One additional detail:
electromagnetic and weak nuclear forces have
been unified, shown to be different faces of
a single force, the electroweak interaction.
QUESTION:
Can gravity slow down the atom ?
ANSWER:
You are asking, essentially, if atoms feel
gravitational forces. Anything which has
mass will feel the gravitational force and
atoms have mass; so, yes, atoms experience
the gravitational force and thus can be made
to slow down. For example, if you shoot an
atom straight upwards, it will slow down
just like a baseball thrown upward with the
same speed. The very fact that we have an
atmosphere is proof that gravity is acting,
keeping all those atoms from speeding off
into space.
QUESTION:
What is the smallest change in frequency
within the visible light spectrum that the
human eye/brain can discern? My new TV says
it can show over a billion colors. Can
humans even see 1 billion colors!
ANSWER:
The way a tv works is by mixing the three
primary colors, red, green, and blue, in
varying amounts. There are an infinite
number of possible ratios. All the colors
you can see are not all the possible
wavelengths you can see (also an infinite
number) but mixtures of many many
wavelengths. When the manufacture refers to
"billions" he is probably referring to how
accurately the electronics can mix the
three. For example, if the relative amounts
of the three were 10, 7, 3, the electronics
would have an uncertainity in how accurately
they could set these numbers; maybe the
numbers could only be set to a 0.01%
accuracy, the ten would be 10±0.0001. Then
you could estimate the number of
possibilities you could have over the whole
visible spectrum.
QUESTION:
I have been thinking about this for
countless hours and though not a physics
student, and haven’t taken any classes of
the sort I have wanted to know this
question: we know black holes are brought
down to a point of singularity. So could and
atom being so small do the same? And if this
is true would that be the last atom we see
on the periodic table?
ANSWER:
The force which holds a star together is
gravity; a normal star does not contract to
a black hole because it has collapsed to the
point where burning hydrogen (fusion) in the
star heats it up so that the pressure
balances the gravity trying to push it
smaller. As the fuel runs out, the star
starts getting smaller. At this stage,
various things can happen, depending on
various conditions, mainly the total mass of
the star; one possible thing is eventually
collapsing all the way to a black hole. An
atom is a completely different kettle of
fish. We should really look at the nucleus
since that is where nearly all the mass is.
Unlike a star, energy is not being produced
in the nucleus, its energy remains constant.
An atomic nucleus is held together not by
gravity but by the strong nuclear force.
Unlike gravity which is a very longrange
force, the strong interaction is very short
range, so each proton and neutron see only
their near neighbors, not all the other
particles in the nucleus. The result is that
the heavier a nucleus becomes, on average,
the less bound it comes so its tendency is
to break apart, not collapse. By the way,
the nucleus does not collapse because the
nuclear force at extremely short distances
becomes repulsive, not attractive.
QUESTION:
How much does a thought weigh? And how
much does one molecule of carbon dioxide
weigh? I want to compare the two.
ANSWER:
A molecule of CO_{2} has a mass of
about 7.3x10^{26} kg (about 1.6x10^{25}
lb). A "thought" is a collection of billions
of electrical impulses and chemical reations
in the brain. Even if these could be
assigned a "weight", which they can't, all
"thoughts" are not the same and would have
different weights.
QUESTION:
Franklin first coined the term negative
and positive in 1750 and Volta invented the
battery in 1800. Did Volta assign a positive
and negative end to the battery? Or did he
think the negative and positive applied to
electrostatics and not "flowing"
electricity. You see where I am going with
this. I need to go up the chain to J.J.
Thompson to see where the goof up was on
adoped circuit diagrams. The explanations I
have read are non sense. They do not
actually describe why the convention had to
be reversed. All the examples I read use
circular reasoning and thus far no smoking
gun of how and when the goof up happened. I
thought if I could start with Volta I might
be able to unravel it.
