QUESTION:
I am a part time instructor and one of the things I teach is rigging. An
important part of the class is calculating sling stress. I was asked a
question that I couldn't quite answer. I don't believe it was my high
school or college physics teachers that failed me rather the more than two
decades since I sat in a physics class. The question relates to vector
forces. If I am hoisting a load of a given weight (say 2000 lbs) and I have
two slings that are at a given angle (say 45 degrees) from the load. Each
sling would be carrying half the load or 1000 lbs straight up. However, the
sling in this example would have 1414 lbs of force (inverse sine if I
remember correctly). The question I was asked was regarding a crushing or
compression force. Since each sling is carrying 1000 lbs vertical load that
means it has 414 lbs of force perpendicular to the vertical. Does that mean
that there is 414 lbs of "crushing" force or is it 828 lbs since each sling
is pulling towards the center of gravity.
ANSWER:
After an email exchange with the questioner, I was able
to determine that "sling" is just a rope or a chain or a
string etc. In the diagram I have shown all the
forces on the load, the tensions in the slings,
T_{1} and T_{2}
and the weight of the weight, W.
The magnitudes of the tensions are equal, T_{1}=T_{2}
=T. Also shown are the horizontal (x)
and vertical (y) components of the tensions;
they are T_{1x}=Tcosθ,
T_{2x}=Tcosθ,
T_{1y}=T_{1y}=Tsinθ.
So the equilibrium equation is 2TsinθW=0;
therefore T=W/(2sinθ).
So, for your example, sinθ=cosθ=1/√(2),
W=2000 lb, and T=1414 lb. Now, your
interest is in the horizontal components; each is Tcosθ=W/(2tanθ)=1000
lb "crushing force" on each side. Your main error is that
you treated the vector forces as scalars, you assumed
T=T_{y}+T_{x} where in fact T=√(T_{y}^{2}+T_{x}^{2}).
QUESTION:
Since everything in the universe is always moving, is the idea of being still/idle/stationary just an illusion?
ANSWER:
I would not put it like that. If you are in a system
where Newton's first law (if the net force on an object
is zero, it will remain at rest or moving with a constant
velocity) is correct, you may think of yourself at rest.
This is called an inertial frame of reference. But the
catch is that this is not the only frame of reference in
the universe where Newton's first law is true, any other
frame which moves with a constant velocity relative to
yours is also an inertial frame. In other words, there is
no such thing as absolute rest. A frame accelerating
relative to yours, however, is not an inertial frame.
QUESTION:
I am a cancer patient about to receive Y90 treatments. If Y90 has a half life of 64 hours: 1. When does that time clock begin running and 2. How can they ever have a full strength dose on hand at any given time?
ANSWER:
^{90}Y is the decay product of ^{90}Sr
which has a half life of about 29 years. 90Sr is an
abundant byproduct of the fission of uranium and
therefore available in the waste of reactors. When the
^{90}Y is needed it can be chemically separated
from the ^{90}Sr. I am sure that the hospital
cannot keep a supply on hand and will have it delivered
when it is needed for your treatment. It does not matter
how long it has been since it was separated, only that
the radiation level be correct for the prodedure. It
probably arrives with a level too high and they wait for
it to be at the needed level.
QUESTION:
I am struggling with the concept of relativity, as it relates to time. Not in terms of the underlying principle or the mathematics. These have been confirmed experimentally and are now in everyday use through technologies such as satellite navigation systems. My concern relates to the way that relativity treats time as a variable rather than an absolute quantity. It seems to me that we can only measure or experience time by reference to some physical process or change, whether that is the oscillation of a quartz crystal or the lifetime of a muon! Any experimental proof of time dilation as a result of the effects of either velocity or gravity really just relates to the way that we experience the passage of time and does not exclude the possibility that for the universe, as a whole, time is absolute and unchanging, passing in a constant fashion regardless of relativity and whether we measure it or not. Therefore, although relativity theory works in practice, this is only because it deals with our limited understanding of, and ability to measure and/or experience, the passage of time. Is this nonsense or is it a possibility? It follows from this reasoning that we can never prove experimentally that time either speeds up or slows down, only that the rate of how we measure or perceive time can vary. It also implies that time (and indeed space) are concepts that we will never fully explain as they ultimately cannot be qualified or quantified by measurement or mathematics.
ANSWER:
You are correct that "…time is absolute and unchanging,
passing in a constant fashion…" but only in your
frame of reference; any clock at rest relative to you
runs at a constant rate. But, what special relativity
tells us is that clocks not at rest relative to us do not
run at the same rate as ours. I have always felt that the
best way to convince someone that this is the case is the
light
clock. In order that this example is convincing to
you, you must accept that the
speed of light is a
universal constant in all frames of reference; but this
is an experimentally wellverified fact and also is a
consequence of the principle of relativity—that the laws
of physics are the same in all inertial frames of
reference.
QUESTION:
SINCE FREE FALL DONT NATURALLY OCCUR ON EARTH BECAUSE FLUID FRICTION IS PRESENT IF YOU
DROP A BOWLING BALL AND A BASEBALL OFF A TOWER OF PISA WHICH WOULD LAND FIRST
ANSWER:
The (upward) drag force on a falling sphere of radius R and speed
v in air
at sea level may be approximated as F=¼πR^{2}v^{2}=0.79R^{2}v^{2}
(this is correct only in SI units); the (downard) force of
the weight W on an object of mass m is
W=mg. The net force is the mass times the
acceleration a (Newton's second law), ma=mg+0.79R^{2}v^{2}.
So as the object falls it goes faster and faster until
v is large enough that it falls with a constant
value speed v_{t} (a=0), v_{t}=[√(mg/0.79)]/R.
For a baseball, m=0.144 kg and R=0.037
m, so v_{t}^{baseball}=9.28 m/s.
For a bowling ball, m=8 kg and R=0.12
m, so v_{t}^{bowling ball}=83.0
m/s, nearly ten times greater than the baseball. So the
baseball stops accelerating sooner than the bowling ball
and loses the race to the ground.
QUESTION:
My 8 year old asked me if the cycle of the universe expanding and contracting will ever end?
ANSWER:
What makes you and your son think that the universe is in
such a cycle? The ultimate fate of the universe is one of
the most important questions in cosmology and nobody
knows the answer. I believe most cosmologists believe
that the universe will continue expanding forever.
Eventually the supply of hydrogen and helium in the
universe will be exhausted (fused into heaviers elements)
and stars will no longer exist. The universe will be
dominated by black holes which will eventually decay away
by Hawking radiation, so the universe will be a very
dilute, cold, dark place filled with radiation. To get
more information, check out the
Wikepedia article on the fate of the universe.
QUESTION:
My son has a question.
A shadow is long or short, wide or narrow. He thinks this seems like area (taking up space). But the real question is... can you have a shadow without matter? Because if you can't have a shadow without matter, then doesn't that mean the shadow is actually matter? (I have to ask because I'm not sure how to answer this; I suppose the shadow is actually the part of the matter that doesn't have light reflecting off of it...? Or how do I answer this for him?)
ANSWER:
A shadow is not matter;
it is not anything, it is a region in which light
illuminating something is blocked by some obstruction.
You might say that a shadow is the lack of anything,
light which might have otherwise illuminated where it is.
And there are degrees of "shadowness". If all the light
illuminating an area is blocked by the obstruction, it is
called an umbra; if only part of the light is
blocked it is called a penumbra. (Umbra is Latin
for shadow.)
QUESTION:
U recently answered a question regards Earth's precession = 26,000 yrs. Can u show what specific equation(s) used to yield this time? I doubt that a single equation was used.
ANSWER:
Just like the precession of a top is due to the torque
exerted by its own weight, the precession of the earth is
due to torques on the earth, mainly by the sun and the
moon. Your equations may be found in this
Wikepedia article.
QUESTION:
On an ordinary day as one goes upward from the surface of the ground the electric potential increases by 100 volts per meter, this means that outdoors the potential of your nose is 200 Volts higher than that at your feet. Then why is it that none of us get a shock when we go out in the street?
ANSWER:
As is often the case, the most lucid answer to your
question can be found in the
Feynman Lectures.
QUESTION:
Why doesn't the earth just stop spinning and orbiting don't you need energy for motion where does the earth gets it's energy and is the earth in perpetual motion?
ANSWER:
You are hitting one of Isaac Newton's most important
discoveries—inertia. It says that you do not need
to exert a force (torque) to keep an object moving in a
straight line (rotating about an axis) to keep it going;
this means, also that the energy it has by virtue of its
motion will not change. So the rotating earth keeps
rotating since there are no torques on it*. The earth is
in a circular orbit† around the sun and so its orbital
motion keeps going since, even though there is a force on
the earth, there is no torque.
*Actually, the moon does exert a torque on the earth
which is causing the spinning to get smaller, but the
change is so tiny that it takes millions of years to be
significant.
†The earth's orbit is
not exactly circular. When it moves closer to the sun it
speeds up, when it moves farther from the sun it slows
down by exactly the same amount, so the orbit just
continues forever.
QUESTION:
With the devastating tragedy that happened in Lebanon, there has been some videos surface of the explosion. One such
video is of some ladies in a shop, the pressure wave hits the shop and the lady outside the shop doesn't get thrown back but inside the other lady does. What is the explanation here??
ANSWER:
Note that the glass doors are closed. Outside the front
of the pressure wave hits, but it takes some brief time
to reach its maximum. However, the pressure difference
between inside and outside is not large enough to break
the glass until it is near its maximum, so the pressure
difference changes much more quickly so the air rushes in
quickly. I will admit that this is just an educated
guess!
QUESTION:
If electricity always take the path of least resistance, why does a lightening discharge form a zigzag path to earth and not straight down which would be the shortest distance?
ANSWER:
Because air is not just a uniform homogeneous medium.
There are local fluctuations in temperature, humidity,
density, wind speed, and composition which makes the path
of least resistance at any instant, not a straight line.
QUESTION:
how momentum is conserved of both objects with same masses moving opposite direction to each other collide to a point and at same time both objects chemically combined to each other?
ANSWER:
In an isolated system, linear momentum is always
conserved; this is because the definition of linear
momentum is designed to be conserved in the absense of
any external force. This is also true in special
relativity where the linear momentum is not p=mv but
rather p=mv/√[1(v^{2}/c^{2})]
where c is the speed of light. In the event of
chemistry going on, there is either energy gained
(endotermic) or lost (exothermic). If lost, the momentum
of the radiation has to be taken into account. If gained,
the energy will be taken from the kinetic energies of the
incident objects but momentum will be conserved.
QUESTION:
Does precession and nutation ( or anything else ) affect the axis of the earth?
ANSWER:
Precession of the earth's axis is caused mainly by
torques exerted on the earth by the moon and the sun. The
period of precession is 26,000 years. But precession does
not cause the actual rotation axis to change. What causes
the axis to change (such that the geographical positions
of the poles would change) is thought to be three things:

