QUESTION:
I learnt about the Rutherford experiment where they disproved the
plum pudding model for atoms where they used a gold leaf which was
apparently 100 atoms thick. I have since read about a gold leaf that
was 2 atoms thick. If atoms are mostly empty space why wouldn't you be
able to see right through said gold leaf.
ANSWER:
Gold looks golden because it
reflects yellow, orange, and red if illuminated with
white light. Interestingly, if an object reflects a
particular color, it is also a very good absorber of that
same color. What that means is that, for gold, yellow,
orange, and red will be exceedingly unlikely to get
through gold leaf; if any light at all gets through it
will be other colors. Indeed, light transmitted through a
gold foil will be bluish in color. Typical gold leaf is
more than ten atoms thick and two atoms is a very recent
achievement, I believe. I would think that the thinner
the foil, the more of the gold colors would also transmit
through.
QUESTION:
The Laws of physics say nothing can travel faster than the speed of
light. Given the universe is expanding, two stars on opposite sides of
the universe would be travelling away from each other at much faster
than the speed of light. From the perspective of planet a, planet b is
travelling faster than the speed of light; how is that possible.
ANSWER:
I presume that you are using
Galilean relativity to deduce the speeds of two objects
relative to each other if you know their speeds relative
to something which you will call "at rest". Let's do an
example: suppose that, relative to earth planet a is
moving away from you with a velocity V_{ay}=0.8c
and, on the other side of the universe, planet b is
moving away from you with a velocity V_{by}=0.8c
where c is the speed of light. You conclude
that the velocity of a relative to b is V_{ab}=1.6c
because the velocity addition formula for Galilean
relativity is V_{ab}=V_{ay}+V_{yb}
and V_{yb}=V_{by} (or,
you might say, just by "common sense"). However, for
objects moving with speeds comparable to the speed of
light, Galilean relativity, and indeed, all of Newtonian
mechanics, are not valid laws of physics. The correct
velocity addition formula is V_{ab}=(V_{ay}+V_{yb}
)/[1+V_{ay}V_{yb}/c^{2}]=1.6c/1.64=0.976c.
QUESTION:
I am emailing you with a specific question my 5 year daughter asked my husband and I about rainbows. I’m hoping you might be able to help us, or point us in a direction that would help us with an explanation. We found you online, and we need a professional.
We showed our daughter a prism refracting the light just like a raindrop would. She asked, if we hung 100 prisms from the ceiling that should make 100 little rainbows, right? I said correct. BUT!!!! she asked "If each raindrop can refract the sunlight into the colors of a rainbow. Then why aren't a million rainbows in the sky? Why do we only see one big rainbow? How do the reds stay together, the orange together etc...Please help. I haven't been able to find any information on the web.
ANSWER:
The prism demonstrates
dispersion, the fact that different colors of light
travel with different speeds in any medium. The prism is
made of glass and the rain drop is water; both
demonstrate dispersion but the details are different
because the rain drop is a sphere. All the light rays
from the sun come in essentially parallel. In my figure,
several rays are shown and one is followed as it refracts
when it enters the water, disperses into the colors, and
strikes the back of the drop; when it reaches the back,
part of it goes back into the air (not drawn here), and
part of it is reflected back through the water as shown;
then it reaches the surface again and part of it refracts
back into the air as shown, part of it reflects into the
water again (not shown). For this particular ray, and all
others from millions and millions of other raindrops for
which light enters at the same place on the surface,
the
emerging light constitutes the rainbow you see. But, note
that the angle between the direction of this incoming ray
and the exiting "rainbow rays" is 42°. But, what if I had chosen one of the other rays? I would get a different path for the ray bouncing around inside the drop and
the rainbow would emerge in a different direction. So, the whole sky would be covered in overlapping rainbows! And, it actually is. But not all of these infinite rainbows have the same brightness; it turns out that at 42° the emerging light is the brightest. You can actually derive this angle of maximum brightness if you know calculus and trigonometry; see the Wikepedia article on
Rainbow.
So, if you are standing in just the right place to see
the colors from this drop, then all other drops which sent their rainbow in your direction will also be seen but as coming from different points in the sky. The locus of those points will be a perfect circle. If you are in an airplane it is possible to see the full circle.
One last issue: it turns out that we often see a second,
dimmer rainbow above the bright one. The reason for this is that some of the incident rays will reflect twice instead of once inside the drop before they come out; these form the second rainbow. I hope this is not too complicated for a fiveyear old, but
hopefully you can digest it and explain it in terms she would understand.
QUESTION:
It says in a lot of places that wavelength is inversely proportional to frequency. Would it not be possible to increase the speed of the wave, (and therefore the frequency) so more waves pass through a point per second, without increasing the wavelength?
ANSWER:
In general, v=fλ
and you can hold any one of the three constant and ask
what happens to the other two. Here is a simple example:
A guitar string of length L vibrating with its
fundamental frequency is a standing wave with λ=2L.
You can change the speed of the wave by changing the
tension in the string (which is what the tuning pegs do).
But the wavelength cannot change so the frequency must.
QUESTION:
Does the gravity of earth comes from molten core.and if so can we create artifital gravity by making molten core simillar to earth by rotating it at high speed.
ANSWER:
Gravity is caused by mass. The
earth's core constitutes approximately 1/3 of its total
mass and rotational speed has nothing to do with it.
Neither does the "moltenness" have anything to do with
gravity. So if you had a molten core alone, rapidly
rotating, its gravitational field would be just the same
as if you had a solid sphere of the same mass; there
would be nothing "artificial" about it.
QUESTION:
what force are acting on a ping pong ball at the top of its bounce is there just gravity or is the kinetic energy from the ball still at play.
ANSWER:
The first thing to note is that
kinetic energy is not a force. And I don't know if when
you say "top of its bounce" whether if is in purely
vertical motion or is in a trajectory during which it
will still have a speed at the top. If it is the former,
it is at rest and the only force on it is its weight
(gravity). If it is the latter there will also be an air
drag force opposite its direction of motion. (Air drag is
quite important for a ping pong ball.)
QUESTION:
My question is about the actual energy consumption that is required to produce a certain amount of electrical output:
Framing the setup for the question:
If I have an electrical generator that can produce 1kw of electrical energy and I spin the armature with NO load, the amount of work I am producing to spin the generator is fairly nominal. It's simply the frictional losses of the bearings, the amount of energy to accelerate the mass of the armature to a particular rpm (lets say 1000 rpm), and perhaps some wind resistance. This will have some numerical value, I am assuming in joules or horsepower. (please correct me if I am wrong).
Now, if I throw a switch that connects 10, 100 watt lightbulbs. The back EMF is tremendous. Anyone who has tried to spin a generator by hand can attest to how difficult it is to try and light just 1 100 watt light bulb.
Assuming the generator is 98% efficient; how much work (or energy) is required to produce the 1kw of electricity to light the bulbs?
I would like to compare the "NO LOAD" power requirement to spin the armature to the "FULL LOAD" requirement to spin the armature.
ANSWER:
I don't think there is much of a
mystery here. Let's forget about your first paragraph
because when you say 98% efficient, that presumably
includes all the losses you enumerate. This generater can
generate 1 kW of power at 1000 rpm. Now you are asking it
to give you 1 kW if you attach 10 100 W bulbs. You ask
how much energy you have to put in but that is not the
right thing to ask; you should ask how much energy per
second (Watts) you need to supply, power. Since the
efficiency is 98%, you need to put in 1 kW/0.98=1.02 kW
of power.
QUESTION:
Hello, i was curious why science hasn't utilized the physics of "slipping" into generating energy...
I understand there are road blocks, but is the potential of slipping to generate energy not extreme enough to warrant overcoming them?
Can you not generate a lot of energy with slipping? or is it just that in comparison to other forms of generating energy, it just doesn't compare and it's not as exponentially capable of generating energy as im assuming. Like when i imagine moving a large object at high speeds once you've propelled it to the speed you want it wouldn't the physics of slip, give it the continuous push it needs to maintain or even exceed the level you had put it too..and would not the physics of slipping then take over the propelling of the object...and if we could create or find..a suitable force that creates slip and resists wear and tear long enough to warrant using would that not enable us to drastically improve our capacity to generate energy??
ANSWER:
I presume you mean friction?
Friction generally takes energy away from a system. For
example, a box with kinetic energy sliding across a floor
stops because friction takes the energy away from it.
That energy which the box had does not really disappear
but shows up as heat (the floor and box will get a little
warmer) and sound (while sliding you can hear it). Are
you suggesting that we could use the heat and sound
energy somehow? Why not just use the energy the box
already had? Usually we try to eliminate friction as much
as we can to maximize other sources. One example is the
brakes of a convential car which transform all the
kinetic energy of the car into heat; electric and hybrid
cars use magnetic induction to put that kinetic energy
into the batteries rather than into heat in the drums or
disks of brakes.
QUESTION:
Hi, today I read an article on Armstrong Limit and I have a question about it, if saliva/tears boil in 36 degrees Celsius in the Armstrong Limit why is it a problem, does it cause any health issues?
ANSWER:
Your body is about 60% water. How
could you ask if it would be any health issues if all
that water started boiling?
QUESTION:
If you could dissassemble a human atom by atom, and then store the human in his or her dissassembled atom state, could you reassemble said human in the future without them aging essentially creating some form of atomical time travel? Also if it were achievable, would they retain their conciousness, memories and all that?
ANSWER:
The answer is no. To read a
discussion of the physics of the impossibility of the
"beam me up Scotty" problem, I suggest The Physics of
Star Trek by Lawrence Krauss, HarperCollins
Publishers, 1996.
QUESTION:
I am wondering the following;
1. Will the electron of atoms always apear when looked for in its cloud, or can it sometimes be "missed"?
2. Will it actually only be in two places at once or could it be where ever it is looked for simultaneously?
3. Is the "quality" or "strength" of an electron compromised when seen simultaneously in its different locations?
ANSWER:
The problem with your questions
is that you have the idea that you can even talk about
the electron being at any particular position. All you
can know, until you make a measurment, is the probablity
of the electron being in any particular arbitrarily small
volume. A measurement means that you look at a particular
small volume and if the probability of finding it there
is 1% you will only see it once every 100 times you look
there. It is never in two places at the same time, it is
really smeared over all places until you observe it and
then it is where you observe it to be. It can never be
observed at the same time in two places; otherwise there
would mysteriously be two electrons even if you know
there can only be one.
QUESTION:
I have a potentially unusual question to which I am currently unable to find an answer and I thought that you may be able to help. I was riding my motorcycle today and turned a corner and the front tyre came into contact with a large patch of gravel on the surface of the road. I had touched the front brake immediately prior to this and, as the front tyre hit the gravel the bike began to slide and to pitch sideways. I put both feet on the ground and steadied the bike, effectively holding it upright and avoiding a spill. Now for the question: My body is very sore after around 5 hours after the event and this has made me wonder what kind of energy did I absorb in my attempt to avoid the accident? The bike weighs 220kg, I weigh 125kg, I was travelling at 20mph (after I had released the brakes )on relatively flat ground and stopped the momentum of the bike within 2.5 yards (after I had released the brake). I am completely unable to fathom a calculation for this, I’m afraid . Can you help me at all?
ANSWER:
So, we have m=345 kg moving with a
speed v=20 mph=8.94 m/s and stopping in d=2.5 yd=2.3 m. The
initial kinetic energy was E_{1}=½mv^{2}=2.73x10^{4}
J and the initial linear momentum was p_{1}=mv=3.08x10^{3}
kg·m/s; the final energy and momentum are zero. Change in
energy is equal to work done, so W=E_{1}=Fd=2.3F
where F is the average force exerted by the
ground on you (frictional drag); so F=1.19x10^{4}
N. You could also say that the change in momentum equals
the Ft where t is the time to stop, so
t=1.19x10^{4}/3.08x10^{3}=3.86
s. The average acceleration over the stop was F/m=34.5
m/s^{2}, about 3.5 times the acceleration due to
gravity, 9.8 m/s^{2}. So why were you so sore?
Because Newton's third law says if the ground exerts a
force on you, then you exerted an equal but opposite
force on the ground. So you were exerting a force on the
order of 10,000 N, about 2000 pounds, for about 4
seconds. That sounds like an awful lot to me. You said
the brakes were not engated as you traveled 2.5 yards but
was the bike skidding sideways? That would have been like
having your brakes on and have lessened the amount of
force you needed to apply to stop the ride.
QUESTION:
My question is about Newton's 3rd law. I understand it pretty good, so I'm making myself questions trying to understand the limits. Here is my question.
Let's suppose I'm in a target range, shooting at the target. Obviously the bullet go through it. In the very precise instant that the bullet hit the card, how can that be explained by the 3rd law. Is there an opposite force, equal in magnitude, opposite in direction, even though the bullet go through the target?
ANSWER:
During the time the bullet and
target are in contact with each other, the bullet exerts
a force on the target; this force is evidently stronger
than the target can withstand and that is why it tears
and the bullet passes through. Now, when that is
happening, Newton's third law states that the target
exerts an equal but opposite force on the bullet. It is
not very hard to punch a hole through a sheet of paper,
so the force is not very big in this situation;
nevertheless, the bullet feels that force and as a result
emerges on the other side with a slightly smaller speed.
If you had 100 sheets of paper to make your target, the
bullet would be going with a considerably smaller speed
when it came out the back; with a thick or strong enough
target the bullet would just stop in the target.
QUESTION:
In this
video, starting at 2:00, a pair of 2liter bottles suspended over 2 loudspeakers are made to orbit about a fulcrum on which they are balanced when excited at their resonant frequency purely by sound waves. But such wave motion is back and forth, so how can it impart momentum in just the forward direction? I must know how this works!
ANSWER:
Yes, the pressure at any point in
the bottle, including the open end, will fluxuate with
the frequency of the air in the bottle. When the pressure
is low, air from the outside is sucked into the bottle;
when the pressure is high, the air is pushed out. But the
two air motions are not analogous. As shown in the
figures below, when air flows in it comes from many
directions so this gives the bottle only a small thrust
to the right. But when it flows out, the neck tends to
align the air motion causing a larger thrust to the left.
A cavity with a narrow, open neck at one end is called a
Helmholz resonator.
QUESTION:
Einstein's theory claims that a large object in space will create a gravitational field that will bend SpaceTime. My question is  if indeed a large body like the Earth does create a bend in SpaceTime do we exist in another aspect of time than that of the surrounding space which is empty and thus does not have a gravitational effect on that around it?
ANSWER:
The rate at which clocks run
depends on the intensity of the gravitational field. A
clock 100 m above you runs at a different rate as yours.
So I guess you do "exist in another aspect of time"
(whatever that means!) from any region where the
magnitude of the gravitational field is different than
yours. This effect is often referred to as the
gravitational red shift. It turns out that gps systems
must correct for this effects since timing between you
and satellites at high altitudes must be extraordinarily
accurate. I should note that the differences in time are
extremely small, even for something as large as the sun
which is incredibly massive.
QUESTION:
If at the start of the universe all the matter was within a cubic meter how didn't it create a black hole
ANSWER:
I state clearly on the site that
I do not normally do astronomy/astrophysics/cosmology, so
take with a grain of salt that I am no expert. I would
guess that there is so much energy in this cubic meter
that whatever forces might be acting are simply
inadequate to reverse the expansion. Also, I used the
word "other" because we do not know what the laws of
physics were in the very young universe.
QUESTION:
got a discusion with a friend, about the riding wind versus side wind.
when driving, the air thats in front of the care needs to move, and puts presure on the car. does that increase when you drive harder, and will a side wind have more or less effect when you drive harder?
ANSWER:
If you are driving into a
headwind there is a force back on you due to the speed of
the wind. You would therefore have to press harder on the
accelerator to keep going with the same speed you would
go in still air. But that does not have a handling the
car, merely holds you back causing you to use more
gasoline. For example, if your car has a speed of 60
km/hr into a 20 km/hr headwind, it would be the same as
driving 80 km/hr in still air. In a crosswind, however,
handling is more of an issue. There is a tendency to push
the car in the same direction as the wind is blowing; if
the road is wet or icy and the wind is strong enough, the
friction of the tires could be inadequate to keep you
from sliding across the road. Even if the road is dry,
the tendency for the car to turn with the wind and you
need to slightly steer into the wind. Also keep in mind
that a strong headwind will not necessarily keep blowing
exactly opposite your direction or else you may need to
do a maneuver where the wind now is partly blowing across
your path. So, always be extra careful on a windy day.
QUESTION:
I respect that Heisenberg's Uncertainty Principle can't be violated. . But a daily lab event seems to say "no you can". Here's the situation. Take the cathode ray tube in an oscilloscope. An electron is ejected from the cathode, deflected by coils around the tube's neck and then impacted at a precise screen location. The UP says I can never accurately know the electron's position and momentum at T. But I do accurately know P, it's relativistic mass is constant and so is it's velocity. In order to place that electron at the precise point on the tube face, the circuit must know its exact location to know when to increase voltages to the neck coils for proper deflection. Why is this scenario not a violation of UP?
ANSWER:
Well, how accurately can you
actually know P? Certainly not better than maybe 0.1%.
And position? Maybe to a micron or so. But, the P
you are talking about and the position you are talking
about are not the proper quantities to be talking about
in terms of the UP. Suppose we call the line between the
electron gun to the center of the screen the zdirection.
Then the UP is ΔzΔp_{z}≥ℏ
but the position on the screen would be perpendicular to
z, could be either x or y,and ΔxΔp_{z}
has no uncertainty principle associated with it.
QUESTION:
So I'm in high school and I'm in physics. I'm also a
musician (piano, singing, etc.). I was wondering about
pendulums, but specifically metronomes. If I wanted to
use a metronome as a model in an energy conservation
project, would I simply apply the same rules as normal
pendulum situations? I guess I'm just saying that I'm
having trouble with the difference between metronomes and
other pendulums. I tried to find the answer online, but
whenever a metronome is used in a physics example is
always is talking about simple harmonic motion or
synchronizing metronomes, which isn't really what I'm
asking. I'm not advanced enough yet to figure out the
physics behind a metronome by myself, either. Sorry about
the long paragraph for what's probably a pretty simple
question! I'm basically just asking if, when discussing
energy transfer/conservation, work, and force in physics
does a metronome abide by the same rules as a normal
pendulum, and if not, how does it differ?
ANSWER:
A metronome is certainly not a
simple pendulum which is basically a weightless stick of
