QUESTION:
In lifting an object to a higher level directly over its original location, the energy I expend increases potential energy. But, does some of the energy used in lifting it also go to accelerating it to higher rotational velocities as the circumference of its "orbit" increases as it is raised over its original position? Does this add kinetic energy and mass to the object whereas increasing potential energy does not?
ANSWER:
(Preface: all my calculations below assume that the height lifted is
much smaller than the radius of the earth. I also neglect the change in
the gravitational force over the distance the mass is lifted. Also, to
simplify things, all my calculations are at the equator.) Yes, work is
done to increase the kinetic energy. As viewed from an inertial frame,
watching the mass M get lifted to h, I estimate that
the kinetic energy changes by ΔK≈2hK_{initial}/R=MhRω^{2}
where R=6.4x10^{6} m is the radius of the earth and ω=7.3x10^{5}
s^{1} is the angular velocity of the earth. For example,
lifting 1 kg a height of 1 m requires 0.03 J of work to increase the
kinetic energy. But wait a minute! Once we acknowledge that the earth is
rotating, we have to recognize that the mass, being in a circular orbit,
has a centripetal acceleration a_{c}=Rω^{2}
and therefore the net force on M is MgMRω^{2}.
Therefore, the net work done is W≈(MgMRω^{2})h+MhRω^{2}=Mgh.
QUESTION:
Can gravity be focused like light thought a lens? Say following the plane of a qalaxy. If so, could stars be traveling around galaxies at the same speed because gravity effects time?
The reason for the question above, IF gravity focuses out along the plane of a galaxy, could it be slowing down the stars closer to the center of the galaxy because gravity effects time?
ANSWER:
Gravitational lensing is light being focused by gravity, not gravity
being focused. So I am afraid that your idea to explain the anomalous
orbiting properties of stars in a galaxy will not work. An example of
gravitational lensing is shown to the right; shown is "Einstein's cross"
which shows four images of the same quasar.
QUESTION:
Why don't liquid pipelines that run downhill rupture from the weight of the liquid in them? For instance, a 14 mile long section of 10" ID pipe carrying crude oil that runs at a 22.5 degree angle has about 301,592.89 gallons of crude in it. That Crude weighs 91,328,360.39 pounds. I calculate that the static weight at the bottom of that run should be about 22,832,090.10 pounds of oil. That exceeds the tensile strength of the pipe wall by a factor of 50 to 100. At first I thought it was because a closed pipe would have sort of a vacuum at the top, but then I realized that would make the pipe crush from atmospheric pressure. I'm missing something simple I am sure, but darned if I can figure out what it is...
ANSWER:
There is something seriously wrong with your numbers. They imply that
the weight of one gallon of crude is about 9x10^{7}/3x10^{5}=300
lb, and 14 miles at 22.5° would take you to 14sin22.5°=5.4
miles, higher than Mount Everest! Also, since we are working with a
fluid, pressure would be a more appropriate quantity to look at than
force. I will start from scratch with new
numbers: the density of crude oil is about ρ=900 kg/m^{3
}and I will take a distance of about y=1000 m between the
top and the bottom, about the height of a small mountain; you need only
the drop, as we shall see below. Like you, I will assume that the oil in
the pipe is static which simplifies things. A good approximate way to
solve this kind of problem is to use Bernoulli's equation, P+ρgy+½ρv^{2}=constant;
P is the pressure, g=9.8 m/s^{2} is the
acceleration due to gravity, and v (=0 for us) is the speed. At
the top, P_{top}=P_{A} and y_{top}=1000
m and at the bottom P_{bottom}=P+P_{A}
and y_{bottom}=0; here, P_{A} is the
atmospheric pressure. Therefore, P_{A}+900x9.8x1000=P+P_{A}+0.
So the gauge pressure is P≈10^{7} N/m^{2}≈100
atm=1450 PSI. The brief research I have done indicates that this pressure
is at the extreme upper limit of specifications for pipelines. To make
the oil move, you need to add a lot more pressure. My use of Bernoulli's
equation is a very crude approximation because it assumes a fluid with
no viscosity and no frictional forces with the walls, obviously not a
very good approximation.
QUESTION:
Energy question. On earth kinetic energy transfer is more or less sound and heat given the little knowledge I have on the subject. In a vacuum sound cannot be heard, but does that mean it does not exist? if it does not exist than where does the energy go? if it does exist, and our ears are simply broken in a vacuum does sound expand faster in the vacuum like light? Every thing I read suggests sound cannot exist in a vacuum, but energy must be conserved, so what form of energy does snapping my fingers in a vacuum take if it cannot take the form of sound?
ANSWER:
First of all, sound, being waves in air, does not exist in a vacuum. By
removing one possible way for energy to be removed from the system
simply means that the energy needs to be taken away by other mechanisms.
Let's do an example. Imagine a plucked guitar string which has a certain
amount of energy. In air, I can think of three ways to take energy from
the string—the radiated sound, the drag of the string
moving through the air, and the friction associated with the string
bending and unbending. A guitar in a vacuum no longer has the first two
ways to lose energy, so only the bending friction would be converting
kinetic energy of the string to heat in the string. It would therefore
take longer for the string to stop than in the air, but the total energy
converted would be the same; if the string did not have a way to get rid
of its energy (it does, it could radiate or conduct to the pegs, but
let's forget that), it would end up hotter than it would in air.
QUESTION:
Is there a law in physics that allows me to calculate the magnetic field at a certain point created by a charged particle moving in a straight line
with constant speed in empty space?
ANSWER:
Yes, there is an equation. I will warn you, though, that this is quite
technical and pretty highlevel. Refer to the figure to the left. The
charge q, with velocity v is at
position x(t) at time t; we
wish to know the field at position r at time
t. However, since the information about the field propogates at
the speed of light, c, the field at time t is
determined by where the charge was at some earlier time t_{ρ};
rx(t_{ρ})≡ρ=ρ=c(tt_{ρ}).
After much calculation (see the detail in Chapter 10 of David Griffith's
book
Introduction to Electrodynamics), the magnetic field is
B(r,t)=(vxE(r,t))/c
where the electric field is
E(r,t)=[q/(4πε_{0})][ρ/(ρ·u)^{3}][(c^{2}v^{2})u];
the vector u is defined as u≡c1_{ρ}v
where 1_{ρ} is a unit vector in the
direction of the vector ρ.
For high speeds, the fields look like the diagrams below.
ADDED
NOTE:
The above expressions for electric and magnetic field are exactly
correct. You can write approximately correct equations for particles
with speed much less than the speed of light, v<<c:
E(r,t)=[q/(4πε_{0})][(rx)/rx^{3}]
and
B(r,t)=[qμ_{0}/(4π)][vx(rx)/rx^{3}].
These are simply
Coulomb's law and the
BiotSavart law.
QUESTION:
Why do moving charged particles respond to magnetic fields? I know that every charged particle has it's own magnetic field and permanent magnet will attract/repel these particles, but the force will be so little that it won't be able to be measured at all, but when it comes to moving, a "magic" happens, and I don't understand what is special with moving charges versus stationary charges.
Does the magnetic field between magnet and charged particle increase proprtional to velocity and the force gets noticable? In turbines, when magnet is rotating, how that makes electrons move?
Why are those electrons affected by magnetic field at all? There is Lorentz law but how was that
equation figured out? Does that equation only depend on experiments? Did they just see that electrons start moving when we rotate magnet next to them and that's all?
ANSWER:
Site groundrules specify single, concise, wellfocused questions so I
should have just thrown this out. Instead I have edited the question a
bit to make it more focused. There is no way I can fully answer the
questions because they really ask that I give you a full course in
magnetostatics. First of all, disabuse yourself of your second sentence.
All charged particles do not have a magnetic field if they are not
moving; true, most elementary particles (electrons, protons, etc.)
have magnetic moments, but these are incredibly weak and do not normally
react to a magnetic field; if you just have an electric charge Q,
it experiences zero force if at rest in a magnetic field. Yes, it is an
empirical fact (experimentally observed) that a charged particle Q
with velocity V in a magnetic field
B experiences a force F=QVxB.
It is now understood that there is only one field, the electromagnetic
field, and electromagnetism is a relativistic theory; it is, though, no
longer a vector field like you are familiar with, but a tensor field
with nine components. What this means is that the answer to your
question about there being electrons at rest with a magnet moving is
that an electron moving in a static field is no different from an
electron at rest and the magnet moving; that, essentially is
relativity—all that matters is relative velocity. Also, once you understand that there is a single field, the
Lorentz force arises naturally.
QUESTION:
If I weigh 200lbs and am riding a kick scooter that weighs 14lbs, and I am riding at a speed of 10 mph and I jump the scooter off a curb, say 6 inches, what is the force in terms of pounds, that I am applying to the scooter as it lands?
ANSWER:
Usually it is not possible to answer this kind of question because what
is needed is to know how long the collision between you and the ground
lasted. In this case, though, we can estimate the time of this
collision. As you may know, paratroupers are trained to not land with
stiff legs, rather to bend at the knees during the landing; the purpose
is to prolong the landing time and this reduces the average force on the
legs during the landing. If a mass M hits the ground with some
speed V and stops in time t, the average force over
the time is F=W+MV/t where W is the weight,
200 lb; to convert the
weight to mass, divide by the acceleration due to gravity, g=32
ft/s^{2}: M=W/g=(200 lb)/(32 ft/s^{2})=6.25
ft·lb/s^{2}. The speed at impact can be determined
from V=√(2gh)=√[2x(32
ft/s^{2})x(½ ft)]=5.7 ft/s. If we approximate that
the distance S over which your legs bend on landing as S=1
ft, the time to stop is t=2S/V=(2x1 ft)/(5.7
ft/s)=0.35 s. So, finally, F=200 lb+(6.25
ft·lb/s^{2})(5.7 ft/s)/(0.35 s)=302 lb. This is
the force the scooter exerts on you which, by Newton's third law, is
equal the the force you exert on the scooter (but in opposite
direction). If you stop in ½ ft, the force would be 404 lb. Keep
in mind that this is a very approximate estimate of the average force.
QUESTION:
If someone spins a stick in a circle,What happens if the stick is scaled up to such a size that the outer edge of the stick would have to break the Speed of Light Limit?
ANSWER:
See an
earlier answer.
QUESTION:
I have been trying to understand this for years. In Einstein's theory of time verses speed I believe he used the situation of a man (lets call him man A) standing in a train yard. A second man is on a train (lets call him man B) and the train is moving. They both drop an object at the same time. If I standing in the yard with man A to me the object man A drops would appear to me in normal time. Man B's object would appear to take longer causing the difference in time. I understand this. My main question is if a man C was on a different train going the same direction as man B's train the time difference between me and man A to man B
& man C. I understand that to man B & Man c they would be the same. What if man C's train was going in the other direction? To me and man A the would seem the same, but man B to man C I believe the difference in the dropped object would be appear to be doubled to them between them. Since I believe it would appear doubled between man C
& D. I think there should be a double time difference. From man A's view the age difference would be the same, but between man B
& man C there should be double the time difference. How can there be doubled the time difference when they are traveling at the same speed?
ANSWER:
Your question boils down to what is called velocity addition.
In classical physics, v_{CB}=v_{CA}+v_{AB};
the notation is "v_{IJ} is the velocity of I relative
to to J". I have written this so that it corresponds to your question—A
is you, B and C are the trains; it is more convenient for you if we
rewrite the equation as
v_{CB}=v_{CA}v_{BA}
which we can do because v_{AB}=v_{BA}.
If the trains have speeds v in the same direction, v_{CB}=vv=0;
in opposite directions, v_{CB}=v(v)=2v—each
sees the other moving with speed 2v. But, this form of velocity
addition is wrong for very high speeds (see an
earlier answer). The relativistically correct velocity addition
equation is
v_{CB}=(v_{CA}v_{BA})/[(1(v_{CA}v_{BA}/c^{2})]
which reduces to the classical equation for the speeds much less than
c. So, for the trains moving in opposite directions, v_{CB}=2v/(1+v^{2}/c^{2});
for example, if v=0.5c, v_{CB}=c/(1+0.25)=0.8c.
Now, you seem to think that if you double the speed, you double the time difference. This is not correct—time dilation goes like the gamma factor, 1/√(1v^{2}/c^{2}). So, for your situation with v=0.5c, t_{A}=1.33t_{B}, t_{A}=1.33t_{C}, t_{C}=1.67t_{B}, and t_{B}=1.67t_{C}. These are confusing, I admit. They are meant to denote, for example, that t_{B}=1.67t_{C }means that B sees his clock tick out 1.67 s when C's clock ticks out 1 s; B observes C's clock to be slow.
QUESTION:
I am having a debate with my brother about climbing on an incline.
I understand the basics of climbing a hill on a diagonal. If you climb diagonally, you can avoid taking larger vertical steps at the cost of more horizontal movement. This makes each step take less energy while increasing the overall work and time needed.
However, it seems as though this rule does not work for stairs. Stairs do not allow for shorter vertical steps (you either make 100% progress on a step or 0%). Do I have this correct? Am I missing something?
ANSWER:
Slaloming up the incline will increase time spent but not increase work
done. This assumes no frictional forces are important, the only work you
do is the work lifting you. Since work done does not change but elapsed
time does, the average power you are generating going straight up is
greater than zigzaging. Going up steps, though, if you go across a step
you do no work, the only work done is lifting you to the next step. The
only way to get the equivalent lowered average power output as you do by
slaloming up the slope is to rest between steps.
QUESTION:
How strong would a man have to be to push a 16,000 lb bus on a flat surface?
ANSWER:
That depends on how much friction there is. And not just the friction on
the bus, but more importantly, the friction between the man's feet and
the ground. Newton's third law says that the force the man exerts on the
bus is equal and opposite the force which the bus exerts on the man (B
in the picture). Other forces on the man are his weight (W),
the friction the the road exerts on his feet (N),
and the force that the road exerts up on him (N).
If the bus is not moving, N=W and f=B, equilibrium.
The biggest that the frictional force can be without the man's feet slipping is
f=μN where μ is the coefficient of
static friction between shoe soles and road surface. A typical value of
μ for rubber on asphald, for example,
μ≈1, so the biggest f could be is
approximately his weight W;
this means that the largest force he could exert on the bus without
slipping would be
about equal to his weight. Taking W≈200 lb, if the
frictional force on the bus is taken to be zero, the bus would
accelerate forward with an acceleration of a=Bg/16000=200x32/16000=0.4
ft/s^{2} where g=32 ft/s^{2 }is the
acceleration due to gravity; this means that after 10 s the bus would be
moving forward with a speed 4 ft/s. If there were a 100 lb frictional
force acting on the bus, the acceleration would only be a=0.2 ft/s^{2}.
If there were a frictional force greater than 200 lb acting on the bus,
the man could not move it.
QUESTION:
While reading about the twin paradox, I've been told at the end of the traveling twin's journey, he begins decelerating in order to land back on Earth, and he, the traveling twin, observes his brother's clock on Earth to SPEED UP. This makes sense to me except for one problem: This suggests that the light pulse in the Earth clock would be percived by the traveling twin to be moving faster than C. Of course, the traveling twin is no longer in an inertial frame. I thought perhaps that since he feels himself moving now, he would also measure himself and the light having a CLOSING SPEED greater than C, even though he would see the light moving across the ground on Earth to equal C? If so, at what rate would a clock in frame S behind the traveling twin run at, faster or slower? Is it also possible that acceleration, from the point of view of the traveling twin, causes length contraction perpendicular to the the ship's vector, shortening the distance the pulse has to travel?
ANSWER:
There is no need to discuss acceleration to understand the
twin paradox.
Acceleration just makes everything harder to understand. Basically, just
assume necessary accelerations (departing, turning around, landing)
occur in a vanishingly short time. But, you are not really interested in
the twin paradox, you are interested in how things appear in an
accelerated frame. I am sorry, but nothing in your question after "…clock
on Earth to SPEED UP…" makes any sense. First of all, the fact
that any observer will measure the speed of light to be c is a
law of physics and the general principle of relativity states the laws
of physics are the same in all (not just inertial) frames.
Second, to measure a speed in your frame of reference you must use your
clock, not somebody else's. And third, how another clock looks is really
irrelevant because how it appears to run and how it
is actually running are not the same.
I would like to address how the clocks look from the perspective of the Doppler effect. The relativistically correct Doppler effect is usually expressed in terms of the frequency of the light; for our purposes, it is more convenient to express it in terms of the periods, T_{observer}=T_{source}√[(1+β)/(1β)] where β=v/c and β is positive for the source and observer moving apart; it makes no difference which is the observer—each twin will see the other's clock running at the same relative rate. Let's illustrate with a specific example, β=0.8, the traveling twin coming in at 80% the speed of light; T_{observer}=T_{source}√[(10.8)/(1+0.8)]=T_{source}/3 so the observer will see the source clock running fast by a factor of three. But special relativity tells us that moving clocks run slow, not fast; 1 second on the moving clock will be 1/√(10.8^{2})=1.67 seconds on the observer's clock. How the moving clock looks is thus demonstrably not a measure of how fast it is actually running. Now, if the incoming twin puts on the brakes such that he slows to β=0.6, T_{observer}=T_{source}√[(10.6)/(1+0.6)]=T_{source}/2 so the observer will see the source clock running fast by a factor of two, apparently slowing down. Now, 1 second on the moving clock will be 1/√(10.6^{2})=1.25 seconds on the observer's clock, speeding up compared to when β=0.8.
QUESTION:
Sir can you please tell me where I am going wrong in this one? (attachment)
ANSWER:
The pendulum bob is not in equilibrium.
FOLLOWUP QUESTION:
I found the second equation in the book "An Introduction to Mechanics" by Daniel Kleppner and Robert Kolenkow.. Which you must be knowing about... So the second equation is definitely correct.. And first equation is just like the second one.. Whereas in second equation I have taken the components of tension force.. In the first one I have taken the components of weight.
ANSWER:
Did you not read my first answer? You are trying to apply Newton's first
law where it does not apply. Is the pendulum in equilibrium? Or will it
accelerate in some way if no other forces are applied? If it is a simple
pendulum (at rest right now), the mass will have an acceleration
perpendicular to the string; as soon as it begins moving it will have a
component of its acceleration along the string. Therefore, both of your
equations will be incorrect. Your second question reveals that this is
not a simple pendulum, but rather a conical pendulum (the mass moves in
a horizontal plane with the string tracing out a cone); you should have
given me that information in your first question. Now I can give you a
more complete answer. The conical pendulum, with m moving in a
circle, has only a horizontal centripetal acceleration toward the center
of that circle, a_{c}=v^{2}/(Lsinθ)
where L is the length of the string; applying Newton's second
law, Tsinθ=mv^{2}/(Lsinθ).
There is no acceleration vertically and so you can apply Newton's first
law,
Tcosθmg=0.
QUESTION:
Suppose I have a 10 tons weight hanging 5 meters up in the air. I want to get electricity by lowering the weight against a dynamo (for example).
How much energy do I get?
A 100 W light bulb needs 100 W of power when it's ON. So, if it stays on for 10 hours it will consume 1 KW, am I right?...
Ok, so my question is... How many 100 W light bulbs can I have ON at the same time with the energy coming from that falling weight?  while the weight is falling, obviously.

if the weight falls for 1 hour

if the weight falls for 2 hours.
What's the formula?
Somebody asked the same question on some forum on the web. His weight was 200 tons falling 100 meters for 1 hour, and someone said that the solution is:
dU=Fdy =
200,000kg * 9.81m/s2 * 100m
=
196.2MJ =
196.2MW/3600 = 54,500KW/h
Is the formula right? If yes, how do I apply it? Because I get extremely small numbers if I change his 200,000 with my 10,000, and his 100 meters with my 5 meters.
Plus, I don't know what KW/h means. All I'm interested is knowing how many Watts are available at any given moment while the weight is falling.
ANSWER:
The first thing we need to get straight is what a watt is. The unit of
energy in SI units is the Joule (J); 1 J is 1 N·m where a Newton
(N) is the unit of force and the meter (m) is the unit length. A Joule
is the kinetic energy which a 2 kg mass moving with a speed of 1 m/s
has; or, it is the work you need to do to lift a 1 kg mass to a height
of 1/9.81 m. A Watt (W) is the rate at which energy is delivered or
consumed, 1 W=1 J/s. Therefore, your 100 W bulb consumes 100 J of energy
every second. Incidentally, if you look at your electricity bill, you
will be billed for how many kW·hr you have consumed; a
kW·hr is a unit of energy, 1
kW·hr=1000x3600 J=3,600,000 J.
The example stated is correct but the units are not. It is fine up to the point where the potential energy of the mass is 196.2 MJ (mega=M=10^{6}). Now, if you let this mass drop over 3600 s, it is losing its energy at the rate of (196.2x10^{6} J)/(3600 s)=5.45x10^{4} J/s=54.5 kW (not kW/hr). For your case, your mass has a potential energy of 10^{4}x9.81x5=4.9x10^{5} J. If you deliver this energy over an hour, the power is 4.9x10^{5}/3600=136 W; You could power one light bulb over this hour and have some energy left over at the end (about ¼ of what you started with). Clearly, the power delivered over two hours would only be half as much, not enough to power even one 100 W bulb.
QUESTION:
Imagine free electron is falling towards the Earth due to the gravitational interaction. In order to prevent the electron from falling, you come up with the idea to fix a second electron below on the ground to prevent the first electron from falling. What is the distance between the two charges, such that the top electron is in balance (the net force is zero)?
ANSWER:
I am going to assume that I can write the gravitational force as mg;
if I find the distance d between the electrons to not be very
small compared to the radius of the earth, I will have to start over
again and use the force as MmG/(R+d)^{2}. The
electrostatic force between the two electrons is ke^{2}/d^{2}=9x10^{9}x(1.6x10^{19})^{2}/d^{2}=2.3x10^{28}/d^{2}=mg=9x10^{31}x9.8=8.8x10^{30};
solving, d=5.1 m. Good, I don't have to start over!
QUESTION:
How far would a 100 pound pig go if you put it in a sling shot with a
500 pound pull back with a 5 mile an hour wind? In feet?
ANSWER:
There is not enough information. How far do you have to pull the sling
shot to reach 500 lb? With that info I can get a good estimate without
including air drag which means without the wind. Air drag adds quite a
bit of difficulty to the problem.
FOLLOWUP QUESTION:
You would have to pull it back 175 feet.
ANSWER:
This is a truly peculiar question! I will first neglect air drag and the
wind; then I will include an approximate calculation including air drag.
I will assume the pig launches at 45º and returns to the
same level from which it was launched. You should know that scientists
prefer to work in SI units, so I will convert all your numbers and
convert back to English units at the end. 100 lb=45.4 kg, 500 lb=2224 N,
175 ft=53.3 m. So the spring constant k=2224/53.3=41.7 N/m. So
the energy stored in the spring is ½kx^{2}=½x41.7x53.3^{2}=59,200
J. This must equal the kinetic energy of the pig at launch, ½mv^{2}=½45.4v^{2}=22.7v^{2}=59,200.
Solving for the launch speed of the pig, v=51.1 m/s. The range
of a projectile launched at 45º is R=v^{2}/g=266
m=873 ft.
If air drag is included, the angle for maximum distance is changed. I used a calculation from a demonstration by Wolfram. Without going into details, the graph shows the maximum distance gone is about 120 m=394 ft and the launch angle is about 36º. The red shows the path with air drag, the blue without. I emphasize that this is a very rough calculation but it is useful to demonstrate that air drag is important. To include the wind would be pointless since the uncertainty in air drag is much bigger than any effects such a wind would have; however, if you were to do it, you would need to specify the direction of the wind.
QUESTION:
if the jeep weighs 2000 kg and tricycle weighs 1000 kg. among them? who will difficult to stop when both travelling same initial speed?
ANSWER:
(The numbers you give are masses, not weights. Weight is the mass times
the acceleration due to gravity, W=mg, where g=9.8 m/s^{2}.
But it doesn't really matter for this question because, as you will see
below, the stopping distance does not depend on what the weight is.) The
maximum force you can get when braking is the maximum frictional force
which, on level ground, is μW where μ is
the coefficient of static friction between the wheels and the road and
W is the weight. The work done by friction is equal to the
change in kinetic energy, μWs=½mv^{2}
where s is the stopping distance, v is the starting
speed, and m is the mass. But, the mass is the weight divided
by the acceleration due to gravity, m=W/g; therefore
s=v^{2}/(2μg). As you can see, the minimum
stopping distance does not depend on weight, only on both having the
same wheels on the same surface, e.g., rubber on asphalt.
QUESTION:
Does amplified sound travel further than unamplified sound,i.e. Two objects emit sound at 65 decibles apiece, one objects sound is a "raw" 65 decibles , while the second object is 65 decibles coming from an amplifier, do the amplified decibles travel further?
ANSWER:
There is no way to unambiguously answer this. A decibel is a measure of
the intensity of sound relative to a standard intensity. It is a
logarithmic measure which means that when dB increases by 10 dB, the
intensity increases by a factor of 10; e.g., 10 db to 20 db increases
the power ratio from 10 to 100. Intensity of sound is energy/area/time
and usually measured in watts/square meter (W/m^{2}) and the dB
is proportional to the logarithm. So, if you want to specify the "dB of
the source" you need to specify some geometry. Suppose we say that we
will measure all the energy passing through a sphere of radius 1 m.
Then, you would measure for both sources that the total energy per
second passing through that sphere would be the same. Using the
definition of the
dB, 65 dB=10^{6.5}10^{12}=3.2x10^{6} W. Now we
come to the tricky part; you should be able to see that how far we will
be able to hear this depends on how this power is distributed over the
sphere. If the source radiates equally in all directions, this power
will be evenly distributed over the sphere; as an example, let's assume
your "raw" is distributed that way. Then the intensity of the sound is
the power divided by the area of the sphere, 3.2x10^{6}/(4π1^{2})≈2.5x10^{7}
W/m^{2}. Now, suppose that your amplified sound comes from a
speaker which only radiates in the forward direction; if you think about
it, the intensity at the 1 m distance will be about 5x10^{7}
W/m^{2}, twice as loud. As you moved farther away, you would
find that the intensity fell off like 1/r^{2} where
r is your distance from the source; at 500 m away, the intensity of
the "raw" will be at 2.5x10^{7}/500^{2}=10^{12}
W/m^{2} and the amplified will be at 2x10^{12} W/m^{2}.
The "threshhold of hearing" is about 10^{12} W/m^{2},
so, in principle, you could barely hear both, the amplified being
louder. Beyond this distance, and out to about 700 m, only the amplified
sound would be heard. These specific distances depend on the assumption
that there are no other damping mechanisims in the air. The bottom line
is that it all depends on the pattern of the radiation of both. You
might be interested in an
earlier
answer.
QUESTION:
How can two people play catch on a moving train. Without the ball zooming past like if you are in a car?
ANSWER:
Because the ball is moving along with the train so if the two are on the train and the train moves with constant velocity, all laws of physics are exactly the same as for somebody standing beside the tracks. So a game of catch will be exactly the same either in the train or beside the tracks because it is the laws of physics which govern how the ball will move.I
QUESTION:
To help fight the fires, the state uses planes to drop water and fire
retardants on the flames. One such plane flies horizontally over a fire at a
speed of 60 m/s and drops a giant water balloon to help extinguish the fire.
It flies at a height of 200 m.
If the plane released its load when right over the flames, it would
overshoot its target. It must release it a little earlier, marked by d on
the drawing. How far before the fire must it release the water?
ANSWER:
No homework. But there are lots of advertisers on this page which will help
with homework.
FOLLOWUP
QUESTION:
Please acknowledge the following question as this turned into a big argument with my son yesterday. This is not a homework question by any means, but rather the values were just manipulated by me and the question is just bare with no values for any measurements provided in the question.
ANSWER:
OK, I'll take your word for it this time. The equation for y
motion is y=0=y_{0}+v_{0y}t½gt^{2}=2000½9.8t^{2}=2004.9t^{2},
so t=6.39 s. The equation for x motion is x=x0+v_{x}t=0+60t=383
m.
QUESTION:
Can you possibly explain to me what the precise nature of the mechanism in matter is which allows it to retain its internal inertial frame of reference even when it is tumbling in its flight? This is something which we take for granted every day, yet I have never been able to see into matter to understand why this is so. I am more than happy to make a meaningful donation if you can possibly explain this to me.
ANSWER:
I do not understand your question. Something "tumbling" "in its flight"
is not an inertial frame of reference. If you are inside an airplane
which is doing violent maneuvers and not buckled in, you will be thrown
around because the frame of reference is not inertial.
QUESTION:
I received a novelty gift that purports to find the "balance point" of a golf ball by spinning it up to 10,000 rpm. After 1020 seconds the ball reaches an "equilibrium" spin and a horizontal line is marked that indicates the "balance axis". The assumption that on tees and greens you orient the ball to put the line vertically along the intended path so the
center of gravity (CoG) is rolling/spinning over the target line and thus minimizing potential effects of the CoG being on the side and potentially causing a "wobble". Putting the ball at different starting orientations in the device doesn't matter. It does tend to find the same "equilibrium" spin after a time so it is consistent.
I decided to mark a dozen balls using the device and then put them in a container of salt water to compare it to finding the CoG using a buoyancy test. Put in enough salt and eventually the golf balls float and
will reorient to put the CoG at the lowest possible position in the solution. Out of 12 balls only 1 had the previously marked axis running through the top of the ball. The others were all off by somewhere in the 4045 degree range. The physics of the buoyancy test seem pretty straightforward and understandable to me. What is happening in the spin device is less clear. Can you explain the difference in the two tests? Is there a different "balance point"/CoG that is being located by the spin device? And to preempt your first obvious statement, Yes, I understand that none of this has a significant impact on my golf game compared to all of the other variables at play.
NOTE
FROM THE PHYSICIST:
This question has a history of numerous exchanges between me and the
questioner. It took, as you can see below, several hypotheses before I
was finally able to understand the results of his experiments. I have
elected to put the final answer first, labeled as "NEW ANSWER". But, I
have left the original answers in, beginning with "OLD ANSWER", because
there is quite a bit of interesting physics there (just not really the
solution to this question) and because it is of interest to see how a
series of ideas eventually can lead to the right idea in science.
NEW
ANSWER:
First, let's consider the physics of the spinning method. Clearly
the idea is that if the CoG is not at the geometrical center of the
ball, if you spin the ball about a vertical axis, the CoG will be
"thrown out" horizontally. But, will it end up in a horizontal plane
passing through the center of the ball? What I show here is that the
answer is no. If the CoG is a distance r from the
geometrical center of the ball, has a mass m,
and is spinning with an angular velocity of ω, the CoG experiences
(in the rotating frame) three forces: the weight mg, the force T
holding it in place, and the centrifugal force C=mrω^{2}.
In spite of all
Now, the questioner found that for most balls, θ≈45º so tanθ≈1. Taking g≈10 m/s^{2} and ω^{2}=(10,000 rpm)^{2}≈10^{6} s^{2}, I find that r≈10^{5} m. This is very small, 1/100 mm, but demonstrates that the spinner will not find the correct plane for very small r. Technically, it never finds the correct plane, but for r>1 mm, θ<0.6º. Since it is my understanding that modern golf balls are very homogenous, this device is not useful for most balls. It would work better if the axis of rotation were horizontal.
Next I should address the question of whether such a small r will be detectable by the floating method. We can use the same figure above except with C=0. The mass of the ball is about 0.046 kg and the radius of the ball is about R=0.021 m. Taking r=10^{5} m, the torque about the center of the ball is τ=mgrcosθ=4.6x10^{6}cosθ N·m. The moment of inertia of the sphere is I=2mR^{2}/5=4.1x10^{5} kg·m^{2}. So, the angular acceleration is α=τ/I=0.11cosθ s^{2}. This means that, if you start at θ=0º, after 1 s the angular velocity would be about 0.11x180º/π≈6º/s, easily detectable I should think.
The bottom line here is that the questioner discovered that only one of the 12 balls was not essentially perfectly "balanced".
All this is truly academic, though, since I am sure nobody really thinks that a ball with its CoG less than 1/10 mm from the center of the sphere will behave in any measureable way differently than a perfect ball. So the CheckGo Pro does no harm to your game, it just does no good unless you happen to have a really offcenter CoG. If you do both measurements, you can locate surprisingly precisely where the CoG is with θ giving you r and the vertical giving you the direction of the line between the CoG and the center of the sphere.
OLD ANSWER:
You have devised a simpler way to find the "balance axis" than the fancy
gizmo you received. And I see no way that your method would not work—the
CoG should be vertically below the geometrical center of the ball. The
spinning gizmo seems like it ought to work also with the CoG being
forced to seek the plane perpendicular to the axis of rotation. So, why
are the two experiments different? I believe that the the two
experiments would yield identical results if the golf ball were a
perfectly rigid
^{ }
body. But look at the picture above. A struck ball experiences a force of about 2000 lb during the collision with a club; the weight of the ball is about 0.1 lb, so this is about 20,000 times the weight, or about 20,000g. (One "g" of force is equal to the weight of the object.) Now, if your gizmo gets the ball spinning with an angular velocity of ω=10,000 rpm≈1000 s^{1}, the centripetal acceleration is a=Rω^{2} where R is the distance from the axis of rotation. A point only 1 mm from the center would have a=1000 m/s=100g and for a point on the surface where R≈2 cm, a=2000g; the resulting forces on the ball will surely cause the ball to deform into an oblate spheroid, flattened along the axis of rotation. (The figure shows a much exaggerated flattening compared to what the actual would be.) However, unless the CoG is at the center of the ball, changing the shape of the ball will change the location of the CoG and after the ball stops spinning the CoG will move back to its original location. The CoG will never be very far from the center, so I would guess that the errors in locating the plane in which it lies could be quite large. I can see no reason why your floating method would not work. Maybe you ought to market it!
FOLLOWUP
QUESTION:
One follow up question (again purely academic) if I might. Since a golf ball spin rate could vary between 25009000 rpm depending on the club, would the "spinner" axis be more accurate/appropriate for a ball spinning at a high rate and the buoyancy axis more accurate for a ball rolling on a putting green?
Also, the buoyancy test wasn't my idea. I wish I were that clever. It's been around for a long time. I'm told that Ben Hogan checked his golf balls that way back in the 1950s when golf balls were quite a bit less uniform than those manufactured today.
ANSWER:
I believe that the driven ball will experience almost no sensitivity to
the location of the CoG. The reason is that the axis about which the
ball is rotating will always pass through the CoG and there will be no
"wobble", unlike the rolling ball. An extreme example of the CoG far
from the geometrical center is shown in the hammer projectile above.
Although parts of the hammer are often far from the trajectory of the
CoG, it moves smoothly overall. It might have a very minor effect on the
ball by virtue of the lift generated by the spin, but I cannot imagine a
ball where the CoG is more than a millemeter from the center and any
such effect would likely be unmeasurably small.
FOLLOWUP
QUESTION:
Your response to the second question makes this even more academic (if possible) because if the rolling ball is the only one affected then clearly the spinner is superfluous.
One last followup question. I promise. After your first response, I wondered if the change in ball shape is measurable at all. I mounted a laser on a level and shot it over the top of the ball at rest so that it barely touched the top of the ball. After spinning the ball up to full speed I couldn't perceive any change at all in the amount of light touching the top of the ball. Is the expected change in shape that small? If so, does that still explain a 45 degree axis differential in the measuring methods?
ANSWER:
I cannot really predict how large the actual deformation will be other
than I expect it to be quite small. Still, the spinning could cause the
mass distribution to change. Suppose that the solid rubber core is
enclosed in a very rigid spherical shell; keep in mind that the core,
made of rubber, is compressible. At high rpm, the ball will act like a
centrifuge and the rubber core will be squeezed out toward the surface
making the density of the core far from the rotation axis larger than it
is closer to the axis. So, even if it does not change shape, the mass
distribution will change and the CoG would not (necessarily) be in the
same plane as it is when not spinning. Again, if the distance of the CoG
from the center is very small, even a very small change of its location
while spinning could introduce a large error in determining the plane in
which it lies. I cannot understand why 11 out of 12 balls would be off
by 40º45º, though.
I have to admit, though, as I get deeper into this problem, that I have some reservations about my answers. Most modern balls use an inner core made of polybutadiene rubber which is what superballs are made of. It turns out that this rubber with a bulk modulus K=1.52 GPa is nearly as incompressible as water with K=2.2 GPa. Assuming your buoyancy results are reproducable, I am puzzled. My final conclusion would have to be that if the ball is very close to homogenous, it is impossible to make a really accurate measurement of the plane in which the CoM resides but not so hard to find the line on which it resides.
There are lots of videos on youtube about this spinning device compared to the saltwater method. Most are promos for a golfball called RealLine. There is one with a result similar to your results. One shows a kid using it to mark a ball and the line is clearly off the equatorial position (i.e. would not cut the ball in equal halves). If you read the comments on the spinner on Amazon, there are some enlightening comments on what might go wrong. Unfortunately, I could not find any comprehensive comparisons of the two methods like you have done.
QUESTION:
If me and my friends went on a camping trip up in the mountains and place a pot on a fire with water in it with one egg in and and cooked it the same length of time we would of if we were home but the egg was not done what would be the cause of that
ANSWER:
At high altitudes the pressure is smaller (less air per unit volume). At
low pressures water boils at a lower temperature.
QUESTION:
Object A (lets call it a train which is moving) has a mass greater than object B (a man named conner who is stationary (and with abnormal strength)) but object B outputs more Force than object A. What would happen to the objects when they collide?
ANSWER:
Your phrase "…outputs more Force…" does not really
mean anything. During the collision, Newton's third law requires that
the force the train exerts on the man must be equal and opposite to the
force the man exerts on the train. However, you could imagine the man
pushing with his very strong arm during the collision time such that he
adds a certain amount of energy to the system; this would result in a
different situation from if he just stood there. Nevertheless, Newton's
third law would still apply. The extremes of his just standing there are

a perfectly inelastic collision where the man sticks to the train afterward and

a perfectly elastic collision where the man flies off with no loss of total energy.
The speeds of each may be found easily and are classic introductory physics problems. Because the mass of the train is much larger than the mass of the man, the results are pretty simple:

For the inelastic collision, the man and train continue with approximately the speed that the train came in.

For the elastic collision, the train continues at approximately the speed it came in with and the man proceeds in the same direction with approximately twice that speed.
If the collsions last some short time t, the average forces on the man during that time will be approximately mv/t for the inelastic collision and 2mv/t for the elastic collision; of course, the forces on the train will be the same.
Now suppose the man (mass m) adds just the right amount of energy to stop the train (mass M) coming in with speed v; I assume that there is no other energy lost or gained. Then the man's speed after the collision would be equal to (M/m)v, a huge speed! In this case the average force on the man would be Mv/t; he will have to be amazingly strong to endure this! The energy he would have to add to stop the train would be approximately (M/m) times the energy the train came in with.
QUESTION:
On a recent BBC program I watched it said that the universe could 'borrow' energy from a vaccum so long as it gave it back quickly enough  that electrons and positrons would spontaniously form in a vaccum, then anialate each other to return the energy.
My question is this. If you had a vaccum inside an incredably strong magnetic field, could you pull these particals appart before they anialate? and if so what would happen, where would the energy for their creation come from
etc.
ANSWER:
A magnetic field would not be a good choice to try to do what you want
because the particles would be nearly at rest and experience little
force. However, a very strong gravitational field around a black hole
could add the energy of one particle's mass and it would escape while
the other was absorbed into the black hole; the black hole would get
lighter by the mass of one particle. This is called
Hawking
radiation.
QUESTION:
Hello, I was wondering about the good old water in the bucket example in a verticle circle and why does the water stay in the bucket at the top of the circle, when it is upside down (at the top).
I'm actually having a bit tough time visualizing the forces acting, i mean there should be a tension acting downwards on the bucket (due to my hand)? Which could be minimum since at the top, the centripetal force could all be provided by mg (weight). (Is there a good reason why mg can provide all of the centripetal force?)
So mg is a constant, as per N3 law, the water should exert a force on the bucket upwards, and the bucket would exert a force on the water downwards, so there is also a normal reaction force pointing downwards towards the centre of the circle, should this be a minimum too? But wait, aren't we supposed to consider the bucket and water as a "one" object?
Moreover, what's the relation of the speed to this? Why should there be a minimum speed for the water not to fall? Also I'm guessing the radius in this case, would be the length of my arm?
ANSWER:
You are making this way too hard. The most important thing in solving
mechanics problems is to focus on one body at a time. You want to
understand when and if the water will fall, so choose to look at the
water as the body. The tension in the rope and the force of your hand
are irrelevant because they are not forces on the water. The only forces
on the water are its own weight, mg down, and the normal force
which the bucket exerts on the water, N. So, Newton's second
law is mg+N=mv^{2}/R where v is the
speed and R is the length of the rope; I have chosen down as
the positive direction. Solving, N=m[v^{2}/R)g].
Now note that if v<√(gR) then N is
negative which means that it would be a vector which points up since I
chose down as positive. But the bucket is unable to exert a force up on
the water and so the water would not stay in the bucket if going too
slowly.
QUESTION:
[It took several exchanges with this questioner to get all
the information I needed, so I have paraphrased the question.]
An 850 pound 4wheeler (including the weight of the occupants) is sitting still stopped, and gets pushed 32 ft from getting rear ended by an
800 lb motorcycle (including the weight of the rider) how fast is the motorcycle going?
After the collision, the motorcycle is at rest and the 4wheeler has its
brakes locked and skids on dry asphalt.
ANSWER:
I can only do a very approximate calculation and will only retain 2
significant figures throughout. I prefer to work in SI units, so the
input data are 800 lb=360 kg, m=850 lb=390 kg, s=32
ft=9.8 m. I will use momentum conservation for the collision, 360v=390u
where v is the incoming speed of the motorcycle and u
is the outgoing speed of the 4wheeler; so u=0.92v. As
the 4wheeler slides, the friction of the brakes does work which takes
the energy of the 4wheeler away. The energy is K=½x390xu^{2}=½x390x(0.92)^{2}v^{2}=170v^{2}
J; the work done by the friction is W=μmgs=0.9x390x9.8x9.8=34,000
J where μ=0.9 is the approximate coefficient of static
friction of tires on dry asphalt and g=9.8 m/s^{2} is
the acceleration due to gravity. So, K+W=0=170v^{2}34,000
or v=14 m/s=31 mph.
FOLLOWUP
QUESTION:
31 mph? That's it, really?
ANSWER:
With the data you gave me, that is the best possible estimate. The place
where there might be a problem is the fact that the motorcycle is at
rest after the collision. Since the two masses are very close, this
implies that the collision was very nearly perfectly elastic, almost no
energy lost in the collision which would be surprising to me; given your
data, only 6% of the energy was lost. If the cycle were going much
faster and tried to brake to avoid the collision and continued on a
little way after the collision before stopping, the answer would have
been faster than 31. In any case, I can tell you that the speed of the
4wheeler immediately after the collision had to be in the neighborhood
of 30 mph.
QUESTION:
I'm confused? One moment I'm reading about "inertial reference frames" and that "acceleration due to gravity" is unaffected by mass. This is followed up with examples such as the Bowling ball and feather. All good. Maths seems clear enough.
But then we start talking about a particular body/objects "acceleration due to gravity" at or near the surface of the Earth as being 9.8m/s^{2} which is calculated using the masses of earth and the "falling" body and Newton's Law. Similarly I read that if I go and stand on the moon the "acceleration due to gravity" will be different because the masses are different? Again seems to be clear.
What am I missing? How is it that the mass of two object does not affect the acceleration one moment but the accelerations of the moon and the earth on a body have differing values due in part (large part) to their mass?
ANSWER:
Look at the figure. Two masses, M and m, are separated
by a distance r. M exerts a force F_{mM}
on m and m exerts a force F_{Mm}
on M; because of Newton's third law, these forces are equal and
opposite, F_{mM}=F_{Mm}.
Because of Newton's law of universal gravitation the forces have
magnitude F_{Mn}=F_{mM}≡F=mMG/r^{2}.
Then, using Newton's second law, each mass will have an acceleration
independent of its own mass of a=[mMG/r^{2}]/m=MG/r^{2}
and A=[mMG/r^{2}]/M=mG/r^{2}.
Note that I have been viewing this from outside the system; this frame
of reference is is called an inertial frame of reference. It is
important that we view the system from an inertial frame, because
otherwise Newton's laws are not correct.
So, what you have been taught is correct only if you view things from outside the twobody system. But what you have also probably been taught is that you measure this constant acceleration relative to the surface of the earth and that is technically incorrect because the earth is accelerating up to meet the falling mass and is therefore not an inertial frame. However, the earth's acceleration is extremely tiny, too small to measure. If M is much much larger than m, which is certainly the case for the earth and the moon, what you have been taught is, for all intents and purposes, correct.
QUESTION:
My question has to do with the conservation of angular momentum and black holes. If a rotating object enters a black hole, would that objects angular momentum not be converted to massenergy increasing the mass and momentum of a black hole? As I see it, this would indeed violate the law of conservation of angular momentum. However an alternative theory had occurred to me. If indeed the angular momentum of the object is destroyed, it might impart a rotation on the universe as a hole, in the opposite direction of the spinning object. I do not know the mathematics or physics well enough to know whether or not that is a feasible hypothesis, but if it is, could it be possible that this universal rotation which was initially essentially non existent has continued to steadily increase such that today the centripetal force resulting from this rotation might explain the effects of dark Energy which opposes gravity over great distances. Please let me know if there is something basic I am misunderstanding, and as for the dark energy hypothesis, I am sure there is some reason I am wrong but if not and that is a valid hypothesis please let me know.
ANSWER:
As I state on the site, I do not normally answer questions on
astronomy/astrophysics/cosmology and I give some links to sites to which
it would be more appropriate to pose such questions. First of all, there
is no doubt that some black holes, perhaps all, have angular momentum.
Second, any angular momentum which an object has once it has entered the
event horizon, must become added to the black hole it already had
because no information may be transmitted from inside that radius.
Regarding whether angular momentum could be transferred to the "universe
as a [w]hole" (interesting typo on your part!), the universe is not a
rigid object, so any angular monentum lost (via interactions with other
nearby objects) before the object passes the event horizon would be
transferred to objects nearby. But the angular momentum of the whole
universe would remain unchanged since the object is also part of the
whole universe and what the rest of the universe gained, it lost. I urge
you to find another more authoritative source!
QUESTION:
Please explain how kinetic energy
affects the human body in flight.
ANSWER:
I do not understand. Kinetic energy of what? In flight?
QUESTION:
My apologies. I was asking if/how kinetic energy in the airplane
affects my human body while in flight and after.
Perhaps I know too little to even pose a question given that I am already assuming that this 'kinetic' energy is freely existing in the 'air' inside the plane.
That I just love the beauty of physics while remaining totally 'illiterate' to the laws
saddens me so I truly thank you for your kindness.
ANSWER:
Kinetic energy is the energy something has by virtue of its motion and is not something which
"affects" you. It is something you have or don't have, depending on your motion. And it is something which is relative, it depends on the frame
in which you calculate your energy; you have kinetic energy relative to
the ground but not relative to the airplane. The kinetic energy may be
written as K=½mv^{2} where m
is the mass and v is the speed. For example, if you have a mass
of 60 kg (about 130 lb) and are in an airplane going 500 mph=224 m/s,
your kinetic energy relative to the ground is K=½·60·224^{2}=1.5x10^{6}
Joules, a million and a half Joules! How did you get that energy?
During the takeoff the airplane pushed on you with some force to speed
you up and give you that energy. Suppose that the time to speed you up
to that speed was 10 minutes=600 seconds; then the power that the plane
had to deliver over that time would have been 1.5x10^{6}/600
J/s=2,500 Watts. The power delivered by the airplane to you would be
enough to power 25 100 Watt light bulbs. During the 10 minutes, the
airplane would have been pushing on you with a force of about 22.4
Newtons=5 lb; that is not much of an effect on you. But suppose that you
had accelerated to 500 mph in 1 second instead of 600; then the force on
you would have been 13,400 N=3000 lb which would have crushed you. So
you see, it is the force when you are acquiring your kinetic energy
which "affects" you, not the kinetic energy itself.

