
moment of inertia of a uniform solid sphere is 2MR 2 /5
and of a hollow spherical shell 2MR 2 /3.
QUESTION:
My question is about how torque and power each contributes to the performance, and acceleration of a car.
The acceleration should be equal to the force the wheels pushes the road with, devided by the mass of the car. The force is determined by the torque of the engine if i'm not mistaken.
On the other hand, acceleration of a car can be described as the accumulation of kinetic energy, given constant mass of course. The rate of that is determined by the power delivered to the wheels.
My hope is that you can give me a better understanding of <torque vs power>.
ANSWER: You are right, the force which accelerates the
car is the frictional force between the road and tires. But,
performance-wise, what matters is how quickly the engine can transmit
energy to the wheels. In physics, think first of force: Force is what you
need to do in order to give some mass an acceleration. Torque is the
rotation analog of force, that which you must do in order to give some
object an angular acceleration—for example to increase the angular
speed of a gear from 5 rpm to 10 rpm. Torque is defined as product of the
force F times the distance from r from the axis of
rotation, T=rF ; if you double the torque in a particular
situation, you will double the rate at which the rotational speed (RPM)
increases. Torque for a combustion engine usually envisions the drive shaft
as the agent for delivering torque from the engine. But you really cannot
just say that a particular engine has a particular torque because the
torque depends on many variables such as the RPM of the engine and also on
the gearing between the drive shaft and the wheels. Now, the engine
delivers energy to the car at some rate. If something is accelerating in a
straight line, the force F times the velocity v is the
power P , P=Fv . For example, suppose you want to see what
force is required to deliver P =200 horsepower (hp)=1491 watts (W)
at the speed v =50 mph=22 m/s; F=P /v =1491/22=68
N=15 lb. Now the same thing can be done for delivering energy to a rotating
object with a torque: The torque required to deliver power P to something
with a rotational velocity ω is given by the equation
P=Tω ;
in this equation P is in N·m/s, T is in N·m, and
ω is in s-1 . I presume you would like to work in RPM and horsepower, so
making those conversions,
P =(2π (ω /60)T )/746,
with ω in RPM, P in hp, and T in
N·m.
T =7114(P /ω ).
So if you want to know how many N·m of
torque you need to generate 200 hp at 1000 RPM, it is 7114x200/1000=1423 N·m=1050
ft·lb.
QUESTION:
If light slows down ( I appreciate it is by very little ) in water what speed does it travel at after leaving the water and entering a vacuum again?
If it returns to c then where does the extra energy come from to speed up again?
If it does not return to c then does that not mean that the only light in the universe travelling at c is light that has never been slowed down by anything else?
This came up in a conversation with some friends ( we have weird conversations )
I have a sneaky feeling that it will be some special case because of the nature of light ( being massless? ) that means conservation of energy does not apply.
Could you explain in terms a 55yo with a degree in maths ( not physics ) can understand?
ANSWER: Let's start with a fundemental fact about
waves: the speed of a wave v equals its wavelength λ times its frequency
f , v =λf . I assume you know what
wavelength and frequency of a wave are. Now imagine a very long string
which we are wiggling at one end with some frequency some frequency f .
So there will be waves of frequency f traveling down the string. The speed of
waves on a string depends on the mass density of the string (grams per
meter, for example) and the tension in the string, how taut it is. Now,
suddenly the mass density gets bigger at some particular location on the
string, so at that point the wave on the new part of the string, while
still vibrating with the same frequency as the lighter part of the string,
moves more slowly. As a result, the wavelength gets shorter. Note that no
energy leaves or enters the wave when the speed of the wave changes. If the
string becomes lighter again, the waves will speed back up again.
Light behaves the same way, slowing down in the
medium and speeding back up at the interfaces with the vacuum. The
difference is the mechanism for slowing the waves down or speeding them up; v =λf
still holds. There is another way you can look at light: You may know that
light waves of some given frequency may be thought of as a swarm of
photons, each having an energy E=hf where h is Planck's constant.
Since f is the same in and out of the medium, the photons always
have the same energy whatever the wavelength of the corresponding waves.
QUESTION:
(the video holds my try to solve a textbook question with my intuition, it's not about how to solve a textbook question but rather it is that whyy my intuition is incorrect. Although it's not necessary to watch a video of me struggling to solve a problem and in the end get the wrong answer.) Sir it's not about the textbook question.. but I am asking that why is equivalent mass of a pulley system (the mass of a block by which we can substitute the whole pulley)
different than, the (m1+m2) where m1 is mass at the left side and m2 is mass at the right side of the massless, frictionless pulley.
Whyy is it not m1+m2 because I only see that much mass there on the pulley
And in the equivalent mass formula, 4m1m2/(m1+m2)
Where does some extra mass come from???
Well, if we measure the weight of the whole pulley system, it should be (m1+m2)g !!
And not 4gm1m2/(m1+m2) right?
see some say that the tension in the upward string that is holding down the whole pulley system will depend on the accelerations of masses happening inside the pulley,
But
Those are internal accelerations, and if one block goes up, the other one goes downand therefore balancing the accelerations.
Can you please solve the confusion on the equivalent weight of the whole pulley being different then the actual weight of the pulley??? Why is it different?
ANSWER: m 1
is at rest. The questioner then was correct in his reasoning that the
tension in the thread connected to m 1 must be m 1 g
and it will be the same on both sides of the upper pulley. He then seems to think that the
whole lower pulley sistem can be thought of as an object of mass
m 2 +m 3 . This is only true if those two
masses are equal. If they are not, then one accelerates down and the other
up and all bets are off, you cannot do the problem that way. (Whoops, I
just did what I said I wouldn't—tell him what he was doing wrong!)
The first thing to do is to look at the lower pulley. There is an upward
force m 1 g and two downward forces which must
be equal in magnitude which I will call T ; the pulley must be in
equilibrium, so T =m 1 g /2. Next I will
look at m 2 . I will choose the +y direction to
be vertically up and so
m 1 g /2-m 2 g =m 2 a 2
where a 2 is the
acceleration of m 2 . Next I will look at m 3 .
I will choose the +y direction to be vertically down and so
-m 1 g /2+m 3 g =m 3 a 3
where a 3 is the
acceleration of m 3 . But the magnitude of the two
accelerations must be the same so a 3 =±a 2 .
Since I have chosen +y to be in opposite directions, a 3 =a 2 =a .
If you now solve the two equations (eliminating a ) you will find
[4/m 1 ]=[1/m 2 ]+[1/m 3 ].
QUESTION:
I'm a recent physics graduate, and I've realized that most of what I've done while studying physics is very theoretical, with the exception of some basic kinematics, circuits, and optics labs.
However, if at all possible, I would like to perform an experiment to show Relativity (or at least the underlying postulates) in the real world for myself. After all, it's different enough from our everyday experience to be difficult to accept on an intuitive level, even if you learn the math and such.
I figured I'd ask around and see if you had any ideas for such an experiment using either materials you can find at home or basic undergraduate lab equipment.
ANSWER: Well, I couldn't think of a table-top experiment
which would meet your criteria. So I did a quick Google search and found a
question basically the same as yours on
Physics Forums . I think you will find the ensuing discussion
interesting but probably nothing which would fit your wishes.
QUESTION:
Hello, I'm another 11th grader and I'm trying to understand how EM waves work.
The electric and magnetic waves are in the same phase. However, I was told that during induction the changing magnetic field induces the electrice one, and vice versa. From this I'd assume that B is shifted by compared to E, supposedly being its derivative, but that's not the case. I looked some things up on ChatGPT and the internet and found the following Maxwell equation:
After getting a rough understanding of gradients and partial differentiation (took a long time), I concluded that the rate of change in the electric field equals the negative rate of change of the magnetic field (divided by the change in time). Which still doesn't make sense! If E and B change alongside each other (beside some constant, or directional difference), then charged objects would gain a constant magnetic field and freshly magnetized pieces of iron would gain an electric field, which they could keep, until their magnetic field declines.
Besides the former, I'm struggling at understanding how energy is stored within photons, if E and B oscillate at the same time. Due to how inductors and capacitors can be charged, I am of the opinion that electric and magnetic fields contain energy, like in this equation about energy density: However, if E and B reach zero at the same time, that would mean the photon has no energy, which it clearly has, when E and B are at their maximum. This goes against the conservation of energy, which means I'm really missing something. Can you help me in this regard?
ANSWER: [This question had equations and a figure which
I have edited out because I do not do LaTex symbol language. Also, because
the questioner is not really prepared to understand electromagnetism at
this level.]
I admire your quest to understand electromagnetism.
You are already in a good place in understanding qualitatively. Until you
have studied vector calculus and differential equations, however, be
satisfied that you have already achieved way beyond your years. I will
briefly give you some more information to help with your pursuit of this
difficult subjece. The first equation you show is
∇xE B t . This is not a gradient of E E E B
The other equation you quote (corrected here) is
energy density of an electromagnetic wave,
U =½(ε0 E 2 +B 2 /μ 0 )
in empty space. What you are missing is that this
equation has been time-averaged. And before it is time averaged any spot in
space would be oscillating between 0 and some maximum value; there is no
violation of energy conservation as you suggest. Regarding the photons,
this is not what the energy density is for them. If the frequency of the
wave is f
then the energy of one photon is hf where h is Planck's
constant. Hence, for photons U photons = Nhf where N is the number of photons per unit
volume. In both cases U has units J/m3 .
QUESTION:
For the past several months, I (an aerospace engineer with an interest in quantum mechanics) have been trying understand Heisenberg's frame of mind when he first came up with his crucial insights in 1925. Everywhere I look I see essentially the same statement: "His leading idea was that only those quantities that are in principle observable should play a role in the theory, and that all attempts to form a picture of what goes on inside the atom should be avoided."
What none of the sources has conveyed to me is the following: granted that at the time Heisenberg developed his theory there was not a microscope powerful enough to see individual particles within the atom, but how did he conclude that seeing the positions and motions of the particles was "in principle" impossible? This was two years before he derived the Uncertainty Principle, which in any case is predicated on his original derivation of quantum mechanics. So without any of that "knowledge," he just arbitrarily decided that some quantities which had not yet been observed were in principle unobservable? I don't understand that. Can you help me?
ANSWER: Your question is strange. What does
"Heisenberg's frame of mind" even mean? Anyhow, that isn't physics, that's
history of physics or maybe psychology of physics. In 1925 Heisenberg
constructed the first mathematically rigorous theory of quantum mechanics;
it was based on matrix algebra. Prior to that there was only the
Bohr-Sommerfeld semiclassical description of what I refer to as quantum
physics. Within a year an alternate theory, based of differential
equations, the root being the Schrödinger equation which is a wave
equation. The latter was generally more accessible to physicists and is the
favored theory to introduce to students for that reason. Still,
Heisenberg's quantum mechanics is very elegant but more difficult to
calculate with. But the two were subsequently shown to be 100% identical in
rigor and both describe the exact same physics. What is different is the
mathematics used by each.
The uncertainly principle does not say some
quantities are unobservable. It says that some pairs variables are
'conjugate variables' and therefore you cannot measure both to arbitray
good accuracy, you cannot know both of them exactly. It is actually built
right into quantum mechanics, so no 'arbitrary' decisions needed to be
made.
I think you should give priorty to learning quantum
mechanics rather than Heisenberg's state of mind.
QUESTION:
Is it possible for a person to jump/drop onto an impact force measuring device and have the initial impact force be less than body weight?
Reference/context: Science world had a kinesiology exhibit, and one of the displays was about impact force measurement.
The person gets weighed and then climbs up to a platform approximately 18-24 inches high. The person then jumps/drops down onto an impact force measurement device, and a display shows impact force by body weight.
After a couple attempts I achieved an impact force reading of 1.2 times my body weight, and I feel I could have done better. Could that number have been below 1.0?
ANSWER: I need to know what the device actually
measures. "Impact force" is not a quantity in physics. A force is a force,
but an impact is the product of a force times a time. For example, if you
have a mass m , hit the device with a speed v and come to
a stop in a time t , your average impulse over that interval is
I=Ft=mv . where F is the average force you feel during the
collision time. So you could write F=mv /t. I suspect that
F is what this device measures. Assuming m is some fixed number,
how could you control how big or small F is? If you increase v
or decrease t , F will be bigger; if you decrease v
or increase t , F will be smaller. An everyday example is
consistent with this: If you fall and hit your head on a cement sidewalk,
you might get badly hurt because your head stops in a very short time; if
instead you fall on a thick foam mat, you would probably not get hurt
because it takes a pretty long time to stop.
So let's see if there is a way to get an F
which is less than your weight. Being a scientist, I must use SI units, but
I will try to convert to Imperial units when it seems helpful. Mass is
measured in kilograms (kg), length in meters (m), and time in seconds (s).
So suppose that your mass is 100 kg, the height is 0.5 m, and the time is
t . ; acceleration due to gravity is about g =10 m/s2
so your weight is W =1000 N (a newton is mass times g and
is 1 N=0.224 lb). So, your weight would be about 224 lb. since v/t is your
acceleration, I finally come up with F /W=a /g
where a is your average acceleration. So you just need to have the
rate at which you are slowing down less than the acceleration due to
gravity. If you drop from 0.5 m your speed when you hit the device is about
v =3.16 m/s. So if you can make v /t =10, F /W= 1.0,
so t =0.316 s. If you make the time greater than about a third of a
second you will get below 1.0; you can do that by watching what
parachutists do when they land—go into a crouch as slowly as you can.
Of course, I do not know how the measuring device works so my numbers may
be off, but anything you can do to increase the time from when you hit and
when you stop will lessen the average force you feel.
QUESTION:
When Hiroshima and Nagasaki where bombed with nuclear weapons in World War 2 there was total devastation. However within a relatively short period of time people were able to move back in and re-build the cities. However when there is a nuclear power plant meltdown like Chernoble or Fukushima the area around the power plant remains contaminated with radioactivity for thousands of years. Why is that?
ANSWER:
I
have previously answered
this question.
QUESTION:
I'm having trouble with Hawking radiation. It seems to me that if you separate a pair produced at the "event" horizon, with one half of the pair falling into the black hole and the other half escaping, you have added mass\energy to the black hole rather than subtracted from it. The problem for me here is the assumption that the mass\energy to produce the pair in the first place is somehow subtracted from the black hole?! To me this seems to presuppose that the mechanism for the pair production is well understood and the reservoir for the mass\energy for this pair production can communicate across the "event" horizon.
ANSWER:
I have answered this
question before.
QUESTION:
Red shift/blue shift—light traveling toward us (Andromeda Galaxy as an example of blue shift) has a shorter wavelength than stars/galaxies that are moving away from us. These light waves are part of the visible part of the electromagnetic spectrum (right?), but what about the parts that are invisible to the eye—ultra-violet, infrared, radio waves, etc? Do these waves change frequency in the above examples? (I know about Doppler radar and how it works, so is the same principle at play here?)
ANSWER:
All frequencies of electromagnetic waves are Doppler shifted.
QUESTION:
Black holes because of their mass and gravitational
'pull' suck things in and there is no escape (at least as I understand it), and this includes light. But light has no mass, so what is it about light that makes it susceptible to the gravity of a black hole?
ANSWER:
Our best theory of gravity is general relativity. In general relativity a
photon, the quantum of the electromagnetic field, has energy. (Of course it
does because the electromagnetic field itself has energy.) The energy E
the photon has is E=hf where h is Planck's constant and
f is the frequency of the field. If you throw a ball straight up
in the air with some initial speed v , it initially has energy E =½mv 2 ,
but the higher it goes the smaller v becomes until it is zero; so
when it runs out of energy it stops and, like any mass, will fall back to
the ground. The photon, however, still loses its energy but in another way:
Because of the red shift of light moving away from a heavy object, the
photon's frequency will get smaller as it rises. For most objects, a star
for example, the photons will end up with a pretty small shift in
frequency. For a black hole, however, the photon loses all its energy
before it can escape, that is, it simply disappears. If you are wondering
where the energy of the photon went, it went to increasing the mass of the
black hole by an amount m , mc 2 =hf or the mass
of the black hole has increased by m=hf /c 2 .