length L frictionlessly pivoted at one end and
attached to a point mass m at the other end; the point
mass oscillates below the pivot. As you have probably
learned, this seemingly simple problem cannot be solved
analytically but can be approximately solved is the angle
is small. The result is that the frequency f (cycles per
second) is approximately f≈[√(g/L)]/(2π)
where g is the acceleration due to gravity.
Note that the mass does not matter, a bit of a surprise
perhaps.
Now the metronome is clearly not a simple pendulum
because the mass is above the pivot point. But, how can
that be? If you just stuck a mass on the end of a stick
and rotated the mass to a point above the pivot, would it
oscillate about the very top? Of course not, it would
still oscillate about the very bottom but with an
amplitude which was very big. So, how does the metronome
do this? Well, I have earlier worked
out how a pendulum works. You may want to have a look
at this but, being in high school you probably are not
ready for the math there. And maybe your physics class
has not even gotten to rotational motion yet so you would
not know about moment of inertia, angular acceleration,
etc. But you find out the nature of the pendulum in a
metronome: there is a bigger mass below and out of site.
So this is really a "double pendulum" and as long as the
hidden mass is bigger, it (the bigger) will oscillate
about the bottom while the smaller oscillates about the
top. The picture here of the metronome made of plexiglas
shows that bottom mass. You can also get the final answer if the "stick" has
negligible mass compared to the two masses, f=ω/(2π)=[√(g(ML_{M}mL_{m})/(ML_{M}^{2}+mL_{m}^{2})]/(2π)
where M (m) and L_{M}
(L_{m}) are the mass and distance from
the pivot of the lower (upper) weight.
Now to your question, whether energy conservation applies
to a metronome. No realworld pendulum as its mechanical
energy conserved because friction of some sort is always
present. However, if you put it in a box from which no
energy can come in or go out, the total energy will be
conserved, even if the pendulum stops, the air in the box
will heat up a little bit and if you were to measure all
the energy contained in thermal energy you would find all
the energy which the pendulum originally had. So the
answer to the "basically
just asking" is yes, the two pendula "abide" by the same
physics rules. Also note that there is a little
springdriven motor you can see inside; when you wind it
up you do work to give it potential energy and it then
gives this energy to the pendulum to replace energy lost
to friction.
QUESTION:
Let's consider crushing some item with scissors blade. It is easier to do it when the item is positioned closer to the pivot(assuming the force of squeezing the scissors is constant). I am asking to check the following explanation:
Is it so because to avoid being crashed the item must generate reaction force impacting the blades of such magnitude that the blades are not moving(theirs torques have to equal zero).Let's assume that force momentum coming from squeezing the scissors is constant(we don't change force of our fingers nor the distance from the pivot) and potential of generating reaction force by the item also. Then the only thing we can do to decrease this torque coming from reaction force(and therefore making the item less "resilient") is to shorten the distance between item and pivot.
My wonder is if it all comes down to considering movement of scissors blades around axis, meaning: blades are moving the item gets crushed, blades are still(torque equals 0) the item is in one piece.
ANSWER:
You can understand the principle
by just considering one half or the scissors; then the
other does just the same but with opposite forces. All
your talk about "reaction force" is wrong; look at my
diagram—the force labelled F_{1}
is the force your thumb exerts on the scissor and the
reaction force is the force which the scissor exerts on
your thumb which is the same magnitude but oppisite
direction. I will assume that there is some small object
you want to "crush" located somewhere along the blade.
This gets a little confusing, so read the next sentence
carefully. If you exert the force F_{1
}on the scissor and the scissor is not closing, the
sum of the torques exerted on the scissor must
equal zero; if the object is located near the pivot, like
where F_{2} is, the
object must exert a force opposite to F_{2}
but of the same magnitude since F_{2
}is the force the scissor exerts on the object which
is what you are interested in. (Ignore F_{3}
for now.) To calculate the magnitudes of the torques you
need to multiply the force times the moment arm (yellow
lines). The moment arm for F_{1} is much
longer than for F_{2} and therefore
F_{2} should be much larger than F_{1}.
Now, consider placing the object out near the end of the
blade where F_{3} is
drawn; its moment arm is larger than for F_{1}
so its magnitude is smaller. I hope it is now clear why
placing the object closer to the pivot results in a
larger force on it than farther away.
QUESTION:
I have a question about the weak nuclear force.
Why do we call weak force a force?
All what's said about the weak force is that it cause decays by changed one particle into another (or at a fundamental level, by changing the flavour of quarks).
But , "what makes it a force?"
Do we call it a force only because it involves W/Z bosons or there's something else.
ANSWER:
It is better named the weak
interaction. The concept of a force is generally not
useful in quantum physics, but, since we all have a gut
feeling what a force is in macroscopic classical physics,
it is natural to think of interactions between objects as
resulting in forces. But, to treat interactions between
objects quantum mechanically we use the concept of fields
to describe those interactions; fields are also used in
classical physics as well—think electric, magnetic,
and gravitational fields. At the quantum level fields
alone are useful, not forces. And when you have a field
it can be quantized and the quanta are often viewed as
the "messengers" of the field; the field quanta are
photons for the electromagnetic field, gluons for the
strong interaction, and W^{±} and Z bosons
(as you note) for the weak interaction.
It is interesting that the fourth "force" in nature is
problematical: no one has successfully quantized the
gravitational field and achieving that is one of the holy
grails of physics.
Also interesting is that the notion of force is not
useful in the theory of special relativity because the
usefulness of force depends on the usefulness of
acceleration. At high speeds where special relativity is
important, different observers will see different
accelerations for the same object; if we think of force
in terms of Newton's second law (F=ma),
different observers would deduce different forces. If,
instead, we thought of Newton's second law as force
equals time rate of change of linear momentum, force
could be invisioned but only if linear momentum (mass
times velocity) were differently defined. But I
digress...!
QUESTION:
I am a 180 lb man competing 180 lb weight class five time world champion powerlift. I bench 500 lb and I would like to know how much force am I executing to lift 500 lbs?
ANSWER:
To hold 500 lb at rest or to move
500 lb straight up with constant speed requires a force
of 500 lb. Since
the weight you lift is initially at rest and momentarily
at rest when your arms are fully extended, you must exert a
force larger than 500 lb to get it started
moving upward and smaller than 500 lb for it to
stop at the top. I have watched videos of the bench press and
it appears that during most of the time the bar is moving
with a constant speed; so most of the time the force you
exert is about 500 lb. To do a rough calculation of the force
necessary to get it moving, I will use a total lift distance of
about 1 m in a time of about 1 s so the speed is roughly
v=1 m/s; 500 lb has a mass of about m=227
kg. (I will use SI units, like scientists prefer, and
convert back to pounds in the end.) Suppose that the time it
takes to accelerate the bar to 1 m/s is t=0.1 s;
so the acceleration is a=(1 m/s)/(0.1 s)=10 m/s^{2}.
So the force F you must exert is calculated
using Newton's second law, ma=(FW)
where W=mg is the weight of the bar and g≈10
m/s^{2} is the acceleration due to gravity;
F=227(10+10)=4540
N=1000 lb, twice the weight you are lifting.
Similarly, if it takes 0.1 s to stop the bar, the force
you would apply would be F=227(10+10)=0, the
weight will stop itself because of gravity. Keep in mind
that these are estimates using reasonable numbers; the
actual numbers depend on the way the individual does the
lift.
MY ANSWER IS WRONG, OR AT LEAST INCOMPLETE. SEE
BELOW
QUESTION:
Recently when cleaning, I opened the cover of our ionisation smoke detector, as I was opening the smoke detector cover, dust fell from the inside of the smoke detector into my eye. The whole inside of the cover was covered in this dusta brown colour dust, not grey colour. I was concerned that this dust has been given off by the Americium due to the slats on the ionisation chamber which allow particles to escape from the radiation. Could this be so? I was concerned as it said online that it was dangerous to ingest or inhale the americium from smoke detectors and this dust entered my eyes?
ANSWER:
I have recently answered a
question about smoke detectors; you should read
that. The americium is in a
sealed container and cannot get out. Also, the color of
dust depends on where it comes from. There is probably
some source of brown dust in your house and you wouldn't
notice its color except when something like your smoke
detector accumulates a relatively large amount over a
long time.
CORRECTION:
I received an
email with additional
information about which I was not aware. Essentially it
said that the recoiling americium nuclei could collide
with and eventually damage the casing around the source.
Since your smoke detector is 27 years old and the rule of
thumb is that the useful lifetime of americium smoke
detectors is about 10 years, you should dispose of this
one and get a new one. (Use the recommendations of the
answer you included in
your question to dispose of it safely.) I still believe
that you need not obsess over the small exposure you
received when you were cleaning, given the very small
numbers I quoted in the earlier answer about smoke
detectors; the unit quoted there,
rem (Roentgen equivalent man), is specifically meant
to indicate exposure to ionizing radiation of human
tissue and is therefore more useful than Curies used by
the author of the
answer which just measures the amount of radiation
emitted without consideration of their health effects.
QUESTION:
A piece of cork is thrown vertically downwards from a sky scraper 300 meters high.
Its initial velocity is 2 m/s. Air resistance produces a uniform acceleration of 4 m/s^{2} until the cork reaches a terminal velocity of 5 m/s.
Why does the cork reach terminal velocity?
ANSWER:
There is something terribly wrong
about this question. I see only two ways I can interpret
this question.