Where happens all the matter that enters a black hole? Is all the matter in a black hole crushed into a "singularity" What are the current theories you suggest I study?

If a black hole can consume matter without limits, could a super massive black hole also consume the mass our universe?

So, could the "Big Bang" be from this "singularity", a singularity that holds all the mass of our universe? (This makes sense to my way of thinking & boggles my mind.)

What are White Holes? Have they been proven to exist yet? Or like "worm holes" are they only theoretical today? And does current thinking say White Hole evolve from Black Holes?
ANSWER:
Well, I certainly appreciate your donation—almost nobody
does, even for my best, most creative answers. I do, however, recommend
that people wait for an answer before donating unless you like my site
so much that you just want to support it and I greatly appreciate that!
(Also, I have no way to "refund" contributions.) In your case, you did not read enough
on the site before submitting your
question(s). Two quotes from the site: "If your question is clearly astronomy or astrophysics, particularly detailed questions about black holes, stellar evolution, dark matter or dark energy, the big bang,
etc., areas in which I am not expert, I may not answer" and "Please
submit single, concise, wellfocused questions". Because of your contribution, I
will cut you a little slack this time.

The simplest idea of a black hole is an object with mass, charge, and angular momentum which has infinite density and where time has stopped. I believe the latest thinking is that it is not really of infinite density. Predictions of singularities are from classical general relativity; if you view the situation quantum mechanically, then uncertertainty ideas do not want zero size or stopped time because spacetime itself should have a granularity. A discussion which you might find accessible is at Physics Forums.

This depends on how things happen to be moving around. You could imagine a much older universe composed of nothing but black holes and the radiation they emit when they undergo Hawking radiation but moving in such a way that they will never encounter each other.

There is a very interesting book by Lee Smolin, Time Reborn, where he reprises an idea originally by Wheeler and deWitt that black holes are the seeds of new universes. There is a nice interview with Smolin on space.com and a lecture by him that you can watch.

White holes are hypothetical and have never been observed. I know nothing about them.
I hope I have earned your donation!
QUESTION:
is there any gravity on space? infomations are random on internet, like gravity is every where, there is no zero gravity concept, but as we go far from earth gravity start decreases as black holes are in space, and they have such a strong gravity that nothing escape of it, so where black hole got such a strong gravity, i know with increasing mass gravity increases, but black holes are not near earth surface so from where they gain gravity???
ANSWER:
Every object in the universe with mass causes gravity. The bigger the
mass, the bigger the gravity. You cause a gravitational field, but it is
tiny compared the the earth's field. The earth's field is tiny compared
to the sun's. The sun's field is tiny compared to a black hole's. The
entire universe is permeated by gravitational fields. And gravity is a
very longrange force; although the force gets smaller as you get
farther away, it extends all the way across the universe.
QUESTION:
I have calculated that a 50 g marble attached to a 1 m string wrapping itself around your finger held above your head at a rate of one revolution every second until it reaches 5 cm from your finger will end up applying a centrifugal force of the equivalent of the force of gravity on 1600 kg. Does this prove that conservation of angular momentum is a fallacy?
ANSWER:
This is a variation of the tetherball problem, a classic in introductory
physics courses. If you want to talk about angular momentum
conservation, you have to ask under what conditions angular momentum is
conserved. The angular momentum of your marble relative to the center of
your finger is conserved if there are no forces which exert a torque on
it. The only force is the tension in the string and, as you can plainly
see, the tension T has a component perpendicular to r
(Tsinθ) and therefore exerts a torque
τ=rTsinθ and therefore angular momentum is
not conserved; this does not make it a "fallacy", it just does not hold
for this particular problem. What is conserved, though, is energy.
Because the displacement (along the velocity vector) is always
perpendicular to the tension, the tension does no work so energy is
conserved, ½mv_{1}^{2}=½mv_{2}^{2}
or v_{1}=v_{2}, the velocity stays
constant. So, if the m=50 g=0.05 kg marble starts at about 1 m
away from your finger with a frequency of ω=1 rev/s=2π
radians/s, its speed is v≈ωL=2π
m/s; then when L=5 cm=0.05 m, T=mv^{2}/L
=0.05x4π^{2}/0.05=39.5 N=4.0 kgforce.
THE
PHYSICIST:
The questioner submitted followup questions. To see these,
link here to the
OfftheWall Hall of Fame.
QUESTION:
Since electrons occupy discrete energy levels in an atom , shouldn't the electrons that are more energetic (are in a higher energy level) need less energy to escape the atom ? When light is shone onto the material and is above the threshold frequency electrons are emitted , but are those electrons from the highest energy level ?? It confuses me , because if that's true , then that would mean that the work function would differ for each energy level which doesn't make any sense , because work function is defined as the minimum energy for an electron to be released.
ANSWER:
You should not look at electrical properties of a conductor by looking
at atomic structure of its constituents. A solid composed of huge
numbers of atoms does not behave the same way as a single atom does. In
a conductor, atoms all interact with their neighbors in such a way that
at least one electron per atom (called conduction electrons) moves
around inside the solid pretty much freely like an electron gas. The
photon strikes the surface and gives all its energy to a single
electron. For that electron to be ejected from the metal, it must have
more kinetic energy than the work you would have to do to just pull it
out of the metal, and this is not how much work you would have to do to
pull it out of a single atom because it is not bound to any atom; the
former is called the work function, the latter is
called the ionization potential. The idea is that when an
electron is removed from the metal, one positively charged atom is left
behind resulting in an attractive force trying to hold the electron in.
A point charge in front of a plane ideal conductor is a classic
electrostatics problem, usually solved using the
method
of images. If you then integrate from the size of an atom (~10^{10}
m) to infinity you can get a ballpark estimate of the work function,
W≈3.6 eV.
QUESTION:
My question is that throughout The Principia Newton uses the fact that for a body continuously moving in a uniform circular motion with constant velocity, the body falls a small distance towards the center of the circle. Had there been no centripetal force, the body would move along a straight line(tangent), but due to the force the body falls towards the center, Due to its velocity the body does not completely fall but moves in a circle. Now as Newton, Chandrashekhar and Feynman have shown in their respective books, and also quite obviously, in a very short interval of time the deviation produced by the force or more precisely the distance fallen by the body will pe parallel to the radius at the initial point and parallel to the central force at the point. Since the displacement will be parallel to the force ,some finite work would be done, because work is defined as the product of the force and the displacement caused by the force in the direction of the force. Important thing is that the displacement to be taken is that which is caused by the force and not due to initial velocity. Therefore some finite , very small amount of work will be done by the centripetal force since the fallen distance and and the force are parallel. When this work be integrated over half a circle or 1/4 or 3/4 of circle it will give an increase in kinetic energy showing that the velocity has increased. This is contradictory. So what is wrong over here or is something correct ?
ANSWER:
A short answer to a long question: It is true that an object in a
circular orbit is constantly "falling". It is not true that it is moving
toward the center, rather that it has an acceleration in that direction.
Although the change in velocity is centripetal, the displacement is
always tangential and the force does zero work. By the way, the body
does not move "with constant velocity" as you state, but with constant
speed. The direction of the velocity vector is constantly changing.
QUESTION:
we have defined momentum as the product of mass and the velocity of a body, we say that photon is a mass less particle,i find it really confusing how can a massless particle still have the momentum?
ANSWER:
Linear momentum p was, indeed, defined as mv before
the advent of the theory of special relativity. If v is much
less than c, the speed of light, this is an excellent
approximation, but not exactly true. The correct expression for p
is m_{0}v/√[1(v^{2}/c^{2})]
where m_{0} is the mass of the object when at rest. The
relation among energy E, momentum p, and rest mass
m_{0} is E^{2}=p^{2}c^{2}+m_{0}^{2}c^{4}.
So, you see, even if a particle has zero mass (like the photon) it still
has momentum if it has energy, p=E/c. I am also often
asked how a photon can have energy because E=mc^{2} and
m=0 for a photon. You can find links on the
faq page to
answers which discuss this question.
QUESTION:
A wheel rolls without slipping with angilar velocity ω and radius r what is the angular velocity of a point in the rim at the same level as the centre ?
ANSWER:
I believe that you are asking the wrong question; you must be asking
what the velocity of the point is, not its angular velocity. It is
important to understand how the wheel is rotating. Since the point of
contact with the ground is at rest at any instant, the whole wheel is
rotating about that point at that instant. The angular velocity ω
about this point is v/R where R is the radius
of the wheel and v is the speed which the center of the wheel
is moving forward with speed v. Now, the point on the rim also
has the same angular velocity ω, but its distance from
the axis of rotation is R'. Because the angle which R'
makes with the ground is 45º, it is easy to show that
R'=R√2. Therefore,
ω=v/R=v'/R'=v'/(R√2),
and so v'=v√2; the direction of the velocity
v' is 45º below the horizontal, as shown above.
QUESTION:
Is it possible to create a gravitational lens without a black hole or dense object? If so, can gravitational lensing be practical for use? I've been thinking about it for quite some time now and tried thinking of possible uses.
ANSWER:
By definition, gravitational lensing is the result of strong
gravitational fields which are caused by large masses. So, without a
large compact mass, appreciable lensing will not occur.
QUESTION:
Gravity Well question that's been puzzling me since I watched Interstellar...If an object is orbiting in the gravity well of a massive planet/star, it is falling (conforming perfectly to the curved space locally). Other than its orbital velocity, why would its depth in the gravity well cause its clock to tick any slower than another object floating more distantly from the same gravity well.Seems to me that gravity well time dilation only applies if you're fighting the pull of that well (e.g., by standing on the planet's surface)Thoughts?
ANSWER:
Gravity does not just warp space, it warps spacetime. The larger the
gravitational field, the more strongly spacetime is warped. Therefore
one finds that clocks run slow in a gravitational field, the stronger
the field, the slower they run.
QUESTION:
What is the physics behind a baseball curving?
ANSWER:
See an
earlier answer.
QUESTION:
If a sun's gravity attracts a planet, is it possible it would repel an anti planet?
ANSWER:
No, that is not possible. There is only one kind of gravitational mass
and therefore only attractive gravitational forces are possible. Every
experiment ever done with antiparticles indicates that the mass is
identical to that of its particle counterpart.
QUESTION:
I am trying to explain to my brother why on a spinning wheel a point farther out is going faster than a point closer to axis, though the wheel is spinning at the same rpms. But he just cant figure it out. Could you give an explanation a 2 year old could understand?
PS My brother is 21.
ANSWER:
I could probably not convince a twoyear old, but if your brother is
just a little smarter than one, I can probably convince him. The speed
of something is defined as the distance traveled divided by the time it takes to
travel that distance. For example, a car going around a circular race
track which has a total circumference of 2 miles takes 2 minutes to go
around once, its speed is (2 miles)/(2 minutes)=1 mile/minute=60 mph.
Suppose a wheel has an angular speed of 10 rpm and has a circumference
of 2 m. Then the distance a point on the rim will go in 1 minute is 20 m
because the wheel goes around 10 times; the speed of that point is
therefore 20 m/min. Now look at a point halfway from the axle to the
rim; it will move in a circle of circumference only 1 m so the distance
it travels in 1 min is only 10 m so its speed is therefore 10 m/min. In
a nutshell, a point near the center travels a shorter distance than a
point far from the center in the same time.
QUESTION:
my question is about speed, if our earth is traveling at 1600KM per hour and the Milky way is more then several Million Km Per Hour etc, how our equmileting speed is not reaching the speed of light, or at least breaking the sound barrier..
and if i jump i move little from my original spot?
ANSWER:
See an earlier answer. Also, the sound barrier
is irrelevant because there is no sound in space. Although not really
related to your main question, if you jump vertically upward in a
rotating coordinate system (like the earth spinning on its axis) you
will indeed not land exactly where you launched but the difference is
very small.
QUESTION:
If gravity is not a force but just a curvature of spacetime then how does a massive object (like earth) affect a much less massive object (like a tennis ball) when they are not in motion relative to each other?
For example, if I were able to travel out to space far enough away from earth and then stop so that I am not in motion with respect to earth and let go of a tennis ball it will stay stationary.
But if I traveled to say 100,000 ft above sea level and were able to hover there so I am not in motion to earth and then let go of a tennis ball it will immediately begin to move towards earth.
In other words, how is it possible for the curvature of spacetime to affect bodies that are not in motion to other?
ANSWER:
You have some misconceptions here. First of all, no matter how far away
you get, there will always be a small force toward the earth. It may be
so small that you would have to wait a millenium to see it move a
millimeter, but it is still there. The curvature of spacetime justs gets
smaller as you get farther away. It is certainly true that if you drop
something from an altitude of 100,000 feet it will accelerate toward the
earth. But, what does that have to do with motion? Even if you gave it
some motion, say throwing it horizonatlly, it would still accelerate
toward the earth just the same as dropping it, but now it would also
have a velocity component parallel to the earth as well. Maybe you have
never seen the "trampoline
model" which illustrates (in a simplistic, not literal way) how
warping the space (the surface of a trampoline) by a massive object
(bowling ball) will cause a light object (marble) to be attracted to the
massive object.
QUESTION:
Engineers in the Bay of Fundy are trying to harness tidal power to generate electricity. It's claimed that they can generate enough electricity to power the entire Atlantic Provinces of Canada.
This got me to thinking, if that much power can be harnessed, where has that energy been going? I'm assuming it goes to heat and warms the water. If that's the case, would harnessing the power therefore cool the water and potentially harm aquatic ecosystems?
ANSWER:
My research shows that all the power projects on the Bay of Fundy will
generate about 20 MW. These will be powered by underwater turbines. I
did a very rough estimate of the total power available by the falling
water: The average rise in sea level is 15 m, area of the bay is 13,000
km^{2},
density of water is 1000 kg/m^{3}. I find that the total volume
to fall is about 2x10^{11} m^{3}, the mass is 2x10^{1}4
kg, and the potential energy of this mass is about 3x10^{1}6 J. If
you deliver that much energy over the course of a day, the average power
is about 350 GW. So, the power plants will only reduce the energy of the
tidal flow by less than 0.01%, a truly negligible amount.
QUESTION:
I recently read Michio Kaku's book that discusses impossible technologies and how they can be achieved and even possible. He discusses that a perpetual motion is next to impossible, because energy is always wasted, attributed to the laws of thermodynamics. Now, when talking about using antimatter as energy, it is described as having 100% efficiency. Wouldn't that violate the laws of thermodynamics, or is it one of the few cases where we see the laws of physics bend?
ANSWER:
The relevant "laws of thermodynamics" are essentially just the
conservation of energy. Energy is always conserved in an isolated
system, it is just that in the macroscopic world some of the energy at
the end is not useful. We normally specify efficiency of a machine as
the relation of energy added to a system to the work we get out; any
difference between the two shows up elsewhere, usually as heat. In
particleantiparticle annihilation energy is conserved just like the for
macroscopic machine, but there is no work, just electromagnetic energy
(two photons); if you think of it as a machine, you need to convert
those two photons into useful work and that conversion would likely not
be 100% efficient. On a microscopic level, any interaction among
elementary particles is "100% efficient" if you compare energy before
with energy after. Shoot an electron at an atom and have it excite the
atom; the electron leaves with less energy than it came with, the atom
recoils because of having been hit by an electron, and then the atom
deexcites by emitting a photon. Add up the energy of the electron, the
photon, and the atom after the collision and you get the same total as
the energy of the electron and the atom before the collision.
QUESTION:
If I were to fall straight through the earth, from one side to the other, how long would that take if it were possible?
ANSWER:
If the earth were a uniform sphere, constant density throughout its
volume, this is a classic elementary physics problem; a solution may be
seen on the
hyperphysics web site. The answer is about 42 minutes. However, as
discussed in an earlier answer, the earth is
by no means a uniform sphere so the answer would be different but not
easy to calculate.
QUESTION:
If there's less gravity on the moon, then why do astronauts move slower? Why don't they have more spring if gravity's pull is weaker? I'm thinking that this should be a matter of resistance, but obviously there's a factor I'm not accounting for.
ANSWER:
Sure, with less gravity (about 1/6 of earth) an astronaut can jump much
higher, but it will take longer to get to the top and back down than on
earth; that would look like slow motion, right? Or, just think about
taking a step. Your center of gravity is forward of your trailing foot
and so you rotate about that foot. The rate of rotational acceleration
is only 1/6 that on earth, again like slow motion.
QUESTION:
Some car drivers reason that since tractor trailers have more tires, they
should be able to stop quicker. How much work does each car tire need to do
to stop it, compared with how much work each truck tire needs to do to stop
it? For ease of reference let's say a car is 4000 pounds and the truck is
80000 pounds while the speed involved is 65 miles per hour. Finally is the
work done proportional to the size difference of truck brakes versus car
brakes?
ANSWER:
I replied that this is not a homework
solving site.
FOLLOWUP
QUESTION:
I didn't realize I had posed this in the form of a homeworkish question. I'm a 47yo truck driver, tired of hearing all the lame@$$ rhetoric of the motoring public. Long, quiet, empty night highway helped me come at this from a different angle. Without too much detail, one truck wheel wrangles slightly more than a single car's worth of mass; is why trucks do NOT stop better despite having more wheels.
ANSWER:
I have never heard this but it is pretty nonsensical. How quickly any
vehicle can stop is determined by the maximum force which can be applied
by braking; if the coefficient of static friction between the tires and
the road is μ, the maximum force on a level surface
is μW. where W is the weight of the vehicle. The
distance s traveled will be determined by the work done by this
force, μWs, which will equal the initial kinetic
energy, ½Mv^{2}=½(W/g)v^{2},
where M is the mass and g=32 ft/s^{2}. So,
s=(1/(2μg)v^{2}. So here is the
interesting thing: the distance traveled does not depend on the weight
of the vehicle, only by the initial speed and the condition of the road.
The force applied does depend on the weight and the force per wheel
would be 0.7x4000/4=700 lb for the car and 0.7x80000/18=3111 lb; but the
larger force is simply because of the larger weight and the only thing
which is important is the distance to stop. Of course, this also depends
on whether you have antilock brakes which let you get the maximum amount
of friction when you are just on the verge of skidding; if you lock the
brakes and you skid, you go farther before stopping. For your example,
v=65 mph=95 ft/s and μ is approximately 0.7 for
tires on a dry road, so s≈200 ft. I have neglected all
other forms of friction like rolling friction and air drag, but these
should be relatively unimportant. Bottom line—stopping distance
does not depend on number of tires nor on weight.
QUESTION:
I am trying to figure out how to calculate the Cubic yards of a relatively flat surface. Specific example: A carpet or rug (let's say it is 1 inch thick and 100 square feet). How many cubic yards (3ftx3ftx3ft) would it be when rolled up? Or what is the equation on how to get the answer for varying sized carpets?
ANSWER:
If the area is A and the thickness is t, the volume is
V=At. Of course, you have to be careful to have A and
t measured in the same units. V is the volume
regardless of how you change its shape. Since you apparently want your
answer to be in cubic yards, you should probably convert everything to
yards at the outset. For the example you state, A=(100 ft^{2})(1
yd/3 ft)^{2}=100/9 yd^{2} and t=(1 in)(1 yd/36
in)=1/36 yd. So V=(100/9)(1/36)=0.309 yd^{3}. You can
get more detailed if you know the length L and width W
of the rug so A=LW and V=LWt. If you roll it up to a
cylinder of length L, its volume will be V=πR^{2}L=LWt
where R is the radius of the cylinder; therefore, R=√(Wt/π).
So, if your rug is 10x10 ft^{2}, it will roll up with a radius
of R=√[(10/3)(1/36)/3.14]=0.172 yd=0.52 ft. All this
assumes that the rug does not compress when you roll it up.
QUESTION:
Kind of an odd question with a potentially obvious answer, but i was wondering, if you were on the moon and you jumped forward in a superman like pose, would you keep going? like, would you gain speed or would you still eventually fall back to the moon? and would the spin of the moon have an affect on this?
ANSWER:
The escape velocity of the moon is about 2.38 km/hr which is more than
5000 mph. I do not think you can jump that fast. A low altitude orbital
velocity is about 3700 mph, so I don't think you can jump that fast
either. The gravity on the moon is about 1/6 that on the earth, so you
could certainly be able to jump farther than on earth; but you would
never "keep going". And you would certainly not "gain speed". And the
spin of the moon on its axis would be of no use since it only goes
around its axis once a month.
QUESTION:
I work with a product by the name of Stereotaxis. It's a magnetic navagation system designed to move catheters inside a patients body by people in a control room (as to shield the operator, or physician, from the effects). Unfortunately, my colleagues
& I (registered nurses) are not as fortunate. We typically sit about 3 feet in front of the Stereotaxis magnet for most of the day (can be 10 hours or more). There are 2 magnets that produce 800 gauss each (we sit in front of one magnet, not both). The other magnet only effects us when it is deployed for use with the patient. Is this dangerous to us? We all work 40 hours per week; at least, sometimes much more.
ANSWER:
I have just done some cursory research on the effects of magnetic fields
on biological systems. It would appear to me that there are virtually no
effects for fields smaller than about 10,000 gauss and your exposure is
a full order of magnitude below that level. And, if the magnets are
rated at 800 gauss the field where you are is probably less than that.
Also, the effects of a magnetic field would not be cumulative like
exposure to radiation would be—when you go home there would have
been no damage to your cells. The reasons for the people being in the
control room would certainly not be to isolate them from these fields. I
have had two atrial ablations myself and navigation was done, I believe,
by xray for which longterm exposure would much more a concern; maybe
the control room is a vestige from earlier days. Also, the control room
likely provides a better environment for concentration by the
professionals doing the procedure.
QUESTION:
If a charged particle is at rest on the surface of the earth, it
emits no light visible in the inertial frame of someone standing on
earth (both accelerated at the same rate, equivalence). Consider a
charged particle at rest on the earth, as viewed by an orbiting
observer, far above the earth, or viewed by an observer in a rocket
hovering thousands of miles above the earth (where gravitational
acceleration is less than on the surface). Will the observer see light
emitting from the charge since it is accelerating (in gravitational
equivalence) relative to the observer?
ANSWER:
Someone standing on earth is not in an inertial frame, but that does not
really matter here because the charge is at rest relative to that
person. And, yes, anyone accelerating relative to the charge will
observe a radiation field. An
earlier question is very similar to yours and you should read that
answer to get more detail.
QUESTION:
i was studying something on the internet where i need to know one question i.e If an object Travelling at about 2500 km per hour or may be higher collides with other object in the ocean around 90 meters below the ocean level is it possible that the debris of that object will go around 2 km approx ahead in the water itself?
ANSWER:
Extremely unlikely. Let me show you a very rough calculation which will
demonstrate this for an extreme case. If one object collides elastically
with another mass whose mass is very small compared to the incident
mass, the collidedwith mass will recoil with a speed twice the incident
speed, in your case that would be 2x2500 km/hr=5000 km/hr≈1400
m/s. The drag force on an object with speed v in water may be
approximated as F_{drag}=½CρAv^{2};
here I will choose C≈1 (order of magnitude for most
shapes) is the drag coefficient which depends only on the shape, ρ≈1000
kg/m^{3} is the density of water, and A≈1 cm^{2}≈10^{4}
m^{2} is the cross sectional area of the object. I want this
object to go as far as possible, so I have chosen A to be
relatively small; an object this small will have a relatively small
mass, certainly no larger than m≈100 gm=0.1 kg. Knowing
all this, one can calculate the velocity v and position x
as functions of time t, v=v_{0}/(1+kt)
and x=(v_{0}/k)ln(1+kt)
where k=½CρAv_{0}/m≈7000
s^{1} and v_{0}=1400 m/s is the
starting velocity. I find that at t=½ s, v≈0.4
m/s and x≈1.6 m; the object will have lost nearly all its
speed after having traveled only about a meter and a half. I can think
of no earthly way that any debris could propogate 2 km!
QUESTION:
Would my weight be the same on the surface of earth and one mile underground? How about one mile in the atmosphere?
ANSWER:
Above the surface of the earth the gravitational force falls off like 1/r^{2}
where r is the distance from the center of the earth. If R=6.4x10^{6}
m is the radius of the earth, W is your weight at the surface,
and W^{+} is your weight 1 mile=1.6x10^{3} m above the
surface, then W^{+}/W=R^{2}/r^{2}=0.9995;
your weight would be about 0.05% smaller. If you assume that the mass of
the earth is uniformly distributed throughout its entire volume, the
gravitational force falls off like r as you go toward the
center. Then W^{}/W=r/R=0.99975;
your weight would be about 0.025% smaller. Given other factors like
local variations in the density of the surrounding earth, these would be
unmeasurably small; one mile (about 1600 m) is, after all, extremely
tiny realtive to the size of the earth. I should also note that
approximating the earth's density as uniform is a very poor
approximation; see an
earlier answer. If you had asked your weight 100 miles below the
surface the answer would be that your weight is nearly the same as at
the surface.
QUESTION:
How is the light red shift, observed in the universe, a factor of distance AND speed and the Doppler affect (in which it is often compared) is only a factor of distance? Galaxies farther away have a greater red shift so we are told they are farther away and moving away faster than nearby galaxies. With Doppler the farther the object is away the greater the shift, but the speed of travel could be constant, say using the historic train example. I could buy into the universe is expanding, I'm not convinced we can predict how fast relative to time.
ANSWER:
The Doppler effect does not depend on distance, only speed. Many years
ago astronomers discovered that distant stars and galaxies exhibited
redshifted spectra which they could use to determine the speed at
which those objects were traveling. Later methods were developed which
allowed an independent measure of the distance of those
objects. When the two separate measurements were compared, it was found
that the velocity of the objects was a linear function of their
distances. The graph shows an example where distance was determined
using the brightness of type Ia supernovae. I do not understand your
question beginning with "…With Doppler the farther…"
QUESTION:
People say that Universe is expanding, right?
We all know that Gravitational force is an attractive force and it is predominant at the planetary level among all other forces. So shouldn't the universe come closer together as time passes? Not at a very fast rate, but at least very slowly.
I don't understand why still people do say it expands, Can you please help me reasoning it out?
ANSWER:
Right after you throw a ball straight up, it is moving upward
("expanding" relative to you); the harder you throw it, the farther it
goes before you and it "come closer together". The universe is still in
the early stages where it is still "going up". But, if you throw it hard
enough (called the escape velocity), it will never come back.
And, if you throw it with a velocity greater than the escape velocity,
it will never stop going. So, the ultimate fate of the universe is that
it will either expand forever or everything will turn around and fall
back together (the big crunch). All that is out of date,
though, because it was discovered almost two decades ago that gravity is
not purely attractive or else there is some other force which is
operative as well (called
dark energy);
this results in the fact that the distant objects in the universe are
not just moving away from us, they are actually speeding up. This
repulsive force is not yet fully understood and the ultimate fate of the
universe is not known.
QUESTION:
I am a volunteer guide at South Foreland historic lighthouse in the UK. We have an optic weighing approximately 2 tons, floating in a close fitting trough containing only approx. 28 litres of mercury. What is the theory which enables this optic to float as it does not appear to fit within the basics of Archimedes principle.
ANSWER:
To float 2 metric tons (2000 kg) you must displace M=2000 kg of
mercury. The density of mercury is ρ=13,600 kg/m^{3},
so the volume you must displace is V=M/ρ=0.15 m^{3}=150
l; this, I presume, is what is bothering you since only 28 l are used.
Suppose that the reservoir for the mercury is a cylinder of radius 1 m
and depth d; to contain 0.15 m^{3}, the depth of the
container would have to be d=0.05 m=5 cm (estimating
π≈3). So, I will make a container 6 cm high for an
extra 20%, 180 l and I will buy that 180 l to fill it up. Now, let's
make the pedestal on which the lens sits be a solid cylinder of radius
99 cm so that when you put it into the reservoir there will be 1 cm gap
all around. So as you lower it into the reservoir, mercury will spill
out the top and you will be sure to capture it. When you have captured
150 l, the whole thing will be floating on the mercury. You return the
extra 150 l and have floated the lens with only 180150=30 l. Of course,
in the real world you would only buy 30 l, put the pedestal into the
empty reservoir, and add mercury until it floats. (I realize that the
shape of the pedestal and reservoir are probably not full cylinders,
since you said "trough", but my simple example wouldn't be so simple
with more complicated volumes. The idea is the same, though.)
QUESTION:
In a controlled environment with no crosswinds, with a tailwind of 10mph (or any speed really), would it be possible for a person to move fast enough to equal the effect the wind has on his clothes, hair, etc...? I.E. get to a point where his clothes are essentially motionless as if a perfectly calm day?
ANSWER:
If you move with a velocity equal to the velocity of the air, you are at
rest relative to the air. Therefore the answer to your question is yes.
An example is a hot air balloon; the balloon will be pushed in the
direction of the wind and attain the speed of the wind. If you are
riding in the balloon, the air around you will be still. This becomes a
problem if you wish to steer the balloon because to change direction of
any aircraft you need to be moving relative to the air.
QUESTION:
If the world was a cube would we fall off?
ANSWER:
No, you would not fall off because the cube would still have a
gravitational field which would attract you. You would weigh the most
(least) at the center of each face (the corners) because you are closest
to (farthest from) the center of the cube. At the centers your weight
would be "straight down", i.e. perpendicular to the ground. Elsewhere
your weight would have a component parallel to the ground as shown in
the figure above. Note that if you released a ball at an edge, it would
roll to the center of the face; or if you were to walk from the center
to an edge, it would be like walking uphill. (By the way, you should use
the subjunctive—If the earth were
a cube… )
QUESTION:
A stone and a feather are falling in a vacuum chamber on earth and if left to fall, which would fall first? I still say
the stone because the gravity force is still a function of mass. so basically more mass, more attraction, if it weren't true, jupiter and feather would reach at the same time.
What is right?
ANSWER:
An object of mass m a distance R from the center of
the earth feels a force F=mΦ where Φ=GM/R^{2}
is the gravitational field and M is the mass of the earth.
But Newton's second law tells us that F=ma where a is
the acceleration of m; therefore a=Φ
regardless what m is. All this assumes that the earth, which
also feels the same force as m, has a mass much larger than the
object, M>>m so that the acceleration of the earth is
negligible. If m were comparable to or larger than M,
the earth would accelerate toward the object so the two would meet
earlier; but the the heavy object's acceleration would be just the same
as the feather's even in this scenario. You are right that the stone
feels more force but it also has a greater inertia which results in
identical accelerations.
QUESTION:
The below question I have found in an old text book. I am soon to be a teacher in training so brushing up on my mathematics, but this one has stumped me. I am not sure if I am going along the right lines.
A uniform rod ab of weight W and which is 20 cm long is suspended by two vertical springs
X and Y attached to the ends of the rod. The upper ends of the springs are attached to a horizontal beam. When the springs are unextended they have the same length. The tension in X is given by
T_{X}=Kx and the tension in Y is given by T_{Y} = 3Ky, where
K is a constant and x and y are the extensions of X and Y respectively. At what distance from A must a body of weight 5W be attached to the rod if the rod is to be horizontal?
ANSWER:
Each spring is stretched by the same amount, so if the tension is X is
T then the tension is Y is 3T. W acts at a
distance 0.1 m from X and 5W acts at a distance d from
X. You now need to apply the equilibrium conditions (Newton's first
law). The sum of the forces is zero, 4T6W=0, so T=1.5W.
The sum of the torques about the left end is zero, (3x0.2)T0.1W5dW=0=0.8W5dW
or d=0.16 m. (The solution which you attached, not shown here,
got the relative tensions in the springs correct, but you never applied
Newton's first law and never thought about torques at all.)
QUESTION:
one, is there anything we know of faster then light?
two, could I say that the light being pulled into a black hole is the speed in which light cant escape therefor making that faster or is that a measurement of some kind of force. or does it not matter the force speed, speed is speed?
ANSWER:
One: no.
Two: This sentence makes little or no sense to me. What I
can tell you is that light falling into a black hole is gaining energy
but it is not gaining speed. Energy of light increases by increasing the
frequency (decreasing the wavelength).
QUESTION:
Is there a way to calculate the weight equivalent of pulling a load of X pounds up and incline of Y degrees? If I tow a 1000 pound load up a 5 % grade, it's like pulling how many pounds on a level?
ANSWER:
There a couple of things ambiguous about this question. First, "weight
equivalent" is not defined; it seems to me that what you want is how
much force you must exert to pull a weight up an incline. Comparing this
force, as I will show below, with how much force it takes to pull it
horizontally is not really meaningful. Second, you really cannot answer
this question without specifying the friction between the weight and the
incline/floor. Suppose that the coefficient of kinetic friction is
μ, the weight is W, and the angle of the incline is
θ. Then
the force to pull it with constant speed up the incline is F_{incline}=W(sinθ+μcosθ)
and to pull it along the floor is F_{floor}=μW.
If there were no friction (like if the weight were on very well
lubricated wheels) and your 1000 lb weight were pulled up a 2.86º
(5%) slope, F_{floor}=0 and F_{incline}=49.9
lb. With μ=0.5, F_{floor}=500 lb and F_{incline}=549
lb. The reason F_{incline }is bigger is that, in
addition to pulling the mass in a horizontal direction, you are also
lifting it.
QUESTION:
i should know this already, but i can't completely understand. in chemical processes, a reaction is either exothermic or endothermic. if you put heat into it, the heat can be extracted. why is fusion and fission not the same? i mean if you break the strong bonds, i can understand a release of energy, but when you create new strong bonds in fusion? gravity provides energy for the fusion in stars, but in reactors or weapons, i don't understand
ANSWER:
There is only one rule you need to know for exothermic/endothermic
reactions: If the sum of the masses after the reaction is smaller than
the sum of the masses before the reaction, it is exothermic; otherwise,
it is endothermic. Chemistry is a really poor source of energy, so the
mass changes are very tiny (essentially unmeasurable), so this was not
known for chemical reactions until we knew that E=mc^{2}
and is not taught in elementary chemistry courses. And, you should avoid
the notion that if you must add energy to make the reaction go it is
endothermic because you might get more energy out than you put in.
Fusion is such a case: in order to fuse two deuterons (^{2}H) to
one alpha (^{4}He) you must overcome their mutual Coulomb
repulsion; but if you compare the masses before and after the fusion,
you will find that their is much less after (an alpha is much more
tightly bound than a deuteron). Similarly, if you compare the masses
before and after a fission of a heavy nucleus, the total mass afterward
is measurably smaller. Fusion (fission) is exothermic for nuclei lighter
(heavier) than iron. That is why stars do not make elements heavier than
iron; heavier elements are created in supernova explosions or other
exotic astronomical events. You might have a look at an
earlier answer.
QUESTION:
Could you please explain why the mass and energy of the Carbon 14 decay into Nitrogen 14 do not seem to add up? Carbon 14 has a mass of 14.003241u, Nitrogen 14 has a mass of 14.003074u, and a beta particle (e) has a mass of 0.000548u. It appears that the mass after the decay is larger than the mass before the decay. In addition, the beta particle is supposed to have energy of up to 156 keV. Doesn't this energy come from the conversion of mass into energy? How is it that there is more mass after this decay as compared to before the decay?
ANSWER:
The masses you use are atomic masses. The ^{14}N atom
left after the decay has only 6 electrons, not the 7 in a ^{14}N
atom, so you do not add the mass energy of the β^{
}(or, if you wish, you add it and subtract the mass energy of the
missing electron).
QUESTION:
In the old days, many types of tags were made out of paper. Nowadays, lots of tags are made out of film, b/c the film is waterproof, grease resistant, and stronger than paper. One application
where film tags are being used is in the grocery store for items like hams, turkey, and fish.
Our company, YUPO, makes film that is used in grocery stores to tag hams, but unfortunately, the grocery store is complaining b/c the tags are failing in the store. I am writing to ask for some help determining how much strength is required from my film to actually work for this application. I hope you can help.
The tag dimensions are 5" long x 2" wide, and 250 microns thick. A key point: of the tag design is that in the middle of small dimension there is a 5/8" wide extension (we call it a neck, b/c it extends out). It extends 2" away from the rest of the tag, so the total length of the tag is 7" long.
The 2" extension gets looped over a string and back onto the body of the tag. This is how the tag attaches to the ham.
Now for the set up to the question. Customers pickup the ham using the tag. They raise it to about shoulder height. They lower it into the cart holding onto the tag. Then, suddenly, at the last second, they jerk the tag upwards so the ham doesn't crash into the other items in the cart.
When the customer jerks it up at the last second, the tag snaps. The hams weigh 10 pounds. I am wondering how much force is required to prevent the tag from snapping. I think this can be explained through physics, but I don't know how to do it. Our lab has equipment that can be used to test tensile strengths, elongation, etc, but I am asking your help figuring out how much strength will be required. Can you help me?
ANSWER:
Every customer is going to lower it differently, so it is, or course,
impossible to give a definitive answer. I have worked out the force
F which the neck would have to exert on an object of weight W
(lb) if the customer simply dropped the object from a height h
(ft) and then stopped it completly in a distance s (in): F=W[1+1.06√(h/s^{2})].
For example, if a 10 lb ham dropped from 2 ft and stopped in 2 in, F=17.5
lb. That would be a pretty extreme case, though, since I would guess
most people would probably lower it at a lower speed than it would
achieve in free falling 2 ft. I think engineers like to insert a factor
of 2 safety factor. Overall, I would guess that the tag should to be
able to handle at least twice the weight of the product.
QUESTION:
I sent you a idea about putting microscopes end to end, or it's called a compound microscope. I thought you could use the scratch proof stuff they put on glasses, to fill in the cracks. But, I think there would still be a seam, so I thought you could make a mold and put the scratch proof stuff on the mold. It still would not be perfect, but it would get rid of the scratches and only make it blurry. What might be new that I thought of is, the blurriness would have a pattern, and a machine could look at the blurry image and draw a picture of what it would look like if it was focused. You could do it if you knew the exact shape of the lens, and look at a colored grid through the lens and that would help see how to focus the picture. Of course you could try a lot of lenses and find one area that is almost perfect. The lenses today are all scratches if you look up close, and I was taught that is the problem with looking at a virus up close. A microscope that enlarges 1000 times could be enlarged 1000 times, and that would be 1000,000 times. I would take like two microscopes.
ANSWER:
The reason that you cannot image a virus with an optical microscope has
nothing to do with the design of the optics (how you arrange lenses like
your idea of using a second microscope to look at the image of the
first). If the size of the object is comparable to the wavelength of the
light diffraction of the light causes the image to be very blurred,
losing all detail. So, if you enlarge a little fuzzy image all you will
get a big fuzzy image.
QUESTION:
Payne Stewart's Learjet was said to have hit the ground at 300400 MPH. If terminal velocity is 120 MPH, how is this possible?
ANSWER:
Where did you get the idea that the terminal velocity of a Learjet in an
uncontrolled dive is 120 mph? I am sure that it is much larger than
that. 120 mph is often given as a typical terminal velocity for a
skydiver, but a jet has much more mass and less drag than a human body
which would imply a much larger terminal velocity.
QUESTION:
I've tried asking Maths folk this question but have yet to receive an answer so I'm thinking perhaps it's more of a physics question:
If I'm standing on Earth how do I calculate the size and distance of the celestial bodies (Moon Sun Planets stars etc) without knowing either the size or distance of any of them?
I've been driving myself crazy.
ANSWER:
Don't go crazy! There are many ways you can make these measurements.
Objects in the solar system are pretty easy. If you know the law of
gravitation, you can relate the period (time of rotation) to the size of
the orbit and thereby deduce where they are relative to earth. Once you
know the distances, you can deduce the size by measuring the angles they
subtend. Outside the solar system is a bit trickier. For stars or
galaxies not too distance, if you measure the position in the sky at one
time and then another time 6 months later (one full diameter of our
orbit apart) you can deduce the distance by
parallax. For more distant
objects, it gets a bit harder and we are getting into the realm of
astronomy in which I am not an expert. By measuring the
redshift of the
spectra of stars we can deduce the distance if we know the
Hubble constant
because the farther away an object is from us, the faster it is going.
Another way to determine distance is to have a certain type of
astronomical object which has the same total lumninosity regardless of
where it is. Such an object is called a
standard candle
and an example is a type Ia supernova; knowing how bright something is
and how bright it looks can be used to determine how distant it is.
QUESTION:
my question is based on newton's 3rd law . An elephant pushes a rabbit with 250 N force , then the rabbit should also apply 250 N force but the rabbit is only capable to apply 25 N force . WHY THIS HAPPEN . according to newton's 3rd law the rabbit should also apply this much of force .
ANSWER:
The rabbit does not need to use his muscles to exert a force. Suppose we
had a toy rabbit made of steel. You cannot ask it to exert a force! But,
if an elephant comes up and pushes down on it with a 250 N force, it
will surely exert an upward force on the elephant. If it were a real
rabbit, the force might squash the rabbit and kill it, but it would
still exert an upward force on the elephant. A classic example is a car
hitting a mosquito, mass of about 2 mg=2x10^{6} kg. If the
collision lasts 10 μs=10^{5} s and the car is going
25 m/s, the mosquito felt an average force of 5 N; the car will feel a
backward force of 5 N and you know a mosquito could not lift a ½
kg mass!
QUESTION:
Please explain in layman's language, what exactly is time dilation? I solve questions on STR and twin paradox but never understood how time can be delayed? According to twin paradox, after completion of journey one will be younger than other.... How can this be possible?? I mean at biological level :\
Also in interstellar movie, 7 minutes in another universes is equal to 23 years in our universe and when that hero reached home, his girl become old while he was still young.
ANSWER:
There are two kinds of time dilation. In the movie Interstellar,
the time dilation is called gravitational time dilation which is a
feature of general relativity, not special relativity; here time runs more slowly because of a strong
gravitational field. In the movie a planet orbits close to a spinning
black hole and time is slowed. It is a bit of a stretch, though. See an
interview with
Kip Thorne who was the science advisor of the movie. The other kind
of time dilation is easier to understand. First, you need to understand
why moving clocks run slow. The best way to understand this is to use
the
"light clock" as an example which I have done in many earlier
answers. In order to find the light clock useful, you must accept that
the speed of light is the same in all reference frames; see my
faq
page for answers which address this. Finally, the twin paradox is
easy to understand; again, see an
earlier answer. Actually, the twin paradox relies more on length
contraction than time dilation to understand because the traveling twin
sees the distance to the distant star as smaller than the earthbound
twin does although both agree on the traveling twin's speed.
QUESTION:
Would it be possible to have a platform in geosynchronous orbit that was tethered to the surface of the Earth, upon which we could build an elevator into orbit?
ANSWER:
A space elevator is not that simple. But it is not impossible either.
See the
Wikepedia article.
QUESTION:
How much force would be needed to tip over a 31 foot tall,22 ton and 12 feet wide.
ANSWER:
You don't want the force, you want the torque. Also, it depends on where
the center of gravity is. I will give you a general solution and you can
calculate the numbers yourself. The green cross is the center of
gravity, h above the ground and L to the right of the
near side of the object. The
smallest force F you can apply is at the top, H above the ground (31 ft
for you, but if you want to solve for F at a different height, just use
a smaller H in the final equation I will give you). The weight
of the object is W, the normal force is N and the
frictional force is f. The object, just about to tip, has N
acting at the right hand lower edge of the object, so N and
f are acting there. Newton's first law tells you that N=W
and f=F. Finally, sum the torques about the right hand lower
edge: FHW(dL)=0 or F=W(dL)/H.
For your case, if the weight is uniformly distributed, i.e.
h=H/2 and L=d/2, then F=Wd/(2H)=22x12/(2x31)=4.26
tons. You must push it until the center of gravity is over the right
hand lower edge and from there it will fall on its own. Also note that
if the floor is too slippery it will slide before it tips.
QUESTION:
If i am traveling along a path with the velocity of 300km/h,and fired a bullet in opposite direction with the speed of 300km/h,what will happen to the bullet?
ANSWER:
There is an old
Mythbusters
episode which perfectly answers your question.
QUESTION:
Imagine a cylinder in space, rotating at the appropriate velocity so that objects objects resting on the rounded surface of the interior experience a centripetal motion as if they were experiencing normal Earth gravity. Except, with Earth level gravity and atmosphere, (ignoring wind) a pressurized balloon with enough lift to rise in the atmosphere will travel "upwards," relative from the surface; while a balloon ascending from the surface inside a rotating "space station" probably wouldn't behave exactly the same, due to the angular momentum of the rotating surface. Would the observer piloting the balloon find himself rising at a vertical angle relative to the surface, rather than straight upwards? Would he eventually collide with the surface at another point? In short, what would be the path of the rising balloon in the centrifuge?
ANSWER:
I love questions about "artificial gravity" in rotating space stations;
there are sometimes surprising results unless you are standing still on
the interior surface and the radius is very large compared to your
height. Rather than going over a lot of preliminary stuff that I have
done before, I recommend that you carefully read through an
earlier answer where I
examine the problem of whether you could play catch in such an
environment. Another
earlier answer addresses what happens if you jump "straight up".
Perhaps the most important thing to take from those answers is that this
is not really like gravity at all because the instant you lose contact
with the surface you experience no forces at all so you move in a
straight line; nevertheless, with a large enough radius, the behavior of
the path as viewed by an observer on the surface can be quite similar to
how it would be on earth, particularly for the straightup jump. Also we
will need to know that the angular velocity ω for the centripetal
acceleration of a cylinder of radius r_{0} to be g=9.8
m/s^{2} is ω=√(g/r_{0}).
Now, why does a hotair baloon rise? It is because of the buoyant force and the buoyant force arises because the pressure on the bottom of the balloon is larger than the pressure on the top. So the first question we need to answer is whether there is a pressure gradient in the space station (i.e. pressure decreases with distance from the surface) and how that compares with the pressure gradient in a gravitational field. In the space station there is certainly a gradient because (as you note) it is simply a giant centrifuge so air will tend to move to larger radii. For an incompressible fluid there is a centrifuge equation for pressure a distance r from the center, P(r)=P_{0}½ρω^{2}(r_{0}^{2}r^{2})=P_{0}½(ρg/r_{0})(r_{0}^{2}r^{2}); here r_{0} is the distance from the center to the outer surface, P_{0}=P(r_{0}), and ρ is the fluid density. Technically, this is not correct because the density varies with r, but we are really only interested in situations where the density is fairly constant; for example, if r_{0}=1000 m, r=900 m, and P_{0}=10^{5} N/m^{2} (approximately atmospheric), P(r)=0.99x10^{5} N/m^{2}. In a uniform gravitational field there is an empirical equation to calculate pressure as a function of altitude h, P(h)=P_{0}[1(Lh/T_{0})]^{gM/(RL)}. Without going into detail, I find that P(h=100 m)=0.98x10^{5} N/m^{2}; we can therefore conclude that the buoyant force in the space station will be very similar to that on earth, at least for the first few hundred meters.
I want to look at what the balloon does from two perspectives—an observer outside the space station and the guy inside who releases the balloon.

