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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 dust-a 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.

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 (4-9.8)=-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 self-consistent.

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 Non-Central 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, well-focused 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
liquid-drop 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 5-6 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
proton-proton 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
$700-1,100 + for a system. The European
Union has developed standards for impact
protection pads for motorcyclist's. EN
1621-2:2014 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 G-forces. The rider was not
injured. Assuming that the rider suffered a
TRANSIMTED force of the no more than 12kN
(as per the EN 1621-2:2014). How much energy
in Kn did the airbag system absorb ? What is
the energy absorption capability difference
between the airbag and a EN 1621-2:2014 pad
with 12 kN transmitted energy ? Problems
1) I have been unable to determine the EN
1621-2:2014 IMPACT test force, I only know
the TRANSMITTED test force is no greater
than 12kN 2) Rider crash data uses
G-force 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 G-forces, 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, well-focused questions..." I can,
however, help you with the single question
of confusion between kN and g -force.
The g -force 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 g -force
is usually expressed in g s; for
example, because of some force you
experience an acceleration of 10 g s,
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 random-motion
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=√(2xP x2x0.0167/0.76).
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 =100-T .
I will neglect any edge or geometry effects
in the bar, in other words I will treat it
as a one-dimensional problem where the heat
flow vector in the bar is in the direction
of the bar and uniformly distributed across
the cross-sectional area. In that case the
equation for the rate R is

R =ΔQ /Δt=kA ΔT /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 steady-state 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 heat-flow 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 1-d 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
one-dimensional 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 )[A sin(sx )+B cos(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:

B exp(-c ^{2} s ^{2} t)+D=T _{0
} B =0, D =T _{0}

exp(-c ^{2} s ^{2} t )[sA cos(sL )]+C= 0
C =s (exp(-c ^{2} s ^{2} t) [A cos(sL )])
s ≠0 so C =0A _{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 )=100-Σ{(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 long-term 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 crash-couch/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
merry-go-round; 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=mg-kv ^{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
β -decay. 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 )(R-d );

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 (M-m )/(2M -m ),

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 anti-particles 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 anti-particle 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 x-rays) 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
1000-3.4=996.6 N, about 0.3% smaller than your weight.
This is a long-winded 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 "hot-water-freezes-faster" 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/anti-quark 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 goobledy-gook 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 (x-vt ). The most commonly used example of
waves is sinesoidal, for example, f=A sin(kx-ωt )
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} =A sin(kx-ωt ),
and an identical shaped wave traveling to the left, f _{left} =A sin(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+T-Mg=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+T-mg=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 (M-m )/(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 β -decay.
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 dyne-centimetres? If so, then that's easy to work with since that's a measure of torque I'm familiar with. So I'm half way there. If it's something else, then perhaps you can help explain.
Then there is the question of gm/cm/angle. The problem is that none of the calculators I've seen have a notion of angle. This is specific to torsion springs. The torque, in this case, is the working load when the spring bends 100°.
Will I ever find the correct spring? Sadly, I have spoken to the manufacturer, and they were unable to help me.

ANSWER: I think the "number" of interest is not a torque, rather
the spring constant associated with the spring for
angular displacement. First, a review of linear springs
which when stretched stretch a distance x when a
force F is applied to the spring. It is
experimentally determined that, to an excellent
approximation for a good steel spring, is that the
distance stretched is proportional to the force applied,
x ∝F or F=kx . The
proportionality constant k is called the spring
constant and is measured, in SI units, as N/m (Newtons per
meter) [dynes/cm in CGS units, lb/ft in Imperial units].
k indicates the stiffness of the spring
and this equation is called Hooke's law. The form of
Hooke's law for angular motion (as in a torsion spring)
is that the angular displacement θ is proportional
to the applied torque τ or τ=κθ ;
I believe the quantity you want is κ
(kappa) and is measured, in SI units, N·m/radian
[dynes·cm/radian in CGS units, lb·ft/degree in Imperial
units]*. I have noted that κ is often called, by
manufacturers and vendors, the rate
of the spring since it is the rate of change of torque
per unit angle.
I am puzzled by your reference to a constant defined as
gm/cm/degree; 1 dyne=1 gm·cm/s^{2} , so κ
should have units gm·cm^{2} /s^{2} /degree;
sometimes the gram is treated as a unit of force (called
a gram force where 1 gram force=980 dynes), in which case
κ
would be gm·cm/degree, not gm/cm/degree. So, there
must be somewhere in the reference you are using a
definition of what the CGS number is. Suppose that it is
0.76 dynes·cm/radian (which would be the likeliest
units if purely CGS system of units is applied) but you want it in
lb·in/degree; then 0.76 (dyne·cm/radian)x(2.25x10^{-6}
lb/dyne)x(0.394 in/cm)/(57.3 deg/radian)=1.18x10^{-8}
lb·in/degree. Or, suppose you wanted it in (gm
force·cm/degree); then 0.76 (dyne·cm/radian)x(gram
force/980 dyne)/(57.3 deg/radian)=1.39x10^{-5} (gm force·cm)/degree).There are lots of unit converters
around, my favorite is
here .

The choice of units for κ which it seems most vendors use
is lb·in/deg. The best calculator I found to get
an idea of the magnitude of these springs can be found
here .
Until you are able to deduce the units of the so-called CGS value of
0.76 you cannot know what it is.

*Sometimes
κ is implicitly given by specifying the
torque at a particular angle.

ADDED
THOUGHTS: In the book by Hans Jendritzki,
Watch Adjustment , he says that in the CGS system
for characterizing the spring in question (which I will
call N _{CGS} ), the unit of force is the
dyne and the diameter of the spring is 1 cm. He then
uses not the restoring torque to calculate N _{CGS}
but rather the
restoring couple associated with the force; the
restoring couple is the restoring force times the
diameter whereas the torque is the restoring force
times the radius . Most vendors use the
restoring torque to characterize springs, so ½N _{CGS
} should be used to convert to the vendor's units.
Although the units of the angle are not given in
Jendritkzi, I have found that the more natural radians
gives values much too small when converting to vendor
units; so N _{CGS} =(restoring couple in
dyne·cm for 1 cm diameter)/degree.
One vendor charactorizes the spring by specifying a
torque given by N gm/cm/100°; this does not
appear to be dimensionally correct unless the gram is
gram force (I will denote it as gmf) where 1
gmf=981 dynes (the weight of 1 gm in dynes) and /cm
means that again this is for a 1 cm diameter spring. So,
assuming that the vendor's torque is indeed torque
(force times half the diameter), ½N _{CGS} =N .
For example, suppose N _{CGS} =2; then
N =½(2) (dyne·cm/1°)(1 gmf·cm/981
dyne·cm)/(100°/1°)=0.102 gmf·cm/100°; so the conversion
from the CGS value to this vendor's value is N =0.0491N _{CGS} .
So, in the case of N _{CGS} =0.76
dyne·cm/1°, N =0.037 gmf·cm/100°.

[Disclaimer: I am
not a clockmaker and I am not familiar with conventions
clockmakers might use or meanings they may attach to
certain words which are not in accordance with
definitions, units, or conventions used by physicists.
In the event that I have misunderstood or that either
Jendritkzi really meant torque or the vendor really
meant couple, the conversion factor would be 0.102
instead. I have done my best to determine conversions
between two numbers representing the same thing but in
different systems of units.]

QUESTION:
If you lined up frictionless gears over an exceptionally long distance, could you effectively communicate faster than the speed of light through one person moving the first gear and a second person reading/interpreting the resulting movement of the last gear?

ANSWER:
I have answered this question in many guises
many times . In your
variation, each gear will experience a force from the
previous gear in the line and exert a force on the next
gear. The time it takes for the force to travel from the
input force location to the output force location is
determined by the speed of sound in the gear. The message
you send would travel at the speed of sound, much smaller
than the speed of light.

QUESTION:
If you are driving side by side at 60mph with another car and throw a can of coke at them would the can be travelling at the same speed so hit the target as if you were parked next to each other or merely disappear behind your vehicle?

ANSWER:
When you throw it out the window it originally has a
forward component of its velocity of 60 mph. At this
speed, the can will experience a very significant force
slowing its forward velocity down because of the air
drag. So the can will not keep pace with the two cars and
will appear, from either car, to be accelerating
backwards. So, no, the can will not hit the other car
where it would have if the cars had not been moving.

QUESTION:
Under relative motion, it's experimented that A person sitting on a back seat in a moving bus has a speed with respect to the earth which is the same as that of the bus. But if he now walks towards the driver of the bus,he has a speed relative to the earth which is more than that of the bus. Suppose the bus is moving at 100km/h and the person walks at 5km/h towards the driver, his forward speed relative to the earth is 100+5= 105km/hr. But when he walks back from the driver to his back seat at the speed of 5km/hr his speed relative to the earth is now 100-5= 95km/hr.
WHY the increase speed of such person when moved forward and the decrease when moved backward

ANSWER:
The problem you are having is that you do not understand
the phrase "relative to the earth". Let's alter your
example by having the train moving at 5 km/hr. If you are
in the train walking 5 km/hr in the opposite direction,
someone standing by the side of the tracks will
see you standing still; the same as running on a
treadmill in a gym with the treadmill and you moving in
opposite directions with the same speed. If you walk in
the same direction as the train is going, you will be
seen by the trackside observer as moving with a
speed of 10 km/hr.

QUESTION:
IF WAVES need a medium to transmit energy but light can travel through a vacuum.
Does dark matter have Anything To Do With Lights Ability To Do This?

ANSWER:
There is absolutely no mystery about light waves;
electromagnetism is arguably the best-understood theory
in all of physics. You may be sure it has nothing to do
with dark matter. Dark matter is arguably the
least-understood theory in all of physics. There are many
features of the universe which do not fit into our
current understanding of physics and introducing some
kind of particle which interacts with the rest of
creation only via gravity would solve a lot of
them; all attempts to directly detect dark matter have
failed and until there is direct observation we
understand nothing. To my mind, failure to observe dark
matter may be indicative that we do not understand
gravity as well as we think we do.

QUESTION:
Bowling ball vs.
billiard ball
What would ultimately travel further, a 12 lb bowling ball or a standard
billiard ball: if thrown/rolled with the same amount of physical strength/athletic ability, at ideal trajectory for each, on a frozen lake (not without small imperfections but as close to perfect as could be found naturally), under normal atmospheric conditions (Wisconsin in the winter), with little to no wind? Would temperature play a roll in determining a winner? Would the small difference in friction coefficients be significant? Any other variables Etc...
My brother and I have been debating this for years while ice fishing.

ANSWER:
(You might not want to read this rather long
introductory paragraph if you are just interested in the
final answer. ) This problem was very interesting to me
because I initially thought it was pretty easy to
understand. The billiard ball starts out with much greater
speed than the bowling ball, jumps into the lead.
Neglecting air drag is usually what is done initially in
considering such problems and there has to be some
kinetic friction for the balls sliding on the ice; but
because the kinetic friction is proportional to the
weight, each ball experiences the same rate of change of
its speed as it slows. Therefore the billiard ball is the
obvious winner. When I initially solved the problem this
way using reasonable numbers, I found that the bowling and
billiard balls had initial speeds of 14 mph and 81 mph,
respectively; the bowling ball went 686 ft and the
billiard ball went 21,700 ft, a little more than 4 miles! When I
was writing the final paragraph I discussed the possible
effects of two things I had neglected, rolling of the
balls rather than sliding and the possibility of air
drag. Just for fun I estimated the effect of air drag and
discovered that at the beginning, particularly for the
billiard ball, the drag force was considerably larger than the
sliding friction. I realized that I had to redo the whole
problem, now much more difficult and mathematically
involved than originally. Because the air drag estimate I
used is valid only if you work in SI units, I will do
that throughout and convert to imperial units when needed.
The mathematics gets very complicated and I will only
sketch my calculations and whoever is interested can fill
in the gaps.

First I will tabulate quantities needed for the problem:

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

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

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

For the air drag
I will use the approximation D =¼Av ^{2} where
A
is the area presented to the oncoming air (πR ^{2} )
and v is the speed. The radii of the balls
are 0.0286 m and 0.108 m. As noted above, this
equation is only valid for SI units because the
factor ¼ includes things like air density, drag coefficients,
etc .

We are now ready to
start doing the physics. Start with Newton's second law

ma=m (dv /dt )=-¼A v ^{2} -μmg

dv /dt=-Cv ^{2} -μg
(C =¼A /m ).

This equation is
integrable. The result is

t =tan^{-1} [v √(C /(μg ))]/√(Cμg )|_{v(0)} ^{0} .

Evaluating
this expression between the limits v=v (0)
to v =0 and inverting to solve for v (t )
is straightforward but messy. The essential
results are illustrated in the two graphs below.
The effect of air drag on the bowling ball is
modest but noticeable, reducing the time to stop
by about 10 s. The effect on the billiard ball,
though, is enormous; because of its high speed
at the start, it is slowed much more than due to
sliding friction alone. Still, it stayed in
motion about 20 s longer than the bowling ball
and was going faster most of the time, so
certainly went farther before stopping. Still,
it would be useful to find the position as a
function of time to determine how far each ball
went.

To find the position, x (t ), is
straightforward but again messy: integrate the
differential equation dx /dt =v (t )
where v is the function being plotted
in the graphs above. The results are shown in
the graphs below. The bowling ball stops at
t =54 s at a position x =158 m=518
ft; the billiard ball stops at t =77 s
at a position x =535 m=1755 ft. The
winner is the billiard ball which goes more than
three times farther than the bowling ball! I
believe that regardless of any details of the
friction and drag forces or how we choose the
quantity Fd , the billiard ball will
always go faster if the initial kinetic energies
are equal.

Finally, I want to discuss rolling vs .
sliding. It is really hard to get an object
rolling without slipping on a very slippery
surface. A sliding ball will not roll until it
is nearly ready to stop, and that would have
very little effect on my calculations. Also,
rolling friction is velocity-independent and
proportional to the normal force so it would
affect both balls the same.

QUESTION:
I have a question about gravity. In order to general relativity mass bends the space. I saw a simulation of gravity that a man put a mass on a surface like silk. that the mass bends the silk. and we put another mass they attract to each other. that 2D silk is an examinition of our #d world. I wanna know why masses attract to each other. is it becuase of bending space?

ANSWER:
That simulation is a good way to understand how space
bending can alter how objects move but keep in mind that
it is only a cartoon to illustrate the warping of
spacetime. See my
faq
page .

QUESTION:
If repulsion forces between protons is high why hasn't the nucleus exploded

ANSWER:
It is true that the Coulomb force experienced by protons
in a nucleus is very strong. However, the nuclear force
at very short distances is stronger and wins the
"tug-of-war" and holds them inside. However, this strong
force is also a very short-range force and if you pull a
proton just outside the nucleus, it will be repelled and
fly off. It is also interesting to ask why the nucleus
doesn't collapse if the nuclear force is so strong. The
nuclear force becomes repulsive if the protons get too
close together; this is called the saturation of nuclear
forces.

QUESTION:
What would be a ballpark estimate of all the mass converted to energy (lost) by all the fusion reactions in all the stars that are or have ever been in the last 13 billion years (lots of assumptions)?
Would this lost mass effect the overall gravitational attraction of the universe?
Would it be significant enough to be a factor in the current acceleration of the universe?