Incidentally, There are two kinds of red/blue
shifts. One is the one you are probably familiar with, generally called
"Doppler shift" and caused by motion of the source or observer. The other,
as discussed in this answer, is referred to as gravitational red/blue
shift.
QUESTION:
I've just generally been thinking about this for a while, and it's difficult to get an answer from Google. I think if I understand things correctly, it's actually vital for our universe that the speed of light is constant, and this constant effectively creates the electromagnetic spectrum we observe. Light can't go faster. It can only increase in frequency as more energy is added. If there was no constant speed of light, it might be the case that all energy in the universe would evaporate instantly? Maybe? I don't know. I'm just trying to conceptualize what's going on.
But what would be the physical mechanism by which the speed of light is constant? It must be a property of space? Do physicists have a solid understanding of what causes this?
ANSWER:
Go to the faq page. You will find "Why
is the speed of light constant to all observers? "
QUESTION:
I am an 11th Grade Student, and my teacher currently teaching us newton's laws of motion, in the above image there is a pulley arrangement, with all surfaces smooth! my teacher was discussing about this in the last few minutes of school and i couldnt understand it. He said that the big 10kg block will move towards the left! what? why would it move towards left? it would need a force in the left direction right? but here, i dont see any force on the 10kg block
which is on left direction... [followed by 364 unnecessary rambling
words!]
ANSWER: Once again I am faced with a question which
violates site ground rules requiring "...single, concise, well-focused
questions." Oh well, I have just edited it down and, rather than doing a
complete solution, I shall only address the student's primary concern: when
released, what happens to the 10 kg block? I always told my students to
"choose a body " when starting a Newton's second law problem. I choose the
pulley which, although not stated explicitly, must be frictionless and
massless. The next step is to find all the forces on that body .
Clearly the string is pulling on the pulley to the left by T 1
and down by T 2 ; because this is a
massless pulley we can assume T 1 =T 2 =T.
But, wait, is there anything else touching the pulley which might
exert a force on it. Yes! This setup would not work if the pulley were not
attached to the 10 kg block. The block must therefore exert some force on
the pulley to hold it on the corner of the block, F g . This is wrong.
QUESTION:
My question is about the way steam and heat works when food is cooking. For instance, if you cook food too long in a steamer, the food generally gets mushy if cooked too long. However if you heat up food in an airtight container in the microwave, the food does not get mushy if cooked for too long. In fact it will burn and get crunchy instead, despite having steam and lots of moisture in the container. Is there a physics-related reason that the two steaming methods vary so drastically in outcome if overcooked?
ANSWER: In a steamer there is a reservoir of water
continually adding stream to the inside of the pot. The cooking happens
from the outside in. If you cook too long it will be overcooked. The same
happens simply boiling the food. When you microwave you heat up the whole
thing uniformly. And any steam that is in there came from the food itself,
not nearly as much as in a steamer. But if you do that too long you will
boil the water out of the food and it will certainly become "crunchy".
QUESTION: I need to move an 80 pound speaker 28.5 inches
above the floor using a dolly on a 7 foot long plank. How much force will
I need to apply? (A very long question heavily edited by me!)
ANSWER:
QUESTION:
You have 4 hot hard-boiled eggs you want to cool down quickly. They are in a pot. You throw in a handful of ice and then add cold water. Do you just cover the eggs with the water or fill the pot? Which is faster?
ANSWER: I want to start out by noting this is an
academic question. The way cooks normally cool hot foods quickly to halt the
cooking is to plunge the hot food into an ice bath, water to which enough
ice has been added to bring the temperature of the water down to 0°C
(ice water).
This can get to be a very complicated system. I
will simplify it to include only what is going on with the water, ice, and
eggs. In other words, imagine that these three are enclosed in an
insulating box through which heat cannot flow. This is probably a good
approximation because cooling the food happens quickly enough that the
interaction with the bowl and air in the room is minimal. What happens is:
ICE: There are two
ways ice can absorb heat: (a) if the temperature is less than 0°C, it
can warm up until it gets to 0°C, and (b) at 0°C it will absorb heat
but not increase in temperature, instead melt. Most home freezers have a temperature of
0°F=-18°C, so initially, the way the ice will absorb the energy is
by warming up; when the ice reaches 0°C it will absorb energy by melting.
WATER: The question seems to ask, mainly, how
much water should we add. I will look at several amounts of water. I
will start the water at 15°C, a typical temperature for tap water.
EGGS: The mass of an egg is about 50 grams and
we have four of them. They have just come out of boiling water at
100°C.
I struggled with answering this question. The main problem is that the
speed at which the temperatures change depends sensitively on things like
how heat diffuses from the centers of the eggs into the water and the size
of the pieces of the "handful of ice"; crushed ice will absorb energy much
faster than a big block of ice. So I cannot answer the question, which
presumes that only the amount of water determines the speed to come to
equilibrium—there are many more factors which play. I think it likely
that it would take longer times for large amounts of water to come to
equilibrium, so I could glibly suggest that the answer is the smaller
amount of water is faster; but the problem has too many variables to know
for sure. What I have decided to do is demonstrate the physics of how the
final temperature T is achieved and work that out for one example:
I will find the minimum amount of ice we must add to 1 liter of water so
that the equilibrium of the ice/water/eggs system will be T =0°C.
Heat is just the flow of energy. If an object has a mass M and a
temperature T 1 and an amount of heat Q is
added its temperature increases to T 2 . Then the change
of temperature is proportional to Q , Q =MC (T 2 -T 1 ),
where C is a proportionality constant called the specific heat of
the material absorbing energy. If the object is instead having energy taken
from it, T 2 -T 1 <0 so Q <0.
If a solid is melting it is absorbing energy but not getting warmer; in
that case one needs a constant called latent heat of fusion, L which is the
energy absorbed per kilogram of material melted, Q=ML. So, the
following is the computation.
and the latent heat of
fusion for:
ice: L =3.33x105 J/kg.
I will calculate the amount of ice which is
needed for the final temperature to be T =0°C
starting with 1 kg (a liter) of water.
M w =1 kg of water at T w =15°C,
M e =0.2 kg of eggs at T e =100°C,
and
an unknown mass M i of ice at
T i =-18°C.
So, the equation (which is simply energy
conservation) to solve is:
Note that this is a linear equation for M i
as a function of M w .; this is shown in the figure. So,
1 kg will about cover the eggs and about a third of a kilogram of ice
will totally melt when T reaches 0°C. If you start with more ice, there
will simply be ice still unmelted in the pot when equilibrium is reached.
QUESTION:
I am currently studying for my physics exam. A student in the group chat asked us all a question and it got us stunned. Can you help?
Please be aware that this is NOT a homework question, we all just share different opinions on the answer.
Question:
A container with very thick/sturdy walls is on a frictionless surface. Inside the container, on the left side, there is a cannon with cannonballs. The cannon fires all its cannonballs against the right wall, where they fall directly to the ground upon contact (they do not bounce back). What is the effect on the container?
a) The container stays in place
b) The container moves to the right
c) The container moves to the left
d) Not enough information
I personally think the answer is A, but a lot of people are saying C, and others also say D. I'm lost.
ANSWER: See the figure. I assume that the box and
cannonball are rigidly attached and have a combined mass of M ; the
cannonball has mass m . Before the cannonball is fired, the whole
system is at rest and so its linear momentum is zero. Afterwards the cannon
ball is moving (relative to the ground which is where the momentum was
previously zero). But the total momentum must still be zero (conservation
of momentum) because there are no horizontal external forces on the system.
So, mv-MV =0 or V=mv /M ; when the cannonball hits
the wall its horizontal linear momentum becomes zero. There are still no
horizontal external forces so the motion of the box must also stop. Clearly
the answer is c).
There is another way you could look the problem.
Since after the the cannon is fired the cannonball, previously at rest, now
is moving. So the cannon/box must have exerted a force F -F on the cannon/box
etc .
QUESTION:
I am a high school senior who writing software for the avionics of our school's model rocket for the TARC rocketry competition. I have taken AP mechanics (Calculus based) and had a difficult time figuring out the projectile path of a rocket.
After burnout the rocket will coast to a target apogee (around 800 ft) at subsonic speeds. The rocket has onboard an accelerometer (plus gyro), a really accurate clock, and an altimeter (with barometer). I was wondering if there was a way to predict the exact apogee of the rocket based soley on the accelerometer readings and the clock (while knowing all the constants: mass, drag coefficient, etc.)?
My team and I have tried several ways to do this but we get stuck in dependency loops. Drag is based on velocity while velocity is based subsequently on drag. We tried using energy, momentum, even modified kinematics (for changing acceleration), none of which worked.
What we need is a single function that takes in current vertical acceleration and the constants and predicts a delta height which we can added to the current height to predict apogee. Also, since air pressure is part of drag and that changes based on height according to a function, it would be nice if we could put that into the predictive function.
I really don't want to make this question too long but if I could add one more layer of complexity. We are using air brakes to control the altitude to achieve a target apogee. We planned to run the above mentioned predictive function in a loop and reverse engineering it to create another function to find out how much more drag area we would need to achieve the target altitude since the airbrakes are radial flaps that just increase our cross sectional area.
ANSWER: First of all, you have not given me nearly
enough information about what you are trying to do. Is the time the rocket
is burning part of your calculation? It is very different (and harder) than
the trajectory after burnout for several reasons. Is the trajectory
one-dimensional, i.e . straight up and back down? Or does the
motion of the rocket also have a horizontal component? Does the
accelerometer give you the direction of the acceleration vector? And, if
not, it is not very useful. Do you know the mass of the rocket before and
after burnout?
All the above is necessary for me to give you any
guidance. But the way you are approaching the problem, common among
newcomers to physics, is all wrong. One of the most important things to
consider when solving a problem in the real world (not the ideal models of
on-the-blackboard problems in a physics classroom) is to determine what is
important and what can be approximated without serious error. Except in
textbook problems, you should never expect to be able to get the
appropriate equation in just one step, you need to make sure you understand
things step by step. You should be starting out by imagining you are doing
a textbook problem—ignore drag and ignore the change of mass during
the burn. Then you might put in drag. Here is a little tip: A fairly good
approximation for air drag D is D ≈¼Av 2 .
Here A is the area presented to onrushing air and v is the speed;
the vector D must be
opposite the vector v D will come out in Newtons. The estimate for D does not look
dimensionally correct; that is because lots of stuff is included in that
little factor ¼, like that the fluid is air, the density is for atmospheric
pressure at sea level, the shape is a generic sphere, etc .
Including D in your calculations will tell you if it is actually
very important for your problem.
That's probably all I should do at this point. If
you answer my questions about what you are trying to accomplish, maybe I
can give you some other instructions.
By the way, if you know the acceleration as a
function of time you should be able to determine the position and velocity
as functions of time if you know the initial position and velocity. It
would require numerical integrations because a will not be constant.
REPLY:
I am only simulating the rocket after burnout. As such, the problem is more like finding the trajectory of a cannonball mid-flight.
Like a cannonball, the projectile will have velocities in the horizontal and vertical direction, however only the vertical component is needed as I am trying to predict apogee.
The problem is that at an arbitrary time during the ascent of the cannonball, we will be given vertical acceleration (drag and gravity with direction), current altitude, and time (if that is needed though it might not be), and we must determine the apogee time and altitude.
Constants like mass (after burnout), cross sectional area, drag coefficient, etc. are known.
Drag is quadratic (v^2)
Without drag, the apogee is simply found with kinematics (constant acceleration), but since drag is a changing force that imparts a changing acceleration unlike the constant 9.81 of gravity, no one on the rocket team could figure it out.
Also, I have found other equations online that include drag and give altitude as a function of time, but the equation we need should just give out a number for max altitude based on the given values stated above. That's why we tried using energy formulas to solve the problem since at the top there is no kinetic energy, only potential (again only vertical components).
ANSWER: There is a good reason why you are having so much trouble: You are high school students who, at most, have had introductory differential and integral calculus. As I will show, this problem
will require differential equations. And not just simple linear
differential equations, but coupled second-order nonlinear differentials.
Such equations can only be solved numerically, almost never analytically. I
have given a little thought and have thought of a way that just might work
for you. It won't be easy. One thing before we get started: You said "only
the vertical component [of the velocity] is needed ." This is
not correct because to get the behavior of the vertical component you must
also know the horizontal velocity.
One thing which I am not clear on is what you are
given. It is imperative that you know the initial velocity, both magnitude and direction, to
solve this problem. When you say "we will be given
vertical acceleration" I think you must mean "we will be given
acceleration" because the acceleration is not vertical. When you say "drag
and gravity with direction" I assume that the drag force, both vertical and
horizontal components are given. The drag force is usually written
=-C|v |v ,
so Dx =-C √[vx 2 +vy 2 ]vx
D y =-C √[vx 2 +vy 2 ]vy v =√(D /C ) andvy /vx =Dy /Dx =tanθ
θ= tan-1 (Dy /Dx ) is the angle v makes with the horizontal.
From the initial drag you can now get the two components of v vx =√(D /C )cosθ and
vy =√(D /C )sin θ .t =0) values of the components of
the velocity. I will choose that the initial values of x and y
are both zero, i.e. I choose the origin of my coordinate system to
be at the initial position of the rocket.
So now we are ready to do some physics. What's key
in classical mechanics? Newton's second law, F =ma m (d2 r t 2 )=-C |v |v -mg 1 y 1 y m is the mass,
r is the vector pointing from the origin to the
rocket at time t , and g is the acceleration due to
gravity. I realize that vector calculus is something you have not yet
studied, but it's not so bad because we just write it in component form:m (d2 x /dt 2 )=-C|v |vx =-C √[vx 2 +vy 2 ]vx
m (d2 y/dt 2 )=-C|v |v y -mg=-C √[vx 2 +vy 2 ]v y -mg. These are second-order (second derivative), non-linear (square root),
coupled (vx and vy both appear in
both equations) equations. But we can make them much more tractable by
noting that (d2 x /dt 2 )=(dvx /dt )
and (d2 y /dt 2 )=(dv y /dt ).
Finally we have vx /dt )=-β √[vx 2 +vy 2 ]vx
andv y /dt )=-β √[vx 2 +vy 2 ]v y -g β=C/m . Now we can't solve these analytically but we can
do it numerically (that's where you come in!). This will just involve
arithmetic. vx ={-β √[vx 2 +vy 2 ]vx }Δt
andvy ={-β √[vx 2 +vy 2 ]vy -g }Δt. So here is your computer programming problem. You know β , g ,
vx , and vy at t= 0, they
all are just numbers. Choose some Δt , say Δt =1 s
and then you can calculate the changes in the velocity components. Now you
know the (approximate) values of the component velocities at t =1 s. Now you
calculate with those and get the values of the velocities at t =2 s.
Continue until you find the time when vy ≈0; going
past this t vy will become negative. You will probably
need to try several values of Δt until it is small enough that the
final answer doesn't change (you are, after all, simulating what you would
get with infinitesimal Δt). After you find the total time to
the top, you still haven't answered where it is. What you need to do is to
integrate into your program calculation and tabulation of Δx=vx Δt and Δy=vy Δt
at each interval, and calculate running sums. The final sum of Δy
is your answer.
I know that this is a lot for a high schooler to
digest. But I have tried to simplify as much as I could. As I warned you at
the beginning, when you have air drag to contend with, you have a much more
complicated problem than with constant forces. If you are able to pull it
off I would love to see your results.