From outside, the instant that the balloon is released it experiences no force except the buoyant force B. At that instant it has a tangential speed of v=r_{0}ω=√(gr_{0}). The buoyant force is constant in magnitude but changes in direction since it always points toward the center of the space station. So as the balloon is carried in the direction of the initial tangential velocity and rises due to B, the direction of both B and v changes and the magnitude of v changes.
If the radius is very large (obviously not the case in the figure above), the direction of B will not change much and the path followed will approximately be a parabola. 
When looking from inside the rotating cylinder there are two fictitious forces (see earlier answer for more detail), the centrifugal force F_{r}=mg and the Coriolis force F_{c}=2mv√(g/r), in addition to B. In the figure below I have shown the balloon shortly after liftoff. As you can see, the Coriolis force will deflect the balloon to the right and F_{c} will get larger and change direction as v turns to the right and gets larger and r decreases. So, if there were a ceiling (space station like a torus) it would definitely not strike directly above the launch point.
Again, if r_{0} were large enough, the effect would be minimal because F_{c} would never get large enough to cause a significant deflection because of the 1/√r term.
Note that I have not considered another possible force, air drag. I believe this would have a small effect and would not alter the conclusions above. You were right to speculate that the result would be different from on earth but angular momentum has nothing to do with it; it is simply due to the Coriolis force.
FOLLOWUP
QUESTION:
What would happen if an observer got into a car and accelerated beyond the rotational velocity of the
cylinder in the same direction, once the vehicle lifted from the surface? Would it then glide for a certain period until, from the perspective of another observer inside the centrifuge, descending back onto the surface at a slightly slower speed? Also, the second part to
the question is, would the driver experience weightlessness if he was slowed enough while airborne? Would any airborne object, for that matter, potentially experience weightlessness if slowed or misdirected?
ANSWER:
I will first look at the situation from inside the space station, the
rotating frame. The diagram below shows the car moving with some speed u
in the same direction as the cylinder is rotating as you stipulated.
There is one real force acting on the car, the normal force N from the "floor". There are two fictitious forces acting, the centrifugal F_{r} and the Coriolis F_{c}, both pointing radially outward (the Coriolis force is F_{c}=2muxω). So, you see, it has gotten "heavier" rather than "lighter" because of the Coriolis force and will not "lift off". So, what happens if you travel opposite the rotation direction as in the figure below?
Now the Coriolis force points radially inward, so there will be a speed where N=0. Now, it gets a little tricky to understand what happens as u increases. The first thing you have to understand is that the car, in the frame of the rotating cylinder, is going around the circle; so the sum of all the forces must add to the centripetal force: mu^{2}/r_{0}=Nmg+2mu√(g/r_{0}). So, when N=0, u^{2}+gr_{0}2uv=0=u^{2}uv. So, when u=v, N=0. This makes total sense because if you look at the car from outside, it is at rest and, since this is taking place in empty space, there are no forces whatever on the car. In this situation, any small force with a radial component could launch the car and, as seen from outside, it would simply move in a straight line with constant speed (no forces on it once it loses contact with the wall) until it encountered the wall; the car and all its passengers are weightless. But there will never be a "lift off" situation; if you view the situation from from the inertial (outside) frame, for any u≠v the car will be traveling in a circle and therefore require N≠0.
I must confess that the fact that all the forces on the car must add to the centripetal force mu^{2}/r_{0} eluded me at first. If you set NF_{r}+F_{c}=0 you find u=½v for N=0 and this just did not seem right to me. I therefore posted a question on Physics Forums and got a great answer from Simon Bridge which set me straight.
FOLLOWUP
QUESTION:
This is my final question about being airborne in a large atmospheric centrifuge in space: If you decided to locate yourself in the very center of the sphere, having no rotation, and you were then pushed toward the rotational equator at a walking pace, what would happen to you before you reached the surface, or rather how do you figure it out? Toward the surface, wind current would be rotating in the centrifugal direction, and the resistance would surely begin to alter your trajectory from "vertically, straight down" to "diagonally," from an observer on the equator, on a curve toward the surface. Assuming that the sphere is large enough that the accelerating effect of wind has enough time, would you ultimately be able to meet the surface on a landing curve that merges into the circumferential curve? And, if you accelerate too much, would you skip like a stone or a soccer ball as you collide with surface, slowing each time until you reach the rotational velocity? Ideally, you would be wearing a protective suit.
(Note from The Physicist: in earlier parts of this discussion I have
edited out "sphere" in the questions many times and replaced with
"cylinder" because a sphere is not a good model for pseudo gravity
because only at the equator of a sphere would the apparent gravity mimic
earth's. My answer below replaces "sphere" by "cylinder" and "center" by
"axis")
ANSWER:
If the initial push gave you a speed of 5 mph=2.2 m/s and the radius of
the cylinder were 1 km, here is what would happen. The speed of the rim
would be about v=√(gr_{o})≈100
m/s=≈220 mph; this would be be the "wind speed" at the rim as
seen by an observer not rotating with the cylinder, including you
starting at the center; this speed would decrease linearly to zero on
the axis. It is easiest to view your gedanken from outside, but
I will discuss both observers. I will also discuss two scenarios, with
and without air.
 If there were no air in the cylinder:
 If viewed from outside, you would move in a straight line away from the axis with a constant velocity 2.2 m/s until you hit the rim (in about 1000/2.2=455 s=7.6 min) moving tangentially relative to you with speed 100 m/s—bad news for you!
 If viewed from inside, the centrifugal force would would be
radially out and the Coriolis force would push you azimuthally
in the direction opposite the direction the cylinder rotates.
You would still hit the rim with a relative tangential speed of
100 m/s and a radial speed of 2.2 m/s (which implies your
landing speed in the rotating cylinder frame is about 100 m/s)
after 7.6 minutes. Since the period of the cylinder's rotation
is about 63 s (T=2πr_{0}/v),
your path would spiral around 7.2 times before landing.The paths
might look something like the picture below, the blue being the
outside view and the black the inside view.

But, there is air, so you would have two real forces acting on you, the buoyant force acting toward the axis and the air drag acting opposite your velocity. Earlier in this discussion it was concluded that the buoyant force would be comparable to that on earth, and you certainly would not expect the buoyant force to have much effect on you here; it would be a very small force pointing toward the axis causing you to slightly slow down vertically. But the wind is a different story.

If viewed from outside, as you started falling air drag would be very small because the wind speed would be small, your speed is small, and because the air density would be low. It is too difficult to try to calculate this with any precision, but clearly what is going to happen is that the outside observer would see you spiraling out from the axis in the same direction the cylinder was rotating. Given the large speeds near the end (more than a category 4 hurricane), it is indeed likely that you would land with a speed close to the speed of the rim; this would be a much better scenario on landing than if there were no air. You might think that the force due to 220 mph winds would hurt you, but as you acquire more and more tangential speed, the wind speed you see becomes smaller and smaller. The total fall time would be much longer than 455 s because of the large tangential velocity you acquire; the average acceleration over the whole trip would be much less than 100/455=0.22 m/s^{2}=g/45 because the time is much longer.

If viewed from the inside, the centrifugal force would give you an acceleration g radially out (toward the rim). Early in the descent the Coriolis force would accelerate you in the direction opposite the direction of rotation; as the velocity acquired an azimuthal component, the Coriolis force would have a radial component radially inward, opposite the centrifugal force. The air drag would always act opposite the velocity vector, so you would never reverse the direction (you never rotate with the cylinder). It is interesting that the inside observer sees you going against the rotation but the outside observer sees you going with the rotation; this would mean that your angular velocity is always smaller in magnitude than the angular velocity of the space station.

QUESTION:
If there are two objects, object a (a ball) and object b (a wall), how is it possible that the ball can get closer to the wall for all infinity and never touch the wall? What i mean...say the ball is ten meters away from the wall, then 5 meters, then 1 meter, then 99 cm, 50 cm, 1 cm, 50 mm, 1 mm, and the cycle can go on and on forever. Because numbers are infinite the ball can get closer to the wall forever without ever touching it. What is interesting about this question to me is that the ball and the wall are moving at different speeds in the same direction. The ball is always moving closer to the wall, so that means that the ball must be traveling faster than the wall. Now if the ball is getting closer to the wall infinitely (1 m, 1 cm, 1mm, and on and on forever, that means that the ball is traveling faster than the wall forever. So how can object A, the ball, travel faster than object b, the wall, for an infinite amount of time in the same direction but never touch or pass the wall?
ANSWER:
First of all, it makes no difference whether the wall is moving or not;
all that matters is the relative velocity, the rate at which the ball is
approaching the wall. Secondly, your question is just a variation of
Zeno's paradox. See an earlier
answer.
QUESTION:
myself and a friend were discussing this one day when we were bored if you were to fire a projectile and suddenly remove all the wind resistance (think of driving your impossibly fast car alongside said projectile at an identical speed with the window down and moving sideways into the bullets path so that it is now inside the car ) what effect would this have upon the projectile??
we came up with two hypotheses
frst being that the reduction of wind resistance that the windshield affords means the projectile suddenly has a lot less holding it back and would speed up and the second and more likely outcome is that the projectile would continue to decelerate but at a much slower rate owing to the fact that it is now travelling in a pocket of air that is traveling at the same speed as itself. and cannot possibly speed up due to it already being fired thus it has no way of gaining any energy from anywhere
ANSWER:
So, I assume your magic flying car is keeping pace with the bullet
before it comes through the window. Due to the air drag on the bullet,
it is slowing down and so you must also be slowing down. When the bullet
gets inside the car is no longer slowing down, but your car is.
Therefore the bullet will appear to you to accelerate toward the
windshield but you and I both know that what is really happening is that
the car is decelerating toward the bullet.
QUESTION:
Four forklifts are parked in a square like configuration, so that each forklift has its forks under the other. If all four were to be commanded to lift simultaneously, what would be the outcome?
ANSWER:
All four would mysteriously levitate above the ground. Just kidding!
Your forklift would feel four forces: its own weight, down, the force
the floor exerts up, the force which one neighboring forklift exerts up
trying to lift yours which is up, and the force which your other
neighboring forklift—whom you are trying to lift—exerts on
yours. Now, what is the direction of that last force? You are exerting
an upward force on him and therefore he is exerting an equal and
opposite force on you, down; this is due to Newton's third law.
You can conclude that all the forces which the four exert on each other
cancel out if you take the ensemble of 4 as the object to focus on. As
long an nobody tips over, nothing will happen.
QUESTION:
What happends to a sound wave after it is created? How does it end up? Or does it last forever?
ANSWER:
Any sound wave loses energy as it travels because of internal friction,
viscosity in fluids. Also, the intensity of the sound decreases
approximately by 1/d^{2} where d is the
distance from the source. Eventually, the intensity will drop below the
sensitivity of the ear. At some intensity the medium will no longer be
able to support the sound wave, but when that happened would be
complicated to understand and depend on the medium properties and the
frequency of the sound.
QUESTION:
I work on a maintenance team in a production facility and i am trying to make a conveyor that moves pallets of cardboard. I have an A/C motor with a sprocket and chain connected to a driven shaft turning the conveyor. My problem is that the pallet is too heavy and my only options are to buy a 3 phase motor or mess with the sprockets to change the gearing. Now to my question, will increasing sprocket size on the driven shaft, therefor decreasing speed, actually increase the torque and therefor increase the amount of weight it can move, or just lower the speed?
ANSWER:
The problem is friction, not weight. If there were no friction, you
could move any pallet with any motor; and, once you got it going (it
might take a while to bring it up to speed for a large mass and small
motor), you could turn off the motor and it would keep moving. The
friction gets larger as the load on the conveyor gets larger. Suppose
that the motor is capable of exerting some torque τ_{motor};
if the radius of the gear or pulley attached to the motor shaft is r,
this results in a tension in the chain or belt of F=τ_{motor}/r.
Now, if the gear or pulley on the drive shaft of the conveyor is R,
the resulting torque on that shaft will be τ_{conveyor}=FR=τ_{motor}(R/r).
So, yes, increasing the size of the drive gear increases the torque.
This is the physics of an ideal situation. I am not an engineer!
QUESTION:
My text book says the direction of displacement and velocity are always the same. that's incorrect isn't it? Cos direction of velocity depends on the direction of movement, and the direction of displacement depends on the initial position right?
ANSWER:
No, that is not incorrect. In fact, the definition of velocity
incorporates this very equality. Start with average velocity. If an
object has a position R_{1} at time
t_{1} and a position R_{2}
at time t_{2}, then the displacement vector
D is defined as D≡R_{2}R_{1}
and average velocity vector is defined as v_{avg}≡D/(t_{2}t_{1}).
As you can see, the average velocity and the displacement are in the
same direction. To get instantaneous velocity you find the limit of the
average velocity as the time difference approaches zero. If you know
calculus, the instantaneous velocity is v=dR/dt;
again, velocity v and displacement dR
are in the same direction.
FOLLOWUP QUESTION:
If a boy moves from position A, towards the East direction 100 m, then turns around and moves 60 m in West direction. The displacement would be 40 m in East direction. The velocity would be (lets say he was moving at 30 ms1 instantaneous velocity) in the west direction right? Or lets say we draw a displacement time and a velocity/time graph each for the movement. The displacement would have a positive value at the end of the movement (+40 m). But the velocity wud have a negative value, (if we considered velocity in East direction as positive). am i correct?
ANSWER:
You are correct but your example is meaningless. It makes no sense to compare the displacement over two different times with the instantaneous velocity at the second time—you could get any answer. Read my first answer carefully. The relationship between velocity and the corresponding displacement over any time interval is clearly defined, even in the limit that the time interval shrinks to zero.
QUESTION:
How do you convert sound energy into electrical energy?
ANSWER:
There are many ways. I suggest you read the Wikepedia article on
microphones.
QUESTION:
I have come across this interesting fact about black holes just about every time i research on them: their gravitational force is so strong that even light cannot escape. However, what i do not understand is that in school, i learnt that gravitational force F= GMm/r^2. if light photons are packets of energy with NO mass how is it that they experience this gravitational pull? is it to do with massenergy conversion?
ANSWER:
What you learned in school is only part of the story. The theory of
general relativity is the modern explanation for gravity, Newtonian
gravity is empirical, originally designed to explain the motion of
objects in our solar system. In general relativity, the presence of mass
actually deforms the space around it and this results in light passing a
heavy object being bent, apparently feeling a force. I would suggest you
read some of the earlier answers about
general relativity linked to
on the faq page. Incidentally, only photons inside the
Schwartzchild radius R=2GM/c^{2}
will be captured.
QUESTION:
I recently saw my son spinning an object on a wire overhead
& wondered how fast is the object at the end moving in mph? Obviously there are many factors to consider, but let's say a very athletic man was spinning an ideal weight (let's say a marble or ball bearing of equal size) on an ideal length of wire (let's say 3 feet) in a circular motion similar to a sling
& a stone of days past, what approximate range or max speed could the marble/ball bearing reach?
ANSWER:
I cannot imagine a frequency of more than about f=5
revolutions/s. If the radius of the circle is R=3 ft, the
velocity is v=2πRf=2x3.14x3x5=94 ft/s=64
mph.
QUESTION:
1 light year equals to how many earth year?
For example one of the orion's belt star is 900 light years away. Is this mean that 1 day in the star equal to 900 years on earth?
I assume that time would be slower on earth? Judging from Einstein's gravitational time dilation?
ANSWER:
You have this all wrong for the following reason: a light year is not a
measure of time, it is a measure of distance. A light year is the
distance that light travels in one year, so if you aimed a light source
toward this star, it would take 900 years to get there. Likely time
would run at about the same rate in the close vicinity of this star
because gravitational time dilation is extremely small unless the
gravitational field is very strong—like in the vicinity of
a black hole or neutron star.
QUESTION:
Rod tied to a string tilts vertically but when it is rotated it becomes horizontal Why ???
Can't seem to find any answer
ANSWER:
The easiest way to see this is to introduce the (fictitious) centrifugal
forces shown in the figure above. As you can see, both exert a torque
about the suspension point which will tend to make the rod horizontal.
When the rod becomes horizontal, the forces are still there but no
longer exert torques.
QUESTION:
Hi, I am a mathematician (graduated some years ago) and I am currently studying physics because I will take admission exams in Physics Departement of my local university (in about 9 months).
While studying (from the book "Physics for Scientists and Engineers" by Serway and Jewet) I found a very interesting problem:
I'm not asking you to solve this. I did it myself. But the physical meaning eludes me. [Here the questioner adds a bunch of mathematics discussion which is not really relevant to the physics, but the crux of what eludes him is, since this is a quadratic equation, what are the physical meanings of the two solutions and does their average have any physical significance.]
ANSWER:
Normally, I discard this kind of question because the purpose
of the site is not as a homework helper or tutoring service. I emailed a
glib answer without really analyzing the problem. But when I looked more
closely I realized that the problem did not really seem to make sense;
in particular, where did the 2.80 m come from and why is the potential
energy term involving x on the same side of the equation as the
kinetic energy term?
The equation clearly is intended to represent modeling the problem of a m=46 kg object (child) moving with a speed v=2.4 m/s when landing on an ideal spring with spring constant k=1.94x10^{4} N/m. This is a standard introductory physics energy conservation problem. If you choose y=0 at the top of the spring and the +y direction as vertically up, the energy conservation equation is ½mv^{2}=½ky^{2}+mgy where y is the position(s) where the mass is at rest. This says that the kinetic energy of m when y=0 (initial energy) equals the potential energy of the spring plus the gravitational potential energy of the mass when the object is at rest. Note that this equation assumes that the mass is always attached to the spring and so the two solutions are how far the mass travels in the negative direction (y<0 solution) and how far it rebounds in the positive direction (y>0 solution). These two solutions are the extremes of the resulting oscillation of the mass and the oscillations are symmetrically about the location where the mass would be at rest if not moving, y_{rest}=mg/k. Since this is the center of the oscillation, ½(y_{1}+y_{2})=y_{rest}. The solutions for the problem at hand are y_{2}=+0.096 m, y_{1}=0.142 m, y_{rest}=0.023 m; note that ½(y_{1}+y_{2})=y_{rest}.
If you solve the equation in the problem as stated, you find that the positive solution is greater in magnitude than the negative solution and that the average is not the equilibrium condition. I believe that what the author intended was that the stack of mattresses was 2.8 m high, the floor was chosen as the zero of gravitational potential energy, and the final gravitational potential energy was moved to the left side for compactness. This would mean that the final potential energy assumed by the author was mg(x) and therefore x must mean the distance that the mass went down. But that is wrong—the final potential energy should have been mg(2.80x). This would then reduce identically to my equation except that x=0 at the position of the unstretched spring and the +x direction is vertically down. The problem as stated is incorrect.
QUESTION:
How high off the ground would a vehicle weighing 6200 lbs have to be to reach a speed of 45 mph before it impacts the ground?
ANSWER:
If you neglect air drag, the height above the ground, h=v^{2}/(2g),
would be about 68 ft. If air drag is included, I estimate that the
terminal velocity (see earlier answer) would be
about 262 mph. Since this is so much larger than 45 mph, I judge that
air drag would not be important and 68 ft would be your answer.
QUESTION:
How is the gravitational force directly proportional to the product of masses and inversely proportional to square of distance between the masses? i just want to know how they proven it ?
ANSWER:
It is simply an experimental fact. This hypothesis describes the
motion of the bodies in the solar system almost perfectly.
QUESTION:
What factors affect light intensity and how?
ANSWER:
I guess you want a definition of intensity. It is simply the
amount of energy passing through an area per second divided by the area.
Physicists prefer to measure this as watts per square meter, W/m^{2}.
But, to discover how convoluted the measurement of intensity can be, see
an earlier answser.
QUESTION:
Why don't oranges being carried in a semitruck get crushed by all the oranges on top?
four layers are shown
ANSWER:
It is because of the way that spheres tend to
pack
together. As you can see by carefully examining the figures above, each
orange is pushed down by three oranges from above, is pushed up by three
forces from oranges below, and is pushed horizontally by six oranges in
the same layer. These twelve forces all are directed toward the center
of the orange and all add up to zero net force, but there is a
net pressure over the surface of each orange approximately trying to
squeeze it into a smaller orange. But an orange is mostly water which is
nearly incompressible so the orange does not get crushed. Think of a
nicely packed snow ball: if you try to crush it into a smaller ball by
squeezing with cupped hands you will fail; to crush it you will have to
flatten it by pushing it with diametrically opposed hands. You might
think the bottommost layer would get crushed because there is just one
force pushing up; but, each orange has six others in the same layer
pushing toward its center and these keep it from getting flattened.
Here is a little more about sphere packing. There are two possible packings which achieve the maximum density of π/(2√3)≈74%: hexagonal closepacked (HCP) facecentered cubic (FCC); these are compared in the figure above (HCP on the left).
FOLLOWUP
QUESTION:
What would happen to the oranges once the force exerted upon them reached a critical strength which they couldn't bear? Does the entire group of oranges burst simultaneously in a flood of juice?
ANSWER:
Because the forces are not spread uniformly over the surface
area, there would be a tendency to be squeezed to a different shape but
still approximately preserving the volume of each orange. And I would
not expect it to happen all at once because the oranges on the bottom
are certainly experiencing greater forces. I would expect the
outoflayer forces (from above and below) to tend to flatten the
lowerlayer oranges but the neighbors in the same layer to cause
the oranges to have a hexagonal shape, so the oranges would tend
toward a hexagonalprism shape. Each orange would occupy about
the same volume, but the amount of empty space (previously 26%)
would decrease and each layer would get thinner. This ignores
the possibility of orange peels rupturing, but I would think
things would tend toward this shift before much juice flowed!
QUESTION:
What is the effect of mass on torque? A wind turbine fan's blades are commonly very long to increase torque and to decrease speed.
How can I decrease speed using MASS? Or, can I increase torque, by increasing of mass, without increasing length of the blade? (without losing the energy.)
What is the formula applicable here?
ANSWER:
You are asking many questions here with no simple answers.
The simplest place to start is your first question: does the mass of the rotor have an effect on the torque on it? Typically, the turbine has three blades. I will just analyze a single one and the same arguments could be made for the other two. Call the length of the blade L and assume that the force on it due to the wind is approximately uniform along the length of the blade (the force on a tiny piece of the blade near the center is the same as the force on an identical tiny piece near the end). Then the total force F due to the wind will depend on the length of the blade, but the force per unit length, Φ=F/L will be more useful because it will depend only on how hard the wind is blowing. It is now pretty easy to show that the torque due to the wind is τ_{wind}=½ΦL^{2}. So, the answer to your question is no, mass does not affect the torque; the torque depends only on how hard the wind is blowing and how long the blade is.
Your second question is how can you decrease speed by changing the mass M. If I model the blade as a uniform thin stick of length L, its moment of inertia is I=ML^{2}/3. If it has an angular velocity ω_{1}, its angular momentum is L_{1}=Iω_{1}=Mω_{1}L^{2}/3. If you increase the mass to M+m, the moment of inertia will increase to I'=(M+m)L^{2}/3 and its angular velocity will change to ω_{2}. But, the angular momentum will not be changed, Iω_{1}=I'ω_{2}; you can then solve this for the new angular velocity, ω_{2}=(I/I')ω_{1}=[M/(M+m)]ω_{1} which is smaller. However, the rotational kinetic energy E of the blade is now lower, E_{1}=½Iω_{1}^{2} and E_{2}=½I'ω_{2}^{2}=½I{(M+m)/M}{[M/(M+m)]ω_{1}}^{2 }or E_{2}=[M/(M+m)]E_{1}. On the other hand, if you wanted to add mass but keep the energy the same, E_{2}/E_{1}=1=I'ω_{2}^{2}/Iω_{1}^{2} or ω_{2}=ω_{1}√[M/(M+m)]; in this case, the angular momentum will have changed.
Your third question is moot since we have established that torque does not depend on mass.
QUESTION:
If I stood beside a small operating hovercraft with a sail built into the front of it and blew air into the sail with a leaf blower I know that the craft would move forward. Now the question.If I sat down on the hovercraft with the leaf blower in hand and we became one with the hovercraft and I then blew air into the sail would we move forward or would action and reaction of the leaf blower neutralize the forward motion?
ANSWER:
It is easiest to understand if you think first of using a stick
instead of a leaf blower. Standing on the ground and pushing with the
stick on the sail, there is an unbalanced force acting on your
hovercraft (the stick). Now, if you stand on the hovercraft, the stick
exerts a backward force on you (part of the hovercraft, now) and the
stick exerts a forward force on the hovercraft and these cancel out. Or,
if you like, the only forces which have any effect on a system are
external forces and by becoming part of the system what you do is no
longer an external force. The leaf blower is a little trickier, but I
believe even worse! The leaf blower will exert a backward force on you
(like a little jet engine) and the stream of air will exert a forward
force on the sail; but some of the force from the stream of air will be
diminished by the air slowing down on its way to sail because of
interaction with the still air. So, the net effect would be for the
whole hovercraft to move backwards; probably not noticible because of
friction and the smallness of the loss of power due to the still air.
QUESTION:
Helium. Where do we get it from if it is lighter than air and doesn't react with any other elements in the normal human tolerant environment?
ANSWER:
Good question. Even though it is the second most abundant
element in the known universe, there is virtually none in the atmosphere
(because it is so light that its average speed is greater than escape
velocity and it shoots off into space) and is not tied up in rocks,
water, or other chemicals (because it is inert) like hydrogen is, for
example. This element was not even discovered until 1868 as a spectral
line in the sun (where untold zillions of tons are being produced every
second from nuclear fusion) and not found on earth until 1895 when trace
amounts were found coming from uranium ore; the source was as a nuclear
decay product in αdecay. The first large amounts were
discovered in 1903 as a byproduct mixed with the methane in natural gas
wells; today large scale amounts come only from helium trapped
underground.
QUESTION:
Why nucleon number is the sum of the number of protons and neutrons instead of the sum of the number of electrons and neutrons or between the number of protons and electrons , explain the logic ?
ANSWER:
It is very simple. Nucleon means either proton or neutron; a
proton is a nucleon and a neutron is a nucleon. Nucleon
number means number of nucleons. Electrons are not
nucleons.
QUESTION:
If basketball (A) weighs 1 lb and is tossed upward to goal (A) at a height
of 8 ft (5 ft above my daughter's head) and basketball B is 6lbs, At what
height should goal (B) be to generate the same force to toss?
FOLLOWUP QUESTION:
Actually this is not homework, let me explain the situation. My daughter is 5 years old and playing Kindergarten basketball. Only one person on her team can toss the ball high enough to score, the ball weighs roughly 1 lb and the goal is roughly 8 ft high. I purchased a weight trainer ball that is exactly the same diameter as the regulation ball she uses but it's a 6 lb ball.
My theory is that I can build her a goal in the house that is not as tall but would require the same energy to make the basket. therefore making her stronger. And when it comes time to shoot the lighter ball in the taller goal, she shouldn't have any problems.
ANSWER:
For the 1 lb ball, the energy which must be supplied is 1x5=5 ft·lb,
assumning that she releases the ball at the level of the top of
her head. The 6 lb ball, if sent vertically with an energy input
of 5 ft·lb, will rise to a height of h=5/6=0.83
ft=10 inches above her head. All this assumes that the ball is thrown
straight up. Note that I have not really answered your question
because you asked for force and the energy input depends both on
force F and the distance s over which it is applied.
The energy input W could be written as W=Fs,
So, if you assume that she throws it the same way and pushes as
hard as she can, the force need not be known.
QUESTION:
I am curious about a topic. In golf, if I hit a ball very hard and then I hit one very softly, is the one
hit very softly more likely to move or sway in its straight path?
ANSWER:
You refer to "its straight path". No golf ball goes in a
straight path, so I presume you mean that it does not curve left or
right; such a ball, if not curving, would have a projected path on the
ground (like the path of its shadow) which is straight. For a
righthanded golfer, a ball which curves right is called a slice
and one which curves to the left is called a hook; these have
opposite spins. Neglecting the possibility of wind, the reason that a
ball curves is because it has spin. But now it gets complicated because:

the hardhit ball is in the air much longer than the softlyhit ball;

the lateral force causing the curve depends on both the rate of spin and the speed of the ball, so the hardhit ball will experience more lateral force than the softlyhit one if they have the same spin;

even if the slow ball has a bigger lateral force, the fast ball is likely to be deflected a greater distance because of its longer flight time;

a lateral wind will exert the same force on both, but the fast ball will be deflected farther because of the longer time.
So, you see, there is no simple answer. To avoid curving, learn to hit the ball without imparting significant spin!
QUESTION:
We had a phenomenon happen recently about 8:30 PM that is inexplicable to us, but there must be some explanation. My wife
& I were in separate rooms when we both heard an extremely loud noise from the living room! The noise sounded like a large glass that just hit a hard tile floor, but loudness was magnified. As it turned out we came into the living room to find a glass platter that we had sitting on the coffee table for about a year just shattered. –It broke completely by itself as there was no one in the room.
Do you know how this may have shattered/blew apart all by itself?We used the platter to put 3 little oil lamps on.
Inside our house the next morning at 6 AM we heard thunder outside so thought it might have had something to do with the barometric pressure.Very low barometric pressure
& the type of glass it was made up combined just right to explode it like that? The temperature was a constant 68 degrees as it was for the months that was on the table.
If you have any idea about this, we would appreciate it.
ANSWER:
Glass, as you know, is manufactured at very high temperatures.
It has a quite large coefficient of theremal expansion (a large
change in size for a small change in temperature) and is a poor
conductor of heat. This means that as it cools it does not all
cool at the same time. This can result in very large stresses
being "frozen in" at some locations. What causes it to
spontaneously break is usually difficult to determine; most
likely it had recently been bumped or your oil lamps might have
caused hot spots on the glass. Such things could have caused a
tiny fracture to begin and the final shattering could easily
come at some unpredictable later time. Unusual but not
unexpected.
QUESTION:
On a skate board going down a .5 mile hill at 45 degrees slopes if I weigh
187 lbs how many mph would I be going by the bottom of the slope.
FOLLOWUP
QUESTION:
This isn't homework I'm a 35 year old heavy equipment operator and my son had a accident on skate board and we are curious how fast he was going when he wiped out.
I just wasn't sure how accurate I was when I said about
3035mph. Please if you don't know just tell me so I can find someone who does—we got bets on it now amongst the family.
ANSWER:
Wow, 45º is pretty darn steep! A half mile would
correspond to his having dropped by about 0.35 miles≈560 m. If
there were no friction at all his speed would have been v=√(2gh)=√(2x9.8x560)=105
m/s=235 mph! Back to the drawing board! There is some friction due to
the wheels and bearings and I estimate that this is probably not more
than about 15 lb; this would only slow him down to about 99 m/s=220 mph.
Back to the drawing board! Finally, since the speed is going to get
pretty big, we need to take air drag into consideration because the drag
force is proportional to the square of the speed. A rough estimate would
be that the force is about F_{drag}≈¼Av^{2}
where A is the area his body presents to the onrushing wind. When F_{drag}
is equal the net force down the incline (component of weight minus
friction, which I estimate to be about 117 lb=520 N), he will stop
accelerating. Taking his area to be about 2 m^{2}, you can then solve ¼Av^{2}=520 to get
v=32 m/s=70 mph. This is all very rough but should give you an
orderofmagnitude estimate. (I still find it hard to believe that he
went down a half mile, 45º slope without braking at all!)
QUESTION:
I am curious about generating power in space. Why do they always use solar instead of the windmill type of generation?
A coil/magnet rotating. It seems to me, once the rotation is started, it would continue forever? Thus if you used a rocket to start the rotating part of the generator, and it kept spinning, could you use the magnetic field to protect say, an astronaut inside the generator? If it was big enough. Would you get perpetual energy if you used the electricity created in say, a microwave rocket engine or electromagnet. Or does the magnetic force alone cause the spin to lose momentum?
ANSWER:
So, you start something rotating in a vacuum and it never stops
because there is no air drag. You could even imagine making extremely
lowfriction bearings so you could mount this on the side of your
spacecraft and it would at least spin for a very long time before
slowing down. But, the minute you hook it up to a generator you are
asking it for energy so it immediately begins to slow down, giving its
kinetic energy to you to power a light bulb, maybe. There is no free
lunch in this universe, and if you want energy you need something to
give it to you and the sun is the most convenient source in our
neighborhood.
QUESTION:
Is dark energy real? If all matter in the universe expanded from a single point the size of an atom (the big bang) wouldn't things be moving faster from each other because of geometry? If you start with a sphere the size of an atom and it expanded outwards over billions years even a Planck length size degree difference would be immense. It would cause things to move farther apart at faster rates as time went on. Are things moving faster away from each other because of geometry and angles, and not from hypothesized dark energy?
ANSWER:
I do not usually answer questions in
astronomy/astrophysics/cosmology but think I can answer this
one. If a collection of objects interact only via an
attractive force (gravity in this case), any one of them can
only speed up when moving toward their center of mass. The
details of the motion would be determined by the initial
conditions. If all the objects were moving away from some common
point at some time, the only possible motions would be

for all to move forever away from each other, but forever slowing down;

for all to slow down and eventually turn around and speed up back together; or

for some to come together and some to keep going.
The simple reason is that the potential energy of such a system increases as the objects get farther apart so the kinetic energy must decrease to conserve energy.
QUESTION:
But I read somewhere that the faster you travel through space, the slower you travel through time, and if you reach light speed, time stops. If that's true, why does light have a speed, instead of just being instant? Does time not stop for light when it travels?
ANSWER:
First, you cannot reach light speed, no object with mass can; so
let's not talk about how fast your clock would be running if you went
light speed. What happens at high speed is that your clock will run
slow when measured by an observer you are passing. To you, time would
seem perfectly normal; however you would observe distances along your
line of travel to be shorter and therefore you would take a shorter time
to get there. Now, if you are traveling at almost the speed of light,
say 99.99% of it, I would still see you traveling at that speed
regardless of what your clock is doing; so, if the photon had a
(nonchanging) clock on it, I would still see it going at the speed of
light. Regarding whether time stops for a photon, my stock answer is
that a photon does not have a "point of view" and it is pointless to ask
how fast a photon's clock is moving because a photon does not carry a
clock with it.
QUESTION:
Does the existence of gravitational waves imply the existence of gravitons?
ANSWER:
No, gravitational waves have nothing to do with gravitons.
Gravitational waves are predicted by general relativity, the
best current theory of gravity. Gravitons would be the quanta of
the gravitational field is a successful theory of quantum
gravity is ever devised. You can look in the faq page for
earlier answers about
gravitons
and
general relativity.
QUESTION:
I've been dealing with a false prophet who says that a comet is coming and is going to skim the earth, as if to skip off of it, like a stone skipping on water. Is this even possible? She says it will skip off of the earth and keep going into space. Please let me know if this is even possible?
ANSWER:
Yes, I believe this is indeed possible. You might recall that
during the Apollo 13 failed moon mission there was concern that if the
spacecraft
reentry angle were too small that they would "skip off" the
atmosphere into space.
QUESTION:
If it's true that oceanic tides can be caused when the
moon's gravity pulls the molecules of ocean water up and away from earth by
a certain distance, and if it's also true that earth's atmospheric tides
can, likewise, also be caused by the moon's gravity pulling the molecules in
the atmosphere up and away from earth by a certain distance, then what stops
the atmospheric molecules, once they have accelerated even just a tiny
distance in the direction away from earth and towards the moon, from
continuing on their journey to the moon? (Let's, for example, say that the
atmospheric molecules in question areto simplify mattersthe ones at the
very top of earth's atmosphere, so that no other atmospheric molecules are
between these particular atmospheric molecules and the moon, as the moon
pulls on them.)
ANSWER:
This can get very complicated because the molecules in the air
have a whole range of speeds from very slow to very fast, but I do not
think that that complication is important to answer your question. The
reason that the molecules do not fall to the moon is the same as if you
throw a ball straight up and it does not fall to the moon—the
earth is pulling on it harder than the moon is. How anything moves is
determined by the net force on it and its weight (the force the earth
exerts on it) is bigger by far than its "moon weight".
QUESTION:
I need to figure out the force of impact from an object weighing 3.25 lbs falling from 2'
I have on reference that a five lb object falling two feet creates a force of impact of approximately 319 lbs.
This is not a school question. I got hit in the head with this object.
ANSWER:
I always try to emphasize that you cannot know how much force an
object exerts when it hits unless you know how quickly it stops (or,
equivalently, how far it goes while stopping). I find that the object
was going about 11 ft/s when it hit your head. Suppose that it stopped
in about 1 inch; in that case, the average force during the time of
stopping would have been about 75 lb. Had it stopped in ½
inch, the force would have been about 150 lb.
QUESTION:
If photons are the charge carriers for the electromagnetic force, then why don't magnets and power lines glow in the dark? Is it because those photons are not in the visible part of the spectrum or is it something else?
ANSWER:
Your terminology is a little off. Photons are the quanta of the
electromagnetic field; we then think of them as the "messengers of the
force" communicating the force among charged particles. However, they
are virtual photons which means that they pop into and out of
existence very quickly, too quickly for you to observe them—hence,
no glow! Also, if you think about it for a minute, if you saw a glow and
the fields did not change, that would violate energy conservation.
QUESTION:
If the planet earth was perfectly smooth and spherical will the water cease to flow?
ANSWER:
Not if everything else stayed the same. If the earth were
completely isolated, not rotating, and without atmosphere, water would
flow until it formed a uniform layer over the earth; eventually any
currents would damp out due to the viscosity of the water. The fact that
the earth is rotating and heated by the sun and has an atmosphere would
mean that the water would try to distribute itself mostly uniformly but
with an equatorial bulge; however heating and cooling of the atmosphere
would cause weather patterns and the resulting winds would move the
water around just like what happens today. Also, the moon causes tides
which are, by definition, motions of the water. You probably could think
of many more reasons the water would not become totally static.
QUESTION:
I am trying to figure this out for a dear Uncle on Vancouver Island as a challenge. I have suffered a concussion so trying is difficult. He used to live in S. Wales and dropped stones done old coal shaft. He was a teacher.
Wonder if you could help me please: "A stationary rock is allowed to drop down an 800 foot shaft. Without compensating for air resistance, how far does it fall during
the sixth second of its descent? This is the formula. Assume gravity value to be 32 feet per second per second.
Please set out your answer clearly showing your thought process, line by line. Use words as well as numbers.
I'm afraid your answer so far is incorrect.
If needed, the formula we used was S = ut + half gt^{2}."
Thank you very much. I am in Gr 5!! I want to by a physicist.
ANSWER:
I will assume that this is not a homework problem (forbidden on
this site!); at least if it is you went to a lot of trouble to
disguise it! Your equation is right except since we will start
the clock (t=0) when you let go of the stone, u=0
because u in your equation is the speed at t=0.
Also this equation assumes that S=0 at t=0 and
that S increases in a downward direction. So, at the
end of 5 seconds (the beginning of the 6^{th} second)
the position is S_{5}=½x32x5^{2}=400
ft; at the end of 6 seconds the position is
S_{6}=½x32x6^{2}=576 ft.
So the total distance traveled is 176 ft. I trust you will not
present this work to your uncle as your own.
QUESTION:
I have a really general question regarding the concept of work in terms of Physics. I'm aware that if work is negative, it means that the displacement and force act in opposite directions. However, does negative work also always imply that the the speed of the object is decreasing, or is this only true when looking at objects moving on a horizontal plane.
ANSWER:
The acceleration of any object is determined by all the forces
on it. If only one force acts on an object and the work done is
negative, it must be slowing down. If any other forces are present, all
bets are off. For example, a box sliding down an incline has friction
f doing negative work and gravity mgsinθ
doing positive work; if f<mgsinθ
the box will be speeding up.
QUESTION:
I'm a physics teacher in South Australia.
My question is related to the HeNe laser and has bothered me for some time as to the actual mechanics.
As He is raised to 20.61eV and then transfers to Ne with 20.66eV for population inversion etc...
where has the extra 0.05eV appeared from? the quanta of energy is lower therefore Ne electrons should not be excited to that state.
ANSWER:
It comes from the kinetic energy of the collision between the He
and Ne.
QUESTION:
Why is water used to cool car engines?
ANSWER:
Because it is cheap, readily available, nontoxic, minimally corrosive, and can be kept from boiling with pressure. Perhaps most important,
though, is that it has a high specific heat which means it can absorb a lot of heat without a large temperature increase.
QUESTION:
What would be the estimated terminal velocity be of a 4,300lb car falling from 30,000ft above sea level be?
ANSWER:
The terminal velocity, v_{t},
does not depend on the altitude from which you drop your car.
This can be a very tricky problem because
v_{t}
does depend on the density of the air which changes greatly from
sea level to 30,000 ft. So to get a first estimate, I will just
assume sea level density everywhere. There is an estimate for
the drag force in sea level air which is good for a rough
estimate, F_{D}=¼Av^{2}
where A is the cross
sectional area and v
is the speed. From this you can show that
v_{t}=2√(mg/A).
In SI units, m=4300
lb=1950 kg, g=9.8 m/s^{2}, and A≈2x4=8
m^{2} (estimating the car as 2 m wide and 4 m long).
Then v_{t}≈100
m/s=224 mph.
I guess we should now ask whether we expect it to reach terminal velocity before it hits the ground. Actually, it will technically never really reach terminal velocity, only approach it—see an earlier answer. I will calculate how far it falls before it reaches 99% of v_{t}. In the earlier answer, I show that the height from which you must drop it for it to reach terminal velocity with no air drag is h^{no drag}=v_{t}^{2}/(2g), and the height from which you drop it for it to reach 99% of terminal velocity with air drag is h^{drag}=1.96v_{t}^{2}/g (derived from the expression v/v_{t}=0.99=√[1exp(2gh/v_{t}^{2})]. So, for your case, h^{no drag}≈510 m and h^{drag}≈2000 m. At 2000 m (around 6000 ft) the air is about 85% the density of sealevel air, so I believe that my approximation assuming constant density is pretty good and the car would probably reach 99% of the terminal velocity by the time it hit the ground. To actually put in the change in density with altitude would make this a much more difficult problem.
QUESTION:
What is after death (AD), birth of christ (bc)? Then how we are caluclate the age difference in between AD
& BC.
ANSWER:
This is not physics, but easily answered. Actually, AD is
anno domini, (in the year of the Lord in Latin) and BC is
before Christ. The dividing line is, supposedly, either the
birth or conception of Jesus. Times before are labeled BC and those
after are labeled AD. There is no year zero, so the first year after
this time is labeled 1 AD and the first year before is labeled 1 BC.
Hence, the time from, e.g., January 1, 10 BC to January 1, 10
AD is 20 years; but, the time from January 1, 1 AD to January 1, 20 AD
is 19 years.
QUESTION:
We're trying to design a vessel for working in the vacuum of space. We have a vacuum chamber that can pull a 0.01 atm partial vacuum, so the question is : How does the force difference compare on the walls of a vessel, with 14 psi inside to outside chamber or space, i.e. force comparison between 0.01 atm in the chamber and the 10^{14} in space?
My guess is that we are capturing 99% of the effective force differential using the chamber, so not much more to expect from the vacuum of space.
ANSWER:
Ok, the 14 psi inside your chamber is about 0.953 atm and the
pressure outside is 0.01 atm making the net pressure difference 0.943
atm. The pressure outside in space is, for all intents and purposes,
zero, so the net pressure difference would be 0.953 atm. So, the percent
difference is 100x(0.9530.943)/0.953=1.05%, about what you guessed. In
other words, the force on any part of the walls of your chamber is
about 1% smaller than it would be in space.
QUESTION:
Not sure if this is physics or not but how can you record silent sound ? I found the patent for it on google it's
5159703
and I wanna know how you can record it because a regular microphone doesn't pick it up.
ANSWER:
The idea here is essentially the same as AM radio where the
highfrequency radio wave is modulated by the audible message. For this
invention the radio wave is replaced with sound of a frequency larger
than is audible but modulated by an audible signal. You could certainly
make a detector (call it a microphone if you like) to detect these
highfrequency sound waves; ultrasound imaging in medicine does just
that. Then you would need some electronics to extract the audible signal
from the carrier, just like you need a radio receiver to extract the
audible signal from the AM radio carrier.
QUESTION:
Does a round and square object, the same weight, fall the same?
ANSWER:
If air drag is negligible, like if you drop them from a few
feet, yes. If they fall fast enough for air drag to be important, they
will fall differently and, if they are about the same size, say a sphere
and a cube, the sphere will fall faster. That may be all you want, but I
will go on and explain in a bit more detail. The drag force F_{D}
on an object may be approximated for every day objects, masses, and
speeds as F_{D}=½C_{D}ρv^{2}A
where ρ is the density of the air, v is the speed,
A
is the crosssectional area presented to the onrushing air, and C_{D}
is called the drag coefficient. C_{D }depends only on
the shape of the object. C_{D}≈0.5 for a sphere
and C_{D}≈0.8 for a cube (falling with one face
to the wind). So, if their areas are about the same, the drag on the
sphere will be smaller and it will go faster. Of course if the sphere
area were ten times bigger than the cube area, that would be more
important than the somewhat smaller drag coefficient and the cube would
win.
QUESTION:
What would happen if light traveled at normal speed?
ANSWER:
I recommend the
Mr Thompkins books by George Gamov.
QUESTION:
Is it correct to say that nuclear fusion violates the law of conservation of mass since a portion of the mass is converted into photons?
ANSWER:
Well, you could say that if there were such a law as
conservation of mass. Since 1905 when Albert Einstein showed us that
mass is just another form of energy, the only valid such law is
conservation of energy. Even in chemistry where conservation of mass
appears to be correct, the ultimate source of energy is mass being
changed into kinetic energy of the chemical reaction products (heat);
chemistry is such an inefficient source of energy that the mass changes
are miniscule.
QUESTION:
Can you tell me why muons are extremely unstable (lasting only fractions of a second) while electrons and neutrinos are pretty much stable? I just don't get why muons are so unstable since they're just leptons like electrons and neutrinos just more massive than the other two.
ANSWER:
If you look at the decay of the muon, you will see that the mass
of the products, an electron and two neutrinos, is much less than the
mass of the muon. This means that the decay is energetically possible,
energy is released by the decay so the decay products have kinetic
energy afterwards. In nature, almost always when a process is
energetically possible it will occur. Only in cases where a decay would
be prohibited by some selection rule will decay not occur. For example,
a proton cannot decay into three electrons because charge conservation
would be violated, even though it is energetically possible.
QUESTION:
I think that it is possible to travel faster than speed of light. I explain it in this
link.
Is it correct?
ANSWER:
No, it is not correct. In your example, an earthbound observer
observes the traveler to travel 4.24 ly in 4.28 years. The
traveler, as you correctly surmise, observes the trip to take
0.604 years. However, you have not taken length contraction into
account. The traveler observes the distance she has to travel to
be 4.24/γ=0.598 ly, so she perceives
her speed to be 0.99c.
QUESTION:
So if black holes suck in everything in including light that must mean everything is getting pulled in as fast or greater then the speed of light. So if light is weightless and it is sucked in. What happens to any mass as it is sucked in. Would the mass of the object then cease to have mass? Because im pretty sure anything traveling at the speed of light has to be massless correct? And how does gravity effect something with no mass? I dunno if it is a good question or not but i couldnt find a whole lot on the subject.
ANSWER:
Nothing ever goes faster than the speed of light and only light
can go at the speed of light. When an object merges with the
black hole, its energy, E=mc^{2}, is not lost
and the black hole becomes more massive by the amount of the
mass energy. When light is swallowed by the black hole, the mass
increases even though the light does not have any mass because
light does have energy and the energy shows up as increased mass
of the black hole, m=E/c^{2}.
QUESTION:
What happens gravitationally when the center of mass can no longer be considered a point but is instead an area? Specifically, suppose the Sun was to "explode" or supernova; ignoring the obvious destruction of the solar system, what would happen to the planetary orbit of Earth? I presume it would be roughly akin to letting go of the string at the end of which I have a ball spinning around me.
ANSWER:
The center of mass is always a point. If the sun were to
"explode", the center of center of mass would continue to be at the
center of where it was before the explosion. A star explodes
approximately isotropically, that is, material goes out at the same rate
in all directions. So, until the material reached the earth's orbit, the
orbit would be unchanged. But, as material gets outside the earth's
orbit, only the material inside would contribute to the force felt by
the earth (this is Gauss's law). So the earth would behave as if there
were a star of constantly decreasing mass at the original center of
mass.
QUESTION:
okay, so i wanted to ask what would be the KE of a tungsten rod of length 70 m with a conical end of height 10 m, 6 m in diameter at one end and 6 cm at the other , weighing 38307.5 tons and falling at terminal velocity which i think is about 2340530 m/s?
ANSWER:
The speed, while large, is still less than 1% the speed of
light, so you can use the classical expression for kinetic energy, KE=½mv^{2}=½x3.83x10^{7}x2.34x10^{6}=4.48x10^{13}
J. Note that the composition, size, and shape are irrelevant. Since
everything else was in SI units, I assumed that ton is metric tonne, 1
tonne=1000 kg. This must be some kind of projectile in a computer game.
QUESTION:
Can an electric current flowing in a wire be stopped by a magnetic field? If so, how?
I need to stop it from distant.
ANSWER:
The magnetic force on a moving charge is always perpendicular to
its velocity. To stop a moving object you must apply a force
antiparallel to its velocity.
QUESTION:
I bought a 400lb gun cabinet and need to pull it on a 2 wheel hand cart up
a 12ft ramp at about 35degrees to the horizontal How much load does 1 or 2
people have to carry and how much is borne by the wheel. I am trying to make
sure we can be comfortably safe!
ANSWER:
I could make a rough estimate but would need to know the dimensions of the
cabinet, if the center of gravity is near the geometrical center. I would
assume that the cabinet was parallel to the ramp when being pulled.
FOLLOWUP QUESTION:
It is 20X29X55. It would not be parallel to the ramp but about 20 degrees from the ramp (which is about 35 degrees to the ground (thus avoiding 4 steps).
ANSWER:
Since only an approximation can be reasonably done here, I will
essentially model the case as a uniform thin stick of length
L with weight W, normal force
N of the incline on the wheel, and a force
F which you exert on the upper end. In
the diagram above, I have resolved F
into its components parallel (x) and perpendicular (y)
to the ramp. Next write the three equations of equilibrium,
x and y forces and the torques;
this will give you the force you need to apply to move it up the
ramp with constant speed.
ΣF_{x}=0=F_{x}Wsinθ
ΣF_{y}=0=F_{y}+NWcosθ
Στ=0=½WLcos(θ+φ)NLcosφ.
I summed torques about the end where you are pulling. Putting in W=400 lb, θ=35º, and φ=20º, I find F_{x}=229 lb, F_{y}=206 lb, and N=122 lb. Note that you do not need to know the length L. The net force you have to exert is F=√[(F_{x})^{2}+(F_{x})^{2}]=308 lb. If someone were at the wheel pushing up the ramp with a force B, that would reduce both F_{x} and F_{y}. This would change the equations to
ΣF_{x}=0=F_{x}Wsinθ+B
ΣF_{y}=0=F_{y}+NWcosθ
Στ=0=½WLcos(θ+φ)NLcosφ+BLsinφ.
For example, if B=100 lb, the solutions would be F_{x}=129 lb, F_{y}=169 lb, and N=159 lb; so your force would be F=213 lb.
QUESTION:
Imagine if you wrapped a rope tightly around the earth. How much longer would you have to make the rope if you wanted
it to be exactly one foot above the surface all the way around?
ANSWER:
I hope you don't think that the rope would spontaneously rise up if
it were longer than the circumference of the earth; you would have a
slightly slack rope laying on the ground. You are specifying the difference
in radii between one circle with a circumference C and another of
circumference C+δ; that
is not really physics. But, it is easy enough to do. If C is the
circumference of the earth, then C=2πR_{ }where R
is the radius of the earth and
C+δ=2πR' where
R' is the radius of the circle your rope would make if δ=1
ft. Then δ=2π(R'R)=2π=6.28 ft.
Note that δ depends only on how high the rope is above ground,
not how big the earth is: if the earth were 1 ft in radius and you increased
the length of the rope by 6.28 ft, the rope would be 1 ft above the surface!
QUESTION:
I'm writing a research paper for my college english class and the topic is Thorium Reactors. My question is "Are thorium based reactors such as LFTR
fusion or fission and based?" I was just wondering because the it seems from what I've learnt that the reactors use the thorium to produce a reaction that makes another element such as uranium 233 which I assume is fusion because I'm sure they're using the energy put off from that initial reaction to power something. But after the uranium 233 is used and to produce energy as efficiently as possible I would think that you would implement a system that would immediately and directly use said produced uranium into some form a fission reactor.
ANSWER:
Fusion always involves light nuclei and there is no such thing
as a fusion reactor, only ideas for them. So a thorium reactor must be a
fission reactor. It would be inaccurate, though, to call thorium the
fuel because thorium is not
fissile. If a thermal neutron is absorbed by a fissile
nucleus, it will fission and result in more neutrons leading to more
fissions and the reaction can be selfsustaining. Thorium, which is 100%
^{232}Th, absorbs a neutron to become ^{233}Th which
βdecays to ^{233}Pa (half life 22 minutes)which
βdecays to ^{233}U (half life 27 days). The ^{
233}U is fissile and is nearly stable (αdecay half
life 160,000 years). Thorium is said to be fertile, absorption
of a neutron results in production of a fissile fuel. So a thorium
reactor is a breeder reactor, a reactor whose purpose is create fuel. At
the startup one needs a starter fissile fuel as well as the
thorium to provide neutrons to create the ^{233}U. As the ^{
233}U builds up, it becomes the fuel.
QUESTION:
Would it be physically possible to create a parachute capable of delivering a main battle tank safely to a theatre of war?
Like how huge would it have to be?
ANSWER:
A reasonable estimate of the force F of air drag on an
object of mass m, speed v, and cross sectional area A is F=¼Av^{2};
this works only in SI units. The speed v when F=mg is called
the terminal velocity. I estimated a reasonable terminal velocity would
be the speed the tank would have if you dropped it from about 10 m,
v≈14 m/s. The mass of the
MBT70 (KPz70)
is about 4.5x10^{4} kg. Putting it all together, I find
that A≈10^{4} m^{2}, a square about 100 m on a
side or a circle of radius about 30 m. You could do a more
accurate calculation but this gives you a reasonable estimate.
QUESTION:
If you stand holding a box but the box is not moving, are you doing any work?
ANSWER:
See an
earlier answer.
QUESTION:
Why is it necessary to have a minimal temperature of 150 million degrees Kelvin for nuclear fusion on earth if the sun does nuclear fusion at a temperature of 15 milloin degrees?
ANSWER:
There is more than one reason that I can think of. The mass of
the sun is 2x10^{30 }kg, quite a bit bigger than the
mass of fuel in a fusion reactor. Therefore the rate of fusion
in the sun can be low but the energy output would still be huge.
Increasing the temperature would increase the rate. The density
of the sun's core is about 150 g/cm^{3}, 150 times more
dense than water. In a fusion reactor, the practical densities
are many orders of magnitude below this. Again, the rate of
fusion would be dependent on the density of the fuel.
QUESTION:
My name is Justin and my friend Richard blew my mind when he told me that nothing doesn't have mass. Again. Nothing doesn't have mass. Is this true? I was taught that everything has an inherent mass. Is he wrong? Am I wrong? Does sheer nothingness have a mass or not.
Please help me. My friend just blew my mind and I'd like it to be back together.
ANSWER:
It is true that the vacuum does not consist of nothing. See a
recent answer for more detail.
However, the vacuum does not contain mass in the sense you
normally think of it—you cannot "weigh" a vacuum.
Also, photons, the quanta of light, do not have mass. Until
recently it was thought that neutrinos have no mass; they
actually do have very tiny masses. All the double negatives in
your question and my answer make this a little confusing; I hope
I have answered what you meant to ask!
QUESTION:
in classical physics angular momentum is a variable factor, but why in quantum mechanics angular momenta are quantized. and how its related to spin.
I have read that Paul Dirac showed, how changing some relativistic factors in Shroedinger's equation can spontaneously lead to the "spin"
concept. but what the spin actually tells us and how we can visualize it.
ANSWER:
The reason why orbital angular momentum is quantized is that,
when you solve the Schrödinger equation for the atomic wave
function, the wave function is not normalizable unless the
angular momentum is
ħ√[L(L+1)] where L
is an integer. (For our purposes here, "not normalizable" means
that the wave function becomes infinite somewhere.) It also
turns out that L is called the orbital angular
momentum quantum number and its being an integer has nothing to
do with spin except that spin is also an angular momentum. To
visualize spin, read two earlier answer (#1
and #2). In relativistic quantum
mechanics, the Dirac equation replaces the
Schrödinger equation. When you write the Dirac equation for
an electron, it turns out that spin is indeed predicted to have
an angular momentum quantum number of ½ as is
observed experimentally.
QUESTION:
I've just found out that a proton isn't made up of two up and one down quark but also contains zillions of other up and down quarks along with their anti matter equivalents. Here's a
link
to an article from the L.H.C people at Cern that shows this.
I'm amazed and confused because I thought that matter and anti matter particles would annihilate each other. Why don't they? And, if they do, how is the 'zillions of other quarks' balance maintained?
ANSWER:
The crux of what is going on here is that a vacuum is not really
a vacuum as we generally think of it—nothing.
Particleantiparticle pairs are continuously popping into
existence and then annihilating back to nothing after a short
time. This is called virtual pair production. Also, if a
particular particle experiences a particular force, the
messenger of that force (gluons for the strong interaction,
photons for the electromagnetic interaction) are continuously
being emitted and reabsorbed. All this is called vacuum
polarization. So, I could give you a similar description of the
hydrogen atom: the hydrogen atom is not really a proton and an
electron, it is a proton and zillions of electrons and positrons
and photons. If you really want to understand the hydrogen atom in
detail, you need to take the effects of vacuum polarization into
account (see the
Lamb shift,
for example). The CERN explanation should have included zillions
of positrons and electrons and photons inside the nucleus also.
Just as a hydrogen atom is pretty well described as a proton and an
electron as a first approximation, a proton is pretty well
described as three quarks as a first approximation.
QUESTION:
Why the whole matter of radioactive sample does not disintigrate at once or Why it always take half life to disintegrate half of initial value?
ANSWER:
Because decay is a statistical process. Whenever you have a
large number N of anything they will, at some time
t, have a time rate of change R=dN/dt.
If R<0, N is getting smaller (as in
radioactivity) and if R>0, N is getting larger
(like bacteria growth). For a great many cases in nature, it
turns out that the rate is proportional to the number, R∝N.
Radioactivity turns out to behave that way: if you measure R
for 2x10^{20} radioactive nuclei it will be twice as big
than when you measure R for 10^{20} nuclei. You
can then solve for what R is for a given situation:
dN/dt=λN where λ
is the proportionality constant, called the decay constant; the
minus sign is put there so that λ will be a
positive number since the rate is negative. If you know
differential equations, you will find that N=N_{0}e^{λt}
where N_{0} is the number when t=0.
The decay constant is related to the half life τ_{½}=ln(2)/λ.
However, this all depends on there being a large number to
start. In the extreme case, if N_{0}=1, they
would all decay at once! You just could not predict when. If
N_{0}=2, they might both go at once or else one
might go before the other but not necessarily at τ_{½}.
QUESTION:
When solving questions involving two identical springs being stretched to points A and B to create a total length and the natural length of the springs are given. What is a consistent way to calculate the amplitude?
Is it half of the length of AB subtract the natural length of the spring?
ANSWER:
I assume that the springs are in series—one attached to
the end of the other. Suppose that you exert some force F
such that the springs are stretched by a distance s.
Then each spring will be stretched by ½s.
If k is the common spring constant, F=½ks.
Therefore the two together behave like a single spring with
spring constant ½k.
QUESTION:
How fast are we moving? Besides Earth rotation and revolving, there's our path in the galaxy, and even greater in relationship to all other galaxies, and please include the expansion of spacetime.
ANSWER:
First you should understand that it only makes sense to ask what
our velocity is relative to some inertial frame of reference.
And, also, it depends on things like what time of day it is,
what time of year it is, what year it is, etc. because
all these things affect your velocity; e.g., relative
to the earth's axis, you are moving in opposite directions at
noon and midnight. There is a nice qualitative discussion of
this question at
Curious about Astronomy; adding up all the various
velocities, she finds a maximum speed relative to the center of
our supercluster of about 900 km/s. Given the roughness of this
calculation, trying to finetune it by adding in the expansion
of spacetime is pointless.
QUESTION:
In my AP Physics class, we have had a classwide debate over a physics
problem for the last few days. The problem asked how much work it would take
to move a satellite that was orbiting Earth at a certain height to a greater
height. Some of us say that the work equals the change in potential energy,
while others say that the work is the change in the total mechanical energy.
The total energy method gives an answer of exactly half of the amount the
potential energy method gives. Who is right?
Both orbits are circular.
ANSWER:
Since the speeds in the two orbits are different, the kinetic
energies will be different, so the correct way to do this is
½mv_{2}^{2}mMG/r_{2}^{2}=W+½mv_{1}^{2}mMG/r_{1}^{2}.
Of course you will have to figure out what the velocities are in
terms of the radii, but that should be a piece of cake for AP
students!
QUESTION:
A car is travelling 45 mph, with the road being at a 58% incline. How
long would it take a car to slow down without using breaks? This is not homework. I am a claims examiner that is trying to get some information as to the rate of speed in which a car decelerates.
ANSWER:
You are a claims examiner and you spell brakes "breaks"?! I could do this if I assumed no friction whatever, but it would not be predictive of the real world because there is plenty of friction acting on a moving car even without brakes applied. I could make a better estimate if you could tell me how far this particular car, starting at 45 mph, traveled on level ground with brakes not applied. Also, is the car in gear? In neutral? Engine running?
If there were absolutely no friction, the car would keep going until it had gone
vertically up a distance of about 20 m≈66 ft
regardless of the grade of the incline. For a 6% incline, the
distance traveled by the car would be about 1100 ft. In the real
world it would be way less than this.
QUESTION:
As we know the earth is revolving around the sun. Why don't we have any feeling of revolving?
ANSWER:
In order to have a "feeling" of motion, there must be an
acceleration. The acceleration of the earth in its orbit is
about 6x10^{6} m/s^{2}. For comparison, if you
are in a car with an acceleration from 0 to 60 mph in 10 s, your
acceleration is about 3 m/s^{2}; this means that you
"feel" pushed back into your seat with a force about one third
of your weight. The force you "feel" due to the orbital motion
of the earth is about 1/600,000 of your weight, far less than
you could "feel". The acceleration due to the earth's rotation
on its axis is larger than this but still way smaller than you
could "feel".
QUESTION:
Trying to understand a relationship between nuclear physics and particle physics.
In nuclear fusion 4x H1 fuses to give (ultimately) 1 x He4 + 1 positron a neutrino and energy.
However, 4 protons have become 2 protons and 2 neutrons and neutrons are more massive than protons  so we have apparently gained mass  which should require an input of energy.
My research finds that the He4 nucleus is less massive than the 4 protons that made it because mass has been converted to 'binding energy'.
Questions:

Since a lot of the mass of a proton comes from energy, doesn't this 'binding energy' add mass to the He4 nucleus?

The protons that formed this nucleus are made up of Quarks held together by energy in the form of the exchange of Gluon particles. To convert to a neutron, an Up Quark has become a Down Quark with more mass. What mass has the He4 Nucleus lost?

Do the Quarks in the He4 nucleus have less mass than those in the H1 since the Quarks appear to be the only mass in the process that can convert to energy? Surely if the energy came from the Gluons it would still just be energy and hence not have the mass/energy conversion ratio?

If the energy did come from the Gluons, are Quarks in He4 less tightly bound than those in H1?

Does this continue to occur as you move up the fusion sequence to higher mass elements?

Has the creation of matter from energy (e.g. the electron produced in this fusion) ever been actually achieved? Or are we working purely on theory based on remote observation of the Sun?
ANSWER:
I should really delete this question altogether because it has
ignored the site ground rules stipulating "…single,
concise, wellfocused questions…" However, I will address at
least parts of the question to clear up some major misconceptions.
Although the questioner asserts that he is "trying
to understand a relationship between nuclear physics and particle
physics", I will ignore all questions relating to quarks and gluons
because if the nuclear physics itself is not understood, then there is
no point in trying to relate it to particle physics. To understand the
fundamentals of nuclear physics, you do not need to even know that
quarks and gluons exist, only that a strong nuclear interaction exists.
In the opening paragraph the mass of the ^{4}He nucleus is
compared to the mass of four protons. This is a meaningless comparison
because the constituents are two neutrons and two protons. (See below*
for how protons are converted into ^{4}He nuclei.) Indeed, the
mass of the ^{4}He nucleus is less than the mass of its
constituents. If one knows E=mc^{2},
the reason for this mass difference is intuitive. Does it take work to
pull a proton or neutron out of the ^{4}He nucleus? Of course it
does, otherwise the ^{4}He nucleus would not stay together. So
energy must be added to disassemble the nucleus and this energy resides
in the greater mass of the constituents. I find this much easier to
understand than to say that the excess mass of the constituents is
"converted" to binding energy.
As you move up to create heavier and heavier nuclei via fusion, the fusions continue to release energy but in decreasing quantities until the final product is iron. Thereafter, if you want to fuse nuclei, you must add energy. So, elements heavier than iron are not produced in stars. Heavier elements are produced in supernova explosions.
You ask if "…creation of matter from energy…" has ever been observed. It happens all the time. But that is really the wrong question because matter and energy are not different things, matter is simply a form of energy. So, for example, if you fused two ^{32}S nuclei to a single ^{64}Ge nucleus, the Ge would have more mass than the two S. Any time you have an endothermic chemical reaction you end up with more mass than you started with (although almost immesurably small because chemistry is so inefficient).
*As can be seen in the figure above, the protonproton cycle requires an input of 6 protons and ends with one ^{4}He and 2 protons. So the net input is 4 protons. However, there are numerous outputs—2 positrons, 2 neutrinos, and 2 photons, all of which carry energy away. That is why comparing the net input with the final product is not meaningful.
QUESTION:
The substance which undergoes deformation with a small force
will be elastic or inelastic?
ANSWER:
You cannot judge the elasticity of something by how easily it
deforms. If it stays deformed after the force is removed, it is
inelastic. If it returns to its original shape after the force is
removed, it is elastic.
QUESTION:
The particles that make up a rock are constantly moving. However a rock does not visibly vibrate.
Why do you think this is?
ANSWER:
The amplitude of the vibration is less than the distance between
atoms in the rock, ~10^{11 }m. The atoms are all randomly
vibrating so that there always as many atoms moving in one direction as
in the opposite direction. The frequencies of the vibrations are ~10^{13
}Hz, far above a frequency which you could feel, hear, or see.
QUESTION:
I was just wondering if you have ever thought that dark energy was actually just gravity? Could the gravitational force act differently over large(r) scales? This seems like something I would like to study at a PhD level as this could be an area to reveal the connection between gravity and quantum mechanics.
ANSWER:
The best theory of gravity is general relativity (GR). GR
already contains a term, called the
cosmological constant, which can add a repulsive force to the
attractive gravitational force. I do not quite see how this could be a
route to quantizing gravity, but who knows? Good luck!
QUESTION:
A quantity of electrical energy is defined by volts x amps x time. A quantity of mechanical energy could be defined (or is) by force x distance which equates to kinetic energy. When electrical energy is converted into mechanical, a force can be created by applying voltage and current (amps). Is this a paradox? Electrical energy is functionally force x time while mechanical energy is force x distance.
ANSWER:
1 V=1 J/C and 1 A=1 C/s, so 1 volt·amp·second=1
Joules=force·distance. No discrepancy, no paradox. Another way to
look at this is that current times voltage is power and power is W=J/s.
FOLLOWUP QUESTION:
I totally understand that 1 amp is 1 coulomb per second. I don't know where 1 volt is equal to 1 joule per coulomb comes from or why that is true.
ANSWER:
The electric field E is defined to be the force F
felt by a charge Q divided by Q. The electric
potential V is defined as E times distance d
over which it acts. V=Ed=Fd/Q=[J/C]
QUESTION:
If gravitons mediate the gravitational force between particles, how does this work for a black holes where gravitons cannot escape the event horizon?
Are gravitons predicted by General Relativity?
ANSWER:
See an earlier answer. Gravitons are
not predicted by general relativity. Physics inside the event horizon is
not well understood.
QUESTION:
Why is there a slightly rough surface on a basketball? Does this affect the static friction acting between your hand and the ball during a shot?
ANSWER:
The dimples on a golf ball and the fuzz on a tennis ball reduce
air drag and thereby allow the ball to go faster and farther. However,
the speeds of these balls are much larger than speeds ever encountered
by basketballs and this cannot be the reason for the bumps on a
basketball. A little research reveals the purpose is just to make the
ball easier to grip and handle, as you speculated.
QUESTION:
Are there any known stable nuclei for witch Z>N? Why are they so
rare?
ANSWER:
The only stable Z>N nucleus is ^{3}He. The
reason that all stable isotopes heavier than helium have the same or
more neutrons than protons is complicated because the nuclear force is
complicated. One important reason is that the repulsive Coulomb force as
well as the attractive nuclear force is present trying to blow the
protons apart and the more protons you add, the more important this
becomes; adding neutrons tends to push the protons farther apart to
reduce the repulsive force. Neutrons are sort of "insulation" for the
protons' electric repulsion from each other.
QUESTION:
Electrons orbit around the nucleus in varying degrees of proximity to the nucleus. Do electrons farther from the nucleus orbit at a different rate of speed than those closer to the nucleus?
ANSWER:
Yes, if you use a Bohr model for the electrons. However, the notion of electrons running around in welldefined orbits is naïve and incorrect.
FOLLOWUP QUESTION:
Would it be correct to apply Kepler's Laws of Motion to the revolution of electrons about the nucleus in the Electron Cloud Model?
ANSWER:
The electrostatic force is a 1/r^{2} force just like the
gravitational force, so if the atom were a classical system Kepler's
laws could be used for an atom. In fact, the
BohrSommerfeld model, the first extension of the Bohr's circular
orbit model, essentially does this by including elliptical orbits and
appropriate quantization. Of course, the atom is not a classical system
and although such models can be instructive, they are not strictly
accurate. What you refer to as the "electron cloud model" would be the
proper solutions to the Schrodinger equation, not having welldefined
orbits.
QUESTION:
I would like to know if one horsepower is equivalent to 33,000 pounds per minute, and for a four cycle engine fires on every 2nd revolution of the crankshaft, and lets just say this 1 hp engine runs at 2800 rpm , so it fires 700 times , is there a formula to calculate how much force was created in the combustion chamber in pounds?
ANSWER:
Power is energy per unit time; some possible units of power are
Watts (W), horsepower (hp), and footpounds/second (ftlb/s). So your
lb/min is not an acceptable unit for power because lb is a unit for
force, not energy. But, you apparently looked this up somewhere because
1 hp=33,000 ftlb/s. The power of the engine does not tell you the force
in the combustion chamber, in fact force is not really what would be of
interest. What you want is the energy delivered by each piston for each
stroke. If the pistons fire, for example, at the rate of 1000 times per
minutes, each piston delivers 3300 ftlb per stroke.
QUESTION:
I knew that when a car which travels very fast and is brake very sudden, the
car will like "fly away". But is there any theory or principle or rule can
explain this?
ANSWER:
I do not know what "fly away" means.
FOLLOWUP QUESTION:
To explain, an illustration is made. Imagine a bicycle that
travels in very high velocity and it is braked suddenly and
hardly, what will happen. The bicycle will like "flying up"
caused by the momentum. It is also related to the principle of
acceleration and deceleration.
ANSWER:
OK, I get it now. Refer to the figure above. The easiest
way to do this is to introduce a
fictitious force. If the car is accelerating with
acceleration a (which points opposite the velocity
v when braking), Newton's first law will be valid in the
car frame if a force F=ma acting at the center of mass
(COM) is introduced. The "real" forces on the car are a normal
forces up on each wheel by the road, the frictional forces
backwards on each wheel by the road, and the weight mg
which acts at the COM. In the drawing I have only labeled the
weight, the normal force on the rear wheel, and the fictitious
force because those are all you need to answer your question. If
you wish, you could find both normal forces and the sum of the
two frictional forces in terms of a; if all wheels were
locked you could find the individual frictional forces if you
knew the coefficients of kinetic friction. Now, I will sum
torques τ about the point where the front wheel
touches the ground, Στ=mah+NLmgs=0 where
h is the height of the center of mass above the ground,
L is the horizontal distace from the front axle to the COM,
and s is the horizontal distance between the wheels.
You can now solve for N, N=(mgsmah)/L.
Now think about N; if N<0, the road would have
to pull down on the back wheels to hold the car from rotating
forward about the front wheels. N will be zero when the
car is just about to "fly up"; therefore, if a>g(s/h),
the car will "fly up".
QUESTION:
If you aimed a high powered laser beam in a tangential direction to the surface of the Earth, would that beam travel in a straight line and shoot out into space, or would Earth's gravity be enough to at least bend the light in a measurable amount?
ANSWER:
See an earlier answer. If you
calculate the angle for earth instead of the sun, φ=4GM/(c^{2}r)=0.002
arcseconds, just about impossible to observe.
QUESTION:
When speaking of particle accelerators,the accelerators keep adding energy to the particles, even though they cannot speed up any further. But where does the energy go?
ANSWER:
Well, they never really get to a speed where they cannot go any
faster because they never reach the speed of light. I have always
thought "accelerator" was a misnomer for very highenergy machines
because the acceleration (rate of change in speed) is almost zero; I
would prefer "energizer". The easiest way to think about it is that the
particles acquire mass and that is "…where…the energy
go[es]". The energy of the particles is E=mc^{2}=m_{0}c^{2}/√[1(v/c)^{2}]
where m_{0} is the mass at rest. For example, if v/c
increases from 0.99 to 0.999, that is only a 1% change in speed; but the
energy increases from 7.1m_{0}c^{2}
to 22.4m_{0}c^{2}, more than triple.
QUESTION:
A cylinder is kept in friction less inclined plane (curve face touching incline). Why it does not roll (transnational motion only) although a torque is working on it with respect to contact point due to component of force parallel to incline
ANSWER:
Because the cylinder is accelerating. The point of contact is
also accelerating and so you cannot use it as a point about which to sum
torques because Newton's laws are not valid in an accelerating
coordinate system. An accelerating coordinate system is called a
noninertial frame. There is, however, one point where you can always sum
torques even if that point is accelerating—the center of
mass. The net torque about the center of mass is zero because the weight
vector acts at that point and the normal force passes through it.
QUESTION:
I am looking for relationship between viscosity of air and pressure.
I want a table listing the viscosity at different pressures. Pressure should start from 0.1 atm to 1 atm.
ANSWER:
The table above comes from an
article
by Kadoya, Matsunama, and Nagashima. Pressure is given in MPa
and 0.10 MPa=0.987 atm≈1 atm. So you want the
variation between 0.01 MPa and 0.1 MPa. As you can see the
variation is very small; the viscosity is much more dependent on
the temperature.
QUESTION:
Two questions for you concerning the recent proof that gravity flows in waves. 1). How fast do these gravity waves move? Do they move at a constant speed or is their speed predicated by the mass of the object that generated the waves? 2). Since the earth is trapped in orbit around the sun by the sun's gravity, how can a wave moving out from an object in turn pull a second object back towards the object that generated the wave? In other words, how can gravity waves moving out from the sun at the same time pull the earth towards the sun?
ANSWER:
The answer to your first question is in a
recent answer. In your second answer, you are confusing gravity with
gravitational waves. Our picture of gravity is that mass (like the sun)
deforms the space around as in the animation above. Something like the
earth orbiting the sun is not really feeling a force, it is following
the contour of the space. (Do not take this
simplistic model too seriously, it is really the fourdimensional
spacetime which is "deformed".) But, if something accelerates, like the
earth going around in its orbit, it will send out ripples and those are
the gravitational waves. On the scale of this animation, the waves the
orbiting earth is sending out are far too tiny to be seen. But, if the
objects orbiting are much more massive, like the two orbiting black
holes which were observed in the recently reported
observation of the waves, the waves are much bigger.
QUESTION:
Ok so you know how they say, "if you look up at the stars in the night sky, you are seeing light that's taken millions of years to travel to earth, so you are seeing Into the past potentially"? Ok so here's the question. If say I'm on a planet in the Andromeda galaxy, and you're on earth looking at me with a super telescope( to where u can actually see me wave at you) and let's say we agreed on a specific time and date to do this. If I start waving at you at the exact time you are observing me with the telescope, do you ever see me wave or does it take millions of years to see me wave at you ?
ANSWER:
How do you plan to agree on a specific time and date to wave if
it takes millions of years to communicate. Just imagine that there is a
star exactly halfway between you and me and we have, by some magic, both
been instructed to do our things (wave and look) when we observed the
star to become a supernova. If you and I are separated by a million
light years, I will not see you wave until a million light years have
passed. Furthermore, the supernova would have occurred a half million
years before either of us saw it.
QUESTION:
Why is it that when I listen to music on my PC through an external speaker set and my cell phone right next to me gets messages, etc... and they cause distortions in how the music sounds?
ANSWER:
Your cell phone is both a receiver and a transmitter of radio
waves. The transmitted radio waves may be detected by your sound system.
QUESTION:
I mean to say the angle derived in Einstein equation φ=4GM/(c^{2}r)=1.75
arcseconds is total deflection when light source is opposite side of the Sun?
ANSWER:
QUESTION:
When a television set is turned on it often generates an electrostatic field. You can still feel this charge if you move your arm near the screen shortly after the set has been turned off. Why does this field not disappear as soon as the television set has been turned off?
ANSWER:
You probably have an old cathode ray tv. The screen is coated
with a phosphor which glows if you strike it with an electron. Electrons
are shot from the rear of the tube to cause a picture to be formed. Some
of these electrons are still on the screen when you turn off the power.
QUESTION:
Ships are often built on ways that slope down to a nearby body of water. often a ship is launched before most of its interior and superstructure have been installed and is completed when a float.
Is this done because the added weight would cause the ship to slide down the ways prematurely?
ANSWER:
Friction can be a tricky business, but the simplest behavior is
that the frictional force increases proportional to the weight. But the
force of the gravity trying to slide the weight down the slope is also
proportional to the weight. Therefore, doubling the weight of the ship
should not increase its tendency to slide down. Besides, if this were a
concern you could always temporarily block the path down the slope like
placing blocks in front of a vehicle on a slope to keep it from rolling.
I suspect the real reason is that the structure of the ramp is probably
not strong enough to support the full weight of the ship.
QUESTION:
If one wanted to turn a cylinder 5 feet in outer diameter and 3 feet inner diameter and the reel is supported via a 2 in bar through the center of it Look like this i suppose (more like a reel) A rope is wound around the outer diameter and pulled (think of a yo yo) how much torque would you have to put on the rope to get it to turn? at 35 rpm? Can you show how one can figure that?
ANSWER:
It would depend on how long you pulled, what the mass of the hollow cylinder is, and how much friction there is.
FOLLOWUP QUESTION:
I have asked several people to give an answer on this and no one has been able to answer it for me. So I attached a pic I just drew out for you...
On the pic you see that the master reel is 5000 lbs... I want to pull the
rope off the reel and put on another reel... I would like to do so at 35 rpm on the small reel... The small reel is just like the large reel but the OD is 24 inches and the center of the reel is 12 inches... with a bar going through it just like the large one... how much pulling torque would be needed to get the large reel to start turning... neglect friction...not enough there to consider...
ANSWER:
First you have a misconception. If friction is truly negligible,
any torque, no matter how small, will start the big reel
turning. It is a matter how long you want to wait until you get
up to 35 rpm on the smaller reel. For several reasons this is a
quite complicated engineering problem. The most important
problem to deal with, I think, is that as rope comes off the
larger reel its weight gets smaller and its radius gets smaller;
at the same time the smaller reel gets larger and heavier. So if
you want to keep the smaller reel going at a constant 35 rpm the
larger reel will have to change its speed if the rope is to
remain taught. So, I will only address the question of how you
start up just to demonstrate how torque determines what happens.
From your drawing, it looks like the rope fills the reel so its
outer radius would be about 3 ft≈1 m and its inner radius
would be about 1 ft≈0.3 m. Since the mass of the rope is
5000 lb≈2300 kg is large compared to the 200 lb reel, I
will neglect the contribution of the reel; this simply says that
the torque you apply mainly has the effect of getting the rope
spinning. Since it is the smaller reel you want to go 35 rpm,
the upper reel would need to have an angular velocity of
(0.5/3)x35=5.8 rpm=0.61 radians/s. (Assuming that your idea is
to use the empty reel to pull the heavier reel via the rope.)
Again, I will neglect the torque necessary to get the smaller
reel going since it will presumably be lighter than the already
neglected weight of the heavier reel. The basic physics
principle to use is the rotational form of Newton's second law,
τ=Iα where α
is the angular acceleration, τ
is the torque, and I is the moment of inertia.
In your case, the angular acceleration would be the final
angular speed divided by the time to get there; so if you want
to get up to speed in 10 seconds, the angular acceleration would
be 5.8/10=0.58 rpm/s=0.061 radians/s^{2}. You see, now,
why there is no answer to your question: the torque will depend
on how quickly you want to get up to speed. The moment of
inertial of the rope is I=½M(a^{2}+b^{2})=½(2300)(1^{2}+0.3^{2})=1250
kg·m^{2}. Finally you can estimate the torque
τ=1250x0.061=76 N·m=56 ft·lb.
A longer spinup time would need a smaller torque.
QUESTION:
I've got a basic question about signal processing within the discipline of management information systems. Today, some basic signals within management information systems, which people come across are electrical, light (fiber optics), and radio waves. Is it true that radio waves are a type of electromagnetic radiation or many types of electromagnetic waves, which can travel at the speed of light? Is that true? I didn't think it was possible for anything to travel at the speed of light. As I understand it, the speed of light in a vacuum, such as space or other, commonly denoted c, is a universal physical constant that's very important in many areas of physics. The speed of light, it travels (approximately 3.00×10^{8} m/s) or 186,000 Miles per second. Can anything you know of travel at the speed of light especially radio waves?
ANSWER:
When a physicist refers to "the speed of light" she is talking
about the speed of electromagnetic radiation in a vacuum. Of all the
possible wavelengths of light, visible light is but a tiny fraction.
QUESTION:
Whether light is a particle or wave? Which is right and why?
ANSWER:
Light is not a particle or a wave, it is a particle
and a wave. This is called waveparticle duality. If you design an
experiment to observe a particle, you will observe a particle; and, if
you design an experiment to observe a wave, you will observe a wave.
QUESTION:
If you have a wind up toy car, meaning that when it starts you can wind it up to maximum and have a constant force pushing it forwards, how would you expect the distance in travels to change as you add mass to it?
ANSWER:
The spring will deliver a certain amount of energy. Giving the
same amount of energy to a large mass and a small mass will
result in the small mass going faster. The small mass will
therefore go farther. This assumes that the wheels never slip.
QUESTION:
If all things with mass have a proportional amount of gravity and gravitational waves were recently observed to have been produced by 2 black holes converging, is it correct to deduce that all things with mass produce gravitational waves proportional to their mass?
ANSWER:
Yes. Any object with mass which accelerates should radiate
gravitational waves. Wave to someone and you cause gravitational waves.
However, gravity is the weakest force in nature so, for the the waves to
have a big enough amplitude to be detectable, the masses must be
extremely large (as in black holes).
QUESTION:
How fast do gravitational waves move?
Is that rate constant?
ANSWER:
The speed of gravitational waves has never been measured. The
speed at which a gravitational field propogates should be the same
but it has never been measured either; this would determine, for
example, if the sun suddenly disappeared how long it would be until the
earth stopped orbiting. The theory of general relativity, which predicts
gravitational waves, say that the speed of gravity should be the same as
the speed of light. The recent
observation of gravitational waves determined the distance to the
source to be about 1.33 billion light years, but the uncertainty was
very large, about 40%, so it really provides no measurement of the
speed. Whatever the speed, there is no reason to think that it would not
be constant everywhere in empty space.
QUESTION:
A car is moving forward through a road.
Which part of the wheel of the car moves fast the upper part or the lower part of the wheel?
ANSWER:
Inasmuch as the part of the tire in contact with the road is at
rest, the answer should be obvious. You should be able to show
that the top of the wheel is moving forward with twice the speed
of the car.
QUESTION:
Has anybody ever done the double slit experiment on a very large scale. By large I mean physically where the electrons are shot at tow slits several feet wide separated by an even greater distance from a large distance away to see if the electrons themselves will still manage to make it through the slits at all? As long as the electrons are unobserved acting as a wave they should still continue to go throught the slits and produce a wave pattern behind the slits regardless of scale. However does this deteriorate at some point and break down to cause the electron to again act as a particle and simply hit the mass between the two slits and if so could this be somehow used as a physical means to measure the energy of the electron? Could an electron with a greater energy then once again act as a wave?
ANSWER:
As far as I know, nobody has ever done a true doubleslit
experiment with electrons at all. Given the extraordinarily short
wavelength of an electron, the spacing between slits would have to be on
the order of 10^{10} m and you just cannot make physical slits
that close because that is like the distance between atoms. What is
actually done is to shoot electrons at a single crystal and you get a
doubleslit like experiment. For diffraction to be observable the slit
spacing and slit width must be small compared to the wavelength of the
wave.
QUESTION:
I am doing an experiment on factors affecting the travel distance of a toy car from down a ramp and thought it would be a good idea to understand how a physicist thinks of things. My question is, how do you think weight of an object affects the distance it will travel after going down a ramp?
ANSWER:
The best discussion I have seen of the physics of pinewood derby
races is this
youtube video. You will see that what matters more than the added
weight is where you put it. (Thanks to my son Andy for pointing me to
this video; his son and my grandson Finn placed second in the Cub Scout
pinewood derby last year on the strength of the tips here!)
QUESTION:
Hello, question in physics. Why would the speed of a rollercoaster with potential energy of 5MJ have less than that predicted for a perfect frictioless track at the bottom of the slope?
ANSWER:
A roller coaster has wheels and some of the kinetic energy would
be in their rotation, not in the speed of the cars. Of course, if the
track were perfectly frictionless the wheels would not go around at all.
Maybe your problem means no energy lost to friction? Or maybe the bottom
of the slope is not at the zero of potential energy?
QUESTION:
What will happen if we fill water in the tyres of our cars instead of air?Does it have any effect?.
ANSWER:
Three important issues I can think of. First, it would add a lot
of weight to the vehicle which would hurt your gas economy. Second, the
moment of inertial of a wheel would be larger requiring a larger torque
having to be exerted for either acceleration or braking. Third, air is
compressible and water is not and so the wheel would not have a
cushioning effect on the ride.
QUESTION:
We say that time slows as we accelerate. Is time some existent entity that can slow? Or is it the accelerated object [or particles it is composed of] that somehow "ages" more slowly because of the energy applied to it? i.e., is it the change in mass, and not time that slows aging?
ANSWER:
You do not need to talk about acceleration. Time dilation says
that clocks which have a high velocity relative to you run more
slowly than your clocks. And, it is not that they look like they
are running slower, in fact they may look to be running faster,
but they actually do run slower; see
faq
page for links discussing this. The best way to see this, I
think, is to consider the light clock discussed in an
earlier answer. To understand the light clock, you must
accept that the speed of light is the same for all observers;
see the faq
page. Also, to help you with understanding time dilation,
read about the
twin paradox.
QUESTION:
What would happen if you threw a baseball at the speed of sound?
ANSWER:
At such a large speed, air drag has an enormous effect on ball.
To see the mathematical details see earlier answers for a
lacross ball and a
baseball. I will make the same assumption that I did in
those answers that the amount by which the ball will fall will
be very small compared to its horizontal distance and the speed
acquired in the vertical direction will be very small compared
to its horizontal speed. So I will ignore the small effect which
the vertical motion will have on the horizontal motion. As
discussed in the
earlier answer, the horizontal distance x and speed
v are given by v=v_{0}/(1+kt)
and x=(v_{0}/k)ln(1+kt)
where v_{0} is the initial velocity and k is a
constant determined by the mass, geometry, and initial velocity
of the ball. For v_{0}=340 m/s (speed of sound)
and a baseball (mass=0.15 kg, diameter=0.075 m) these become
v=340/(1+2.8t) and x=121ln(1+2.8t)
and are plotted below.
In one second the ball goes about 160 m and slows down to less than 100 m/s. During the same time, the ball will fall approximately 5 m and so, if launched horizontally from a height of 5 m will hit the ground in one second as shown below. Be sure to note the difference in vertical scales; an insert shows the trajectory drawn to scale. The small distance fallen is the justification for my approximations above.
QUESTION:
Hi! I am new to Quantum Mechanics and I am a little confuse about the representation of the spin. So if I have a 1/2 spin particle I usually put spin up in the positive direction of Z axis and spin down in the negative one and the probability of getting spin up or down is proportional to the cos^2 of half of the angle between spin and the certain direction. But if I have multiple options for the spin (e.g. in spin one i have spin 1,0,1) how do I visualize them? I mean If i put upward and downward in the z axis, and at z=0 for spin of 0 I don't get the probability one if I apply the above formula.
ANSWER:
I do not know what you mean about the "…cos^{2}
of half of the angle…", but you seem to not know what, e.g.,
spin
½ means. When you say spin s, this means that
the spin angular momentum quantum number is s; this means that
the magnitude of that spin angular momentum is
S=ħ√[s(s+1)]
where
ħ
is the rationalized Planck constant. But, there is another quantum
number, m_{s}=s, s+1, s+2, …,
s1, s which is the quantum number for the zcomponent
of S, S_{z}=m_{s}ħ.
Hence, the angle θ which the vector makes with the
positive zaxis is θ=cos^{1}(S_{z}/S).
For s=½, θ=cos^{1}(½/√(3/4))=54.7^{0}
and θ=cos^{1}(½/√(3/4))=125.3^{0};
similarly, for s=1, θ=45^{0}, 90^{0},
135^{0}.
QUESTION:
Does a uniformly charged ring rotating at constant angular velocity about its axis perpendicular to its plane radiate electromagnetic waves?
Because the magnetic field produced is constant hence there is no changing magnetic field and hence no em waves.is it ?
ANSWER:
Any accelerated charge radiates. The radiation from a charge
moving in a circle is called synchrotron radiation. The theory
of synchrotron radiation is very difficult and not appropriate
to do here. I can tell you, though, that the energy which the
charges have is so low that the power radiated would be
immeasurably small.
QUESTION:
I just saw a
music video in which a group of performers appear to be in an aeroplane cabin in freefall for 2 minutes and 40 seconds. The choreography is spectacular, and it appears to have been done in ONE take!
How far would the plane have had to descend to maintain zero gravity conditions for 160 seconds?
ANSWER:
This video was shot in Russia in a reducedgravity jet provided
by S7 Airlines. Weightlessness is not achieved by falling but by
following the parabolic path which a projectile would follow.
Imagine that someone shot you from a cannon with a speed of 500
mph (the typical speed of a commercial jet). If there were no
air drag, you would follow a parabolic path. The plane which
contains you now follows that exact path and that is how you
appear to be weightless. An alternative way would be for the
plane to simply go straight down with an acceleration of 9.8 m/s^{2},
but I think you can see that this would not be a very safe
situation. Anyhow, back to your question, I could not find
reference to any such parabolic flight lasting longer than 30 s,
so the video must have been shot in more than one take. Because
this is such an unfamiliar environment, I cannot believe that,
even with a lot of practice, it could be done in a single take
without errors. And, if the plane were simply falling for 160 s,
it would have to have started at an altitude of about 80 mi (far
higher than a plane can actually fly) and would end with a speed
of about 3600 mph (far faster than the plane could fly without
disintegrating).
QUESTION
(submitted by my daughter!):
Is
this exciting for you?? I dont really understand it but I'm trying to...it seems cool!
ANSWER:
Observing gravitational waves has been a holy grail of physics since
before I was an undergraduate (like 55 years ago, gasp!). So yes, it is pretty exciting news. This is actually just the first
direct evidence for gravitational waves.
Indirect evidence was found when observing a pair of stars orbiting each other and spiraling in toward a collision. The loss of energy turned out to be exactly
equal to the amount of energy they would lose if radiating gravitational waves. A nobel prize in physics was awarded in 1993 for this observation.
The animation below shows the predicted gravitational waves from
two neutron stars orbiting each other.
QUESTION:
A bit off beat from what you are usually asked. Jews are not allowed to drive a car on the Sabbath. The problem is the internal combustion engine which creates a
spark in the piston igniting the gasoline.
In a Tesla automobile does the magnetic field around the motor ultimately create a spark driving the wheels?
ANSWER:
It is natural, if you have a little knowledge about
electric motors, to think of the magnetic field driving the motor. In
fact, a magnetic field
never does work because of its very nature. Although the presence of
magnetic fields is imperative, work is always done by electric fields
they produce. Most
electric motors have "brushes", contacts which slide on the rotating
armature, and small sparks are inevitable when electrical contact is
made or broken. The Tesla electric car, however, employs a type of motor
called an
induction motor which is brushless and is therefore probably
sparkless.
QUESTION:
Hi this may be a hard question but If I wanted to run 2200 gpm through a 2500 foot run with 50 feet of fall what size pvc pipe would this take?
This is all gravity.
ANSWER:
At first I just did a calculation with no corrections for
viscosity or drag. I found the velocity had to be about 17 m/s
and the diameter of a pvc pipe would have to be about 0.1 m≈4
in. But then I worried about the fact that a pipe that long is
likely to have significant drag over its length. It is a pretty
complicated engineering calculation and I was unfamiliar with
many of the parameters. But I did find a
web site which seems to have made it easy for me by
including a calculator. Frankly, I have no idea what the
roughness coefficient is, but it suggests a value of 150 for
plastic. The result is below. As you can see, to get a flow rate
of about 2200 gpm would require a pipe with diameter of about 10
in.
QUESTION:
What if Einstein's General relativity wrong? Would all physics need to be rewritten?
ANSWER:
Actually, the theory of general relativity, essentially the
theory of gravity, has very little effect on most of physics.
Furthermore, like all theories, it is an idealization and approximation
and is already "wrong" at some level. It is also incomplete because
there is no theory of quantum gravity and it does not address the issues
of dark matter and dark energy.
QUESTION:
If I place a liquid filled container on a scale and suspend a mass with greater density than the liquid within the liquid and then release the mass, will the scale register the full weight of the mass while the mass is in motion (falling) as compared to when the mass has settled on the bottom?
Will the scale read the same while the mass is accelerating as when it has achieved terminal velocity?
ANSWER:
Your second question indicates that you understand the answer
will be different depending on whether the falling mass is accelerating
or not. The figure shows that the weights of the fluid and the container
will act down on the scale. Now look at the falling object. In addition
to its weight there are two upward forces, the buoyant force
B and the drag force D;
these are both forces which the fluid exerts on the object. But Newton's
third law says if the fluid exerts a force on the object, the object
exerts an equal and opposite force on the fluid. Therefore the scale
will read W_{f}+W_{c}+B+D. If
the object is falling with constant speed, it is in equilibrium and so
B+D=W_{o} and the scale reads the total weight of
container, object, and fluid. If the object is accelerating down,
B+D<W_{o} and the scale reads less than the total weight of
container, object, and fluid. There is an
earlier question
similar to yours except the object is rising instead of
sinking.
QUESTION:
Have scientists done experiment on what is the value of gravity below the earth surface as depth increases? if done pl. provide chart g vs depth.
ANSWER:
The
deepest hole ever drilled is only about 12 km deep. I could
not find any reference to attempts to measure g at various depths
down
this hole. Since the radius is about 6.4x10^{3} km, you would
only expect about a 0.2% variation over that distance. There are models
of the density of the earth, though, which have been determined by
observing waves transmitted through the earth during earthquakes or
nuclear bomb tests; these are believed to be a pretty good
representation of the radial density and can be used to calculate g.
The two figures above show the deduced density distribution and the
calculated g.
Usually in introductory physics classes we talk about the earth as having constant density, but as you can see, that is far from true—the core is much more dense than the mantles and crust. If it were true, g would decrease linearly to zero inside the earth. Instead, it increases slightly first to around 10 m/s^{2} and remains nearly constant until you are at a depth of around 2000 km. There is little likelihood that g will ever actually be measured deep inside the earth because the temperature increases greatly as you go deeper, already to near 200ºC at 12 km. However, if you have detailed information on density distribution, there is really no need to measure g.
QUESTION:
I'm so confused, my question is that if i hit the chair and chair change it's position from it's original position, then where is the reaction of chair and how according to third law of motion
ANSWER:
The force which you exert is on the
chair. The reaction force is the force which the chair exerts
on you. Only forces on the chair
determine how the chair will move.
QUESTION:
I have a question thats driving me nuts.
Say on a circle that spins, there are two points, one on the outer rim and one close to the rim.
Both are in line with each other and are traveling at the same time but the distance of the point on the outside making one revolution is longer than the distance of the one on the inside.
So the speed are the same but the distance of the inner point is shorter.
The time they take to complete one rotation is also the same as the two points stay in line with each other.
So speed is equal to distance divided by time as far as i know.
Something have to give, don't understand it.
ANSWER:
The two points have the same angular speed but different
translational speeds. For example, if the distance from the center is 1
m to the rim and 0.5 m to the other point, and the angular speed is 1
revolution/second, the distance traveled in one second is 2π
m for the rim and π m for the second point; the second point
has half the speed (π m/s) of the first (2π
m/s). Don't go nuts!
QUESTION:
Scientists say that centrifugal force is the only thing
stronger than the pull of a black hole. What i want to know is what
would happen if somehow the material around a black hole stopped
spinning around it what would happen and how might the spinning be
stopped?
ANSWER:
Funny, I never heard scientists say this. All that I can imagine
you are thinking about is that in the right circumstances an object
might orbit around a black hole just like the earth orbits around the
sun without falling in. But, the black hole exerts such a large force
that even light can orbit at a certain radius which is outside the
Schwartzchild radius; inside the orbiting light, nothing with mass could
have a stable orbit. See an earlier answer.
Obviously, if you stopped this orbiting, anything would fall into the
black hole.
QUESTION:
As part of our business we bag wrap passengers bags / suitcases prior to flying at the major UK airports.
We use and have used for many years a power pre stretch cast film – 17 micron nano with a 300% capability.
Recent feedback from Heathrow airport suggests some of the passengers bags are sticking to the conveyer belts and are being missdirected. I am being asked for the “coefficient of friction” for the film we are using.
I have advised our supplier of this, they have sent through the data spec sheet but there is no mention of COF, on speaking with them they have never had this question raised before. Personally, I do not think this Is an issue with our film but more where customers themselves are wrapping their own bags with home use film. However, I need to provide proof that the film we are using does not have any adhesive properties.
My question is – would the COF affect this and how do I get the actual information on the film?
ANSWER:
The force of friction f depends on only two things:
what the surfaces which are sliding on each other are (conveyer material
and your plastic film) and the force N which presses the two
surfaces together; normally, on a level surface, the force N is
simply the weight of the object (suitcase in your case). There are two
kinds of COFs, kinetic and static. The kinetic coefficient, μ_{k},
allows you to determine the frictional force on objects which are
sliding. In that case, f=μ_{k}N.
The static coefficient, μ_{s}, allows you to
determine how hard you have to push on the suitcase in order for it to
start sliding; in this case f_{max}=μ_{s}N
where f_{max }is the greatest frictional force you
can get. Since you are being asked to prove that it is not too "sticky",
it is the static, not the kinetic, coefficient which you need; measuring
μ_{s }is quite easy. The only problem is that μ_{s
}depends on the surfaces so you must have a piece of the material
from which the conveyer belts are made to make a measurement. Once you
have that, use it as an incline on which to place a wrapped suitcase.
Slowly increase the slope of the incline until the suitcase just begins
to slide. Your COF (μ_{s}) is equal to the tangent
of the angle of the incline which (see diagram above) is simply μ_{s}=H/L.
QUESTION:
I have a hill approximately 25 degree slope and 300 feet in length. Want to install a tow rope, what size of motor would I need at 240V, 3/4" rope and a 3 to 1 gear box. The anticipated speed is approximately 4 feet per second and at most I will have only two people on the tow rope at one time.
ANSWER:
I
can only give you a rough answer. I will calculate the power needed to
lift two large people (total mass 200 kg) up the hill. I will work in SI
units because that is the system in which the watt is the unit of power.
Your speed is v=4 ft/s=1.22 m/s so the rate at which the load
is rising is vsin25^{0}=0.516 m/s. The rate at which
the energy of this mass is increasing is mgv=1011 J/s≈1
kW. I guess I would throw in a safety factor to account for friction and
other energy losses, so maybe a 2 kW motor would do you. The voltage
(240) and gear box are not really relevant in determining the power.
Disclaimer: I am not an engineer, so you should get a second opinion!
QUESTION:
A 1500lb 8'x8' box that is 3'6" tall is lifted at one of the four side so that the opposite side acts as a pivot on the ground (like a strong man flipping a giant tire in a a world strongest man competition). How much actual weight is being lifted? assuming that the weight distribution of the box is perfectly even.
ANSWER:
Well, that depends on how you lift it. Let's assume that you
lift it so that you cause it to rotate with uniform speed. One way that
you could accomplish this is to push in a direction perpendicular a line
drawn from where you are pushing and the edge on the box remaining on
the floor. Referring to the figure above, the equilibrium conditions are
N+FcosθW=0, fFsinθ=0,
and ½LWLFcosθ=½WFcosθ=0;
here F is the force you exert, W is the weight, N
is the force the floor exerts vertically, and f is the
frictional force exerted on the floor. Solving these I find that N=½W,
F=½W/cosθ, and f=½Wtanθ.
I suspect that the case you are interested in is when you first lift it
off the ground, θ=0. F=750 lb, half the total
weight. Note that this analysis is valid as long as the floor is not too
smooth, that is the box does not start sliding at some angle; the angle
is less than θ=tan^{1}(L/H)
because at that angle the center of gravity is directly over the pivot
side.
Of course there are lots of other ways you could lift it which would be more efficient if your aim was to tip it over; for example, you could start pushing horizontally once you got it off the ground so that the floor would hold up all the weight rather than half the weight. There is an old answer very similar to yours that you might be interested in.
QUESTION:
If a tennis ball and a football is thrown from a certain hieght then which ball will land first?
ANSWER:
I will assume that you do not mean an American football, rather
a soccer ball. Also, it is not unambiguous what "…thrown
from a certain height…" means. Imagine that we just drop each
from some height. If air friction is ignored, they hit the ground
simultaneously; this would approximately be the case if they were
dropped from a few meters. However, air drag becomes more important as
the speed increases. The drag force f may be approximated as
f≈¼Av^{2} where A is the
cross sectional area (πR^{2}) and v is the
speed (you must use SI units). So, for a ball, fmg=ma where
m is the mass, a is the acceleration, and g
is the acceleration due to gravity. Since f gets bigger as
v gets bigger, eventually f=mg and the ball will stop
accelerating and fall with a constant speed. The velocity at which this
happens is called the terminal velocity v_{t}=√[4mg/(πR^{2})].
