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During a discussion with my 6th grade class about the Law of Inertia and space travel, a student asked: "If a spacecraft leaves a solar system, for example Voyager 1, will its velocity increase due to the lack of gravity from the sun." Fairly certain the answer is no; however, it did spark a rather interesting debate. Can you explain?
Perhaps the key is to note that "leaves the solar system"
does not mean that there is no longer any gravity from the sun.
Rather there is a boundary where the solar wind, particles like protons and
electrons, stops; this is called the heliopause and has been definitely
observed by the Voyager 1 spacecraft as shown in the graph to the right.
This shows the amount of solar wind the Voyager detected as it moved during
the months of 2011-2. In September 2012 it dropped to near zero. This is
where we define the edge of the solar system to be and it is about 50
Astronomical Units (AU) from the sun; the earth is one AU from the sun. But
the gravity from the whole solar system is still present and the craft will
continue slowing down, but ever so slightly since as you get farther away
from a mass the gravity gets weaker. This craft has enough speed that, if it
never encountered any other mass it would keep going forever. I did a rough
calculation and found that the acceleration of Voyager is about a=-0.01
mi/hr2 which means that it loses about 1/100 mph per hour; but
the speed is about 40,000 mph, so I think we could agree that it is moving
with an almost constant speed. As it gets farther away, the acceleration
will get even smaller (physicists call slowing down negative acceleration,
not deceleration). Until it gets close to something else, like some other
star, it will keep going with an almost constant speed. When it does
approach another star it will begin speeding up. Your students should
appreciate that the only thing which can change the speed of something is a
force, a push or a pull.
I am building a cold frame to keep veggies alive in the winter. It will be 3' x 6' with two "doors" (called lights) each 3' x 3' that will be hinged to the frame. The frame probably weighs about 10 - 15 lbs. The doors will be quite lightweight, possibly only 3lbs each. I wanted to use magnets to keep the doors closed at night or when I am not venting the cold frame. We often get very windy days with 40mph wind speeds and gusts to 60 mph. From what I've read, magnets have different "pull force" properties. I'd like a way to figure out what pull force the magnet for each door needs to have to withstand the winds we get. Please don't tell me to just hook the doors closed -- shockingly, it appears magnets are a more cost effective solution.
where ρ is the density of the
fluid (ρair≈1 kg/m3), P is the pressure,
g is the acceleration due to gravity, and h is the height
relative to some chosen h=0. For your situation both surfaces are at
essentially the same height so PA=P+½ρv2
where the pressure inside your frame is atmospheric pressure (PA)
and the velocity inside is zero. So, PA-P=ΔP=½ρv2=½∙1∙26.82=359
N/m2=7.5 lb/ft2. This would be the pressure trying to
lift the door. So the total force on each door would be F=AΔP=9∙7.5=67.5
lb where A=3∙3 ft2=9 ft2 is the area of the
door. But, this is not the answer since we want to keep it from swinging
about the hinges, not lifting into the air. So, assuming that the force is
distributed uniformly over the whole area, you may take the whole force to
act in the middle, 1.5 ft from the hinges, so the torque which is exerted is
1.5∙67.5≈100 ft∙lb. But, the weight of the door also exerts a torque, but
opposite the torque due to the wind (the weight tries to hold it down)
-3x1.5=-4.5 ft∙lb. So, the net torque on the door about the hinges is about
95 ft∙lb. To hold the door closed, one needs to exert a torque equal and
opposite to this. To do this, it would be wisest to apply the force at the
edge opposite the hinges to get the maximum torque for the force. The
required force from your magnets would then be F=95/3=32 lb. Note
that this is just an estimate. Fluid dynamics in the real world can be very
complex. Also note that, if my calculations are anywhere close to correct,
you should probably be sure the whole thing is attached to the ground or the
side of your house since the total force on the whole thing would be
67.6+67.5-3-3-15=114 lb, enough to blow the whole thing away in a 60 mph
wind! Also, once the door just barely opens, the wind will get under it and
simply blow it up, Bernoulli no longer makes any difference.
I must say that I cannot believe that you could not buy a
couple of simple hooks/latches for under $5, but I will do a rough
calculation for you to estimate the force you would need to apply at the
edge of the doors to hold it down in a 60 mph=26.8 m/s wind. When a fluid
moves with some speed v across a surface, the pressure is lower than
if it were not moving; this is how an airplane wing works and why cigarette
smoke is drawn out the cracked window of a car. To estimate the effect,
Bernoulli's equation is used: ½ρv2+ρgh+P=
If a particle has a lifetime of say 100 years then what are the chances that that particle will decay in 1 year. How will one go about calculating particle decay probability?
N=exp(-0.01)=0.99. If you want to interpret that as a 1% probablility
that the particle has decayed, I guess that that would be ok. But, of
course, there is no such thing as 99% of a particle. Keep in mind, though,
that this is not linear; for example, if t=50=τ/2,
N=exp(-0.5)=0.61, not 0.5.
Lifetime has no meaning for a single particle. It is a
statistical concept as I will show. The rate at which a large ensemble of
particles decays is proportional to how many particles there are, dN(t)/dt=t/τ
where N(t) is the number of particles at some time t
and τ is the mean lifetime. The solution of this differential
equation is N(t)=N0exp(-t/τ)
where N0=N(t=0). So, when t=τ,
N=0.37N0, there are 37% of the original particles
left. For your question, you might want to ask what is the situation when
t=1 if N0=1 and τ
where does electron go that created by beta emission?