ADDED THOUGHT: To check that I have not made an algebra error
somewhere I have written a PYTHON code to do the calculations I suggested
above. I randomly chose reasonable boundary conditions, β =0.001, v x t =0)=100 m/s,
v y t =0)=200 m/s, g =9.8, and Δt =0.25 s.
Shown are the components of the velocity, the components of the position,
and the path of the rocket.
QUESTION:
We have read for years that the universe is composed of certain percentages of visible matter, dark matter, and dark energy.
I can understand that there are ways of reasonably estimating the amounts of visible and dark matter. They are described for us
in published articles. How does anybody estimate the amount of dark energy?
And what is it anyway? Is dark energy expected to have all the characteristics types that regular energy has - kinetic, thermal, etc.?
All of the articles that I have seen simply present a percentage associated with dark energy. There don't seem even to be some
typical characteristics of dark energy that would allow us in the public to form a concept of what it is. Falling back on the Classical
Physics model, dark energy has to be the ability to do work on dark matter, if that helps!
Regardless of what dark energy actually, is one still wonders: how is it estimated?
ANSWER: Site ground rules clearly state that I do not do astronomy/astrophysics/cosmology. I can tell you though that it is not clearly understood what either dark matter or dark energy actually are.
FOLLOWUP QUESTION:
But as a last thought, they tell us that dark matter can be evidenced by its observed effect on real matter. And then they quote actual percentage estimates for the
amounts of dark matter and energy.
And yet dark energy, I think, has not even been detected. There must be a process used to arrive at these numbers. That's what I am really after. Can you direct me to anyone or any organization that I can communicate with to find out just how it is done? My thought is that, if I can understand the process, I can more clearly
Understand what "they" think these things are. Maybe not.
ANSWER: Dark matter is thought by most astrophysicists to be some kind of particle which interacts only via gravity. It is very hard to "observe" something if you cannot interact
with it using electromagnetism. There are numerous anomalies in the
universe which seem to not be described by our current "best" theory of
gravity, general relativity. You may make a model of hard-to-detect
particles with mass but which only interact with the rest of the universe
via gravity, but if that model works it does not mean that you have
"detected" dark matter, you have only fitted the data, not necessarily one
which tells you what dark matter is ; at this level it is
empiricism. It may be that dark matter is not stuff, but an indication that
we don't understand gravity as well as we think we do.
Dark energy is a different kind of thing. You say
that it "…has not even been detected…"
That is not correct, it has been "detected" just as surely as dark matter
has. The evidence for it is that very distant galaxies are observed to be
not just moving away from us but accelerating. In other words you can
assume that there must be some universal repulsive force pushing the
universe apart. In my opinion, dark energy is in just as detected as dark
matter is.
There is an interesting back story about dark
energy. Circa 1918, when Einstein was putting the finishing
touches on his theory of general relativity, it had not yet been discovered
that the universe is actually expanding; it was assumed to be static. That
presents a big problem since gravity is a long-range attractive force which
should have been causing the universe to be collapsing. So he added a term
to general relativity to hold everything in place and it became called the
cosmoligical constant λ . Much later in his life, when it was discovered that
the universe is not static, he said that λ was his "biggest blunder".
In 1998 when acceleration of distant galaxies was observed, new life was given to
λ ! If you are interested in delving into this more deeply,
check out this
link.
QUESTION:
I was wondering if a feather and a bowling ball actually pull towards the earth at the same rate? Why would the gravitational fields produced by the feather and the bowling ball not increase the rate of their individual attraction towards the earth?
ANSWER: First we need to specify that this notion
requires that there be no forces (like air drag in the real world) other
than gravity acting, the earth must be assumed to be enormously more
massive than the feather and bowling ball (which is true), and the
gravitational force which the earth exerts on the objects is uniform (which
is true only close to the surface of the earth). Also, you seem to be a bit
muddled as to what is going on. The reason that the two masses accelerate
is that each feels a force due to the gravitational field of the earth, it
has nothing to do with the gravitational fields of the objects themselves.
You seem to be thinking that the bowling ball feels a bigger force and that
is true; the force anything feels (which is called the weight of that
thing) is proportional to (depends on) its mass. But the effectiveness of
that force, what acceleration (the rate at which its speed will change
while falling) it falls with is inversely proportional to its mass. So, if
the mass of the ball is 1000 times bigger than the feather's mass, the
force on it will be 1000 times bigger. But, the more massive an object is,
the more force required to give it a certain acceleration is. It just so
happens (because of Newton's second law that the force necessary is
proportional to the mass. Therefore although the ball has 1000 times the
mass, it takes 1000 times as much force to accelerate it as it takes to
accelerate the feather. The acceleration is g =9.8 m/s2 .
If you want a more mathematical discussion, see an
earlier answer .
QUESTION:
One of the issues YouTube physicists discuss regarding interstellar travel is that the high speeds we would achieve would render the spaceship more vulnerable to constant bombardment by particles, gases, and dust, requiring a thicker hull in the direction of travel. Assuming interstellar travel is possible with current or near-future tech (not a given), is it true that more particles and gases would impact in the opposite direction of travel? That would seem to assume that most particles are more or less at rest in the frame of reference we are coming from with Earth. Wouldn't they all be going at random velocities up to near the speed of light relative to us?
Within the solar system, are particles mostly at rest relative to Earth's frame of reference? Within the gravitationally-bound Local Group?
ANSWER: Unfortunately, the space between planets, stars galaxies are not true vacuums as we often think of them. So, inevitably, atoms and molecules will be present in
varying densities, sometimes just moving around in relative vacuum, sometimes congregated in clouds. And you are right, they will be moving in random directions with a distribution of speeds, but the
speeds are all very small compared to the speed of light, c . But a ship traveling to distant stars would have to have speeds which are not so tiny compared to c . And
to someone on that ship, all the particles will be moving mainly at speeds
close to the speed which the ship has. When you say "mostly at rest" you do
not define what that means. What is true is that their speeds are very slow
compared to c ; if that's what you mean by "mostly at rest", then yes most
particles which you encounter will end up coming toward you with a speed
very close to the speed you are traveling.
FOLLOWUP QUESTION:
Thank you for answering my question!! I have refined it to clarify the point you brought up.
We are measuring the velocity of the spacecraft relative to Earth's frame of reference. Therefore, we are measuring the velocities of the particles it might encounter from the same frame of reference - Earth's. You can't compare any particular velocity to the speed of light without a frame of reference. You must have an observer with its own frame of reference to measure the speed.
Why should we assume that particles outside the Solar System or Local Group travel at very low speeds compared to the speed of light from our frame of reference? Shouldn't we see the same number of high-speed particles bombard the ship from all directions, regardless of its velocity relative to Earth?
Does the idea that the spacecraft will encounter more particles in the direction of relativistic travel rely on the fact that all particles in the Local Group are gravitationally bound together?
ANSWER: My bad! I should have specified my reference
frames. Let's call the earth our "rest" frame. First some numbers:
Typical atom/molecule densities most places in
our galaxy is 10-100 particles/cm3
Typical atom/molecule densities in intergalactic
space is 1 particle/m3
Typical temperatures of most places in our
galaxy is 10-100 K
Typical RMS speed of molecules is slow, e.g. ,
CO2 at 10 K is about 300 m/s
So if we say that probably particles with speed
10 v RMS are the fastest with any appreciable
number, those have speeds 3000 m/s=10-5 c .
Since the biggest speeds relative to the earth are
so tiny compared to the speed of light, they look like they might as well
all have been at rest compared with the motion of the ship. Therefore, on
the ship with some velocity U every atom or molecule will be seen moving
with a velocity very close to -U
Suppose that you are driving in a forest at 50 mph.
Every tree from your frame of reference is moving backwards at 50 mph, not
just trees in front of you.
QUESTION:
Since the discovery of the Higgs Boson, which is said to be what makes matter essentially makes things be physical. Then wouldn't it make sense that there is indeed a graviton? In where the particle is in a wave function until check and collapsed into a particle? Not unlike the double slit. Couldn't this be the case? And it is my opinion that this could fit nicely into an current theory.
ANSWER: I don't understand hows Higgs or double slits have anything to do with whether "…there is indeed a graviton…".
I will assume that the main issue is you want to know is whether "…there is
indeed a graviton…". As you may know (see the next question) general
relativity is generally depicted as a geometric model, deformation of
spacetime by mass. So gravity is not a force, merely geometry? If that is
the case, why seek a quantum theory of gravity where the quanta (gravitons)
would be "messengers of the force" when there is no force, only geometry?
But a few years ago I came across a short
essay which made the whole thing clear to me. A theory does not have to
have only one framework of characterizing it. For example, when quantum
mechanics was in its infancy, two competing mathematical frameworks were
used, both of which seemed to come to the same result when calculating:
Matrix mechanics was formulated by Heisenberg, Born, and Jordan;
differential equation formulation of quantum mechanics was introduced by
Schrodinger. The two were eventually shown to be identical. Similarly
general relativity has the properties of a field theory and can be
interpreted as having forces and being quantized. Hence, there should be a
theory of quantum gravity. If you want to read more see my
earlier answer
and be sure to look at the
essay linked there.
QUESTION:
Hi. I've been pondering gravity lately. Why is it that -- although Einstein's general relativity best explains what we call
"gravity"-- one still sees statements such as "all masses attract each other—you and your pet, the earth and the moon . . ." etc., etc. What magical
"attraction" are they talking about? Also, could you recommend a
good illustration of what actually happens when an object near the earth
"falls"?
ANSWER: The theory of general relativity (GR) is quite complicated mathematically.
To understand a complicated physical theory requires that we have a
qualitative explanation of some effects predicted to visualize. GR is a
theory which uses advanced geometry formalism, a four-dimensional spacetime
(time being the fourth dimension). It therefore becomes a theory we think
of as a geometrical theory and envision massive objects to deform the
spacetime around them resulting in attraction. But, there's a problem—since
we live in a 3-dimensional world we really can't visual a 4-dimensional
world. So what is usually done is to imagine a 2-dimensional plane which,
if we put a bowling ball in it, gets deformed. If there is a marble nearby
it will roll toward the bowling ball; I often refer to this as the
trampoline model and it is a cartoon not to be taken too seriously but to
provide an example of an object deforming some space around it. You can
also give the marble a little shove and it will orbit the bowling ball as
shown in the figure. The marble also deforms the space around it but since
it has so little mass compared to the bowling ball, it is not apparent in
the figure.
QUESTION:
Recently I've got interested in one question. Lets assume we have 2 balls with same mass in space on some distance from each other (no other objects around). At the initial moment of time they are staying still (velocities = 0). Then, due to gravitational attraction, they start moving toward each other. At some moment they collide and collision is elastic—so they move away from each other, slow down and then again start moving toward each other (then collide ets.)
The question is—will they bounce forever (because of conservation of energy) or they eventually, like, stick to each other?
ANSWER: The problem with your question is that you ask for an answer in an ideal situation, perfectly elastic collisions and perfectly empty space. Under those conditions, the two balls would bounce
forever. The fact is, though, that
no macroscopic balls can collide perfectly elastically and no vacuum is
perfect. So in the universe we live in, the balls would lose a little
energy every time they collided and quickly stop, "…stick[ing]
to each other…". And even if you had perfectly elastic collisions,
actual "empty space" (intergalactic space) contains about one molecule per
cubic meter; it might take eons, but eventually the balls will collide with
those molecules giving some of their energy to them and eventually sticking
together. And don't forget, there are lots of photons zooming around which
could add or subtract energy when they hit the balls.
QUESTION:
A while back I was reading an article about the electromagnetic spectrum, the visible light spectrum specifically. And a question raced through my mind. If I understand correctly the colors we perceive are based on reflective properties of objects in our environment. Certain reflective properties reflect different wavelengths of light and we perceive them as colors. My question is in the absence of all light does anything truly have a color? And if so what do you think that looks like? An infrared photo is the closest that I can imagine.
ANSWER: If a tree falls in the forest and there is nobody nearby to hear it, does it make any sound?
Your question is similar—both are about semantics, not physics. If
you say that sound is your perception of it, the tree makes no sound if
there is nobody to hear it; on the other hand if you say sound is the
presence of compression waves traveling through the air, there is sound.
Color is trickier, though, because the color you perceive depends on how it
is illuminated. We usually think of white light as a distribution of all
the wavelengths of sunlight. So just saying something is red is a grossly
undetermined description because something red in one illumination may look
pink in another. So physics avoids this confusion and simply uses the word
color in a qualitative sense, just like you might describe something as
warm to denote temperature. So if you want to know what "color" something
is you specify the distribution of wavelengths emitted from an object when
it is illuminated by a different distribution of wavelengths.
QUESTION:
I was trying to bounce my Walmart ball at the ground and have it bounce off the ground, hit the ceiling, bounce off the ground and then hit the ceiling again. I noticed
that no matter how hard I threw, it would never hit the ceiling twice. I assume it is impossible for a physics reason and would like to know specifically what. The ceiling is about 8 ft high, ball is about 42 inches around.
ANSWER: The "physics reason" is that your arm is
incapable of giving the ball enough energy. Each time the ball has a
collision with ceiling or floor it loses a fraction f of its
energy, so if its incoming energy is K , its outgoing energy is
K (1-f ); as it travels from the floor to the ceiling it loses
energy V=mgh where m is the mass of the ball, g
is the acceleration due to gravity, and h is the distance the
center of the ball travels; as it travels from the ceiling to the floor it
gains energy V . What I did is to calculate the energy step-by-step
to find the minimum initial energy K 0
needed for the final kinetic energy to be zero, just as it reaches the
ceiling for the second time. Then I solved the equation I got for K 0 .
The result was
K 0 =V (F 2 -F +1)/F 3
where F =1-f , the fraction of energy retained in a
collision.
The energy is just the kinetic energy,
K 0 =½mv 2 =mgh (F 2 -F +1)/F 3
so v =√[2gh (F 2 -F +1)/F 3 ].
So let's look at your case, the radius of the ball
is about 6" so h is about 7'=2.1 m, g is 9.8 m/s2 , and I will
guess that F is about 0.5 which means half of the energy is lost
in each collision (I dropped a similar ball and it bounced back about
halfway from where I dropped it). Therefore, v =15.7 m/s=35.2 mph. I doubt that you
could give it that velocity.
QUESTION:
I'm doing my level best to finish my first speculative fiction novel, and this is a question to which I'm unable to have anywhere close to 100% confidence in my answer. I'm hoping your much more vast breadth of knowledge might help give me a bit more certainty with it; even if you have to do some educated guessing - which is more than I'd be able to do (though I've shared my guess below). So here goes:
I would like to have as close to an accurate sense of how water would move on other worlds; specifically, were all atmospheric conditions identical to Earth's, how would water (and most other liquids then) move on the Moon ("The Expanse" showed liquids moving much more slowly on the Moon) and on Mars?
My guess is, if they were right - I assume Astronomer/producer Naren Shankar has an pretty solid sense of it - and atmospherics are essentially the same, the biggest factor is the strength of the gravity; more than the speed of rotation (but this is a factor I'm uncertain about). Stronger gravity will make liquids move more quickly, I believe, but if they were right on that show, then twice as fast as on the Moon - or if rotation speed is a bigger factor than I'm thinking, that's something I'll have to educate myself about.
ANSWER: The speed v in deep water is v deep =√(gL /2π ) where
L is the wavelength of the wave and g is gravitational acceleration; 'deep' means
h ≥½L . So the speed is proportional to the
square root of g and to the square root of L . For example, a water wave on
the moon would have a speed of about v moon =√(1.6/9.8)v earth =0.4v earth .
On Mars, v Mars =√(3.7/9.8)v earth =0.61v earth .