glacial ice
melting and sending its water to the oceans,

glacial rebound
which is the raising up of the land previously
supporting the melting glacier, and

tectonic motion,
the drift of large land masses.
QUESTION:
The momentum transmitted to the golf ball comes from the club head and,to a lesser extent, from the club shaft.
Golfers are instructed to hold the club very loosely. Thus, it seems, no momentum is transmitted from the arm.
What if the golfer instead holds very tightly to the club? Will the momentum of the arm be transmitted to the golf ball?
ANSWER:
The physics of a golf swing is much more complicated than
you might think. From the little research I saw, your
"hold the club very loosely" is an oversimplification.
The club needs to accelerate very quickly on the way to
the ball and that acceleration is only going to come from
the torque you provide. What should happen is that, just
before impact, your wrists must relax. There is an
excellent discussion on the golf swing which, for the
first half is fairly conceptual and the second half goes
into the detailed physics analysis of the swing.
FOLLOWUP QUESTION:
I understand that the standard golf swing depends on uncocking the wrists to generate velocity at the moment of impact.
Moe Norman, a Canadian golfer now deceased, was the best ball striker in the history of golf. Moe used a completely different swing that he developed. Among other things, it involved holding the golf club tightly.
Thought experiment: The shaft of the golf club is fused to the golfer's arm and runs all the way up to the shoulder. That is, there is no wrist play at all. I understand that such a setup will generate less velocity than a hinged wrist. Therefore, the 'v' in 'mv' will be lower. My question is whether the 'm' will be higher. Given that the golfer is now swinging a much more massive club, albeit from the shoulder, I posit that the mass striking the ball is greater.
Take the center of gravity of the clubarm, calculate its velocity, multiply by the mass of the clubarm, and one has the momentum that is transmitted to the ball (I think). This may or may not be as great as the momentum of the faster wristhinged swing.
I just want to know whether the 'm' will be greater with this fused club.
ANSWER:
I believe that there is no way to understand the club as
a point mass, so to talk about its mass in a collision is
meaningless. Think of it as a machine which imparts
momentum to a small mass by exerting a force on it over
some short time. I think it probably boils down to having
the machine achieve the greatest possible velocity when
it hits the ball; if you can achieve that by holding the
club more tightly, go for it! It was interesting to learn
a little about Moe Norman of whom I had never heard.
QUESTION:
When a golf club hits a golf ball, part of the momentum of the golf club transfers to the golf ball. The golf ball, being light, takes off at a velocity greater than the club head speed.
What if the golf club is accelerating at the point of impact? Will the ball take off faster (and go further) than when it is hit at the same impact velocity by a club that is not accelerating?
Note: amateur golfers tend to reach maximum velocity at the point of impact. Pros reach max velocity after impact. Pros do swing faster, which imparts more velocity to the ball. I would like to know whether their acceleration adds even more velocity.
ANSWER:
What will determine the exit speed of the ball is the
speed of the club at impact. The impact time is
approximately Δt=0.0005 s. How big would the acceleration
need to be in order to significantly change the speed
over this time? A typical speed of the club head is about
150 mph, about v=70 m/s. Suppose that the speed increased
over the impact time by Δv=1 m/s. Then the acceleration
would have to be a=Δv/Δt=1/0.0005=2000 m/s^{2};
this is about 200 times greater than the acceleration due
to gravity. We could estimate the acceleration by
guessing that the club took about 0.1 s to go from zero
to 70 m/s, a=700 m/s^{2}. So the speed
of the club might increase by 700x0.0005=0.35 m/s and the
average speed during impact would be about 70.2 m/s. My
feeling is that this would have a negligible effect on
the speed of the ball.
FOLLOWUP QUESTION:
I did some more research and I have a follow up.
This article states that Force equals mass times acceleration.
It indicates that applying a force to an object over time creates momentum.
Therefore, it seems to me that a club head that is accelerating at the point of impact transmits Force and, therefore, momentum to the golf ball. And that this momentum would be in addition to the momentum of the club head.
A club head that is not accelerating will transmit no force to the golf ball.
It seems to me that a club head accelerating is going to hit the ball further, velocity at impact being equal.
ANSWER:
Sorry, but you have this all wrong. (This is what comes
from using formulas when you do not really understand
what they mean or when they are applicable!) F=ma
is Newton's second law and simply says that if you push
or pull on a mass it will accelerate. Indeed, if you
apply a force over a time you will accelerate it and
therefore change its momentum. But these do not
imply that the source of the force itself needs to be
accelerating. Here is what happens: when the club strikes
the ball, it exerts a force on the ball; Newton's third
law says that if the club exerts a force on the ball, the
ball exerts an equal and opposite force on the club. For
the short time that they are in contact (after which the
forces disappear), the ball will speed up and the club
will slow down if there are no other forces on it (which
may be the case*). If the average force they exert on
each other is F and the time of contact is t,
the ball will acquire a linear momentum of p=Ft;
the ball will experience a change in its momentum but by
how much is not so simple since other forces act on it.
Note that F=p/t=mv/t, so the
smaller t, the larger F, all else being
the same. In my answer to your original question, using
v=70 m/s and t=0.0005 and noting that
the mass of a ball is about m=0.046 kg,
F=0.046x70/0.0005=6,440 N=1448 lb.
*If the club is accelerating when it strikes the ball
there must be some force on it—ultimately that
comes from you.
QUESTION:
I am writing a video response to Dr. William Lane Craig, a well known apologist, and so I am answering every position he states from a debate between him and Christopher Hitchens at Biola university some years ago.
I write to you now because Dr. Craig says something that I don't understand. He states,
"First, when the laws of nature are expressed as mathematical equations, you find appearing in them certain constants, like the gravitational constant. These constants are not determined by the laws of nature. The laws of nature are consistent with a wide range of values for these constants."
I don't understand this at all.
ANSWER:
(This is not the whole question submitted, but gets to
the heart of what the questioner wants to know. Reminder
that site ground rules specify "…concise, well focused
questions…") I consider that the most important
"laws" of nature are not equations, they are
proportionalities. In order to keep this discussion
focused on my basic description of nature, I will focus
solely on classical physics to make my points. In physics
we begin with three fundamental concepts to lay the
foundation of describing nature: mass, length, and time.
In the SI unit scheme we choose kilograms (kg), meters
(m), and seconds (s). With the chosen units we can define
velocity as the rate of change of position with units of
m/s; then we can define acceleration as the rate of
change of velocity with units (m/s)/s=m/s^{2}. So
if an object changes its speed from 2 m/s to 6 m/s in a
time of 2 s, its acceleration is 2 m/s^{2}. Mass
is trickier to understand but it is a quantity which
measures how resistant an object is to being accelerated
if you push on it; it is harder to accelerate an object
of mass 1000 kg than it is one of 2 kg. Note that we have
not yet quantified that push (or pull) which we normally