The mass and radius of a tennis ball are m=0.059 kg and R=0.069
m; the mass and radius of a soccer ball are m=0.43 kg and R=0.22
m. So, I find v_{t}^{tennis}≈50 m/s for
the tennis ball and v_{t}^{soccer}≈11 m/s
for the soccer ball. The tennis ball will easily win the race because it
continues accelerating long after the soccer ball stopped accelerating.
QUESTION:
I love hearing about new discoveries from particle accelerators, but one aspect of them confuses me. If we accelerate particles to 99% of the speed of light, shouldn't that make the matter appear to go slower to us, a la the twins paradox? Why don't these particles approach infinite mass and compress time from our perspective?
ANSWER:
Stable particles, those usually accelerated, do not carry clocks
with them; but, if they did, those clocks would run slowly compared to
yours. Some particles do carry clocks, those which are unstable. Suppose
that the average lifetime of some particle were 1 s and its speed was
0.99c. You would observe it to live for 1/√(1.99^{2})=7.1
s. The accelerated particles do indeed approach infinite mass but they
have a really long way to go to get there even in the most powerful
accelerators; see an
earlier answer.
QUESTION:
My question relates to my own voluntary time involved with the rope/rigging community. It is not of any commercial interest to me, but the answer to my question has eluded me for a long time despite research and attempts to calculate it ! If you could at least explain to me the correct method of attempting to work our the answer I would greatly appreciate it.
Question background  horizontal restraint line.
A 7.6 m long horizontal rope is anchored at both ends. It is 2m horizontally above ground level (AGL). It is pretensioned to 50kg.
A 20kg mass is suspended vertical above the horizontal rope and attached to the rope by a leash of 0.7 m length.
The object/leash attachment point is midway along the 7.6 m horizontal rope i.e.
3.8 m from either anchor.
The 20 kg object is then dropped.
At the time of peak impact force, the sag in the horizontal tension line is 1.0 m. Despite my best efforts I do not seem to be able to calculate theoretical (i) peak impact forces on the anchors (ii) peak impact force on the object.
I do however have peak impact force load cells that have recorded average (i) peak impact forces on the anchor 220
kg and (ii) peak impact force on the object 115 kg.
Is it possible for you to detail the correct formula/method for at least working the theoretical approach to calculating these forces ?
QUERY:
It would be helpful if you could tell me the sag and forces when the mass is
hanging at rest on its leash. I take it that the 115 kg measurement is the
tension in the leash.
REPLY:
ANSWER:
Let me first try to understand the data for the "loads"
(essentially the tension in the rope, T, and
the tension in the leash, F)
in terms of simple physics; then I will try to generalize it. Normally,
physicists do not like to use kg to measure force, but I will go ahead
and do that here. Focusing on the point where the leash is tied to the
rope, the force which the rope pulls up is 2Tsinθ=F.
In the case of the hanging mass,
sinθ=s/√(L^{2}+s^{2})=0.41/√(3.8^{2}+0.41^{2})=0.11,
so F=20=0.22T; thus T=91 kg which is in
reasonable agreement with the measured value of 100 kg. In the case of
the maximum s,
sinθ=0.25 and F=0.51T.
Taking the measured value of F=115 kg, T=225 kg, again
in good agreement with the measured value of T=220 kg.
At this point what a physicist normally does is to try to understand the situation in terms of a simple spring model. If the rope is like a simple spring, i.e. its tension is proportional to its stretch, you can usually make approximations which would result in the simple model that F≈ks where k is a constant. For the equilibrium situation, then, 20≈0.41k or k≈49 kg/m. This would then predict that the force when s=1 m would be F=49 kg which is too small by more than a factor of 2.
The previous try indicates that the rope is probably not approximated as an ideal spring. My last attempt is to try to treat the rope more explicitly as an ideal spring, not using small approximations used above. So, looking at one half the rope, I will write T=50+kδ where δ=√(L^{2}+s^{2})L is the amount by which this half of the rope is stretched relative to L. Note that k is not the same as in the approximation above where s was the stretch parameter rather than δ. Calculating δ for the at rest situation where T=100 kg, k=2270 kg/m. Then, calculating T for the maximum s using this value of k, T=50+2270x0.129=344 kg which is too large by nearly 60% compared to the measured value of 220.
I conclude that the rope is poorly approximated as an ideal spring and that to do any more detailed analysis of this problem would require measuring s as a function of the load by varying the load for the equilibrium situations.
FOLLOWUP
QUESTION:
If possible, without the aid of the load cell values which I get by practically performing the drop test. How would you derive the theoretical values for the tension/force at the both the anchor and on the body in the drop test scenario. I understand how to calculate forces etc for the system at rest, but it is the dynamic falling system, deceleration etc that I am unable to clarify.
To summarise, if you have the load/object, placed at any height above or below the horizontal line (depending on the leash length), without loading the horizontal line, and then you drop the object, what would be the correct formula method to determine the approximate theoretical peak impact forces on the anchors and body, without the luxury of any load cell info. when it comes to rest after being decelerated by the horizontal line.
ANSWER:
You apparently did not understand my answer, particularly the second and third paragraphs. The first paragraph shows that the data are consistent with the simple model for the tensions in the ropes. I found these very encouraging. From the data for the dynamic point you could, of course, calculate the acceleration of the weight at that point. 20a=(11520)g so
a=4.75g=46.6 m/s^{2}.
Then I tried to model the rope as a simple spring, the tension is proportional to the stretch. I failed to do so in both attempts. You do not know the force which this rope will exert given a certain amount of stretch. Physics may be powerful, but you cannot do any predictive calculation if you are ignorant of the force. I have two data points but that is not enough if the force is not linear. The last paragraph suggests the only hope for having a predictive analysis: you need to measure the tension in the rope as a function of what you call sag. To do this you need to use many weights, say at 2 kg steps, and measure load#1 and sag. This would give me the information necessary to do predictive calculations. Life in the real world is not always simple and analytical. It occurs to me you do not really need to hang varying weights because you have the load#2 device; just pull down on it until it reads 2 kg, 4 kg, 6 kg, etc.
It is, of course, simple to calculate the speed the weight has just before the leash goes taught: v=√(2gh) where is the distance fallen.
QUESTION:
A conductor, even though it is carrying a current, has zero net charge. why then does a magnetic field exert a force on it?
ANSWER:
Magnetic fields exert forces on moving charges. In the
conductor, negative charges are moving but positive charges are not. A
magnetic field exerts no force on charges at rest.
QUESTION:
Do all five balls on a Newton's Cradle have to weigh the exact same amount in order for it work correctly?
ANSWER:
No. The balls on the ends need to be the same mass. However, you
could no longer do the demonstration where if you send 2, 3, or 4 balls
in that there would be 2, 3, or 4 balls out.
QUESTION:
If F=Q1.Q1/R^2, does it mean that two unlike charges touching each other have infinite forces of attraction between each other? If so, how are we able to separate 2 oppositely charges objects which are stuck together with our bare hands?
ANSWER:
As often happens, you are applying an equation (Coulomb's law)
without asking whether it applies. This law is true for two point
charges or for two sphericallysymmetric charge distributions; in the
second case, R is the distance between the centers of
the spheres. So, if you have two charged insulating spheres of radii 1
cm and 2 cm which are touching, R=3 cm.
QUESTION:
What is inertia? From what I understand, the inertia for a sphere is: I = (2/5)MR^2. Why is the radius included? Why is the volume of the sphere related to the resistance to change motion? As long as any 2 objects have the same mass, why should they have different inertias?
ANSWER:
Inertia means the inherant ability of something to resist being
accelerated. There are two kinds of acceleration, translational
acceleration a (like a car speeding up or slowing down), and
rotational acceleration α like the rotational speed of an
engine's flywheel speeding up or slowing down. Inertia for resisting
a is simply mass which you can see from Newton's second law,
a=F/m. If you double the mass on which a given force acts,
you halve the acceleration; mass is sometimes called inertial mass for
this reason. For rotational acceleration, the rotational inertia depends
not just on how much mass there is, it also depends on how it is
distributed; for example, it requires much more effort to get a wheel
with radius 1 m spinning than it does to spin a wheel with the same mass
but a much smaller radius. The rotational form of Newton's second law is
α=τ/I where τ is the torque
applied to the object whose moment of inertia is I. In
rotational physics torque plays the role of force and moment of inertia
plays the role of mass.
QUESTION:
If friction acts perpendicular to the direction of motion, if i place an object against a wall and let go, since the object is falling downwards, will friction from the wall cause the object to move away from the wall? Also if I am standing still, gravity results in a downwards force so friction should act leftwards or rightwards. Which one is the correct one (assuming that i am not moving)?
ANSWER:
Friction does not act perpendicular to the direction of the
motion but opposite it. When the surface is vertical there is no
frictional force. This ignores irregularities in the wall and air drag.
If you are standing on a horizontal surface and not moving, the
frictional force on you from the floor is zero.
QUESTION:
I am an engineer working on the reconstruction of a traffic accident where it is alleged that a car traveling over a railroad crossing became airborne at a speed lower than the posted speed of the road. The information that I have available includes the type and make of the car and the geometry of the road. Nothing in the engineering literature that I can find addresses this issue. Despite the five quarters of physics that I took long ago, I am having trouble finding the information that I need to model this.
Any suggestions?
ANSWER:
This is sort of a classic introductory physics problem. The idea
is: when does an object lose contact with the surface on which it is
moving, usually taken to be a segment of a circle. I will do an
approximation that the shape is a circle and that the car is a point
mass. You can then generalize to your case from there or else give me
more information. In the sketch above, the radius of the circle is R,
the weight of the car is mg, the angle specifying the current
position of the car is θ, the force which is
causing the car to move with some constant speed v is F,
and the normal force of the road on the car is N. Note that
although the car has a constant speed, it has a centripetal acceleration
a=v^{2}/R toward the center of the circle.
Applying Newton's laws, Fmgcosθ=0 and mgsinθN=mv^{2}/R.
The first equation tells you what force you need to keep it moving at a
constant speed, F=mgcosθ which is really not of
interest to you; note that the force is in the direction of v
on the way up and opposite on the way down since the cosine changes sign
at 90^{0}. The second equation tells you what N is for
any position of the car, N=mgsinθmv^{2}/R;
note that if N is negative it points toward the center but the
road cannot pull down, only push up, so the car could not stay on the
road at that speed and angle. What is really of interest is under what
conditions would N=0=gsinθv^{2}/R
or v^{2}=Rgsinθ; this would
tell you the speed (angle) at which a car with a particular angle
(speed) would leave the road. Note that it is independent of the mass.
Notice also that v^{2}/Rg<1 because the sine
function cannot be larger than 1.0. For example, for what speed will the
car leave the road at 45^{0} if gR=300 (m/s)^{2}
(R≈30 m)? v=√(300x0.707)=14.6 m/s=32.7
mph. Or, at what angle will a car with speed 35 mph=15.6 m/s leave the
road? θ=arcsin(15.6^{2}/300)=54.2^{0}.
One thing which occurs to me though is, since you know the car and railroad crossing, why don't you just do the experiment and drive it over the crossing?
QUESTION:
When i place an object in between my 2 palms, why does it not fall? My palms only supply horizontal force, where does the vertical force come from to hold the object between my palms without falling?
ANSWER:
You are wrong to say that your "…palms only supply
a horizontal force…" When surfaces are in contact, the forces
they exert on each other have components both perpendicular (normal) and
parallel (friction) to the surfaces. It is the friction which holds the
object. If the object were very heavy, say 100 lb, and your hands were
greased, you would not be able to hold it up because the frictional
forces would not be as large as the weight.
QUESTION:
Say there is a cylinder on a ramp and the friction force from the ramp cancels out the parallel component of gravity. Therefore, the cylinder should be in linear equilibrium. However, from the reference point of the center axis of the cylinder, there is a net torque exerted by the friction force. Additionally, there is also a net torque exerted by the gravitational force from the reference point of the point of contact of the cylinder and the ramp. Therefore, it is not in rotational equilibrium and should start to rotate, correct? How is this possible, because if the cylinder starts to roll how can it also be in linear equilibrium?
ANSWER:
There is a simple answer to your question: the frictional force
is not equal to the component of the weight along the incline. Rather,
f+mgsinθ=ma where θ is
the angle of the incline and a is the acceleration of the center of mass
down the incline.
FOLLOWUP QUESTION:
Thank you very much for your response. However, I think you may have misunderstood my question. I was asking what would happen in a case where the frictional force is set to cancel out the parallel component of weight. It seems as if the center of mass cannot move, but the cylinder needs to rotate. Therefore, it would appear as if the only outcome of this situation would be a cylinder rotating in place on a ramp, which does not seem possible. I think that the cylinder would have to roll down the ramp, but I can't see how this would be consistent with linear equilibrium.
ANSWER:
I did not misunderstand your question. You cannot simply adjust
the friction to be what you want it to be. You can, however, simulate
what you want to happen by wrapping a string around the cylinder and
pulling up on the string with a force mgsinθ
where
θ is the angle of the incline and m is the mass
of the cylinder; imagine that the incline is smooth (frictionless). Now
there will be a net torque about the center of mass of mgRsinθ=Iα
where I is the moment of inertia and
α is the angular acceleration of the cylinder. The
cylinder will spin in place. Your hand will have an acceleration of
a=mgR^{2}sinθ/I; for a uniform
solid cylinder with I=½mR^{2}, a=2gsinθ.
ADDED
THOUGHT:
If the coefficient of kinetic friction is exactly equal to
μ_{k}=tanθ,
the cylinder will slide down the incline with constant speed because the
frictional force will be f=μ_{k}N=(tanθ)(mgcosθ)=mgsinθ.
So, if you start it sliding and not rolling, it will begin spinning
about its center of mass because of the torque due to the friction and
have an angular acceleration α=fR/I; it will continue
sliding down the plane with constant speed, though.
FOLLOWUP QUESTION:
Thanks again but I am still a bit confused. It makes sense that the center of mass will move at a constant velocity while the cylinder is rolling, but how did it acquire that velocity in the first place if I start the cylinder at rest and not sliding as you wrote in the additional thought. In other words, what happens if the coefficient of static friction is equal to mgsinθ and the cylinder starts with no velocity of any kind?
ANSWER:
If you start the cylinder at rest on the incline and
μ_{k}=tanθ
(coefficient of static friction will be larger),
the cylinder will roll without slipping. If you solve the dynamics for
the cylinder rolling without slipping you will find that (see the figure
above) f=(mgsinθ)/3 and a=(2gsinθ)/3
where f is the frictional force and a is the
acceleration of the center of mass. Since f<mgsinθ,
if you simply let it go, it will roll without slipping. So, if you want
it to slide down with constant speed, you must give it a shove to start
it slipping.
QUESTION:
Does mass of an object increase its mass exponentially as it approaches infinitely close to the speed of light?
ANSWER:
If df/dq=Aq for some function f(q),
where A is a positive constant, f is said to be
exponentially increasing with q. For your question, f
is m and q is v. The expression for m(v)
is m=m_{0}/√[1v^{2}/c^{2}]
where m_{0} is the rest mass. Calculating the
derivative, dm/dv=mv/[c^{2}v^{2}].
Thus, although m increases without bound as v
increases, it does not increase exponentially.
QUESTION:
Can sustaining enough angular momentum help us stay in a black hole for longer(near the event horizon)? If yes, how much energy would be required to sustain it for ten minutes.
ANSWER:
What do you mean by "in a black hole"? A black hole is a
singularity so if you were "in" it you could not have any angular
momentum. In any case, I am assuming that your idea is to orbit the
black hole so as not to fall in. If you were able to do this, no energy
would be required; the earth orbits the sun in a stable orbit with no
energy input. However, it is not possible for any object to orbit a
black hole anywhere near the Scwartzschild radius because the speed
would be too large. At a radius of 1.5 the Schwartzschild radius, light
would orbit a black hole in a circular orbit (see
earlier answer). This is a result of general relativity. No stable
orbits inside this exist and anything inside would fall into the black
hole.
QUESTION:
Hi! I have been told that hypothetically speaking when a metal bar or something other object travels in space at the speed of light it shrinks.
So my question is if it is correct and why yes or no?
ANSWER:
First of all, nothing can travel at the speed of light. However,
the fact that moving lengths are shorter is not hypothetical, it is
simple fact which has been verified by measurements. It is the result of
the special theory of relativity. And, it is not because they appear
to be shorter, they actually are shorter. You might look at an
earlier answer about length contraction.
QUESTION:
We have material that generates electricity when exposed to light, or force.
Why haven't not found one that does this when exposed to atomic radiation. Imagine almost permanent batteries.
ANSWER:
There are
batteries which get their primary energy from radioactive decay.
Atomic batteries are routinely used in heart pacemakers and lowpower
requirements in space probes.
QUESTION:
My question is about the theory of relativity and time dilation. It appears to only work in one direction and I'm not sure why. For example, take a person on earth and compare to a person in a rocket going near the speed of light. The person on earth observes the rocket man going near light speed and aging slower. The man in the rocket observes earth moving past him near the speed of light except the earth man ages quicker.
ANSWER:
Congratulations, you have discovered the twin paradox! It is
true that the earth man's clock runs slow in the rocket man's frame. It
also is true that the rocket man's clock runs slow in the earth man's
frame. But to make a definitive comparison, they must somehow bring
their clocks together to compare. In other words, the rocket man has to
come back to compare his clock with the earthbound clock. To get the
full explanation of the twin paradox, look on the
faq page.
QUESTION:
My question is about the theory of relativity and time dilation. It appears to only work in one direction and I'm not sure why. For example, take a person on earth and compare to a person in a rocket going near the speed of light. The person on earth observes the rocket man going near light speed and aging slower. The man in the rocket observes earth moving past him near the speed of light except the earth man ages quicker.
ANSWER:
Congratulations, you have discovered the twin paradox! It is
true that the earth man's clock runs slow in the rocket man's frame. It
also is true that the rocket man's clock runs slow in the earth man's
frame. But to make a definitive comparison, they must somehow bring
their clocks together to compare. In other words, the rocket man has to
come back to compare his clock with the earthbound clock. To get the
full explanation of the twin paradox, look on the
faq page.
QUESTION:
Can a ball bounce higher than the height it was dropped? I know that air resistance would slow it down but is it possible?
ANSWER:
Even if you bounced a ball in a vacuum where there would be no
air resistance, it could not bounce higher than the height from which
you dropped it. The only way to achieve this would be to add more energy
to the ball, either throw it down rather than drop it or maybe have a
little "kick" from its interaction with the floor.
QUESTION:
An icehockey player throws his stick on the ice. The stick translates and
rotates. Before it stops, it always rotates and translates  never only
rotates or translates. Why do both motions occur simultaneously like this?
ANSWER:
It actually depends on how it is thrown
FOLLOWUP QUESTION:
Suppose it isn't thrown intentionally in such a way so as to just make it spin or just make it translate.
ANSWER:
The motion of a rigid body moving in two dimensions (the ice,
call it the xy plane) may be broken into two components,
motion of the center of mass (translation) and motion about
the center of mass (rotation). There are, in your example, no forces in
the x or ydirection once the stick has been thrown
if you neglect friction. If the force you throw it with is directed
through the center of mass, it will not rotate because there is no
torque about the center of mass to get it rotating. If the net force you
throw it with is zero (you would need to use two hands to do this) it
will not translate. If you just randomly grabbed it and threw it, there
would be a net force which would result in the center of mass
accelerating during the time you were throwing it; and there would be a
net torque which would result in the stick having a rotational
acceleration during the time you were throwing it. Once you let go,
there is no force and no torque on the stick and so both linear and
angular momentum would be conserved meaning it would continue moving the
way it was moving when you released it.
QUESTION:
Suppose a container full of water is kept in a certain area. Let no external force
be applied. Now the question is: Will there be an overall circulation of molecules of water?
Or will the rest position of water perfectly 100% at rest?
ANSWER:
There is a difference between no "circulation of molecules" and
"100% at rest." The molecules in a glass of water are never at rest,
they are always moving around. However, in a glass which has been
sitting undisturbed for a long time the average velocity is zero, there
are always as many moving in any direction as there are moving in the
opposite direction. However, you could have a fluid which has a zero
average velocity of molecules but they could be circulating (like a
whirlpool). If you start such a circulation, it will often last for a
relatively long time but, due to viscosity and fluid drag forces, it
will eventually die out.
QUESTION:
This is a brain teaser I have been having trouble with. Lets say a clock is moving towards me at a 50% of the speed of light. When the clock is moving towards me, do I observe its hands to be ticking slower, faster, or at the same rate compared to the rate of ticks that I observe when it's moving away from me?
ANSWER:
A clock will appear to run faster than yours when
moving toward you. A clock will appear to run slower than yours
when moving away from you. But that clock is actually running
more slowly than yours in both directions (and not by the same factor as
it appears to run when going away from you). I think you will understand
why is you look at earlier answers to a particular
faq on the faq page. Keep in
mind that moving clocks run slow, they do not just appear to run slow;
how things are and how things seem are often not the same.
QUESTION:
Many Science Fiction ships have for protection a shield that stops projectiles and saves the ship from a lethal impact. Such an example would be the energy shields possessed by the faction known as Covenant from Halo.
So given the laws of Newton and the third law about there always being an equal force countering the force that was exerted first, wouldn't this just mean that whenever the energy shield takes a hit by a projectile, the force would ''travel'' from the shield to whatever generator generates the shield.
Like the UNSC Frigates shoot a 600 ton slug at 30km/s. They're 30ft long and around 7ft wide. So they have a lot of force and momentum behind them.
So wouldn't an impact like this just cause the shield generator to be violently thrown off its attachments and ''fly off'' through the compartments of the ship destroying a lot of the ship?
ANSWER:
"So they have a lot of force and momentum behind them." Yes, these
projectiles have a lot of momentum, but saying that "…there is a
lot of force…behind them…"
has no meaning. The momentum a slug has is p=mv=(6x10^{5}
kg)(3x10^{4} m/s)=1.8x10^{10} kg m/s. When this hits
something, it will certainly exert a force, but the magnitude of that
force will depend on how long it took to stop. I have no idea how big it
is, but suppose it has a radius of 10 km from the ship; I will think of
it as very flexible and suppose that it stretches inward just stopping
before it hits the ship. I also will assume that the ship is much more
massive than the slug; (elsewise how could a comparablysized ship carry
a bunch of them and fire them without huge recoil?) If it stops in 10
km=10^{4} m with uniform acceleration, I can apply simple
kinematics, 3x10^{4}=at and 10^{4}=3x10^{4}+½at^{2},
and find that t=1.5 s, the time for the slug to stop. The
average force felt by the slug is (Newton's second law) the rate of
change of the momentum, F=1.8x10^{10}/1.5=1.2x10^{10}
N. You are certainly right, this very large force will be felt by the
ship because of Newton's third law. But suppose that the ship is 100,000
times more massive than the slug, 6x10^{10 }kg; in that case,
the final velocity of the ship after being hit will be found from F=mv/t=6x10^{10}v/1.5=1.2x10^{10}
N, so v=0.3 m/s, not so bad! If the shield were very rigid, it
would be catastrophe for the ship. I have never played these games but I
expect the shield is shown as stopping the slug almost instantly.
A little should be said about the other end, launching the slug in the first place. In this case, unless the cannon has a 10 km barrel, the recoil force on the ship will be huge. You would be interested in similar Q&As along the same line as yours.
QUESTION:
If I were to drop two round balls of different mass under water would they both fall to the bottom at the same velocity or would one reach the bottom first?
ANSWER:
The forces on a ball are its own weight mg down; the
buoyant force B up which would be equal in magnitude to the
weight of an equal volume of water; and the drag force f up
which would depend on the size the ball and its speed. The net force
F would be F=mg+B+f; as the ball went faster and faster,
f would get bigger and bigger until eventually F=0 and
the ball would go down with a constant speed called the terminal
velocity. The larger the mass, the larger the terminal velocity. Without
specifying the the sizes of the balls, your question cannot be answered.
If they had identical sizes, the heavier ball would reach the bottom
first because it would have a larger terminal velocity. Of course, if
B>mg the ball would float!
QUESTION:
If you performed the double slit experiment in outer space and were observing the electron particles would you get a diffraction pattern or two rows of particles. Does the earth's atmosphere have any effect on the experiment?
ANSWER:
Any electron diffraction experiment is always performed
in a vacuum. The range of the electrons in air is short enough that it
would entirely ruin any experiment you tried to do. Electrons interact
strongly with any atoms in their vicinity. In outer space you would have
a vacuum so you would see a diffraction pattern.
QUESTION:
I'm trying to find a fast/easy way to test whether a sealed, consistentlydimensioned rectangular box is sufficiently "stable" for transport on a (small) 2wheeled bicycle trailer. The box is pretty tall. If it's overweighted and topheavy, it'll flip the trailer around turns (which are sufficiently tight/quick). I figure there might be a quick, static "tip test" with a combination pull gauge, inclinometer and scale, but my math skills are primitive. Is there a simple way to ascertain whether, for a given object, a target stability threshold is met?
ANSWER:
There is an
earlier
answer about a bicycle making a turn. It would be helpful for you to
read that first. The easiest way
to do this problem of your trailer turning a curve is to introduce a fictitious
centrifugal force which I will call C, pointed
away from the center of the circle; the magnitude of this force will be mv^{2}/R where
m is the mass of the box plus trailer, v
is its speed, and R is the radius of the curve. The picture above shows all the forces on the box plus trailer: W is the weight and the green
x is the
center of gravity (COG) of the box plus trailer; f_{1} and f_{2} are the
frictional forces exerted by the road on the inside and outside wheels
respectively; N_{1} and N_{2} are the normal
forces exerted by the road on the inside and outside wheels respectively;
the center of gravity of the box plus trailer is a distance H above the road and the wheel base is
2L (with the center of gravity halfway between the wheels). Newton's
equations yield:

f_{1}+f_{1}=C for equilibrium of horizontal forces;

N_{1}+N_{2}=W for equilibrium of vertical forces;

CH+L(N_{1}N_{2})=0 for equilibrium of torques about the red x.
If you work this out, you find the normal forces which are indicative of the weight the wheels support: N_{1}=½(WC(H/L)) and N_{2}=½(W+C(H/L)). A few things to note are:

the outer wheel supports more weight,

if C=0 (you are not turning), the inner and outer wheels each support half the weight,

at a high enough speed C will become so large that N_{1}=0 and if you go any faster you will tip over, and

if the road cannot provide enough friction you will skid before you will tip over.
Now we come to your question. You first want the maximum speed without tipping. Solving for v in the N_{1}=0 equation gives
v_{max}=√[RWL/(mH)]=√[RgL/H]
where g=9.8 m/s^{2}=32 ft/s^{2} is the acceleration due to gravity. For example, suppose that R=7 ft, L=17 in=1.42 ft, H=30 in=2.5 ft. Then v_{max}=√[7x32x1.42/(2.5)]=11.3 ft/s=7.7 mph.
Be sure to note that the assumptions of a level road (not banked) and wheels not slipping are used in my calculations. Also be sure to note that W is the weight of both box and trailer and 2L is the wheel base, not the box width.
One more thing is that you might not know how to find the COG of the trailer plus box. If the COG of the trailer is H_{trailer }above the ground (probably close to the axle) and the COG of the box is H_{box} above the ground, then H=(H_{box}xW_{box}+H_{trailer}xW_{trailer})/W.
ADDED
THOUGHT:
When just about to tip, all the weight is on the outer wheel and
so N_{2}=W and f_{2}=μN_{2}=μW,
where μ is the coefficient of static friction. If
you work it out, the minimum value the μ must have
to keep the trailer from skidding is μ_{min}=L/H.
For the example worked out above, μ_{min}=0.57.
For comparison, μ for rubber on dry asphalt
is about 0.9, so the trailer would not skid.
QUESTION:
If a 20 pound rock and 100 pound rock or drop from 1000 feet which will hit the ground first.
ANSWER:
If air drag is neglected, they would hit simultaneously. If air drag is
considered, it would depend on the geometry of the two objects. If they have
identical shapes and sizes, the heavier rock would hit first; otherwise, you
would need to know the shape and size of each to calculate the times to
fall. See the faq page.
QUESTION:
A starship pilot wants to set her spaceship to light speed but the crew and passengers can only endure a force up to 1.2 times their weight. Assuming the pilot can maintain a constant rate of acceleration, what is the minimum time she can safely achieve light speed?
ANSWER:
This question completely ignores special relativity. It is impossible to go
as fast as the speed of light. Furthermore, acceleration is not really a
useful quantity in special relativity and you must use special relativity
when speeds become comparable to the speed of light. I have
earlier worked out the
velocity of something which would correspond to occupants of your spaceship
experiencing a force equal to their own weight due to the acceleration which
I will adapt to your case later. (See the graph above.) First, though, I
will work out the (incorrect) Newtonian calculation which is presumably what
you want. The appropriate equation would be v=at where v=3x10^{8}
m/s, a=1.2g=11.8 m/s^{2}, and t is the
time to reach v; the solution is 2.5x10^{7} s=0.79 years.
For the correct calculation, you cannot reach the speed of light; from the
graph above (black curve), though, you can see that you would reach more than 99%
of c when
gt/c=3. To make this your problem, we simply replace g by 1.2g
and solve for t. I find that t=7.7x10^{7} s=2.4
years.
QUESTION:
Why do wavefunctions need to be normalized?
ANSWER:
So that the absolute square of the wave function can be interpreted as a
probability density. The integral of ψ*ψ over all
space is then the probability of finding the particle somewhere which, of
course, must be 1.
QUESTION:
Will a heavier ball roll down a small slope faster than a lighter ball?
Or will a lighter ball roll down a small slope faster than a heavier ball?
ANSWER:
If both are solid balls of the same radius but of different masses, they
will take equal times if air drag is neglected as might be appropriate for a
"small slope". If they get going fast enough that air drag becomes
important, the heavier ball will win the race.
QUESTION:
During refraction why does a light ray bend...and how we can account for conservation of momentum?
ANSWER:
This is not simple. See
this article.
QUESTION:
The following situation occured with my son, would you mind sending me the
formula or the resulting impact force?
Car A 1430kg rearends car B 1143 kg. Car B (Automatic gear is on park) is
shoved 2 meters (6.56168 feet) before stop.
What is the impact force on Car B Tons?
ANSWER:
No way to calculate this without much more information. See the FAQ page for
other impossible to calculate the force situations.
FOLLOWUP QUESTION:
Thank you very much for your reply.
I went to your site faq and understand what you mean (speed + stop time),
but on this site
it gives me force on the car 7. Tones with just the weight of car one and 2
meters of course it refers to 1 car hitting a tree, but the force of impact
must be close.
Does it sound reasonable to you?
ANSWER:
Maybe I could make a very rough estimate if I knew whether the incoming car
was skidding, if the pavement was dry asphalt, was there any notable
compression of the cars (like a bumper moved in by 2 cm)? Would a speed of
about 15 mph be reasonable?
FOLLOWUP QUESTION:
Thank you very much for tryng to help find the force of impact.
It was a mass pile up 6 cars (cars 1 to 4 Totalled)
01 Car 1950kg (estimated speed 55/60mph) >02 Car 1172kg no brakes>(((Car 03 A 1430kg no brakes >Car 04 B 1143 kg in Park gear)))>C 05 1270Kg (+ 4 occupants + 280Kg no brakes)> Shoved 1 meter to Car 06 Irrelevant
 To simplify the calculations I reduced it to Car A and Car B
 It was on dry asphat. The incoming (A) car was shoved 1meter (neutral gear no brakes) before hitting B and Car B in Park was shoved 2 meters into Car C which was shoved 1 meter into Car D
 B car rear bumper was compressed by 25cm (lower body frame steel bent 10cm)
 The estimated speed 25 mph of impact (Car 03 A to Car 04 B)
ANSWER:
Good grief! If I understand things correctly, it really is impossible now to do any meaningful calculations. I had thought from your earlier question that car B was at rest after going 2 m but now there are far too many unknowns. With two cars there was a possibility that I could have done a very rough estimate, with this situation it is truly hopeless. There is no reasonable way to estimate car A’s speed or car B’s speed after 2 m.
FOLLOWUP QUESTION:
Yep. That's why I simplified it to 2 cars.
According to the lady in car A 1430kg she thinks her car was travelling at about 40Km / (I would say 25 miles) and shoved car B 1143 kg for 2.5 meter until it stop. (Stop means B hit C but this is irrelavant to a rough estimate)
Dry asphalt
Car B 25 cm rear bumper compression
Without making it complicated with the compounded forces from other cars etc, all I would like to have a rough idea is, what aproximatly was the force of impact exerted from Car A onto car B.
Do you that is possible to roughly calculate the force of impact Tons?
ANSWER:
OK, I give in! It is not irrelevant that car B was moving when it hit car C,
but we can make a rough estimate now that I have the speed of A and the
compression of the damage on B without any information about after the
collision. I will assume that the time of the collision (compression of 25
cm=0.25 m) is very short so that I can ignore friction of the road, the
collision is perfectly inelastic (A and B stick together), the speed of car
A was 40 km/hr=11 m/s, and the compression of car B is negligible. Because
this is a very rough estimate, I will retain only two significant figures
throughout. Then the speed v after the collision has finished of
the cars may be estimated from momentum conservation: 1400x11=(1400+1100)v,
so v=6.2 m/s. The average force on car B is the rate of change in
its momentum F=(6.2x1100)/t=6800/t where t
is the time of the collision. Since we know the acceleration a is
6.2/t, we can use the kinematic equation for distance to get the
time: 0.25=½at^{2}=3.1t, so the time
of collision is t=0.081 s; therefore F=84,000 N=19,000
lb=8.5 ton.
QUESTION:
Why
must the trajectory of the reducedgravity airplane to be parabolic. Why not arc of the circle or anything else?
ANSWER:
The reason is that the path a projectile follows is a parabola. If you were
in an airplane and it suddenly disappeared, you would follow a parabolic
path determined by what your velocity was when the plane disappeared. So, if
the plane flies to follow that path, you would be in free fall. For more
information about this, see an
earlier
answer.
QUESTION:
I am writing a blog and had my friends to ask of what they would want to talk about.. I get this one. I have no clue with physics, so, I am asking. Explain why a photon particle which is a very small bundle of energy and travels at the speed of light seems to defy the laws of physics by never losing speed or velocity?
ANSWER:
There is no law of physics which says an object naturally loses velocity. To
change the velocity of something you have to apply a force to it, it must
interact with something. If you had a bowling ball which had no forces on it
it would continue going with a constant speed in a straight line forever.
This is just Newton's first law. However, Newtonian mechanics is not valid
for a photon, but it behaves like any other particle when it experiences no
interactions which would change it. There is one important difference—regardless
of how it interacts with something else, it never speeds up nor slows down;
the speed of light in a vacuum is a constant of nature. If you look on my
faq page you will find links to discussions regarding
why the speed is constant and why it has the value it does. When you throw a
ball up in the air, it slows down; you throw it down from a tall building,
it speeds up. Photons don't do that but they do change their energies by
changing to a longer, redder (shorter, bluer) wavelength when going up
(down) in a gravitational field.
QUESTION:
Gasoline contains 40 megajoules of energy per kilogram and gasoline trucks have around a hundred tons of it. So how does a gasoline truck exploding not produce an explosion similar to a small nuke?
ANSWER:
Since it is always a little ambigous what is meant by a ton, I did my own
calculation using the volume of a tanker truck of about 10,000 gallons and
the density of gasoline of about 2.7 kg/gallon. I got the total energy
content of about 10^{12 }J. The energy of the Nagasaki bomb, a small
bomb by today's standards, was about 10^{14} J, 100 times bigger.
Two things to consider are:

The bomb number represents total energy delivered whereas I would guess that likely less than half the energy content of the gasoline would actually be delivered.

The time over which the energy is delivered is likely much longer for the gasoline explosion than the nuclear explosion. As an example, just to illustrate the importance, suppose the bomb exploded in 1 ms=10^{3} s and the tank in 1 s. Then the power delivered by each is 10^{8} GW for the bomb and 10^{3} GW for the tank. As a result the destructive power of the bomb would be much bigger.
QUESTION: Centrifugal force seems so much confusing to me. I read on the books that centripetal and centrifugal force never act on the same object (HallidayResnicks' "physics" and several of our intermediate text books) and centrifugal force never acts on the moving object. But in many cases (effects of earth's rotation on g, roller coaster's loop etc) the centrifugal force are said to act on the moving object (as our text books say!). So what's right and what centrifugal force actually is?
ANSWER:
Newton's laws are true only in what we call an inertial frame of reference.
Any frame which is accelerating is not an inertial frame of reference. Here
is an example: suppose you are in a car going 80 mph and the driver slams on
the brakes. If you are not wearing your seatbelt, you suddenly experience an
acceleration toward the windshield. But, there are no forces on you which
suddenly impel you forward, so Newton's first law is false for you because
you are accelerating in your frame even though there are no forces on you.
Of course someone watching this from the roadside has no problem
understanding what is happening: because there are no forces on you, you
continue moving forward with a constant speed while the car is stopping.
There is a very neat trick you can use to make Newton's laws work in an
accelerating frame: you just add a force which is the negative of the
acceleration of the frame times the mass of the object you are calculating;
such a force does not really exist and we call it a fictitious force.
So, for you in the accelerating car we add a force, let's call it the
"impulsive force" in the forward direction, which is your mass times the
magnitude of the car's acceleration. You can then say that you accelerate
forward with the same acceleration as the car is accelerating backward
because you are being pushed by the impulsive force.
Centrifugal force is a fictitious force. If you are on a merrygoround you feel a force trying to push you off; there is no such force. But if you pretend there is, you can do use Newton's first law: there is a centrifugal force outward which is mv^{2}/R and you are at rest; the force which is keeping you from accelerating outward is friction with the floor, so it must be equal to mv^{2}/R. If someone on the outside analyzes the problem she will say that you have an acceleration v^{2}/R inward so there must be a force (centripetal) inward to provide that acceleration, F=mv^{2}/R which is the same frictional force we found on the merrygoround. Clearly the centrifugal and centripetal forces in these two cases are acting on the same body—you—so your claim that texts say otherwise is strange. In 40 years of teaching I have never heard such a claim. It is true that both centripetal and centrifugal forces never act in the same solution to a problem; if you choose a person on the merrygoround to analyze, you never have both a centripetal and centrifugal force acting on that person. Perhaps you misunderstood and that is what is meant by never acting on the same object.
You can learn more about fictitious forces in an earlier answer and the links it leads to.
QUESTION: Two balloons, one with He gas is attached to the lower part of car and one with ordinary air is attached to roof hanging downward. When brakes are applied in what direction both balloons will move?
ANSWER:
The helium balloon will move backward; see an
earlier
answer for an explanation. (Note that the earlier answer has
acceleration rather than braking.) The air balloon will move forward because
it feels no buoyant force and therefore its inertia will cause it to want to
crash into the windshield.
QUESTION: If a man could travel with a near speed of light, what would happen to him if he'd run though an elephant. Would that man be demolished or could he survive that impact?
ANSWER:
Are you kidding? A man going 200 mph, far below the speed of light, would be
instantly killed in this scenario. A 100 kg man going at 80% the speed of
light would bring in about 5x10^{18} J of energy to the collision
which is equivalent to about 10,000 Nagasaki atomic bombs. Not only would
the man and elephant be obliterated, also anything within miles would be.
QUESTION:

What exactly is momentum and how is it different from force?

I understand that p = mv and F = ma but they seem so similar in application that I haven't fully wrapped my head around it.

Additionally, what does it mean for something to have momentum in the first place, and why must it be conserved?

How does light have momentum if it, by nature, has 0 mass (I presume E =mc^2 comes into play somewhere here)?

Finally, what is the significance in the fact that light does have momentum (if it does)?
ANSWER:
You have lots of questions, really. I have rearranged your question to
delineate it into parts:

For your first question, just say that p=mv. My answer to #2 should clarify how force and linear momentum are related. Force and momentum have to be different because they are not even measured in the same units—momentum is mass*length/time and force is mass*length/time^{2}.

What is acceleration? It is rate of change of velocity, a=(v_{2}v_{1})/t where t is the time to change speed from v_{1} to v_{2}. So one could write that F=(mv_{2}mv_{1})/t=(p_{2}p_{1})/t; so force may be thought of as the rate of change of momentum. Newton actually stated his second law this way, not as F=ma. It is the second law which is a fundamental law of physics, momentum is just defined because of its simple and natural relationship to the second law.

It doesn't "mean" anything for something to have momentum, it is just a definition. However, consider the second law if there is no force acting; then F=0=(p_{2}p_{1})/t. In other words, p_{2}=p_{1 }which simply means that momentum does not change (is conserved). The condition for conservation of momentum of a system is that there be no external forces on it. For example; suppose you look at an isolated galaxy which has billions of stars in it all interacting with each other and there are negligible forces from the outside; if you sum up all the momenta of all the stars today and in 10 years from now, you would get the same answer even though the shape and orientation of the galaxy would change.