For shallow water, h ≤0.04L , the speed is approximately
v shallow =√(gh ); here the speed is
proportional to root h rather than root L . There is even
an approximate expression for the intermediate depths, v inter =v deep tanh(kh )
where k =2π /L . This is probably more than you
need to know, but if you want to see the whole reference which I used, go
here .
QUESTION:
I understand that time slows down for objects moving at relativistic speeds. However, I don't understand why the slow-down in time directly affects things. Why are biological reactions slowed down (so I age more slowly). Or why are chemical or nuclear reactions slowed down? Even though they call time the
'fourth dimension', it seems to have a much more direct effect of things than space does. If space where increased or decreased by 50% in extent, I wouldn’t become 50% bigger or smaller. If physical objects were directly affected by the growth or reduction of space dimensions, how would you ever be able to measure that change?
ANSWER: You misunderstand time dilation. If you are
moving with some large speed relative to someone else, your clock runs at
some rate which is totally the same as you would at any speed. You
do not see that clock running at a different rate because you were moving
relative to someone else. If your ship had no way to look outside, you
would not even know you were moving. What is different is the rate which
you clock runs is different than the clock which is at rest relative to you
runs. You should read my description of the
twin paradox .
QUESTION:
So I am having problem in understanding if the acceleration must always be constant when we think of Free Fall. For example, imagine if I were about 70,000km above Earth's surface and the only object in my vicinity is Earth, then I’d only feel gravitational force of earth. Suppose I am falling towards Earth, then by definition what I experience is free fall. But when I get closer to Earth, say when I am 1,000km above the surface, the acceleration I'd feel would not be the same as what I felt when I was about 70,000km above the surface.
Having said that, can I conclude that the only reason we experience constant acceleration when we experience free fall on Earth is because the height for the motion is significantly less than the radius of Earth and therefore, we can ignore the minute changes in gravitational acceleration?
ANSWER: The gravitational force F on a mass
m at some distance r from the center of a spherically
symmetric sphere of radius R and mass M is F=MmG /r 2
where G is Newton's universal gravitational constant.
Because of Newton's second law, a=F /m , so a=MG /r 2 .
If you set M and R to be the mass and radius of the earth
you will find that a is about 9.8 m/s2 . That is the
only r which has this as the value of a ; in other words,
the acceleration is never uniform. But, as is often the case in
science, a good approximation can yield perfectly adequate approximate
results provided that you understand the limitations. It is clear from your
question that you already understand that very well. I will give you a
quantitative way of estimating how high h in altitude you can go
before the approximate acceleration differs from the ground-level
acceleration:
First, a (r=R )=g =MG /R 2 a (r=R+h )=g'=MG /(R+h )2
Next, write (R+h )2 =R 2 [1+(h /R )2 ]
so that we can write g'=g /[1+(h /R )2 ]
or g /g' =1+(h /R )2
Suppose we want there to be only a 1% error at
some h ; what is h ?
(g-g' )/g =g (1-g /g' )=0.01g =g (h /R )2
so h =√(0.01R 2 )=0.1R
QUESTION:
If the Laws of Thermodynamics and the Conservation of Matter state thate neither energy nor mass can be created nor destroyed, why do theoretical physicists fixate on the big bang as the creation of the universe?
To me it seems logical that both energy and mass (being interchangeable) are eternal, and that the big bang that caused the present universe may just be the latest of the likely cycle of eternal big bangs. What do you think, is it logical?
ANSWER: There is no such thing as "conservation of matter" because matter can be created or destroyed. The total amount of energy in the universe cannot be changed although the amount of mass may change. However, I see no reason
why the "eternal" amount of energy in our universe should preclude "fixating" on the big bang.
QUESTION:
But energy cannot exist without matter and matter cannot exist without energy.
As both would be stagnant in the absence of the other.
The laws state that in an isolated system neither mass nor energy can be created or destroyed. Mass is a body of matter of indefinite shape or size.
Therefore mass is matter and cannot be created or destroyed.
How then, do you validate your proposition that it can be?
ANSWER: This is utter nonsense. "The laws" state that in
an isolated system the total energy must remain constant. Mass is just a
form of energy, E=mc 2 . You are not alone in believing
that conservation of mass is a law of physics; I often get questions based on this assumption. So. where did this "law"
come from? Modern chemistry was developed about 400 years ago. Experiments
showed that if you measured all the mass which contributed to a chemical
reaction and then measured all the mass after the reaction, the two, within
the accuracy of measurement, would be equal. The catch is that the accuracy
of the measurement is too poor to say that the two are precisely equal.
When special relativity was developed, it became clear that mass could be
converted into energy because E=mc 2 . In an exotheric
reaction, energy is released in the form of heat and the origin of that
energy is a tiny decrease in the mass. Indeed, even with the most sensitive
instruments it is virtually impossible to measure this; chemistry is a very
inefficient source of energy. But if you look at nuclear reactions, the
mass change is easy to measure; e.g ., the mass of a helium nucleus
(two protons and two newtrons) is about 3% lighter than the summed mass of
two protons and two newtrons. Stars, like the sun, are using nuclear
reactions to release energy from mass. If the sun were burning using
chemistry, it would have been totally reduced to ashes millions of years
ago.
QUESTION:
I was wondering how the oceans can absorb CO2 from the atmosphere if they have achieved equilibrium.
ANSWER: Lot's of reasons. First, the CO2 doesn't just swim around but iteracts chemically so that
the water dynamically reduces levels allowing more CO2 to replace that decrease.
If the amount of CO2 in the atmosphere is increasing which
causes the equilibrium to be undone whether the CO2 is
decreasing or not. So the real equilibrium is dynamic, not static. By the
way, the chemistry causes the ocean to become more acidic, not desireable.
QUESTION:
I have come across this paper written by a person called John Mandlbaur. He claims:
"My papers are rejected because they come to a conclusion which the editor doesn't believe. That is what rejected without review means.
Numerous physicists have described my maths as perfect and nobody can defeat it."
Now I do not believe this because although I am no physicist, I know physics well enough that papers are not rejected without good reason so I am sure his is rejected because the calculation and/or reasoning is fatally flawed but being no physicist myself I am not qualified to comment on the paper. If however you identified the relevant flaws in straightforward English, I could post your reply on Quora and hopefully prevent others from being taken in by these claims of Mr Mandlbaur's.
If this is not for you, but you can recommend someone else to explain why the paper is refuted in plain English please let me know.
ANSWER: I
interacted with this fellow about 7 years ago.
We had a very lengthy back-and-forth discussion of his "ideas" which
eventually became so outrageous that I relegated him to the
"Off-the-Wall" Hall of Fame ". I
don't want
to call names, but there has never been a questioner more worthy of
off-the-wall status than him. It is very sad that he has held on to these
delusions for such a long time. I think it was Bill Murray who said "It’s hard to win an argument with a smart person, but it’s damn near impossible to win an argument with a stupid person."
I am sure that if you go over these ramblings you will have plenty of
examples to convince yourself that Mr
Mandlbaur is a crackpot.
QUESTION:
I am writing a science fiction story, and I'd like the science to be accurate. So: if a space ship starts from Earth and steadily accelerates at 2 G's, how far will it be from Earth
in 3 months? I'm writing a love story intertwined with the difficulties of
interstellar commerce.
ANSWER: Before you do anything you should read the four
previous answers
which you can find in the faq page. If you don't care to get into the
details you can just go to
this link which will let you know how I did the calculations for you. You first need to understand that acceleration is not a very useful quantity in physics if velocities are comparable to light speed;
this is discussed in the previous answers I refered to above. Since you are
interested in high speeds, we need to first determine whether classical
Newtonian physics can be used or whether we need to use special relativity.
We first need to specify who is measuring the acceleration since it is not
the same as seen by everyone with high relative speeds; I assume that you
are interested in where the ship is and how fast it is going after three
months. The figures show calculations I have done assuming g =10
m/s2 , for both relativistit (red) and Newtonian (black)
mechanics, and an acceleration observed from the rest frame of 2g .
and essentially velocity as a function of time. The first figure calculates
up to about t= 23 months, the second to 3 months. Note that the
first calculation has the relativistic calculation approaches c at
large t whereas the Newtonian calculation increases linearly
forever, already at about twice light speed at about one year. But, as
shown in the second figure, the two calculations are approximately equal at
3 months. Therefore at 3 months you could say that the speed of the ship is
about half of light speed and the distance traveled was about d =½at 2
=10x(7.78x106 )2 =6x1014 m=0.06 light years.
It would be much more complicated to calculate the distance
relativistically.
QUESTION:
My question is about spinning black holes. Or rather spinning singularities. Like singularities are one dimensional objects right? A point of infinite density.
Anyway. don't you need like...another dimension to be capable of spinning? Like what axis does the point spin around? How can you rotate when there is no geometry? I Thought it was a point. How does a point rotate? It wouldn't make sense though. The thing would suddenly stop spinning because there was no more geometry to rotate.
Anyway, everytime I try to think about it I sorta reach this block in my head.
ANSWER: Not one dimension, zero dimensions. Still your questions are pertinent. The first thing you should accept is that we do not know what the physics of a black hole is. The laws of physics are
not known inside the Schwarzschild radius. But reasonable calculations which can be done seem to lead to a true mathematical singularity.
I feel that most physicists are uncomfortable with the literal idea of
infinities. That said, there is no reason why a point particle cannot have
angular momentum. When black holes initially form from the gravitational
collapse of a dying star and the star originally had angular momentum, that
property will be conserved. From that point of view, one would expect just
about every black hole to be "spinning". You are thinking classically, but
most elementary particles have intrinsic angular momentum (spin) but if you try to interpret it as the
physical spinning of some mass distribution about some axis you will fail.
Electrons are thought to be point particles and they certainly have spin.
QUESTION:
I am trying to figure out how quickly (approz) a tangential velocity (I think) something would have to have, rotating around the earth at an altitude of about 400kms, to have a centripetal force of 1g ie 9.82 newtons. I think that's the right terminology?
Here's my thinking:
I think escape velocity at that distance is (very) roughly 27,360kph?
So if you're going at just or just under that speed you're in constant freefall round the earth, ie experiencing zero gravity, in a vertically constant orbit.
(First hurdle: I don't even know if that's correct!)
FYI, I'm imagining a hypothetical band looping entirely around the earth at ~400kms altitude, for example, creating its own centripetal force.
If you stood on the inside of that band, head towards the Earth, and this is where my thinking gets v v doubtful,
would you then have to spin whatever the whole artefact (I think it comes out at something like a tangential velocity of 29,366kph to create a 1g centripetal force?) on top of the originally calculated escape velocity in order to create 1g's reactionary force?
So, is the end tangential speed the band would need to rotate approx 56,726kph?
ANSWER: I am sorry to tell you that most of the things you say are nonsense.
Let me just tell you how things are rather than picking your question
apart. First, qualitatively: A circular orbit of radius r (r
being the distance from the center of the earth) will have a particular
velocity associated with it; you seem to think that you can choose the
velocity you would like to have in that orbit. The gravitational force on
any orbiting object gets smaller the farther away from the center of the
earth you are; as you know, the gravitational force on an object of mass
m at r=RE is
F (RE )=9.82m
where RE 6
m is the radius of the earth. So the acceleration is g (RE )=9.82
m/s2 .
But, in general g (r )=ME G /r 2 where
G= 6.67x10-11 N·m/kg2 is Newton's universal gravitational constant
and ME = 6x1024 kg. So now you can see that every orbit is in free fall
except the acceleration of that fall is dependent on the radius of the
orbit and is only approximately equal to what you think of as g at
near-earth orbits. We can also calculate the speed at any orbital radius
r
from Newton's second law
F=mg (r )=mv 2 /r =m ME G /r 2 .
So, solving, v (r )=√(ME G /r ).
You are interested in the altitude h =4x105
m so r =RE +h =6.41x106
m. Doing the arithmetic, g (r )=9.74 m/s2 and v (r )=7.9x103
m/s=28,440 km/hr.
ADDED THOUGHT: Rereading your question it appears that you might want to be making "artificial gravity" like the big donut-shaped space ships which you see sometimes in sci-fi movies like
2001 A Space Odyssey . But I don't know why you would want to do it
around the earth where there is already gravity present. If you are
interested in rotating space ships to create the appearance of gravity
because of the centrifugal force, I have answered questions along those
lines in the past. Find links on the
faq page .
QUESTION:
I suppose this is an off the wall question, but it is quite practical, and I'm not sure where to ask it. A web search turns up nothing of interest.
I've convinced myself that my coffee cup is interfering with the 2.4GHz signal between my mouse and its receiver, which is plugged into the back of my display. The spatial orientation of the three objects is not ideal, but my desk setup is a bit limited, basically thus:
2.4GHz wireless mouse on pull-out keyboard tray, mouse on the right.
My ceramic coffee mug on the surface of the desk, just above and beyond the mouse, also on the right.
My display is just above and beyond that on the desk, with the mouse receiver plugged into a USB port in the middle of the back of the display.
Ordinarily, these three things tend unconsciously toward an unfortunate syzygy*. I suspect the coffee cup is interfering with the signal, causing severe mouse lag.
So far unexceptional. I don't see why the ceramic mug or its mostly water contents might not be reflecting the signal.
But there doesn't seem to be any interference when the mug is full. Only after drinking half or more do I notice the lag. If I move the mug way over to the left, that seems to eliminate the lag.
So my question is, does the water level in the mug change the overall resonant frequency of the mug such that it can interfere with the signal at some levels but not others? Is it like filling a jug to different levels to get different tones when you blow across the opening? And considering that microwave ovens work on a similar frequency by heating water, would water be an especially problematic interloper at 2.4GHz?
ANSWER: Interference and blockage of electromagnetic
waves are tricky to understand and track down. Like sometimes when I am driving and come to a stop there will be a specific spot where a local
radio station will suddenly almost be totally blocked. All the things you have tried are reasonable: I salute you because physics is, at its root, an experimental science. If you could tolerate moving your cup to the
left side, that is likely your best solution. The 2.4 GHz seems like a
likely culprit here, particularly since there seems to be negligible
blockage once the water is removed. Cups of different construction would be
useful to look at, e.g. , a metal cup with and without water. I
doubt that it is anything to do with interference like the blown over
bottle because sound waves are of the same order-of-magnitude wavelength as
the resonating object whereas the GHz waves are nowhere close to the size
of the cup. Good luck!
*If you, are puzzled, as I was, by the funny word
syzygy, it means when three or more objects line up as in eclipses for
example.
QUESTION:
Sir,My doubt is in POTENTIAL ENERGY .I am really struggling in understanding this topic .please help me out.Sir my question is that -"is potential energy just a term to make sums look easier?" Sir, taking an example - if we throw a ball with v velocity perpendicularly upward then at the highest point the velocity becomes zero therefore the kinetic energy becomes zero but at this point we say that the energy has been transferred into potential energy but acutually where did the energy go ? And thus can we say that we defined potential energy so that our famous rule does not gets violated i.e. "energy can neither be created or destroyed it can only be transferred from one form to another " is it so ??? What does it mean that energy is stored???(potential energy)?????
ANSWER: My, you really are struggling. It is clear that you are just beginning your physics education, so I will keep my answer addressed only at that level. In fact, I will only discuss
problem involving a uniform gravitational field. First, suppose we know nothing about potential energy. So, the example you give is taking a ball of mass m and throwing it up vertically with some velocity v .
Then it starts with energy E 1 =½mv 2 and rises to some height h where all its energy is gone, E 2 =0. So energy was not conserved; why not? Because some force must have done work on the mass to take its energy.
Of course, that force is gravity which is mg . Because the force is constant and pointing down and the displacement is positive is positive, the work is negative, W=-mgh . The work done is equal to the
change in energy W=E 2 -E 1 =-mgh =0-½mv 2 . You could use this to find out how high the ball will go,
h=v 2 /(2g ); you have probably done such a problem
before you knew anything about potential energy.