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

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

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

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

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

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

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

T(0,t)=T_{0}

∂T(x,t)/∂x_{x=}_{L}=0

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

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

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

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

You have a ball
in your hand;

you throw the
ball with some speed;

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

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

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

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

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

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

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

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

Electric charges cause electric fields.

Electric currents cause
magnetic fields.

Changing electric fields cause magnetic
fields.

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

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

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

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

If 0<T<½(M+m)g
you will accelerate downward with an
acceleration smaller in magnitude than g. Of
course T<0 is not possible for a rope.
*It is important to
note that, because of Newton's third law, if the rope
exerts a force up on you, you exert an equal and opposite
down on the rope. T is a measure of how hard you
are pulling. How large T can be depends only on
how strong you are and how strong the rope is.
QUESTION:
We define image as the intersection point of reflected rays and the focal point is the point at which light will meet after reflection.
So, why is it that we have different positions of image when object is placed at different distances from concave mirror and not always at focal point except when object is placed at infinity?
ANSWER:
Because only rays which come in
parallel to the optic axis are reflected through the
focal point. This occurs only (approximately) for objects
very far from the mirror.
QUESTION:
Does radiation from your phone stay on your hands after you use it?
ANSWER:
Absolutely not. Any radiation
from the phone is electromagnetic, essentially radio
waves. It is transient and does not "stay" anywhere. Even
when you hold the phone the radiation is not "on your
hands" but harmlessly passing through them just like the
waves from some radio station is passing through your
whole body.
QUESTION:
When a uranium or plutonium atom is fissioned, energy is released visa E=MC2. I believe this energy is in the form of photons. What I do not understand is how these photons (light) create such high temperatures in an uncontrolled nuclear explosion. Can you please explain?
ANSWER:
Actually, emitted photons account
for only about 2.5% of the total fission energy. Emitted
neutrons account for 3.5%. Most of the energy is
contained in the kinetic enerngy of the fission
fragments, 85%. The other approximately 9% of the total
energy shows up later when the fission fragments, which
are not stable, decay radioactively, mainly from βdecay.
The high temperatures are due to the kinetic energy of
the fission products which have speeds typically 3% the
speed of light.
QUESTION:
I'm embarking on a horological project. I need to buy a very particular type of spiral spring. I guess it's a type of torsion spring. The only supplier I can find who sells these springs classifies them with two numbers. The overall diameter in mm
and the torque in gm/cm/100°.
(As noted below, I assume that this means the torque on the spring when
the angle is 100° and it should be gm·cm, not gm/cm.) I know the spring I need. But I only know in terms of the CGS System. I'm following an old book and I've ascertained that the spring I need has a
"CGS number" of 0.76.
So, my challenge is to find a way to convert between these two unfamiliar units of torque measurement. A
"CGS number" and gm/cm/angle.
There are plenty of calculators for converting between different units of torque. But I don't think it's as simple as that. I have two questions.
I know a bit about the CGS system and there is loads of info online. But I'm not clear what it is in the context of classifying a torsion spring. The book only says
"the CGS number indicates the restoring couple of the spring when its diameter is 1cm." So what is the CGS number exactly? Is it dynecentimetres? If so, then that's easy to work with since that's a measure of torque I'm familiar with. So I'm half way there. If it's something else, then perhaps you can help explain.
Then there is the question of gm/cm/angle. The problem is that none of the calculators I've seen have a notion of angle. This is specific to torsion springs. The torque, in this case, is the working load when the spring bends 100°.
Will I ever find the correct spring? Sadly, I have spoken to the manufacturer, and they were unable to help me.
ANSWER:
I think the "number" of interest is not a torque, rather
the spring constant associated with the spring for
angular displacement. First, a review of linear springs
which when stretched stretch a distance x when a
force F is applied to the spring. It is
experimentally determined that, to an excellent
approximation for a good steel spring, is that the
distance stretched is proportional to the force applied,
x∝F or F=kx. The
proportionality constant k is called the spring
constant and is measured, in SI units, as N/m (Newtons per
meter) [dynes/cm in CGS units, lb/ft in Imperial units].
k indicates the stiffness of the spring
and this equation is called Hooke's law. The form of
Hooke's law for angular motion (as in a torsion spring)
is that the angular displacement θ is proportional
to the applied torque τ or τ=κθ;
I believe the quantity you want is κ
(kappa) and is measured, in SI units, N·m/radian
[dynes·cm/radian in CGS units, lb·ft/degree in Imperial
units]*. I have noted that κ is often called, by
manufacturers and vendors, the rate
of the spring since it is the rate of change of torque
per unit angle.
I am puzzled by your reference to a constant defined as
gm/cm/degree; 1 dyne=1 gm·cm/s^{2}, so κ
should have units gm·cm^{2}/s^{2}/degree;
sometimes the gram is treated as a unit of force (called
a gram force where 1 gram force=980 dynes), in which case
κ
would be gm·cm/degree, not gm/cm/degree. So, there
must be somewhere in the reference you are using a
definition of what the CGS number is. Suppose that it is
0.76 dynes·cm/radian (which would be the likeliest
units if purely CGS system of units is applied) but you want it in
lb·in/degree; then 0.76 (dyne·cm/radian)x(2.25x10^{6}
lb/dyne)x(0.394 in/cm)/(57.3 deg/radian)=1.18x10^{8}
lb·in/degree. Or, suppose you wanted it in (gm
force·cm/degree); then 0.76 (dyne·cm/radian)x(gram
force/980 dyne)/(57.3 deg/radian)=1.39x10^{5} (gm force·cm)/degree).There are lots of unit converters
around, my favorite is
here.
The choice of units for κ which it seems most vendors use
is lb·in/deg. The best calculator I found to get
an idea of the magnitude of these springs can be found
here.
Until you are able to deduce the units of the socalled CGS value of
0.76 you cannot know what it is.
*Sometimes
κ is implicitly given by specifying the
torque at a particular angle.
ADDED THOUGHTS:
In the book by Hans
Jendritzki, Watch Adjustment, he says that in
the CGS system for characterizing the spring in question
(which I will call N_{CGS}), the unit of
force is the dyne and the diameter of the spring is 1
cm. He then uses not the restoring torque to calculate
N_{CGS} but rather the
restoring couple
associated with the force; the restoring couple is the
restoring force times the diameter whereas the
torque is the restoring force times the radius.