The electron is not "created" in a vacuum, a proton and a
neutrino are also "created" and a neutron "destroyed." The net effect is
that there is still the same amount of charge, zero, after the decay as
there was before. One possibility is that the electron never leaves the
source and, since the atom it left is now positively charged, the electrons
in the source all move around until all the atoms have the correct number of
electrons. Or, if it leaves the source, it will encounter other matter with
which it will interact, lose energy and either find a positive ion somewhere
to join or stick to a neutral atom which becomes a negative ion. Charge can
build up to some limit on anything.
Galileo was punished by the Church for teaching that the sun is stationary and the earth moves around it.His opponents held the view that the earth is stationary and the sun moves around it.If the absolute motion has no meaning, are the two viewpoints not equally correct or equally wrong? i.e, By the concept of relative motion can't we say both?if so, then why do we usually say that earth goes round the sun,and the other way round?
If the sun and earth were both just moving with constant
velocity and not interacting in any way, you would have a point. However,
because of their gravitational interaction and the fact that the mass of the
sun is hugely bigger than the mass of the earth, there is no way you can
sensibly argue that the sun orbits the earth. Each exerts a gravitational
force on the other (equal and opposite) but because the sun is so massive,
the force has almost no effect on it. If the earth were not orbiting but
simply released from rest, it would fall into the sun. There would be no
question which object had the greater acceleration—it would be obvious to
any observer that it was the smaller mass, the larger mass practically
unaffected. So, if any frames are accelerating, they are not equivalent to
those not accelerating. What you call "the
concept of relative motion" applies only to unaccelerated frames.
Two spherical bobs, one metallic and other of glass, of same size are allowed to fall freely from the same height above the ground. Which of the two would reach earlier and why?
Two objects with identical geometries (same size and shape)
experience identical air drag force while moving with a speed v.
While falling, the force has a greater effect on the less massive causing it
to slow down more. Therefore the mass, not whether glass or metal,
determines which gets to the ground first; the winner is whichever has more
mass. You can find links to many old answers about air drag on the
Suppose you have radiation detectors fixed on the ground on Earth. Will they detect radiation coming from a charged particle in free fall near them?
The first answer that comes to mind is: Yes, they will detect radiation because the particle is accelerated, and electrodynamics predicts that accelerated charges must radiate in this situation.
According to the Equivalence Principle, this situation is equivalent to detectors fixed on an accelerated rocket with acceleration g moving in the outer space and far away from the influence of other bodies. If the answer to the previous question is yes, then the detectors on the rocket should also detect radiation coming from a charge in free fall as observed by the reference frame of the rocket. But a charge in free fall in this reference frame is at rest in the inertial reference frame fixed with respect to the distant stars, and a charge at rest in an inertial frame should not radiate.
Is it possible that detectors fixed on the rocket detect radiation but detectors at rest in the inertial frame do not? Is radiation something not absolute, but relative to the reference frame?
This is a fascinating question and points to an experiment
which would seemingly violate the equivalence principle. The answer to your
first question is an unequivocal yes, an electric charge accelerating in
free fall in a gravitational field radiates electromagnetic waves, an
electric charge not accelerating does not radiate. But, suppose that you are
falling along with the charge; relative to you the charge is not
accelerating and therefore not radiating. Or, equivalently, suppose that you
are in a spaceship in empty space with your rockets turned on. If you
release an electric charge inside, it will "accelerate" toward the rear of
the ship and therefore radiate because the equivalence principle states that
there is no experiment you can perform which can distinguish between the
accelerating frame and a static gravitational field. However, the charge
will move with constant speed relative to an inertial observer nearby and
therefore not radiate. In both cases we have an electric charge both
radiating and not radiating, a seeming paradox. Although I had not heard of this
paradox before, apparently it has been a topic of many articles. The most recent of
Almeida and Saa,
has evidently laid the paradox to rest. They demonstrate in this article
that observers for whom the charge is not accelerating "…will not detect any radiation because the radiation ﬁeld is conﬁned to a spacetime region
beyond a horizon that they cannot access…" and "…the electromagnetic ﬁeld generated by a uniformly
accelerated charge is observed by a comoving observer as a purely electrostatic ﬁeld."
Like all "paradoxes" in relativity, there is not really a paradox; rather a
radiation field in one frame may be a static field in another. Basically,
you nailed it when you said "radiation
[is] something not absolute, but relative to the reference frame."
What would happen if a needle that weighed as much as the Earth and everything on it was placed on the surface of the planet? Where would the pin end up? What would be the impact on the Earth? Would the outcome be different if it was set on land or water?
This is a wacky question but I guess I can do a wacky
question now and then. I will take "placed on" as being synonymous with
"suddenly appears at the surface". It will also be easier to think about if
the needle appears at the equator. I will also assume that no matter what
happens, the needle retains its size and shape. If the needle just sat
there, the effect would be that the earth would now rotate not about its
axis but about a parallel axis which is halfway between the center of the
earth and the needle; the length of the day would increase to about 5 times
the current 24 hours. But, it would certainly not stay on the surface of it.
The force which the needle would exert on the surface of the earth would be
about 6x1025 N. But this force is over an area of a needle, very
tiny, so the pressure would be astronomical. This would cut into the earth
and the needle would end up, probably after oscillating back and forth for a
while, at the center of the earth. The length of the day would return to
about 24 hours and everything on the surface would weight twice as much. If
the needle were to appear at a pole, there would be no effect on the length
of the day at all.