call force. However we can certainly imagine figuring out
a way to push or pull on something twice as hard, or
three times as hard, etc. Whatever force is,
suppose we push on something with a constant force F
and vary the mass; we will find that if we double the
mass we halve the acceleration, if we halve the mass we
double the acceleration, if we make the mass 10 times
larger, the acceleration is 1/10 of its original value,
etc.
We have found that, F being constant,
acceleration a is inversely proportional to mass,
a∝1/m.
Now suppose we vary F but keep m constant; we find that
if we double the force we double the acceleration, etc.,
i.e. a∝F. We can now put
the two together and get a∝F/m or F∝ma.
This is what I consider to be Newton's second law; it
tells us an important connection of how quantities depend on
each other, it is a law of physics; and, you really did
not have to have defined the meter and second and
kilogram for it to be true, you just had to understand
conceptually what mass, length, and time are. If you know
even a little physics, you do not recognize this as
Newton's second law—any textbook will tell you that
it is F=ma. Now, how can one turn a
proportionality into an equation? It is simple, just
introduce a proportionality and voila! F=Cma
where C is an arbitrary constant since we
have not defined how we will measure F. Don't
think of it as a "fundamental constant" because if we had
already defined how to measure force, we would have to
measure C. For example, if we had defined a unit
of force to be a pound but measured m and a
in SI units (kg and m/s^{2}), F=ma would not be a valid
statement of Newton's second law.
Now let's discuss an example of when a constant is a
fundamental constant. Newton's law of gravitation
expresses the magnitude F of the equal and
opposite forces exerted on each other for two point (or
spherical) objects of masses m_{1} and
m_{2} separated by a distance of r:
F∝m_{1}m_{2}/r^{2}.
Now, to make this an equation, we must introduce a
proportionality constant G: F=Gm_{1}m_{2}/r^{2}. But can we choose it to be whatever we like? No, because
there is nothing in this proportionality which is
undefined; we can take two 1 kg masses, separate them by
1 m, and measure the force between them (a really hard
experiment!). In other words, we must measure the
constant G. This is a true
fundamental constant of nature, it measures the strength
of gravitation. Of course, its numerical value depends on
how we measure mass, length, and time, but it is still a
universal constant. In SI units it is 6.67x10^{11} N·m^{2}/kg^{2}.
You might be interested in two earlier answers,
1 and
2, along similar lines as
yours.
QUESTION:
I wondered what happens if one would make a magnet spin really fast. Could it be theoretically possible that it could create visible light.
ANSWER:
The frequency range of visible light is about 48x10^{14}
Hz, so you can be sure that you cannot do this with any
macroscopic magnet. One way to get magnetic radiation is
to fabricate something called a
splitring resonator (SRR) but technical problems
limit SRRs to about 2x10^{14} Hz, near infrared.
Another technique uses spherical silicon nanoparticles
and these can generate visible magnetic light. The method
is too technical to discuss here, but you can read about
it at
this link.
QUESTION:
i had a doubt, to detect an inertial frame we need to define 0 force, but to define 0 force we need an inertial frame,0 force is defined as the sum of real forces should be 0, and we can identify real forces if we believe that the real forces drop with distance,and this could be a possible solution to the above problem, but it based on an assumption, and even if we consider that we cannot detect an inertial frame, and all the physics upto special relativity is just based on if there is an inertial frame, then how come we could do experiments to confirm the theory and how did physics work uptill general relativity, if we cannot solve this basic conceptual problem.
ANSWER:
Technically, there is no such thing as an inertial frame.
No matter where you go in the universe there are always
electromagnetic and gravitational fields. That does not
mean we cannot define an inertial frame as one in which
Newton's laws are true at low velocities. In fact we can
do better by defining an inertial frame as one in which
the total linear momentum (relativistic
definition) remains constant; then you are not
confined to low velocities. Then we will find that very
good approximations to inertial frames can be found in
nature. There is nothing wrong with basing a theory on an
idealized (fictional) concept as long as the ultimate
theory agrees with experimental results in the real
world.
QUESTION:
The idea of dark matter and dark energy is puzzling me. It is understood that dark matter acts like glue to hold galaxies together. On the other hand, dark energy is responsible for expansion of universe. Scientists say inside galaxies dark matter overcomes dark energy so gravity wins and keeps galaxies together but when it comes to intergalactic space, it is said that dark energy overcomes dark matter and wins the fight, so galaxies are moving away. How do you explain this contradiction? Why dark energy wins in intergalactic space but not inside the galaxies.
In other words, what makes dark matter overcomes dark energy in galaxies but loses in the space between galaxies.
ANSWER:
Even though I state on the site that I do not do
astronomy/astrophysics/cosmology, I will take a stab at
this one. Of course, keep in mind that nobody knows what
dark matter or dark energy are in any detail. Dark
matter, as the name implies, is thought to be some kind
of mass and therefore subject to gravitational forces. It
therefore would be not surprising to find it more
concentrated in the vicinity of large masses like
galaxies; so its density is likely to be smaller in
intergalactic space. Even less is known about dark energy
but, if it is anything like uniformly distributed, it is
more likely to lose a "tugofwar" with dark matter
inside a galaxy than outside.
QUESTION:
What happens to the gravitational effect of mass when it transforms into energy, as in e=mc^2?
ANSWER:
In general relativity, our best theory of gravity, the
bending of spacetime, which is what gravity is, is not
caused only by mass. Any local energy density will cause
gravitation.
QUESTION:
I would like to ask about gravitational mass.
I know inertial mass is changing by motion (speed) according to m=mo/(1v2/c2)^0.5 And also that is inertial mass which sits in E=mc2.
If the statements above is correct, now how about gravitational mass? Does it change with motion (speed)? And what mass should be used for general gravitational formula F=GmM/r2? should we use mo (rest mass) regardless of speed of the object? Or should we use m=mo/(1v2/c2)^0.5 to substitute in F=GmM/r2?
In other words does mass equivalence principle (inertial mass=gravitational mass) hold in hight speeds?
ANSWER:
Read the answer to the previous question. The moving mass
has more energy than the stationary mass and therefore
has a stronger gravitational field as measured by a
stationary observer. Newton's gravitational equation is
amazingly accurate for speeds small compared to the speed
of light. However, if you are at very high speeds you
should not use it at all. The whole notion of force is,
itself, not useful in relativistic circumstances. We
might ask why force is a useful concept in classical
physics. Newton's second law, the central concept in
classical mechanics, is F=ma. Why? If someone is
moving past you and measures the acceleration a'
of something in your frame, she will measure exactly the
same value as you. Therefore you will both conclude the
same force is being applied to m, if a=a' then
F=F'. But if relative speeds are not small,
a≠a' so Newton's second law is no longer true.
Instead we rely more on conservation principles more in
relativity theory. In particular, we want to
redefine momentum so that it is conserved in a closed
system.
QUESTION:
I have recently learnt about the LIGO detector and watched a few videos on how it works, I understand how they detect the stretching and contracting of space time due to the gravitational wave only stretching and contracting a few hundred times per second and the new photons entering this space must travel farther or shorter deviating from the normal time travelled.
I would like to ask how these new photons enter this stretched space time since the gravitational wave itself is travelling at the speed of light does that not mean the stretched space will either move with a certain packet of photons, and since the wave travels at light speed, there wont be enough contractions to allow detection to occur.
ANSWER:
You just make this too hard to visualize thinking of photons.
Think instead as
electromagnetic waves. (Although
the speed of the gravity wave does not matter for the
interferometer to work, it is just as if somebody were
jiggling the mirror ends, I want to note that it is only
assumed that the speed of gravity is c; the
speed of gravity has never been measured, we just know
that it is very fast. Since the gravitational field has
never been quantized, we can only assume that the
graviton has no mass. Note that for decades it was
assumed that neutrinos had no mass, something we now
understand is not true.)
QUESTION:
Who first derived the relativistic energymomentum relation
E^{2}+p^{2}c^{2}=m^{2}c^{4} and in what publication? Wikipedia says that it was Dirac in 1928, but I cannot find evidence of this in his famous 1928 paper, "The Quantum Theory of the Electron."
ANSWER:
As far as I can tell, Dirac's 1928 paper used the
energymomentum relation as a jumping off point for his
derivation of the Dirac equation. Certainly this equation
was known long before 1928 since special relativity
itself was introduced in two papers by Einstein in 1905. The
first paper did not mention momentum at all. In the
second paper he shows that if a particle of mass m
radiates energy L that the mass changes its
kinetic energy by L/c^{2},
hence E=mc^{2}, but does not mention
momentum. Finally in 1907 he writes an
article in which
he shows that if relativistic momentum is p=mv/√[1(v/c)^{2}],
then it is a conserved quantity as in classical Newtonian
physics. From here, and knowing E=mc^{2}/√[1(v/c)^{2}],
it is
straightforward to get the energymomentum
relation.
QUESTION:
Why would lifting my feet off the ground make me heavier. Let's say I were to do a push up while my friend is on my back for extra weight. If the friend was to plant
her foot on the ground, why would the pushup become easier.
ANSWER:
In the first figure I show all the forces on the man in
yellow. They are: N is the force which the
ground must exert up on the man, W is his
weight, and F is the force which the woman
exerts down on the man. (I have drawn N as one
vector, but it is really distributed among his two feet
and two hands. Nevertheless, the size of N is
proportional to the force he must exert to lift himself.)
The man is in equilibrium, so the forces on him must add
to zero, NWF=0 or N=W+F. In white are
the forces on the woman: w is her weight and
f is the force which the man exerts up on the woman
which by Newton's first law equal and opposite to the
force which she exerts down on the man so F=f.
Since she is also in equilibrium, the sum of the forces
on her must add to zero, so fw=0 or f=w=F.
So, finally, we see that the total weight the man must
cope with is the sum of his and the woman's weight,
N=W+w.
So now I put a blue brick under the woman's foot so that
it rests on the brick. This is a new force n up
on her, shown in red. Each person's weight is the same as
before, your weight is just the force down due to
gravity. But now there are three forces on the woman
adding up to zero: n+f 'w=0 or f '=wn.
The force f ' exerted up on her by the
man, which has the same magnitude of the force F
' which she exerts down on the man, is smaller than
before. So now,
N 'WF '=0
or N '=W+F '=W+wn. He has a smaller total force
to contend with than before.
The mistake you made is that you assumed that the force
down on the man was the woman's weight. But the woman's
weight is not a force on the man, it is a force on the
woman. This is accidentally correct in my first example
because if only the man is supporting her, it just
happens that the force she exerts down is equal to her
weight in magnitude. But in the second example the brick
is supporting part of her weight so the man supports
less. She does not become lighter!
QUESTION:
I am writing a fantasy novel currently, and at the ending a giant Dragon, with scales made out of stone is falling into the sea infront of a city and I'm imagining a tsunami destroying the city. The dragon weighs around 2,000 tons, how high would the wave be and how far would it reach inland? (the city is basically a flat plain)
ANSWER:
Well, I have to make some wild approximations to address
this at all. I will perhaps get you within an order of
magnitude of how big the wave might be. Maybe I should sort of do a rough
general case first and then I can throw in some
reasonable numbers. If the dragon drops from a height
h and has a mass m, then the energy she
starts with is E_{1}=mgh, where
g is the acceleration due to gravity. When
she hits the water she will have a smaller energy because
of air drag falling down and when she comes to rest she
will lose some energy to heat and sound and damage to her
body; I will say that a fraction f of the original energy
goes into the energy E of the wave created,
E=fE_{1}. Now I will assume that all this
energy goes into a circular pulse which has a width of
w
and a length of 2πR (circumference of a
circle) where R is the radius of the circle at
the time of interest, namely when the wave hits the
shore; so R is the distance from shore where the
dragon hit the water. Now, I found an expression on
Wikepedia which relates the energy of a wave to its
height: E/A=ρgH^{2}/16
where A=2πRw is the total
horizontal area of the wave, H is the height of the wave,
and ρ=1000 kg/m^{3} is the density
of water. If you put it all together you will find
H=√[8fmh/(ρπRw)].
Suppose h=5000 m, m=2x10^{6} kg (2000 metric
tons), f=½, and w=20 m; then I
find H≈25 m. This is in the same ballpark as the
largest tsunami wave ever observed which was about 100
ft. There is no way I could estimate how far inland the
wave would travel; it would depend a lot on the terrain,
obstacles, etc.
QUESTION:
This question is in reference to quantum entanglement.
When a Super positioned photon is measured to determine spin does it's spin stay in that orientation as long as they a measuring it or does it immediately go back to a Super positioned state?
In other words if you determined the spin of a quantum entangled particle at say 12:00 pm and constantly measured it, would it's spin continue to point in the same direction at say 12:02 pm. or is the measurement only good for the exact moment it was initially measured?
ANSWER:
When you observe it, you put it in a single state (and
simultaneously, put the other entangled photon in a
single state). Unless it interacts with something else
later, they remain in states you put them, they do not
reëntangle.
QUESTION:
What is the maximum distance between two electrons where they will notice each other? They would both be at rest to each other in a vacuum with no external stimuli or forms of energy. At what distance do the electrons "See" each other.
ANSWER:
This question has no meaning because "notice each other" is not quantified. In principle, the Coulomb
force extends all the way to infinity.
You could, for example, ask at what distance would the force
each electron feels would result in an acceleration of 1 mm/yr^{2}=[1x10^{3 }
m]x[(1
yr)x(365 day/yr)x(24 hr/day)x(3600 s/hr)]^{2}≈9x10^{22}
m/s^{2}.
In general, the force is F=ke^{2}/r^{2}=ma,
so r=√[ke^{2}/(ma)]
where k=9x10^{9} N·m^{2}/C^{2},
e=1.6x10^{9} C, and m=9x10^{31} kg. I
calculate, approximately, that r≈5x10^{11}
m≈300,000,000 miles which is about 3 times the
distance to the sun. Of course, you could never do such
an experiment since you would need to be in a universe
which contained nothing but these two electrons.
QUESTION:
Aren't "constants" simply a "fudge number" to make equations work? It seems to me, constants are values of a concept we have yet to identify or understand, but need to make equations with known concepts and values work out. Since these equations work correctly with a variety of known variables, they are generally accepted. It seems to me, nonelectrical magnetism (i.e. that which is retained in magnetic material like iron) and gravitation have a variety of descriptions and equations which predict their effectsyet we still are devoid of actual cause.
ANSWER:
Constants are not just "fudge numbers" as you suggest.
They can be, but in general constants play a very
important role in physics. I will give you a few
examples.
The laws of physics are stated most elegantly as
proportionalities rather than equations. For example,
Newton's second law states that the acceleration (a)
of an object is directly proportional to the applied
force (F) and inversely proportional to the mass
(m), a∝F/m; so you double the
acceleration if you push twice as hard and halve it if
you double the mass. Now, to calculate we prefer to have
equations rather than proportionalities, so we introduce
a proportionality constant, call it C, a=CF/m.
What does C mean? Since we know how to
measure a (m/s^{2}) and m (kg),
your choice of C determines how we will measure
force. I would like the force to have the magnitude of 1
if a=1 m/s^{2 }and m=1 kg, so I
choose C=1. The constant C is not a fudge
factor, rather its choice defines how we measure forces.
(We call the unit kg·m/s^{2 }a Newton, N.)
Newton also discovered the universal law of gravity: if
two masses (point masses or spheres) m_{1} and
m_{2} are
separated by a distance d, they exert equal and opposite
attractive forces on each other the magnitude of which is
F; then F∝m_{1}m_{2}/d^{2}.
So the equation for universal gravitation would be of the
form F=Gm_{1}m_{2}/d^{2}.
where G is some constant. But we know how to measure mass
and distance and force, so we cannot choose G to
be any old thing we want as we did for Newton's second
law, we must measure it. It turns out to be G=6.67x10^{11}
N·m^{2}/kg^{2}. This is one of the
most fundamental constants in all of nature and most
certainly not a "fudge factor".
Constants are often used to quantify a particular
situation. For example, the frictional force f
of one object sliding on another is found, approximately,
to be proportional to the force which presses one object
to the other (often called the normal force, N).
This is not a physical law, just an approximation of
experimental measurements, and it obviously involves only
the magnitudes of the forces since their directions are
perpendicular to each other. So, including a
proportionality constant μ, called the
coeficient of kinetic friction, f=μN. This
constant tells you how slippery the surfaces are; its
value for rubber sliding on ice will be much smaller than
for rubber sliding on asphalt.
Sometimes, as you suggest, constants are added to fit
data and their physical meaning is unknown, you might
label such constants as "fudge factors".
You may be interested in learning about Planck units
discussed in an earlier answer.
where new units are expressed by combinations of the five
fundamental constants of nature.
QUESTION:
In a example of a electrical generator. Central in the generator are rotating magnets. I understand that as the magnet moves the cores of the surrounding coils will react producing a magnetic field. Also the copper wire winding around the coil will react.
My question is does the copper wire winding react to magnetic field according to the field produced by the core or the moving magnet ? In a lenz law sense.
Plus would I be correct in thinking if there is no circuit on the copper winding there no organised field would form on the copper wire winding itself.
ANSWER:
The best way to understand the priciples is to look at
the simplest possible generator. The figure shows a
generator consisting of a single loop rotating in a
magnetic field which is approximately uniform. There
would be no difference in the results if it were the
magnets rotating around the coil as you describe. Since
the induced EMF in the loop is proportional to the time
rate of flux through the loop and the flux is
proportional to the cosine of the angle θ between the field
and a normal to the area of the loop, the induced EMF
will be sinesoidal. The angle may be written as θ=ωt where
ω is the angular velocity of the loop (or magnets)
and t is some arbitrary time, so the
voltage is proportional to cos(ωt). If
there is a load across the terminals, and therefore a
current through the loop, there will be a field due to
the current which does not affect the output voltage if
the source of power maintains a constant angular
velocity. If you are drawing power from the generator, it
is harder to "turn the crank" than if there is no current
flowing. If you need detailed information on "reallife"
generators, it is more an electrical engineering question
than physics.
QUESTION:
How did Einstein derived general relativity math when he understood that the gravity is just the space bending around an object how did he applied this to the world of math?
ANSWER:
You can learn about how Einstein struggled with the
mathematical methods necessary to fully describe general
relativity and with which mathemeticians he consulted in
his biography by Walter Isaacson,
Einstein: His Life and Universe.
QUESTION:
I have been thinking about clocks lately and found myself in a sort of a pickle over the velocity of a dial. (I am not native so sorry if my description makes it very complicated). Suppose we have a disc with a dial starting at its center. The dial draws two circles on that disc as it turns. The first circle has its radius equal to r, whereas the second circle has a radius of 2r. Now suppose that 1 full turn takes 1s (which would be the same for both circles since they are being drawn by the same dial). Now because of the difference of radii, there is also a difference in their circumference
c. And so we've got t=1s, c=2πr and c'=4πr (2π2r). Now here's the question  if the time is the same, but the circumference or distances are different does that mean the velocity of these two points on the same dial is different? How is that possible since the dial turns with a set velocity which in my understanding would be the same for each point on that dial?
ANSWER:
It is simply because all points on a rotating rigid body
have the same angular velocity, (revolutions per second,
e.g.) but not the same translational velocity (miles per
hour, e.g.). In general, v=rω
where v is the translational speed, r is the distance
from the axis of rotation, and ω is the
angular velocity in radians per second.
QUESTION:
I was just wondering why does the centripetal acceleration formula work a=v^2/r?
ANSWER:
Why does it work? It "works" because it is correct. Maybe
you wanted to ask "where does it come from?" or "how is
it derived?" I will give you a brief standard derivation.
On the left of the figure I have drawn the particle
moving with constant speed v around a circle of radius R.
In some time interval Δt the particle moves through
an angle theta and has velocities v_{1}
and v_{2} but both
speeds are equal to v, v_{2}=v_{2}=v.
On the right I have placed the velocities tailtotail
and drawn the difference vector v_{2}v_{1}=Δv.
So now I see two isoscoles triangles, each with the same
angle between their equal sides; they are therefore
similar triangles and so we can write v/Δv=R/Δs.
If I now rearrange and divide each side by Δt,
I find Δv/Δt=(v/R)Δs/Δt.
Now, Δv/Δt is the magnitude
of the average acceleration; to get the instantaneous
acceleration, we must take the limit as Δt approaches zero.
But, when Δt approaches zero, Δs/Δt approaches
v. So, finally, a=v^{2}/R.
QUESTION:
Is Planck time a concept derived from the theory of relativity or its premises?
ANSWER:
No, that is not where the concept
came from. The origin of
Planck units came from the
desire to have a system of units which are based only on
constants of nature, not on the specific units like the
meter, the second, etc. You take the five
universal constants:

the speed of
light in vacuum, c,

the universal
gravitational constant, G,

the rationalized
Planck's constant, ℏ
,

the Boltzman
constant, k_{B}, and

the Coulomb
constant, k_{e}=1/(4πε_{0}).
You now form
combinations of these constants which have the correct
dimensions of the units you want:

length,

mass,

time,

temperature, and

electric charge
I have copied the table from
Wikipedia showing the values of these Planck units in SI
units:
It is hypothesized that, at
distances smaller than l_{P}=1.6x10^{35}
m, the usual physics as we know it will not work. I think
you might say that the origin of the idea of Planck units
is quantum field theory.
QUESTION:
A toy top is a diskshaped object with a sharp point and a thin stem projecting from its bottom and top, respectively. When you twist the stem hard, the top begins to spin rapidly. When you then set the top's point on the ground and let go of it, it continues to spin about a vertical axis for a very long time. What keeps the top spinning?
ANSWER:
There is a property of spinning
objects called angular momentum. For example, the angular
momentum of your top is a vector approximately equal to ½MR^{2}ω
where M is the mass of the top, R its radius, and ω
is the rate at which it is spinning; the direction of the
angular momentum vector is straight up if it is spinning
counterclockwise if viewed from above, straight down if
clockwise.
There is a physical law which states that if there are no
torques on a spinning object, its angular momentum never
changes. There are no torques on the top if its spin axis
is vertical and so it does never stops.
QUESTION:
The third dimension is 3D (spheres, etc.). The first dimension is a dot. Everything that has a front and back, has an edge. So isn't a dot 3D?
ANSWER:
A dot has zero dimensions. The
formal definition of the dimensionality of a space is an
answer to the question "how many numbers does it take to
specify the location of a point in that space?" You may
heard of Cartesian coordinates, (x,y,z);
they were named in honor of the 17th century philosopher
and mathematician
René Descartes. Legend has it that he
was once watching a fly buzzing around his room and asked
"what is the minimum number of numbers I must specify to
describe the position of the fly at any time?" He figured
it was three: the distance up from the floor, the
distance from the back of the room, and the distance from
one of the side walls; his room is said to have three
dimensions. One dimension is a line; here you simply
measure the distance from one point on the line to
wherever the particle is. The line need not be a straight
line. A circle is onedimensional and the location of a
point on the circle is usually specified by an angle
relative to some point we call zero. Two dimensions is a
surface. The surface of the earth is twodimensional, the
position being angles, latitude and longitude. One
surface of a flat sheet of paper is twodimensional and
the two numbers are usually Cartesian coordinates (x,y).
The spatial universe in which we live is
threedimensional, just like Decartes' room. A solid
sphere has three dimensions usually measured by the
distance from the center and two angles (longitude and
latitude).
QUESTION:
Why there is only time dilatation and no time shrinkage? For me I think there should be time dilatation regarding objects traveling away from each other and shrinkage regarding those closer to each other. As what makes one gets older and the other still young for both of them are moving with same speed relative to each other.
ANSWER:
Thinking how you think something
should be does not make it true! You should read my
answer about the
twin paradox. How fast a clock runs and how fast it
appears to actually runs are two different things; see
earlier answer.
QUESTION:
While we use the combination of rear and front brakes to stop a motorcycle, I mean we can chose between the the front and rear without any effort, why do cars primarily use the front brakes to stop? Why does the brake pedal engage only the front brake? The rear brake is only seem to be used when the car is parked
ANSWER:
All modern cars have fourwheel
braking. So where did you get the idea that the brake
pedal engages only the frontwheel brakes? Parking brakes
engage only rear wheel brakes. Your confusion may be that
the braking causes the frontwheel tires to wear more
than the rearwheel tires; this is due to the car
slightly "rocking forward" when stopping and also because
in a front engine car more of the weight is likely to be
supported by the front tires.
QUESTION:
This is a kinematics question about
the motion of wheels; specifically, about how friction between a wheel
and a surface causes forward motion of the entire wheel. There's
something unsatisfactory to me about how this forward motion is
typically explained.
Consider the rear wheel of a bicycle; it is driven by torque applied to the wheel (via chain drive connected to pedals, etc.). In the classical treatment of this scenario, it is claimed that the friction between the wheel and the road surface, which is a
forwardpointing force, causes forward motion of the wheel. But this frictional force is tangential! So wouldn't this tend to be a rotational force instead of one of displacement? It would seem to me that somehow the tangential rightward force of friction at the point of contact needs to be transformed into a rightward force applied at the *center of mass* (i.e., at the axle) of the wheel in order to explain an induced forward/rightward motion, but no source that I have found has attempted to explain this gap.
ANSWER:
The figure shows all forces on
the rear wheel of a bicycle:

C,
the force exerted by the chain on the sprocket
(radius r),

W,
the weight of the wheel (radius R),

N,
the vertical force exerted by the ground on the
wheel,

f,
the frictional force exerted by the ground on the
wheel, and

F,
the force exerted by the rest of the bike on the
wheel.
Note that I have
resolved F into its horizontal,
H, and vertical, V,
components. For any body which cannot be approximated as
a point, Newton's second law (N2) has two parts, translational
and rotational:

The sum of all
forces on a body is equal to the product of the mass
(m) and the acceleration (a) of the
center of mass of the body;

The sum of all
the torques on the body about any axis is equal to
the product of the moment of inertia (I) of and the
angular acceleration (α) about that axis.
You seem to think
that a force which exerts a torque about the axis you
have chosen (I will choose the axle for the bike wheel)
only can cause angular acceleration; in fact, every force
on the object contributes to the translational
acceleration. Suppose now that you are standing on the
pedal but also applying the front brake such that the
bike remains at rest; both translational and angular
acceleration are zero. So the translational N2 yields two
equations, NWV=0 and C+fH=0; the
rotational N2, choosing the axle as the axis, yields
CrfR=0. (F, W, and N
exert no torque about the axle.) If you now release the
brake, H will get much smaller
and the wheel will start moving forward and rotating.
If you are
interested in the entire bike, not just the rear wheel,
the force C has no contribution to the motion because it
is an internal force; the chain exerts a force
C on the rear sprocket but a force
C on the front sprocket
netting zero net force.
FOLLOWUP QUESTION:
You say, "in fact, every force on the object contributes to the translational acceleration" but the force, f, in your answer, is not directed at the center of mass of the wheel.
I guess my question is simply, "why is f treated as a force directed at the center of mass and not as a torque on the wheel"? I don't understand where the forward acceleration of the *center of mass* of the wheel is coming from.
ANSWER:
You did not read my answer
carefully. The translational N2 is "The sum of all
forces on a body is equal to the product of the mass
(m) and the acceleration (a) of the
center of mass of the body". Note that it does
not say
…the sum of all forces which are directed through the
center of mass… Let me give you a simpler
example. Suppose that you are in the middle of empty
space and you have a wheel which has a string wrapped
around its circumference. You pull on the string so that
the tension in that string is the only force on that
wheel and it is tangential. Surely you do not
think that your pulling will cause the wheel to only spin
and not accelerate toward you too. Here is another
example, a yoyo. There are two forces on it, its own
weight which passes through the center of mass and points
down and the tension of the string which you are holding
which points up and does not pass through the center of
mass. If the string does not affect the translational
acceleration, the only force on the yoyo would be its
weight so it would fall with acceleration due to gravity,
g, which it clearly does not do.
SECOND FOLLOWUP QUESTION:
In classical billiards problems, for example, when a force is applied to a ball, the force has to be decomposed into its normal component directed at the center of mass (call it f_n) and the tangential component (call it f_t). Only f_n will cause translation, right? It's my understanding that f_t will only contribute a torque, but not to translation. So why is the bike case any different? Contact friction between the wheel and the surface is a tangential force, so, according to my understanding, it can not cause translation, because it has no component directed at the center of mass of the wheel...
ANSWER:
Look at the figure*. It is much
better to resolve F into its
horizontal and vertical components rather than normal and
tangential components. It is wrong to insist that a force must be
directed toward the center of mass for that force to
contribute to tranalational motion; my examples above
should have convinced you of that. Now, a billiard ball
is confined to move on the table, so there will be no
translational motion in the vertical direction. So the
ball is in equilibrium in the vertical direction, so
NVW=0. But in the horizontal direction there will
be an acceleration, ma=Hf. Only f
and F will exert torques and it
will be a little tricky to calculate the torque due to
F but just geometry. I have
drawn the line of force to be below the center of mass;
if it is through the center of mass, it will exert no
torque and only f will cause
angular acceleration.
Here is the takeaway for you: just use N2 as I have
given it to you, always look at all forces on the object,
and do not continue insisting that a force which exerts a
torque does not affect translation.
*I have drawn
the forces on the ball, which are