I next want to generalize this problem. The ball is
thrown vertically from a height y 1 with a
velocity v 1 and at a later time it is at different
height y 2 and now has a
velocity v 2 . So now the work done is -mg (y 2 -y 1 )
and the change in energy is
½mv 2 2 -½mv 1 2 .
So, rearranging, I can now write (½mv 2 2 +mgy 2 )-(½mv 1 2 +mgy 1 )=0.
If there are no forces other than gravity which are doing work on the ball,
the quantity ½mv 2 +mgy never changes! So
now I get the brilliant idea (just one way of looking at this problem) to
call K =½mv 2 the energy due to motion
(kinetic energy) and V=mgh the energy due to position (potential
energy). So you can call gravitational potential energy a clever trick
which allows you to never worry about calculating the work done by gravity.
Another great reason to use potential energy is that energy is a scalar
quantity whereas force is a vector quantity.
There are some things you have to be careful of. If
you are using V=mgy for gravitational potential energy, you must
choose a coordinate system where +y is up; e.g ., if the
top of a cliff is at y =0 m, the bottom of the cliff cannot be at
y =100 m. You cannot use potential energy for just any old force,
it has to be a conservative force. A conservative force is one which does
zero total work over a closed path (ending where you started). E.g.,
gravity does negative work on the ball and positive work on the way down.
If you push a book across a table you must do work against friction; but
now if you push it back it will be the same amount of work so you cannot
have a potential for friction. Your "famous rule" is not very useful. I
think you should think in terms of the energy equation, E final =E initial +W ext ,
the energy you end up with is the energy you started with plus what you
added. W ext is the work done by all external
forces, forces for which you have not introduced a potential energy; forces
represented by potential energy functions are not external, you have
internalized them.
QUESTION:
I saw an older question on your site regarding how fast Flash could run a quarter mile.
Using similar assumptions, in Science Fantasy - verse Warhammer 40000, there are Space Marines who are biologically enhanced super warriors with 19-22 implanted extra organs that make them able to overwhelmingly outperform a regular human. Space Marines are roughly 7-8.5t tall (depending on if it is earlier variant with 19 organs or later one with 22 organs then the height is on the taller end of the spectrum) with a mass of 750 pounds unarmoured.
Space Marines have been decribed running 87 km/h taking six meter strides.
Seeing that much shorter and less massive Flash could run quarter mile in ten seconds giving a velocity of 144 km/h, using the formula, how fast would you put the maximum velocity a Space Marine could run given their mass of 750 pounds?
Can you give one estimate using the rubber boot sole, and one for ceramite boot sole that their armour is made of?
And when a Space Marine wears armour, his mass is increased to 500-1000kg, so how does wearing armour affect the maximum velocity when affected by laws of friction, gravity, etc...?
ANSWER: Oh my, you missed the whole point of that
caluclation. The mass does not matter because the frictional force he can
exert without slipping is proportional to his mass. And his acceleration is
inversely proportional to his mass; just the same reason why the
acceleration of a falling mass (in vacuum) is independent of mass. And
strength doesn't matter either if he can exert the maximum frictional force
without slipping, μ static mg.
QUESTION: How an ellipsoid object, say an American football...ball is thrown through the air. It shows projectile motion, but the other interesting observation is that the front of the ball points in the direction of travel. I have thought that air resistance possibly applies unequal torques on different sides of the ball perhaps, but I read somewhere that it has to do with gyroscopic precession - that this axis is different to the one for angular frequency, and somehow that explains it. I'm not sure though, any simpler way to understand this?
ANSWER: If we were to pass the football with spiral in a
vacuum, the spin axis would remain pointing in the same direction because
of conservation of angular momentum; this is the case for the lower
trajectory shown in the figure. In air, though, the angular momentum mainly
remains pointing in the same direction as the velocity of the ball. I always am surprised to find an everyday phenomenon the physics of which is unexplained or explained only recently. This is such a case and was explained only recently (2020)
in a paper in
American Journal of Physics . A more accessible description of the
phenomenon can be found at
this link . Your hunch that the air has something to do with it was
spot-on.
QUESTION: so, tell me if I'm wrong but if an object is energy it doesn't have mass yet it is still affected by gravity, but you need mas to be affected by gravity 1. why is energy still affected by gravity 2. is there any way whatsoever for something without mass that isn't energy be affected by gravity.
ANSWER: No, you do not "need mass to be affected by gravity".
The only object we know which has no mass is the photon (a quantum of
light). And because it moves, it has kinetic energy which is
well-known; you shouldn't say that a photon is energy though,
rather it has energy. When a photon finds itself in a strong
gravitational field it does not go in a straight line (as you see things)
which you might have expected, but rather it follows the curved lines of
the space where the gravity is; in other words, in a curved space a curved
line may be the shortest distance between two points. Your second question
refers to something which has neither mass nor energy; that isn't
really something , is it?
QUESTION: I hope this question isnt stupid, also English isnt my first language, but here I go: People always say that you can't travel faster than light speed, but what is the point from which you count the speed (0 m/s)? Because the earth is already moving through space, so would we need to accelerate less when accelerating in the same direction and more if we accelerate in the direction it comes from?
Follow up, may be explained by the answer to the first question: does the light emitting from let's say a LED travelling at 60mph reach a target 100 miles away faster than a staionary LED (assuming we turn both on the second they are next to each other, so when the moving one passes it)?
ANSWER: One of the triumphs of the theory of relativity is that its success validates
the principle of relativity: The laws of physics are the same in all frames
of reference. From this principle we can deduce that the speed of light in
a vacuum is independent of the motions of the observer or the source. So
that answers your 'followup': a bystander would see the two light flashes
arrive at the same time. Your main question asks which frame of reference may a moving object not exceed the
speed of light? Since all observers see the same speed of light, the answer
is any frame.
QUESTION: Dad I have a (possibly) physics question: how come tempered glass just spontaneously combusts without any apparent cause? I remember my old roommate's shower fully falling apart during the pandemic when no one was even there
ANSWER: Hi Hannah! I don't think you mean "spontaneously combust" because that would mean burning/flaming. Glass does spontaneously shatter or break in unexpected circumstances,
however. I have answered
variations of this question several times ( e.g. ,
1 ,
2 ,
3 );
1 is probably the most complete.
QUESTION: I have been thinking about a question, like how we can write the change in unit vectors over a finite time interval, now it
may seem like a homework problem but trust me it's not it's a genuine question that I have been thinking about. I know about the change in unit vectors in an infinitesimally small-time interval 'dt' and that it's given by the magnitude of the really small angle that the unit vector has turned about, but what about a change in the unit vector over a finite time interval how can we write that.
I tried to find the answer in many sources, but I was unable to find any.
ANSWER: It is not really clear to me what you are asking. If you are talking about Cartesian coordinates, they never change. But if you have a coordinate system like spherical polar coordinates, the directions
of the unit vectors are very dependent on where a point happens to be. I will use a two dimnensional polar coordinate system where the vectors have both radial and tangential components
to keep the answer from being too complicated. The figure shows a red dot
moving along some path (red dotted line) which shows the path taken by the
particle between two times t 1 and t 2 .
The unit vectors are shown in green. If you know those two locations you can see that the unit vectors all
rotate through an angle θ 2 -θ 1 in the clockwise direction.
The path taken by the particle will be described by some vector function
f fr and fθ
as shown in the figure. So the average rate of change of direction of unit
vectors is ω avg =Δθ /Δt =(θ 2 -θ 1 )/(t 2 -t 1 ).
QUESTION: A ship at the speed of light turns on headlights, understand that a person on that ship sees light go out at the speed of light, relativity. What does a person "see" from outside that area of relativity? Can they see that, is that currently possible, what if, if not current? Do they see a ship with no lights, light but no ship, a spectrum of both that equals the speed of light? I am, and with more questions, being as specific as possible: I find it hard to articulate a concept from the written word, please accept and understand my ignorance/lack of understanding!
ANSWER: First of all, no ship can travel "…at the speed of light…" and I do not answer questions which stipulate at
or faster than the speed of light. So I will answer your question assunming
v =0.99c , 99% the speed of light. As you say, an observer
on the ship sees the light in front of him having a speed of c
regardless of how fast the ship is moving. An observer relative to whom the
ship is moving with speed 0.99c also would measure the speed of
the light to be c . This is one of the postulates of the theory of
special relativity: The speed of light in a vacuum is independent of the
motion of the source or the observer. However he would not
"see"
the same "color" of the light because of the doppler shift. Suppose that
the light as seen by the ship had a wavelength of 450 nm=4.5x10-7
m, blue; for 0.99c speed of the source the doppler shifted wavelength would
be about 4.5 nm, low-energy x-rays. Hence the stationary observer would see
nothing but if he could measure the speed of the x-rays coming out the
front of the ship he would see them as having a speed of c .
QUESTION:
ANSWER: Imagine an object to the left of a spherical lens
as seen in the figure. There is an imaginary line drawn perpendicular to
the lens and through its center call ed the optic axis. There is some
object consisting of many colors. I have chosen to look at just three
points in this object, one red, one blue, and one green. Each rays from these
points illuminates exactly one point where the image is formed; although I
have only drawn three rays for each point, every ray passes through the the
same single point on the image. Apart from size and whether the image is
inverted or not, the image is an exact copy of the object. Your statement
that "…light itself is not coloured…" is not correct--the
color of light is determined by its wave length.
QUESTION: Assuming no engineering challenges like parasitic losses in materials etc…. Is it theoretically possible to take:
Speaker A. Play a tone. 440hz. Digitally. Using EXACTLY 100 watts of power.
Speaker B. Play a tone. 440hz. Digitally. Using EXACTLY 100 watts of power.
Have speakers A and B facing each other head on. EXACTLY 180 degrees.
The 2 speakers are playing the same note. Same signal, timbre, frequency amplitude pitch etc….
Would these two “sounds” cancel out perfectly such that “in between” there would be zero “sound”. I understand that small changes in angles and frequencies will have enormous effects on there and how the sound will dissipate if this system is not perfectly balanced from the get go. Very loud. But if the experiment were set up perfectly, is it theoretically possible to zero out the wave? Full 100% destructive interference? Chaos? Feigenbaum?
ANSWER: No, the result is a standing wave of the same frequency.
In the animation the original waves from the left (blue) and right (green)
combine to make the black waves which have the same frequency and
wavelength but zero velocity. This is due to the superposition principle
which states that if two or more waves are moving in the same medium, the
net displacement is equal to the sum of the individual displacements at
each time and location. Note that the black curve is sometimes totally
zero, when the two traveling waves are out of phase with other. If two
identical waves are traveling in the same direction they could completely
cancel if they were 180° out of phase. In the second (nonanimated) gif,
imagine the green and blue are moving in the same direction. The black
would be the resultant wave.
QUESTION: I am trying to understand heat, that is, mostly the idea of hot
and hotter. My understanding is that heat is basically motion, and
heating something up (cooking, let's say) is about transferring motion
from one place to another. So that the changes that take place in
cooking are really caused by the molecules of stuff rubbing against
each other. So hotter really means MORE motion, because the actual
temperature at which things burn, like natural gas, can't be
increased just because you burn more of it.
Does that make some sort of sense to you? Clearly my mind wanders
in the kitchen, and this has been nagging at me for a long time.
ANSWER: I believe that your question is answered by an
earlier answer .
You refer to motion; in my earlier answer the kinetic energy referred to is
K =½mv 2 where m is the mass of
some atom or molecule in your flame or pot or food, and v is its
speed (motion).
QUESTION: I have a question about fluid velocity/acceleration and its effects on tubular friction.
Is the relationship linear? Meaning , if the fluid velocity is being increased, is there a moment or a few moments where the friction increases more than what it would’ve been if the relationship was linear?
ANSWER: From what I have been able to learn while researching your question is that "friction" as in two solids sliding on their surfaces is not applicable
to fluid inside a pipe. The fluid exactly on the surface is not sliding but
rather adheres to the surface and is at rest. The motion of remainder of
the fluid has some velocity just outside the surface increases to a maximum
at the center but certainly not linearly. Everything about this velocity
profile is due to the viscosity of the fluid, sort of the fluid version of
friction. More detail can be seen in
this link and links which it contains. The figure shows a figure from
that link and shows the shape of the velocity profile, v (r ).
Eventually, at high enough speeds the fluid becomes turbulent and "all bets
are off".
QUESTION: How should I get to calculating the force that needs to be applied to a piston to compress a container of gas? Suppose the temperature is constant, and by default, as the volume decreases the pressure increases. Let's suppose the volume gets halved.
ANSWER: The force depends on the area A of the piston. And, if the gas cannot be well approximated as an ideal gas,
you would need detailed properties about the gas. For an ideal gas, PV /T
must remain constant and if T is constant, PV is
constant. Now, it seems easy, like if you halve the the volume you need to
double the pressure; that would give you the answer for the force to hold
the piston there once you got it there, F=P 2 A= 2P 1 A .
Be sure that you understand that this is what you mean because if you
simply started at the beginning with this force and continued pushing with
that force, when you got to the half-volume position you would go right by
it and it would start slowing down until finally it would turn around and
start coming back.
QUESTION: For reference I'm 31.
Am I closer to the age 60 than I am 20?
I think I'm closer to the age 60 because time is moving forward and we cannot move backwards to my age at 20. So wouldn't I be closer to 60 because time can only move in a linear motion towards 60 and we are constantly distancing ourselves from the age of 20?
ANSWER: You are 11 years older than 20 and 29 years younger than 60. You are closer to 20 by almost a factor of 3.
If you had asked which you would be closer to being, you would be closer to
being 60 than 20; but you didn't.
QUESTION: Is a mirror VERY different to how people perceive me in reality?
I am bothered by the fact that I never realised before but I guess there's a reason…
How impactful is the reversal effect?
I always thought that's how I look to others.
ANSWER: The image you see is not how others see you, but a "mirror image". If you look at the image of yourself in the mirror, the image of your right hand is the left hand of the mirror you. If you want
to see an image of yourself which is what others see, you need to use a corner mirror.
However, since most of us are pretty much left-right symmetrical, there is
seldom much difference.
QUESTION: My eight year old has a question.
Get ready.
If two particles are in entangled, and one is on the earth and the other has passed the event horizon of a black hole.
Would the entanglement be broken or even if it's unmeasurable still be entangled.
Does the medium "the ether" transcend the event horizon?
Silas Massicotte (8) and Malachi Massicotte (6).
ANSWER: Seriously? Your 8-year old understands quantum mechanics and cosmology?! That would be pretty amazing. I usually don't do, as stated on the site, astronomy/astrophysics/cosmology,
but I did a little research and think I can give you an answer. When one of the two entangled particles is totally destroyed, the other one will not know at all because there is no communication between the two.
Since the wave function of the destroyed one simply disappeared, the wave function of the other will remain in the mixed state it was in originally, but it will no longer be entangled with anything.
If a measurement is made on the survivor, there will be no change in
whatever remains of the other particle. There is a pretty good explanation
at this
link . There is no such thing as the ether.
QUESTION: how can electromagnetic waves travel in vacuum?
how can something without any physical body and dimensions travel in nothing?
it should need something to mediate its propagation
like sound waves require air molecules.
if its pure energy then what is pure energy?
ANSWER: This is exactly the way physicists were thinking
at the end of the 19th century. Serious searches for the medium required
for light to be able to propogate were being made. The best known search
was the Michaelson-Morley experiment which failed to find it. It was called
the luminferous æther. The punch line is that it doesn't exist at
all. Light can travel through empty space. One way to understand this is
through electro-magnetic theory, quite well understood at the begining of
the 20th century. James Clerk Maxwell unified all that was known about
electromagnetism into four equations, today known as
Maxwell's equations .