If a dogs bark measures approximately 60-80 decibel, how much will it reduce traveling in air (average 20 Celsius) over 500 meter.
Could you please brake down the formula for me so I can calculate other distances.
it will be below the "threshold of hearing" and you will no longer hear it;
that corresponds to 0 dB.
You should first read an
earlier answer for a detailed
explanation of what a decibel (dB) is. The main reason for loss of sound
intensity I (measured in watts per square meter, W/m2) is
that the sound waves spread as they get farther away so the energy per
second (power) striking your ear gets smaller. The intensity falls off like
1/R2 where R is the distance from the source. You
do not specify the distance from the dog that the 60-80 dB level is
measured, so I will arbitrarily put it at 1 m. Therefore the ratio of
intensities at 500 and 1 m would be I500/I1=1/5002
or I500=4x10-6I1. But,
the catch is that dB is a measure of the level L (a logarithmic
scale), not the intensity. So we need the equations to convert between L
and I. These are L=10∙log10(I/10-12)
and I=10-12∙10(L/10) where I is
in W/m2. So, as an example I will choose L=70 dB, so I1=10-12∙10(70/10)=10-5
W/m2 and I500=4x10-6∙10-5=4x10-11
W/m2. Finally, we find what the dB level of I500
dB. I believe that this will the main source of quieting with distance, not
any absorption of the sound by the air. Wind can also have an effect on the
intensity if it is across the direction the sound travels to reach you. By
the way, when the intensity reaches 10-12 W/m2
My question is that whether gravitational field an be reversed
in Earth`s atmosphere? it would bring revolutionary changes in Aerospace
but space organisations train their astronauts on how to deal in zero gravity in camps. then they have to create artificial zero gravitational field. how is it possible??
You can simulate zero gravity but you certainly do not do it
by reversing the gravitational field. We are stuck with the field we have
and to simulate zero field there are two ways. The first is to put the
astronaut trainee in a free-fall situation. If you are in a freely falling
elevator, for example, it will seem like there is no gravity. The more
practical way to do this is to ride in an airplane which is moving over a
parabolic path it would follow if it were simply a projectile; inside that
airplane it will seem like there is no gravity. The plane which does this is
affectionlately known as the "Vomit Comet". A second way you can simulate
weightlessness is to build a giant tank of water and then the buoyancy will
provide a force opposite to your weight making the net force on you equal to
Finally, it should be noted that when you are in
orbit you are also not truly "weightless". In orbit you are constantly in
free fall so, although you still have your weight, you are in free fall just
as you were on the Vomit Comet.
If we are given a graph showing pressure versus distance for a sound wave, does a higher pressure amplitude indicates a louder sound produced?
Sound waves are longitudinal waves of pressure as shown in
the animation at the right. You are right that the greater the pressure
difference between the lowest and highest pressures, the louder the sound.
You should not think of the regions of high pressure being the regions of
loudest sound, though. Because the elapsed time between low and high
pressure hitting your ear is so short, your ear/brain averages the square of
deviation of the pressure in the wave from normal atmospheric pressure to
determine loudness. The information from the time difference is also
processed by the brain and interpreted as the frequency or pitch of the
Let's say that we started mining for metals and such somewhere other than earth. What if any would the effects of large scale increase (and I mean really really big) in planet mass be? Also what would happen to the other planets we were robbing of minerals as their mass decreased?
As long as the mass of the orbiting body is much less than
the mass of the orbited body, the orbital motion is independent of the mass
of the orbiting body. So, the length of the year would be unchanged.
However, increasing the mass and radius of the planet would increase its
moment of inertia, so its rotational speed would decrease and the day would
get longer. Keep in mind, though, that a huge amount of stuff would have to
be added for any real difference to occur.
In case of string, when only a pulse is sent then why do the particles vibrate just above the mean position and not below the mean position?
Please answer in a way that i could understand. If your answer include some mathematics esp the calculus please simplify so that i could understand
There is an equation, called the wave equation, which to a
fair approximation applies to waves on a string. One of the properties of
this equation is that the wave retains its shape as it propogates. Hence, if
you start with a pulse above the string, it will remain that way.
I am trying to determine the buoyancy of a 10 kg polyethylene tray (density = 0.95 g/cm3=950
kg/m3) that is used in lobster pounds to hold lobsters. The tray holds 60 kg of lobster, and fills up with water up to the lid (which floats at the top of the water). The volume of the tray is 0.1223 m3. The lobsters are held in seawater (density = 1026 kg/m3). So, my question is how much air is required to be placed in cavities in the lid to keep the tray floating at the top of the water?
I will take the "volume of the tray" to be the inside volume,
that is the volume occupied by lobsters and water. The density of a lobster
is sure to vary from animal to animal, but it certainly must be larger than
the water since the creatures dwell on the bottom. I actually found a
measurment of a lobster's density to be 1187 kg/m3, so I will use
that as an approximation for all lobsters. So, the volume occupied by 60 kg
of lobsters is Vlobsters=60/1187=0.0505 m3. So,
the volume of water needed to fill up the tray is Vwater=0.1223-0.0505=0.0718
m3; so the mass of the water is Mwater=1026x0.0718=73.7
kg. The volume of the polyethylene is Vpoly=10/950=0.0105 m3.