the ball's
weight W,

the normal force
N the table exerts
vertically on the ball,

the frictional
force f the table exerts
horizontally on the ball, and

the force
F exerted on the ball.
QUESTION:
A hammer hitting an anvil creates energy (sound) which moves in all directions at about 700 mph in atmosphere. Sound energy is also created by a large fire. In space, vaccuum, that same energy must also be present if a hammer hits an anvil in space, same as a star (a large fire ball) should also create sound energy, along with all the other energies it creates, so if sound energy in space, a hammer hitting an anvil, does not have atmosphere to slow it down, or to make noise, how would one measure that energy, its speed and pressure. Could this be what is called "Dark Energy".
ANSWER:
Sound is a wave that exists in a
solid, liquid, or gas. If you are in a vacuum, there is
no sound. The energy which the sound carried away in air
is instead retained as sound in the anvil (and hammer).
But this sound eventually damps down and where that
energy goes is an increase in the temperature of the
anvil (and hammer). This energy eventually radiates away,
not as sound but as electromagnetic waves (mainly
infrared) until thermal equilibrium is achieved. It
certainly has nothing to do with dark energy.
QUESTION:
So my question is a bit out of reality question as its not possible to do this in real life. But if you were able to turn a cannon ball into a musket ball and fire it from a flintlock, having the ball turn back into a cannonball apon exiting the gun, would the cannonball retain its velocity after changing or would it lose all energy and drop to the ground?
ANSWER:
Essentially what you are asking
is if an object with a certain mass m and speed
v suddenly loses or gains mass, what happens?
The mass either disappears without trace or appears
without a source. Suppose your musket ball has a mass
m and speed v and your cannon ball has a
mass M. Since there are no external forces on
the ball, linear momentum must be conserved, that is
mv=MV where V is the speed of the cannon
ball. So V=mv/M, much smaller than v.
However, the energy is not conserved: E_{1}=½mv^{2} and
E_{2}=½MV^{2}=½mv^{2}(m/M)=(m/M)E_{1};
a large amount of energy is lost.
QUESTION:
How many pounds per inch does a lacrosse ball exert when it hits a surface? For info: Ball is traveling 90mph. A lacrosse ball weighs 8 ounces. The ball is 5.25 inches in circumference. As a sidebar, does it matter if the surface the ball hits is stiff like a wall or springy like a bounceback? If so, could you expand on the difference in pressures against the two different surfaces?
ANSWER:
If it is pressure you want, you
should ask for pounds per square inch. But I do not think
that is what you really want; it is more straightforward
to estimate the average net force the ball exerts on
whatever it hits. Then you could think about pressure,
how that force is spread over the area which the two
surfaces have touching; pretty hard to do that, though.
If something with mass m with a velocity v_{1
}collides with a much more massive object at rest and
rebounds with velocity v_{2}, the
average force F can be estimated as F≈m(v_{2}v_{1})/t
where t is the time the collision lasts. ("Much more
massive" means the struck object does not recoil
significantly.) You seem to want to get answers in
imperial units (lb, ft, s); without going into details,
m=0.5 lb/(32 ft/s^{2})=0.016 lb·s^{2}/ft=0.016
slug
and v_{1}=90 mph=132 ft/s. Suppose that
v_{2}=0, called a perfectly inelastic
collision; then F=132x0.016/t=(2.1
lb·s)/t. So, the smaller the time, the
larger the force. (The fact that the force is negative
means that this is the force the struck object exerts on
the ball because I have chosen the incoming velocity to
be positive. The force the ball exerts on the object is
positive and equal in magnitude.) That addresses your
second question: the ball will certainly stop in a much
shorter time when hitting a hard surface than a soft one.
Of course that should not be too surprising to you
because you know that if you fall onto a mattress it
hurts much less than if you fall on a concrete floor. So,
if t=1 s, F=2.2 lb whereas if t=0.01
s, F=220 lb. Now, if the ball hits a hard
surface it does not penetrate very far and therefore the
force is spread over a small area and pressure is large.
If the ball hits a soft surface, it penetrates deeper and
therefore the force is spread over a larger area and
pressure is smaller.
QUESTION:
Since this isn't technically a homework question and was a challenge put out to the class to see if anyone can do it, I hope you will consider it.
I'm attempting to do some fun physics maths as a challenge from our professor. It may be a trick question though since she's done that in the past. The problem assumes that c is only 30000 kmph. The problem was to determine the time to reach 29999 kmph from 25000 kmph with a total of 6.48MN of force in a vehicle that weighs 168000kg (168t) in the vacuum of space, too far away from any celestial body to have gravity noticeably affect it.
So just to clarify, there is no air resistance or gravity or time dilation. Relative mass increases when reaching said 30000kmph, and the goal is to find the time it takes to accelerate to 29999kmph from 25000kmph.
ANSWER:
I don't see the point of this odd
choice of c; it makes the problem neither easier
nor harder, just unphysical! You will still have to do some
work! I am going to point you in the direction of solving
it yourself. First, go to an
earlier answer; you will see that I have solved your
problem but starting at rest. If you start at some
initial velocity v_{0}, the result is
that Ft=pp_{0}. Here, p is the
relativistic linear momentum, p=mv/√[1(v/c)^{2}]
where m is the rest mass, 1.68x10^{5} kg
in your case; similarly, p_{0}=mv_{0}/√[1(v_{0}/c)^{2}].
So now you know everything except t, so just
solve for it; this is just algebra/arithmetic. I would
advise you to convert everything to SI units (kg, m, s)
before doing your calculation. Then t will work
out to be seconds.
I do not understand what you mean by "…there
is no…time dilation…" since time dilation
is simply not of any interest here but would come into
play if the whole thing were seen by an observer in a
different frame. Also, I almost never apply the idea of
relativistic mass because I view the linear momentum as
being the quantity redefined in relativity, not mass.
Read my discussion of this in
one of the links in the earlier answer.
QUESTION:
Atoms are mostly empty space. How can they form solid objects?
ANSWER:
First of all, atoms are not
mostly empty space. See an
earlier
answer. Although almost all the atom's mass is in a
volume much smaller than the volume of the whole atom,
the electrons fill the rest of the space in a "cloud".
Bohr's model of the atom, tiny electrons in tiny orbits,
is wrong. If you need a simpler explanation without
reference to the "electron cloud" think of all the atoms
connected to their nearest neighbors by tiny springs.
Each atom connects to its neighbors by some interaction
which can be approximated as a spring. The atoms close
enough to interact with each other will stick together
and thus form a solid object.
QUESTION:
2 balloons, one filled with 1 cubic meter of air, and the other with 1 cubic meter of Helium, are lowered to a depth of 50 feet under water and simultaneously released.
Will they reach the surface at the same time?
ANSWER:
With the information you gave me,
I will have to put stipulations on the problem. The
baloons are rigid so they do not get compressed under the
water, they have identical shapes, their masses (without
being filled) are the same, and both are filled to
atmospheric pressure. There are three forces on each
balloon: its own weight mg, the buoyant force
of the water B, and the drag force opposite the
velocity v
which has the form Cv^{2} where C is
some constant which depends only on the shape of the
balloon. So the net force on each balloon is mg+BCv^{2}=ma
where I have put this equal to the mass times the
acceleration, Newton's second law. When you release each
balloon, each will accelerate up and, as v increases,
eventually a=0 and it will have a constant speed
v_{t},
called the terminal velocity, thereafter. It is easy to
show that v_{t}=√[(Bmg)/C]
The only difference between the two is m which
is smaller for the helium which therefore has the larger
terminal velocity. The helium balloon reaches the surface
first.
QUESTION:
does a bouncing ball possess simple harmonic motion?
ANSWER:
A bouncing ball is not an example
of simple harmonic motion (SHM). By definition, the
ball's motion must be describable by simple sinesoidal
functions of time t, i.e. like y(t)=Asin(ωt)+Bcos(ωt)
where A, B, and ω are
constants and y is a variable describing the motion, for
example the height above the ground in your case. If the
ball were perfectly elastic, the motion would be
harmonic, but not SHM, because y(t) would be a train of
identical downwardopening parabolas. If it were a real
ball, it would not even be harmonic because each
subsequent bounce would be a little smaller that the
previous bounce.
QUESTION:
What exactly is the 4th dimension? How do we know it's real? How are space and time related?
ANSWER:
I have
recently answered this question in some detail.
QUESTION:
As I understand it, electricity moves at the speed of light. Conductivity is where I'm getting stuck on. Are we able to measure speed from it? I tried searching "siemens/meter to kph" with no results.
My questions here are these: If silver has a conductivity of 63x10^6 siemens/meter, and copper has a conductivity has a conductivity of 59x10^6 siemens/meter, does that mean energy in silver is travelling faster than copper, or something else? But, silver can't go, say 101% the speed of light. So, there's obviously something else I'm not understanding.
I guess the point I'm getting at is this: What does conductivity mean, and can you convert siemens/meter into a speed calculating how many particles move through 1 meter of wire in x amount of time, thus making copper's speed x time/meter, while silver is y time/meter?