One of those equations says that time-varying magnetic fields cause
electric fields; another says that time-varying electric fields cause
magnetic fields. Using all four you can show that one possible solution is
a wave equation for both electric and magnetic fields. Since waves are
time-varying, these waves will "keep each other going" which has nothing to
do with any medium. If you go to the link above, you can follow other links
telling you more.
QUESTION: I have a question about a calculation in the USPSA (United States Practical Shooting Association) rulebook. The power factor of ammunition is measured by the weight of the bullet in grains times the velocity in feet per second, divided by 1000. A 165 grain bullet at 1000 fps would be 165 power factor.
This is intended to keep recoil the same between competitors. Felt recoil is subject to a lot of variables that have more to do with the firearm than anything else. But the rule is to ensure that everyone has a minimum level of recoil to manage.
Does a 110 grain bullet at 1500 fps, or 165 power factor exert the same force backwards as a 250 grain bullet at 660 fps, or 165 power factor? All else being equal, is power factor an accurate measure of recoil? The muzzle energy is much higher for the lighter bullet.
ANSWER: The quantity power factor is nothing more than linear momentum
(mass times velocity) in physics but in unusual units, milligrain feet per second:
mgr·ft/s); physicists usually express it in
kg·m/s. But the units you use don't affect the physics so I will use
yours. The thing about linear momentum is that if there are no external
forces on a system, it never changes its magnitude; this is called
conservation of linear momentum. So, if the rifle has a mass of M =3 lb=21,000 gr=21,000,000
mgr and the
two bullets both have equal power factors of 165 you can calculate the
speed V of the rifle after the gun is fired: Before the gun is
fired momentum is zero, so it must be zero afterwards: 0=165-21,000·V
or V =0.00786 ft/s=0.094 in/s. Now, you asked for the force on the
shooter. There is not enough information to calculate this because that
depends on how quickly the rifle takes to stop—if you have a bony
shoulder it would hurt a lot less if you stopped it more slowly by having a
squishy pad between you and the rifle butt. Suppose it takes 1 ms to stop;
then the average force on you over that time would be [3 lb x 0.00786 ft/s
/ 0.001 s] = 24 lb. But you said "all else being equal" so I presume that
the recoil force would be the same for both scenarios because the masses of
the guns and the recoil velocities are the same.
QUESTION: When text book gets to quanta, it would say E=hf. It would say f = n * fundamental frequency. Ok, I can understand there are 1Hz and 2Hz and so on. But couldn't one build a circuit that makes 1.5Hz? At the end, it is a voltage completing a cycle within some time.
And if f can be decimal, then it is continuous. So E is continuous.
I must be missing some concept here.
ANSWER: Not everything is quantized, but
even if everything were in some way quantized, you could only detect with
your senses in some situations. Usually we think of quantum physics as what
happens in very small systems, systems on the general order of atomic
sizes. The thing which is bothering you, I think, is that something which
is macroscopic which shows no signs of being quantized means that an a
microscopic version of that system would not be quantized either. Take, for
instance, a mass m hanging on a spring. So if you pull it down to some
amplitude A , say 1 cm and let it go, it will oscillate up and down
with some frequency f determined by the stiffenss of the spring
k and its energy will be E =½kA 2
and its frequency will be f =2 π √(k /m ).
You could quadruple the energy by doubling the amplitude. But, of course,
you would be able choose any amplitude you liked? Like 1.2379 cm? Of
course, assuming you could actually measure that accurately. So quantum
mechanics is nonsense, right? No, I could contend that the macroscopic
oscillator is also quantized (theoretically at least). What is the spacing
of the energy levels of this quantum system? hf , of course. what
is hf if f =1 cycle/second? ΔE =6.6x10-34
J! Suppose that the spring in this oscillator had a spring constant k =1
N/m; then, since E =½kA 2 , going to the
next allowed level would require a change of amplitude of ΔA =√[2x6.6x10-34 /1]=3.6x10-17
m. That is about 1/10 the size of an atomic nucleus! I don't think you
would notice this constraint!
Your intuition tells you that quantum mechanics can't be
right because you have never noticed its effects. What is your intuition
based on? Your experience! Do you have any experience regarding such tiny
sized things?
QUESTION: I am working on a piece of science-fiction centred around the mid-career physicist, upon whose desk objects appear for significantly timed short periods, a significant pattern. At some point in the peace, the working assumption becomes That these objects are being transmitted from
"Parallel universe"/other reality in the Multiverse.
This may seem to be an odd question if such objects were assumed to be capable of being transmitted between parallel universes or other reality planes in a Multiverse, do you think they would be capable of being photographed and or scanned in some other forms such as x-ray?
ANSWER: Of course, this is all
speculative, but I would say that if your eye can see them a camera could
see them because the eye is, essentially a camera. But you could also say
that maybe the eye is not really seeing them but rather only your brain was
seeing them, like a halucination. I remind you that I do not usually do
cosmology, the realm in which multiverse is included.
QUESTION:
I have a theory (or just an idea) concerning faster than light travel that I would like you to answer. My idea is that if FTL travel is possible, then the energy required to make a warp drive, wormhole, etc. work is exactly the same amount of energy as would be required based on Newtonian physics if there were no law of relativity.
For example, suppose you have a 1000 metric ton FTL spaceship that can travel four light years in four days. Please calculate how much energy it would take to accomplish this if only Newtonian physics were involved.
Then look up some FTL scheme. Here's one you can use https://newatlas.com/physics/ftl-warp-drive-no-negative-energy/ Or just pick another than you prefer. Then calculate how much energy it would take to move the 1000 metric ton spaceship four light years in four days. My idea is that perhaps the universe is designed in such a way that the energy required is the same. In other words even if FTL travel is possible, you don't save any energy by using it. In other words, there is no “free lunch.”
I donated $20 for your answer, since I suspect this will take a lot of work.
REPLY: (Other
readers note: I am only answering this question because the questioner has
sent a donation.)
ANSWER: 8 =1.1x1011
m/s. The energy which a your spaceship would have at that speed is
mv 2 =½x106 x(1.1x1011 )2 =6.2x1027
J=1.7x1021 kW·h.
Relativistically, the distance 1 light year (ly) would have to
be length-contracted to 1 light day (ld) would be 1=365√(1-(v 2 /c 2 ))
or v =0.999992494c or β 2 =0.999984988.
Now, total energy is E =γmc 2 where γ =1/√(1-β 2 )=258.
But we don't want the total energy, we want the kinetic which is
K=E-mc 2 =(γ- 1)mc 2 =257x1.x106 x(3x108 )2 =2.31x1025
J=6.42x1018 kW·h.
So you see, contrary to your idea, it takes less
energy to go relativistically than classically.
FOLLOWUP QUESTION: I read of some warp drive scheme in which the author pointed out you'd have to convert the mass of Jupiter into pure energy via E=MC^2 in order to make the warp drive work.
Would you calculate this energy and compare it to the amount needed in my previous question regarding sending a 1000 metric ton ship four light years in four days? I would also like to know whether you think humans on the ship could survive the g forces of accelerating and decelerating on such a trip, once again assuming just Newtonian physics applied.
ANSWER: M J c2 =1.9x1027 x3x1016 =5.7x1043
J, many orders of magnitude larger than previous calculations. Humans can
survive accelerations of about 10g and not for very long. You
would have to plan your trip carefully to avoid accelerations much greater
than g .
QUESTION: I am a new and upcoming math and
physics teacher at a local high school. A more senior physics teacher
posed this question to me before he retired, and I am rather
embarrassed that I cannot solve it. Apparently, he said this system is
accelerating and that acceleration varies with the angle.
[COMMENT: m =50 kg, M =100 kg, θ =13°, μ s =0.41, and
μ k =0.35. I chose to do it in general and we can
stick those numbers in later.]
ANSWER: m and m must have
the same magnitudes, t 1 =t 2 =t ,
so we have
ΣFy =0=-T +2t sinθ .
The x -equation just tells us what we already know, that the x -components
cancel each other out. We now have one equation with two unknowns, T
and t , so we need to choose another body. I choose m on the left.
ΣF y =0=N-mg-t sinθ
ΣFx =ma =t cosθ -f
f=-μ k N if m is moving or
f=-t cosθ if t cosθ <μ s N
Now we have five unknowns, T, t, N, a, and
f but only 4 equations. The other m is not going
to give
us any new information because of the symmetry, so we are left with only
M . This gives another equation,
ΣF y =MA=Mg-T .
Note that I have chosen +y to be down so
that m and M are moving in their positive directions. The sixth equation will be the relationship between a
and A,
a=A ·tanθ.
I also calculated that the net force on
m goes to zero
at 55° so the static coeffecient of friction would kick in and the
system would stop.
t=Mg /(2sinθ )
f=μ k (mg+Mg /2)
t cosθ-f =0
and therefore,
θ =tan-1 (500/350)=55°.
I leave it to the reader to do the algebra above.
When I did the calculation
using WolframAlpha A =0
at 55°.
So I now had 4 equations with 4 unknowns (I
incorporated the
a-A and the f-N relations to reduce the number of equations and unknowns to 4.) I put
in the value of θ =13° and the solution is shown below,
calculated in WolframAlpha θ =13°.
The results of calculations over the whole range 0-55 are shown in the
graph below.
ADDED THOUGHTS: a=A tanθ although I did enough qualitative thought to the relation that I feel pretty confident. Also, the behaviors of all
the variables are just what I expect them to be using that relation. It is
interesting that at θ =0, because when we usually think if a weight hanging on a horizontal string,
that the string would have to have infinite tension to not deviate from the
horizontal; that situation, though, requires the (massless) string to be tied between two immovable walls. The equations here have a kinetic frictional force acting on the "walls" so they are not immovable.
Since the string at 0° has no vertical component, T must be 0
N,
as the calculation shows; M must therefore be in free fall for the
instant that θ =0°. Since the frictional force is 0.35x490=-171.5
N, the solution finds that t =171.5 N also; I guess technically
t could be anything and still have a= 0 because of the
tangent factor, but small angle nonzero θ will have
whatever t is required. At θ= 55° everything has
stopped and must be in equilibrium: both N and T are at
980 N and t =980x0.35/cos(55)=598 N. Because all these accurately
describe what we know they should actually be, I am pretty confident that
a=A tanθ is correct.
QUESTION: My understanding is that atomic clocks circling the earth run
1) slower because they are moving fast
but
2) faster because they 'feel' less of earth's gravity
than clocks on the surface of the earth.
Which wins out, speed or gravity?
How much is the time dilation/contraction due to gravity and how much due to speed?
Could there be a planet of some mass that you could put a satellite in orbit around at a certain speed so that the two would balance out so that the orbiting clock would measure time at the SAME rate as one on the planet's surface?
And, how much math would I need to figure out the answer to that last question?
ANSWER: r .
I suggest that you look at another Wikipedia article
on time dilation .
Near the end of the article the gravitational time dilation is discussed.
The location of the zero dilation is at 1.497 earth radii, about half an
earth radius altitude.
Everything below in this
answer is incorrect (except for the red curve). I was misled by an AI error, several incorrect
internet posts, undue confidence in myself to be able to do the GR
calculation! It is an excellent example of how incorrect an internet post
can be and how AI probably learns from its memorizing everything on the
web. I did more research because it seemed extremely odd that the
geosynchronous orbit and the zero dilation orbit would be the same and the
fact that my incorrect answer was about 10% smaller than the actual
geosynchronous orbit.
I am leaving it posted as an example of how wrong
the internet can be!
Wikepedia article on gravitational time dilation; I would recommend it
because it gives the form of the gravitational dilation (I leave it to the
reader to understand the notation here. t 0 is at the
surface of the earth and t f is at some distance r
from the center of the earth):
t 0 =t f √[1-(2GM /(rc 2 ))]
What needs to be plotted is (t 0 -t f )/t f
which must be zero on the surface. So I conclude that
(t 0 -t f )/t f =√[1-(2GM /(rc 2 ))]-√[1-(2GM /(Rc 2 ))]
The kinematical time dilation turned out to be
easier for a circular orbit of radius r. Without any details except to note
that GM /r=v 2
(t 0 -t f )/t f =-1/√[1-(GM /(rc 2 ))]+1,
ensuring that an orbit at r=∞ have zero speed.
So, here's my calculation. The kinetic calculation
agrees very well with the Wikepedia graph but the gravitational one does
not. The result is that the zero net dilation is at about 6 earth radii,
but the Wikepedia calculation was at about 1.3 earth radii, a really big
discrepancy. I stared and stared and couldn't find an error. So I decided that
I would ask AI about this; here is my interaction:
Six earth radii is approximately where the
geosynchronous orbit, where all the communications satellites are. It would
appear that my calculations are right. I did check by googling and the
usual answer was the same as ChatGPT's.
I think AI got it backward, though: kinematic
dilation is slower, gravitational dilation is faster.
QUESTION: Picture an attic that covers a house and its driveway. The area of the driveway and the house are the same, so they are each covered by 50% of the attic. In the center of both the house and the driveway are 2 identical circular holes into the attic above.
The temperature in the house is 68°F, attic is 86°F and driveway 104°F. Since warmer air flows upwards, will the driveway air get into the attic, or will the colder air in the house work as a barrier to keep the attic air from coming down?
ANSWER:
The zeroth law of thermodynamics is that heat
always flows from a higher temperature to a lower temperature.
The zeroth law is true for an isolated system.
If there is some other agent which acts to add, subtract, or
redistribute the heat energy, it may not be true.
Particularly for gases, the notion of "warm air
rises" is a result of the fact that if you increase the temperature the
volume increases and so the gas becomes less dense; it rises because of
gravity so the warm "floats" on the cold just like wood floats on
water.
There are three ways that heat is transferred—radiation,
conduction, and convection.
First, lets consider these systems are totally
isolated—no gravity, perfect insulation all over the outsides, no
furnaces, ACs, etc . Looking at all three at once would muddy the
water, so I will imagine there is a perfect insulating wall in the middle
of the attic. Heat would flow from the hot to the cold by convection
through the holes and conduction through the ceilings. Eventually all the
air would be at the same temperature, somewhere between the two starting
temperatures, which would depend on the amount of air in each of the two
volumes. The same thing would happen for the garage/attic except heat would
flow up, not down. Next, turn on gravity; then there would be a continuous
increasing temperature from bottom to top. The ceiling would have the same
temperature top and bottom. Similarly for the garage/attic half. You might
want to imagine that the ceilings are perfectly insulating; the net result
would be the same but only convection through the holes would transfer the
heat.
Finally, your scenario, all three volumes connected
through the holes and no wall in the middle of the attic. With no gravity
everything would end up with the same temperature. With gravity added,
there would then be a continuous increase from floors to the roof of the
attic. Don't forget that this is the real world and any sources or sinks of
heat will change the details.
QUESTION: I have searched the internet and not found an answer.
If there are 2 identical blocks of ice.
One is at -1 Degrees C
One is at -50 Degrees C
Do they melt at different rates? Will the colder piece take longer to melt?
I'm assuming the colder piece will take longer, but I don't actually know.
ANSWER: T somewhere above the freezing point or below the boiling point of water.
Heat flows from a higher temperature environment to a lower temperature environment. When the heat flows into the lower tempeature environment (the ices in your case) it is absorbed and can do
two things: it increases the temperature or it can cause a phase change (melting, freezing,
condensing, or evaporating) without a change of temperature.