Most vendors use the restoring torque to characterize
springs, so ½N_{CGS }should be
used to convert to the vendor's units. Although the
units of the angle are not given in Jendritkzi, I have
found that the more natural radians gives values much
too small when converting to vendor units; so N_{CGS}=(restoring
couple in dyne·cm for 1 cm diameter)/degree.
One vendor
charactorizes the spring by specifying a torque given by
N gm/cm/100°; this does not appear to be
dimensionally correct unless the gram is gram force
(I will denote it as gmf) where 1 gmf=981 dynes (the
weight of 1 gm in dynes) and /cm means that again this
is for a 1 cm diameter spring. So, assuming that the
vendor's torque is indeed torque (force times half the
diameter), ½N_{CGS}=N.
For example, suppose N_{CGS}=2; then
N=½(2) (dyne·cm/1°)(1
gmf·cm/981 dyne·cm)/(100°/1°)=0.102 gmf·cm/100°;
so the conversion from the CGS value to this vendor's
value is N=0.0491N_{CGS}. So,
in the case of N_{CGS}=0.76 dyne·cm/1°,
N=0.037 gmf·cm/100°.
[Disclaimer: I am
not a clockmaker and I am not familiar with conventions
clockmakers might use or meanings they may attach to
certain words which are not in accordance with
definitions, units, or conventions used by physicists. In the event that I have misunderstood
or that either Jendritkzi really meant torque or the vendor
really meant couple, the conversion factor would be
0.102 instead. I have done my best to determine
conversions between two numbers representing the same
thing but in different systems of units.]
QUESTION:
If you lined up frictionless gears over an exceptionally long distance, could you effectively communicate faster than the speed of light through one person moving the first gear and a second person reading/interpreting the resulting movement of the last gear?
ANSWER:
I have answered this question in many guises
many times. In your
variation, each gear will experience a force from the
previous gear in the line and exert a force on the next
gear. The time it takes for the force to travel from the
input force location to the output force location is
determined by the speed of sound in the gear. The message
you send would travel at the speed of sound, much smaller
than the speed of light.
QUESTION:
If you are driving side by side at 60mph with another car and throw a can of coke at them would the can be travelling at the same speed so hit the target as if you were parked next to each other or merely disappear behind your vehicle?
ANSWER:
When you throw it out the window it originally has a
forward component of its velocity of 60 mph. At this
speed, the can will experience a very significant force
slowing its forward velocity down because of the air
drag. So the can will not keep pace with the two cars and
will appear, from either car, to be accelerating
backwards. So, no, the can will not hit the other car
where it would have if the cars had not been moving.
QUESTION:
Under relative motion, it's experimented that A person sitting on a back seat in a moving bus has a speed with respect to the earth which is the same as that of the bus. But if he now walks towards the driver of the bus,he has a speed relative to the earth which is more than that of the bus. Suppose the bus is moving at 100km/h and the person walks at 5km/h towards the driver, his forward speed relative to the earth is 100+5= 105km/hr. But when he walks back from the driver to his back seat at the speed of 5km/hr his speed relative to the earth is now 1005= 95km/hr.
WHY the increase speed of such person when moved forward and the decrease when moved backward
ANSWER:
The problem you are having is that you do not understand
the phrase "relative to the earth". Let's alter your
example by having the train moving at 5 km/hr. If you are
in the train walking 5 km/hr in the opposite direction,
someone standing by the side of the tracks will
see you standing still; the same as running on a
treadmill in a gym with the treadmill and you moving in
opposite directions with the same speed. If you walk in
the same direction as the train is going, you will be seen by the trackside observer
as moving with a
speed of 10 km/hr.
QUESTION:
IF WAVES need a medium to transmit energy but light can travel through a vacuum.
Does dark matter have Anything To Do With Lights Ability To Do This?
ANSWER:
There is absolutely no mystery about light waves;
electromagnetism is arguably the bestunderstood theory
in all of physics. You may be sure it has nothing to do
with dark matter. Dark matter is arguably the
leastunderstood theory in all of physics. There are many
features of the universe which do not fit into our
current understanding of physics and introducing some
kind of particle which interacts with the rest of
creation only via gravity would solve a lot of
them; all attempts to directly detect dark matter have
failed and until there is direct observation we
understand nothing. To my mind, failure to observe dark
matter may be indicative that we do not understand
gravity as well as we think we do.
QUESTION:
Bowling ball vs.
billiard ball
What would ultimately travel further, a 12 lb bowling ball or a standard
billiard ball: if thrown/rolled with the same amount of physical strength/athletic ability, at ideal trajectory for each, on a frozen lake (not without small imperfections but as close to perfect as could be found naturally), under normal atmospheric conditions (Wisconsin in the winter), with little to no wind? Would temperature play a roll in determining a winner? Would the small difference in friction coefficients be significant? Any other variables Etc...
My brother and I have been debating this for years while ice fishing.
ANSWER:
(You might not want to read this rather long
introductory paragraph if you are just interested in the
final answer.) This problem was very interesting to me
because I initially thought it was pretty easy to
understand. The billiard ball starts out with much greater
speed than the bowling ball, jumps into the lead.
Neglecting air drag is usually what is done initially in
considering such problems and there has to be some
kinetic friction for the balls sliding on the ice; but
because the kinetic friction is proportional to the
weight, each ball experiences the same rate of change of
its speed as it slows. Therefore the billiard ball is the
obvious winner. When I initially solved the problem this
way using reasonable numbers, I found that the bowling and
billiard balls had initial speeds of 14 mph and 81 mph,
respectively; the bowling ball went 686 ft and the
billiard ball went 21,700 ft, a little more than 4 miles! When I
was writing the final paragraph I discussed the possible
effects of two things I had neglected, rolling of the
balls rather than sliding and the possibility of air
drag. Just for fun I estimated the effect of air drag and
discovered that at the beginning, particularly for the
billiard ball, the drag force was considerably larger than the
sliding friction. I realized that I had to redo the whole
problem, now much more difficult and mathematically
involved than originally. Because the air drag estimate I
used is valid only if you work in SI units, I will do
that throughout and convert to imperial units when needed.
The mathematics gets very complicated and I will only
sketch my calculations and whoever is interested can fill
in the gaps.
First I will tabulate quantities needed for the problem:

We need to
quantify what is meant by "…the same amount of
physical strength/athletic ability…" I will
assume that the thrower exerts a constant force F
over a distance d on both balls. Physics says
that the result is that a kinetic energy K=½mv^{2}
is acquired where m is the mass of the ball and
v is the speed it has when it leaves the
thrower's hand; this kinetic energy is equal the the work
W done which is W=Fd; that is, Fd=½mv^{2}.
If you solve this equation for v you find the speed,
v=√(2Fd/m). I will use
Fd=110 J (about 25 lb over 1 m) because this
value would give the billiard ball a speed of about 80 mph
which seemed reasonable to me.