I'm sorry if this is a bit confusing, or not even your field. But, I figured electricity is, in some part, in the field of physics.
ANSWER:
I am going to give you a
qualitative overview since you seem to not have any
understanding of what goes on in a conducting wire. If
you want a more detailed discussion of the simple model
of electric currents, you can find it in any introductory
physics text book. A conductor has electrons which are
moving around at random in the material, not really bound
to any of the atoms; these are called conduction
electrons and they can be thought of as a gas of
electrons. But there is no flow of current because the
electrons move randomly so there are just as many going
in one direction as there are going in the opposite
direction. When this wire is connected to a battery an
electric field is established inside the wire which
points from the positive terminal to the negative
terminal. When you say "electricity moves at the speed of
light", what that means is that the establishment of the
field moves with speed c; the electrons do not
move at that speed. So all electrons see the field turn
on almost immediately. All electrons, having
negative charge, begin accelerating in the direction
opposite the field. If the only force seen by the
electrons was due to the field, they would accelerate
until they reached the positive terminal and left the
wire. But, even though they are not bound to atoms, they
certainly see the atoms and when they hit one they are
momentarily stopped or scattered. So each electron is
constantly being accelerated, then stopped, then
accelerated and the net result is that, on average, all
the electrons participating in the current move with a
very small velocity, on the order of one millimeter per
minute; this is called the drift velocity. All the
electrons are continually gaining and then losing kinetic
energy and that lost energy shows up in the heating up of
the wire. What determines the conductivity is the number
of conduction electrons per unit volume of the material
as well as its crystaline structure and how individual
atoms scatter the electrons.
QUESTION:
The classic escape velocity formula takes in consideration an uniform acceleration of gravity, g. What would be a mathematical model for escape velocity where the force of gravity varies as the body distances from the planet?
ANSWER:
"The classic escape velocity formula"
that I know does not assume a uniform gravitational
field, it assumes the correct field. Let's see where it
comes from. The potential energy for a particle of mass
m a distance r from the center of a large
spherical mass M is, choosing zero potential
at r=∞ is V(r)=mMG/r
where G is the universal gravitational constant.
If m has a speed v at r=R, where
R is the radius of the
earth, the total energy is ½mv^{2}mMG/R=E.
Now, suppose we want v to be big enough that m will go
all the way to r=∞ before it finally
comes to rest; then E must be zero. Therefore
v_{e}(R)=√(2MG/R);
but, MG/R=gR, so v_{e}(R)=√(2gR).
That is the escape velocity at the surface of the planet
(ignoring any other forces, notably air drag) where g
is the acceleration due to gravity. Now, you want it
everywhere else. That's easy, just replace R by
r, v_{e}(r)=√(2MG/r).
You could use g=MGr if g now means the
acceleration due to gravity at an altitude of rR.
(Incidentally, this does not work anywhere inside the
planet.)
QUESTION:
I have a bit of a debate with colleagues and looking for a professionals take. I'm a plasterer and when mixing up the plaster introducing too much air to the mix makes it less workable. Some people I work with think mixing counter clockwise introduces less air to the process than clockwise, I think it would make no difference.
ANSWER:
Well, I just so happen to have a
plaster/paint stirring tool which is powered by a drill.
Examine it carefully and you will see that clockwise and
counterclockwise are not the same in their effect; if the
direction is clockwise (as viewed from above), the blades
push down on the mixture and if the direction is
counterclockwise, the blades push up. I do not know for
sure which, if either, of these would introduce more air.
My best guess would be that for the clockwise rotation
there would be a whirlpool into the surface, whereas for
the counterclockwise rotation there would be more of a
hump on the surface; I would think the whirlpool would
pull in more air. I must also note that not all stirring
tools are necessarily of the same design as mine.
QUESTION:
Why does this toy move in a straight line? Shouldn't the toy rotate due to gyroscopic precession?
https://youtu.be/hs4iv7IHvGg
ANSWER:
Well, it
does not move in a
straight line. One thing I noticed is that the toy starts
moving immediately when released; this indicates to me
that the surface it is on is not level. Just because it
is a gyroscope does not mean it will precess. There must
be an external torque acting on it; imagine an axis
running between the two legs about which you would
calculate torque. In the figure that I have clipped from
the video you can see the motor and battery and the cds
which are rotating; the toy is leaning such that I would
certainly expect the center of gravity to be beyond the
being above the axis and, were the discs not spinning the
toy would certainly fall. If you could put the center of
gravity directly above the axis, it would not precess.
You can see several instances in the video where the toy
is moving in a curved path, and in one case the path is a
pretty tight circle; these are due to precession because
of the torque due to the weight.
QUESTION:
Would you be marginally taller when on top of a tall building/mountain or at the bottom of the ocean. Does the pressure/gravity change and if so which one makes you taller?
ANSWER:
If you are in the the vicinity of
the earth there are two forces which are different
between your head and feet, tidal force and pressure
force due to the fluid you are in. The tidal force is due
to the fact that the gravitational field at your feet is
slightly larger than at your head, both pointing down;
therefore the net force on you tends to stretch you
taller. We can estimate the tidal force because the
gravitational field falls off like 1/r^{2}: F_{feet}/F_{head}=(r+h)^{2}/r^{2}=[1+(h/r)]^{2}≈1+(h/r)^{2}
where r is the distance to your feet from the
center of the earth and h is your height; I have
used the binomial expansion because h/r<<1. So, F_{feet}F_{head}≈F_{head}(h/r)^{2}.
Clearly, either force is going to be approximately equal
to your weight, so the tidal force is about mg(h/r)^{2};
if I take h=2 m, mg=1000 N, and r=6.4x10^{6}
m, this is about 10^{10} N (2.2x10^{11}
lb).
The pressure force is
down on your head and slightly larger and up on your
feet; so the pressure force tends to compress you
shorter. The net force on you is simply the buoyant force
which equals the weight of the displaced fluid. If I take
your volume to be about 0.02 m^{3}, the buoyant
force force is about 1000 N in water and 1 N in air.
So, as you can see, the stretching due to the tidal force
is negligible to forces due to fluid pressure. So, you
will be shorter than your "real" height, but more so in
water than air. Therefore you are taller up high in the
air than down low in the water.
QUESTION:
I know that at sea level, atmospheric pressure is 14.7 pounds per square inch. I know that the higher in elevation you go, the thinner the air gets and a corresponding decrease in the amount of oxygen such that trying to survive above 1618k feet is not very easy. My question is this: If you where to drain all of the oceans so that the existing atmosphere before draining was now occupying the void left by the water, would you be able to survive at what was sea level? My thought is that the atmosphere would become so diluted with the oceans gone that you would actually have to travel way below what was sea level to able to survive the atmospheric pressure change and the oxygen dilution. Wondering about this has always bugged me lol.
ANSWER:
The volume of all the oceans is
about 1.4x10^{18} m^{3} and their average
depth is about 3700 m (about 12,000 ft). Now, dealing
with the volume of the atmosphere is tricky because the
density of the atmosphere gets smaller with altitude so
you have to be careful how you describe volume. The mass
of the whole atmosphere is pretty wellknown, though, to
be about 5.15x10^{18} kg; so you can calculate
the volume if the whole atmosphere were at sea level
pressure where the density is 1.225 kg/m^{3},
5.15x10^{18}/1.225=4.2x10^{18} m^{3}.
So, if the atmosphere were uniform, about 1/3 of it would
flow into the emptied oceans. But, it is not uniform so
less than 1/3 would come; there would be some base
pressure 3700 m below the sea level (since there is no
sea now, when I say sea level I mean what it was) which I
would estimate to be about the same as the pressure at
sea level was because 3700 m is small compared to the
radius of the earth, 6.4x10^{6} m (0.06%
smaller). Since the oceans occupy about 70% of earth's
surface, sea level pressure would be about what the
current pressure is at about 12,000 ft, the height of a
modestly high mountain and certainly habitable. The
highest habitable place on earth is about 16,000 ft,
so lots of places (like Denver) habitable now would not
be. And certainly nobody would be climbing Mt. Everest in
this new world!
QUESTION:
What are the signs that show scientists that some particles have a quark structure and others do not?
ANSWER:
Quarks are something which were
hypothesized after most of the elementary particles and
their properties were known. There are three kinds of
particles, hadrons, leptons, and field quanta. Leptons
are few, electrons, muons, and neutrinos. Field quanta
are also few, photons, gluons, Z bosons, W bosons, and
the Higgs boson. Hadrons are many, the proton and neutron
being the best known, but what has been referred to as a
"zoo" of other hadrons. To try to systemize and explain
this array of particles, first group theory and then,
building on that, introducing hypothetical particles
called quarks was extremely successful. There is no way I
could answer "what are the signs" because the "signs" are
all the hadrons and their properties.