(You might want to read a recent answer to a question similar to yours.) At
the beginning of the experiment each piece of ice will heat by increasing
its temperature. The 1°C piece will get to 0°C in some short time, t 1 , then energy coming in is used to melt the ice until it is fully melted in a time t 2 . As the ice is melting, the water from the melting
will start warming while the remaining ice will continue to melt without warming. Eventually all the water will be heated to the temperature of the room; call the time from when all the ice was melted
until all the water
was at T to be t 3 . It will take the first piece t 1 +t 2 +t 3
to get to the end of the experiment. The only difference between the two
pieces is that it will take the colder piece fifty times longer to get to 0°C.
So it gets to the end of the experiment in a time 50t 1 +t 2 +t 3 .
The times for the two pieces to completely melt will be the same, but it
will take the colder one longer before it starts melting.
QUESTION: In the known universe, is there anything else in existence besides matter, waves, forces, movement and change?
In other words what are the very basic essential physical things and the effects, excluding any measures that we seem to give a reality to quite often?
Maybe waves could exclude matter or maybe waves are just matter in motion? Space itself seems not to be truly part of existence?
Time to me is a measure of forced motion and change and the same with energy.
What are the bare essentials of the known universe that I can detect through my senses, not just concepts alone?
Would you include fields?
What are the indisputables?
ANSWER: etc ..
The answer is that the universe is ENERGY.
REVISED ANSWER:
FOLLOWUP QUESTION: What is energy and where has it come from?
Does the ability to do work come from energy or the forces or both?
It seems to me that of all the forms, shapes, colours, sizes, amounts of energy that can be stored or transferred but not created or destroyed.. energy is a quantitative measure of the movement that forces create on an object or particle.
Could energy be dispersing waves of just motion through fields since the big bang.
Anyway, if not a measure, what is it? Please not the ability to do work.
ANSWER: site ground rules , are not "…single, concise, well-focused questions…"
You are expecting me to give you a complete tutorial on energy, something
which takes a couple of years of classes to fully understand. Essentially
we invent a quantity which, for any isolated system, is conserved,
i.e . remains constant no matter what happens inside it. By being
isolated we mean that there are no external forces doing work on the system.
[Sorry, there is no way to talk about energy without mentioning work.]
Conceptually, isolated means any system to which energy is neither
being added to nor taken from the system. Assuming that the entire universe
is all that there is (or not interacting with other systems outside the
universe), the total energy of the universe never changes.
QUESTION: While water is freezing, energy is being released in the form of heat- how does this jive with a temperature plateau during the freezing phase change?
ANSWER:
Well, I see that my little story has not answered
your question, it is the reversal. So my story continues: The water and the
air are now at some temperature, say +10°C. Now the power comes back on
and the water and air start cooling, the freezer removing heat from both.
When the water gets to 0°C it starts freezing and the heat being
removed results in freezing, not cooling; but the air continues to cool.
Finally when all the water is frozen the ice now cools because heat is
being removed by the freezer. Eventually everything is at -10°C again.
QUESTION: I have a question about time. I am a nursing student, taking anatomy courses and learning the atomic physiology of bodily processes. For example, how ions flow in and out of heart cells to make it contract. The whole process takes so much time and thought to track it all out, when in reality, it takes .8 seconds to actually complete a cycle.
My question is, how fast are these atoms moving in order to move at the speed to complete our bodily processes in the small fraction of a time we experience as humans? Or, is it that atoms are so small, they experience time different than we do? or is that not an accurate thought? thanks for the help!
ANSWER:
Hearts are not conducting wires but they can conduct
electric currents because flesh is an electrolyte which plays the roll of
conductor and the charge carriers are ions (atoms or molecules either
missing an electron or carrying an extra electron). So when some action in
the heart requires a message to be sent from point A to point B, a voltage
between A and B is turned on at A resulting almost immediately in the pulse
arriving at B; an ion created at A does not get sent to B. I
believe that calcium ions are important charge carriers in the heart; I
take a drug which is called a Ca channel blocker but I couldn't tell you in
any detail how it works.
So the answer to your question is that the ions move
very slowly but the information which they carry arrives very, very
quickly.
*An electric field is the thing which causes there
to be a force on the charge carriers which drives the electric current.
QUESTION: When light goes from, say, air to glass or glass to air at a non-90° angle, it's refracted. Does that use energy? Are the air and the glass heated? And if so, is that what's happening when a microwave oven heats things the most at the interface between different things (like the ceramic plate and the ham).
ANSWER:
QUESTION: Concerns the speed of light in a medium. I understand the speed of light varies slightly with hot and cold earth atmospheres, primarily (apparently) with density, as it does in other mediums. Intergalactic space is a medium, however tenuous, and not a true vacuum. It would therefore seem that the speed of light would vary slightly even in it. The standard speed of light is calculated as in a true vacuum. Is this an answerable question?
ANSWER: 3 . The likelihood of
there being an interaction which would measurably change the average speed
of a single photon are about as close to infinitesimal as we could
imagine.
QUESTION: Air is moved downwards, how can momentum be conserved if the helicopter remains stationary? Should there not be an equal and opposite momentum upwards?
ANSWER: W L L W
PRELUDE:
QUESTION: My question is with regard to an apparent mathematical disagreement with inverse square law. Let's say you have point source in 3D space positioned at (x_0, y_0, z_0). It emits photons in uniform 3D directions. --- I just want to exclaim that properly picking uniform 3D directions is not obvious! See Wolfram's Sphere Point Picking page ---. Anyway, you have an infinite length/height vertical y/z plane detector at x=1. One can derive the governing distribution of hits on this detector and it turns out to be the multi-variate Cauchy distribution given by:
p(y,z) = 1/(2*pi) * (1-x_0)/sqrt((1-x_0)^2 + (y-y_0)^2 + (z-z_0)^2)^3
This density does not seem to follow the inverse square law. It has a 3/2 power in the denominator! It would need to be 2/2. I have derived densities for other 3D geometries (e.g. cylinder) that are not exactly inverse square law either. I am not a physicist but I am struggling with a physicist who claims it must be inverse square law. It is not. It is similar, but agrees less than the Cauchy density when compared to Monte Carlo simulation. What am I missing?
ANSWER: r dependance of what ever the "source" is emitting? By definition, a point source of a vector force field creates a field pointing radially and having a magnitude which decreases like 1/r 2 . I don't know what a multi-variate Cauchy distribution is, or if I ever did I don't remember. You seem to be talking mathematics here and not physics.
REPLY: The Cauchy distribution (maybe you know it as Lorentz?) is the distribution of where rays hit a vertical or horizontal line when emitted in uniform directions from a 2D point source (x_0, y_0). The bivariate one I wrote is the distribution of where rays hit a y/z plane (at x=1 in this case) when rays are emitted in uniform directions from a 3D point source (x_0, y_0, z_0).
ANSWER: r 2 and the total amount of stuff hitting the surface of the sphere does not change (because there are no other sources or sinks), it follows that the density of the stuff must decrease like 1/r 2 .
REPLY: I think I have an answer for you! Though it has stumped me for more than a month now. If I look at the behavior of probabilities on a small area of the y/z detector plane and I allow the plane to move... say x=1,2,3,etc... Then the ratio of probabilities of detection in this small area will approximately follow the inverse square law. A PDF is not a probability itself. You have to integrate over an area of the detector plane to calculate probability of detection within that area. So if I calculate the probability of detection in, say, [-0.1,0.1]^2 for the y/z detector plane positioned at x=2, then it's approximately 1/4th the probability of detection in [-0.1,0.1]^2 for x=1. If I do the same for the y/z plane positioned at x=3, then it's approximately 1/9th the probability of detection relative to the x=1 scenario. And so forth and so forth.
I think this other physicist assumed you could just arbitrarily define a density function with inverse square law. This seems to only be true for histograms! The bins of a histogram can be thought of as the integral of some PDF over a small area.
Still, that Cauchy/Lorentz density is an inverse cube where the calculated probabilities behave like inverse square law. That's pretty weird and unintuitive to me!
OK. I'm a poor communicator at this stuff, I'm sorry. So the way this physicist thinks is that if a point source (x_0,y_0,z_0) emits a number of photons in 3D uniform directions, then the number of photons that should hit a specific location on a surface that is a distance r away from the source should be proportional to 1/r^2. So he asserts that you can just define a probability density of photons hitting a detector surface directly using the inverse square law. For example, if it's a y/z plane at x=1, then r=sqrt((1-x_0)^2 + (y-y_0)^2 + (z-z_0)^2) and you get your inverse square law probability density function of p(y,z) = C*1/r^2 where C is a normalizing constant. That seems reasonable to me! And it's quite elegant, right? If you know how to calculate the distance to the detector surface, you can trivially describe the probability density of photons hitting any location of the detector surface. Well... but there's this Cauchy distribution that defines a distribution of photons hitting a planar surface when emitted from the point source. That one is proportional to 1/r^3... and it lines up almost perfectly with Monte Carlo simulation. So how can that be 1/r^3?
The physicist asserts inverse square law must be true. It's well-tested and known to be true. I understand that. But the Cauchy distribution describes the simulation better than the inverse square law-based density. From a math point-of-view, it's so straightforward to derive the Cauchy distribution that it would be baffling if it wasn't the correct probability density. But the darn thing really has 1/r^3 in it!
ANSWER: r 3 function
appears quite naturally. I will look at an infinite sheet occupying the xy plane and a
point charge Q , a distance d from the sheet. The
electric field has a magnitude E=kQ /r 2 and
points radially away from Q . Although your calculations were in
Cartesian coordinates, cylindrical coordinates are much more natural. At the
surface of the sheet the electric field has no azimuthal component and so
the area element to consider is dA =2πρdρ . My aim here is
to get the total flux through the infinite sheet, but I will look at other
things as I go along. A few things we need are:√ (ρ 2 +d 2 )r 2 =kQ/ (ρ2 +d2 )ρ2 +d2 )z =EkQd /(ρ 2 +d 2 )3/2 sinθ=ρ /√(ρ 2 +d 2 )Eρ =E sinθ=kQρ /(ρ 2 +d 2 )A =2πρ dρ
The first thing to note is that E z is
where your 1/r 3 is; just because you r 3
behavior on the x,y plane does not contradict what the behavior of the
field is. Ez * is what is important here if we are interested in something
spread over the entire sheet because for every Eρ on the area
element, there is an oppositely-pointing Eρ 180° away which cancels it.
dA·E cosθ is what physicists call electric flux and, taken over the area dA , the flux is dΦ =2πkQρ dρ/ (ρ 2 +d 2 )3/2 .
Finally we can calculate the total flux through the sheet:
Φ =2π 0 ∫∞ Ez ρ dρ= 2πkQ
* The graph shows Ez
normalized to Ez (ρ= 0, z= 0) as a
function of ρ in units of d.
QUESTION: I recently learned that at CERN they produced streams of protons by applying large magnetic fields to hydrogen gas.
This removes the electrons and these only the nucleus specifically protons.
This made me wonder as alpha particles are merely the nucleus of the helium gas, were you to apply a large magnetic field to helium gas would you merely be left with alpha particles?
ANSWER:
Ionize hydrogen gas to make a plasma. This is
done with a radio frequency electric field.
You then extract the protons from this plasma
using electric fields and inject those into an accelerator which speeds
them up.
Then you repeatedly speed them up, again using
electric fields and magnetic fields to steer and focus the protons
being accelerated.
Most accelerators, including for alpha particles,
work the same way. In fact, magnetic fields cannot ever exert the forces
necessary to accelerate charged particles. The reason is that the force
which a magnetic field B v v
QUESTION: So I just read an article about virtual particles and Hawkins radiation, and how a black hole can take virtual particles, separate them, and take one of them while the other Hawkins radiation is ejected out and that's how a black hole can evaporate, which I still find hard to believe. If that is the case, and a virtual particle pops existence and one is taken away and the other is ejected what happens with that energy? So, if there's energy between the virtual particles and that energy can't be destroyed only transferred could that energy end up as dark energy?
I'll explain it again in the hopes of clarifying, near a black hole, virtual particle, can pop into existence, be separated, with one particle, being devoured by the black hole, and the other one being ejected out into space, if there is energy between the two particles at the moment of them being separated could that energy simply turn into dark energy or would it stay with the Hawkins radiation being pushed into outer space?
ANSWER:
QUESTION: When a golf putt hits the edge of the hole, sometimes the ball will "lip out," whipping around the edge and heading off in a new direction. Some lip outs can look quite violent and travel surprisingly far from the hole.
My question: can hitting the edge actually increase the distance to the hole? Formally, if point H the center of the hole, is it possible for a lip out to finish further from H than an identical putt would if there were no hole punched in the ground?
Or does conservation of momentum make this impossible, and this is all just an optical illusion caused by the brain misinterpreting the abrupt change in direction?
ANSWER: earlier question I discussed
the motion of a putted ball in great detail. If you want to understand my
answer to your question, you must read this
Q&A first to see how I use
the frictional force which is what I used there. I will also assume that
the parameter
μ R =0.093 and the mass of a golf ball is m= 0.046
kg. So the frictional force of a rolling golf ball f =-mgμ R =-0.045x9.8x0.093=-0.042
N if the green is level which I will assume to be the case. The
acceleration of the ball along a straight-line path will be a=f/m =-0.91
m/s2 . The figure
shows the two paths you stipulate, one rolling a distance D 1 ; the second, rolling a distance D 2 +D 3 +D 4 .
D 3 , not labeled in the figure, is the path during
which the ball is "lipping out", moving on the rim aroundθ ;
so D 4 =Rθ where θ is in radians
and R= 0.054 m is the radius of the hole. I will assume* that the
acceleration along the curved path is the same as along the straight
paths. Now, we can find D 1 because, from the earlier answer, D 1 =0.55v 0 2 =where
v 0 is the velocity which the ball has as it leaves the
putter. I will choose v 0 =1.5 m/s which yields D 1 =2.05
m. Since I have assumed all paths have the same acceleration along all
paths, each of the two putts must travel the same distance, soD 2 +D 3 +D 4 =D 1 =2.05
m D 4 =2.05-D 2 -0.054θ
Since we are trying to find out whether or not it is
likely that the ball which lipped out is farther from the hole or not,
knowing D 4 will likely tell us. D 2 is just the distance from the hole
center where the lie was; θ will probably not be bigger than 90°=1.6 radians
and, besides, it is very small in the scheme of things. If we choose D 2 =1
m and θ= 90°, D 4 =2.05-1-0.085=0.965
m. It makes more sense to look at a straight shot which just missed getting
lipped with the one that lipped by 1.6 radians. The second figure shows how
far each are from the center after stopping. The lipped one is closer by
about 8%.
*The question about what the rolling friction should
be in the arc was what worried me most when I started analyzing this
problem. But the normal force on the ball by the ground has to be bigger
than it is when going straight because of a centrifugal force acting
outward. But the conclusion would be the same that the lipped ball will end
closer to the center of the hole would would not be changed; if the normal
force is bigger, the friction is bigger, so it loses more speed and would
be even closer.
QUESTION: I was trying to measure the bounce of a golf ball to determine whether its lifespan was up. In order to control the ball's bounce as to not get away I thought bouncing it in a tube with viewing hole would be a good environment. Unfortunately I noticed the balls bounce approx 50% less in the tube than outside of it.
I would like to know why the golf ball bounces higher outside than inside a 2 inch PVC tube from 5 feet off the ground? Is there anything I can do to alter the tube to get the same results (ie drill holes in it, cut out sections in the bottom to release air)?
ANSWER:
Now your situation is not like this but it is very
similar except it is leaky; the air will leak out as the ball falls, but
the pressure in the air below the ball will still increase. All the air
which the ball is pushing through has to squeeze through a much smaller
space than if the pipe weren't there. That is going to considerably impede
the fall of the ball. If you used a pipe much larger to keep from having to
chase the ball it would probably better match the freely falling ball.
QUESTION: This is the 90 year old man in a nursing home and would like to know if it is possible to have a total vacuum in a closed space.