ANSWER:
I would not presume to make even the roughest estimate
you ask for. However, it is not relevant to the questions
you seem to be interested in. According to general
relativity, gravitation is caused by energy density, not
specifically mass. (Don't forget, mass is just a form of
energy, E=mc ^{2} .) So changes of mass
will affect the details of the gravitational field in
some volume of space but if you get very far from that
volume, the gravity you see originating from that volume
will not change.

QUESTION:
Our 5-year-old niece wants to know why it's so hard for her to ride her Big Wheel uphill (3% slope) on her family's driveway. My BSN-RN wife would like a friendly, informative way to break that down for her (Daphne), including whether her weight (50 pounds), her position on the Big Wheel, the coefficient of friction on said drive wheel, and other factors play a part. Said aunt would also like a kid-friendly way to calculate the horsepower necessary for her to ascend that grade on her Big Wheel for perhaps 100' (the driveway's length). No, this is in no way homework. Except it's at their home. And, well, it is work.

ANSWER:
A little tough for a
five-year old! Here, maybe, are some suggestions:

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

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

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

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

Friction, I
would opine, is an unnecessary complication.

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

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

Friction has
little to do with it because the friction is not
significantly different on level ground vs .
the incline. Position on the trike plays no role
since any change would be the ease of pedaling which
would be the same in either case.

QUESTION:
what would a light/sound wave look like in 3 dimensions, all diagrams are in 2d and it just came to mind that it wouldn't be like that in real life

ANSWER:
Here is a cylindrical wave. The blue surfaces represent
wave fronts.

QUESTION:
I am currently doing an experiment on the metronome pendulum. As you know a metronome pendulum has two types of pendulum which is the inverted pendulum and simple pendulum. It is called as double-weighted pendulum. This type of pendulum has two weight which is one above the pivot and one below the pivot. In my experiment I want to find the relationship between the varying mass of the top mass of metronome pendulum and its angular frequency but I cant seem to find any formula that describes my metronome pendulum. What formula for angular frequency should I use because from my data, I will be looking for its period of oscillation. There are too many angular frequency formula for different types of pendulum so I am confused on which formula I should be using.

ANSWER:
I will show you how to get a general answer. Then you can
apply it to the conditions of your experiment. Newton's
second law for rotational motion is τ=Iα=I (d^{2} θ /dt ^{2} ) where
τ is the net torque about the pivot (the blue cross in the figure),
I is the moment of inertia about the pivot, α is the angular acceleration, and
θ is the angle of the system at some time
t . If the system is a pendulum with no forces
exerting torques except weights (vertically down), all torques are
proportional to sinθ where here
θ is the angle relative to the vertical.
Now our equation may be written as

(d^{2} θ /dt ^{2} )=-(|τ| /I )sinθ

where
|τ| is the torque about the pivot when
the pendulum is horizontal (θ=π /2). The
negative sign results from the fact that the angular
acceleration and the angle are always opposite, a
restoring torque. This equation is extremely difficult to
solve, but if the angle is small (much smaller than 1
radian), you can use the small angle approximation, sinθ≈θ ,
which gives us

(d^{2} θ /dt ^{2} )≈-(|τ| /I )θ

This is the simple harmonic oscillator equation for an oscillating
system for which we know the angular frequency, ω =√(|τ| /I ).
So, let's apply it to your situation as illustrated in my
figure. There are two point masses and the mass of the
rod itself which I will take as m . The rod has a center
of gravity (orange cross) a distance of L _{3}
from m _{1} as shown; if your rod is
uniform, L _{3} =L /2. (If the
mass of the rod is much smaller than both the point
masses, you can approximate m as zero.) Now,
since all weights are of the form W=mg , you can
write |τ| :

|τ|=g (m _{1} L _{1} -m(L _{3} -L _{1} )-m _{2} L _{2} ).

The moments of inertia for the two point masses are

I _{1} =m _{1} L _{1} ^{2}
and I _{2} =m _{2} L _{2} ^{2} .

The moment of inertia for the rod is trickier, so to
avoid a lot of messy algebra I am going to assume a
uniform rod where L _{3} =L /2. I
find

I _{rod} =(m /(3L ))(L ^{3} -L _{1} ^{3} )

The total I is just the sum of the three moments
of inertia. So, finally,

ω =√[(g (m _{1} L _{1} -m (L /2-L _{1} )-m _{2} L _{2} ))/(m _{1} L _{1} ^{2} +m _{2} L _{2} ^{2} +(m /(3L ))(L ^{3} -L _{1} ^{3} )]

And, if m can be neglected,

ω =√[g (m _{1} L _{1} -m _{2} L _{2} )/(m _{1} L _{1} ^{2} +m _{2} L _{2} ^{2} )]

QUESTION:
So I have just thought about this, let’s just say that hypothetically you were able to fall through the earth completely. I was wondering what the gravity would do to you if you went in one side and came out the other

ANSWER:
See the
faq page .

QUESTION:
If everything is expanding, and every galaxy is moving away from the Milky Way, why is the Milky Way and Andromeda predicted to colllide in 3.5
billion years?

ANSWER:
See an
earlier answer . See a simulation
video .

QUESTION:
this is for a horror book I am writing. how fast can a bat fly, if it weights 4500 pounds and is 25 feet 6 inches tall? and in 1 flap of its wings, can rise a rise a quarter, of a mile? also how wide would its wingspan be?

ANSWER:
The best I can do is to estimate how the wings would have
to be if the ratio of weight (W ) to wing area (A )
is the same as a real bat. I find that a typical value
for that ratio for bats is about R =W /A =20
N/m^{2} =0.42 lb/ft^{2} .
So, for your case, A=W /R =4500/0.42≈10,000
ft^{2} or one wing has an area of about 5000 ft^{2} .
If I model a wing to be a triangle with base B =25 ft and
a width H , 5000=½BH =12.5x25xH
or H =400 ft, a wingspan of about 800 ft.
Since it has the same R as for a real bat, it
should be able to fly as fast as a real bat which can be
as high as 100 mph.

QUESTION:
I am a part time instructor and one of the things I teach is rigging. An
important part of the class is calculating sling stress. I was asked a
question that I couldn't quite answer. I don't believe it was my high
school or college physics teachers that failed me rather the more than two
decades since I sat in a physics class. The question relates to vector
forces. If I am hoisting a load of a given weight (say 2000 lbs) and I have
two slings that are at a given angle (say 45 degrees) from the load. Each
sling would be carrying half the load or 1000 lbs straight up. However, the
sling in this example would have 1414 lbs of force (inverse sine if I
remember correctly). The question I was asked was regarding a crushing or
compression force. Since each sling is carrying 1000 lbs vertical load that
means it has 414 lbs of force perpendicular to the vertical. Does that mean
that there is 414 lbs of "crushing" force or is it 828 lbs since each sling
is pulling towards the center of gravity.

ANSWER:
After an email exchange with the questioner, I was able
to determine that "sling" is just a rope or a chain or a
string etc . In the diagram I have shown all the
forces on the load, the tensions in the slings,
T _{1} and T _{2}
and the weight of the weight, W .
The magnitudes of the tensions are equal, T _{1} =T _{2}
=T . Also shown are the horizontal (x )
and vertical (y ) components of the tensions;
they are T _{1x} =T cosθ ,
T _{2x} =-T cosθ ,
T _{1y} =T _{1y} =T sinθ.
So the equilibrium equation is 2T sinθ-W= 0;
therefore T=W /(2sinθ ).
So, for your example, sinθ =cosθ =1/√(2),
W =2000 lb, and T =1414 lb. Now, your
interest is in the horizontal components; each is T cosθ=W /(2tanθ )=1000
lb "crushing force" on each side. Your main error is that
you treated the vector forces as scalars, you assumed
T=T_{y} +T_{x} where in fact T =√(T_{y} ^{2} +T_{x} ^{2} ).

QUESTION:
Since everything in the universe is always moving, is the idea of being still/idle/stationary just an illusion?

ANSWER:
I would not put it like that. If you are in a system
where Newton's first law (if the net force on an object
is zero, it will remain at rest or moving with a constant
velocity) is correct, you may think of yourself at rest.
This is called an inertial frame of reference. But the
catch is that this is not the only frame of reference in
the universe where Newton's first law is true, any other
frame which moves with a constant velocity relative to
yours is also an inertial frame. In other words, there is
no such thing as absolute rest. A frame accelerating
relative to yours, however, is not an inertial frame.

QUESTION:
I am a cancer patient about to receive Y-90 treatments. If Y-90 has a half life of 64 hours: 1. When does that time clock begin running and 2. How can they ever have a full strength dose on hand at any given time?

ANSWER:
^{90} Y is the decay product of ^{90} Sr
which has a half life of about 29 years. 90Sr is an
abundant byproduct of the fission of uranium and
therefore available in the waste of reactors. When the
^{90} Y is needed it can be chemically separated
from the ^{90} Sr. I am sure that the hospital
cannot keep a supply on hand and will have it delivered
when it is needed for your treatment. It does not matter
how long it has been since it was separated, only that
the radiation level be correct for the prodedure. It
probably arrives with a level too high and they wait for
it to be at the needed level.

QUESTION:
I am struggling with the concept of relativity, as it relates to time. Not in terms of the underlying principle or the mathematics. These have been confirmed experimentally and are now in everyday use through technologies such as satellite navigation systems. My concern relates to the way that relativity treats time as a variable rather than an absolute quantity. It seems to me that we can only measure or experience time by reference to some physical process or change, whether that is the oscillation of a quartz crystal or the lifetime of a muon! Any experimental proof of time dilation as a result of the effects of either velocity or gravity really just relates to the way that we experience the passage of time and does not exclude the possibility that for the universe, as a whole, time is absolute and unchanging, passing in a constant fashion regardless of relativity and whether we measure it or not. Therefore, although relativity theory works in practice, this is only because it deals with our limited understanding of, and ability to measure and/or experience, the passage of time. Is this nonsense or is it a possibility? It follows from this reasoning that we can never prove experimentally that time either speeds up or slows down, only that the rate of how we measure or perceive time can vary. It also implies that time (and indeed space) are concepts that we will never fully explain as they ultimately cannot be qualified or quantified by measurement or mathematics.

ANSWER:
You are correct that "…time is absolute and unchanging,
passing in a constant fashion…" but only in your
frame of reference; any clock at rest relative to you
runs at a constant rate. But, what special relativity
tells us is that clocks not at rest relative to us do not
run at the same rate as ours. I have always felt that the
best way to convince someone that this is the case is the
light
clock . In order that this example is convincing to
you, you must accept that the
speed of light is a
universal constant in all frames of reference; but this
is an experimentally well-verified fact and also is a
consequence of the principle of relativity—that the laws
of physics are the same in all inertial frames of
reference.

QUESTION:
SINCE FREE FALL DONT NATURALLY OCCUR ON EARTH BECAUSE FLUID FRICTION IS PRESENT IF YOU
DROP A BOWLING BALL AND A BASEBALL OFF A TOWER OF PISA WHICH WOULD LAND FIRST

ANSWER:
The (upward) drag force on a falling sphere of radius R and speed
v in air
at sea level may be approximated as F =¼πR ^{2} v ^{2} =0.79R ^{2} v ^{2}
(this is correct only in SI units); the (downard) force of
the weight W on an object of mass m is
W=mg . The net force is the mass times the
acceleration a (Newton's second law), ma =-mg +0.79R ^{2} v ^{2} .
So as the object falls it goes faster and faster until
v is large enough that it falls with a constant
value speed v_{t} (a =0), v_{t} =[√(mg /0.79)]/R .
For a baseball, m =0.144 kg and R =0.037
m, so v_{t} ^{baseball} =9.28 m/s.
For a bowling ball, m =8 kg and R =0.12
m, so v_{t} ^{bowling ball} =83.0
m/s, nearly ten times greater than the baseball. So the
baseball stops accelerating sooner than the bowling ball
and loses the race to the ground.

QUESTION:
My 8 year old asked me if the cycle of the universe expanding and contracting will ever end?

ANSWER:
What makes you and your son think that the universe is in
such a cycle? The ultimate fate of the universe is one of
the most important questions in cosmology and nobody
knows the answer. I believe most cosmologists believe
that the universe will continue expanding forever.
Eventually the supply of hydrogen and helium in the
universe will be exhausted (fused into heaviers elements)
and stars will no longer exist. The universe will be
dominated by black holes which will eventually decay away
by Hawking radiation, so the universe will be a very
dilute, cold, dark place filled with radiation. To get
more information, check out the
Wikepedia article on the fate of the universe.

QUESTION:
My son has a question.
A shadow is long or short, wide or narrow. He thinks this seems like area (taking up space). But the real question is... can you have a shadow without matter? Because if you can't have a shadow without matter, then doesn't that mean the shadow is actually matter? (I have to ask because I'm not sure how to answer this; I suppose the shadow is actually the part of the matter that doesn't have light reflecting off of it...? Or how do I answer this for him?)

ANSWER:
A shadow is not matter;
it is not anything, it is a region in which light
illuminating something is blocked by some obstruction.
You might say that a shadow is the lack of anything,
light which might have otherwise illuminated where it is.
And there are degrees of "shadowness". If all the light
illuminating an area is blocked by the obstruction, it is
called an umbra ; if only part of the light is
blocked it is called a penumbra . (Umbra is Latin
for shadow.)

QUESTION:
U recently answered a question regards Earth's precession = 26,000 yrs. Can u show what specific equation(s) used to yield this time? I doubt that a single equation was used.

ANSWER:
Just like the precession of a top is due to the torque
exerted by its own weight, the precession of the earth is
due to torques on the earth, mainly by the sun and the
moon. Your equations may be found in this
Wikepedia article .

QUESTION:
On an ordinary day as one goes upward from the surface of the ground the electric potential increases by 100 volts per meter, this means that outdoors the potential of your nose is 200 Volts higher than that at your feet. Then why is it that none of us get a shock when we go out in the street?

ANSWER:
As is often the case, the most lucid answer to your
question can be found in the
Feynman Lectures .

QUESTION:
Why doesn't the earth just stop spinning and orbiting don't you need energy for motion where does the earth gets it's energy and is the earth in perpetual motion?

ANSWER:
You are hitting one of Isaac Newton's most important
discoveries—inertia. It says that you do not need
to exert a force (torque) to keep an object moving in a
straight line (rotating about an axis) to keep it going;
this means, also that the energy it has by virtue of its
motion will not change. So the rotating earth keeps
rotating since there are no torques on it*. The earth is
in a circular orbit† around the sun and so its orbital
motion keeps going since, even though there is a force on
the earth, there is no torque.

*Actually, the moon does exert a torque on the earth
which is causing the spinning to get smaller, but the
change is so tiny that it takes millions of years to be
significant.

†The earth's orbit is
not exactly circular. When it moves closer to the sun it
speeds up, when it moves farther from the sun it slows
down by exactly the same amount, so the orbit just
continues forever.

QUESTION:
With the devastating tragedy that happened in Lebanon, there has been some videos surface of the explosion. One such
video is of some ladies in a shop, the pressure wave hits the shop and the lady outside the shop doesn’t get thrown back but inside the other lady does. What is the explanation here??

ANSWER:
Note that the glass doors are closed. Outside the front
of the pressure wave hits, but it takes some brief time
to reach its maximum. However, the pressure difference
between inside and outside is not large enough to break
the glass until it is near its maximum, so the pressure
difference changes much more quickly so the air rushes in
quickly. I will admit that this is just an educated
guess!

QUESTION:
If electricity always take the path of least resistance, why does a lightening discharge form a zigzag path to earth and not straight down which would be the shortest distance?