So, the total mass is 60+10+73.7=143.7 kg and the total volume is
0.1223+0.0105=0.1328 m3. The density of the full tray is then
ρ=143.7/0.1328=1082 kg/m3; since this is greater than the
density of the water (1026 kg/m3), it will sink, hence presumably
the point of this question. In order for it to "just" float, volume must be
increased until the density is equal to 1026 kg/m3, i.e.
1026=143.7/V or V=0.1401 m3; so, the volume of the
air should be Vair=0.1401-0.1328=0.0073 m3=7300
cm3. This would be a volume of a cube about 20 cm on a side. I
guess I would increase that by at least 50% as a safety factor. (Note that I
have neglected the density of the air which is about 1 kg/m3.)
What happens to gamma rays when they are blocked by materials like lead, and how come they're not blocked by some other materials?
For all intents and purposes, gamma rays interact only with
electrons in the material. The two most important ways they interact are the
photoelectric effect where the photon gives all its energy to an electron
and Compton scattering where the photon scatters from an electron, giving
some of its energy to the electron. Which of these is more important depends
on photon energy and other factors. But, as you would think, the density of
electrons in the material is very important. To a very rough approximation,
atoms are all about the same size. That means that there are about the same
number of lead atoms in a cubic centimenter of lead as there are aluminum
atoms in a cubic centimeter of aluminum. But every lead atom has 82
electrons whereas every aluminum atom has 13 electrons. There are therefore
roughly 6.3 times more electrons in a cubic centimeter of lead than in a
cubic centimeter of aluminum; therefore lead will be much more effective in
shielding against gamma rays. Regarding "what happens" to the gamma rays,
they disappear completely if photoelectric effect occurs or become gradually
less energetic and changed in direction if Compton scattered.
To tighten the loose head of a hammer, the base of the handle is sometimes struck on a hard surface. Explain the physics behind this maneuver.
This sort of sounds like homework. The physics is inertia.
The head of the hammer is pretty massive and so it has a lot of inertia, it
wants to keep going. When the handle suddenly stops, the inertia of the head
keeps it moving so that it moves down a bit on the handle before stopping,
thereby becoming more tightly bound to the handle.
What would happen to a beam of proton particles if the pass through a coil (perpendicularly) imagine a closed circle shaped loop. Currwnt is passing through this coil hencr generating a magnetic field. Will this B field make the particles vibrate? Accelerate them?
I assume that you have the beam of protons very narrow and
moving along the axis of the loop. As you can see from the figure to the
right, the on-axis field is along the axis. Therefore, a beam of protons on
the axis would pass through unaffected because the force on a charged
particle is zero if the velocity and field are parallel. If the proton were
off axis, there would be a small force which, if the proton were in the
plane of the page here, would point either into or out of the page.
Isnt the gravitational pull of an object determined SOLELY on the mass of that object and NOT its size or density? obviously distance from said object plays a role just as size and density affect mass.... however, if what i believe is true, there can be no such thing as black holes, which even today their existence is only theoretical, not proven... anyway my point is this... i believe that if our sun was the size of a basketball but still had the same mass it has today, (giving it almost infinite density)..(what many scientists today consider a 'black hole') but the Earth was still the same 93,000,000 miles from the actual surface as it is today, the orbit of the Earth and the planets would remain, and the sun would not be considered a 'black hole', nor would it behave like one, (sucking in all its orbiting masses, and even light itself).. it would be much smaller, but given the exact same mass and distances surface to surface, little would change as far as orbits go... could this be correct theoretically?
If the object is spherically symmetric and you are totally
outside the mass distribution, then you are right—only the total mass
matters. But, this does not mean that black holes do not exist. It is
dramatic to say that nothing can escape a black hole, not even light, if you
are inside a critical distance called the Schwartzchild radius, but that
does not mean that objects could not orbit the black hole, either inside or
outside that radius. If the sun were a black hole, its Schwartzchild radius
m; so all the planets would be outside and would orbit just fine; they could
even be dragged away if you wanted. Only objects inside 3000 m would be
"trapped" but even they could still orbit the black hole. The figure to the
left shows some orbits of stars around the supermassive black hole at the
center of the galaxy which have been observed by astronomers.
I don't understand the hydrogen spectrum..the theory given in my book says that Balmer series will be observed when electron falls from n=3,4,5......(up to infinity) to n=2 but isn't n=2 for an electron of the hydrogen also an excited state,so it must fall back to its ground state?? Or am i getting something wrong here??
Yes, it must get down to the ground state again once it
reaches n=2. But that is part of a different series, called the Lyman
Series. The radiation from that transtion is called the Lyman alpha line and
is in the ultraviolet region of the spectrum. The rest of the series is
n=3,4,5… going to n=1; all the lines in this series are in the ultraviolet.
Every level has a
terminating on it.
if i create vaccum on earth ,like in a spherical shell,if pierced it will suck air around it, its not the case with atmosphere of earth,why it is not that space vaccum suck it?gravity is indeed much stronger near earth and atmosphere much more denser ?
The reason that air does not leak into space is that gravity
holds it to the earth. Think of the air as a collection of molecules all
moving around with different velocities. The velocities are distributed
according to the Maxwell-Boltzmann distribution.
At normal temperatures, most molecules (e.g., O2, N2,
H2O, CO2,…) have almost no molecules have velocities
large enough to escape the
earth's gravitational pull. Lighter molecules, in particular H2
and He, have much higher velocities and do escape into space; you will find
almost no hydrogen or helium in our atmosphere.