ANSWER:
I really don't understand your perception of
"goof up". There are two different "kinds"of
electric charge. Coulomb's law shows that
the magnitude of the force F
which a point charge q feels due to
the presence of another point charge Q
is proportional to the product of the
charges Qq and inversely to the
distance r between them, F∝Qq/r^{2}.
Also, if the charges are of the same "kind"
they repel and of different "kind" they
attract; since attract/repel result in
exactly different directions for the vector
force if we randomly assign signs to the two
"kinds", we can write F∝1_{r}Qq/r^{2}
where 1_{r}
is a unit vector in the direction of the
vector r from
Q to q, 1_{r}=r/r.
This proportionality does not
depend on which "kind" we call positive and
which we call negative, there is no right
and wrong choice for the sign for an
electron. Franklin made a random choice
which resulted in electrons as having
negative charge; he might have made a
different choice if he knew electrons
existed and were usually the charge carriers
in currents (nobody knew then), but it was
not a wrong choice. Later, when currents and
magnetic fields were studied, all of
electromagnetism simply continued the
original choice with no problem whatever.
All this means that positive electric
currents in wires actually are almost
electrons moving in the opposite direction
which is endlessy confusing to students. I
really don't know anything about Thompson
correcting "a goof up" in circuit diagrams
other than pointing out that currents shown
moving in one direction (from + to  battery
terminals) is actually electrons moving in
the opposite directions. No physics was
changed.
QUESTION:
What will be the motion of the earth if
the sun disappears after t=0 sec?. I was
asked this question in an exam and the
answer is not tangential to the motion
ANSWER:
It is assumed, although never very
accurately measured, that the speed of
gravity is the same as the speed of light.
Therefore, the earth would continue in its
orbit until t=8.3 minutes, the time
it takes light from the sun to reach the
earth. At that time the earth would go dark
and proceed in an approximately straight
trajectory.
QUESTION:
How do gravitons that exit a certain
mass cause other objects to be attracted to
that mass. I can imagine Einstein's idea of
curved space being the cause for the seeming
attraction, but what does that have to do
with gravitons?
ANSWER:
There is no successful theory of quantum
gravity; therefore gravitons are
hypothetical particles, never observed, just
an idea. Gravity is the only one of the four
fundamental forces of nature which has not
been quantized. See an
earlier answer and the links in that
answer. You will see that the theory of
general relativity is not unambigously
described by the notion of warped spacetime
but is also a field theory. Any field theory
should be able to be quantized; in a
quantized field theory the "messenger" of
the field would be called a graviton, just
as the messenger for the electromagnetic
field is the photon, and is the gluon for
the strong nuclear field.
QUESTION:
Why does a rounded pencil I put on a
flat ground make a small harmonic movement
before stay steady?
ANSWER:
One side of the pencil is heavier than the
other. For example, if you had a uniform
ball and glued a weight to its surface, it
would always rotate so that weight would go
to the ground. With some pencils the lead in
the middle is not exactly on the axis. The
"lead" is actually graphite (carbon) with a
density of 2.26 g/cm^{3}; wood has a
density of about 0.6 g/cm^{3}. So
the lead will seek its lowest p
ossible
point it can possibly find. The figure
illustrates this situation. The weight of
the lead will cause the pencil to roll to
the right; but when it gets there it will be
moving and so the pencil will continuing to
roll until the lead stops and then begins
rolling to the right. It will oscillate
forever if there is no friction; but there
is so it will quickly damp out with the lead
directly below the center axis of the
pencil.
QUESTION:
I am a glass artist. When I fume gold
(vaporize in the torch flame) and it
collects onto the borosilicate glass it
appears pink/purple. If I encase this gold
with clear, it then looks green. I can
throughly manipulate the glass and maintain
the green color. I read that Faraday
obtained green gold in a thin film with
stress on the film. I also fumed gold onto a
rod and stretched it at a fairly cold glass
working temperature to stress the gold and
avoid annealing the gold. The gold took on a
green appearance. I have been doing this for
over 25 years and have not been able to find
out why gold turns green when encased (or
stressed like with the stretch) I am asking
a physicist because this has to do with
light and gold, plasmon resonance etc are
pretty heavy topics. If you can answer this
question I’d love to gift you one of my art
works! Why does a thin film of gold turn
green when encased in glass?