ANSWER:
Because air is not just a uniform homogeneous medium.
There are local fluctuations in temperature, humidity,
density, wind speed, and composition which makes the path
of least resistance at any instant, not a straight line.

QUESTION:
how momentum is conserved of both objects with same masses moving opposite direction to each other collide to a point and at same time both objects chemically combined to each other?

ANSWER:
In an isolated system, linear momentum is always
conserved; this is because the definition of linear
momentum is designed to be conserved in the absense of
any external force. This is also true in special
relativity where the linear momentum is not p=mv but
rather p=mv /√[1-(v ^{2} /c ^{2} )]
where c is the speed of light. In the event of
chemistry going on, there is either energy gained
(endotermic) or lost (exothermic). If lost, the momentum
of the radiation has to be taken into account. If gained,
the energy will be taken from the kinetic energies of the
incident objects but momentum will be conserved.

QUESTION:
Does precession and nutation ( or anything else ) affect the axis of the earth?

ANSWER:
Precession of the earth's axis is caused mainly by
torques exerted on the earth by the moon and the sun. The
period of precession is 26,000 years. But precession does
not cause the actual rotation axis to change. What causes
the axis to change (such that the geographical positions
of the poles would change) is thought to be three things:

glacial ice
melting and sending its water to the oceans,

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

tectonic motion,
the drift of large land masses.

QUESTION:
The momentum transmitted to the golf ball comes from the club head and,to a lesser extent, from the club shaft.
Golfers are instructed to hold the club very loosely. Thus, it seems, no momentum is transmitted from the arm.
What if the golfer instead holds very tightly to the club? Will the momentum of the arm be transmitted to the golf ball?

ANSWER:
The physics of a golf swing is much more complicated than
you might think. From the little research I saw, your
"hold the club very loosely" is an oversimplification.
The club needs to accelerate very quickly on the way to
the ball and that acceleration is only going to come from
the torque you provide. What should happen is that, just
before impact, your wrists must relax. There is an
excellent discussion on the golf swing which, for the
first half is fairly conceptual and the second half goes
into the detailed physics analysis of the swing.

FOLLOWUP QUESTION:
I understand that the standard golf swing depends on uncocking the wrists to generate velocity at the moment of impact.
Moe Norman, a Canadian golfer now deceased, was the best ball striker in the history of golf. Moe used a completely different swing that he developed. Among other things, it involved holding the golf club tightly.
Thought experiment: The shaft of the golf club is fused to the golfer's arm and runs all the way up to the shoulder. That is, there is no wrist play at all. I understand that such a set-up will generate less velocity than a hinged wrist. Therefore, the 'v' in 'mv' will be lower. My question is whether the 'm' will be higher. Given that the golfer is now swinging a much more massive club, albeit from the shoulder, I posit that the mass striking the ball is greater.
Take the center of gravity of the club-arm, calculate its velocity, multiply by the mass of the club-arm, and one has the momentum that is transmitted to the ball (I think). This may or may not be as great as the momentum of the faster wrist-hinged swing.
I just want to know whether the 'm' will be greater with this fused club.

ANSWER:
I believe that there is no way to understand the club as
a point mass, so to talk about its mass in a collision is
meaningless. Think of it as a machine which imparts
momentum to a small mass by exerting a force on it over
some short time. I think it probably boils down to having
the machine achieve the greatest possible velocity when
it hits the ball; if you can achieve that by holding the
club more tightly, go for it! It was interesting to learn
a little about Moe Norman of whom I had never heard.

QUESTION:
When a golf club hits a golf ball, part of the momentum of the golf club transfers to the golf ball. The golf ball, being light, takes off at a velocity greater than the club head speed.
What if the golf club is accelerating at the point of impact? Will the ball take off faster (and go further) than when it is hit at the same impact velocity by a club that is not accelerating?
Note: amateur golfers tend to reach maximum velocity at the point of impact. Pros reach max velocity after impact. Pros do swing faster, which imparts more velocity to the ball. I would like to know whether their acceleration adds even more velocity.

ANSWER:
What will determine the exit speed of the ball is the
speed of the club at impact. The impact time is
approximately Δt =0.0005 s. How big would the acceleration
need to be in order to significantly change the speed
over this time? A typical speed of the club head is about
150 mph, about v= 70 m/s. Suppose that the speed increased
over the impact time by Δv =1 m/s. Then the acceleration
would have to be a =Δv /Δt= 1/0.0005=2000 m/s^{2} ;
this is about 200 times greater than the acceleration due
to gravity. We could estimate the acceleration by
guessing that the club took about 0.1 s to go from zero
to 70 m/s, a =700 m/s^{2} . So the speed
of the club might increase by 700x0.0005=0.35 m/s and the
average speed during impact would be about 70.2 m/s. My
feeling is that this would have a negligible effect on
the speed of the ball.

FOLLOWUP QUESTION:
I did some more research and I have a follow up.
This article states that Force equals mass times acceleration.
It indicates that applying a force to an object over time creates momentum.
Therefore, it seems to me that a club head that is accelerating at the point of impact transmits Force and, therefore, momentum to the golf ball. And that this momentum would be in addition to the momentum of the club head.
A club head that is not accelerating will transmit no force to the golf ball.
It seems to me that a club head accelerating is going to hit the ball further, velocity at impact being equal.

ANSWER:
Sorry, but you have this all wrong. (This is what comes
from using formulas when you do not really understand
what they mean or when they are applicable!) F=ma
is Newton's second law and simply says that if you push
or pull on a mass it will accelerate. Indeed, if you
apply a force over a time you will accelerate it and
therefore change its momentum. But these do not
imply that the source of the force itself needs to be
accelerating. Here is what happens: when the club strikes
the ball, it exerts a force on the ball; Newton's third
law says that if the club exerts a force on the ball, the
ball exerts an equal and opposite force on the club. For
the short time that they are in contact (after which the
forces disappear), the ball will speed up and the club
will slow down if there are no other forces on it (which
may be the case*). If the average force they exert on
each other is F and the time of contact is t ,
the ball will acquire a linear momentum of p=Ft ;
the ball will experience a change in its momentum but by
how much is not so simple since other forces act on it.
Note that F=p /t=mv /t , so the
smaller t , the larger F , all else being
the same. In my answer to your original question, using
v =70 m/s and t =0.0005 and noting that
the mass of a ball is about m =0.046 kg,
F =0.046x70/0.0005=6,440 N=1448 lb.

*If the club is accelerating when it strikes the ball
there must be some force on it—ultimately that
comes from you.

QUESTION:
I am writing a video response to Dr. William Lane Craig, a well known apologist, and so I am answering every position he states from a debate between him and Christopher Hitchens at Biola university some years ago.
I write to you now because Dr. Craig says something that I don't understand. He states,
"First, when the laws of nature are expressed as mathematical equations, you find appearing in them certain constants, like the gravitational constant. These constants are not determined by the laws of nature. The laws of nature are consistent with a wide range of values for these constants."
I don't understand this at all.

ANSWER:
(This is not the whole question submitted, but gets to
the heart of what the questioner wants to know. Reminder
that site ground rules specify "…concise, well focused
questions…") I consider that the most important
"laws" of nature are not equations, they are
proportionalities. In order to keep this discussion
focused on my basic description of nature, I will focus
solely on classical physics to make my points. In physics
we begin with three fundamental concepts to lay the
foundation of describing nature: mass, length, and time.
In the SI unit scheme we choose kilograms (kg), meters
(m), and seconds (s). With the chosen units we can define
velocity as the rate of change of position with units of
m/s; then we can define acceleration as the rate of
change of velocity with units (m/s)/s=m/s^{2} . So
if an object changes its speed from 2 m/s to 6 m/s in a
time of 2 s, its acceleration is 2 m/s^{2} . Mass
is trickier to understand but it is a quantity which
measures how resistant an object is to being accelerated
if you push on it; it is harder to accelerate an object
of mass 1000 kg than it is one of 2 kg. Note that we have
not yet quantified that push (or pull) which we normally
call force. However we can certainly imagine figuring out
a way to push or pull on something twice as hard, or
three times as hard, etc . Whatever force is,
suppose we push on something with a constant force F
and vary the mass; we will find that if we double the
mass we halve the acceleration, if we halve the mass we
double the acceleration, if we make the mass 10 times
larger, the acceleration is 1/10 of its original value,
etc .
We have found that, F being constant,
acceleration a is inversely proportional to mass,
a ∝1/m .
Now suppose we vary F but keep m constant; we find that
if we double the force we double the acceleration, etc .,
i.e. a ∝F. We can now put
the two together and get a∝F /m or F∝ma .
This is what I consider to be Newton's second law; it
tells us an important connection of how quantities depend on
each other, it is a law of physics; and, you really did
not have to have defined the meter and second and
kilogram for it to be true, you just had to understand
conceptually what mass, length, and time are. If you know
even a little physics, you do not recognize this as
Newton's second law—any textbook will tell you that
it is F=ma . Now, how can one turn a
proportionality into an equation? It is simple, just
introduce a proportionality and voila! F=Cma
where C is an arbitrary constant since we
have not defined how we will measure F . Don't
think of it as a "fundamental constant" because if we had
already defined how to measure force, we would have to
measure C . For example, if we had defined a unit
of force to be a pound but measured m and a
in SI units (kg and m/s^{2} ), F=ma would not be a valid
statement of Newton's second law.

Now let's discuss an example of when a constant is a
fundamental constant. Newton's law of gravitation
expresses the magnitude F of the equal and
opposite forces exerted on each other for two point (or
spherical) objects of masses m _{1} and
m _{2} separated by a distance of r :
F ∝m _{1} m _{2} /r ^{2} .
Now, to make this an equation, we must introduce a
proportionality constant G : F=Gm _{1} m _{2} /r ^{2} . But can we choose it to be whatever we like? No, because
there is nothing in this proportionality which is
undefined; we can take two 1 kg masses, separate them by
1 m, and measure the force between them (a really hard
experiment!). In other words, we must measure the
constant G . This is a true
fundamental constant of nature, it measures the strength
of gravitation. Of course, its numerical value depends on
how we measure mass, length, and time, but it is still a
universal constant. In SI units it is 6.67x10^{-11} N·m^{2} /kg^{2} .

You might be interested in two earlier answers,
1 and
2 , along similar
lines as yours.

QUESTION:
I wondered what happens if one would make a magnet spin really fast. Could it be theoretically possible that it could create visible light.

ANSWER:
The frequency range of visible light is about 4-8x10^{14}
Hz, so you can be sure that you cannot do this with any
macroscopic magnet. One way to get magnetic radiation is
to fabricate something called a
split-ring resonator (SRR) but technical problems
limit SRRs to about 2x10^{14} Hz, near infrared.
Another technique uses spherical silicon nanoparticles
and these can generate visible magnetic light. The method
is too technical to discuss here, but you can read about
it at
this link .

QUESTION:
i had a doubt, to detect an inertial frame we need to define 0 force, but to define 0 force we need an inertial frame,0 force is defined as the sum of real forces should be 0, and we can identify real forces if we believe that the real forces drop with distance,and this could be a possible solution to the above problem, but it based on an assumption, and even if we consider that we cannot detect an inertial frame, and all the physics upto special relativity is just based on if there is an inertial frame, then how come we could do experiments to confirm the theory and how did physics work uptill general relativity, if we cannot solve this basic conceptual problem.

ANSWER:
Technically, there is no such thing as an inertial frame.
No matter where you go in the universe there are always
electromagnetic and gravitational fields. That does not
mean we cannot define an inertial frame as one in which
Newton's laws are true at low velocities. In fact we can
do better by defining an inertial frame as one in which
the total linear momentum (relativistic
definition ) remains constant; then you are not
confined to low velocities. Then we will find that very
good approximations to inertial frames can be found in
nature. There is nothing wrong with basing a theory on an
idealized (fictional) concept as long as the ultimate
theory agrees with experimental results in the real
world.

QUESTION:
The idea of dark matter and dark energy is puzzling me. It is understood that dark matter acts like glue to hold galaxies together. On the other hand, dark energy is responsible for expansion of universe. Scientists say inside galaxies dark matter overcomes dark energy so gravity wins and keeps galaxies together but when it comes to intergalactic space, it is said that dark energy overcomes dark matter and wins the fight, so galaxies are moving away. How do you explain this contradiction? Why dark energy wins in intergalactic space but not inside the galaxies.
In other words, what makes dark matter overcomes dark energy in galaxies but loses in the space between galaxies.

ANSWER:
Even though I state on the site that I do not do
astronomy/astrophysics/cosmology, I will take a stab at
this one. Of course, keep in mind that nobody knows what
dark matter or dark energy are in any detail. Dark
matter, as the name implies, is thought to be some kind
of mass and therefore subject to gravitational forces. It
therefore would be not surprising to find it more
concentrated in the vicinity of large masses like
galaxies; so its density is likely to be smaller in
intergalactic space. Even less is known about dark energy
but, if it is anything like uniformly distributed, it is
more likely to lose a "tug-of-war" with dark matter
inside a galaxy than outside.

QUESTION:
What happens to the gravitational effect of mass when it transforms into energy, as in e=mc^2?

ANSWER:
In general relativity, our best theory of gravity, the
bending of spacetime, which is what gravity is, is not
caused only by mass. Any local energy density will cause
gravitation.

QUESTION:
I would like to ask about gravitational mass.
I know inertial mass is changing by motion (speed) according to m=mo/(1-v2/c2)^0.5 And also that is inertial mass which sits in E=mc2.
If the statements above is correct, now how about gravitational mass? Does it change with motion (speed)? And what mass should be used for general gravitational formula F=GmM/r2? should we use mo (rest mass) regardless of speed of the object? Or should we use m=mo/(1-v2/c2)^0.5 to substitute in F=GmM/r2?
In other words does mass equivalence principle (inertial mass=gravitational mass) hold in hight speeds?

ANSWER:
Read the answer to the previous question. The moving mass
has more energy than the stationary mass and therefore
has a stronger gravitational field as measured by a
stationary observer. Newton's gravitational equation is
amazingly accurate for speeds small compared to the speed
of light. However, if you are at very high speeds you
should not use it at all. The whole notion of force is,
itself, not useful in relativistic circumstances. We
might ask why force is a useful concept in classical
physics. Newton's second law, the central concept in
classical mechanics, is F=ma . Why? If someone is
moving past you and measures the acceleration a'
of something in your frame, she will measure exactly the
same value as you. Therefore you will both conclude the
same force is being applied to m, if a=a' then
F=F' . But if relative speeds are not small,
a≠a' so Newton's second law is no longer true.
Instead we rely more on conservation principles more in
relativity theory. In particular, we want to
redefine momentum so that it is conserved in a closed
system.

QUESTION:
I have recently learnt about the LIGO detector and watched a few videos on how it works, I understand how they detect the stretching and contracting of space time due to the gravitational wave only stretching and contracting a few hundred times per second and the new photons entering this space must travel farther or shorter deviating from the normal time travelled.
I would like to ask how these new photons enter this stretched space time since the gravitational wave itself is travelling at the speed of light does that not mean the stretched space will either move with a certain packet of photons, and since the wave travels at light speed, there wont be enough contractions to allow detection to occur.