With respect to special theory of relativity, light speed c is an invariant, but wouldn't rest mass also be an invariant? If true, why is it not one of the postulates?
Velocity and acceleration are not invariants, but are there other properties that are invariants?
Light speed is not usually referred to as an invariant; it is
simply a universal constant. Furthermore, as I have
stressed before, I see no need to call
the constancy of c a postulate; it is simply a
result of the principle of relativity
which states that the laws of physics be the same in all inertial frames.
Rest mass is the inertia something has in its own rest frame, so of course
it must be a constant. The word invariance usually refers to any
quantity which remains constant when a Lorentz transformation is performed.
For example, the total energy of an isolated particle E=√(m02c4+p2c2)
is an invariant.
I've been trying to calculate how much impact force an object weighing 2,000,000 lbs traveling at 120,000mph will generate. I've run calculation conversions that tells me that the Newtonian Force is a whapping 48665704243.2 but what does that mean? I read about how an 80 pound 1 foot long object traveling at 52,000mph hit the surface of the moon with a force equivalent to 5 Tons of TNT exploding, creating a 65 foot wide crater!
What would my 1,000 Ton 157.64 Mach object do?
tons=0.66 Megatons of TNT. For comparison, the bomb dropped on Nagasaki in
WWII was about 0.02 Megatons.
For the umpteenth time, you cannot calculate a force by
knowing the mass and velocity of something; see my
FAQ page. What you can do, as your
example does, is calculate the energy which the object carries in. I will
not do what I usually do and convert to SI units since you seem to like
English units. The kinetic energy something has is K=½MV2
where M is the mass and V is the speed. Using your example,
KTNT=5 if K=½∙80∙52,0002=1.1x1011
lb∙mph2, so to convert energy in lb∙mph2, to energy in
tons of TNT you need to multiply by a factor of 5/1.1x1011 lb∙mph2
Why are loops provided for transporting oil/water for longer distances?
When the temperature of the pipe changes it changes length.
In the figure to the left, the pipe will expand if you heat it up and
contract if you cool it down. If it were just a straight pipe, the resulting
forces on the pipe along its length could be large enough to cause it to
buckle and fail. Inserting loops allows the length changes to be taken up by
the size of the loop as shown.
Can u get pure energy? Can you make it visible with power modulation?
Since mass is just another form of energy, everything you can
see or feel is pure energy. Maybe you mean energy without mass? In that
case, the answer would be electromagnetic radiation like light or radio
waves. I have no idea what your second question means.
My question relates to time dilation and the speed of light. When we consider the passage of time, and the speed of an observer or traveler on Earth, how does the speed of our planet around the sun, our arm of the galaxy and the speed of the galaxy itself play into the speed of time for someone on Earth? We talk about a spaceship travelling at a certain speed, but isn't the Earth itself like a spaceship for its inhabitants? And isn't the Earth rotating around the sun at a certain speed, and the spiral arm of our galaxy also rotating at a certain speed? So isn't the Earth also?
Time dilation is meaningful only as a relative thing. One
observer can measure the rate at which the clock of another observer runs.
Relative to a clock on the sun, an earth clock would run slowly. Relative to
the center of the galaxy, an earth clock would run slowly differently.
Relative to a clock on Andromeda galaxy, an earth clock would run slowly
differently. This is the thing about relativity—it is only relative
measurements that mean anything; that's why they call it relativity.
Underlying your question is the misconception that there exists some
absolute rest relative to which all clocks in the universe will run. There
is no such thing as "absolute rest"; any inertial frame may be thought of as
being at rest. Incidentally, for all but the most accurate measurements, the
time dilations I have alluded to are negligibly small because the relative
speeds are small compared to the speed of light.
I don't understand angular momentum but as a first step here is my question. If you hold out weights and turn on a pivot and bring the weights in; are the weights moving at the same velocity and you are spinning faster so that the weights move at the same velocity or are the weights actually moving faster?
The angular momentum L of a mass M a distance
R from the axis around which it is rotating is of L=MVR where V is its speed. It requires no torque to change
R, so L will stay the same if R is changed because
angular momentum of a system on which there are no torques is constant
(conserved). So, for example, if you change R to ½R, V will double. If there were no
person spinning with the weights and no external torques (like friction),
this would be your answer. It is more complicated if you have anything other
than the weights. You probably want to ignore the rest of the answer below,
but I want to do it for completeness.
If you take into account the presence of the
person you need to introduce the moment of inertia I. The angular
momentum now is written as L=Iω where ω is the angular
velocity (in radians per second) of the system. If a point in the system
is a distance R
from the axis of rotation and moving with a speed V, the angular
velocity is ω=V/R. So, comparing with the idealized situation above
you can see that the moment of inertial of a point mass must be Ipoint-mass=MR2.
Now, the man holding the weights has some moment of inertia which I
shall call Iman; I will ignore the contributions from
his arms, so his moment of inertia does not change when he pulls in the
weights. The total angular momentum at the beginning is now L1=(Iman+2MR2)ω1=(Iman+2MR2)V1/R.
So let's again reduce the distance to ½R. The new angular
momentum is L2=(Iman+2M(½R)2)ω2=(Iman+½R2)V2/(½R);
this needs to be the same as the original angular momentum, so (Iman+½MR2)V2/(½R)=(Iman+2MR2)V1/R.
Note that if Iman=0, you get the same as above, V2=2V1.