It turns out that if v is very large, comparable to the speed of light c, Newtonian mechanics is incorrect. (You could say that Newtonian mechanics is wrong but a superb approximation for low speeds.) If you say that p=mv it turns out that momentum is not conserved for an isolated system at very large speeds. However, since momentum conservation is such a powerful way to solve problems, we redefine momentum (in the theory of special relativity), to be p=mv/√(1(v^{2}/c^{2})), momentum is conserved again and we still have p≈mv for small v. It also turns out that, as a result of this new definition of p, we can write that E=√(p^{2}c^{2}+m^{2}c^{4}) where E is the energy of a particle of mass m. So if the particle is at rest, p=0 and E=mc^{2}; if the particle has no mass, p=E/c.

No more significant than if a billiard ball has momentum—it just does.
I have deleted your question about angular momentum—it is off topic.
QUESTION: I am a high school student interested in relativity and I recently read an article about relativity . The article stated that the "The combined speed of any object's motion through space and its motion through time is always precisely equal to the speed of light." For an object moving at 90% the speed of light, it should only be moving at 10% the speed of light through time. I assume that means that time should pass only at 10% the actual speed of time for the object (correct me if I am wrong). However based on the Lorenz factor, time passes at 0.43 the actual speed of time for the object. Why is this so? Furthermore when time dilation occurs it is only seen by someone else in a stationary frame of reference. In the moving object's frame of reference time does not slow down at all. Does this mean that the combined speed of the moving object's motion through space and time can be more than the speed of light in the moving object's frame of reference? Can the moving object go beyond the speed limit in its frame of reference?
ANSWER:
Look closely at the example given: "To get a fuller sense of what Einstein found, imagine that Bart has a skateboard with a maximum speed of 65 miles per hour. If he heads due north at top speed—reading, whistling, yawning, and occasionally glancing at the road—and then merges onto a highway pointing in a northeasterly direction, his speed in the northward direction will be less than 65 miles per hour. The reason is clear. Initially, all his speed was devoted to northward motion, but when he shifted direction some of that speed was diverted into eastward motion, leaving a little less for heading north."
The speed in the northward direction will now be v_{N}=65cos45^{0}=46
mph; his velocity in the eastward direction will be v_{E}=65sin45^{0}=46
mph. His total velocity will be √(v_{N}^{2}+v_{E}^{2})=65
mph, not (v_{N}+v_{E})=92
mph. In relativity N and E now are x and t for the
stationary observer and x' and t' for the position and
time of the moving observer as seen by the stationary observer. The catch,
though, (which I presume is why the author of your article avoided these
details) is that time and space are slightly different from space (N) and
space (E) in that to get the length of the vector you calculate the square
root of the difference of the squares instead of the sum. So the speed
through spacetime would be √(c^{2}v^{2}).
So, the stationary observer will see a "spacetime speed" of v_{spacetime}=√(c^{2}v^{2})
and the "space time distance traveled" by the moving observer as seen by the
stationary observer would be d_{spacetime}=v_{spacetime}t'
where t'=t/√(1(v^{2}/c^{2})).
If you do your algebra you will see that d_{spacetime}=ct
indicating that the stationary observer sees the moving observer having
a speed of c through spacetime.
QUESTION:
Bodies of water bend with the curvature of the planet. How large would a body of water have to be in order to measure a difference of 1 inch from one end to the other.
ANSWER:
I
often get questions like this. I have used this figure many times before.
Here R=6.4x10^{6} m is the radius of the earth, d=1
in=0.024 m is your 1 inch, θ is the angle which
subtends the arc length, call it s, you seek.
From the triangle you can write cosθ=R/(R+d)=[1+(d/R)]^{1}≈1(d/R)+…
where I have done a binomial expansion of [1+(d/R)]^{1
}because (d/R) is extremely small. Now, because
θ is also very small, I can represent cosθ by
the expansion cosθ≈1½θ^{2}+…
so (d/R)≈½θ^{2}.
Finally we can write that θ=s/R. If you now solve
for s you will find s≈554 m.
QUESTION: Hey! So, somebody is standing on Earth, and they measure the distance from Earth to some star to be D. Now, they hop in a spaceship, and travel 90% the speed of light towards the star. How will they measure the distance between Earth and the star now? <D, D, or >D? I initially thought they would measure to be <D, because I've heard of how length contraction works. After I thought about it more, I changed my guess to be D, because I thought ONLY the spaceship would experience length contraction, but I could be wrong there (I'm not 100% familiar with the phenomenon of relativistic length). Now I'm stuck between D and <D, and I'd greatly appreciate if you could clear this up!
ANSWER:
Length contraction says that a length which is moving is shortened. From the
perspective of the spaceship the distance between earth and the star is a
length which is moving backwards relative to the the spaceship. It is
therefore <D. A meter stick on the spaceship, however, will be seen by the
passengers to be 1 m in length.
QUESTION: magnetic force on charge q moving with velocity v =qV x B if i observe this charge from a car moving with same speed and direction as that of q than it velocity as observed by me will be 0 so the force will be 0.i am not able to understand this dilemma at one time force non zero and at other time it is 0
ANSWER:
The problem is that the electric and magnetic fields in one frame of
reference are not the same as in another moving frame. (This is special
relativity.) In your case you first start with a magnetic field and zero
electric field. Suppose that the magnetic field is in the ydirection,
B=jB, E=0,
and the velocity of q is in the xdirection, v=iv.
Then the force would be F=kqvB
in the zdirection. In the moving frame the new fields would be
B'=jγB and E'=kγvB,
where γ=1/√(1(v/c)^{2}).
Note that
E'=vxB';
therefore the force, as seen in the moving frame is F'=qE'=qvxB'=kγqvB,
as you would expect. Note, however, that F'≠F
, they differ by a factor of
γ; this is because force is said to be not
Lorentz invariant
and it is not really a useful quantity in relativity.
QUESTION:
We can see things billions of lightyears from Earth. This means that light photons have traveled for billions of years at the speed of light to reach us. Why don't those photons slow down or stop? Is there no energy consumed over that vast distance in order to keep that photon moving forward?
ANSWER:
It is a law of physics that photons in a vacuum can only exist moving with
the speed of 3x10^{8} m/s. So, nothing has to "…keep that
photon moving forward…" But, space is not really a vacuum and not
100% of photons make it over such vast distances. Some interact with the
occasional molecule they might encounter and are changed or scattered. Some
find themselves striking a star or planet and being removed from the stream
coming from the source. Some are deflected by the strong gravity near
galaxies. Some encounter a black hole and disappear altogether. Because
space is so empty, though, the majority come through unchanged.
QUESTION:
The light emitted from a light bulb is thermal radiation. Probably the rushing electrons cause the atoms to vibrate and this vibration will cause photons to be emitted with a Gaussian distribution in the EM spectrum.
Still i wonder. Why is the vibrating of the atoms causing EM radiation in the visible spectrum and not lets say microwave or xray part of the spectrum.
Also, I still not sure where the photons pop into existence. Is it from all the electrons , in all shells and perhaps also the vibrating nucleus?
ANSWER:
All objects are always radiating and absorbing thermal radiation. It is
electromagnetic radiation. Many objects are wellrepresented by blackbody
radiation which you might want to research a little bit. As you can see from
the plot of blackbody radiation at 5000 K above, all wavelengths are
present, not just visible light. It may appear that there is no very short
wavelength radiation, but at this temperature it is just too small to
register on the graph. (Incidentally, this is not a gaussian curve.) Atoms
or molecules in a solid or liquid have electric charge distributions (nuclei
and the surrounding electron clouds), most importantly dipole moments, which
oscillate in ways which depend on the temperature. An oscillating charge
distribution looks like a tiny antenna and radiates the energy. These
"antennas" are the source of the radiation. Historically, trying to
understand the blackbody radiation classically was what was the first
indication of quantum physics. The oscillators must be quantized which means
that only discrete packets of radiation are emitted—your
photons.
Q&A OF THE WEEK, 2/52/11 2017
QUESTION:
I work in a high school where this question was posed by one of the pupils in a class I support. The question is this....
If you could attach a rope to a rocket, which would also be attached to earth, and sent it into space (out of our atmosphere) until the rope went taught and then cut the string. Would it stay in space or would it fall back to earth?
ANSWER:
This is one of those problems which I had fun with and I hope it will not be
too exhaustive an answer.
I will make the following assumptions:

the rocket always goes straight up;

the rocket stops moving vertically when the rope is taught;

the rope is cut the instant that the rocket stops;

fuel and the weight of the rope are not issues; and

the launch is from the equator. This makes things much simpler and I will briefly talk about a similar launch from some latitude at the end.
My view of this problem, therefore, is the same as if the rocket were on top of a very long stick vertically straight up and the stick is suddenly removed.
The thing to appreciate is that even though the rocket goes straight up, it will have the same angular velocity ω as the earth so its speed will be ω(L+R) where L is the length of the rope and R is the radius of the earth. The angular velocity is ω=[(2π radians)/(24 hours)]x[(1 hour)/(3600 seconds)]=7.27x10^{5} s^{1}. If L is just right, the rocket will assume an orbit like the geosynchronous communication satellites; this turns out to be if L=5.6R.

So, if the rope happens to be 5.6 times larger than the radius of the earth, the rocket will remain (apparently) stationary above its launch point; it is actually going in a circular orbit with a period of 24 hours. (The animation above illustrates this, although not to the correct scale.) For any other L the orbit will be an elipse with the center of the earth being at the focus. Visualize.

If L>5.6R, the starting point of the rocket will be the perigee (closest point to the earth) of its orbit. Visualize.

If L<5.6R, the starting point of the rocket will be the apogee (farthest point from the earth) of its orbit. Visualize.

For L<5.6R, though, there will be some critical distance L=L_{c} when the perigee of that orbit is exactly equal to R; in that case the orbit will just skim the surface of the earth. After some laborious algebra I found that L_{c}≈3.7R, about 2.2 earth radii inside the geosynchronous orbit. Visualize.

For L<L_{c}, the rocket will crash back into the earth but not where it was launched from because it is a projectile which has a horizontal speed greater than that of the earth's surface. Visualize.

Finally, if the horizontal speed of the rocket ω(L+R) is greater than or equal the escape v_{e}=√[2MG/(R+L)], where M is the mass of the earth, the rocket will escape the earth and never come back. I calculated this to be when L=7.6R, two earth radii beyond the geosynchronous orbit. Visualize.
For any other latitude θ the speed of the satellite will be ω(L+R)sinθ. The resulting orbits, while all elliptical, are much more difficult to visualize and maybe we should save that for another day! However, the simplest launch of all would be from the north or south pole (θ=0^{0} or 180^{0}) because it acquires no horizontal velocity (ω(L+R)sinθ=0) where the rocket would fall straight back down regardless of how high it went.
QUESTION:
Assume there are 2 black bodies, 1 with a mass double of the other. If both black bodies reach a temperature of 5000K, do both emit blue light as black body radiation, with the heavier body emitting a brighter light. Or does the heavier body emit white light while the lighter body emits blue light? Why?
ANSWER:
A
black body does not radiate one color but rather a spectrum of colors. The
spectrum for 5000 K is shown above. The most intense color of this spectrum
is at 580 nm
which is the green color shown to the right. The intensity of
the light emitted does not depend on the mass, it depends on the surface
area of the black body; if you double surface area, you double the amount of
energy being radiated.
QUESTION:
What force causes your hands to
warm up when you rub then together?
ANSWER:
Friction. But the energy to warm your hands comes from you. When you rub
your hands together you are exerting a force so you are doing work. But the
energy derived from your work does not speed up your hands, rather it is
lost to friction and shows up as heat.
QUESTION:
I'm a science geek and an IT person who just found out that there are sixteen types of water ice. I've been googling the phenomena, and can only find discussions of the physical makeup at the atomic level. Can you help me with some discussion of the various forms of ice at a macro level? Like, what does it look like, how does it act, etc. Everything I've found is in science speak, which I really can't envision myself.
ANSWER:
Refer to the figure above which came from the Wikepedia article on
ice. You can get a clearer
picture there. The thing to notice is that all the ices except for ice I and
ice XI occur at extremely high pressures—1000 atmospheres or
higher. So, you cannot really "look" at them and say what they look like
since they are formed in a containment vessel of some sort. Ice XI occurs at
atmospheric pressure but only at very low temperatures. If you look at the
Phases section of the
article, you will find links to separate articles for all 16 forms of ice
where you can get information about their properties. Although the
terminology for the crystal structure is pretty "science speak" as you say,
you will find usually pretty comprehensible pictures of these crystal
structures. I think you would find these articles about as understandable as
you will find.
QUESTION:
If a good absorber is a good emmiter also then why don't we prefer to wear black clothes in summer or in hot conditions?
ANSWER:
Because we are better off if we do not absorb any heat in the first place.
We wear a poor absorber.
QUESTION:
Can electric field and magnetic field collide to each other? If yes what will happen when they collide to each other.
ANSWER:
Two fields "colliding" really does not have any meaning. Certainly an
electric field and a magnetic field can exist at the same time and location.
If either is changing at that time, it will cause the other to change.
QUESTION:
How much physical energy is exerted in a "typical" gym and how does that compare to the amount of energy necessary to operate that same facility. I am wondering if there have been any investigations into redirecting this type of activity towards more productive ends.
ANSWER:
A person doing vigorous exercize may expend about 10 kCal/hr≈2.8
W. The gym would probably consume a few kilowatts so you would probably need
a couple of thousand folks to generate that much power. I don't think you
are on to something here!
QUESTION:
If vision works when we receive the reflected wavelength of white light from a particular object. Also we knew angle of reflection depends upon incidence angle, how so many people could possibly see the same object unless they stood in the path of reflected wavelength.
ANSWER:
Because the surface is not smooth like a mirror, the angle of reflection is
not equal to the angle of incidence.
QUESTION:
Why is it easy to accelerate tiny subatomic particles like electrons close to the speed of light compared to large macroscopic objects like spaceships?
ANSWER:
There are lots of ways you could explain this. For example, Newton's second
law states that a=F/m where a is the acceleration
of an object of mass m when acted on by a force F. Suppose
you exert a 1 N force on an electron (m≈10^{30} kg)
and a space ship
(m≈10^{6} kg). Then the acceleration of the
electron (initially) is 10^{30} m/s^{2} and that of the
space ship is 10^{6}
m/s^{2}.
QUESTION:
If atoms at room temperature move around at roughly the speed of a jet airplane, how fast do atoms that make up a jet airplane move when said airplane is moving (at room temperature)?
ANSWER:
I assume you are talking about molecules in a gas. It is much more
complicated to talk about molecular speeds in solids. So let's talk about
the air in the airplane cabin. The speeds are distributed from very low to
very high, but the most probable speed would be around 1000 mph, about twice
the speed of a commercial jet. But, that does not mean that the most
probable speed of molecules in a jet flying by would be around 1500 mph
because the molecules measured inside the airplane would be moving in all
directions so the result of adding 500 mph to all the molecules going 1000
mph would yield anything from 500 mph to 1500 mph. Measured inside the
airplane, the average velocity of all the molecules would be zero because
for every molecule going in one direction with some speed v there
is always another going in the opposite direction with speed v. The
average velocity of the air in the airplane as measured by an observer on
the ground would be 500 mph. It could be thought of as like a wind in which
there is a net flow of air in the direction the wind is blowing.
QUESTION:
Hi, my question relates to time dilation and the twin paradox/moving clock theory. A worked example of time dilation with a moving clock (one clock taking a relative tonitself 1 hour round trip at near the speed of light and one clock staying as a reference) shows it seemingly loosing time due to time dilation. Wouldn't this only seem to be correct until the observation of the moving clock ended? For example, the observation point which watched the moving clock is getting its informatiom from the trail of light from the journey, which shows slower time due to the speed at which it is travelling (as light is effectively taking longer to reach the observation point). However, when the travelling clock lands back on earth after an hour (relative to itself) and you were to end the observation of its journey and simply look at its landing spot, it should be there. Not only that but it should be showing the same time as the reference clock. So an hour long trip at the speed of light would only seem to take longer if you only seen it through your observation of its journey. Which would make any worked examples true up until observation ended. If this would in fact be true, wouldnt it be a more suitable explanation that our time is relative to light and that time dilation is no more than a lighttrick as we only see what we see when light reaches us? Or that our time is only relative to the centre of our universe?
ANSWER:
I think you can clear up your confusion by reading the faqs for the
twin paradox and for how fast
clocks appear to run.
QUESTION:
I am trying to understand how to predict stopping distances at other speeds if I know a stopping distance at one speed.
Let's say a car going 25 mph with studded tires on packed powder stops in 20 meters. (I got that from a study on various types of tires on various surfaces it's realistic). What would be a good approach to finding the the stopping distance of the same car with the same tires on the same surface, if it is going 20 mph?
We do not know the mass of the car, but perhaps it is accurate to create an acceleration constant that will apply for other speeds?
a=v/t and v=d/t
so therefore by combining these we get
a=v^2/d
25mph is 11.176 m/s
deceleration constant:
a = (11.176m/s)^2 / 20m = 6.245 m/s^2
then can we apply the deceleration constant to the other question? What will be the stopping distance at 20 mph?
20mph = 8.94 m/s; stopping distance is 12.8m
Is this valid?
My applications include predicting future stopping distances at other speeds if one has experienced a stopping distance at a particular speed, and also, finding a reasonable speed for a traffic situation near where I live, where there is limited visibility.
ANSWER:
Your final answer of 12.8 m is right, but your method is wrong. Your answer
for the acceleration (and stopping time if you had calculated it) is wrong.
Your principle error was that d/t is the average velocity,
not the starting velocity; the velocity is changing the whole time. In this
case, uniform acceleration, the average velocity is ½v_{0
}so your acceleration is wrong by a factor of 2; a=3.12 m/s^{2}.
The correct expression for the stopping distance is d=½v_{0}^{2}/a
where v_{0} is the initial velocity and a is the
magnitude of the acceleration. The force which stops the car depends only on
the nature of the rubbing surfaces and the weight of the car. But the weight
of the car cancels out when you calculate the acceleration, so the
acceleration is independent of both the weight and the initial speed.
Therefore your guess that you can use one datum to determine the
acceleration and then use that for all other speeds is correct; the added
bonus is that it is also independent of how heavy the car is as long as it
is the same road surface with the same tires.
QUESTION:
I am trying to understand how to predict stopping distances at other speeds if I know a stopping distance at one speed.
Let's say a car going 25 mph with studded tires on packed powder stops in 20 meters. (I got that from a study on various types of tires on various surfaces it's realistic). What would be a good approach to finding the the stopping distance of the same car with the same tires on the same surface, if it is going 20 mph?
We do not know the mass of the car, but perhaps it is accurate to create an acceleration constant that will apply for other speeds?
a=v/t and v=d/t
so therefore by combining these we get
a=v^2/d
25mph is 11.176 m/s
deceleration constant:
a = (11.176m/s)^2 / 20m = 6.245 m/s^2
then can we apply the deceleration constant to the other question? What will be the stopping distance at 20 mph?
20mph = 8.94 m/s; stopping distance is 12.8m
Is this valid?
My applications include predicting future stopping distances at other speeds if one has experienced a stopping distance at a particular speed, and also, finding a reasonable speed for a traffic situation near where I live, where there is limited visibility.
ANSWER:
Your final answer of 12.8 m is right, but your method is wrong. Your answer
for the acceleration (and stopping time if you had calculated it) is wrong.
Your principle error was that d/t is the average velocity,
not the starting velocity; the velocity is changing the whole time. In this
case, uniform acceleration, the average velocity is ½v_{0
}so your acceleration is wrong by a factor of 2; a=3.12 m/s^{2}.
The correct expression for the stopping distance is d=½v_{0}^{2}/a
where v_{0} is the initial velocity and a is the
magnitude of the acceleration. The force which stops the car depends only on
the nature of the rubbing surfaces and the weight of the car. But the weight
of the car cancels out when you calculate the acceleration, so the
acceleration is independent of both the weight and the initial speed.
Therefore your guess that you can use one datum to determine the
acceleration and then use that for all other speeds is correct; the added
bonus is that it is also independent of how heavy the car is as long as it
is the same road surface with the same tires.
QUESTION:
An interesting question came up amongst my friends and me resulting in differences of opinion that I hope you can settle.
Question: If an object is dropped from a fixed position while travelling in a moving jetliner(say 300+ MPH) would the same object dropped from the same position while not moving land on precisely the same spot? ie not a micron different. Of course all variables being equal.
ANSWER:
Absolutely they would not land in the same place. The way you describe it is
a little confusing: "dropped from a fixed position while travelling" is kind
of ambiguous isn't it? I assume that it is like dropping a bomb from a
plane; the plane and bomb are both moving forward. Neglecting air drag, the
bomb will land directly under the plane if the plane continues moving with
constant speed in a straight line. The object dropped from rest exactly
where the plane was when it dropped the bomb will land directly below the
drop point. If air drag is taken into account, the bomb will land somewhat
behind where the plane is. See an
earlier
answer.
FOLLOWUP QUESTION:
The scenario involves an object dropped from a fixed point inside a moving airliner and hitting a point under it but still inside the aircraft. No bomb dropping. Then a repeat of this operation while the airliner is sitting still. Does it hit the same spot in each case?
ANSWER:
Sorry, I misunderstood. In an elementary physics class, it is assumed that
the only force which acts on the object is its own weight, a force toward
the center of the earth. And actually that is true. However, it is also
assumed (although not always stated) that the plane is moving in a straight
line with constant speed; that is not true because if the plane maintains a
constant altitude it is actually moving in a circular path above a spherical
earth and the earth itself is spinning on its axis. If the observer is in
the airplane (which she is), Newton's laws are not true because in either
case the experiment is being done in an accelerating frame. However, there
is a neat trick to make Newton's laws work in an accelerating frame by
introducing fictitious forces; the best known fictitious force is the
centrifugal force. The force (real+fictitious) which acts on the falling
object depend on the direction the plane is flying, its speed, and its
latitude. These are different for your two experiments and therefore the two
objects would not land precisely at the same spot. To read more detail on
these fictitious forces, see an
earlier answer.
Finally, there are two places on earth where the fictitious forces are zero and therefore the two objects will move identically—at the north and south poles.
QUESTION:
An object travels at a very high speed relative to some planet. Its velocity is normal to the gravitational field lines of the planet at a particular instant. At that instant, does the gravitational attraction between the two bodies increase due to the object's relative motion or does it stay the same as if the object was not moving at all?
ANSWER:
The object moving at high speed has an increased mass and therefore feels an
increased gravitational force.
QUESTION:
What causes the curvature of spacetime/gravity to be so strong around black holes? Is it due to their density (mass density/energy density)? If not what is it due to? And why dont other extremely dense stars like the sun exhibit properties like black holes
ANSWER:
For any spherically symmetric mass distribution the gravitational field is
proportional to the mass m and inversely proportional to the square of the
distance from the center 1/r^{2}. Because a black hole is a
singularity (zero size) it is possible to get very close to the center where
the field will be huge. Also, many black holes are very much more massive
than a typical star. A regular star might have a very big mass, but it is
spread out over a very large volume; inside the star itself the field
decreases linearly to zero at the center.
QUESTION:
If you were able to drive your car with forward acceleration in outer space and you rolled the window down and stuck out your arm, would it be pushed back like it would here on earth? This isn't a homework or test question, it is a question that came up in a fun discussion at work the other day...
ANSWER:
You had better have your space suit on because outside the car there is no
air in outer space. And the fact is that the main thing which causes your
arm to be pushed back here on earth, particularly for high speeds, is the
air drag, the apparent wind you perceive to be blowing from in front of you.
That is totally absent in space. However, you have specified that the car is
accelerating. When you accelerate forward, it feels as if there is a force
pushing you (and your extended arm) backward. There is no such force (which
is why we call it a fictitious force), but it feels like it because your
accelerating arm needs a forward force (your shoulder provides it) to
accelerate it and your brain interprets this as there is something pushing
your arm backward which your shoulder counters. Here on earth, unless your acceleration is
very large and/or your speed is very small, this force is pretty small
compared to the air drag. So the answer to your question is that your arm
will be pushed back but not "…like it would here on earth…"
QUESTION:
I was wondering what would happen if an object, say a cube, were travelling toward you at a relativistically interesting speed. Its motion is strictly in the x component in an XYZ coordinate plane. Would you see a length contraction in the dimensions of the cube in all three dimensions or just the x component?
Also I was curious about the effects of the length contraction on the observed density of the cube. Observed mass goes up as velocity does, and you observe a length contraction in one or more dimensions (depending on previous answer). Therefore the density would go way up. Is this a proper interpretation of relativity?
ANSWER:
As I have said before, saying that mass increases with speed is a matter of
taste. It turns out that in special relativity the linear momentum p
is not conserved if momentum is defined as p=m_{0}v
where m_{0} is the inertial mass at rest. If you
redefine p to be p=m_{0}v/√(1(v^{2}/c^{2})),
momentum of an isolated system is constant. You can interpret this as simply
a redefinition of p or that the mass has increased so that
m=m_{0}/√(1(v^{2}/c^{2})).
That said, let's answer your question assuming that mass increases.
Length contraction occurs only along the direction of motion. Since mass increases by a factor 1/√(1(v^{2}/c^{2}))) and volume decreases by a factor of 1/√(1(v^{2}/c^{2})), density increases by a factor 1/(1(v^{2}/c^{2})).
QUESTION:
if a cat fell from 100 feet how fast was the cat in mph falling when it hit the ground
ANSWER:
The terminal velocity of a cat is about 60 mph and that speed is achieved
after falling about 50 ft. Therefore falling from 100 ft the speed would be
about 60 mph. See earlier answers (1
and 2)
for more details. Keep in mind that all cats are not the same size, weight,
or fuzziness, so this can only be an approximate answer. It is interesting
that about 10% of cats falling from 5 stories are killed but fewer from
higher stories because, once reaching terminal velocity, they can relax,
spread out (to slow down) and get ready to hit.
QUESTION:
E=MC^2. It implies that energy and mass are exchangeable and consistently so. My problem comes with c^2. How is this possible? c= the speed of light, if I am not mistaken and one of the fundamental laws of the universe is its speed limit, that of c. So how is c^2 possible?
ANSWER:
Consider the following: you have a few hundred square tiles, and they have
varying sizes. I will use d to specify the length of the side on
any particular tile. However, due to some fundamental law of the universe, none of
these tiles is allowed to have to have d>d_{max}=4 cm. Now
suppose that I write an equation for the area of any of these tiles, A=d^{2}.
Now one of these tiles has a size of d=3 cm (not violating our
fundamental law) and so its area is A=9 cm^{2}. But wait,
how can that be? Since d^{2} is more than twice as large as
d_{max}, this must violate our fundamental law of the universe, right? Of
course not. The area of a square is simply a different thing from the length
of the side of a square and does not even have the same units (cm vs.
cm^{2}).
QUESTION:
I am a weight lifter and want to calculate how much "work" I do when performing the deadlift. I did the math and want to ensure my concepts and calculations are correct, as it has been a while since I took physics. Here is what I did:
I am lifting 225lb (1000.85 newton) a distance of 0.613m. So the work (in joules) = 1000.85n X 0.613m = 613 joules. I did 32 repetitions of this in total, so the total work would be 19,632.64 joules. So, if I convert that to watt hours, it is about 5.45 watt hours. Does this mean that to do that same amount of work, 5 watts for 1 hour would be required? Or 20 watts for 15 minutes? Or 300 watts for 1 minute?
Did I use the right calculations?
Ultimately I want to compare several of my lifts to see how much total work I performed (for example 10 sets of 5 reps on Bench Press at 150lbs VS 6 sets of 7 reps on Deadlift at 225 lbs)...
ANSWER:
All your calculations are correct (allowing rough rounding in "__ watts for
__"). The problem is that simple physics like this is not really a good
approximation to how biological systems are actually working. If you simply
hold a weight with a horizontal arm, physics says no work is being done, but
you and I know that sugar is being burnt to provide energy to hold the
weight there. Likewise, physics says you are doing negative work when you
lower the weight (cancelling out the positive work you did lifting it) but I
suspect this lowering also requires you to actually use energy, that is do
positive work. The explanation of how work is being done is given in an
earlier answer.
QUESTION:
This is quite for a long time that I'm thinking about a question in Galilean
kinematics which brings me to a paradox, which I cannot solve myself.
Consider a body with a mass m and velocity V_{1} in one reference system and
V_{1}
+ V' in another one. Now let's say this body starts slow down and its
velocity change is ΔV in both systems. Simple algebra shows that the
kinetic energy change in the 1st system will be
ΔE_{1} = mΔV/2(2V_{1}+ΔV)
and in the 2nd system
ΔE_{2} = (mΔV/2)(2V_{1}+ΔV+2V')
i.e. its different in different reference systems. Now if we imagine that
this slow down and kinetic energy change was due to the friction it follows
that the amount of heat released is different for different reference
systems! Nonsence! It should be a mistake somewhere.
ANSWER:
The first thing to appreciate is that energy is not invariant under
Galilean transformation. This is easy to appreciate since a mass with
speed v has kinetic energy K=½mv^{2}
but zero kinetic energy in a frame moving with it. Therefore, the change
in kinetic energy will be different in different frames. However, the
workenergy theorem (ΔK=W where W is work done by
external forces) will be true in all frames, just not the same numbers in
all frames.
In the first frame in your example,
where m has speed v_{1}, the floor is, I assume,
at rest. To make the algebra simpler and the explanation easier to grasp,
I am going to let m come to a complete stop, Δv=v_{1}.
Then, when m moves some distance s and the frictional
force is f, the work done is W=fs and so ΔK=½mv_{1}^{2}=fs.
You can deduce that the heat generated is ½mv_{1}^{2};
all other frames will see this amount of heat. Now, in a frame with speed
v' moving in the same direction as v, you can easily
show that ΔK'=½mv_{1}^{2}+mvv'.
You already know that ½mv_{1}^{2}
is the heat and mvv' is simply the amount that ΔK
is different in this frame. If you calculate the work done in this frame,
it is different because the distance m slides is different; but,
if you do that you will find, as expected that
W=½mv_{1}^{2}+mvv'.
One more special case. Suppose v'=½v. You should be able to convince yourself that m originally is catching up with the moving frame but then stops and turns around and ends up where it started from; thereafter it moves with constant speed v'. The total distance it has traveled in the moving frame is zero and so W=0 and therefore ΔK'=0. So the heat is not anywhere evident here even though you would feel it. In this case you would find W=½mv_{1}^{2}+½mv_{1}^{2}. Now you can see where that heat is hiding, right?
So, if you want to most conveniently determine the amount of heat generated, you apply the workenergy theorem in the frame where the surface is at rest. This is because to the moving observer the surface is moving as well as m. The moving observer has to correct for that motion by calculating m’s motion relative to the surface. m’s motion relative to the second observer is not really relevant in calculating the heat generated.
Thanks to R. M. Wood and A. K. Edwards for helpful comments.
QUESTION:
I'm trying to come up with an equation at work on how to figure out how much a parent roll of toilet paper weighs just by measuring the distance from the core to the outside of the roll without taking it out of the machine. For instance 1inch of paper on the out of the roll will not weigh the same as 1 inch of paper from the core. The measurements I know are the weight of the parent roll when we start using it, the core diameter, core weight, and distance from the edge of the core to the outside edge of the roll before it's started. As it is running we can measure the current distance from the core edge to the outside of the roll while it is running. My question is how do I calculate a equation to determine the weight of the roll at any given time based on the distance from the edge of the core?
ANSWER:
Not really physics, but I can do it. Notation:

W_{1} is the weight of the parent,

W_{2} is the weight of the core,

W_{3}=W_{1}W_{2} is the weight of the paper in the parent,

R_{1} is the radius of the parent,

R_{2} is the radius of the core,

L is the length of the roll,

V_{1}=πL(R_{1}^{2}R_{2}^{2}) is the volume of the paper on the parent,

W is the weight of the paper on the partially used roll,

R is the radius of the partially used roll, and

V=πL(R^{2}R_{2}^{2}) is the volume of the paper on the partially used roll.
You may be interested only in the final answer, but I will outline the solution for anyone interested.

The density of the paper is ρ=W_{3}/V_{1}=W/V.

Therefore, W=W_{3}(V/V_{1})=W_{3}(R^{2}R_{2}^{2})/(R_{1}^{2}R_{2}^{2}).

What you want, is the weight of the whole partially used roll, W+W_{2}=W_{2}+W_{3}(R^{2}R_{2}^{2})/(R_{1}^{2}R_{2}^{2}).
So, just
plug in your measured values and you will have the weight of the partially
used roll. Incidentally, you say "distance from the edge of the core". If
this (call it D) is more conveniently measured you can replace R
in the equation above by (D+R_{2}) which results in
W+W_{2}=W_{2}+W_{3}(D^{2}+2DR_{2})/(R_{1}^{2}R_{2}^{2}).
QUESTION:
I'm a writer. I'm working on a story. It's sold actually, but they want some changes and I'm trying to figure out the science..
IDEA: A distant planet picks up broadcasts from Earth, so the inhabitants have some knowledge of us, and know that life is out there. A "man" on the alien planet gets part of his consciousness encoded into a light beam that is sent out into space. When it hits earth It is able to project itself and appear to be human. When the light fades the man vanishes, leaving a message that he is long dead. (It's a love story.) But the question is what broadcasts would he have received time wise. His knowledge of earth would be limited, ending at a certain time...
This is the end/explanation. Thank you in advance for your help!
"My people developed a holographic beam so powerful it could be projected through space. Through lightyears. It is more advanced than anything on Earth, so advanced that some actual particles of my being were intertwined with the particles of light being projected. So it was like I was actually here, next to you, talking to you, laughing with you. But I was not. I am not. "When we sent the beam of light traveling into space my particles were woven into it, not all of them, of course. But enough particles of my consciousness were intertwined with the particles of light so that I could interact with anyone I discovered. ...So that I could learn more about the world we'd previously only known through television broadcasts. .. "I traveled in a spaceship made of light. Now that the light has died so have the particles of my consciousness that were interwoven with the beam.... The real 'me, ' the corporeal 'me,' died HOW MANY? light years ago." Long ago, men went to sea. And women waited for them, peering out into black waters, searching for a tiny speck of light on the horizon. Now I too wait, looking out into the vast blackness of space, searching for my love, my heart, HOW MANY? lightyears dead.
ANSWER:
For starters, a light year is a measure of distance, not time, so you want
to say 'years' not 'light years' in your concluding sentences.
(A light year is the distance light will travel in one year.) It
would be better, I think, if the traveler could have learned about earth
from radio broadcasts as well as television because that could take us back
about 100 years ago compared with only about 50 for tv alone. His
consciousness travels here at the speed of light, so he could not have come
here from a distance of greater than 100 light years away from earth which
is a pretty small region of the universe. (The diameter of our galaxy, the
Milky Way, is about 100,000 light years.) So, the most extreme case would be
that it takes 100 years for the signals to first reach his planet, then it
takes his consciousness another 100 years to get here; so when his 'consciousness' arrives
on earth, the stayathome him would be at most 100 years older and he
arrives on earth 200 years after the first radio signals he received were
sent.
One other detail which probably isn't that important would be you could not really carry 'particles' of consciousness (whatever that might be!) on a light beam if they had any mass at all. Nevertheless you could envision the information regarding his 'consciousness' could be carried by light.
ADDED
THOUGHTS:
I did not answer one of your questions. How much information he had would
simply depend on how long he waited after first receiving signals. Since
the light beam would not be able to receive and interpret later signals,
he would have no further information. However, if you were to put his
'consciousness particles' aboard a spaceship going 99.999% of the speed of
light, all the broadcasts could be intercepted and interpreted en
route and incorporated into his consciousness somehow; other
timing would be pretty much the same. Or, if you simply put him on that
spaceship, he would age only about half a year during the trip which the
earth would see as 99.999 years. But then you would not have your dramatic
sad ending!
QUESTION:
I've been given a query by a friend, and I'm having trouble finding answers!
Consider the following:
The tallest building in the world is 250m tall (completely fabricated stat).
At the top of this building is a light detector and a clock.
At the bottom of the building is a laser and identical clock.
At t = 0 the laser fires a single beam of light at the light detector atop the building.
Exactly 5 seconds later the laser fires another beam of light at the light detector atop the building.
Therefore, at the bottom of the building 5 seconds has passed in between laser shots.
How much time has passed between the light detector detecting the first and second laser beams at the top of the tower?
The same, more or less?
I've ben given a query by a friend, and I'm having trouble finding answers!
Consider the following:
The tallest building in the world is 250m tall (completely fabricated stat).
At the top of this building is a light detector and a clock.
At the bottom of the building is a laser and identical clock.
At t = 0 the laser fires a single beam of light at the light detector atop the building.
Exactly 5 seconds later the laser fires another beam of light at the light detector atop the building.
Therefore, at the bottom of the building 5 seconds has passed in between laser shots.
How much time has passed between the light detector detecting the first and second laser beams at the top of the tower?
The same, more or less?
ANSWER:
Since you acknowledge that the height of the building is immaterial, let me
make it 1.0 light seconds tall. (A light second is the distance light will
travel in one second.) So the first pulse is observed at t=1 s and
the second at t=6 s, an elapsed time of 5 s.
FOLLOWUP QUESTION:
the identical clock statement simply means that if they were switched around and the experiment performed again, the result would be the same.
The answer lies in general relativity.
ANSWER:
OK, I did not realize you wanted to get so exacting! I am still going to use
a building 1 ls=3x10^{8} m high to keep the numbers from becoming
impossibly small. Indeed, general relativity tells us that the stronger a
gravitational field is, the more slowly a clock will run. This is called
gravitational time dilation (and is directly related to gravitational red
shift). The rate at which a clock runs slower in a gravitational field (MG/r^{2})
is √[1(2MG/r)] where M is the mass of the
source (the earth in your case), G=6.67x10^{11} N⋅m^{2}/kg^{2}
is the universal gravitation constant, and r is the distance from
the center of the source. If some time t elapses on a clock in zero
gravitational field, the time which will elapse for other clocks is t/√[1(2MG/r)].
For our purposes I will assume that the zerofield clock will measure 1 s
for a pulse traversing your building.

After doing some arithmetic I find that if t elapses on the zerofield clock, t'=t[1(4.4x10^{3}/r)] elapses on another clock. So, for the first pulse where both clocks have begun at t'=0, the elapsed time of the two clocks will be t'_{bottom}=(16.9x10^{10}) s and t'_{top}=(11.5x10^{11}) s. The upper clock will will measure a shorter time by about 6.8x10^{10} s=68 ns.

Now, we need to specify who determines when 5 s have elapsed. The most sensible choice would be the guy at the bottom of the building who will be triggering the laser. When his time is t'_{bottom}=5 s, the zerofield time is t=5/(16.9x10^{10})≈(5+3.5x10^{9}) s and (this is tricky) the top clock will measure t'_{top}=(5+3.5x10^{9}6.8x10^{10}x5)=(51.0x10^{10})s.

We know from the first pulse the additional time on both clocks, so we can now calculate what each clock reads when the second pulse hits: t'_{bottom}=(5+16.9x10^{10})=(66.9x10^{10}) s and t'_{top}=(51.0x10^{10}+11.5x10^{11})=(61.2x10^{10}) s.
The top clock will see 57 ns more time elapse than the bottom clock. I believe that if you use the 250 m tall building, things would simply scale and time differences would be 250/3x10^{8}=8.3x10^{7} times smaller.
QUESTION:
Now, I know the centre of gravity can change depending on how close to uniform the gravitational field is, and it is where the weight appears to act from. But, what is the centre of mass? Ok, it is a constant; doesn't change; and in an uniform gravitational field the two centres overlap; but how do you define the centre of mass without referencing centre of gravity? An online source said it is the point from which the mass is 'equally distributed in all directions' for an object; but I can have irregular shapes. What does that source means? Or is the centre of mass defined in another way? I am really confused!
ANSWER:
Actually, the center of mass is the fundamental quantity, not center of
gravity. The the position r_{cm} of the
center of mass for a collection of point masses is defined as r_{cm}=[Σm_{i}r_{i}]/Σm_{i}
where r_{i} is the position of the point
mass m_{i}. If you have a continuous mass distribution with
mass density ρ(r)
you have to integrate over the whole volume to get
r_{cm}: r_{cm}=[∫rρ(r)dr]/∫ρ(r)dr.
Notice that in both cases the denominator is just the total mass.
QUESTION:
When I was very young, I used a handheld fish scale to try and measure my
own weight by attaching the hook (normally placed in the gills of the fish
just caught) to my belt and then pulled up on it. I'm interested in the
physics behind my folly and, expanding it further, if the scale was attached
to a light (not heavy) bar (that would support more than my weight) and the
scale could measure a fish weighing more than me, and I was strong enough to
lift more than my weight above my head, how close could I expect to get to
seeing my weight on the scale (attached to the bar I'm pushing up, and a
beltdevice around my waist)?
ANSWER:
The way you always solve a statics problem like this is to first choose a
body, then apply Newton's first law which states that the sum of all forces
on the body must equal zero; I choose the scale as the body.
What are the forces on the scale itself? Your hand pulls up with a force
F_{hand}, your belt pulls down with a force F_{belt},
and the earth pulls down on the scale (its weight) with a force W_{scale}.
These three have to add to zero which means that F_{hand}=F_{belt}+W_{scale}.
Now, the scale is calibrated so that it reads zero when F_{belt}=0,
so the scale will read whatever force your hand exerts up. Note that when
you do this analysis, the desired quantity, your weight, never
appears; everything would be just the same if you weighed 1000 lb. Regarding
your second question, it is really no different from the first question:
simply replace 'belt' by 'bar' everywhere. In either case you would observe
the scale reading your weight when you pulled up with a force equal to your
own weight.
FOLLOWUP QUESTION:
Thank you! So it sounds like it wasn't a folly after all (except for the fact that the scale I used wouldn't support my weight at that age).
ANSWER:
Of course it was folly, total folly! Doing this experiment gives
you absolutely no information about your weight. Only if you already knew
your weight and were able to pull with a force that hard would the scale
read your weight.
QUESTION:
I'm having a discussion on a website that came up because of another post. That post asked the question (paraphrased) If you connected a wire to your mouth and another (unstated) orifice, how fast would you have to travel around the earth to create enough current to electrocute yourself?
One of the other posters wanted to know if the voltage would be the same "going" through your body as through the wire. And wouldn't your body have current induced also?
The answer for the induced current in the wire I can deal with, But haven't enough electro/biology (I just made that up) to explain why ones body wouldn't have current induced in it. I know intuitivly why but can't explain it satisfactorily to him.
Can you help me out here (in colloquial English)? I'd appreciate it if you can and willl
ANSWER:
The body is certainly not a perfect insulator or else you would never have
to worry about getting electrocuted. The physiology is very complex and
variable, and it doesn't seem you want to get into too much detail either,
so I will just talk qualitatively. Two important features:

A voltage large enough to drive a large enough current through your heart will kill you.

A conductor moving through a magnetic field will have a voltage induced across its ends. (Details depend on direction of the field and the direction of the velocity of the conductor. Just assume that you have a straight wire moving perpendicular to the field.)
So the wire will have a voltage across its ends and will therefore look like a battery. Similarly, the body will have a voltage of the same polarity (say positive at the mouth and positive at the end of the wire in your mouth) but with a much smaller voltage than the wire; this would create a circuit looking like two batteries, one weak and one strong, which would drive a current through the weaker battery (your body). So, the idea works if you move fast enough. But, the earth's magnetic field is really weak and the speeds would be impossible to achieve without burning everything up in the atmosphere I would bet. Let's just do a rough calculation. The voltage V is about V=BLv where B is the field, L is the length, and v is the velocity. I will take L=1 m, v=18,000 mph≈8000 m/s (low earth orbital speed), and B≈5x10^{5} T. So the voltage would be less than half a volt! Have I been colloquial enough for you?
QUESTION:
Would it be possible physically to encapsulate a black hole with solar panel type devices and use its energy to power a civilization? Like when we reach transgalactic civilization status and run across a black hole and we want to utilize it.
ANSWER:
Well, that is a pretty crazy idea because a black hole is an energy sink,
not an energy source!
QUESTION:
im struggling to get a grasp of this. say car A and car B are travelling at a constant speed, maintaining same distance in between. form car A you shoot a gun 90 degrees up. which vehicle is the bullet likely to fall in. is car A or B?
ANSWER:
Before you fire the gun, the bullet in it is already traveling along with
you in car a with the same speed as the car. When you fire it up it goes up
with some speed but it continues moving parallel to the ground with car A
also. Since there are no forces in the horizontal direction acting on it
(neglecting air drag), it will go straight up and straight back down as seen
by car A. Someone on the ground will see it go in a parabolic path beginning
and ending in car A.
QUESTION:
I don't quite understand escape velocity. I know rockets have to reach 7mi/sec to leave earth's gravity but If I set off in my hypothetical spaceship and I moved at a constant 7 miles per minute on a trajectory perfectly perpendicular to the earth and directly above its launch pad (until I hit a geosynchronous orbit).
If I can maintain this velocity away from and perpendicular to the earth what could possibly be stopping me from escaping its gravity?
ANSWER:
The escape velocity is defined as the speed an object must have at the
surface of the earth (neglecting air drag) to escape. Obviously, as you
note, if you could could continue pushing on something to keep it moving a
speed of 1 cm/s, it would eventually get as far away as you liked. For more
on escape velocity, see the
FAQ
page.
QUESTION:
I was wondering if the charge of a particle increases as its mass does (at relativistic velocities). For example, if a quark is going very fast, does its charge increase the same proportion as its increase in mass?
ANSWER:
No, electric charge is strictly conserved.
QUESTION:
When a photon loses energy (Redshifts) climbing up thru a gravitational field, does the photons decreased energy go into the mass of the gravitational field itself? According to GR, the gravitational field itself contributes to the mass of the system. For earth, it is very small. When I elevate a massive object in a gravitational field, I can say the "potential energy" is in a very small increase of the mass of the object. But photons have no mass! And are never at rest! The only thing I can think of is that the photon "transfers" energy to the gravitational field itself, which appears as a small increase in mass of the field.
ANSWER:
Let's think of a star emitting a photon of frequency f. Initially
an energy of hf is removed from the star. But, by the time that
the photon is very far away the star, it has lost some amount of energy Δhf,
so the net loss to the star and its field is
hfΔhf. At the instant the photon is created, the
mass of the star is reduced by hf/c^{2}. But when the
photon, which is losing its energy to the field, is far away, the energy
of the field will have increased by Δhf/c^{2};
I would not use the terminology "mass of the field" since the field has
energy density, not mass. But now, it seems to me (not a cosmology/general
relativity expert) that there is a bit of a paradox: we always associate a
given mass with a particular gravitational field, so the field should have
the energy content associated with a mass M'=Mhf/c^{2}.
But, in fact, the field would have energy content associated with a mass
M'=Mhf/c^{2}+Δhf/c^{2}.
I am guessing that the field and the mass somehow "equilibrate" so that the
final mass of the star is consistent with the energy of the final field. (I
could easily be wrong! Perhaps it is only meaningful to look at the total
energy of the star and its field. Whatever the case, I would not talk about
mass of the field.)
QUESTION:
Considering a lack of strong ambient light sources, such as the sun, is the exterior glass of a spacecraft, the part that comes into contact with the vacuum of space, cold, neutral, or hot? Would it depend on the temperature inside the craft itself, or would the vacuum have enough of an impact on it to change it?
ANSWER:
It is just the same as a window in your house in winter. The inside surface
will be in equilibrium with the air in the room and the outside surface will
be in equilibrium with the air outside. Then heat will be conducted from
inside to outside at a rate proportional to the temperature difference. If
the sun is shining directly on the window, absorption of the radiation would
raise the temperature of the outside surface somewhat.
QUESTION:
Why isn't
Hawking radiation trapped in a black hole's gravitational pull and eventually sucked right back into the black hole?
ANSWER:
What happens is that pair production occurs near the
Schwartzschild
radius. One particle of the pair is outside that radius and can
therefore escape.
QUESTION:
"IF" we violate the laws of conservation, what would happen?
ANSWER:
There are so many such laws that no single answer could cover it all.
FOLLOWUP QUESTION:
But is it possible to violate the laws even tho we don't know how since its just a law and some laws are meant to be violated...sorry for my curiosity, I just some opinions from a Physicist
ANSWER:
I would say that if a "law" is properly stated, it is not "meant to be
violated". Again, there are far too many conservation laws to discuss. But
let me give you just a couple of examples of how a law can be misstated but,
if properly stated, it cannot be broken.