ANSWER: 3 .
In intergalactic space there is typically 1 molecule per m3 . We
now understand that even if we had a volume with absolutely no molecules
in it, it would still not be truly empty because there are constantly
particle-antiparticle pairs which pop into existence and pop back out,
called virtual particles; this is called vacuum polarization.
QUESTION: I am in a state-run nursing home that has toilet facilities in a small enclosure with two doors measuring 5ft by 8ft and height 10ft. The temp rises to 6 degrees above the adjacent rooms. Is this due to the heating of the particles in the atoms creating energy in the form of heat?
ANSWER:
QUESTION: I just really confused myself with a strange hypothetical situation during the impulse/momentum unit at my school. If a kid is riding a sled on completely frictionless ice with negligible air resistance, and then separates himself from the side of the sled in a way such that both the sled and the kid are moving in the same direction, would the sled still speed up? I know the sled's mass would seem to decrease from the sled's perspective, so it should speed up to conserve momentum, but if the kid is still moving at the same velocity as before, wouldn't any increase in speed for either thing not conserve momentum for the system? I tried to explain this to my teacher, and we clearly didn't understand each other.
ANSWER: external forces
acting on a system the linear momentum will be conserved; most problems
you are likely to encounter have momentum conservation. If there are no
external forces doing work on a system the kinetic energy would be
conserved; if it is (which it usually isn't), it is called an elastic
event. I shall look at two scenarios:
What is the relation between the two final
velocities if we demand that linear momentum is conserved? If
the subscripts labeled 1 and 2 are for the boy and the sled, thenm 1 +m 2 )u =m 1 v 1 +m 2 v 2
v 1 =u [1+(m 1 /m 2 )]-(m 1 /m 2 )v 2
where u is the speed before
separation and v i are the velocities after.v 2 =u /2, then
v 1 =u [1+(½m 1 /m 2 )].
As you can see, there are an infinite number of possible velocities
after the collisions depending just on how the kid got off the sled. Of
course, that is not surprising because we have only one equation for
two unknowns.
Now, suppose that we impose the condition that
energy must also be conserved. The derivation of the solutions for a
perfectly elastic event can be found in any introductory physics
textbook or go to the
Wikepedia article on elastic scattering in one-dimension. Now we
have two equations for two unknowns which means of course there will
only be one solution. If both objects have the same speed
u before the collisions, then these
equations becomev 1 =u
and v 2 =u .
It looks to me that you did not give me all the
information about the problem. If it had been something like "Suppose that
after the separation the sled had a speed of 3u /2, how fast is the
boy moving?" you could find out that the boy was moving more slowly, v 1 =u [1-(½m 1 /m 2 )].
QUESTION: I am a new adjunct instructor for a introductory physical science online class. My background is in chemistry. Anyway, I ran across a conceptual problem in OpenStax Physics 2e, that I am confused about. In the chapter reading, Coulomb's law is introduced followed by electric fields which are derived from Coulomb's law. Neither of which I find confusing. Here is the problem I am confused about: Compare and contrast the Coulomb field and the electric field.
ANSWER: r 2 . The prototypical
Coulomb field originates in an isolated point charge q :
E =r o kq /r 2
where r o is a dimensionless vector of
magnitude 1 which points directly away from the point charge, called a
unit vector; the constant k is sometimes written as 1/(4πε o ).
This Coulomb field plays a crucial role in the determination of the field
of any charge distribution which is not a point charge. If you have a
chemistry background, you surely know calculus, so I will give you the
whole general picture so you can understand what goes on. If the course
you are teaching is "physical science" you surely will not teach it at
this level, probably the electric field due to 2 or more point charges. In
the figure a tiny (infinetesimal, really) piece of the blob of charge has
been focused on. The field at the point labeled P due to the tiny
(point) charge labeled qi is a Coulomb field; now you
have to integrate over the whole volume for every point in 3-D space.
Except for simple shapes like a sphere, this is an extremely difficult
calculation to do analytically. That is why your course will probably not
talk about anything much beyond point charges. You might also touch on the
field of a large uniformly charged plane which is pretty easy to
conceptionalize and introduces a uniform electric field.
QUESTION: Einstein, in his theory of General Relativity, stated that Gravity is not a force. So, why the phisicists are still trying to develop a theory to explain Gravity as a force next to the other three fundamental forces of nature?
ANSWER: earlier
Q&As on this topic; be sure to see links in that Q&A.
QUESTION: I am writing you to seek your expertise in the field of physics.
As you know, Newton's law of universal gravitation demonstrates that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.
The use of r2 instead of r arouses my curiosity.
Any further explanations would be highly appreciated.
ANSWER: T 2 /a 3 was the same
for all the known planets (T is the period and a is the
semimajor axis of the elipse); also, the line from the center of the sun
to the orbit swept out equal areas
r , but if he did it
would not explain the data—Kepler's three laws were the data. But it
would not work. Nor would 1/√r , 1/r 3 , etc., but a 1/r 2 force did work.
There is another way you can look at forces. A force
usually may be thought of as a field where we think of lines of force
directions in the direction a mass would be attracted; the closer the
lines are together, the stronger the field; and total number of lines is
proportional to the mass. The red lines in the figure show the earth's
gravitational field but only in one plane. The blue lines show spheres
again only in two dimensions. You should imagine the lines coming in from
all directions and many imaginary spheres. The strength of the field would
be the number N of lines per unit area A. But every sphere is punctured by
exactly the same number of lines as all the others. The area of a sphere is
4πr 2 so the number of lines per unit area, strength
of field, is N /(4πr 2 ). Eureka! There's your
1/r 2 .
QUESTION: I am a student in Belgium. Near my school, there is a wind turbine that has caught my interest. I have been wondering if it's possible to determine the wind speed by measuring the time it takes for the turbine blades to complete one full rotation, given the radius of the wind turbine. I am assuming a constant wind speed for this calculation.
I believe that the wind speed (Vw ) can be expressed as Vw = (3.6 * 2 * π * Rturbine ) / T (in kilometers per hour), where
Rturbine is the radius of the wind turbine, and T is the time it takes for the blades to complete one rotation. However, this formula does not seem to yield realistic results. Despite conducting some research on this matter, I haven't been able to find a satisfactory answer.
ANSWER:
The blades must move slowly, usually around
10-20 rpm, because of structural limits and noise polution.
So there are mechanisms which keep the
rotational speed within those limits.
Because this is too slow for the generator to
work there is a gear box to increase rotation to the generator by
a factor of about 100.
The blades do not work like paddles or sails,
rather like the wings of an airplane or rotor of a helicopter.
The speed of the tips of the blades is more
important than the rotational speed of the blades themselves for
maximizing efficiency.
Don't forget that, unlike the pinwheel, energy
is being constantly taken away from the blades and converted into
electrical energy which will be sent out to the world; so there must
some energy balance where the energy taken out is equal to the energy
being supplied by the wind minus energy lost to friction, inefficiency,
etc .
There is a very good
webpage going into the details of wind turbines; be sure to look at the
video too. There is another
video which is
also pretty detailed.
QUESTION: if pi measures a circle and and it never repeats it self, but its used to measure a circle doesn't that mean the numbers would eventually go around the circle and repeat.
ANSWER: π does not "go around the circle",
whatever that means; it is a measure of the ratio of two lengths, the radius R and circumference C of a circle, π =C /(2R ).
QUESTION: For matter that is falling into a black hole, even before it gets to the event horizon, it would be impacted by both gravitational and velocity time dilation in accordance with standard and generally relativity. This time dilation can be very extreme at these extremes. So would this mean that through the perspective of the matter falling in, it's still hasn't reached the center or even the event horizon? Because from our perspective outside the black hole it is going to take them billions or even trillions of years to get there. So while for them it's a quick decent, for us it's a very long time. And since we are here now could that possibly mean that all that matter in black holes has yet to reach the center because they are nearly frozen in time dilation? If I'm wrong what aspect is not working the way I think?
ANSWER: link to get a
tutorial about falling into a black hole.
QUESTION: Is infinite energy possible in a frictionless surface
ANSWER:
QUESTION: I was learning about a STEAM activity for my children. This activity involved cutting up pool noodles and putting toothpicks in them to stick the noodle pieces together. I was wondering, why doesn't the toothpick fall out? What force(s) are acting on it to keep the two together? It seems strange to me, because the pool noodle is porous. I want to know so I can teach my children the physics behind the activity.
ANSWER:
QUESTION: If sound like light travels on earth, does the sound like light just continue on continuously.
ANSWER:
QUESTION: As a solid object (ex a box) travels in a circular path, the side of the box closest to the inside of the curve moves a shorter distance than the side of the box towards to the outside of the curve. Yet, the sides of the box remain in the same relative position to each other............
How can one side of the box travel a greater distance than the other and the box still remain intact?
ANSWER: r from the axis has a speed v=rω where ω is the angular velocity in radians per second. So the farther out
any point is the faster its speed is. But if the spinning object is not a rigid body it will not remain intact. If you start with a spinning pancake made of soft putty it will spread out. If the box you were
thinking about was made of rubber it would not retain the same shape if you started to spin it.
FOLLOWUP QUESTION: Thank you so much for replying to my question, but....
I'm sorry, even tho your answer makes perfect sense, it did not answer my question .
How can one side of the box travel a greater distance than the other and the box still remain intact?
Imagine the box is going around a circular race track (like a race car). The track has a diameter of 1000' at the center of the track - the circumference of the center of the track is 3141.59' . Say the box is 4' long and 2' wide. In one trip around the track, the center of the box travels 3141.59'. The outside of the box (at a diam of 1001') travels 3144.73'. The inside of the box (at a diam of 999') travels 3138.44'.
So, the outside of the box has travelled 6.29' farther than the inside of the box....
How can one side of the box travel a greater distance than the other side and the box still remain intact?.
I have wondered about this for many years of my life and have never been able to find someone to explain it to me.
(straight-line physics is so much easier than curved-line physics....)
Thank you so much for your attention to this question.
ANSWER:
linear velocity v of any object rotating about some axis with
angular
velocity ω and at some point a distance d from the
axis: v=ωd. The angular velocity must be measured in
radians/second (s-1 ); since there are 2π radians in a circle,
an angular velocity of 1 rotation per second has ω =2πd.
I have drawn a diagram of a stick rotating around one end with angular
velocity ω showing one end having a speed v but the middle having a speed of
v /2 and the end it is rotating around is at rest; v=Lω .
Your instinct that if it wasn't strong enough to have those differing speeds, remain a rigid body, it will break. A good example of
that is when a very tall chimney falls, it breaks before hits the ground.
QUESTION: I know that electrical current is in units of coulombs/s. Each coulomb contains a bazillion electrons. For a electrical transmission line with an AC signal, I think of the mean value of electrons alternating back in forth at the AC rate, since the current is alternating. What happens to these electrons when they feed an antenna and the power is radiated? Or do I have all my assumptions completely wrong?
ANSWER:
QUESTION: I suspect that Elon is not producing trucks because of the way electric vehicles brake.. Am I correct in saying that there is a transfer or energy to the road surface on EVs, not on normal vehicles that rely on pads.a big rig, heavy.. those wheels, the weight, the method of slowing, I believe that e trucks will not suit our roads. Do you get me?
ANSWER: "…a transfer of energy
to the road surface…" when EVs brake. Let's discuss what braking
does. When a vehicle is moving it has energy by virtue of its motion
called kinetic energy. In order to stop the vehicle from moving, you must
get rid of that kinetic energy. The traditional way to do that is to use
friction of brake pads (or shoes) rubbing on a metal disc (or drum). Where
does the kinetic energy of the vehicle go? Into heat energy because the
brakes get very hot. Wouldn't it be nice if you could capture some of that
energy and store it to use to change back to kinetic energy later rather
than get the energy from the burning fuel. There is a way other than
friction that could be used; that method is called regenerative braking
(regen) and involves using electromagnetism to create an electric current
which could be stored by sending the current to a battery (not to the road
surface!) Before recent times there were no batteries to store all this
energy except the usual 12 V battery which gets recharged but not by a
braking system but by the alternator. But with hybrids or EVs, there are
big batteries which are hungry to grab any energy they can to lengthen the
time before they next need recharging. But no vehicle relies solely on
regen because regen might not slow the vehicle down fast enough or it
might fail altogether; all hybrids and EVs have both conventional brakes
and regen brakes and usually you can adjust what fraction of the
braking is done by each. Furthermore, if the vehicle is not braking fast
enough with just regen, conventional brakes would jump in. Therefore, as
you can see, braking is not an issue at all in whether the vehicle is
feasible. But adding regen to any vehicle is to your advantage if you have
the ability to store the electric energy.
QUESTION: I read that quantum spin is intrinsic angular momentum. I do not understand what intrinsic angular momentum is. Could you please tell me? In ignorance, I guess that it is some number which quantum physicists derive from the positions, movements and interactions of electrons, other particles and photons, and which predicts their positions, movements and interactions, but does not refer to classical rotation because the number is so high that under the equation E = MC2 the electron, other particle or photon would have infinite mass, which is wrong.
ANSWER:
etc . RPM); instead they can only have certain discrete values which is
called quantization of AM. Elementary particles have one fixed AM but the
total angular momentum of a quantum system in some particular state which
has quantum numbers either integers n=0, 1, 2, 3… or half-odd half integers,
n=1/2, 3/2, 5/2… The first set are called bosons and the second set
are called fermions. The angular momentum J of a system with
quantum number n is J is given by J =ℏ√(n(n+1)) where ℏ is the rationalized Planck's constant.
IOM is often referred to as spin. But if you try to make a semiclassical
model of an electron as a little spinning ball you get impossible results,
like the surface of the ball traveling faster than the speed of light. So
we can think of it like a spinning ball but have to keep in mind that it
is not a classical object and just has a property that behaves sort of
like classical objects do.
QUESTION: I am trying to understand something about gravity. To it seems that the gravitational pull equations I was taught is flawed. It is stated that it is directly proportional to the mass of the object and inverse to the distance from the object. The distance makes sense. The mass does not because gravity itself is needed to make the mass. Without it, mass would not exist. It is like trying to describe the color red using crimson.
ANSWER: vice versa . Also,
gravitational force is inversely proportional to the square of the
distance.
QUESTION: Light is able to be bent by both gravity and it easy bent when traveling through clear objects like a prism. Could it be that the secrets to gravity and anti-gravity might be found in light?
ANSWER:
QUESTION: I'm not sure if this counts as astrophysics, but if an object was falling towards something, like the earth (no air resistance), and then the earth disappeared, would it keep the inertia from the fall? Because if considering gravity as just space-time curvature, it disappearing seems like it would leave the object motionless, as it always was motionless in freefall, because standing on the earth is the same as accelerating up?
ANSWER:
QUESTION: My question is does it matter if we have two Dynamos which produce
the same voltage but different frequency does the frequency matter and how?
ANSWER:
QUESTION: Is it theoretically possible for a black hole to contain other smaller black holes?
ANSWER:
QUESTION: How fast would a missile be flying if it was fired from an SR-71 blackbird flying mach 3.2?
ANSWER: the blackbird has a velocity
v with respect to the ground and the speed of this particular missile
has a velocity u with respect to the ground
if fired from the ground , and the
velocities are in the same direction, then the missile has the speed
u+v with respect to the ground if fired
from the blackbird .
QUESTION: So let's say my buddy and I are driving along the highway in our Pontiac Trans Ams at 99% the speed of light maybe a couple hundred yards apart. I'd like to chat with him on the CB radio. Are we traveling faster than the radio waves can be received?
ANSWER: It sounds like you and your buddy are traveling in the same direction with the same speeds. In that case, according to the principle of relativity, there is no difference from your both
sitting at rest separated by "a couple of hundred yards apart."