If Iman>>2MR2, there will
be negligible change in angular velocity and so V2=½V1
because V=Rω and ω did not change. Finally, there
must be a case where Iman is just right that V2=V1;
that would be if Iman=MR2 and would
correspond to ω2=2ω1.
Where can I find the assumptions for conservation of linear momentum? I'm specifically looking for a source which says that both bodies in a linear momentum problem have to attain a common velocity. The assumption or condition I'm looking for is so basic I can't find it.
You can read my
earlier answer on
momentum conservation. I am afraid I have no idea what you mean by "both
bodies…have to attain a common velocity". In fact, that does not even sound
right to me for a general case.
The question, "Why does water freeze at 32 deg. F?" has been proposed all over the internet. There is not an adequate answer to be found. The answers will describe how it freezes, and what takes place, but never why it freezes. The freezing point of hydrogen is -241.746 degrees celcius and the freezing point of oxygen is -218.79 °C. So why does water freeze at 0 deg C?
Iron freezes around 12000C, mercury at -390C,
etc. The temperature of the phase transition between solid and liquid
depends on the material; it is not a simple thing to predict exactly where
that transition will occur, but it can be done in some cases. It is also
dependent on the pressure. But I am curious: Why do you particularly ask
about water? Why didn't you ask me "why does molybdenum freeze at 26200C?"
This question is "all over the internet"? Guess i missed that.
Hello,when we look at objects with triangular prism instead of of eye glasses we see spectrum or rainbow on objects edges ,why this spectrum is always formed on the edges.
I mean the edges of objects for example chair,wall etc. I mean things i look at through prism instead of eyeglasses.
The ideal situation for seeing a spectrum from a prism is if
you have white light, all coming from the same direction and preferably
through a narrow slit. If you just hold the prism up and look through it,
light of many colors is coming from all directions. When you have an
obstruction, like the edge of a table leg, for example, it lis like "half a
slit" and light coming from the edge is restricted to come mostly from one
direction. If that light has a fairly large fraction of white light, you are
likely to see a rainbow spectrum.
Hi I'm a high school student and having hard time understanding something about electric potential energy. Do electrons moving in a circuit have potential energy because battery has done some work moving them against electric field (from low potential to high potential)? If yes, suppose that I just put the wires around the battery and think about the moment when it took 1 electron from low potential to high potential. That electron now has a potential energy equal to the energy spent when moving the charge from low to high potential. But since a battery doesn't generate electrons (and just providing force to move them) and all other electrons were already there in the copper wire, how did other electrons get their potential energy? The battery hasn't done any work for them (or did it?).
The electric current in a conducting wire is not the best way
to learn about potential energy of charges because it is a rather complicate
process. What happens is that the potential difference across the ends of
the wire cause there to be an electric field inside the wire. Electrons see
this field and therefore each believes that it is in an approximately
uniform field and therefore it accelerates and gains kinetic energy as it
loses potential energy it has by virtue of the field. But what next happens
is that little electron almost immediately encounters an atom in the wire,
collides with it, and loses some or all the kinetic energy it has just
acquired and has to start all over again. So each electron bounces slowly
along the wire, repeatedly gaining and then losing kinetic energy. On
average, there is a net drift of electrons down the wire but it is really
quite slow and we consider the average electron to move with a constant
drift velocity. So any electron, moving from the negative terminal of the
battery to the positive terminal of the battery of potential difference V
moved, on average, with no change of kinetic energy but it has lost eV
Joules of potential energy. Where did that energy go? Put your hand on the
wire and you will see that it has warmed up. If you want a more lucid
example of electrons and potential energy, imagine a uniform electric field
of strength E (like in the gap between plates of a parallel plate
capacitor) with an electron released at some point. Then as the electron
moves, accelerating along the direction opposite the field (because it is a
negative charge), it loses potential energy eEz where z is the
distance it has traveled. After it has gone a distance z, it will have
acquired a kinetic energy ½mv2=eEz.
I am reading the ABC of Relativity. This is what I don't understand. If I get on a train in London and travel to Edinborough why is it equally true that Edinborough is travelling towards me? It am the one who initiates the motion. If Edinborough is really moving towards me, how is it also moving towards someone approaching it from the North, at the same time?
The important thing to understand is that there is no test
you can do which determines who is "really" at rest—there is no such thing
as one frame of reference which is absolutely at rest. The way this is
expressed in the theory of special relativity is the laws of physics are
the same in all inertial frames of reference. (An inertial frame is one
which moves with a constant velocity in a straight line, no acceleration
allowed.) This means that there is no experiment you can do inside your
train which will have different result if you were in Edinborough or any other
train. All velocities are relative and it does not matter whom you consider
to be at rest to do physics. You, before you left London, were in the same
frame as Edinborough, and nobody would argue that it was you which caused
the change to a different frame since you were the one who accelerated and
consumed diesel fuel; that does not change the fact that the two frames are,
in all ways, equivalent once you are done accelerating.
By the way, it turns out that the laws of physics
have to be the same in all frames, including accelerating frames. This
is the basis for the theory of general relativity. But, you need not
worry about this is you are just studying special relativity. If you are
interested, see my FAQ page
I was playing a game known as ''Fallout 3'' and in the game there are laser weapons. The laser weapons are powered by a marshmallow sized microfusion cells that are basically miniature nuclear reactors that fuses hydrogen atoms. In the game they produce enough power to turn a 500 kilogram bear into ash in one second. So could a reactor that small produce enough power for the gun and how energy much would a marshmallow sized blob of fused hydrogen produce? A normal microfusion cell in the game has enough energy to fire 24 of these shots. So would it be possible in any way for these laser weapons to be able to be this powerful with an energy source like the microfusion cell?