Energy conservation:

Incorrect: the total energy of any system is conserved. In fact you can change the energy of a system by doing work on it.

Correct statement: the total energy of any system is conserved if no external forces do work on it.


Electric charge conservation:

Incorrect: The total electric charge of a system must remain constant. In fact, you can always add or subtract charge from a system.

Correct statement: The total electric charge of an isolated system must remain constant.

These laws are laws because they are always found to be true and it is rather pointless to ask "what if". If energy conservation were not true, then energy could suddenly appear out of nothing.
Only if there is some overriding physical law could another be broken, but only within certain constraints. For example, the Heisenberg uncertainty principle says that it is impossible to precisely know the energy of a system for a short enough time, ΔEΔt~10^{34} J⋅s where ΔE is the amount by which you change the energy of a system and Δt is the time during which the energy is changed by that amount. For example, 10^{20} J could appear out of nothing for as long as 10^{14} s; after the time had elapsed, though, it would have to disappear again.
QUESTION:
Are there any similarities between light waves and sound waves?
For example,
When I pluck a guitar string while placing the guitar parallel to the tv, I can see waves that I previously couldn't see under visible light. The color blue seems to work best.
ANSWER:
The picture on your TV flickers at a rate of 60 Hz (times per second).
Therefore it is like a strobe light. If the string is vibrating with a
frequency of near 60, 120, 180,…it will appear to be frozen; sound
has frequencies around these frequencies. So, you are probably seeing this
strobe effect. The frequencies
of light are enormously larger, hundreds of THz (10^{12} Hz).
QUESTION:
What was the first thing Einstein did for the Mathematics of General Relativity? I just mean to say how did he start General Theory of Relativity?
ANSWER:
There is a story about how Einstein got the idea for general relativity,
possibly not true. He was sitting at his desk in the Swiss patent office
watching a workman across the square on a ladder painting a building. The
workman fell off the ladder and Einstein thought to himself, "There is no
experiment he can do which could distinguish whether he is in free fall in a
gravitational field or is at rest in empty space." This is a statement of
the equivalence principle which, along with principle of relativity (the
laws of physics must be the same in all reference frames) form the basis of
general relativity.
QUESTION:
If i am walking around the earth. Gravity keeps me on the earth but i'm moving around a sphere so the question is. By how much does the curve of the earth drop to keep me on the surface of the earth after 8000 meters? So that i don't just fling out into space.
ANSWER:
I think you are just asking about geometry. Certainly, the curvature of the
earth does not keep you from being flung "out into space"! You seem to be
asking about the distance d in the figure here which is easy to calculate.
If you go 8000 m along the surface, that is the arc length subtended by the
angle θ. Therefore, θ=8000/R=8000/6.4x10^{6}=1.25x10^{3}
radians=0.0716º. Finally, you can write cosθ=R/(R+d)
or d=R(1cosθ)=5
m.
QUESTION:
What role do gravity or the weak nuclear force have in keeping the human body alive? Clearly, the strong nuclear force keeps us from dissolving into particles, but if you were suited up in between galaxies and gravity "disappeared", would the strong force be enough to keep us intact and alive?
Also, electromagnetism is necessary for brain function (and heat transfer?), but do we use the weak nuclear force when making energy from food or anything like that?
ANSWER:
We evolved in a gravitational environment, so the proper functioning of many
of our biological systems is dependent on gravity. Astronauts come back
after long missions with weakend bones and muscles, for example. The
electromagnetic force is probably the most vital because all of chemistry is
electromagnetic. I cannot think of any biological functions which are
directly dependent on the weak interaction.
QUESTION:
If
g=GM/d^{2} then why does acceleration due to gravity decreases as we go deeper into earth?
ANSWER:
That equation is only valid for the acceleration outside a
sphericallysymmetric mass distribution. If you go inside the mass
distribution, the equation is g=GM'/d^{2} where
M'
is the mass inside d.
QUESTION:
I did read one of your answers. I didn't quite get it. Electricity is electrons flowing through a electric field right? I had the problem, if electrons lose energy as heat etc. then some of the electrons energy must be lost. I thought it must be the electrical potential energy, since the electron is not losing kinetic energy.
Because if it did lose kinetic energy, then the speed would reduce and that would reduce the current in different places of the circuit, which does not happen. Current in all parts of the circuit is the same. But you say, that the electron keeps bumping and has to start again and again, and that he moves real slowly. I thought the electrons were moving at light speed?
ANSWER:
I'm not sure which answer you read.
This one is probably the best. Another good one to read is
this one. So, to understand current you have to appreciate that it is
statistical; there are such a huge number of electrons moving that all it
makes sense to talk about is the average velocity of the electrons which is
referred to as the drift velocity. When a collision with an atom
occurs, kinetic energy is turned into heat energy. On average, electrons
have constant kinetic energy but lose potential energy as they move along in
the electric field. This energy lost to heat is supplied by the power source
which is why you have to change batteries now and then or burn coal in a
power plant somewhere to supply the energy. The drift velocity is, as
explained in the second reference above, extremely slow, not the speed of
light. When you turn on the potential difference, the electric field
establishes itself along the wire at about the speed of light which is why a
light seems to come on instantaneously.
QUESTION:
Why it is important for fission reactions to emit neutron ?
ANSWER:
Because a neutron can cause a fission to happen in another nucleus and
therefore you have the possibility for a chain reaction.
QUESTION:
Do all weak interaction produce neutrinos ?
ANSWER:
Here is one which does not produce a neutrino. It does, however, involve a
neutrino.
QUESTION:
Will two objects traveling in the same direction ever collide? Assume the objects are on earth, unmanned and their mass, volume, weight, density and speed are the same. All variables that can come in to play should be assumed that they are equal, for example no hills, curves bumps and no change in surfaces to have any effect on the coefficient of friction. Just a question my wife and I were wondering about.
ANSWER:
I'm
not really sure what you are getting at here. If you are just wondering
whether two parallel lines ever intersect, the answer is no. But you seem to
want to know about material objects originally moving on parallel lines. To
make the situation simpler, let's just have the two objects, originally with
parallel velocities, move in otherwise empty space. If the original
velocities are equal and they are traveling side by side, they will
eventually collide because there is a gravitational attraction between them
which will eventually bring them together. However, if they are originally
side by side and traveling with unequal speeds, they would not collide if
their relative velocities were greater than the escape velocity. The escape
velocity for equal masses m originally separated by a distance
r is v_{escape}=2√(mG/r) where G=6.67x10^{11}
N⋅m^{2}/kg^{2} is the gravitational constant.
QUESTION:
A 400lb person jumps up 2inches on earth. If same person jumps up on the
moon, how high would the jump be?
ANSWER:
I will assume that whatever the jumper does will add the same energy on both
earth and the moon. The gravitational potential energy U at the
highest point must be equal to that added by the jump, and U=mgh
where m is the mass (400 lb), h is the height (2 in on earth), and
g is the acceleration due to gravity (9.8 m/s^{2}
on earth, 1.6 m/s^{2} on the moon). The mass is the same both
places, so h_{earth}g_{earth}=h_{moon}g_{moon}. Solving,
h_{moon}=12.25 in.
QUESTION:
Hi, I was wondering what are the chances of survival from falling from the ninth floor of a building, going over the science of that how does surface affect the fall, body weight and trajectory. What is the difference from falling from a third story window as opposed to a higher up one?
ANSWER:
Someone else also asked this question; apparently it refers to a recent actual
incident of a student falling out a dorm room window about 85 ft≈26
m high; the student survived without serious injury. The second person also
wanted to know if I could estimate the force experienced on impact. First I
will calculate the speed he would hit the ground if there were no air drag.
The appropriate equations of motion are y(t)=26½gt^{2}. and
v(t)=gt where y(t) is the height above the ground
at time t, and g is
acceleration due to gravity which I will take to be g≈10
m/s^{2}. The time when the ground (y=0) is reached is found
from the y equation, 0=265t^{2} or t=√(26/5)=2.3
s. Therefore v=10x2.3=23 m/s (about 51 mph). The terminal velocity of a falling
human is approximately 55 m/s, more than double the speed here, so the
effects of air drag are small and can be neglected for our purposes of
estimating. (If there is air drag, terminal velocity is the speed
which will eventually be reached when the drag becomes equal to the weight.)
Estimating the force this guy experienced when he hit the ground is a bit trickier, because what really matters is how quickly he stopped. Keep in mind that this is only a rough estimate because I do not know the exact nature of how the ground behaved when he hit it. The main principle is Newton's second law which may be stated as F=mΔv/Δt where m is the mass, Δt is the time to stop, Δv=23 m/s is the change in speed over that time, and F is the average force experienced over Δt. You can see that the shorter the time, the greater the force; he will be hurt a lot more falling on concrete than on a pile of mattresses. I was told that his weight was 156 lb which is m=71 kg and he fell onto about 2" of pine straw; that was probably over relatively soft earth which would have compressed a couple of more inches. So let's say he stopped over a distance of about 4"≈0.1 m. We can estimate the stopping time from the stopping distance by assuming that the decceleration is constant; without going into details, this results in the approximate time Δt≈0.01 s. Putting all that into the equation above for F, F≈71x23/0.01=163,000 N≈37,000 lb. This is a very large force, but keep in mind that if he hits flat it is spread out over his whole body, so we should really think about pressure; estimating his total area to be about 2 m^{2}, I find that this results in a pressure of about 82,000 N/m^{2}=12 lb/in^{2}. That is still a pretty big force but you could certainly endure a force of 12 lb exerted over one square inch of your body pretty easily.
Another possibility is that the victim employed some variation of the technique parachuters use when hitting the ground, going feet first and using bending of the knees to lengthen the time of collision. Supposing that he has about 0.8 m of leg and body bending to apply, his stopping distance is about eight times as large which would result in in an eight times smaller average force, about 5,000 lb.
Falling from a third story window (about 32 feet, say) would result in a speed of about 14 m/s (31 mph) so the force would be reduced by a factor of a little less than a half.
ADDED NOTE:
A rough estimate including air drag would have his speed at the ground be
about 21 m/s rather than 23 m/s as above. Given the rough estimates in all
these calculations, this 10% difference is indeed negligible.
QUESTION:
Does the gravity of the earth work through the moon? I.e. If you are on the "dark" side of the moon, do you weigh more than you would standing on the side facing the earth?
ANSWER:
Yes, the net gravitational field you see is due to all masses, not just the
one you are closest to. On the far side of the moon the earth's field is
weaker than on the near side, but it points in the same direction as the
moon's field; so, you are right, you would be slightly heavier on the far
side.
QUESTION:
Gravity makes Earth orbit sun, Milky Way and Adromeda collide with each other,etc...
My question is, what can provide the necessary energy for a system like this for billions of years?
ANSWER:
The earth orbiting the sun has energy, but it takes zero energy to keep it
orbiting. The Milky Way and Andromeda galaxies have energy but no additional
energy is added as they move toward each other and eventually collide.
QUESTION:
If I use some magnetic bars, cut them perfectly so that I can put them together to form a globe, with the same pole pointing outwards, and the other pole pointing inwards, do I get a Magnetic monopole object?
ANSWER:
Think of your bars as dipoles of positive and negative magnetic charges
(monopoles) separated by a distance d. The magnitude of the magnetic
field B of a monopole is inversely proportional to the square of the
distance r from the charge, B=kq/r^{2} where
k is some constant. In the drawing above the field at point p is B=B_{q}+B_{+q}=kq[(1/(rd)^{2}(1/r+d)^{2}]=4kqrd/[(r+d)^{2}(rd)^{2}].
Now, look at the field when r>>d: B≈4kqd/r^{3}.
The field does not look like a monopole because it falls off like 1/r^{3},
not 1/r^{2}.
QUESTION:
It might look like a homework question, but it is not. Please help me. I have asked this question everywhere I could, but everybody seems to ignore it.
So, the problem is: Let us say we have two bodies A and B in contact with each other, with A lying at the back of B, and the system is on a frictionless horizontal surface. Let A have mass 5 kg and B 10 kg. Now let's say I apply a force of 45 N on A with my hand, then the system begins to accelerate at 3 m/s^2 and the net force on B by A is then 30 N, and B in reaction applies a net force of 30 N on A. Thus, the net force on A is 15 N. What I do not understand is why A is not applying a force of 45 N on B? If it is due to the reaction of B on A, how does A know in the first place that it is to exert a force of 30 N on B so that it receive a reaction of 30 N from B? Is not the reaction force of B on A some kind of a function of the action of A on B, and if it is, then how is the magnitude of the action of A on B is first determined? What is it that I do not understand about Newton's Third Law of Motion?
ANSWER:
OK, I will take your word for it that it is not homework. It is important to
be able to solve these kinds of problems. My method is to choose a body and
look only at that body. I choose first (as you did) to choose both masses as
the body, so M=15 kg and F=45 N and therefore a=F/M=3
m/s^{2}. Next I choose B as the body. The only force on it is the
force which A exerts on it, F_{BA}. Since we know that m_{B}=10
kg and a_{B}=3 m/s^{2}, F_{BA}=m_{B}a_{B}=30
N. Finally choose A as the body. Two forces act on A, F=45 N and the
force which B exerts on it, F_{AB}=F_{BA}=30
N; its mass is m_{5}=5 kg and therefore a=(F+F_{AB})/m_{5}=(4530)/5=3
m/s^{2}. The reason that there is not a 45 N force on B is because
your finger is not touching B, only A is touching B. The reason A "knows" to
exert a force on A is that it has no choice since A's acceleration and mass
are already fixed. Once you know the "reaction" force you automatically know
the "action" force because of Newton's third law, F_{AB}=F_{BA}=30
N for this problem. You could also have chosen A as the body before you
chose B as the body. Two forces act on A, F=45 N and the force which
B exerts on it, F_{AB}. Its mass is m_{5}=5 kg
and its acceleration is 3 m/s^{2}; therefore (45+F_{AB})=m_{5}a=15
N or F_{AB}=1545=30 N.
QUESTION:
I'm bothered by the seeming interchangeability in physics of the terms 'measurement' and 'observation'.
Scenario: Doing a double slit experiment while firing one electron at a time  in a darkened room with a person "observing the experiment" and one without an observer. And also doing another set of experiments with and without a detector  are the results different? if so, how?
ANSWER:
There can be confusion but it is mainly semantic. If you simply "observe"
the double slit experiment by counting electrons which hit the screen and
determining where they hit, no "measurement" has been made regarding the
doubleslit experiment itself, you have simply observed how the results play
out after the electrons pass through the slits. However, if your observation
includes some apparatus to determine through which slit each electron
passed, that is a "measurement"; in that case, the measurement will destroy
what you previously observed—the
interference pattern.
QUESTION:
Are star wars blaster weapons possible? i know lasers wouldnt work because its just light but maybe plasma energy or energy projectiles could work? i just need to know if the blaster rifles/pistols as pictured in star wars and the way they function could EVER work in the distant future and what way you think they could possibly work, assuming we had all the technology necessary to do so. thank you in advance for your knowledge.
ANSWER:
First of all, they are fictional with not much information available
regarding their power. However, a quick search of the web reveals that many
Star Wars enthusiasts have tried to make estimates of their properties from
detailed examination of the movies. First observation is that they are not
lasers because you can see their trajectories and Han Solo is seen to jump
out of the way in time to not get hit; light is just a lot faster. (You
should not write them off just "…because
it's just light…"
light, though; lasers can melt steel.) Also, if the pulses are any kind of
material (superheated plasma, charged particle beams, etc.), they could not
propogate very far in air which they do. But the thing that makes them, in
my view, impossible is the energy each pulse carries on the highest setting;
one site I found estimated the energy of a single pulse to be on the order
of several megajoules. Suppose that a single pulse lasts 1
ms=10^{3} s and carries an energy of 1 MJ=10^{6} J; the
power carried by that pulse would be 10^{6}/10^{3}=10^{9}
W=1 GW. This is the power output of a typical power plant. Where are you
going to get this kind of power in some compact form you can carry around
with you? No way!
QUESTION:
When in the early universe did life become possible? I'm assuming it would have had to have been at some point when the temperature had cooled down enough so that the protons and neutrons of basic elements of life like carbon and oxygen could bond and come together. Also, what do you believe is the purpose for intelligent life in the Universe?
ANSWER:
I usually do not answer questions about astronomy/astrophysics/
cosmology as
stated on the site. I can give you a little information here, though. The
universe just after the big bang was almost exclusively hydrogen. No
elements which could be used to form planets where life might develop or the
material for life itself. The first stars started forming about 100 million
years after the big bang, basically pure hydrogen stars. These had to be
much larger than the sun (perhaps 3001,000 times heavier) because of the
lack of heavier elements and consequently their lifetimes were too short (a
few million years) for life to evolve (it has taken about 5 billion years to
reach our stage of evolution, about half the sun's lifetime). Inside these
stars nuclear fusion caused heavier and heavier elements up to about iron to
form; then when those stars used up all the fuel, they exploded (super
novae) scattering the heavier elements to eventually be part of clouds of
dust and hydrogen which subsequently formed new stars with the possibility
of planets. (As you can see, it is not "cooling down" which is responsible
for heavier elements which are created in stars which are super hot.) So, we
are now at least several billion years into the age of the universe. More
detail on early stars can be read in a
Scientific American article. The "purpose" for anything is not within
the purview of physics.
QUESTION:
I want to design a homebuilt experiment that will physically display the actual curvature of the Earth, and I would appreciate validation that my methods are (at least theoretically)  scientifically accurate.
This test is based on the geometric relationship of a tangent line (globally 'straight' line) when it is perpendicular to a radial line of the sphere.
.
I am going to set 51 fenceposts fifty feet apart for a total length of 2500 feet. I'm going to use a water linelevel to mark a "locally' level line on the first two posts. Then, using a laser light as a straightedge, I will extend this straight line for the full 2500 feet. Then I will take the water linelevel to the last two posts, hold the level to the straight line marks, and note that the globally straight line is now an inch and 3 quarters out of "local" level, due to the Earth's curvature.
.
This inch and 3 quarters divergence is derived from the formula for the Earth's curvature that is eight inches per mile squared: 2500 feet is .4734 of a mile. Squaring that, then multiplying by 8 inches, gives 1.7931 inches.
.
Any advice or criticism or insight you could offer would be well appreciated. For brevity's sake I kept this desription quite short, but I will be happy to elaborate in greater detail if you want me to.
ANSWER:
Here is the problem: You assume that the earth is a perfect sphere. It most
certainly is not, particularly at the halfmile level. You can see some
detail in a calculation I did in an
earlier answer about the
Bonneville Flats where a similar question to yours was looking to observe
curvature at the 10mile level. It takes several hundred miles to be able to
do an accurate measurement of the earth's curvature; the
ancient Greeks did this around 200 BC.
QUESTION:
If I am running at average sprinting pace for an 17 year old male and
I jump off of a 80 meter drop, how far forward will I land from the jumping point? Assuming I am around 11 stone.
ANSWER:
80 m is pretty high, so you will be going very fast when you hit;
therefore, neglecting air drag might introduce significant error. But
air drag is pretty tricky to calculate and I will neglect it; the answer I
get will be somewhat bigger than what would really happen. I will take your
speed to correspond to running a 100 m dash in 15 s, about v ≈6.7
m/s. The equations of motion are x=6.7t and
y=4.9t^{2}
where x and y are the horizontal and ver tical positions
relative to the edge of the cliff and t is the time after jumping.
Solving the y equation for t when y=80 m (the ground),
t=√(80/4.9)=4.04 s, so x=6.7x4.04=27.1 m. That is the answer
neglecting air drag. Note that it is independent of the mass m.
QUESTION:
We represent waves by crest and troughs like ( ups and downs ) but in the slit interference process we represent it in the form of other figures like ))))))) . can u tell me how crests and troughs are represented in this form ?
ANSWER:
"...figures like )))))..." are called wave fronts. Each front is meant to
represent the locus of points at some instant where the wave crests are and
halfway between are the wave troughs.
QUESTION:
According to Einstein,gravity is the bending of spacetime by matter which is unlike Newtonian gravity.But there are boson particles called gravitons which are thought to be the cause of gravity. Aren't these theories conflicting?How can gravitons exist if there is no gravitational force?
ANSWER:
It is generally believed that gravity, like any other field in physics,
should be quantized. So far, there is no successful theory of quantum
gravity. If there were such a quantized gravitational field, there would be
a boson "messenger" like the photon for the electromagnetic field or the
gluon for the strong interaction field; that would be the graviton. Since
there is no theory, there is no graviton, only speculation.
QUESTION:
Many explanations in physics which I have seen proceed with a 3D Euclidean space, empty but for a population of objects. Lines representing electromagnetic radiation are drawn from emitter to sensor to explain various phenomena. The apparently discreet nature of the representations leaves me wondering 
When I see a star, I see it "everywhere", there seem to be no black spots . Spokes radiating from a wheel grow farther apart as they proceed, why not light from a star? Be it particle or wave or some probabilistic combination, does not some m/e travel from the star to strike my eye? Why does this not put the burden of infinite m/e on any light source?
ANSWER:
Those "rays" you see drawn in geometrical optics are not meant to represent
the only place light is. There is no rule about how many rays to draw but
the direction of a ray is always the direction that light is traveling at
any point on the ray and the intensity of the ray is implicit in the density
of the drawn rays. The rule for how to draw a ray between two points is that
you find the path which will take the shortest time. For a point source of
light (like a star) in a vacuum (approximately what is between you and the
star) that will be a straight line. In the picture above you could have
drawn twice as many rays and that drawing would still have conveyed the same
information about the light. An alternative way to represent the waves is to
draw wave fronts which are concentric about the point source and equally
spaced; these are meant to represent the location of the crests of the light
waves at some instant and are imagined to be expanding outward from the
source. The rays are always normal (perpendicular) to the fronts. This
drawing is really a cutaway because the fronts are really spheres, so you
see that wherever you stand you will see light. Because a wave front is
expanding, its area (4 πR^{2})
is getting bigger so the energy of that wave is getting more and more spread
out meaning that the light is getting dimmer.
QUESTION:
I have a question
about the properties of light. If I am traveling at 99.99999... The speed of
light and shine a light in the exact opposite direction, how would that
light behave to an observer? Would it stay still? Would the light still
travel at c?
ANSWER:
Every observer measures the same speed of light, c. That does not
mean that the light is identical. If the source has a huge speed, the light
has a huge red shift, its wavelength is greatly larger than that observed if
at rest relative to the source. It would likely be invisible to the unaided
eye.
QUESTION:
I have 300 grms of baked beans that I want to heat in a microwave. Wil it take longer or more energy to cook these beans in one dish or in separate dishes of 150grms each?
ANSWER:
The rate at which the beans will be heated by the radiation (R_{H})
is proportional to their volume. But, as soon as they start heating up, they
start cooling off by radiating from their surfaces; the rate of cooling (R_{C})
is proportional to the area. Roughly speaking, the volume is proportional
the the L^{3} and the area is proportional to L^{2
}where L is a measure of the size of the pile of beans. The
ratio R_{H}/R_{C}=KL^{3}/L ^{2}=KL
determines the net rate of heating; K is some constant. For example,
suppose your beans were contained in a spheres of radii L and 1.26L;
I chose those because the larger radius sphere would have twice the volume
of the other. Then, the larger sphere will heat up 1.26 times faster than
the smaller (two spheres). Assuming that the microwave energy is constant
with time, the longer time it takes the more energy is used.
The problem with my analysis, though, is that it depends on the power of the microwave. If it has a relatively low power, the radiation will be mostly absorbed in the beans and the assumption that R_{H} is proportional to the volume becomes untrue. Only if a small fraction of the radiation is absorbed by the beans will my analysis be approximately correct. You should do an experiment, try it!
QUESTION:
Why only magnetic moment due to spin is considered in proton nmr spectroscopy. what about magnetic moment due to orbital motion??
ANSWER:
You are right, a proton in a nucleus will have both spin and orbital
magnetic moments. But the whole idea of NMR is to put magnetic moments in a
magnetic field and they tend to align. To flip the spin requires a certain
amount of energy. So you send photons of the right energy, typically radio
frequency radiation) and they will flip the proton spin. You need not worry
about other moments because the photons are tuned only to the right
frequency for spin.
QUESTION:
In understanding that two objects (no matter their phisical properties) will accellerate at the same rate towords the same gravitational force in a perfect vacuum, do we take into consideration the gravity of the falling objects themselves? Would the larger gravitational force of the larger falling object be perfectly offset by the larger ammount of inertia or tendency to resist motion?
ANSWER:
The force F between masses M and m whose centers are
separated by a distance R is F=GMm/R^{2}. But,
the acceleration a of m is a=F/m=GM/R^{2}.
So, as you ask in your last sentence, the inertia (m in a=F/m)
cancels out the mass (m in F=GMm/R^{2}) in the
acceleration. This is actually a profound finding for the following reason.
In Newton's second law, mass is, as you say, inertia (inertial mass, m_{I});
in Newtons universal law of gravitation, mass is the ability to feel or
create gravity (gravitational mass M_{G}). The fact that
gravitational accelerations are the same implies that m_{I}=m_{G},
not something you would necessarily expect. Why these are equal was not
really understood until the general theory of relativity was proposed by
Einstein.
QUESTION:
i am a metal worker and maybe it's dumb but i have thought up an idea for renewable energy. I need to build a huge tank capable of holding 10000 of water, it will have a coned base with a hose coming off it and running back up into the tank, returning the water to it's point of origin....question is...with 10 tons of water down force, 1 meter from the ground....would a decent amount of water travel up that hose pipe and back into the tank???..
ANSWER:
It makes no difference if you have a whole ocean of water, the water at the
bottom can never push water higher than the surface of the tank, can never
cause a continual current like you envision. There is a very detailed
answer for a very similar question to yours that
you should look at.
QUESTION:
In the event horizon of the black hole, a pair production of a particle and an anti particle and the falling of one into the black hole and the releasing of the other from it gives us the Hawkins radiation, but my question is , isn’t it possible for both the particle and the anti particle to fall into the black hole. If that happens defiantly the vacuum energy of the space decreases how can the universe compensate this decrease in energy to validate the law of conservation of energy.
ANSWER:
There are two ways you can imagine the creation of the
particleantiparticle pair
(PAP):

Vacuum fluxuations where a particleantiparticle pair is spontaneously created just outside the event horizon, violating energy conservation. However, the uncertainty principle (UP) allows this violation but only for a very short time. If one escapes, it adds energy to the universe so the captured particle must have negative total energy thereby decreasing the mass of the black hole and maintaining energy conservation. If both particles are captured, they are required by the UP to recombine thereby keeping the mass of the black hole constant.

When a virtual pair is created just outside the event horizon, the intense gravitational field of the black hole can provide the energy for the PAP to become real rather than virtual. In this case, the energy acquired by the pair will be lost by the black hole reducing its mass. If only one of these is captured, the energy of the escaped particle will be added to the universe but the energy of the captured particle will be added to the (already reduced) energy of the black hole resulting in a net loss of energy of the black hole equal to the energy of the escaped particle.
Energy of the whole system is conserved.
QUESTION::
If the speed of sound is inversely proportional to the density of a material, why does sound travel faster in solids (it is the most dense).
I have read that it takes more energy for sound to travel in dense materials so it takes longer but then neighbouring molecules are closer so sound does not have to travel that far, making it faster. This doesn't make any sense because it says the more dense a medium is, sound is both faster and slower.
Also, how does bulk modulus affect the speed of sound.
ANSWER:
You are oversimplifying because the density
ρ
is not the only thing which determines the speed of sound. In general, the
speed of sound v may be written as v= √(K/ρ)
where K is a parameter which specifies the "elasticity" of the
material. (Note that v is inversely proportional to √ρ, not ρ
as you state.) For example, just qualitatively, you might say that steel is
much denser than air but it is also much more elastic. The way K is
specified for gasses is quite different than for solids. For an ideal gas,
K=γP where P is the pressure of the gas and γ=5/3 for
monatomic gasses and 7/5 for diatomic gasses. On the other hand, for solids
K=B+4G/3 where B is the bulk modulus and G is
the shear modulus. For solids K is enormously bigger (~10^{11}
N/m^{2}) than γP for gasses (~10^{5} N/m^{2}).
That answers your question. So, let's just do a rough calculation to see if
I get reasonable values for air and steel:

Taking air as mainly diatomic γ≈ 7/5 , atmospheric pressure P≈10^{5} N/m^{2}, ρ≈1 kg/m^{3}, v ≈374 m/s; a little high, but close.