ANSWER:
I am not sure why it is pink/purple when you
first deposit it. One video suggested that
has to do with the amount of oxygen in the
film; it is also possible that it is a
thinfilm interference effect where the
thickness of the film is an integral
multiple of wavelengths of the color you are
seeing and the light of that wavelength
which is reflected from the front and back
of the film add up to become the dominant
light you see (the same thing going on in
the pretty colors you see on an oil slick).
There such things which use thinfilm
interference, for example the magenta
interference filter shown in the figure.
Now, to
the green issue. Why is gold the color we
see with just a block of gold? The reason is
that when white light strikes the surface,
red/yellow/orange are preferentially
reflected back making the gold look golden.
But some of the light will continue into the
bulk of the gold, and, coincidentally,
red/yellow/orange are preferentially
absorbed. This means that once you get to a
certain depth in the block the only light
surviving is the blue/green part of the
spectrum; eventually all the light gets
absorbed and you will see no light coming
from the back side which originated on the
other side. However, if you have a "block"
which is thin enough, smaller than the
distance where everything is absorbed but
thick enough that all the red/yellow/orange
has been absorbed, the film will look
blue/green. I am quite sure that this is
what is going on but, since I don't know the
geometry of your art, I do not have a
detailed explanation for your particular
situation; e.g., does it look green
when seen from any direction? Maybe if you
sent a photo or two I could go farther.
I have
a question for you. I am a
stainedglass hobbiest and find that red
glass is often more expensive than other
colors. I have been told that the reason is
that gold is required to get a rich, pure
red glass. Do you know if that is true
and/or why?
QUESTION:
A thought occurred to me. From what I
know about particle accelerators, scientists
speed up accelerator particles of matter and
crash them into slower moving or stationary
particles of matter so that they can examine
the breaking apart of all into other
elementary particles. This then allows them
to find more and more basic particles of
matter. I have read that the speeds these
accelerator particles are hoped to reach is
the speed of light, hence the effort to
build more powerful particle accelerators.
So, why not simply have photons of light
(that are travelling at the speed of light
since they are the elementary particles of
light) crash into those other particles of
matter? Why use so much energy trying to
push accelerator particles to reach the
speed of light, when we have light in the
first place and photons already moving at
186,000 miles per second?
ANSWER:
It is not the speed of the incident
particles which matters, it is their energy.
There are no good available sources of
highenergy, highintensity photons. Also,
photons interact only via the
electromagnetic force and we often want
projectiles which interact via the strong
nuclear interaction. So the accelerators are
actually poorly named because they do almost
no change in velocity when they increase
their energy by a large amount; I have
maintained that they would better named
energizers. For example, a proton with speed
99.999% the speed of light accelerated to
99.9999% the speed of light increases its
speed by 0.0009% but its energy increases by
216%.
QUESTION:
A photon hits an atom perpendicularly to
its speed v and it is absorbed as is known
immediately. So it can not act after the
time when it is not perpendicular (e.g. the
force is 0 after that). Certainly all its
energy goes to the atom when it is
perpendicular to v.
But the
impact (absorption) applies a force on the
atom and it is postulated that a
perpendicular force can not do work. So work
has not been done on the atom. Consequently
its energy can not change. If all photon
energy goes to increase internal energy of
an orbit electron from Eo to E1 how then is
the DIRECTION of velocity of the atom
changed? Isn’t this a contradiction?
ANSWER:
Your first paragraph really makes no sense.
Talking about "perpendicular" in a collision
really has no meaning; I think what you mean
is a headon collision, one where the recoil
of the atom is in the same direction as the
incident photon. In a case like this you do
not want to think about force and work.