ANSWER:
You just make this too hard to visualize thinking of photons.
Think instead as
electromagnetic waves . (Although
the speed of the gravity wave does not matter for the
interferometer to work, it is just as if somebody were
jiggling the mirror ends, I want to note that it is only
assumed that the speed of gravity is c ; the
speed of gravity has never been measured, we just know
that it is very fast. Since the gravitational field has
never been quantized, we can only assume that the
graviton has no mass. Note that for decades it was
assumed that neutrinos had no mass, something we now
understand is not true.)

QUESTION:
Who first derived the relativistic energy-momentum relation
E ^{2} +p ^{2} c ^{2} =m ^{2} c ^{4} and in what publication? Wikipedia says that it was Dirac in 1928, but I cannot find evidence of this in his famous 1928 paper, "The Quantum Theory of the Electron."

ANSWER:
As far as I can tell, Dirac's 1928 paper used the
energy-momentum relation as a jumping off point for his
derivation of the Dirac equation. Certainly this equation
was known long before 1928 since special relativity
itself was introduced in two papers by Einstein in 1905. The
first paper did not mention momentum at all. In the
second paper he shows that if a particle of mass m
radiates energy L that the mass changes its
kinetic energy by L /c ^{2} ,
hence E=mc ^{2} , but does not mention
momentum. Finally in 1907 he writes an
article in which
he shows that if relativistic momentum is p=mv /√[1-(v /c )^{2} ],
then it is a conserved quantity as in classical Newtonian
physics. From here, and knowing E =mc ^{2} /√[1-(v /c )^{2} ],
it is
straightforward to get the energy-momentum
relation.

QUESTION:
Why would lifting my feet off the ground make me heavier. Let’s say I were to do a push up while my friend is on my back for extra weight. If the friend was to plant
her foot on the ground, why would the push-up become easier.

ANSWER:
In the first figure I show all the forces on the man in
yellow. They are: N is the force which the
ground must exert up on the man, W is his
weight, and F is the force which the woman
exerts down on the man. (I have drawn N as one
vector, but it is really distributed among his two feet
and two hands. Nevertheless, the size of N is
proportional to the force he must exert to lift himself.)
The man is in equilibrium, so the forces on him must add
to zero, N-W-F =0 or N=W+F . In white are
the forces on the woman: w is her weight and
f is the force which the man exerts up on the woman
which by Newton's first law equal and opposite to the
force which she exerts down on the man so F=f .
Since she is also in equilibrium, the sum of the forces
on her must add to zero, so f-w =0 or f=w=F .
So, finally, we see that the total weight the man must
cope with is the sum of his and the woman's weight,
N=W+w .

So now I put a blue brick under the woman's foot so
that it rests on the brick. This is a new force
n up on her, shown in red. Each
person's weight is the same as before, your
weight is just the force down due to gravity.
But now there are three forces on the woman
adding up to zero: n+f '-w=0 or f
'=w-n . The force f ' exerted
up on her by the man, which is the same
magnitude of the force F '
which she exerts down on the man is smaller than
before. So now,
N '-W-F ' =0
or N '=W+F '=W+w-n. He has a smaller
total force to contend with than before.

The mistake you made is that you assumed that the
force down on the man was the woman's weight.
But the woman's weight is not a force on the
man, it is a force on the woman. This is
accidentally correct in my first example because
if only the man is supporting her, it just
happens that the force she exerts down is equal
to her weight in magnitude. But in the second
example the brick is supporting part of her
weight so the man supports less. She does not
become lighter!

QUESTION:
I am writing a fantasy novel currently, and at the ending a giant Dragon, with scales made out of stone is falling into the sea infront of a city and I'm imagining a tsunami destroying the city. The dragon weighs around 2,000 tons, how high would the wave be and how far would it reach inland? (the city is basically a flat plain)

ANSWER:
Well, I have to make some wild approximations to address
this at all. I will perhaps get you within an order of
magnitude of how big the wave might be. Maybe I should sort of do a rough
general case first and then I can throw in some
reasonable numbers. If the dragon drops from a height
h and has a mass m , then the energy she
starts with is E _{1} =mgh , where
g is the acceleration due to gravity. When
she hits the water she will have a smaller energy because
of air drag falling down and when she comes to rest she
will lose some energy to heat and sound and damage to her
body; I will say that a fraction f of the original energy
goes into the energy E of the wave created,
E=fE _{1} . Now I will assume that all this
energy goes into a circular pulse which has a width of
w
and a length of 2πR (circumference of a
circle) where R is the radius of the circle at
the time of interest, namely when the wave hits the
shore; so R is the distance from shore where the
dragon hit the water. Now, I found an expression on
Wikepedia which relates the energy of a wave to its
height: E /A=ρgH ^{2} /16
where A =2πRw is the total
horizontal area of the wave, H is the height of the wave,
and ρ= 1000 kg/m^{3} is the density
of water. If you put it all together you will find
H =√[8fmh /(ρπRw )].
Suppose h =5000 m, m =2x10^{6} kg (2000 metric
tons), f =½, and w =20 m; then I
find H ≈25 m. This is in the same ballpark as the
largest tsunami wave ever observed which was
about 100 ft. There is no way I could estimate
how far inland the wave would travel; it would
depend a lot on the terrain, obstacles, etc .

QUESTION:
This question is in reference to quantum entanglement.
When a Super positioned photon is measured to determine spin does it's spin stay in that orientation as long as they a measuring it or does it immediately go back to a Super positioned state?
In other words if you determined the spin of a quantum entangled particle at say 12:00 pm and constantly measured it, would it's spin continue to point in the same direction at say 12:02 pm. or is the measurement only good for the exact moment it was initially measured?

ANSWER:
When you observe it, you put it in a single state (and
simultaneously, put the other entangled photon in a
single state). Unless it interacts with something else
later, they remain in states you put them, they do not
reëntangle.

QUESTION:
What is the maximum distance between two electrons where they will notice each other? They would both be at rest to each other in a vacuum with no external stimuli or forms of energy. At what distance do the electrons "See" each other.

ANSWER:
This question has no meaning because "notice each other" is not quantified. In principle, the Coulomb
force extends all the way to infinity.
You could, for example, ask at what distance would the force
each electron feels would result in an acceleration of 1 mm/yr^{2} =[1x10^{-3 }
m]x[(1
yr)x(365 day/yr)x(24 hr/day)x(3600 s/hr)]^{-2} ≈9x10^{-22}
m/s^{2} .
In general, the force is F=ke ^{2} /r ^{2} =ma ,
so r =√[ke ^{2} /(ma )]
where k =9x10^{9} N·m^{2} /C^{2} ,
e= 1.6x10^{-9} C, and m =9x10^{-31} kg. I
calculate, approximately, that r ≈5x10^{11}
m≈300,000,000 miles which is about 3 times the
distance to the sun. Of course, you could never do such
an experiment since you would need to be in a universe
which contained nothing but these two electrons.

QUESTION:
Aren’t "constants" simply a "fudge number" to make equations work? It seems to me, constants are values of a concept we have yet to identify or understand, but need to make equations with known concepts and values work out. Since these equations work correctly with a variety of known variables, they are generally accepted. It seems to me, non-electrical magnetism (i.e. that which is retained in magnetic material like iron) and gravitation have a variety of descriptions and equations which predict their effects – yet we still are devoid of actual cause.

ANSWER:
Constants are not just "fudge numbers" as you suggest.
They can be, but in general constants play a very
important role in physics. I will give you a few
examples.

The laws of physics are stated most elegantly as
proportionalities rather than equations. For example,
Newton's second law states that the acceleration (a )
of an object is directly proportional to the applied
force (F ) and inversely proportional to the mass
(m ), a∝F/m ; so you double the
acceleration if you push twice as hard and halve it if
you double the mass. Now, to calculate we prefer to have
equations rather than proportionalities, so we introduce
a proportionality constant, call it C , a=CF /m.
What does C mean? Since we know how to
measure a (m/s^{2} ) and m (kg),
your choice of C determines how we will measure
force. I would like the force to have the magnitude of 1
if a =1 m/s^{2 } and m =1 kg, so I
choose C =1. The constant C is not a fudge
factor, rather its choice defines how we measure forces.
(We call the unit kg·m/s^{2 } a Newton, N.)

Newton also discovered the universal law of gravity: if
two masses (point masses or spheres) m _{1} and
m _{2} are
separated by a distance d , they exert equal and opposite
attractive forces on each other the magnitude of which is
F ; then F ∝m _{1} m _{2} /d ^{2} .
So the equation for universal gravitation would be of the
form F=Gm _{1} m _{2} /d ^{2} .
where G is some constant. But we know how to measure mass
and distance and force, so we cannot choose G to
be any old thing we want as we did for Newton's second
law, we must measure it. It turns out to be G =6.67x10^{-11}
N·m^{2} /kg^{2} . This is one of the
most fundamental constants in all of nature and most
certainly not a "fudge factor".

Constants are often used to quantify a particular
situation. For example, the frictional force f
of one object sliding on another is found, approximately,
to be proportional to the force which presses one object
to the other (often called the normal force, N ).
This is not a physical law, just an approximation of
experimental measurements, and it obviously involves only
the magnitudes of the forces since their directions are
perpendicular to each other. So, including a
proportionality constant μ , called the
coeficient of kinetic friction, f=μN . This
constant tells you how slippery the surfaces are; its
value for rubber sliding on ice will be much smaller than
for rubber sliding on asphalt.

Sometimes, as you suggest, constants are added to
fit data and their physical meaning is unknown,
you might label such constants as "fudge
factors".

You may be interested in learning about Planck
units discussed in an
earlier answer . where new units are
expressed by combinations of the five
fundamental constants of nature.

QUESTION:
In a example of a electrical generator. Central in the generator are rotating magnets. I understand that as the magnet moves the cores of the surrounding coils will react producing a magnetic field. Also the copper wire winding around the coil will react.
My question is does the copper wire winding react to magnetic field according to the field produced by the core or the moving magnet ? In a lenz law sense.
Plus would I be correct in thinking if there is no circuit on the copper winding there no organised field would form on the copper wire winding itself.

ANSWER:

The best way to understand the priciples is to look at
the simplest possible generator. The figure shows a
generator consisting of a single loop rotating in a
magnetic field which is approximately uniform. There
would be no difference in the results if it were the
magnets rotating around the coil as you describe. Since
the induced EMF in the loop is proportional to the time
rate of flux through the loop and the flux is
proportional to the cosine of the angle θ between the field
and a normal to the area of the loop, the induced EMF
will be sinesoidal. The angle may be written as θ=ωt where
ω is the angular velocity of the loop (or magnets)
and t is some arbitrary time, so the
voltage is proportional to cos(ωt ). If
there is a load across the terminals, and therefore a
current through the loop, there will be a field due to
the current which does not affect the output voltage if
the source of power maintains a constant angular
velocity. If you are drawing power from the generator, it
is harder to "turn the crank" than if there is no current
flowing. If you need detailed information on "real-life"
generators, it is more an electrical engineering question
than physics.

QUESTION:
How did Einstein derived general relativity math when he understood that the gravity is just the space bending around an object how did he applied this to the world of math ?

ANSWER:
You can learn about how Einstein struggled with the
mathematical methods necessary to fully describe general
relativity and with which mathemeticians he consulted in
his biography by Walter Isaacson, "Einstein: His Life and Universe ".

QUESTION:
I have been thinking about clocks lately and found myself in a sort of a pickle over the velocity of a dial. (I am not native so sorry if my description makes it very complicated). Suppose we have a disc with a dial starting at its center. The dial draws two circles on that disc as it turns. The first circle has its radius equal to r, whereas the second circle has a radius of 2r. Now suppose that 1 full turn takes 1s (which would be the same for both circles since they are being drawn by the same dial). Now because of the difference of radii, there is also a difference in their circumference
c. And so we've got t=1s, c=2πr and c'=4πr (2π2r). Now here's the question - if the time is the same, but the circumference or distances are different does that mean the velocity of these two points on the same dial is different? How is that possible since the dial turns with a set velocity which in my understanding would be the same for each point on that dial?

ANSWER:
It is simply because all points on a rotating rigid body
have the same angular velocity, (revolutions per second,
e.g .) but not the same translational velocity (miles per
hour, e.g .). In general, v=rω
where v is the translational speed, r is the distance
from the axis of rotation, and ω is the
angular velocity in radians per second.

QUESTION:
I was just wondering why does the centripetal acceleration formula work a=v^2/r?

ANSWER:

Why does it work? It "works" because it is correct. Maybe
you wanted to ask "where does it come from?" or "how is
it derived?" I will give you a brief standard derivation.

On the left of the figure I have drawn the particle
moving with constant speed v around a circle of radius R.
In some time interval Δt the particle moves through
an angle theta and has velocities v _{1}
and v _{2} but both
speeds are equal to v , v _{2} =v _{2} =v .
On the right I have placed the velocities tail-to-tail
and drawn the difference vector v _{2} -v _{1} =Δv .
So now I see two isoscoles triangles, each with the same
angle between their equal sides; they are therefore
similar triangles and so we can write v /Δv =R/Δs.
If I now rearrange and divide each side by Δt ,
I find Δv /Δt =(v /R )Δs /Δt.
Now, Δv /Δt is the magnitude
of the average acceleration; to get the instantaneous
acceleration, we must take the limit as Δt approaches zero.
But, when Δt approaches zero, Δs /Δt approaches
v . So, finally, a=v ^{2} /R .

QUESTION:
Is Planck time a concept derived from the theory of relativity or its premises?

ANSWER: No, that is not where the concept
came from. The origin of
Planck units came from the
desire to have a system of units which are based only on
constants of nature, not on the specific units like the
meter, the second, etc . You take the five
universal constants:

the speed of
light in vacuum, c,

the universal
gravitational constant, G,

the rationalized
Planck's constant, ℏ,

the Boltzman
constant, k _{B} , and

the Coulomb
constant, k _{e} =1/(4πε _{0} ).

You now form
combinations of these constants which have the correct
dimensions of the units you want:

length,

mass,

time,

temperature, and

electric charge

I have copied the table from
Wikipedia showing the values of these Planck units in SI
units:

It is hypothesized that, at
distances smaller than l _{P} =1.6x10^{-35}
m, the usual physics as we know it will not work. I think
you might say that the origin of the idea of Planck units
is quantum field theory.

QUESTION:
A toy top is a disk-shaped object with a sharp point and a thin stem projecting from its bottom and top, respectively. When you twist the stem hard, the top begins to spin rapidly. When you then set the top's point on the ground and let go of it, it continues to spin about a vertical axis for a very long time. What keeps the top spinning?

ANSWER: There is a property of spinning
objects called angular momentum. For example, the angular
momentum of your top is a vector approximately equal to ½MR ^{2} ω
where M is the mass of the top, R its radius, and ω
is the rate at which it is spinning; the direction of the
angular momentum vector is straight up if it is spinning
counterclockwise if viewed from above, straight down if
clockwise.
There is a physical law which states that if there are no
torques on a spinning object, its angular momentum never
changes. There are no torques on the top if its spin axis
is vertical and so it does never stops.

QUESTION:
The third dimension is 3D (spheres, etc.). The first dimension is a dot. Everything that has a front and back, has an edge. So isn't a dot 3D?