The masses of the balls are
0.17 kg=6 oz for the billiard ball and 5.4 kg=12 lb for the bowling ball.

I will use
a coefficient of sliding
friction for each ball which is very small, μ=0.01.
Each ball experiences a frictional force of f=μmg=ma
(Newton's second law) where a is the
acceleration and g=9.8 m/s^{2} is the
acceleration due to gravity.

For the air drag
I will use the approximation D=¼Av^{2} where
A
is the area presented to the oncoming air (πR^{2})
and v is the speed. The radii of the balls
are 0.0286 m and 0.108 m. As noted above, this
equation is only valid for SI units because the
factor ¼ includes things like air density, drag coefficients,
etc.
We are now ready to
start doing the physics. Start with Newton's second law
ma=m(dv/dt)=¼Av^{2}μmg
dv/dt=Cv^{2}μg
(C=¼A/m).
This equation is
integrable. The result is
t=tan^{1}[v√(C/(μg))]/√(Cμg)_{v(0)}^{0}.
Evaluating this
expression between the limits v=v(0) to v=0
and inverting to solve for v(t) is
straightforward but messy. The essential results are
illustrated in the two graphs below. The effect of air
drag on the bowling ball is modest but noticeable,
reducing the time to stop by about 10 s. The effect on
the billiard ball, though, is enormous; because of its
high speed at the start, it is slowed much more than due
to sliding friction alone. Still, it stayed in motion
about 20 s longer than the bowling ball and was going
faster most of the time, so certainly went farther before
stopping. Still, it would be useful to find the position
as a function of time to determine how far each ball
went.
To find the position, x(t), is
straightforward but again messy: integrate the
differential equation dx/dt=v(t)
where v is the function being plotted in the
graphs above. The results are shown in the graphs below.
The bowling ball stops at t=54 s at a position
x=158 m=518 ft; the billiard ball stops at t=77
s at a position x=535 m=1755 ft. The winner is
the billiard ball which goes more than three times
farther than the bowling ball! I believe that regardless
of any details of the friction and drag forces or how we
choose the quantity Fd, the billiard ball will
always go farther if the initial kinetic energies are
equal.
Finally, I want to discuss rolling
vs. sliding.
It is really hard to get an object rolling without
slipping on a very slippery surface. A sliding ball will
not roll until it is nearly ready to stop, and that would
have very little effect on my calculations. Also, rolling
friction is velocityindependent and proportional to the
normal force so it would affect both balls the same.
QUESTION:
How can a photon have the properties of a wave if it is not in an element? It is said to be in a vacuum so ostensibly it is not in an element, but if it is a wave it must have an element. Has science explained this or is it still an unsolved question?
ANSWER:
You have struck on one of the most important realizations
in physics: light does not need a medium to move through,
it moves through a perfect vacuum. Other waves we know
need a vacuum: sound through air (or just about any other
medium), water waves could not propogate without water,
waves on a guitar string could not exist without the
string, etc. But light is unique and does not
require a medium. The photon is just the smallest
possible piece of light and has wave properties. No
unsolved question here!
QUESTION:
I have a question about gravity. In order to general relativity mass bends the space. I saw a simulation of gravity that a man put a mass on a surface like silk. that the mass bends the silk. and we put another mass they attract to each other. that 2D silk is an examinition of our #d world. I wanna know why masses attract to each other. is it becuase of bending space?
ANSWER:
That simulation is a good way to understand how space
bending can alter how objects move but keep in mind that
it is only a cartoon to illustrate the warping of
spacetime. See my faq
page.
QUESTION:
If repulsion forces between protons is high why hasn't the nucleus exploded
ANSWER:
It is true that the Coulomb force experienced by protons
in a nucleus is very strong. However, the nuclear force
at very short distances is stronger and wins the
"tugofwar" and holds them inside. However, this strong
force is also a very shortrange force and if you pull a
proton just outside the nucleus, it will be repelled and
fly off. It is also interesting to ask why the nucleus
doesn't collapse if the nuclear force is so strong. The
nuclear force becomes repulsive if the protons get too
close together; this is called the saturation of nuclear
forces.
QUESTION:
What would be a ballpark estimate of all the mass converted to energy (lost) by all the fusion reactions in all the stars that are or have ever been in the last 13 billion years (lots of assumptions)?
Would this lost mass effect the overall gravitational attraction of the universe?
Would it be significant enough to be a factor in the current acceleration of the universe?
ANSWER:
I would not presume to make even the roughest estimate
you ask for. However, it is not relevant to the questions
you seem to be interested in. According to general
relativity, gravitation is caused by energy density, not
specifically mass. (Don't forget, mass is just a form of
energy, E=mc^{2}.) So changes of mass
will affect the details of the gravitational field in
some volume of space but if you get very far from that
volume, the gravity you see originating from that volume
will not change.
QUESTION:
Our 5yearold niece wants to know why it's so hard for her to ride her Big Wheel uphill (3% slope) on her family's driveway. My BSNRN wife would like a friendly, informative way to break that down for her (Daphne), including whether her weight (50 pounds), her position on the Big Wheel, the coefficient of friction on said drive wheel, and other factors play a part. Said aunt would also like a kidfriendly way to calculate the horsepower necessary for her to ascend that grade on her Big Wheel for perhaps 100' (the driveway's length). No, this is in no way homework. Except it's at their home. And, well, it is work.
ANSWER:
A little tough for a
fiveyear old! Here, maybe, are some suggestions:

Get her to
understand Newton's first law: an object at rest or
moving with constant speed has all the forces on it
cancelling each other out.

Get her to
understand Newton's second law. If the forces are not
cancelling each other out, the object will start
moving and continue to speed up in the direction of
the uncancelled forces.

For example,
have her imagine sitting on the trike on the
incline with her feet off the pedals or the
ground. She will start moving down the incline.
What is happening is that her force, which is
still straight down, must have a little bit of
itself pushing her down the incline. (This is
basically a component of the weight but the
concept of vectors and their components is likely
too hard for a fiveyear old to comprehend.)

Finally, if, on
the incline, there is a piece of the weight which she
will have to compensate for in order to keep that
force from making her go down the hill: she does this
by pedaling to create a force up the incline: the
force she generates must be greater or equal to the
piece of the weight pointing down the incline. On
level ground there is no unbalanced weight so the
pedaling is much easier.

Friction, I
would opine, is an unnecessary complication.

If she were
heavier, the piece down the incline would be bigger
so it would be harder to pedal.

(This part is
for you, the little girl will not get it.) Your
question about the horsepower is incomplete because
to calculate the power you need to know the time it
takes her to go the 100 ft. Suppose she maintains a
speed of about 2 ft/s. Now, the component of a 50 lb
weight on
a 3% grade is about 50x0.03=1.5 lb so the work done
is 1.5x100=150 ft·lb. The time it takes her is
(100 ft)/(2 ft/s)=50 s, so the power is 150/50=3 ft·lb/s=0.0055
hp.