Rather you should ask if energy and momentum
are conserved; if so, you should solve your
problem that way. Suppose that the system we
are interested in is the photon (mass zero,
energy E=hf, and momentum p=hf/c)
plus the atom (rest mass M, energy
E=Mc^{2}, and at
rest p=0). Because there are no
external forces on this system, its total
energy and momentum must remain constant.
After the absorption, the mass of the
excited atom is M', its kinetic
energy is K, and its momentum is
P'; there is no need to think about
what goes on microscopcally inside the atom.
So we have two equations:
E_{1}=E_{2}
or
hf+Mc^{2}=M'c^{2}+K
p_{1}=p_{2}
or
hf/c=P'.
The
notions of force and work are seldom useful
when looking at atomic sizes. In this case,
the photon carries energy into the atom so
the atom's energy has to increase.
QUESTION:
If l video an energy source such as a
manmade Ultraviolet Curing light, can
watching the video hurt your eyes? And is
there a way of capturing this light to make
it even though it is invisible?
ANSWER:
The video camera is not sensitive to UV
light and therefore records none. The
monitor is designed to emit almost entirely
visible, not UV light and therefore radiates
none. So no, the video will not hurt your
eyes. There are certainly detectors which do
detect UV radiation but it is not really
'capturing' it in the sense that you could
'make' it.
QUESTION:
This question has resulted from
considerable backandforth communications
between me and the questioner. The essence
of the question is this: Two cylinders are
rigidly attached to each other. The smaller
has a radius R and its center axis
(B in the figures below) is separated by a
distance d from the center axis (A)
of the larger. A torque is applied about the
axis of the smaller cylinder but exerts no
net force. Find the resulting torque about
the axis of the larger cylinder.
ANSWER:
The figure shows one possible way of
exerting a torque about point B. The four
tangential forces all have magnitude F/4
so that the torque is τ_{B}=RF.
There is a net force of zero. The
torques about the point A are:
τ_{1}=RF/4
τ_{2}=(d+R)F/4
τ_{3}=RF/4
τ_{4}=(dR)F/4
When
all four torques are added,
τ_{A}=RF=τ_{B}.
A
more general approach to the problem would
be to have a uniform tangential force all
around the perimeter of the smaller cylinder
as shown in the second figure. If we write
the magnitude of the vector dF
as dF=[F/(2π)]dθ,
the integral ∫dF from 0 to 2π
will be F and τ_{B}=RF.
Now the torque due to dF
about A will be dτ_{A}=r×dFdθ=[RF/(2π)][(dsinθ/R)+1]dθ.
Integrating this from 0 to 2π
yields, again, τ_{A}=RF
because ∫sinθdθ=0 over one
whole cycle.
There are two
important provisos here: there must be no
other forces present which exert torques
about axis A and the forces exerting the
torque about axis B must sum to zero net
force.
Sometimes I like to
interject examples of how science is done
and sometimes progress hits snags. Most
often preconceptions are the biggest
stumbling blocks. When I first started to
work on this problem I assumed that the
answer had to depend on d. As you
can see, this is not a particularly
difficult problem. I quickly tried the
general approach and got the answer as I did
above; but I was so sure that d
should be in there somewhere that I assumed
that I had done something wrong. So I did
the calculation for each quadrant
individually; these are tedious
calculations, not difficult but certainly
prone to algebra and arithmetic errors. I
spent hours! Finally, when I was sure that
everything was right I found
τ_{A}=RF.
That was when I tried the simplified
torque with four forces (above) to convince
myself that it was correct.
QUESTION:
I am trying to clarify in my head an
idea regarding electromagnetic force. It is
usually explained in very superficial terms.
One way to explain my question is this: I
understand that two positively charged
particles will repel each other, for example
two protons. They both have a positive
charge, but if they are at rest there will
be no magnetic field. Do electrical charges
repel like magnetic charges? Are they the
same thing? We measure magnetic strength in
one set of units such as Teslas but
electrical potential in another set of
units. I understand the strength of a
permanent magnet to be somehow related to
the arrangement of its electrical charges.