ANSWER: A dot has zero dimensions. The
formal definition of the dimensionality of a space is an
answer to the question "how many numbers does it take to
specify the location of a point in that space?" You may
heard of Cartesian coordinates, (x ,y ,z );
they were named in honor of the 17th century philosopher
and mathematician
René Descartes . Legend has it that he
was once watching a fly buzzing around his room and asked
"what is the minimum number of numbers I must specify to
describe the position of the fly at any time?" He figured
it was three: the distance up from the floor, the
distance from the back of the room, and the distance from
one of the side walls; his room is said to have three
dimensions. One dimension is a line; here you simply
measure the distance from one point on the line to
wherever the particle is. The line need not be a straight
line. A circle is one-dimensional and the location of a
point on the circle is usually specified by an angle
relative to some point we call zero. Two dimensions is a
surface. The surface of the earth is two-dimensional, the
position being angles, latitude and longitude. One
surface of a flat sheet of paper is two-dimensional and
the two numbers are usually Cartesian coordinates (x ,y ).
The spatial universe in which we live is
three-dimensional, just like Decartes' room. A solid
sphere has three dimensions usually measured by the
distance from the center and two angles (longitude and
latitude).

QUESTION:
Why there is only time dilatation and no time shrinkage? For me I think there should be time dilatation regarding objects traveling away from each other and shrinkage regarding those closer to each other. As what makes one gets older and the other still young for both of them are moving with same speed relative to each other.

ANSWER:
Thinking how you think something should be does not make
it true! You should read my answer about the
twin paradox . How fast a clock runs and how fast it
appears to actually runs are two different things; see
earlier answer .

QUESTION:
While we use the combination of rear and front brakes to stop a motorcycle, I mean we can chose between the the front and rear without any effort, why do cars primarily use the front brakes to stop? Why does the brake pedal engage only the front brake? The rear brake is only seem to be used when the car is parked

ANSWER:
All modern cars have four-wheel braking. So where did you
get the idea that the brake pedal engages only the
front-wheel brakes? Parking brakes engage only rear wheel
brakes. Your confusion may be that the braking causes the
front-wheel tires to wear more than the rear-wheel tires;
this is due to the car slightly "rocking forward" when
stopping and also because in a front engine car more of
the weight is likely to be supported by the front tires.

QUESTION:
This is a kinematics question about
the motion of wheels; specifically, about how friction between a wheel
and a surface causes forward motion of the entire wheel. There's
something unsatisfactory to me about how this forward motion is
typically explained.
Consider the rear wheel of a bicycle; it is driven by torque applied to the wheel (via chain drive connected to pedals, etc.). In the classical treatment of this scenario, it is claimed that the friction between the wheel and the road surface, which is a
forward-pointing force, causes forward motion of the wheel. But this frictional force is tangential! So wouldn't this tend to be a rotational force instead of one of displacement? It would seem to me that somehow the tangential rightward force of friction at the point of contact needs to be transformed into a rightward force applied at the *center of mass* (i.e., at the axle) of the wheel in order to explain an induced forward/rightward motion, but no source that I have found has attempted to explain this gap.

ANSWER: The figure shows all forces on
the rear wheel of a bicycle:

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

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

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

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

F ,
the force exerted by the rest of the bike on the
wheel.

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

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

The sum of all
the torques on the body about any axis is equal to
the product of the moment of inertia (I ) of and the
angular acceleration (α ) about that axis.

You seem to
think that a force which exerts a torque about the
axis you have chosen (I will choose the axle for
the bike wheel) only can cause angular
acceleration; in fact, every force on the object
contributes to the translational acceleration.
Suppose now that you are standing on the pedal but
also applying the front brake such that the bike
remains at rest; both translational and angular
acceleration are zero. So the translational N2
yields two equations, N-W-V =0 and
C+f-H =0; the rotational N2, choosing the axle
as the axis, yields Cr-fR =0. (F ,
W , and N exert no torque about
the axle.) If you now release the brake,
H will get much smaller and
the wheel will start moving forward and rotating.

If you are
interested in the entire bike, not just the rear
wheel, the force C has no contribution to the
motion because it is an internal force; the
chain exerts a force C
on the rear sprocket but a force -C
on the front sprocket netting zero net force.

FOLLOWUP QUESTION:
You say, "in fact, every force on the object contributes to the translational acceleration" but the force, f, in your answer, is not directed at the center of mass of the wheel.
I guess my question is simply, "why is f treated as a force directed at the center of mass and not as a torque on the wheel"? I don't understand where the forward acceleration of the *center of mass* of the wheel is coming from.

ANSWER: You did not read my answer
carefully. The translational N2 is "The sum of all
forces on a body is equal to the product of the mass
(m ) and the acceleration (a ) of the
center of mass of the body". Note that it does
not say
…the sum of all forces which are directed through the
center of mass … Let me give you a simpler
example. Suppose that you are in the middle of empty
space and you have a wheel which has a string wrapped
around its circumference. You pull on the string so that
the tension in that string is the only force on that
wheel and it is tangential . Surely you do not
think that your pulling will cause the wheel to only spin
and not accelerate toward you too. Here is another
example, a yoyo. There are two forces on it, its own
weight which passes through the center of mass and points
down and the tension of the string which you are holding
which points up and does not pass through the center of
mass. If the string does not affect the translational
acceleration, the only force on the yoyo would be its
weight so it would fall with acceleration due to gravity,
g , which it clearly does not do.

SECOND FOLLOWUP QUESTION:
In classical billiards problems, for example, when a force is applied to a ball, the force has to be decomposed into its normal component directed at the center of mass (call it f_n) and the tangential component (call it f_t). Only f_n will cause translation, right? It's my understanding that f_t will only contribute a torque, but not to translation. So why is the bike case any different? Contact friction between the wheel and the surface is a tangential force, so, according to my understanding, it can not cause translation, because it has no component directed at the center of mass of the wheel...

ANSWER: Look at the figure*. It is much
better to resolve F into its
horizontal and vertical components rather than normal and
tangential components. It is wrong to insist that a force must be
directed toward the center of mass for that force to
contribute to tranalational motion; my examples above
should have convinced you of that. Now, a billiard ball
is confined to move on the table, so there will be no
translational motion in the vertical direction. So the
ball is in equilibrium in the vertical direction, so
N-V-W =0. But in the horizontal direction there will
be an acceleration, ma=H-f . Only f
and F will exert torques and it
will be a little tricky to calculate the torque due to
F but just geometry. I have
drawn the line of force to be below the center of mass;
if it is through the center of mass, it will exert no
torque and only f will cause
angular acceleration.

Here is the take-away for you: just use N2 as I have
given it to you, always look at all forces on the object,
and do not continue insisting that a force which exerts a
torque does not affect translation.

QUESTION:
A hammer hitting an anvil creates energy (sound) which moves in all directions at about 700 mph in atmosphere. Sound energy is also created by a large fire. In space, vaccuum, that same energy must also be present if a hammer hits an anvil in space, same as a star (a large fire ball) should also create sound energy, along with all the other energies it creates, so if sound energy in space, a hammer hitting an anvil, does not have atmosphere to slow it down, or to make noise, how would one measure that energy, its speed and pressure. Could this be what is called "Dark Energy".

ANSWER:
Sound is a wave that exists in a solid, liquid, or gas.
If you are in a vacuum, there is no sound. The energy
which the sound carried away in air is instead retained
as sound in the anvil (and hammer). But this sound
eventually damps down and where that energy goes is an
increase in the temperature of the anvil (and hammer).
This energy eventually radiates away, not as sound but as
electromagnetic waves (mainly infrared) until thermal
equilibrium is achieved. It certainly has nothing to do
with dark energy.

QUESTION:
So my question is a bit out of reality question as its not possible to do this in real life. But if you were able to turn a cannon ball into a musket ball and fire it from a flintlock, having the ball turn back into a cannonball apon exiting the gun, would the cannonball retain its velocity after changing or would it lose all energy and drop to the ground?

ANSWER: Essentially what you are asking
is if an object with a certain mass m and speed
v suddenly loses or gains mass, what happens?
The mass either disappears without trace or appears
without a source. Suppose your musket ball has a mass
m and speed v and your cannon ball has a
mass M . Since there are no external forces on
the ball, linear momentum must be conserved, that is
mv=MV where V is the speed of the cannon
ball. So V=mv/M , much smaller than v .
However, the energy is not conserved: E _{1} =½mv ^{2} and
E _{2} =½MV ^{2} =½mv ^{2} (m /M )=(m /M )E _{1} ;
a large amount of energy is lost.

QUESTION:
How many pounds per inch does a lacrosse ball exert when it hits a surface? For info: Ball is traveling 90mph. A lacrosse ball weighs 8 ounces. The ball is 5.25 inches in circumference. As a sidebar, does it matter if the surface the ball hits is stiff like a wall or springy like a bounce-back? If so, could you expand on the difference in pressures against the two different surfaces?

ANSWER: If it is pressure you want, you
should ask for pounds per square inch. But I do not think
that is what you really want; it is more straightforward
to estimate the average net force the ball exerts on
whatever it hits. Then you could think about pressure,
how that force is spread over the area which the two
surfaces have touching; pretty hard to do that, though.
If something with mass m with a velocity v _{1
} collides with a much more massive object at rest and
rebounds with velocity v _{2} , the
average force F can be estimated as F≈m (v _{2} -v _{1} )/t
where t is the time the collision lasts. ("Much more
massive" means the struck object does not recoil
significantly.) You seem to want to get answers in
imperial units (lb, ft, s); without going into details,
m =0.5 lb/(32 ft/s^{2} )=0.016 lb·s^{2} /ft=0.016
slug
and v _{1} =90 mph=132 ft/s. Suppose that
v _{2} =0, called a perfectly inelastic
collision; then F =-132x0.016/t =(-2.1
lb·s)/t . So, the smaller the time, the
larger the force. (The fact that the force is negative
means that this is the force the struck object exerts on
the ball because I have chosen the incoming velocity to
be positive. The force the ball exerts on the object is
positive and equal in magnitude.) That addresses your
second question: the ball will certainly stop in a much
shorter time when hitting a hard surface than a soft one.
Of course that should not be too surprising to you
because you know that if you fall onto a mattress it
hurts much less than if you fall on a concrete floor. So,
if t =1 s, F =2.2 lb whereas if t =0.01
s, F =220 lb. Now, if the ball hits a hard
surface it does not penetrate very far and therefore the
force is spread over a small area and pressure is large.
If the ball hits a soft surface, it penetrates deeper and
therefore the force is spread over a larger area and
pressure is smaller.

QUESTION:
Since this isn't technically a homework question and was a challenge put out to the class to see if anyone can do it, I hope you will consider it.
I'm attempting to do some fun physics maths as a challenge from our professor. It may be a trick question though since she's done that in the past. The problem assumes that c is only 30000 kmph. The problem was to determine the time to reach 29999 kmph from 25000 kmph with a total of 6.48MN of force in a vehicle that weighs 168000kg (168t) in the vacuum of space, too far away from any celestial body to have gravity noticeably affect it.
So just to clarify, there is no air resistance or gravity or time dilation. Relative mass increases when reaching said 30000kmph, and the goal is to find the time it takes to accelerate to 29999kmph from 25000kmph.

ANSWER:
I don't see the point of this odd choice of c ;
it makes the problem neither easier nor harder, just
unphysical! You will still have to do some
work! I am going to point you in the direction of solving
it yourself. First, go to an
earlier answer ; you will see that I have solved your
problem but starting at rest. If you start at some
initial velocity v _{0} , the result is
that Ft=p-p _{0} . Here, p is the
relativistic linear momentum, p=mv /√[1-(v /c )^{2} ]
where m is the rest mass, 1.68x10^{5} kg
in your case; similarly, p _{0} =mv _{0} /√[1-(v _{0} /c )^{2} ].
So now you know everything except t , so just
solve for it; this is just algebra/arithmetic. I would
advise you to convert everything to SI units (kg, m, s)
before doing your calculation. Then t will work
out to be seconds.

I do not understand what you mean by "…there
is no…time dilation…" since time dilation
is simply not of any interest here but would come into
play if the whole thing were seen by an observer in a
different frame. Also, I almost never apply the idea of
relativistic mass because I view the linear momentum as
being the quantity redefined in relativity, not mass.
Read my discussion of this in
one of the links in the earlier answer.

QUESTION:

Atoms are mostly empty space. How can they form solid objects?

ANSWER:
First of all, atoms are not mostly empty space. See an
earlier answer . Although almost all the atom's mass
is in a volume much smaller than the volume of the whole
atom, the electrons fill the rest of the space in a
"cloud". Bohr's model of the atom, tiny electrons in tiny
orbits, is wrong. If you need a simpler explanation
without reference to the "electron cloud" think of all
the atoms connected to their nearest neighbors by tiny
springs. Each atom connects to its neighbors by some
interaction which can be approximated as a spring. The
atoms close enough to interact with each other will stick
together and thus form a solid object.

QUESTION:
2 balloons, one filled with 1 cubic meter of air, and the other with 1 cubic meter of Helium, are lowered to a depth of 50 feet under water and simultaneously released.
Will they reach the surface at the same time?

ANSWER: With the information you gave me,
I will have to put stipulations on the problem. The
baloons are rigid so they do not get compressed under the
water, they have identical shapes, their masses (without
being filled) are the same, and both are filled to
atmospheric pressure. There are three forces on each
balloon: its own weight -mg , the buoyant force
of the water B , and the drag force opposite the
velocity v
which has the form -Cv ^{2} where C is
some constant which depends only on the shape of the
balloon. So the net force on each balloon is -mg+B-Cv ^{2} =ma
where I have put this equal to the mass times the
acceleration, Newton's second law. When you release each
balloon, each will accelerate up and, as v increases,
eventually a =0 and it will have a constant speed
v _{t} ,
called the terminal velocity, thereafter. It is easy to
show that v _{t} =√[(B-mg )/C ]
The only difference between the two is m which
is smaller for the helium which therefore has the larger
terminal velocity. The helium balloon reaches the surface
first.

QUESTION:
does a bouncing ball possess simple harmonic motion?

ANSWER: A bouncing ball is not an example
of simple harmonic motion (SHM). By definition, the
ball's motion must be describable by simple sinesoidal
functions of time t , i.e. like y (t )=A sin(ωt )+B cos(ωt )
where A , B , and ω are
constants and y is a variable describing the motion, for
example the height above the ground in your case. If the
ball were perfectly elastic, the motion would be
harmonic, but not SHM, because y (t ) would be a train of
identical downward-opening parabolas. If it were a real
ball, it would not even be harmonic because each
subsequent bounce would be a little smaller that the
previous bounce.

QUESTION:
What exactly is the 4th dimension? How do we know it’s real? How are space and time related?

ANSWER: I have
recently answered this question in some detail.

QUESTION:
As I understand it, electricity moves at the speed of light. Conductivity is where I'm getting stuck on. Are we able to measure speed from it? I tried searching "siemens/meter to kph" with no results.
My questions here are these: If silver has a conductivity of 63x10^6 siemens/meter, and copper has a conductivity has a conductivity of 59x10^6 siemens/meter, does that mean energy in silver is travelling faster than copper, or something else? But, silver can't go, say 101% the speed of light. So, there's obviously something else I'm not understanding.
I guess the point I'm getting at is this: What does conductivity mean, and can you convert siemens/meter into a speed calculating how many particles move through 1 meter of wire in x amount of time, thus making copper's speed x time/meter, while silver is y time/meter?
I'm sorry if this is a bit confusing, or not even your field. But, I figured electricity is, in some part, in the field of physics.