For steel, B ≈1.6x10^{11 }N/m^{2}, G≈7.93x10^{10} N/m^{2}, ρ≈7900 kg/m^{3}, v≈5800 m/s; this is exactly the speed of sound in stainless steel!
QUESTION:
The earth's radius varies depending on latitude. How many times greater is the acceleration of gravity at the poles of the earth than the equator if the radius of the earth is 99.5% smaller at the poles than the equator?
ANSWER:
To calculate exactly would be impossible because the earth is not a perfect
spheroid and its mass is not uniformly distributed. But you can get a good
idea by calculating the difference in gravitational force for spheres of the
same mass but having the two different radii; these radii are 6378 km
(equator) and 6357 km (poles), about 0.33% larger at the equator. Since the
gravitational force is inversely proportional to the square of the radius,
this would give about a 0.11% smaller g at the equator. This effect
is smaller than the effect due to the earth's rotation which results in
about a 0.34% reduction at the equator (due to centrifugal acceleration).
QUESTION:
What is the purpose of utilizing a percentage of body weight to determine how much weight to bench press/push/whatever? I know this seems like a fitness question and not a physics question, but what I am interested in is WHY weight would be used to determine how much one could (or should be able to) lift/push?
For example: A gym teacher wants to grade his students on their strength. He decides to use abililty to push a weighted sled across the floor as the measure. He wants to make the task equally difficult for every student in order to make the grading fair. So, he decides that each student will push 2x his/her body weight for 5 minutes and the grade will be based on how FAR the student is able to push.
So, Student A weighs 170 pounds and pushes 340 pounds (including the weight of the sled) for a total of 160 yards. Student B weighs 240 pounds and pushes 480 pounds (again, including the weight of the sled) for 80 yards. Student A pushed farther and gets a better grade, but Student B complains that he had to push much more weight so he should not get a worse grade.
Does Student B have a legitimate complaint or does his heavier weight contribute somehow to his ability to push that doesn't have anything to do with his strength? As in, does his weight help push the sled in some way?
Sorry, I don't know enough about physics to ask this question using proper physics terms like force, mass, etc. I hope you will still answer my question!
ANSWER:
I cannot comment on the rationale for correlating weight to strength. I can
certainly comment on the physics of your particular example of sled pushing.
I would first of all comment that this example is certainly not one solely
of strength because, since it is a timed activity, endurance as much as
strength is being tested; if one student, for example, were a heavy smoker,
he would likely become exhausted more easily. As a physicist, I would equate
"strength" with force. The specific example you give, though, seems to me to
be more related to energy (work done by the student) or power (rate of
energy delivered) than strength; purely in terms of strength, the heavier
student exerts more force. The force F which each student must exert
depends on the weight w he is pushing and the coefficient of friction
#956;
between the sled and the ground, F= μw.
The work W done in
pushing the sled a distance d is W=Fd= μwd.
The power generated if W
is delivered in a time t is P=W/t=μwd/t .
Both students have the same
μ
and
t , so W_{A}/W_{B}=P_{A}/P_{B}=d_{A}w_{A}/d_{B}w _{B}=(160x340)/(80x480)=1.42.
So student A did 42% more work, generated 42% more power, than student B.
From a physics point of view, B demonstrated more strength, A demonstrated
more power. I would judge that this is not a fair way to assign a grade. It
would be interesting to see if A (B) could move B's (A's) sled 80 (160)
yards.
QUESTION:
Say I have 2 factories producing an object, we'll be arbitrary and say bottle caps. Each produces them at a rate of 1 per second. Now ignoring things like raw materials etc...
One stays on Earth, producing its bottle caps constantly.
The second one (somehow) gets accelerated to (again being arbitrary) 90% the speed of light and remains at that velocity until completing a roundtrip 100 years long.
When the moving factory returns, 100 years later from the perspective of the factory on Earth...will there be a difference in the total number of bottle caps each has produced?
ANSWER:
This is just a variation of the twin paradox.
The number of bottle caps produced by the moving factory will be N √(10.9^{2})=0.44N
where N is the number produced on earth.
QUESTION:
Okay, so for a while I've been wondering why bass travels through certain materials that higher frequency sound doesn't. I have a theory which could be wrong but what I thought is that bass, having a larger wavelength, passes through particles that are bigger, and that shorter wavelengths can't get through. Somewhat like red light and blue light traveling through the atmosphere. Okay, so I looked it up everywhere (almost) and couldn't find a straight answer, even my physics professor didn't know surely why this happens. So, maybe you could. Why can I hear just the bass when someone blasts music in their car?
ANSWER:
The reason is that the attenuation of sound intrinsically depends on the
frequency of the sound, the higher the frequency the greater the
attenuation. The theory of attenuation is rather complicated, so I will just
sketch the results. For a
Newtonian fluid,
Stokes showed that A(d)=A_{0}e^{} ^{ αd }
where
A(d) is the sound amplitude a distance d into the
material, A_{0}=A(d=0), and
α= 8 π ^{2} η f^{2} /(3ρv^{3});
here
η
is viscosity of the fluid, f is frequency of the sound, ρ is
the density of the fluid, and v is the speed of sound in the fluid.
The important thing to notice is that the amplitude decreases like exp(βdf^{2})
for any Newtonian fluid. For example suppose you compare f_{1}=20
Hz with f_{2}=1000 Hz for a thickness of d=1 cm=0.01 m
and A_{1}(0.01)=јA_{0}=A_{0}e^{β�0.01�400}=A_{0}e^{4β}
which can be solved for β, β=0.35 s^{2}/m; then, for
f_{2}, A_{1}(0.01)=A_{0}e^{0.35�0.01�1,000,000}=A_{0}e^{3500} = A_{0}x 9x10^{1521}≈0.
Finally, for any material, Stokes' law of attenuation still may be written
as
A(d)=A_{0}e^{} ^{ αd }
except α=βf
^{ε }where 0<ε<2. Water and
most metals have ε≈2. But even if ε is smaller, it is always
positive which means that the attenuation is always greater for a higher
frequency.
QUESTION:
We know the relative size of the solar system measuring thru the distances of the planets circling our star.
Do we know the thickness of this rotational disc?
Is the gravity of our star concentrated only on this single orbital plane as our star rotates?
If not, why do planets not circle verticaly as well as on horizontal planes?
As far as I know we have only sent space craft thru our orbital disc towards other planets. Do you believe if directed a flight up from our orbital plane would the suns gravity lessen where our space craft could fly faster?
ANSWER:
I should first address your speculation that gravity is stronger on the disc
near which the planets orbit. In fact, gravity depends only on how far you
are from the source, so if you went 100,000,000 miles from the sun in any
direction you would find the same gravitational force. There is not really
any welldefined disc whose thickness you can measure; all the planets are
nearly all orbiting in the same plane and the inclinations of the orbits
relative to say, the plane of the earth's orbit, are all less than 3 Ѕ^{0}.
But that does not mean that everything is near that plane. Many of the dwarf
planets are much out of the plane of our orbit; Pluto, for example, is about
18^{0} inclined relative to our orbital plane. Outside the most
distant planets is the
Oort cloud, a collection of rocky and icy bodies which is distributed
roughly equally in all directions; comets are thought to originate in the
Oort cloud and when they enter the inner solar system they come from all
directions relative to the orbital plane. The reason that the planets are
all nearly on the same plane is that when the sun first formed from a huge
cloud of dust and rocks, it had angular momentum and as the system collapsed
under its own gravity, it spun faster and faster and the resulting
centifugal force caused it to flatten like a giant pizza dough.
QUESTION:
An image is formed
when the reflected ray meet each other at a point. But why does an image is
not formed by the intersection of incident rays? After all they are the same
light rays and it's not like there's some kind of reaction which takes place
or something.
ANSWER:
The light rays from each point on the object all diverge away from that
point. Unless there is something to change their directions, e.g. a
mirror or a lens, they never meet.
QUESTION:
If I were to throw a pencil up into the air, would it come down at the same speed as I threw it up?
ANSWER:
It would if there were no air. However, air drag takes energy away from the pencil and so it would arrive back at your hand going more slowly. For speeds you are likely to be throwing the pencil, though, this effect is small and often neglected in elementary physics classes.
QUESTION:
The demonstration wherein one pours water from a clear vessel and one shines a laser pointer (a red laser pointer in my case) through the water and it appears the laser beam bends with the water pouring out of the clear vessel;
What causes the beam to bend with the water? Would this same type of effect work with other fluids than water, such as air or glass?
ANSWER:
It is caused by
total
internal reflection. If the angle which the light hits the surface
between the water and the air is glancing enough, the light will be
reflected back into the water rather than be refracted into the air. Total
internal reflection can happen whenever light strikes the interface with a
material of smaller index of refraction; the minimum angle for which it can
occur depends on the relative indices of refraction.
QUESTION:
I was reading a introductory book on particles. I read that neutrinos only respond to weak and gravitational forces, isn't that fact that it responds to gravitational forces prove that it has mass? So why was this years Nobel Prize awarded for figuring out that neutrinos had mass?
ANSWER:
No, responding to gravity does not imply mass. Photons, which are known to
be massless, are deflected when passing near a large mass. See
FAQ page.
QUESTION:
If a train runs at 100km/h and if in this train somebody drops a rubber ball and catch it, let's say it drops down 1m and up in one second, it means that the ball has travelled 2 meters in one second for the person in the train.
If I am at the station looking at the train when it passes me by and let's say it is made of glass so I can see what is happening, I can see the ball dropping down and up but for me as the train is moving the ball will have travelled 2 meters plus the distance travelled by the train in one second so more than 2 meters.
I don't think that space time dilatation plays at this sort of speed so how is it explained?
ANSWER:
Actually, you have not calculated the distance traveled correctly. But
because the ball is speeding up on its way down and slowing down on its way
up, you have set up a problem with unnecessarily difficult mathematics to
calculate the distance seen by the person at the station. I will alter your
problem somewhat to make it easier to calculate: the ball is made to first
move down at constant speed and then to move up with the same constant
speed. If the time is still 1 s, the necessary speed is v_{y}=2
m/s. The train is moving with speed v_{x}=100 km/h=27.78 m/s
so the ball and train move a distance 27.78 m horizontally during the one
second the ball is moving. But the distance the ball moves is not
27.78+2=29.78 m but rather
2√( 13.89 ^{2}+1^{2} )=27.85
m because the ball will take a zigzag path. But the ball also has a
different speed as seen from the station, v=√ (v_{x}^{2}+v_{y}^{2})=27.85
m/s. We therefore conclude that the time it takes the ball to travel 27.85 m
is 1 s, so everything hangs together. Your error was to ignore the fact that
the ball has a horizontal component of its velocity in the frame of the
station. You are right that relativity plays no role here.
QUESTION:
Suppose I have a charge +q and there is a point P , Suppose I place a conductor between the charge +q and P . Since there are free electrons in it , Negative electrons move towards +q and equal positive charge inside the conductor near P , The conductor has charge distribution like a dipole. So If I want to calculate E field at P . I could use superposition principle to find E at P due to +q and E due to dipole. But Gauss's law says that dipole doesn't contribute anything to E field at P. Can you explain me 'intuitively' (Not in equations) why the dipole wont contribute anything to the field at P ?
ANSWER:
Gauss's law does not say that the dipole contributes nothing at P. If you
put a spherical Gaussian surface enclosing both q and point P the net
charge enclosed is q but that does not mean that the field due to the
dipole is zero everywhere on that surface. All you can say is that the net
electric flux passing through that surface is q/ ε_{0}.
Gauss's law is usually useful in determining a field only if the field can
be argued to have constant magnitude and normal to the surface everywhere.
Superposition, not Gauss's law, should be used for this problem.
QUESTION:
If I was in a spaceship
moving at half the speed of light (in normal space) and I measured the time
it took the beam of light from a laser pointer to travel from its source to
two points inside my ship, one in the direction of travel and one in the
opposite direction, wouldn't C demand that I observe different time
measurements? If the time measurements are the same, then when I pointed in
the direction of travel, the light would have to be traveling at C plus the
speed of the ship from observations made outside of the ship. If the
measurements are different, then shouldn't we be able to observe the same
effect on earth because of the direction of travel of our planet, solar
system, Galaxy, and group of galaxies?
ANSWER:
I am not sure what you are asking here. Does this
earlier answer help you? One thing is clear is that you do not really
appreciate that the speed of light in vacuum is independent of the motion of
either the observer or the source. This is one of the main postulates of
special relativity. See the FAQ page for questions related to this.
QUESTION:
Sir,I am a mechanical engineer by profession but very interested in reading physics fundamentals. Recently I went through the fundamentals of electro magnetism and I got this doubt. Consider two charges each of charge +q rigidly fixed in a train moving with a constant velocity, V. Let the train speed be negligible compared to speed of light so that we can treat the problem in nonrelativistic terms. The fixity condition ensures that it overcomes electrostatic forces and remain motionless. A traveler in train sees both of them at rest and there wont be any magnetic forces developed between the charges.
Now consider an observer in platform. For him, both charges are moving with a constant velocity, V equal to the train velocity. Each charge will develop magnetic field according to this observer as per BiotSavart law and there will be mutual attraction as each charge is moving under the magnetic field of other.Thus, an observer in train sees no magnetic forces whereas an observer in platform sees mutual magnetic attraction. How do you explain this?
ANSWER:
There is no rule which says that the either the electric or magnetic field
must be the same in all frames of reference, even slowly moving frames like
you specify. The real root of your problem is that electromagnetism is
intrinisically relativistic, even at slow speeds; the electric and magnetic
fields of classical electromagnetism are really both components of the
electromagnetic field which is a tensor and when you change inertial frames,
you cause a transformation of that tensor into another where both the
electric and magnetic field pieces of it are different. In your second case
you would also find that the electric fields were slightly different from
their original values but the differences would be very tiny; the magnetic
fields, though, are nonzero but small, but small is very big compared to the
original magnetic field of zero.
If you are interested, I will give here the electric and magnetic fields for one of the charges moving with velocity v in the +x direction. E'=iE_{x}+ γ(jE_{y}+kE_{z}) and B'=(vxE')/c^{2 }where γ=1/√[1(v/c)^{2}] and i, j, k are unit vectors; the vector E is in the frame where q is at rest and E' and B' are when q is moving.
QUESTION:
I see the equations for computing the slowing of time, based on speed compared to light speed. Do we always use c=3^10 cm/sec? I am asking since light speed is slower, sometimes much slower, in mediums other than air or space. If we use a smaller value for c, the time dilation factor is enhanced.
ANSWER:
The significant thing in the theory of special relativity is the speed of
light in vacuum. What matters in the theory is the universal constant
c, not the speed in the medium you happen to be looking at.
QUESTION:
Is it true that a person flying in an airplane is actually living a shorter amount of time, than a person standing on the ground?
ANSWER:
You must specify " … living
a shorter amount of time … "
with respect to whom . The
person in the airplane sees time progressing at a normal rate. However, the
person on the ground will see the clock of the person on the airplane run
slowly, so he will perceive the traveling person to be " living
a shorter amount of time", i.e. she will have aged less when she
returns to earth. You should read FAQ Q&As on the
twin paradox,
how clocks run, and the
light clock.
QUESTION:
If I was travelling in a bus and the bus was travelling at around 60 mph and
I jumped in the air, how come I don't fly to the back of the bus. Like with flying, if
I was in an airplane travelling at around 500 mph, and I was floating without an aid in the cabin why do
I travel with the plane not just get smashed to the back?
ANSWER:
It is impossible for you to be " floating without an aid ",
so let's assume that you are standing on the floor of both the bus and the
plane. In both cases it is because of Newton's first law. Someone watching
you from the ground would see both you and the vehicle as moving with the
same speed. From that point of view you are in equilibrium because the only
forces on you are your weight down and an equal force up from the floor.
Newton's first law says that if you are moving in a straight line with
constant speed you will continue to do so if all forces are in equilibrium.
When you jump up, the floor no longer exerts a force on you, so you are no
longer in equilibrium vertically and you will fall back to the floor. But
you are still in equilibrium horizontally (there are no forces on you
horizontally) so you will continue, right along with the bus or plane, to
move horizontally with the same speed. Actually, I did this the hard way
because Newton's third law will still be valid in the moving bus or plane,
so when you jump up you will not move horizontally because there are no
forces which act horizontally on you to accelerate you (make you start
moving horizonatlly).
QUESTION:
If I have two soccer
balls the same size, yet one is heavier than the other, which one will go
farther with the same kicking force involved? How does momentum and Newton's
Second Law come into play here?
ANSWER:
I will assume the balls fly through the air, not roll or slide on the
ground. This question has no single answer. Before getting too deeply into
this, I need to modify your question a little. The force does not determine
what happens to the soccer balls (see FAQ
page), so let us change " same kicking force" to "same kicking
impulse"; impulse is essentially force multiplied by the time it was
applied. That is also more convenient because the change in linear momentum
(mv) is equal to the impulse (J). So, with equal impulses, the
lighter ball has a larger initial speed: v_{light}=(m_{heavy}/m_{light})v_{heavy}.
So, if both balls are launched at the same angle and if air drag can be
neglected, the lighter one will go farther because it started with a
larger speed. Maybe that is all you wanted. But, it is altogether possible
that air drag will not be negligible. The force of air drag may be roughly
approximated as F=јAv^{2 }for a sphere where A
is the crosssectional area of the balls. (This is true only for SI units
since ј is not dimensionless.) So now you see that the ball launched the
fastest experiences the larger air drag slowing it down. But the heavy ball
has even more advantage since Newton's second law says a=F/m so even if the two balls
experienced the same force, the less massive ball would have a larger
magnitude of acceleration, slow down faster. Now it becomes complicated
because you have to know the masses, the impulse, and the angle of the
launch to calculate which would go farther.
QUESTION:
Let’s say you have a very long strong tube uncapped at both ends, and laying at sea level. It would have one atmosphere pressure inside. Put a cap on one end and stick that end down to near the bottom of the Mariana Trench. Now pop the cap off. Wouldn't this create a perpetual geyser from the end at the surface? It seems to me that it would come out with enough pressure to run a generator. Would it be considered perpetual motion or just hydraulics? Why wouldn’t this setup solve the world’s clean energy needs? Please pop my bubble. I can't be the first to think of this. What am I not accounting for? Inside the tube I don’t think gravity would be a player. Neither would the pressure fading outside the tube.
ANSWER:
Once the water reached the top of the tube the column of water would be in
equilibrium, just the same as if you had put the tube down without capping
the bottom. You might get a spurt at first but then you would just be left
with a column of water in equilibrium. I guess you could analyze it roughly
by ignoring viscosity and the drag of water moving on the pipe. Suppose the
crosssectional area of the tube were A, the depth of the bottom is
D, and the density of water is
ρ .
At some time after uncapping, the distance up to the current level of the
water in the tube I will call z. The force up on the column of water
would be F_{bottom}=(P_{atm}+ρg D) A,
the weight of the column of water is W=ρgzA,
and the force down on the top is
F_{top}=P_{atm}A. The net force on the
column is F= ρgA (Dz)=ma= ρAza
where a is the
acceleration and m= ρAz
is the mass. So, a= ρgA (Dz)/ ρAz=g [(D/z)1].
Suppose we take g≈10 m/s^{2} and D≈10^{4} m;
then a≈10[(10^{4}/z)1] m/s^{2}. The graph of
the acceleration is shown to the right; note the logarithmic scale for
acceleration to show the huge variation from bottom to top. Note that it is
huge, about 10^{5} m/s^{2} for 1 m and falls by four orders
of magnitude when the column of water is halfway to the surface, finally
falling to zero when the pipe is full. Suppose we start with z=1 m
and the column at rest (where a≈10^{5} m/s^{2}).
After 10 milliseconds (10^{2} s) the column will have acquired a
speed of approximately v=at=1000 m/s and the water level would have
risen approximately to z=1+Ѕat^{2}=6 m. Now, the speed
is almost immediately huge and so the assumption of no frictional forces
becomes impossible. The drag of due to the water moving in the tube for the
column of water going 1000 m/s will be huge
and keep the water from speeding up so rapidly. Also, the air above the
column of water will not just get pushed up will get compressed and
therefore push down on the column with a pressure greater than atmospheric.
It is therefore clear that my simple calculation above is only a best case
scenario and the drag and added pressure will take energy away. The drag
might get big enough that the water in the tube would boil and release steam
to futher impede its progress. I believe that you will end
up with column of water slowing down as it approaches the surface of the
ocean and it will just come to rest there. In any case, when the tube was
filled, even if still moving, the column of water would have zero net force
on it. If you tried to have it do work on your turbine, it would soon loose
all its kinetic energy and stop. Alas, no clean energy solution!
ADDED
THOUGHT:
Robert M. Wood pointed out to me that the most energy you could possibly get
out of the rising water would be the amount of work you performed to get it
down there. If you neglect the weight of the tube itself (or, imagine first
lowering the tube open at both ends and then pushing the water down with a
piston), this work would be
W=ЅρgAD^{2}.
Neglecting any energy loss, this would have to be equal to the kinetic
energy of the column of rising water when it got to the top, K=ЅρDAv^{2},
so v=√(gD). For the case above, v=316 m/s, pretty
modest. And keep in mind that all the water in the tube is now in
equilibrium, so as the water goes above the surface, it is slowing down as
it gains potential energy. And, if you imagine the tube projecting high
above the surface, you can calculate how high h the water will rise
before falling back. The center of mass of the column of stopped water is
y=Ѕh and its mass is ρhA, so mgy=Ѕh^{2}ρgA=ЅρDAv^{2},
so h=v√(D/g)=D. Therefore, with no friction, the
water would rise up and oscillate back and forth for ever. But, as argued
above, friction will be extremely important so this oscillation will quickly
damp out and will never rise up to a height equal to the depth of the ocean.
See the video in the answer to the followup question below as an example for
a miniature of this.
FOLLOWUP
QUESTION:
Thank you for finding interest in my question and for being able to let the nuts and bolts slide. The calculations you created from my questions were like a body punch to my self esteem. I didn't understand a lick of it. And now I'm feeling like the Dickens character asking for more gruel. Your response knocked me for a loop. You pointed out two things I wasn't accounting for. The weight of the water in the tube and the air resistance. The air resistance I've resolved/incorporated.
The equilibrium (if it occurs) from the weight of the water in the tube is a stopper and I hate it. Because it seems so common sense. Then I thought of this,(1000 ft.depth= 30 atmospheres = 420 psi.~), in your model if you bring the tube up at a 70 degree angle instead of the assumed 90, because of the longer tubing, there wouldn't be enough pressure to fill the tube. Also, if you put an uncapped tube from the surface to depth at a 70 degree angle as soon as you reached depth the tube would have too much water in it for the pressure and water would have to be expelled from the bottom until it reached the length of the 90 degree tube length. (I can't even do the baby geometry needed to figure out how much longer the tube would be.) Neither of these seems right to me. What do you think?
What I picture happening is the moment the cap is removed at depth the whole ocean rushes to fill the void at the top and the water pressure at the outlet would be the same as at the intake. The entire tube would become an extension of the 420 psi zone at 1000 ft. If inside the tube was a vacuum and capped at the top wouldn't this be the case? Wouldn't the pressure be 420 psi at the cap?
Here is my main reason for not being able to let this go. (this is my first version of this train of thought) Same long strong tube only this time it is assembled and capped in the vacuum of space 500 miles above earth and the capped end brought down to the surface. When the cap is removed I expect the atmosphere at that point to start rushing up the tube, through it, and out the end in a very long term geyser. And even though the tube is 500 miles long equilibrium would never be reached.
Aren't both examples using the same principal? It seems to me that it is the same principal and that the water at depth would view this one atmosphere pressure at the surface as the next best thing to a vacuum and continue trying to fill it. Just as the tube from space would leak away our air.
Doesn't pressure equal stored energy?
This thought is like a song stuck in my head. I'm tired of thinking about it. It's been months now. Validation would be best for all but, at this point I would settle for a kill shot. Can you help (either way)?
ANSWER:
You have so many misconceptions that I am not even going to try to
convince you with the physics arguments. A couple of comments: The reason
that the pressure is so high deep in the ocean is that the water deep down
must hold up the weight of all the water above it; to to try to ignore
weight of the water in the tube misses the whole point. Suppose you had a
tube 10 km long full of water (not underwater). The pressure at the bottom
would equal to the weight of the tube of water divided by the area of the
tube (+P_{atm}). Also, I have
previously answered your
question about the air in a tube in the atmosphere—it would not suck out all
the air in the atmosphere for the identical reason your water tube would not
create a perpetual geyser. Since you do not know any physics, let me
convince you another way; remember physics, at its heart, is an experimental
science, so let's do an experiment. Your idea should work equally well if
the ocean were only 5000 m deep, maybe not just so robust a geyser, right?
That is, your idea should scale. So, try the following experiment. Take a
straw and put a little cork to plug the bottom and push it down into a glass
of water; pop that cork out some way and see what happens. Does the water
come up the straw and continually squirt out the top (tiny geyser) or does
it come up to the level of the water surface and stop? Maybe it will rise a
bit above the surface but it will settle back to the level of the water in
the glass.
QUESTION:
Suppose we have a horseshoe magnet. Now we bend it in such a way that it becomes doughnut shaped and poles remain in contact with each other. In this situation what will happen to the magnet? Will it behave as a magnet? Where would be its poles? What will happen to domains inside the magnet?
ANSWER:
A horseshoe magnet is just a bent bar magnet, so let's start there. I have
shown a short stubby bar magnet, but you can imagine that as it gets longer
relative to its width the field inside will get more and more uniform. Now,
bend it around into a circle and it will look just like a toroid. The
magnetization inside will remain pretty much the same as before you bent it
and there will just be field inside the torus, not outside. The domains will
all stay pretty much the same.
QUESTION:
the Coriolis effect..
on a freely falling body... is to east...?
motion of earth is towards east right.? so we must feel that falling body deflects to the west isn't it..???
ANSWER:
I cannot give the full derivation of the motion of a particle in a rotating
coordinate system, it is much too involved. I can give you the results,
though. The coordinate system we will use is the coordinate system (x',y',z')
shown to the right. The Coriolis force is given by 2mv'x ω
where v' is
the velocity of m and
ω is the angular
velocity of the earth ( ω≈ 7.3x10^{5}
s^{1}). Now, for a body dropped from some height h, the
direction of the velocity is in the negative z' direction, so the
direction of
v'x ω
is east. This is not what
you would intuitively suspect (as you note), but it is true. An expression
for how far eastward it would drift before hitting the ground is x'=[( ω�cosλ)/3]√(8h^{3}/g)
where λ is the latitude. For example, at the equator where λ=90^{0},
and you drop it from 100 m, the deflection would be x'=2.2x10^{2}
m=2.2 cm.
QUESTION:
What is impulse? My text book explains this as force acting for a short interval of time. Can you please explain this concept?
ANSWER:
What is the effect of applying a constant force F to a mass m?
Newton's second law tells you that you cause the particle to accelerate,
that is if you exert the force over some time t the velocity will
change from v_{1} to v_{2}. Writing Newton's
second law, F=ma=m(v_{2}v_{1})/t.
Rearranging this equation, Ft=m(v_{2}v_{1});
so you can see that the amount which the product mv (which is called
the linear momentum p) changes is numerically equal to the product
Ft which is called the linear impulse J. If the force is
constant, the time need not be short; t could be an hour and J
would still tell you the change in p. I hope you can appriciate that
J=p_{2}p_{1} is simply an alternative
way to write Newton's second law. If F is not constant, you can still
calculate J but it is just harder. In essense what you do is
calculate Ft for many vanishingly small values of t all along
the path the particle takes and sum them to get the total J. (If you
know integral calculus, this is just integrating Fdt over the
time interval.) Of course, Newton's second law is a vector equation, so J
and p are vector quantities, J=p_{2}p_{1}.
If you have a system of particles which interact only with each other, the
net force on the whole system is zero because of Newton's third law, so
J=0 and p does not change as the system moves
around; this is called momentum conservation. Here p means the
momentum of the whole system, the vector sum of all the momenta of all the
particles.
QUESTION:
I've started reading an elementary particle textbook and it doesn't explain why a moving charged particle radiates an EM field. I was hoping you could help with that.
ANSWER:
The semantics of "radiates an EM field" is somewhat ambiguous. A charged
particle at rest creates a static electric field. A charged particle
moving with constant velocity (constant speed in a straight line) creates
both an electric field and a magnetic field and both change with time. But
neither of these situations are said to radiate electromagnetic
fields. Electromagnetic radiation propogates through space as waves; visible
light, for instance, is an electromagnetic wave. Electromagnetic fields can
be created when a charged particle accelerates. E.g., an antenna radiates
radio waves when electrons in the antenna are made to oscillate back and
forth (accelerating). The derivation of how accelerating charges radiate is
a topic in an intermediate E&M course and beyond the scope of this site.
QUESTION:
In the famous book by H.G Wells  "The Invisible Man"  the invisible man explains that his invisibility is due to light getting passed through a membrane (his skin/tissue) clearly , i.e not reflecting or refracting. Is this true? Can a object me made invisible by this method? Further, has there been some actual research in this topic and any advancements?
ANSWER:
First, it is not "true" —this
is a book of fiction! More pertinently, is it possible? I would say, given
what we know today, it is not possible because for the light to not be
refracted the speed of light in the body would have to be the speed of light
in a vacuum, and there is no material for which this is true. Research
regarding invisibility is focused on "invisibility
cloaks".
QUESTION:
A charged parcticle when accelerated radiates electromagnetic radiation or light.
All atoms on earth are accelerated therefore the electrons and protons which are charged are also accelerated so why isnt light radiated?
ANSWER:
First of all, atoms are neutral, have zero net charge. Second, the
accelerations involved because of the earth's motion are very small.
QUESTION:
Why does centripetal force do what it does?
It is a force so why does the object start moving in a circular path and not simply in the direction of the force?
Also it should provide acceleration...so even if some how the object move in circle why doesnt it spiral in??
ANSWER:
If an object is at rest and you exert a force on it, that force will not
cause the object to "start moving in a circular path". You cannot simply
label something a centripetal force. If an object is moving in a circle with
constant speed, it is accelerating because the velocity is constantly
changing. The velocity vector can change either by changing its magnitude
(speed) or its direction. An object moving in a circle therefore requires
some force to accelerate it, for example the sun's gravitational force on
the earth, which points toward the center of the circle. If you stopped the
earth in its orbit, the gravitational force would indeed cause the earth to
crash into the sun. You need to simply look up a derivation of centripetal
acceleration in any elementary physics text you would understand this
better.
FOLLOWUP QUESTION:
unhuh...
how can it change direction without changing its magnitude??
therefore in time dt
the velocity in downward direction becomes adt
where v' is equal to the vector sum ie v'=((v)^2+(adt)^2)^1/2,
so how is the velocity supposed to be constant when the object is not
decelerating tangentially?
ANSWER:
Like I said in my original answer, look this up in an elementary physics
text — I
am not a tutor. The reason your argument is invalid is that dt is
infinitesimal so if adt is perpendicular to the
velocity vector v, the result is to change direction. (Limit(as
dt —> 0){ √[v^{2} +(adt)^{2}]}=v.) Just use
your common sense: a car going in a circular path has a constant magnitude
of velocity, right? And nonconstant direction or velocity, right?
QUESTION:
Two bodies A and B are at rest and in contact with each other. Now if some arrangements made body A exerts a pressure of 10 NEWTON on body B then according to NEWTON'S 3rd law of motion body B will also exert an equal force of 10 NEWTON in opposite direction so the resultant force should be zero, while practically we see that there will be a resultant force of 10 NEWTON acting on body b and if mass and other conditions of body B are such that it moves by applying 10 NEWTON force on it. Then it will start moving. HOW?
ANSWER:
You have this all confused. But, you are not alone! When doing this kind of
problem, you must choose a body to focus on; only forces on that body
affect its motion. So, if you are interested in what body B will do, you
look only at that body. The force which B exerts on A is not a force on B.
So, if B has a mass of, say, 2 kg and there are no other forces on B, it
will have an acceleration of a_{B}=10/2=5 m/s^{2}.
Since the 10 N force continues, A and B remain in contact, so a_{A}
must also be 5 m/s^{2}. One force on A is the 10 N force from B
which is in the opposite direction as the acceleration. But the net force on
A must surely be in the direction of the acceleration, so there must be some
other force (probably due to you pushing on A) on A which is bigger than 10
N and points in the direction of the acceleration. For example, if the mass
of A is 4 kg, F10=4x5=20 N, so F=30 N. Finally, you could
look at A and B together as the body. In that case the forces they exert on
each other do cancel, the acceleration is 5 m/s^{2}, and the total
mass is 6 N; therefore, there must be some external force (you again)
causing this 6 kg to have an acceleration of 5 m/s^{2}: F=ma=6x5=30
N. All three ways of looking at this problem are consistent with each other.
QUESTION:
Can I make a simple bike reflector using 20 thru 40 degree angles?
Everything I have read uses 45 and 90 degree angles.
ANSWER:
I do not know how 45^{0} would work. The point is for the incident
light to be reflected in exactly the opposite direction from the direction
it came in. The only way I know to achieve this is with
corner reflector
which is three mirrors connected as in the inside corner of a cube.
QUESTION:
Why simple harmonic motion is called simple?
ANSWER:
Take a look at the figure above. The red curves are all pure sine waves, just
having different frequencies. These are all called simple harmonic motion
because they may be expressed as a single sine wave. Now, take all four of
these and add them together. The result is the black curve. This curve is
still harmonic (which means that it is periodic, it repeats itself after
some time, called the period, has elapsed) but it is not "simple" because to
describe it you must use four different sine waves.
QUESTION:
I am a Martial Arts instructor who also happens to have a BS in mathematics (well, statistics, sorta like math). I like to explain to my students the science of the art in order to help ground them in reality from a lot of the BS involved in Eastern Mysticism, such as board breaking.
My question is this, wooden weapons, such as nunchaku are usually either round/cylindrical or octagonal. A common technique is to place the weapon lengthwise down the arm and block an incoming attack with the wood. Traditional tonfa are octagonal, but modern police batons (with the side handle) are round. Which would protect the blocking arm from a baseball bat strike (for example) to a greater degree? Would an octagonal shape disperse the incoming force more than a round shape? I believe the octagon would spread the applied energy to a greater degree than a cylindrical/round shape and allow less energy to travel through the wood and into the arm, but I cannot find the math to back this up.
ANSWER:
The only reason that I would say that the octagonal shape would be better is
that it has a larger area of contact with the arm and therefore would
distribute the applied force over a larger area. For example, you would not
just want a point in contact with the arm since all the force would be
applied at a single place. Even better in this regard would be a rectangular
shape. This is probably a negligibly small effect, though. I cannot see how
the shape would have any effect as you suggest to " … allow
less energy to travel through the wood …"
QUESTION:
How much energy is required to move the electron sufficiently far away from the proton such that it does not experience the proton's electric field. I am kind of confused by this question. The effect of electric field will never end as far as we take the electron from the proton. And if we use the relation V=E.L then if E=0, then
L will be infinity?
ANSWER:
Technically, yes the field will be exactly zero only at L= ∞.
But,
that is only if your proton and your electron are the only things in the
universe! You can calculate the ionization energy, the energy necessary to
move an electron from the ground state of hydrogen to infinity; it is 13.6
eV=2.18x10^{18} J. You can also calculate the energy necessary to
move it to any other distance. For example, the energy to move it to about
0.5 μ=5x10^{7} m (which is about 100 times the radius of the
hydrogen atom) is 13.599 eV, essentially the same as to infinity. So you can
see that most of the energy is supplied in close to the proton because that
is where the field (and therefore the force) is strongest.
QUESTION:
So I've been trying to figure this for a while, if a positively charged and a negatively charged quark orbit each other or actually come in contact, also could they form a functional "atom" and be able to have electrons interact with them?
ANSWER:
Twoquark particles are called mesons (one quark plus one antiquark). If you
had a positively and negativelycharged quark you would have an uncharged
meson and so it would not bind an electron and you could not have a
meson+electron atom. However, if you had a positively charged pion, it could
bind an electron. For example, a positive pion
π^{+} is
composed of an up quark (charge +2e/3) and a down antiquark (charge +e/3)
and would have a charge +e and be able to form an "atom" which you might
call pionium.
ADDED
NOTE:
Whoops! Pionium is an "atom" composed of one
π^{+} and one
π^{}. I cannot find any reference to this pion+electron
system, so I guess you can call it whatever you like.
QUESTION:
i love physics and this question i asked to my teacher and principal but they couldn't answer it so my question is about third law of motion "every reaction has an equal and opposite reaction" so when a truck moving with constant speed hits a stationary car so according to newton's 3rd law of motion truck shoud be stopped after collision because car applies equal force on truck which it have during collision.
ANSWER:
Why should it be stopped? Certainly the truck experiences the force which
the car exerts on it, but every force does not have the effect of stopping
the object which experiences a force. If you are running toward a fly which
is hovering at rest in your path, you feel the force of the fly but it does not
stop you.
QUESTION:
Hi ...here is a link to Felix Baumgartner's freefall jump from space 
https://www.youtube.com/watch?v=vvbNcWe0A0
135,890 feet  or, 41.42 km (25.74 mi)
In the video you see speed at which he is supposedly travelling, 729 mph (1173km/hr), 46 seconds after he jumped. The footage is cut then seconds later speed strangely falls to 629 mph...anyway after 4 mins and 18 seconds of free falling he releases parachute. My questions are What speed would he be going at a 4 mins and 18 seconds into freefall? Also how do you explain the deacceleration when it was at 729 mph then down to 629 mph?
ANSWER:
(You might be interested in an
earlier answer.) There is a better video at
https://www.youtube.com/watch?v=raiFrxbHxV0; this video shows data
acquired by instruments on Baumgartner. Here some clips from that video:
The first three are the times also showing speeds; these are roughly the speeds you refer to in your question. The second three show speeds and altitudes. The third graph shows the whole history of altitude, speed, and mach speed. The speeds are all larger than indicated in your video. The speed at 4:20, 113 mph, answers your first question. The speed at 0:50 is about the maximum, 847 mph or mach1.25. The speed at 1:01 has dropped to 732 (about the same change as your video); I suspect this deceleration is due to two factors: there is getting to be much more air resulting in higher drag and, probably after he broke the record he oriented his body to increase drag and slow down. The official maximum speed after everything was calibrated was 843.6 mph.
QUESTION:
Will the drag coefficient be the same in air and water?
ANSWER:
There is no simple answer to this question. The drag coefficient C_{D
}is a constant characterized by the geometry of an object used to
calculate the drag force F_{D} if it is proportional to the
square of the velocity v: F_{D}= Ѕ C_{D} ρAv ^{ 2 }
where A is the area
the object presents to the fluid flow and
ρ
is the density of the fluid.
Whether or not this equation is true depends on a quantity called the
Reynolds number Re=L ρv / η
where L is a length along the direction of flow and
η
is the viscosity of the fluid. It turns out that only if Re>1000 is
the velocity dependence approximately quadratic. If Re<1 the drag
force is approximately proportional to v. Anywhere between these
extremes the drag force is a combination,
F_{D} ≈Ѕ C_{D} ρAv ^{2}+kv.
C_{D} ≈constant
only for the case Re>1000; the drag coefficient would then be the
same for identically shaped objects. It is interesting, though, to get a
feeling for what the relative speeds in air and water corresponding to Re=1000
are. The necessary data at room
temperature are
ρ _{air}≈1
kg/m^{3},
ρ _{water } ≈1000
kg/m^{3},
η _{air}≈1.8x10^{5}
Pa�s, and
η _{water}≈1.0x10^{3}
Pa�s. Then v_{water}/v_{air}≈(1.0x10^{3}/1000)/(1.8x10^{5}/1)=0.056
(which is true for any Re). Taking a sphere of diameter 0.1 m as a
specific example for the critical Re=1000, C_{D}=0.47,
A=7.9x10^{3} m^{2}, L=0.1 m, v_{air}=1000x1.8x10^{5}/(0.1x1)=0.18
m/s and v_{water}=1000x1.0x10^{3}/(0.1x1000)=0.01
m/s. Since the speeds are relatively low, you can conclude that for many
cases of interest the quadratic drag equation holds for both water and air.
Therefore, for many situations you can approximate that the drag coefficient
in air and water are about the same. For a specific case you should check
the Reynolds numbers to be sure that they are >1000. The bottom line is that
if the drag force depends quadratically on velocity, the drag coefficient
approximately depends only on geometry, not the properties of the fluid.
Keep in mind that all calculations of fluid drag are only approximations.
ADDED
COMMENT:
As I said above, for 1<Re<1000, F_{D} ≈av+bv^{2
}where a and b depend on Re; e.g., a≈0 for
Re>1000 and b≈0 for Re<1. So, for Re<1, a
is just a constant and F_{D}≈av≡Ѕ C_{D} ρAv ^{2}
so C_{D}=2a/( ρAv )=2aL/(A ηRe )≡c/Re
where c is a constant. If F_{D} is proportional to
v, C_{D }is inversely proportional to Re; this is
called Stokes' law. As we saw above, Re>1000 leads to C_{D}=constant.
Ferguson and Church have derived an analytical expression which very
neatly reproduces data for the transition from Stokes' law (1/Re) to
constant C_{D}. Calculations for a golf ball are shown at the
left (constant C_{D}, the green line, is called turbulent in
the legend). Note, as we have stated, that the transition occurs in the
region
1<Re<1000. Note the sudden drop in the data around Re=500,000.
I believe that this must be due to the
dimples on the golf ball which
reduce drag.
QUESTION:
In induced magnets, why
does the end nearer the magnet have the opposite polarity to that of the
magnet?
ANSWER:
Think about a piece of iron. It is like a whole bunch of tiny bar magnets,
each of which has a N and S pole; but the iron is usually not a magnet
macroscopically because all the tiny magnets are oriented in random
directions. But, if you bring the north pole of a permanent magnet up close
to the iron, the south poles of all the tiny magnets turn to point toward
the north pole of the permanent magnet.
QUESTION:
I think if an object is turning, it has more gravity than an object which is not turning
ANSWER:
You are right. However, it really has nothing to do with the turning, per
se. When something is turning it has rotational kinetic energy and
therefore a spinning planet has more energy than an otherwise identical
planet not spinning. In general relativity you usually hear about gravity
being caused by mass warping spacetime. However, mass is just the most
obvious source of gravity and what it is which really warps spacetime
is energy density and mass has a lot of energy (E=mc^{2}). The energy due to the turning
is infinitesmal compared to the mass energy of the object so you would never
be able to distinguish the difference due to the turning by looking at the
gravity.
QUESTION:
I've heard that a bullet falls to the ground
at the same speed [in the same time] no matter the charge pushing it forward. More powder and it moves faster but vertical speed is the same. I imagine two equal weight gliders of 20/1 and 30/1 lift/drag ratios would also express the same vertical speed?
My question is, wouldn't this gravity effect also hold true underwater? An underwater glider expressing 30/1 L/D would travel 50% further, faster than a 20/1 L/D in a given vertical?
ANSWER:
I will assume that we are talking about horizontally fired bullets. All
bullets do not hit the ground with the same speed, rather they all take
(approximately) the same time to get to the ground; I have corrected your
question because I believe that is what you meant. Your comparison to
gliders with different lift/drag ratios is not appropriate, because the
thing which results in the times of vertical fall being the same is the
assumption that there is no lift at all. In other words, the vertical speed
of a bullet will always be much less than the terminal velocity so there is
no appreciable component of the drag in the vertical direction. (This is not
true if the gun is fired from a very high altitude since there will be
enough time for the verticle component of the velocity to approach the
terminal velocity.) Underwater (I am just talking about bullets, not
gliders) the bullet will not behave the same because the terminal velocity
is much smaller than in air, in other words the drag is much larger.
QUESTION:
Quantum mechanics has popularized the idea of very small particles existing in two or more points in space during the same instant in time. Is this to be taken at face value as a fundamental aspect of quantum particles, or is this simultaneous existence just a concept used to express our inability to observe these particles without inadvertently altering their speed or location?
ANSWER:
See an earlier answer.
QUESTION:
I read from my history book that when nukes were first created we didn't know that the radiation their explosions produce is harmful to one's health, even to one's life maybe. So my question is, how come neither physicists or medical personnel ever figured out that radiation can be dangerous until we had first victims of radiation?
ANSWER:
Ever hear the expression "hindsight
is 2020"? It is unrealistic to expect that when something new is
discovered that we should somehow know all the effects that might have on
anything else. At the time radioactivity was discovered, nobody even knew
what atoms were composed of or what their structure was. And it was found to
be very tiny bits of matter (e.g. what we know as electrons today)
and who would have thought that getting hit by something trillions of times
lighter than a speck of dust could be harmful? It took experience before it
was appreciated how dangerous it could be. One of the best known examples of
such experience is the case of
radium watch dials.
Radium, mixed with a phosphor, was painted by women onto watch dials and it
would glow in the dark. The workers were encouraged to point their brushes
with their lips to make the fine lines required. Subsequently many became
ill with cancer and other radiation sickness. It seems stupid now, but the
dangers were not known until they were discovered. Also many of the health
effects were longterm effects, the effects not appearing until years or
decades after exposure. One example I know about from personal experience is
the effect of radiation treatment for acne which was popular in the 1950s. I
very much wanted to have this done but my father forbade it, making me
furious with him. Years later people who had had this treatment started
coming down with cancer —thanks,
Dad! Yet another example is the shoefitting xray machine. When I was a kid
it was really fun to buy new shoes because you could look down at an xray
of all the bones in your foot and how well the shoe fitted you; again, when
more was learned about radiation, these machines were all sent to the scrap
heap. When it began to be appreciated that there were dangers, studies began
to be done to try to set allowable dosage levels. But, it is unethical to do
such experiments on people, so lab animals had to be used which always
presents problems with scaling and other variables. Much of what we know
today was learned afterthefact by doing longterm statistical analyses
over decades of medical records.
QUESTION:
Magnetic flux according to my book is total no. of magnetic field lines passing through a given area in magnetic field. ok but why there are not infinite no. of magnetic field lines, because magnetic field line are defined as the path that a magnetic north monopole would take if left in north part of magnet, so if i take a monopole and leave one atom away from the previous position then it should take a slightly different path and that path should be considered as magnetic field line,so in this way i can draw millions of line.
ANSWER:
Magnetic flux is well defined:
Φ_{M}≡∫∫B�dA,
the area integral of the magnetic field. If the integral is over a closed
surface, the flux is zero; this is the famous situation which tells you that
there are no point sources (called magnetic monopoles) of magnetic fields
like there are for electric fields. If you like, you may interpret this as a
number, but that is not really fundamental. If you talk about uniform
magnetic fields which are perpendicular to a plane surface, Φ_{M}=BA
, is the flux through an area A. Then, if you interpret
Φ_{M}
as a number, you would simply say, for example, there were 10 lines through
an area of 1 m^{2} if the magnetic field is 10 T. You could ask the
same question about electric flux,
Φ_{E}≡∫∫E�dA .
This is perhaps a little easier to understand because you can have point
charges. For example, if you have a 1 C point charge, the electric field 1 m
from it is E=Q/(4 πε_{0}r^{2})=1/(4x3.14x8.85x10^{12}x1^{2})=9.27x10^{9}
N/C. The flux at a distance of 1 m from the charge would therefore be Φ_{E} =EA= 9.27x10^{9}x4x3.14x 1^{2}=1.17x10^{11}
Nm^{2}/C. Incidentally, this is the flux regardless of where you
measure it because the area of the sphere
(4 πr^{2})
surrounding the point charge appears both in the denominator of the field
and in the numerator of the flux and cancels. Hence, you could say that
there were 1.17x10^{11} lines of
electric field emanating from a 1 C point charge. You can see this from
Gauss's law,
Φ_{E}=∫∫E�dA=Q/ε_{0}
if the integration is over a closed surface enclosing the charge Q.
QUESTION:
Why degree of monochromaticity is always non zero?
ANSWER:
The practical reason is that you cannot make a laser which is perfectly free
from electronic instabilities and noise. But even if you were able to make a
perfect laser, you could not know its frequency perfectly. The more
fundamental reason is that monochromatic light means light with a single
frequency. But, the linear momentum of the photons is proportional to the
frequency of the light. Because of the Heisenberg uncertainty principle, the
only way you can know momentum of the particle perfectly is to be completely
ignorant of its position; you need an infinitely long wave to know its
frequency perfectly.
QUESTION:
This question is about relativity. If person leaves earth and heads toward a star that is 10 light years away, and he travels at near the speed of light, time will slow down for him. The round trip to the star may take him only one year, but when he returns, maybe 100 years have passed on earth. When he started out to the star, he knew it was
50 [I think you meant 10,
right? The Physicist] light years away. It seems that for him, he will have traveled 20 light years in one year. That is faster than the speed of light. What don't I understand?
ANSWER:
First of all, time for the traveler will not slow down; rather, her clock
will run slower as observed by an observer on the earth. To her, time will
run perfectly normally. However, she will observe the distance to the
destination shortened because of length contraction. You have rounded things
to what they would be if she were going the speed of light, but let's do the
whole problem as if she were going with a speed v=0.999c. The
earth observer would see an elapsed time of 100x0.999=99.9 years (your 100
years). The traveler will see the distance shrunk to D'= 10 √(1.999^{2})=0.447
light years, so the time of travel is t=2x0.447/0.999=0.895 years
(your 1 year). You should read the earlier answer on the
twin
paradox.
QUESTION:
Supposing a weightless container is filled with water. I am sure the pressure at the bottom of liquid, P1 = atmospheric pressure + height of liquid x density of liquid x g = Patm + hdg, where Patm is atmospheric pressure and d is density of liquid. We can calculated this pressure as if liquid in region A and Region C does not exist.
But how about the the pressure at the base of the container, that is P2. Is P2 same as P1? For P2, do we need to consider the whole weight of the liquid, that is inclusive of the weight of water in region A and C?
ANSWER:
It depends on what the force on the bottom is. If the container is in
equilibrium, imagine it sitting on a table. The table would exert an upward
force equal to the weight W of all the water, so P_{2}
would be W/A_{bottom} where A_{bottom}
is the area of the bottom of the container. This assumes that atmospheric
pressure is the same everywhere in the vicinity of the container; in other
words, I have ignored the buoyant force due to the air on the whole
container because it will surely be much smaller than W.
FOLLOWUP
QUESTION:
Indeed the container is resting on the table. For P1, I use h x d x g. But for P2 you use W / Abottom, So can I say P1 not equal to P2 ?
ANSWER:
Yes, but I have to admit that my answer was misleading in that I gave you
the gauge pressure, the pressure above atmospheric. So I should have said
that P_{2}=P_{atm}+W/A_{bottom}.
There is no problem that P_{2} ≠ P_{1}
because the force which the container exerts on the table is not P_{1}A_{bottom}.
Think about it —the
sides of your container exert a downward force on the bottom of the
container.
QUESTION:
Think of words (incapital letters) that can be read properly both with a mirror and without a mirror.
What are these words?
ANSWER:
Well, this is not physics and I usually do not answer such questions.
However, it is a cute puzzle. First, any letter in the word which is
symmetric upon reflection about a vertical axis will look the same in the
mirror: A, H, I, M, O, T, U, V, W, X, Y. Second, any such word must be a
palindrome composed of the allowed letters, e.g. AHA, OTTO, WOW,
YAY, MOM, TIT, TUT, … .
But, wait, it is somewhat more difficult than that! Look at the pictures of
the Camel cigarette package and its reflection, in particular, the word
CHOICE on the side. So words composed of letters with symmetry upon
reflection about a horizontal axis will also reflect the same: B, C, D, E,
H, I, K, O, X will have a reflection which is the same as the original, but
only if turned upside down. There is no restriction for these words to be
palindromes; besides CHOICE, some others would be BED, HEX, BOX, BIKE, HIKE,
CHIDE, …
QUESTION:
What is meant by virtual image?
ANSWER:
It is an image formed by light which appears to be coming from
somewhere but is not actually coming from there. When you look into a
mirror, it appears that the light from the image of what you are looking at
is coming from behind the mirror, but it is actually coming from the mirror
itself.
QUESTION:
Supposing that initial
velocity is 0 Is a displacement vs. time (squared) equivalent to a Velocity
vs. time graph? I tried to find the slope of displacementtime^2: d/t^2 =
V/t which equals acceleration But what I read everywhere is that: d/t^2 =
a/2 according to the equation d= (initial V)(time)+ (1/2)(at^2) I don't what
is wrong with my calculations?
ANSWER:
If you plot d vs. t^{2}, the slope of the line
will be Ѕa.
You know that d=Ѕat^{2}, so suppose that I call t^{2}=u.
Then clearly the slope of the line d=Ѕau is Ѕa. Your
equation d/t^{2}=V/t is clearly
wrong: d/t^{2}= Ѕat^{2} /t^{2}= Ѕa
and Ѕa≠v/t.
QUESTION:
Two of us disagree on part of a sol'n given by two people with Physics background, and I want to know if I am correct, or if I am missing something in the analysis of the problem...in case I have to explain it to a student.
Question concerning Forces/impulse..... 50kg person falling @15m/s is caught by superhero , and final velocity up is 10m/s.
Find change in velocity.
Find change in momentum .
It takes 0.1 sec to catch them.....ave Force is ?
answers are: vel = 25 m/s change in mom.. 1250 kg*m/s, and ave Force = 12,500 N.
Here's where we disagree: Person B says that 12,500 N is equiv. to 25 g ?????
They try to explain that 250 m/s^2 accel. corresponds to 25g.... I said it makes No sense at all, [ I know the accel. is 250, but that doesn't in any way imply a 25 g "equivalence" to me ]. They then went further to "prove" their point........Here is their argument...
500 N/g = 12500N / ( )g ..... I agree the ( ) = 25, but say there is No justification for the 500 N / g in the first place...... any ideas where it comes from , or how to justify that value ?
BTW I teach physics on and off at the HS level.... person B is an Engineer , I think
ANSWER:
Person B is wrong but has the right idea. (As you and your friend have
apparently done, I will approximate g ≈10
m/s^{2}.)
We can agree that the acceleration is a=250 m/s^{2} and that
is undoubtedly 25g. Now, we need to write Newton's second law for the
person, mg+F=ma=500+F=12,500, so F=13,000 N. This is
the average force by the superhero on the person as she is stopped, so the answer that the
average force is 12,500 N is wrong. When one expresses a force as "gs of
force", this is a comparison of the force F to the weight of the
object mg, F(in gs)=F(in N)/mg=13,000/500=26
gs; this simply means that the force on the object is 26 times the
object's weight. So neither of you is completely right, but if there is any
money riding on this, your friend should be the winner because the only
error he made was to forget about the contribution of the weight to the
calculation of the force. I am hoping that superman knows enough physics to
make the time be at least 0.3 s so that Lois does not get badly hurt!
QUESTION:
It is said gravitons are the
expected particles to exchange the gravitational force. Do these particles
exist in real?
ANSWER:
There is no successful theory of quantum gravity, so gravitons are "expected
particles" but I would not call them "real" at this stage; hypothetical
would be a better word.
QUESTION:
Why does it take more time to go to
Dubai from Mumbai than to come from Dubai to Mumbai?
ANSWER:
There are winds at high altitudes called
jet streams. The
prevailing direction of these winds is easterly and their speeds are as
large as 100 mph. Therefore when you travel east (as from Dubai to Mumbai)
you have a tail wind, and when you travel west (as from Mumbai to Dubai) you
have a head wind. Since the greatest speed an airplane can fly, say 600 mph
for commercial jets, is airspeed, that is relative to the air, the ground
speed with a 100 mph wind would be 700 mph going east and 500 mph going
west.
QUESTION:
Why is the direction of angular displacement along the rotational
axes?
ANSWER:
There are two ways in which you can rotate something about some axis, either
clockwise or counterclockwise (as viewed from one side or the other). So, an
angular displacement has two possible directions only. It is like a
displacement in one dimension, like a bead on a long straight wire. You then
call one direction on the wire plus and the opposite direction minus. The
only way to get a similar directional assignment for a rotation is to note
that the rotation axis is just like a onedimensional axis, one way being
assigned to be positive, the other negative. Usually this is done with the
righthand rule such that counterclockwise is a positive vector on the
axis.
QUESTION:
Recently I came across such an article that Gravity isn’t actually a Force, It's the bending of spacetime that cause objects to move on curve paths.
If two objects of equal shape and masses are there they will bend the spacetime in same way by same amount.
Then according to that there shouldn't be any kind of force or something acting between them which isn't true i guess. Gravity forces will act b/w those two. Could you please
explain me this contradiction.
ANSWER:
I do not understand your logic that because they both have the same effect
on spacetime, they will feel no force. Although it is just a cartoon to help
you qualitatively understand warping of spacetime, in this case it is
illustrative. If you place a bowling ball on a tranpoline, what happens? The
surface of the trampoline is warped. Now, suppose that you place two bowling
balls a few inches apart from each other. What happens when you let go of
them? They will behave as if there were a force between them and roll toward
each other but what they are really doing is responding to the shape of the
trampoline.
QUESTION:
I have a question about gravity, or more specifically how to calculate it for the purpose of a scifi fanfiction I am making.
So Earth has the density of 5.5 grams per cubic centimeter, total mass of 5.97219Ч10^{24} kilograms and radius of 6371 kilometers. So how do I put these together in a formula to get the 9.81m/s gravity?
ANSWER:
The density is irrelevant. You need to use Newton's universal law of
gravitation for two point masses. The force F which each feels is
F=GMm/r^{2} where G is the universal gravitational
constant, G=6.67x10^{11} N � m^{2}/kg^{2},
M and m are the masses of the two masses (take M as the
mass of the earth), and r is the separation between M and m.
This also works if the masses are spherically symmetric. But, of course, you
also know in your case that F=mg. So, if you solve for g, g=MG/r^{2} ≈ 6.7x10^{11}x6.0x10^{24}/(6.4x10^{6} )^{2}=9.8
m/s^{2}.
QUESTION:
Why do electrons fall back to the ground state after jumping into a higher orbital?
ANSWER:
The simple answer is that a system will always seek a way to move to a lower
energy, like a ball rolling down a hill and not up it or sitting still. The
more complicated answer addresses whether there is, indeed, "a way" to
achieve this. For atoms, the radiation of photons in the transition is
usually electric dipole radiation and there is a corresponding
quantummechanical operator, let's call it O_{E1}. Then, you
have to look at what is called the transition matrix element for this
operator, < Ψ O_{E1} Ψ'>=∫Ψ� O_{E1} �Ψ'dτ
which is an integral over all space and Ψ' and Ψ are the
excited and ground states, respectively. If this is nonzero, then the decay
will decay with some half life determined by the value of the matrix
element. If it is zero, the state will still probably decay to the ground
state but via a different kind of transition.
QUESTION:
We all know about second law of motion.
The example is about the coin which is placed above the cardboard and cardboard is placed above the beaker... if the cardboard is quickly pulled then the coin falls in the beaker but if the cardboard is slowly pulled then why does the coin come with the cardboard?
Why does this happen?
ANSWER:
This is usually used to demonstrate inertia, Newton's first law; since you
are interested in why it does not work if you move the cardboard slowly, the
second law also must be used. To move the cardboard from rest it must
experience some acceleration. If the coin is to experience an acceleration,
that is move with the cardboard, some force must be exerted on it. The only
force on the coin horizontally is the static friction and the greatest it
can be is μmg
(where μ is the coefficient of static friction) so the greatest
acceleration the coin could have is μg; for example, if μ=0.3,
and you caused the acceleration of the cardboard to be any greater than
about 3 m/s^{2}, the cardboard would slip under the coin. But, if
you gave the cardboard an acceleration of 0.2 m/s^{2} the coin would
move with it.
QUESTION:
How can it be said that the universe is only 13.7 billion years old? This seems ridiculous to me that anyone can make this assumption. Taking in the fact that there are black holes almost as old as the universe itself how is this possible? Stars live billions of years so the first group of black holes would be billions of years after the "big bang". Also due to the fact that the universe is expanding way faster and not slowing down like predicted it would seem to me the current model is wildly inaccurate. Can you please explain this to me? My understanding is that stars form and die out all the time so to be able to accurately determine the age of the universe another method would need to be devised. It is my understanding that galaxies form with the help of super massive black holes.(Which aren't even holes at all but super compacted spheres of matter so dense light cannot escape created by countless other bodies of mass.) Which attracts matter into tight clusters of stars we call galaxies. These galaxies would have to take billions of years to form after the creation of the first stars ever created and after the first black holes were created from some of those stars. Does the current model explain this?
ANSWER:
As clearly stated on the site, I do not do astronomy/astrophysics/
cosmology.
I can at least say a little about your question, though. The age of the universe is not an "assumption", it is based on careful analyses of many measurements. The stars in the
early universe were much more massive than the sun; this
resulted from the fact that the early universe was essentially all hydrogen and a bit of helium.
Stars began forming only 100200 million years after the big bang. Very large stars burn much hotter than smaller stars and therefore have much shorter lifetimes, a few million years rather than
a few billion years for less massive stars (e.g. the sun with an
expected lifetime of about 10 billion years) which would explain why many
black holes are very old. The rate of expansion of the universe is not
inconsistent with the model of a big bang about 14 billion years ago (in
fact, the rate is one of the measurements used to estimate the age of the
universe) even though we do not understand all the details (e.g. dark
energy and accelerating expansion).
QUESTION:
I would like to ask this question because I do not agree to this but it is written in my school handout. It says that work needs only two things to be calculated which is weight and height of displacement.
Now , there is this question that says that two persons who have the same mass went to the top floor of eiffel tower. One used the stairs and the other one used an elevator. The thing is that our handouts states that they both exerted the same amount of work done because they have the same weight and they both have the same height distance covered.
I do not agree simply because the man using the stairs carried his own weight up the distance while the the other one was being carried by an elevator and I immediately thought that the elevator was the one doing the work and not the person which is why they cannot have the same work done.
If I am wrong then I would humbly accept it. However , if our physics teacher is wrong, I would need a legit source stating about the problem and saying that it was wrong or similar problem like the one given with an answer so I can correct my grade. I would really appreciate anyone's help right now.
ANSWER:
The point is that the net work done on each person was the same.
Where the energy came from is different. On the stairs, the energy is
supplied by what the guy ate for lunch. In the elevator, the energy is
supplied by the motor lifting the elevator. Regardless where it came from,
mgh of energy must be somehow supplied.
QUESTION:
How surface tension overcome insect weight and help it to stay on liquid surface, as surface tension is along liquid surface?
ANSWER:
The weight of the insect causes the water to be depressed where the legs
touch the surface, so the surface tension has a vertical component.
QUESTION:
Can we make something like antigravity on Earth, or something similar to that? And i was wondering if we could make something like antigravity with magnets, to actually reduce the weight of some object. I was thinking to make a chamber with a magnet at the bottom, and take magnetized objects and put them on the top, so it will levitate.
ANSWER:
Your device will not work. The magnet on the bottom can be made to exert a
repulsive force on a magnet at the top which tends to lift your chamber.
However, Newton's third law requires that the top magnet exerts an equal and
opposite force on the bottom magnet, so the net force on the chamber is
zero.
QUESTION:
Is rotation a reason for earth being round?
ANSWER:
No. The reason that all large astronomical objects are nearly spherical is
because the force which holds them together is spherically symmetric. If you
have a point mass, the gravitational force another mass feels depends only
on how far apart they are, not on the direction. If you want a lot more
detail, see an earlier
answer about
gravity of a cylinder which shows how the tendency is toward a spherical
shape.
QUESTION:
Using a spring balance, weigh (separately) a piece of wood and a container with water in it. Then weigh them again, but with the wood floating in the water. Does the reading on the balance change? There will be an upthrust on the wood (Archimedes' Principle), causing it to weigh less. I guess there will be an equal downforce exerted by the wood, resulting in the same weight being shown in both cases. This seems a fairly easy question, but I can't find the answer clearly stated anywhere.
ANSWER:
The measured weight will be the sum of the individual weights. The buoyant
force is the force which the water exerts on the wood; Newton's third law
requires that the wood exert an equal and opposite force on the water. As
you correctly "guess", these two forces cancel out.
QUESTION:
Good afternoon. I started learning quantum mechanics and I have a question about Heinsenberg's principle. I was thinking about a photon created by an electronpozitron annihilation. As I understand (and I hope I am not wrong here) all observers, no matter their state, can specify the point (and all will specify the same point) where the photon appears, and I think in a Wilson chamber we can do this thing. Anyway, we can specify its position with an error less than infinity. But at the same time we know it's speed with absolute precision, as, assuming we measure it in vacuum, it's speed is invariably c. So the product between (delta)x and (delta)p will be 0, as (delta)x is not infinity and (delta)p is 0. In other words we can predict it's position with a certain accuracy knowing at the same time it's speed (I am not sure, but i think that the (delta)p is not a vector but a scalar quantity). What is wrong about this as it seems the Heinsenbrg's principle is violated?
ANSWER:
The mass of a photon is zero but it has momentum. Therefore your notion that
linear momentum p is mass m times velocity v must be
wrong; the relativistically
correct expression for momentum is p=mv/ √[1(v/c)^{2}]
where c is the speed of light. For a photon, this is a little tricky
because p=0/0, but you can also write that E^{2}=p^{2}c^{2}+(mc^{2})^{2}
where E is the energy of the a particle of mass m, so p=E/c
if m=0; since E=hf for a photon, where h is Planck's
constant and f is the frequency, the linear momentum of a photon is
p=hf/c . Therefore the frequency of the photon must be
uncertain according to the uncertainty principle. Since the energy of a photon is
hf, there will be an uncertainty in the energy of the photon.
QUESTION:
The units of the Hall coefficient is m^3/C that's, cubic meter per coulomb. How so we get this?
ANSWER:
You just need to know the
definition of the
Hall coefficient R_{H}, R_{H} ≡V_{H}t/(IB)
where V_{H} is the Hall voltage, t is the thickness of
the sample, I is the current, and B is the magnetic field. The
Hall voltage may be
shown to be V_{H}=IB/(nte) where n
is the charge carrier density (units [m^{3}]) and e is the
charge (units [C]]) of a charge carrier. Therefore R_{H}=1/(ne)
so the units of R_{H} are m^{3}/C.