Is that correct? Is a proton at rest a
permanent magnet? Is there some simple
relationship between a static electrical
charge and a permanent magnet? I’m sorry the
question is so long but it is all part of
one big question.
ANSWER:
Your question violates site ground rules for
"single, concise, wellfocused questions" in
spite of your saying it is "one big
question"; even so, it is neither concise
nor wellfocused. I am sorry, but I cannot
give you an entire disquisition on
electromagnetic theory in a single concise
answer. You have a number of misconceptions.
First, there is no such thing as a magnetic
charge. I always like to like to refer to an
answer to a question I answered long ago:
Answer: The laws of
electromagnetism are perfectly symmetric: a
changing magnetic field causes an electric
field and a changing electric field causes a
magnetic field. The first of these is called
Faraday’s law and the second is part of
Ampere’s law. You seem to think that only a
permanent magnet is magnetism. In fact, any
moving electric charge causes a magnetic
field. The most common source of magnetic
fields is simply an electric current. Here
are some facts about electric and magnetic
fields:
• electric charges cause
electric fields,
• electric currents
(moving charges) cause magnetic fields,
• changing magnetic fields cause electric
fields, and
• changing electric fields
cause magnetic fields.
Magnetic
fields are not caused by a charge like
electric fields are, they are caused either
by electric currents (moving charges) or
changing electric fields. So then, what is a
permanent magnet? On the atomic level, think
of electron orbits as little currents and
electrons as tiny spinning charges, also
currents. In a material where all these tiny
magnets align with each other results in a
sum of all their magnetic fields. Regarding
your writing “but if they are at rest there
will be no magnetic field”, this has nothing
to do with any force they exert on each
other by virtual of their charges. Indeed
protons do have a tiny magnetic moment
because they can be roughly be thought of as
spinning, but those moments are very weak.
If you want to really understand all this,
you need to seriously study
electromagnetism.
QUESTION:
As solar energy is becoming a major,
maybe eventually the main source of
electricity, I wonder what the effects of
capturing and retaining significantly more
of the Sun's energy in the system of our
planet could be. Would the overall effect be
comparable to the warming effect of
greenhouse gases, where less energy is
radiated back to Space? What about other
forms of renewable energy, that ultimately
just harness the Sun's energy?
ANSWER:
In greenhouse warming, radiation is absorbed
by the earth and then radiated back into the
atmosphere; some of that energy goes back
into space and some bounces around in the
atmosphere, heating it. Factors, including
the amount of CO_{2} in the
atmosphere, determine the fraction of
reradiated energy which is trapped. Now, if
we capture this energy and store it in
batteries, it stores the energy in a form
which is not heat; then this energy is used
to drive cars, run our utilities, lift
material to build buildings, etc.,
all of which will convert a fraction of the
energy to heat, but a much smaller amount
than if all that energy were absorbed by the
ground and then was reradiated into the
atmosphere. And, using this energy instead
of fossilfuel burning will further help the
situation by not adding to the CO_{2}
in the atmosphere.
QUESTION:
I am trying to calculate the energy
consumed by an electric bike when it is
charging at home. It has a 48V, 14Ah battery
and takes about 6 hours to charge. Is the
electricity in my power bill going to be 672
watt hours (48 x 14) or ~4,000 watt hours
(48 x 14 x ~6)?
ANSWER:
I got exactly the same answer as you but
took a different path to get there. Since an
ampere is a Coulomb/second (C/s), 1 A·h=3600
C, so 14 A·h=5.04x10^{4} C. The
energy to move that charge across a
potential difference of 48 V is E=QV=48x5.04x10^{4}=2.4192x10^{6}
J=2.4192x10^{6} W·s(1h/3600 s)=672
W·h.
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 (1) the temperature of
the bar is at some constant value T_{1},
(3) 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(½nπx/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 equatòr (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 second’s 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:
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:
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:
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.