ANSWER: I am going to give you a
qualitative overview since you seem to not have any
understanding of what goes on in a conducting wire. If
you want a more detailed discussion of the simple model
of electric currents, you can find it in any introductory
physics text book. A conductor has electrons which are
moving around at random in the material, not really bound
to any of the atoms; these are called conduction
electrons and they can be thought of as a gas of
electrons. But there is no flow of current because the
electrons move randomly so there are just as many going
in one direction as there are going in the opposite
direction. When this wire is connected to a battery an
electric field is established inside the wire which
points from the positive terminal to the negative
terminal. When you say "electricity moves at the speed of
light", what that means is that the establishment of the
field moves with speed c ; the electrons do not
move at that speed. So all electrons see the field turn
on almost immediately. All electrons, having
negative charge, begin accelerating in the direction
opposite the field. If the only force seen by the
electrons was due to the field, they would accelerate
until they reached the positive terminal and left the
wire. But, even though they are not bound to atoms, they
certainly see the atoms and when they hit one they are
momentarily stopped or scattered. So each electron is
constantly being accelerated, then stopped, then
accelerated and the net result is that, on average, all
the electrons participating in the current move with a
very small velocity, on the order of one millimeter per
minute; this is called the drift velocity. All the
electrons are continually gaining and then losing kinetic
energy and that lost energy shows up in the heating up of
the wire. What determines the conductivity is the number
of conduction electrons per unit volume of the material
as well as its crystaline structure and how individual
atoms scatter the electrons.

QUESTION:
The classic escape velocity formula takes in consideration an uniform acceleration of gravity, g. What would be a mathematical model for escape velocity where the force of gravity varies as the body distances from the planet?

ANSWER: "The classic escape velocity formula"
that I know does not assume a uniform gravitational
field, it assumes the correct field. Let's see where it
comes from. The potential energy for a particle of mass
m a distance r from the center of a large
spherical mass M is, choosing zero potential
at r =∞ is V (r )=-mMG /r
where G is the universal gravitational constant.
If m has a speed v at r=R , where
R is the radius of the
earth, the total energy is ½mv ^{2} -mMG /R=E .
Now, suppose we want v to be big enough that m will go
all the way to r =∞ before it finally
comes to rest; then E must be zero. Therefore
v _{e} (R )=√(2MG /R );
but, MG /R=gR , so v _{e} (R )=√(2gR ).
That is the escape velocity at the surface of the planet
(ignoring any other forces, notably air drag) where g
is the acceleration due to gravity. Now, you want it
everywhere else. That's easy, just replace R by
r , v _{e} (r )=√(2MG /r ).
You could use g=MGr if g now means the
acceleration due to gravity at an altitude of r-R .
(Incidentally, this does not work anywhere inside the
planet.)

QUESTION:
I have a bit of a debate with colleagues and looking for a professionals take. Iï¿½m a plasterer and when mixing up the plaster introducing too much air to the mix makes it less workable. Some people I work with think mixing counter clockwise introduces less air to the process than clockwise, I think it would make no difference.

ANSWER: Well,
I just so happen to have a plaster/paint stirring tool
which is powered by a drill. Examine it carefully and you
will see that clockwise and counterclockwise are not the
same in their effect; if the direction is clockwise (as
viewed from above), the blades push down on the mixture
and if the direction is counterclockwise, the blades push
up. I do not know for sure which, if either, of these
would introduce more air. My best guess would be that for
the clockwise rotation there would be a whirlpool into
the surface, whereas for the counterclockwise rotation
there would be more of a hump on the surface; I would
think the whirlpool would pull in more air. I must also
note that not all stirring tools are of the same design
as mine.

QUESTION:
Why does this toy move in a straight line? Shouldn't the toy rotate due to gyroscopic precession?
https://youtu.be/hs4iv7IHvGg

ANSWER:
Well, it does not move
in a straight line. One thing I noticed is that the toy
starts moving immediately when released; this indicates
to me that the surface it is on is not level. Just
because it is a gyroscope does not mean it will precess.
There must be an external torque acting on it; imagine an
axis running between the two legs about which you would
calculate torque. In the figure that I have clipped from
the video you can see the motor and battery and the cds
which are rotating; the toy is leaning such that I would
certainly expect the center of gravity to be beyond the
being above the axis and, were the discs not spinning the
toy would certainly fall. If you could put the center of
gravity directly above the axis, it would not precess.
You can see several instances in the video where the toy
is moving in a curved path, and in one case the path is a
pretty tight circle; these are due to precession because
of the torque due to the weight.

QUESTION:
Would you be marginally taller when on top of a tall building/mountain or at the bottom of the ocean. Does the pressure/gravity change and if so which one makes you taller?

ANSWER: If you are in the the vicinity of
the earth there are two forces which are different
between your head and feet, tidal force and pressure
force due to the fluid you are in. The tidal force is due
to the fact that the gravitational field at your feet is
slightly larger than at your head, both pointing down;
therefore the net force on you tends to stretch you
taller. We can estimate the tidal force because the
gravitational field falls off like 1/r ^{2} : F _{feet} /F _{head} =(r+h )^{2} /r ^{2} =[1+(h /r )]^{2} ≈1+(h /r )^{2}
where r is the distance to your feet from the
center of the earth and h is your height; I have
used the binomial expansion because h /r <<1. So, F _{feet} -F _{head} ≈F _{head} (h /r )^{2} .
Clearly, either force is going to be approximately equal
to your weight, so the tidal force is about mg (h /r )^{2} ;
if I take h =2 m, mg =1000 N, and r =6.4x10^{6}
m, this is about 10^{-10} N (2.2x10^{-11}
lb).

The pressure force is
down on your head and slightly larger and up on your
feet; so the pressure force tends to compress you
shorter. The net force on you is simply the buoyant force
which equals the weight of the displaced fluid. If I take
your volume to be about 0.02 m^{3} , the buoyant
force force is about 1000 N in water and 1 N in air.
So, as you can see, the stretching due to the tidal force
is negligible to forces due to fluid pressure. So, you
will be shorter than your "real" height, but more so in
water than air. Therefore you are taller up high in the
air than down low in the water.

QUESTION:
I know that at sea level, atmospheric pressure is 14.7 pounds per square inch. I know that the higher in elevation you go, the thinner the air gets and a corresponding decrease in the amount of oxygen such that trying to survive above 16-18k feet is not very easy. My question is this: If you where to drain all of the oceans so that the existing atmosphere before draining was now occupying the void left by the water, would you be able to survive at what was sea level? My thought is that the atmosphere would become so diluted with the oceans gone that you would actually have to travel way below what was sea level to able to survive the atmospheric pressure change and the oxygen dilution. Wondering about this has always bugged me lol.

ANSWER:
The volume of all the oceans is about 1.4x10^{18}
m^{3} and their average depth is about 3700 m
(about 12,000 ft). Now, dealing with the volume of the
atmosphere is tricky because the density of the
atmosphere gets smaller with altitude so you have to be
careful how you describe volume. The mass of the whole
atmosphere is pretty well-known, though, to be about
5.15x10^{18} kg; so you can calculate the volume
if the whole atmosphere were at sea level pressure where
the density is 1.225 kg/m^{3} , 5.15x10^{18} /1.225=4.2x10^{18}
m^{3} . So, if the atmosphere were uniform, about
1/3 of it would flow into the emptied oceans. But, it is
not uniform so less than 1/3 would come; there would be
some base pressure 3700 m below the sea level (since
there is no sea now, when I say sea level I mean what it
was) which I would estimate to be about the same as the
pressure at sea level was because 3700 m is small
compared to the radius of the earth, 6.4x10^{6} m
(0.06% smaller). Since the oceans occupy about 70% of
earth's surface, sea level pressure would be about what
the current pressure is at about 12,000 ft, the height of
a modestly high mountain and certainly habitable. The
highest habitable place on earth is about 16,000 ft,
so lots of places (like Denver) habitable now would not
be. And certainly nobody would be climbing Mt. Everest in
this new world!

QUESTION:
What are the signs that show scientists that some particles have a quark structure and others do not?

ANSWER:
Quarks are something which were hypothesized after most
of the elementary particles and their properties were
known. There are three kinds of particles, hadrons,
leptons, and field quanta. Leptons are few, electrons,
muons, and neutrinos. Field quanta are also few, photons,
gluons, Z bosons, W bosons, and the Higgs boson. Hadrons
are many, the proton and neutron being the best known,
but what has been referred to as a "zoo" of other
hadrons. To try to systemize and explain this array of
particles, first group theory and then, building on that,
introducing hypothetical particles called quarks was
extremely successful. There is no way I could answer
"what are the signs" because the "signs" are all the
hadrons and their properties.

QUESTION:
What is the difference between a force and weight

ANSWER:
A force is a push or a pull. Weight is the name for a
specific kind of force, one which is caused by gravity.
For example, your weight is the force which the earth's
gravity pulls you down; the gravity on the surface of the
moon is smaller and therefore your weight is smaller
there. All weights are forces but all forces are not
weights.

ADDED
COMMENT: You should be aware that a few textbooks
define weight as being what a scale measures. So if you
are in an upward accelerating elevator you have a larger
weight. I find this to be total nonsense.

QUESTION:
If time slows down near massive objects, how fast is time in the space between Galaxies ?

ANSWER: Suppose that your clock ticks off
a time t if you are in a region of space where there is
essentially zero gravitational field. Now imagine bringing
your clock a distance R from a spherical,
nonrotating object of mass M . The same time interval will
now be given by t'=t √[1-(MG /(Rc )]
where G =6.67x10^{-11} m^{3} kg^{-1} s^{-2}
is the universal gravitational constant and c =3x10^{8}
m/s is the speed of light. For example, let the location
be on the surface of the earth, M =6x10^{24}
kg, R =6.4x10^{6} m; then, t'=t √[1-7x10^{-10} ]≈t [1-3.5x10^{-10} ].
So, for example, the clock in space would click off 1
second while the clock on the earth would tick off
1-3.5x10^{-10} seconds. To see significant
gravitational time dilation requires a larger M ,
for example a black hole; for a black hole with a
Schwartzschild radius R_{S} , the time
dilation equation becomes t'=t √[1-R_{S} /R ]
so time stops when R=R_{S} . If R= 2R_{S} ,
t' =0.7t .

FOLLOWUP QUESTION:
Previous answer did not address this question :
If time slows down near massive objects, HOW FAST IS TIME IN THE SPACE BETWEEN GALAXIES?
Could you do the math at a distance of 10,000 light years from any mass, please.

ANSWER: There is no answer to that
question because time depends on the observer, it is
relative. Any observer in any frame of reference and in
any gravitational field will say that it takes any clock
in his frame 1 s to advance 1 s. Let's take the case of
the black hole example above. The observer in empty space
sees his clock advance 1 s but observes the clock of the
girl near the black hole to advance in 0.7 s in that
time. The girl near the black hole sees her clock advance
1 s but observes the clock in empty space to advance 1.4
s in that time.

QUESTION:
In the theory (?) of quantum entanglement, maybe you could clear up some basic things for me: a) in order for two particles to be entangled - no matter how far apart in space they are - must those same identical two particles have been physically 'partnered' up at some point in the past?
b) If a particle in a spin up/spin down position is OBSERVED, it chooses (?) to become spin up or spin down. WHY does the act of observation force it to make that choice? Or is that even the right question? I understand that this idea of observation causes change goes to the heart of quantum physics, but I'm still trying to get my head around the WHY.

ANSWER: The
answer to your first question is yes, the particles
become entangled by interacting with each other. So, for
example, if you had two electrons, each with a wave
function which is half up spin, half down, unless they
have been put into those states by a mutual interaction,
they are not entangled. For the second question, the
particle does not "choose" anything. Rather, the
measurement puts the particle into the state
which you observe. The result of a measurement is to
observe a single state, not an admixture. So if you have
an electron whose wave function is half up, half down,
your measurement will either give one or the other, not
both; if you observed a large number of electrons, all
with the half and half wave functions, you would observe
spin up for half your measurements and spin down for the
other half.

QUESTION:
The longer the garden hose the less the pressure at the end - Is this a true statement, if so, how much less?
It's not a homework problem, it's a heated discussion with spouse, we're 76

ANSWER:
There is a pressure drop for a fluid flowing through a
pipe. The main reason is that there is frictional loss
due to viscosity of the water and the rubbing of the
water on the walls of the hose. The loss depends on the
size and length of the hose, the density of water, the
flow rate, the viscosity, a constant characterizing the
friction for the material, and the change in height if
the hose is not horizontal.
The physics is very complicated, but there is a handy
on-line calculator where you can get at least a
qualitative estimate of how big an effect it is. I did a
calculation for water in a 1 inch diameter, 100 foot
rubber hose, and for a flow rate of 3 gallons per minute;
I assumed that the hose is horizontal and straight. The
results are shown in the figure. The calculated pressure
drop is 0.38 psi which you can compare with a typical
in-city water pressure of about 40 psi. So the drop is
only on the order of 0.1%. If the water flow were much
faster, say 20 gallons per minute, the drop would be
about 10 psi, a 25% drop.

QUESTION:
I have a question, if the pulling force of mass or acceleration or electromagnetism can cause time dilation... would a pushing force of the repelling force of electromagnetism or the expansion rate of the universe cause reverse time dilation?

ANSWER:
Only speed (magnitude of velocity) matters for
time dilation. It is irrelevant how that speed
was acquired, attractive or repulsive force.

QUESTION:
What is the meaning behind multiplication in physics? Is multiplication in physics purely mathematical or there is a physical explanation to it? How do we explain the product for example, s=v.t? Is there any meaning behind this? For example, I can say that "Distance is defined as the product of velocity 'times' time"? But what does this even mean?

ANSWER: Suppose you are in 6th grade.
Multiplication is explained as simplified
multiple addition, the product of 5 and 3 is 5
plus 5 plus 5. Then we learn all our
multiplication tables to not have to figure out
all the simple products we might need. This is
arithmetic and, indeed, we are taught that it
often means something, like in the example
above, if we have three baskets, each containing
5 apples, we have 15 apples. And then it starts
touching on physics. The area of a rectangle is
the height times the width; or, like your
example (but not exactly) the distance traveled
is the product of the speed times time. But the second
example is sort of ambiguous if you think about
it. What does "distance traveled" mean? It might
mean the distance between the starting and
ending points; but if you get there on a winding
path, that is not the really the distance you
traveled. The problem is that before you get to
physics, there is only one kind of number,
called a scalar quantity, which you can specify
using only a number. Examples of scalars are
time, speed, length, area, etc . But, in
the physical world, some quantities require more
than one number to adequately describe them. The
simplest are vectors which require a
specification of both magnitude and direction.
Examples are force (e.g. 10 lb straight
up), velocity (e.g. 25 mph due north),
displacement (e.g. 5 ft straight down).
Vector quantities are denoted as bold-face, for
example F =10 lb
straight up. Now multiplication takes on new and
different meanings, and the ways we multiply all
have a meaning. The easiest to grasp is how to
multiply a scalar times a vector: as in
arithmetic, multiplication is just repeated
addition, so 3F =F +F + F .
But, what is the multiplication of two vectors?
There are two ways we could imagine multiplying
vectors, namely the product is also a vector or
the product is a scalar. The scalar product or
dot product is defined as
A·B =AB cosθ _{AB}
where θ _{AB}
is the angle between the vectors
A and
B and A and B
are the magnitudes of the vectors. An example of
a scalar product is the work done by a force
F which acts over a displacement
d , W =Fï¿½d .
The scalar product is commutative, i.e .
A·B =Bï¿½A .
The vector product or cross product is defined
as
A×B =u AB sinθ _{AB}
where u is a
dimensionless vector of magnitude 1 which is
perpendicular to the plane defined by
A and B .
You can do additional research on your own if
you want to understand whether u is "up" or
"down" relative to the plane, but vector
products are not commutative,
A·B =-Bï¿½A .
An example of a vector product is torque
τ caused by a
force F a displacement
r from an axis,
τ =r×F .