I have no way to estimate the "power
to turn a 500 kilogram bear into ash in one second". I am sure you realize
that, with today's technology, the possibility of there being such a power
supply is zilch. Let's just do a few estimates to show how hard this is. One
gram of hydrogen fuel (deuterium + tritium), if fully fused into
helium+neutrons, releases something on the order of 300 GJ of energy; so, if
released in 24 one second pulses, each pulse would be about 10 GW. That is
probably way more than your bear burning would need, so let's say 100 MW
would do it; so, we would need about 10-2 g of fuel. I calculate
that to confine that amount of gas in a volume of 10-5 m3
(about 1 in3) would require a pressure of about 5,000,000
atmospheres! That, in itself, should be enough to convince you that this
machine could probably never be possible. If you need more convincing,
consider shielding: 80% of the energy produced is in the kinetic energy of
neutrons. How are you going to harvest that energy in such a small volume
and how are you going to protect yourself from the huge neutron flux? And
surely there needs to be some sort of mechanism to control the process and
convert the energy into usable electrical energy to power the laser; all
that is supposed to fit into 1 in3? This truly is a fantasy game
with no connection to reality!
I read that when no external forces act on a system, the internal forces are paired to balance the momentum.
So if the Universe is considered as such a system where momentum is indeed conserved, every action will have equal and opposite reaction. Is this a sufficient explanation to consider Newton's third law to be a special case of Newton's second law?
Suppose that Newton's third law were not true. For any single
particle in a system of particles Newton's second law would be true but you
would not be able to apply it to the system as a whole. Without Newton's
third law, the total linear momentum of an isolated system of particles
would have no physical significance, it would not be a conserved
quantity. Newton's second law refers only to a single particle and is useful
for systems of particles only because Newton's third law is also true. I
fail to see how that makes the third law a "special case" of the second.
If you were to take a 2 liter bottle of air, down to 10 meters and were to let a slight amount of water into the bottle and then sealed the bottle, what would happen to the bottle when you brought the bottle back to the surface?
It is always hard for a scientist to deal with amounts like
"a slight amount"! If you did not let any air out of the bottle, the
pressure inside would be "a slight amount" higher. If you let in as much
water as would go in without any air escaping, the gauge pressure of the
contained air would be about 1 atmosphere, that is, twice atmospheric
If two deuterium atoms collide at an ideal velocity and at an ideal angle, are there conditions where fusion will not occur and some sort of scattering will take place? If so, why or how would scattering occur?
I do not know what you mean by "ideal velocity" or "ideal angle". I can tell you that whenever the two deuterons come close enough together to interact, there are a great many things which can happen, all with specific probabilities ("cross sections" in the usual parlance). Essentially any possible reaction allowed by conservation laws and selection rules can happen. Nearly always, the most likely thing to happen is elastic scattering where two deuterons exit
after the interaction.
Assuming I had a piece of string that was a light year long and I was holding it on 1 end and my friend was 1 light year away holding the other end, with no slack in the string. If I pulled it would they feel the immediately, or no? If not, why?
You would not believe how many times I have gotten variations
of this question. See my faq page for a
similar question. In a nutshell, the information would travel at the speed
of sound in your string so it would take far longer than one year before
your friend felt your tug.
What would happen to a fly while an airplane takes off assuming that the fly was already flying forward while the plane was at rest?
The effect on the air due to the acceleration of the airplane
is quite small; the air pressure at the rear of the plane would
increase very slightly.
In other words, the air stays almost at rest relative to the airplane. Since
the fly flies relative to the air, he is mostly unaffected. He is, however,
accelerating along with the air and the plane, so he would have to fly a
little harder in the forward direction of the plane to continue moving the
same as he was; since a typical acceleration of a commercial jet is about 3
m/s2, this additional forward force would be about 1/3 of his
why are soap bubbles colorful?what colors are observed when a soap bubble is illuminated by monochromatic light?
The reason is that a soap film has two surfaces from which
reflection of light can occur and they are very close to each other.
Therefore, light of certain colors will interfere constructively while light
of other colors will interfere destructively. A pretty clear explanation as
well as a calculator may be found on the
hyperphysics site. If you illuminate the bubble with monochromatic
light, you will only see that color or no light reflected at all.
I just watched a nature document about the largest snake that ever lived on Earth that was the Titanoboa that could exert a pressure of 400 pounds per square inch and that lived in the water mostly. So my question is that could a Titanoboa destroy a submarine by coiling around it with the pressure of 400 pounds per square inch assuming that the submarine is small enough for the 48 foot snake to coil around it?
The pressure under water increases by about 0.44 psi/ft. To
get to 400 psi, therefore, a submarine must go to a depth of 400/0.44=909
ft. WWII U-boats had collapse depths of 660-920 ft. Modern submarines have
collapse depths of around 2400 ft. So, it depends on the submarine, but most
modern ones could withstand 400 psi. The only proviso is that they are
designed to have the pressure uniformly distributed over the whole surface,
not a narrow band where a snake would squeeze; a very localized pressure
could cause a structural failure at a lower pressure.
I came across
this just now.