Friction has
little to do with it because the friction is not
significantly different on level ground vs.
the incline. Position on the trike plays no role
since any change would be the ease of pedaling which
would be the same in either case.
QUESTION:
what would a light/sound wave look like in 3 dimensions, all diagrams are in 2d and it just came to mind that it wouldn't be like that in real life
ANSWER:
Here is a cylindrical wave. The blue surfaces represent
wave fronts.
QUESTION:
I am currently doing an experiment on the metronome pendulum. As you know a metronome pendulum has two types of pendulum which is the inverted pendulum and simple pendulum. It is called as doubleweighted pendulum. This type of pendulum has two weight which is one above the pivot and one below the pivot. In my experiment I want to find the relationship between the varying mass of the top mass of metronome pendulum and its angular frequency but I cant seem to find any formula that describes my metronome pendulum. What formula for angular frequency should I use because from my data, I will be looking for its period of oscillation. There are too many angular frequency formula for different types of pendulum so I am confused on which formula I should be using.
ANSWER:
I will show you how to get a general answer. Then you can
apply it to the conditions of your experiment. Newton's
second law for rotational motion is τ=Iα=I(d^{2}θ/dt^{2}) where
τ is the net torque about the pivot (the blue cross in the figure),
I is the moment of inertia about the pivot, α is the angular acceleration, and
θ is the angle of the system at some time
t. If the system is a pendulum with no forces
exerting torques except weights (vertically down), all torques are
proportional to sinθ where here
θ is the angle relative to the vertical.
Now our equation may be written as
(d^{2}θ/dt^{2})=(τ/I)sinθ
where
τ is the torque about the pivot when
the pendulum is horizontal (θ=π/2). The
negative sign results from the fact that the angular
acceleration and the angle are always opposite, a
restoring torque. This equation is extremely difficult to
solve, but if the angle is small (much smaller than 1
radian), you can use the small angle approximation, sinθ≈θ,
which gives us
(d^{2}θ/dt^{2})≈(τ/I)θ
This is the simple harmonic oscillator equation for an oscillating
system for which we know the angular frequency, ω=√(τ/I).
So, let's apply it to your situation as illustrated in my
figure. There are two point masses and the mass of the
rod itself which I will take as m. The rod has a center
of gravity (orange cross) a distance of L_{3}
from m_{1}as shown; if your rod is
uniform, L_{3}=L/2. (If the
mass of the rod is much smaller than both the point
masses, you can approximate m as zero.) Now,
since all weights are of the form W=mg, you can
write τ:
τ=g(m_{1}L_{1}m(L_{3}L_{1})m_{2}L_{2}).
The moments of inertia for the two point masses are
I_{1}=m_{1}L_{1}^{2}
and I_{2}=m_{2}L_{2}^{2}.
The moment of inertia for the rod is trickier, so to
avoid a lot of messy algebra I am going to assume a
uniform rod where L_{3}=L/2. I
find
I_{rod}=(m/(3L))(L^{3}L_{1}^{3})
The total I is just the sum of the three moments
of inertia. So, finally,
ω=√[(g(m_{1}L_{1}m(L/2L_{1})m_{2}L_{2}))/(m_{1}L_{1}^{2}+m_{2}L_{2}^{2}+(m/(3L))(L^{3}L_{1}^{3})]
And, if m can be neglected,
ω=√[g(m_{1}L_{1}m_{2}L_{2})/(m_{1}L_{1}^{2}+m_{2}L_{2}^{2})]
QUESTION:
So I have just thought about this, let's just say that hypothetically you were able to fall through the earth completely. I was wondering what the gravity would do to you if you went in one side and came out the other
ANSWER:
See the
faq page.
QUESTION:
If everything is expanding, and every galaxy is moving away from the Milky Way, why is the Milky Way and Andromeda predicted to collide in 3.5
billion years?
ANSWER:
See an
earlier answer. See a simulation
video.
QUESTION:
this is for a horror book I am writing. how fast can a bat fly, if it weights 4500 pounds and is 25 feet 6 inches tall? and in 1 flap of its wings, can rise a rise a quarter, of a mile? also how wide would its wingspan be?
ANSWER:
The best I can do is to estimate how the wings would have
to be if the ratio of weight (W) to wing area (A)
is the same as a real bat. I find that a typical value
for that ratio for bats is about R=W/A=20
N/m^{2}=0.42 lb/ft^{2}.
So, for your case, A=W/R=4500/0.42≈10,000
ft^{2} or one wing has an area of about 5000 ft^{2}.
If I model a wing to be a triangle with base B=25 ft and
a width H, 5000=½BH=12.5H
or H=400 ft, a wingspan of about 800 ft.
Since it has the same R as for a real bat, it
should be able to fly as fast as a real bat which can be
as high as 100 mph.
QUESTION:
I am a part time instructor and one of the things I teach is rigging. An
important part of the class is calculating sling stress. I was asked a
question that I couldn't quite answer. I don't believe it was my high
school or college physics teachers that failed me rather the more than two
decades since I sat in a physics class. The question relates to vector
forces. If I am hoisting a load of a given weight (say 2000 lbs) and I have
two slings that are at a given angle (say 45 degrees) from the load. Each
sling would be carrying half the load or 1000 lbs straight up. However, the
sling in this example would have 1414 lbs of force (inverse sine if I
remember correctly). The question I was asked was regarding a crushing or
compression force. Since each sling is carrying 1000 lbs vertical load that
means it has 414 lbs of force perpendicular to the vertical. Does that mean
that there is 414 lbs of "crushing" force or is it 828 lbs since each sling
is pulling towards the center of gravity.
ANSWER:
After an email exchange with the questioner, I was able
to determine that "sling" is just a rope or a chain or a
string etc. In the diagram I have shown all the
forces on the load, the tensions in the slings,
T_{1} and T_{2}
and the weight of the weight, W.
The magnitudes of the tensions are equal, T_{1}=T_{2}
=T. Also shown are the horizontal (x)
and vertical (y) components of the tensions;
they are T_{1x}=Tcosθ,
T_{2x}=Tcosθ,
T_{1y}=T_{1y}=Tsinθ.
So the equilibrium equation is 2TsinθW=0;
therefore T=W/(2sinθ).
So, for your example, sinθ=cosθ=1/√(2),
W=2000 lb, and T=1414 lb. Now, your
interest is in the horizontal components; each is Tcosθ=W/(2tanθ)=1000
lb "crushing force" on each side. Your main error is that
you treated the vector forces as scalars, you assumed
T=T_{y}+T_{x} where in fact T=√(T_{y}^{2}+T_{x}^{2}).
QUESTION:
Since everything in the universe is always moving, is the idea of being still/idle/stationary just an illusion?
ANSWER:
I would not put it like that. If you are in a system
where Newton's first law (if the net force on an object
is zero, it will remain at rest or moving with a constant
velocity) is correct, you may think of yourself at rest.
This is called an inertial frame of reference. But the
catch is that this is not the only frame of reference in
the universe where Newton's first law is true, any other
frame which moves with a constant velocity relative to
yours is also an inertial frame. In other words, there is
no such thing as absolute rest. A frame accelerating
relative to yours, however, is not an inertial frame.
QUESTION:
I am a cancer patient about to receive Y90 treatments. If Y90 has a half life of 64 hours: 1. When does that time clock begin running and 2. How can they ever have a full strength dose on hand at any given time?
ANSWER:
^{90}Y is the decay product of
^{90}Sr
which has a half life of about 29 years. 90Sr is an
abundant byproduct of the fission of uranium and
therefore available in the waste of reactors. When the
^{90}Y is needed it can be chemically separated
from the ^{90}Sr. I am sure that the hospital
cannot keep a supply on hand and will have it delivered
when it is needed for your treatment. It does not matter
how long it has been since it was separated, only that
the radiation level be correct for the prodedure. It
probably arrives with a level too high and they wait for
it to be at the needed level.
QUESTION:
I am struggling with the concept of relativity, as it relates to time. Not in terms of the underlying principle or the mathematics. These have been confirmed experimentally and are now in everyday use through technologies such as satellite navigation systems. My concern relates to the way that relativity treats time as a variable rather than an absolute quantity. It seems to me that we can only measure or experience time by reference to some physical process or change, whether that is the oscillation of a quartz crystal or the lifetime of a muon! Any experimental proof of time dilation as a result of the effects of either velocity or gravity really just relates to the way that we experience the passage of time and does not exclude the possibility that for the universe, as a whole, time is absolute and unchanging, passing in a constant fashion regardless of relativity and whether we measure it or not. Therefore, although relativity theory works in practice, this is only because it deals with our limited understanding of, and ability to measure and/or experience, the passage of time. Is this nonsense or is it a possibility? It follows from this reasoning that we can never prove experimentally that time either speeds up or slows down, only that the rate of how we measure or perceive time can vary. It also implies that time (and indeed space) are concepts that we will never fully explain as they ultimately cannot be qualified or quantified by measurement or mathematics.