QUESTION:
Something that confuses me is that scientists say it takes approx. 8min for light to travel from the sun to earth. So if the sun were to explode we wouldnï¿½t know till 8min after. However Einstein gathered that the faster an object travels the more time slows down so as you approach the speed of light time stops. But isnï¿½t this just relativity? If you are moving that fast everything seems like itï¿½s not moving cause your moving so fast. If time were to actually stop how could it take 8min for us to realize the sun exploded wouldnï¿½t we realize it instantly? So why canï¿½t you move faster than the speed of light? Wouldnï¿½t that just mean things relative to you would seem to be stopped for longer, meaning time doesnï¿½t stop when you reach the speed of light?

ANSWER: See the
faq page
for an explanation why you can't move faster
than the speed of light. Suppose that someone
was moving with speed 99% of the speed of light
from the sun to the earth, a distance of about 8
light minutes. Because of length contraction the
distance would, for her, be shortened 8x√(1-0.99^{2} )=1.13
light minutes and the time for her to get to
earth would be 1.1/0.99=1.14 minutes; so she
would observe the sun's demise to hit earth in a
time of 1.13 minutes, and she would arrive 0.01
minutes later. But, just because she observed
the time to be 1.13 minutes does not mean
someone on earth does; the earth-bound do not
see the distance to the sun to be shorter but
see it as 8 light minutes.

QUESTION:
Suppose you have two persons watching universe expanding, one of them is on earth, the other is moving fast towards the end of universe. Both of them are watching it all the time. Does this mean, that the "end" of universe exists twice and each is in a different "position"?

ANSWER: You miss the whole point of
special relativity, that the time and position
of some event do not have absolute values. That
the traveler will observe the end of the
universe to be in a different location and at an
earlier time than you do does not imply that the
universe ends twice at two different locations;
rather, it implies that time and position are
different for him than for you.

QUESTION:
I was telling my kids that time is space and explaining the block universe and they asked me, if time is space, then how many minutes are in a mile? It sounds stupid, like it does not have an answer, but the more I think about it, the more it seems like if space is time, it should have a specific answer, and if not, why not?

ANSWER: I will first give the simple
answer without much detail since you are dealing
with kids. But your kids must be fairly
sophisticated to be thinking about space-time at
all, so I will then give a little more
background on how space-time (four dimensions)
arises. When special relativity is mathematically
represented as suggesting a four-dimensional
space, the fourth dimension is not represented
by time (t ) but by x _{0} =ct
where c is the speed of light. So the
fourth dimension has units of length just like
the original three, x _{1} =x ,
x _{2} =y , x _{3} =z .
So, if x _{0} =1 mi=t ·186,000 mi/s,
then t =1/186,000 s=5.38x10^{-6}
s=9x10^{-8} minutes, approximately one
nanominute.

Before Albert Einstein developed special
relativity, it was generally assumed that time
and length are universal and independent. One
second for me is the same for anyone else in the
universe; similarly, one meter is the same
regardless of the position or motion of the
meter stick I am measuring. If someone is moving
with a speed v along my x axis and I measure the
length of his meter stick, I find it is shorter;
similarly, if I measure the rate his clock runs,
I find that it is slower than mine. We are
talking about meter sticks and clocks which, if
located in my reference frame, are identical to
mine. Mathematically this is expressed in what
we call a Lorenz transformation,

x' _{0} =γ (x _{0} -βx _{1} )

x' _{1} =γ (x _{1} -βx _{0} )

where the
primed coordinates are in the moving frame as
measured by the stationary frame, β=v /c ,
and γ =1/√(1-β ^{2} ).
So, two of the coordinates in our 4-space get
mixed up together if the other frame is moving.
Does anything analogous happen in everyday
3-space? The answer is yes; if you take a
coordinate system with three axes, (x,y,z )
and rotate it through an angle θ
about the z -axis, the new x'- and
y' -axes
both depend on what θ is and they
depend on both x and y are:

x'=x cosθ+y sinθ

y'=-x sinθ+y cosθ.

Hence the Lorenz transformation may be
interpreted as a rotation in the x-t
plane. This is the impetus of mathematically
working in a 4-dimensional space (space-time).

QUESTION:
I teach high school physics, and I recently performed a lab to investigate Lenz's law. Given the current state of things, this was done in my own home, with students on video chat timing my experiment with their phones.
My experiment involved dropping a cylinder of neodymium magnets (about 3 cm long and 1 cm in diameter) down a 1.5 m long copper pipe about 2 cm in diameter. I marked out different lengths on the outside of the cylinder, and used a steel nail to hold the magnet in place at these various ï¿½starting lines.ï¿½ So I could directly drop the magnet 153 cm from the top, and 130, 110, 90, 70 and 50 cm using the nail. For each distance, we did three trials. For each of those trials, seven students recorded through the Google Meet. I expected a bit of a hectic disaster, but the precision of the results was amazing.
My idea was to use the graph of distance vs. time to discuss what was happening (they havenï¿½t learned about Lenzï¿½s law yet) in terms of forces. I talked about different types of graph shapes for different situations and how they relate to what is happening with forces: uniform acceleration, uniform velocity and free-fall with air resistance (terminal velocity). My fingers were crossed that our data would approach linearity like the latter, and it did! A valuable learning experience for all. The one hitch is that the
z-intercept of our linear trendline is positive, not negative. When an object goes from rest to terminal velocity, I had shown them that the
z-intercept of the linear portion of the graph is negative, because the speed only increases to this section of the graph. With our results, I am forced to conclude that the magnet slows down before achieving linearity, because the average slope of the graph is steeper prior to the domain of our results.

ANSWER: The questioner said in his
initial message that the graph of the data would
be included but it was not. I have not been able
to get him to reply to any messages, but I spent
some time working this all out, so I thought I
would go ahead and post it. In doing my
calculations I approximated some of the
variables with reasonable guesses. I had three
neodymium approximately 1 cm magnets which when
stacked were about 1 cm; the mass of three was
10 grams so three cm would be about m =30
grams=0.03 kg. I took the acceleration to be
about g =10 m/s^{2} . I made a
wild guess at the terminal velocity by watching
videos on the internet and chose v =20 cm/s=0.2
m/s. I also note that many videos
seem to indicate that terminal velocity is
achieved very quickly. The resistive force is
given by F=-bv (not proportional to
v ^{2} as most air drag problems
are) where b is a constant to be
determined. I choose the coordinate z to
increase in the positive vertical direction and
the initial position to be at z =0 and
the initial velocity to be v =0. This is
a classic problem and I will just state the
results which result from solving the Newton's
second law equation, -mg-bv=m (dv /dt ):

v =(mg /b )[e^{-bt/m} -1]

=0.2(e^{-50t} -1)

z =(mg /b )[(m /b )(e^{-bt/m} -1)-t ]

=0.2[0.02(e^{-50t} -1)-t ]

Note that, for large bt /m , v≈-mg /b=- 0.2
which is the terminal velocity. This can be
solved for b , b =1.5 N/(m/s).

The plot above for v shows that it
takes only about 0.1 s to approximately reach
terminal velocity. The plot for z shows z (t )
in black and z (t ) if the
magnet moved with the terminal velocity the
whole time; the linear portion of the curve is
shifted up by 4 mm, possibly responsible for the
"positive z-intercept" observed by the
questioner. Of course these plots are not what
is being plotted by the questioner because each
new datum starts at rest; but every datum
plotted has the same behavior as the one I have
shown, so all the individual runs are shifted by
the same 4 mm. The shift is so small compared
with the total fall distances, the non linear
time is so small compared with the total fall
times that the experiment should be able to
extract a reasonably good measurement of the
constant b . Unfortunately, I do not
have the times of the various data points, so I
cannot be more specific; the time of fall from
the top if the terminal velocity is 20 cm/s
would be 7.5 s.

QUESTION:
Is following the line of steepest topographic descent (as stream would do) an example of conservation of energy?

ANSWER:
Let's take a stream as the specific example to
discuss the question. The reason the water flows
downhill is that there is a force acting on it,
the downhill component of the force of its own
weight. Generally a stream of water will tend to
follow the path where it finds the greatest
force impelling it down, that is the steepest
terrain. Now, if that were the only force on the
water, it would accelerate and the water would
be going faster at the bottom. But, in order to
do this, the stream would have to get narrower
as you went down because water is basically
incompressible and if you have, say, 100 gallons
per minute flowing at the top of the hill you
need the same flow rate at the bottom where it
is flowing faster. (Think of a hose nozzle where
the water flows much faster out the nozzle than
in the hose.) But this is not what happens
because there are drag forces which impede the
flow. What you generally find is that, if the
stream is about the same depth and width as it
moves down the hill, the speed is pretty
constant. Therefore the kinetic energy at the
bottom is the pretty much the same as at the top
but the potential energy has decreased;
mechanical energy has decreased, not been
conserved. However, if you consider the whole
system of the stream, the stream bed, the banks,
etc ., the energy is conserved because
the lost potential energy of the of the water
will show up as heat in the environment and the
water. However, this would be true regardless of
whether or not the water flowed down the
steepest path. So, I would not use the choice of
steepest path to be an example of energy
conservation.

QUESTION:
I tried posting this question on a
different website and didn't get much of a response.
For some reason there doesn't seem to be a lot of
information available about this subject, or maybe I just
can't find it. Anyway, I am hoping you can help shed some
light on the subject. Please forgive the formatting -
I've posted a couple of different questions, but if you
need a single, concise question to answer, the main one I
have is this: What's going on here?

I have a bottle of water that I put in a freezer for a
period of time, until the water becomes super-cooled and
is still liquid. Now we know that when we shake or
disturb the bottle, some of it crystallizes. But it
doesn't turn into solid ice. Instead it turns into slush
- a mixture of tiny ice crystals and liquid water. Here
are my questions: What determines what parts of the
water will become ice, and which will stay liquid?
What is the temperature of the slush mixture? Is the
water part slightly warmer than the ice part? or is
it all pretty much the same exact temperature? If I
poured out the liquid water into a different bottle, and
left the slush / ice in the original bottle, would there
be any chemical differences between the two? Is the
explanation that the liquid water has impurities such as
minerals, and the frozen water is pure H20? If that
is the case, then if we performed this experiment with
pure water, then it should turn directly into solid ice,
and there'd be no slush, right? I have read about
Fractional Freezing which seems relevant. But this
doesn't seem to fully describe what is happening. I am
hoping for more of a general explanation rather than a
list of individual answers.

ANSWER:
I recommend that you google
supercooled water videos and watch a few. It seems
pretty clear to me that your problem with slush is that
your water is not pure enough. Possibly not cold enough
either so that generated heat warms the water just above
zero before it freezes. This is a very touchy sort of
experiment, one of the videos has the teacher admitting
it took him a number of tries to get a good enough take
to post. By the way, I thought the answer on Reddit was
pretty good.

QUESTION:
Hi, I have a bee in my bonnet about a real world home improvement matter concerning wall tiles and the surfaces they are stuck to.
There is much ado about the supposed weight of tiles per square metre that the sub surface can support and arguements are made for using concrete backer boards which can take greater weights than plasterboard (or drywall if you prefer). So plasterboard can take 30kg/m2 and backer boards 50kg/m2 or more.
The base of wall tiling always finishes at a horizontal surface, such as a floor etc. So given there are no gaps between tiles and that the wall is truly vertical, what is the relevance of the supporting weight figures for the subsurface board?
I can't see that there would be any force acting that would cause any board to fail assuming the tiles end on a horizontal surface that can take the weight. I follow that there will be compressive force down through the tiles and thus through the backing board by virtue of the adhesive to some degree. But surely this compressive force can't affect the backing board as the weight is carried down through the tiles to the horizontal surface that they connect with.

ANSWER:
Well, this isn't really physics, but having had
a recent experience (a backsplash behind our
range), maybe I can make a few comments. If the
wall is sheetrock and sealed or painted, it is
best; this is because the surface is just paper
and if it is necessary to provide vertical
force, it could be too weak. But, as you state,
if the tiles extend all the way to the floor (or
other support like a cabinet as in my
backsplash), there should be no reason to apply
the concrete boards. On the other hand, I would
not put tiles on a bathroom sheetrock wall
because if moisture gets behind the tiles you
would have a mess. The same thing goes for bare
wood. I once had laid large terra cotta tiles on
my kitchen floor on top of an old pine floor; it
turns out I had a slow leak in the ice maker
which I did not discover until the tiles started
coming up from the saturated spongy wood floor.
I had to rip up half the floor and leave it to
dry for several weeks before reinstalling (too
late for a subfloor because it would have raised
the floor too far). When I tiled our laundry
room I did use the backing board and it is still
in pristine condition after 40 years.

QUESTION:
This is really not a question about astrophysics but something simple I am missing. I know that as two massive objects orbit each other they lose energy by emitting gravitational waves and that eventually the orbit decays and they will collide. However as the Earth and Moon rotate around each other and the system loses energy by tidal forces the Moon moves AWAY from the Earth. Why the difference?

ANSWER:
It is understandable that you would think of
this because it would appear that the moon is
gaining energy rather than losing energy. In
fact, it doesn't just appear to gain energy, it
is gaining energy. The reason is rather subtle.
The moon causes a bulge in the oceans which is
responsible for tides. The bulge is actually two
bulges, one toward and one away from the moon as
shown in the figure (greatly exaggerated); this
is because of the
tidal force . Because there is friction
between the oceans and the ocean floors, the
rotation of the earth drags the bulge forward,
so it is slightly ahead of the moon as shown in
the figure (again greatly exaggerated). Now the
bulge exerts a force on the moon which causes it
to accelerate a tiny bit; that's where it gets
the added energy to increase its orbital radius.
Similarly, the moon pulls on the bulge which
causes the earth to slow down. If this were the
whole story, the energy would be conserved, the
earth losing energy and the moon gaining the
same amount. But, as noted above, there is
friction between the earth and the oceans which
takes energy away from the system, giving a net
loss of energy but the moon ends up with more
than it started with.

QUESTION:
If subject A is traveling through space at a high rate of speed and subject B is traveling at a much slower rate of speed would you expect to see a difference in the rate at which either subject can process information. Example: Would it be easier for subject A to process information than subject B or vice versa?

ANSWER: This is too vague. What does
"process information" mean? What do you mean by
"easier". Now, whatever the answers to these
questions are, if each subject is presented with
the same information stream in his frame of
reference and has the same equipment to process
it, they will be equally easy to process. But if
you mean that A sends an information stream
which he processes to B to process (or vice
versa ), they will not process the
information at the same rate. For simplicity's
sake, I will choose the "slow" frame B to be at
rest; in relativity, only the relative velocity
is important. Let's choose a simple concrete
example of information stream: the information
is a stream of ten pulses separated by 1 second
as seen in the reference frame B; B will process
them in 10 seconds. Now, how does A process that
same information which B processed in 10
seconds? Suppose A is moving away from B; then
the time between pulses will be longer than 1
second because he moves some distance away while
waiting for the next pulse to catch up with him—the
information is processed more slowly. You should
be able to see that if you interchanged A and B
everything else would be the same so B would
process the information more slowly. Suppose A
is moving toward B; then the time between pulses
will be shorter than 1 second because he moves
some distance toward B while waiting for the
next pulse to arriveï¿½the information is
processed more quickly. You should be able to
see that if you interchanged A and B everything
else would be the same so B would process the
information more quickly. An
earlier answer would be of interest to you.