It implies a balloon full of air weighs more than the same balloon empty. That doesn't feel right to me as an inflated balloon is surely buoyant by the amount of air it contains. i.e. The extra weight of the air is cancelled out by the buoyancy. The site above is a respected organisation and I would be surprised if is recommending an experiment based on a false assumption. However, I cannot see how the instructions they give could possibly be used to determine the density of air.
You are right, the experiment ignores the buoyant force on
the balloon; if the density of the air in the balloon were identical to the
density of the air outside the balloon, the experiment would fail to find
any weight of air. However, the pressure inside the balloon is greater than
the pressure outside and therefore the density of the air inside is larger
than normal atmospheric density. The experiment then measures the difference
between the mass inside the balloon and the mass of an equal volume of air
at atmospheric pressure. This would still be a reasonable order-of-magnitude
measurement of the density of air. A more accurate experiment would be to
measure the pressure to which the balloon was inflated; with that
information you could do a better measurement of the air density.
I'm interested in why we loose our signal when two walkie talkies get too far apart even there is no obstacle between us. Range of reach of EM waves is infinite, so what's the reason?
You are right, in a vacuum electromagnetic waves last
forever. You are not in a vacuum, but absorption by air is probably very
small for your walkie talkies. The problem is intensity of the waves. If you
are far away compared to the size of the unit, you can say that the signal
spreads out from the sending unit like a sphere. But, there is only a
certain amount of energy in that wave and so the amount of energy
intercepted by the receiving unit gets smaller and smaller the farther away
you go. Eventually, the receiving unit is not sensitive enough to detect the
What should be the temperature of a gas molecule if it needs to escape out of earth's gravitational pull suppose if we take the case of oxygen at what temperature its average velocity will be enough to escape earth's gravitational pull?
Temperature of a gas is
a statistical quantity, no single gas molecule has a temperature. A gas of a
particular temperature has a distribution of speeds called the
Maxwell-Boltzmann distribution which contains all possible speeds. The
figure to the left shows this distribution for N2 for several
temperatures. The escape velocity from the surface of the earth is about
11,000 m/s, so you can see that a heavy molecule like nitrogen has almost no
molecules going that fast at normal temperatures (300 K), or even if T=1000
K. The picture to the right compares the distribution of speeds for N2
with that for H2, about 14 times lighter. You see that the most
likely speed of H2 is about 4 times faster than N2.
Comparing with the figure to the left, 300 K hydrogen gas would have a most
probable speed of more than 1200 m/s. Although there are still very few with
speeds higher than 11,000 m/s, there are still a few which escape.
Eventually, as the slower molecules speed up to fill in the distribution,
they would essentially all leak out of the atmosphere. That is the reason
why there is almost no hydrogen or helium in the atmosphere. Most hydrogen
on earth is locked up in water and other molecules, but since helium is
inert, the only source of it is from underground, usually as a byproduct of
natural gas wells. The form of the Maxwell-Boltzmann distribution is given
Assume we're in a very large hollow sphere. (let's say r = 1 light year).
Then we take a very powerful laser with extremely low diffraction and we fire it while effectuating a full rotation with it.
Wouldn't the laser point travel faster than light on the inside surface of the sphere?
Or does it fall as into the immaterial category and is exempted from the rule?
You do not need such an extreme condition to do what you
want. The distance to the moon is about R=3.8x108 m, so if
you swept your laser beam across the surface of the moon the speed of the
spot would be equal to the speed of light c=3x108 m/s if c=Rω
where ω is the angular velocity you are rotating the laser. So,
per second—really easy to do. But that spot is not anything, really, because
the spot a second from now will not be "made of" the same photons that it is
made of right now. The most important thing is that there is no way that
this spot could be made to carry information from one point on its path to
another. And you are right, it has no mass, but even massless light cannot
travel faster than the speed of light.
As a chemistry teacher, I often get questions from students that are best asked of a physicist. Is there a "short" answer to explain the nature of charge? Why are there only two charges? We understand that the assignment of the negative to the electron could be completely arbitrary, but what exactly is a/the "negative charge"?
For every force field in nature there is a corresponding
source of that field: gravitational mass is the source of gravitational
fields, electric charge is the source of electric fields, electric currents
are the source of magnetic fields, quarks are the source of nuclear fields,
etc. Further, if there are two such sources in proximity of each
other, each experiences a force due to the field of the other. We first are
aware of gravitational fields and conclude after some experimentation that
the property something must have to be a source of that field is mass.
Inspired by our success in understanding the gravitational field, we start
looking around for other forces. Combing our hair one morning, we notice a
new kind of force which is obviously not gravity. After some experimentation
we discover that some objects in nature cause electric fields if they have a
property which we call electric charge. So, putting two electric charges
near each other, we find, not unexpectedly, that both experience a force due
to the other. But now there is something different. In gravity, the force
between two masses is always attractive, never repulsive. So mass is
a relatively simple thing because there is only one kind of it. But with
electric charges, sometimes the force is attractive and sometimes repulsive,
so we conclude that there must be two kinds of electric charge which are
most easily distinguished from each other by assigning a sign, + or -. Of
course, further experimentation reveals that opposite charges attract and
like charges repel. Why are there two? Well, just because that is the way
nature is. You might just as well ask why there is only one kind of
gravitational mass. The way that quarks interact with each other is more
complex than just attractive or repulsive and three kinds of the quark
property called "color charge" are required. So the answer to "what exactly
is charge?" is that it is that property which something has which allows it
to both cause and feel an electric field.
To see questions and answers from longer ago,