ANSWER:
You are correct that "…time is absolute and unchanging,
passing in a constant fashion…" but only in your
frame of reference; any clock at rest relative to you
runs at a constant rate. But, what special relativity
tells us is that clocks not at rest relative to us do not
run at the same rate as ours. I have always felt that the
best way to convince someone that this is the case is the
light
clock. In order that this example is convincing to
you, you must accept that the
speed of light is a
universal constant in all frames of reference; but this
is an experimentally wellverified fact and also is a
consequence of the principle of relativity—that the laws
of physics are the same in all inertial frames of
reference.
QUESTION:
SINCE FREE FALL DONT NATURALLY OCCUR ON EARTH BECAUSE FLUID FRICTION IS PRESENT IF YOU
DROP A BOWLING BALL AND A BASEBALL OFF A TOWER OF PISA WHICH WOULD LAND FIRST
ANSWER:
The (upward) drag force on a falling sphere of radius
R and speed
v in air
at sea level may be approximated as F=¼πR^{2}v^{2}=0.79R^{2}v^{2}
(this is correct only in SI units); the (downard) force of
the weight W on an object of mass m is
W=mg. The net force is the mass times the
acceleration a (Newton's second law), ma=mg+0.79R^{2}v^{2}.
So as the object falls it goes faster and faster until
v is large enough that it falls with a constant
value speed v_{t} (a=0), v_{t}=[√(mg/0.79)]/R.
For a baseball, m=0.144 kg and R=0.037
m, so v_{t}^{baseball}=9.28 m/s.
For a bowling ball, m=8 kg and R=0.12
m, so v_{t}^{bowling ball}=83.0
m/s, nearly ten times greater than the baseball. So the
baseball stops accelerating sooner than the bowling ball
and loses the race to the ground.
QUESTION:
My 8 year old asked me if the cycle of the universe expanding and contracting will ever end?
ANSWER:
What makes you and your son think that the universe is in
such a cycle? The ultimate fate of the universe is one of
the most important questions in cosmology and nobody
knows the answer. I believe most cosmologists believe
that the universe will continue expanding forever.
Eventually the supply of hydrogen and helium in the
universe will be exhausted (fused into heaviers elements)
and stars will no longer exist. The universe will be
dominated by black holes which will eventually decay away
by Hawking radiation, so the universe will be a very
dilute, cold, dark place filled with radiation. To get
more information, check out the
Wikepedia article on the fate of the universe.
QUESTION:
My son has a question.
A shadow is long or short, wide or narrow. He thinks this seems like area (taking up space). But the real question is... can you have a shadow without matter? Because if you can't have a shadow without matter, then doesn't that mean the shadow is actually matter? (I have to ask because I'm not sure how to answer this; I suppose the shadow is actually the part of the matter that doesn't have light reflecting off of it...? Or how do I answer this for him?)
ANSWER:
A shadow is
not matter;
it is not anything, it is a region in which light
illuminating something is blocked by some obstruction.
You might say that a shadow is the lack of anything,
light which might have otherwise illuminated where it is.
And there are degrees of "shadowness". If all the light
illuminating an area is blocked by the obstruction, it is
called an umbra; if only part of the light is
blocked it is called a penumbra. (Umbra is Latin
for shadow.)
QUESTION:
U recently answered a question regards Earth's precession = 26,000 yrs. Can u show what specific equation(s) used to yield this time? I doubt that a single equation was used.
ANSWER:
Just like the precession of a top is due to the torque
exerted by its own weight, the precession of the earth is
due to torques on the earth, mainly by the sun and the
moon. Your equations may be found in this
Wikepedia article.
QUESTION:
On an ordinary day as one goes upward from the surface of the ground the electric potential increases by 100 volts per meter, this means that outdoors the potential of your nose is 200 Volts higher than that at your feet. Then why is it that none of us get a shock when we go out in the street?
ANSWER:
As is often the case, the most lucid answer to your
question can be found in the
Feynman Lectures.
QUESTION:
Why doesn't the earth just stop spinning and orbiting don't you need energy for motion where does the earth gets it's energy and is the earth in perpetual motion?
ANSWER:
You are hitting one of Isaac Newton's most important
discoveries—inertia. It says that you do not need
to exert a force (torque) to keep an object moving in a
straight line (rotating about an axis) to keep it going;
this means, also that the energy it has by virtue of its
motion will not change. So the rotating earth keeps
rotating since there are no torques on it*. The earth is
in a circular orbit† around the sun and so its orbital
motion keeps going since, even though there is a force on
the earth, there is no torque.
*Actually, the
moon does exert a torque on the earth
which is causing the spinning to get smaller, but the
change is so tiny that it takes millions of years to be
significant.
†The earth's orbit is
not exactly circular. When it moves closer to the sun it
speeds up, when it moves farther from the sun it slows
down by exactly the same amount, so the orbit just
continues forever.
QUESTION:
With the devastating tragedy that happened in Lebanon, there has been some videos surface of the explosion. One such
video is of some ladies in a shop, the pressure wave hits the shop and the lady outside the shop doesn't get thrown back but inside the other lady does. What is the explanation here??
ANSWER:
Note that the glass doors are closed. Outside the front
of the pressure wave hits, but it takes some brief time
to reach its maximum. However, the pressure difference
between inside and outside is not large enough to break
the glass until it is near its maximum, so the pressure
difference changes much more quickly so the air rushes in
quickly. I will admit that this is just an educated
guess!
QUESTION:
If electricity always take the path of least resistance, why does a lightening discharge form a zigzag path to earth and not straight down which would be the shortest distance?
ANSWER:
Because air is not just a uniform homogeneous medium.
There are local fluctuations in temperature, humidity,
density, wind speed, and composition which makes the path
of least resistance at any instant, not a straight line.
QUESTION:
how momentum is conserved of both objects with same masses moving opposite direction to each other collide to a point and at same time both objects chemically combined to each other?
ANSWER:
In an isolated system, linear momentum is always
conserved; this is because the definition of linear
momentum is designed to be conserved in the absense of
any external force. This is also true in special
relativity where the linear momentum is not p=mv but
rather p=mv/√[1(v^{2}/c^{2})]
where c is the speed of light. In the event of
chemistry going on, there is either energy gained
(endotermic) or lost (exothermic). If lost, the momentum
of the radiation has to be taken into account. If gained,
the energy will be taken from the kinetic energies of the
incident objects but momentum will be conserved.
QUESTION:
Does precession and nutation ( or anything else ) affect the axis of the earth?
ANSWER:
Precession of the earth's axis is caused mainly by
torques exerted on the earth by the moon and the sun. The
period of precession is 26,000 years. But precession does
not cause the actual rotation axis to change. What causes
the axis to change (such that the geographical positions
of the poles would change) is thought to be three things:

glacial ice
melting and sending its water to the oceans,

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

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

the speed of
light in vacuum, c,

the universal
gravitational constant, G,

the rationalized
Planck's constant, ℏ
,

the Boltzman
constant, k_{B}, and

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

length,

mass,

time,

temperature, and

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

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

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

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

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

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

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

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

the ball's
weight W,

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

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

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