QUESTION:
Matter wave of a particle depends on its momentum,doesn't that mean different observers with different speed will measure different wavelength?

ANSWER:
Yes. This is called the Doppler shift.

QUESTION:
If I am driving in a vehicle in a straight line at a constant speed and activate a drone in the vehicle so the drone is hovering inside the vehicle, open the back of the vehicle and increase my speed so the drone drifts out the back of the vehicle, what then happens to the drone once it is not longer inside the vehicle?

ANSWER: There will be a period when the
drone is leaving the car when there will be lots
of turbulence and other currents of air; I am
answering neglecting this transition period and
also assuming the air outside the car is still.
You have adjusted the drone so that, even though
it is moving forward with the car at speed v ,
it hovers in still air. When it suddenly finds
itself outside it sees a headwind of v ,
not still air. This headwind will exert a force
which will tend to slow the drone down so that
eventually the drone will be hovering at rest
relative to the ground.

QUESTION:
Once a spacecraft breaks free from earth's gravity, how much propulsion does it actually need to reach a destination, say Mars? Would the craft have to continually accelerate through the vacuum of space, or once it reaches its maximum velocity, would its propulsion be turned off?

ANSWER: Technically, you never "break
free from earth's gravity". If you acquire a
speed greater than the escape velocity, then you
will never fall back to earth, but you will slow
down continually forever (assuming you have no
other objects exerting forces on you).
Nevertheless, the force gets very small pretty
quickly. If you are 100 earth radii away from
earth, the weight (force of earth gravity) is
1/100^{2} =10^{-4} times smaller
than on earth. And 100 earth radii is not all
that far in the whole scheme of things; the
distance to the moon is 60. Anyhow, a spacecraft
always coasts nearly all the time on a mission
somewhere in the solar system. Propulsion is
used mainly for getting up to speed and then
slowing down to orbit or land on your
destination; a much smaller use is to steer,
changing direction.

QUESTION:
I have had this question in my mind for over 10y, and I would be very grateful if you could help me to get an answer.
The theoretical experiment goes as follows:
1) Bring to outer space 1 round glass chamber, 1 pure iron ball, and a source of heat.
2) Place the iron ball in the center chamber, and heat it until it gets bright red.
3)Go out of the chamber, close it, and generate a total vacuum.
As far as I have been able to go: Q1: Can we see the bright red iron bar?
A1: Yes, and considering that the ball is in an absolute vacuum the light must be behavior as a particle. *Quantum mechanics dictate that given the duality of light a media is not necessary for light to travel, thus radiation.
Q2: If the light behavior as a particle it must have a mass. Where does this mass come from?
A2. Given that the iron bar is in an absolute vacuum, my logic dictates that the only source of mass for the photons must be the mass of the iron bar itself... Is this correct? Q3. After an X amount of time when the iron bar has finally cooled down. Will the iron bar be lighter than at the beginning of the experiment?
** Consider that the energy in the iron bar can only be released in the form of light.
A3: I have been stuck in here for several years.....
Thank you very much for your time, please notice Iï¿½m not a physicist but a Biologyï¿½s with a deep interest in physics.

ANSWER: I do not see why all the stuff
about a glass chamber, heat source, etc .
is necessary to ask your question which is,
simply: if an object is hot and cools soley
by radiation, does the object have less mass
after cooling?

First, your A2 is wrong. Photons do not have
mass even though they do have momentum and
energy. Therefore, your conclusion that the
object will be lighter because the photons carry
away mass is wrong. However, because mass is
just a form of energy, we can say that the
photons carry away energy (even though they have
no mass). But, where did that energy come from?
The object had to provide it. But there are just
as many atoms after the cooling as before, so
the object evidently changed mass into energy.
This change in mass is very difficult to observe
in the case of a red hot iron object cooling down.
Inasmuch as E=mc ^{2} , the change in mass would be
Δm =ΔE /c ^{2}
where ΔE is the energy carried
off by the radiation. Suppose that ΔE
is a real big number, say a 10,000 Joules; then
Δm =10^{4} /(3x10^{8} )^{2} =10^{-13}
kg=0.1 nanograms! This is approximately the mass
of a single yeast cell.

QUESTION:
For a space traveler that leaves Earth headed for a distant star @ 0.9c, wouldn't his onboard clock (that's @ rest relative to him) read the time passes normally......i.e. as if he were still on Earth? He would see no time dilation nor length contraction!

ANSWER: Wrong. the distance to the star
is measured from the earth; think of it as a
long stick. The stick is moving with speed 0.9c
as seen by the traveler, so he sees the distance
shorter by a factor √(1-0.9^{2} )=0.44.
Since, as you correctly stated, her clock runs
normally, it only takes her 4.4 years to get to
a star 10 light years away. It would be worth
reading my explanation of the twin paradox which
is linked to from the
faq
page.

QUESTION:
Which is the correct term to refer to the speed of light (scalar) or the velocity (vector) of light. They are both used in books & literature. I think its should be the "speed of light", but just asking.

ANSWER: It depends on the context. One
context is "the speed of light is a constant in
all frames of reference and independent of the
velocity of the source or observer". Another
would be "when a ray of light passes close to a
massive galaxy, it is bent and so its velocity
changes although its speed does not."

QUESTION:
How do I calculate the kinetic energy of an object moving submerged under water? Specifically if water pressure is involved. Like if a cannonball was fired at 110 m/s under 9,000 feet underwater. Or If a submarine is moving 10 m/s at a depth of 800 feet. I know kinetic energy is factored in and pressure at that depth and possibly water drag. I'm just not sure if there is a specific equation for this. Can you please help me with this?

ANSWER: The kinetic energy of something
depends only on its mass m and its speed
v , K =½mv ^{2} .
It has nothing to do with its pressure or its
drag force. If there are any forces on the
object, those forces do work which might change
the kinetic energy. Because of the near
incompressibility of water, the depth will have
almost no effect in how something moves. The
effect of the pressure is simply the buoyant
force which is a force equal to the weight of
the displaced water which is nearly the same 100
ft down as 10,000 ft down. Similarly, the drag
depends only on the density, speed, and
geometry, not pressure, the drag will be about
the same at any depth.

QUESTION:
What's the physics behind
"walking ladders" that move down slopes by themselves?
I'm thinking maybe it's the same for slinkys that 'walk' down steps. Gravity and the weight of the ladder being moved around the different legs.
I tried Googling for an answer but can't find any scientific answers.

ANSWER: This was a fun problem to figure
out. First, let's think about all the forces on the
ladder when it is slightly tipped so that two legs are
off the ground. There is the weight W ,
normal forces (not shown) N _{front}
on the front leg and N _{rear}
on the rear leg, and frictional forces (not shown)
f _{front} on the front
leg and f _{rear} on the
rear leg. The weight acts at the center of mass of the
ladder which is centered relative to the two sides, but
is closer to the front than the rear because the front is
heavier. That means that the normal force on the front
leg is greater than on the rear leg; this is important to
remember as we go further. Now, I want to examine the
torque which the weight causes. I have resolved
W into components down the incline
(W _{x} ) and normal to the incline (W _{y} ).
W _{y} causes a torque about the
axis I have labelled DROP AXIS causing the currently-off
legs to drop to the ground. The inertia causes the ladder
to keep going lifting the other legs off the incline; if
none of the legs ever slipped (lots of friction), the
ladder would rock back and forth without going forward.
Now, the other component W _{x} exerts a
torque about the axis I have labelled TWIST AXIS; if no
legs slip, this torque has no effect on the motion.
However, for a couple of reasons, the rear leg may slip:

Because the normal force is smaller for the rear
leg, the maximum static frictional force will be
smaller than for the front leg.

The material contacting the ground may be
different for the rear legs than for the front,
e.g. the rear legs may have a smaller
coefficient of static friction and could slip
even if the normal forces were the same.

So, for the video, the rear legs must slip and
the ladder, with each "step" twists about a
normal axis. There are frictional losses which
would cause the ladder to lose energy and
therefore eventually stop, but since it is going
down an incline, this energy is replenished by
the work done by W_{x} as it
moves down the incline. It seems to me that it
would be pretty tricky to get this set up since
everything must be just right for it to work. ADDED
THOUGHT: A different possibility occurred
to me. I have treated the ladder as if it is
rigid. However, if it is kind of rickety the
lifted legs will tend, if they can, to move
forward because of gravity, even if the two down
legs do not slip. So when they come back down
they will have moved forward a bit.

QUESTION:
My Dad loves physics and always referred to the speed of light as 186,000 miles per second, per second or 186,000 miles per second sq. that is how he always spoke of it. Is that speed of light an excelleration? (even though nothing goes faster than light.)

ANSWER:
The dimensions for speed are length/time, not
length/time/time. Therefore, 186,000 mi/s/s is
wrong.

QUESTION:
E=m(c)2
What is the reason for twice the speed of light? Einstein tried everything and found it?

ANSWER: Oh dear, it is not twice
c it is
c squared. E=m·c·c=mc ^{2} .
And Einstein did not find it by trial and error;
how could he, since it was not really
appreciated at the time that mass was, in fact,
just another kind of energy? It was a natural
result of his theory of special relativity.

QUESTION:
According to Enistine space time is like a fabric and all masses round around the sun so why Moon revolve around Earth not sun?

ANSWER: Not just the sun, but every
object with mass, warps spacetime. It is much
easier to think about this in terms of Newtonian
gravity for which the equivalent statement is
that all objects with mass have a gravitational
field. The force which a mass M exerts
on a mass m a distance r away
is F=MmG /r ^{2} where
G is the universal gravitational
constant. You can calculate the ratio of the
force the sun exerts on the moon to the force
the earth exerts on the moon:

F _{sun} /F _{earth} =(M _{sun} /M _{earth} )(R _{earth} /R _{sun} )=(2x10^{30} /6x10^{24} )(3.8x10^{8} /1.5x10^{11} )^{2} =2.1

So the sun exerts a force more than twice the
force the earth exerts on the moon. But if you
think about it, the moon does revolve around the
sun since it moves right along with the earth;
but because of the force the earth exerts, it
also revolves around the earth.

QUESTION:
according to law of gravitation less mass object attract toward heavy mass object so why we don't attract toward wall because wall have greater mass then a human mass ?

ANSWER: Actually, the law of gravitation
states that two masses attract each other with
equal and opposite forces. So the wall does
exert a force on you and you exert an equal and
opposite force on the wall. However, since
gravity is such a weak force, you do not feel it
at all. For example, if you and have a mass of
100 kg and the wall has a mass of 10,000 kg, the
force you experience, if you are 10 m away, is
about 10x10,000x6.67x10^{-11} /10^{2} =6.67x10^{-8}
N=1.5x10^{-8} lb.

QUESTION:
I'm trying to calculate what effect hitting a 1,500 lb water barrel at 30 mph in a 15,000 vehicle would have upon the truck's momentum?
Also, if I increased the truck's weight to 65,000 and drove it at 50 mph, what difference would this make on the momentum too?
I'm working on a road safety prototype in my shed and to be honest this kind of math is way beyond me, so any help would be much appreciated.

ANSWER: I will assume that the barrel and
vehicle move off together after the collision;
this is called a perfectly inelastic collision.
In a collision the total linear momentum, the
product of the velocity times the mass, remains
constant. Although lb is not a unit of mass, it
is a measure of mass. So, before the first
situation you state, the momentum before the
collision is 15,000x30=450,000 lbï¿½mph and after
it is 16,500v where v is the
speed after the collision; equating the two and
solving for v , I find v =27.3
mph. For the second case, 65,000x50/66,500=v =48.9
mph.

QUESTION:
I have watched
Human Universe with Brian Cox. If you haven't seen it, in
part one he goes to Russian to meet with the Soyuz spacecraft. He tells his
viewers that only knowing two of Newton's equations (F=ma and the law of gravitation
F=GmM /r ^{2} ) he can calculate that the spacecraft only
needs to reduce velocity by 128 m/s to safely reenter orbit. I have looked
and tried to figure out how the calculations work but can't. Do you have
any clue how he would have done the math?

ANSWER: I will assume that the Soyuz is
returning from the International Space Station. First I will give some
information needed later:

R =6.4x10^{6} m
(radius of the earth)

h =4x10^{5}
m (altitude of the space station)

G is
the universal gravitational constant (will not need
it)

M is the mass of
the earth (will not need it)

m is the mass of
Soyuz (will not need it)

g =9.8
m/s^{2} =MG /R ^{2}
(acceleration due to gravity at r=R )

√(MG /R )=√(gR )
(this will be useful later and is why I do not need
to know G or M )

MmG /r ^{2}
is, as you note, the gravitational force on m a
distance r from the center of the earth.

If orbits are
circular, a=v ^{2} /r and you can
show that v =√(MG /r ). I
am thinking that we need to find the difference of speeds
between orbits with r=R+h and r=R :

v_{R} -v_{R+h} =Δv

=√(MG )[√(1/R )-√(1/(R+h ))]

=√(MG /R )[1-(1+(h /R ))^{-1/2} ]

≈√(MG /R )[½(h /R )]

=½h √(g /R )

=247 m/s

Note that, because
h /R= 0.0625 is small compared to 1, I have used the binomial
expansion (1+x )^{n} ≈1+nx .
My answer is about twice what Green
got. Of course, in the real world, air drag when entering
the atmosphere would significantly reduce the required
speed change due to the additional braking it would
cause; but you cannot estimate that using only those two
equations. So, I am at a loss to understand the
difference.

ADDED
THOUGHT: Maybe the Soyuz dropped to a lower orbit
before beginning its reentry, h ~200 km, after
leaving the ISS, but I could find no reference to its
altitude in the video.

QUESTION:
i am currently designing a fully adjustable boom arm for a microphone my prototype is fine and it is functional but i am interested on getting more weight suspension from my joints the joints im using are friction based and use clamping force to lock the joints in the desired position what material should i to use as washers to create more friction in my joints

ANSWER:
it is hard for me to be specific since i do not
know what the joints are made of what you need
to do is find a material with a larger
coefficient of static friction with the material
currently in the joints i think the very best
way would simply apply trial and error try
different materials to see how they work i would
start with obvious choices like rubber why dont
you use punctuation or capitalization it is very
annoying reading your question

QUESTION:
In a fire where does the photons come from? Are photons in a superposition of wave and quanta particles?

ANSWER: Fire is many chemical reactions
going on. The reactions are, for the most part,
exothermic and most of the energy comes out as
heat and light. Photons are particles which have
wave properties.

QUESTION:
I am confused in the formula of orbital angular momentum of electron. According to quantum physics the formula of orbital angular momentum is L=l√(l+1)h/2π B.M. while according to Bhor's postulate the formula of angular momentum is L= nh/2π. Both are giving different result. For example for hydrogen putting n=1 in Bohr formula and l=0 in quantum mechanics formula gives h/2π and 0 respectively.

ANSWER: I believe there is an error in
your question, L =√(l (l +1))h /2π.
The answer to your question is that the
Bohr model is simply incorrect. As luck would
have it, it describes the energy levels of the
hydrogen atom well, but it is incorrect
regarding angular momentum of the levels.

QUESTION:
Why would a heavy object like a bowling ball not float on top of water, but a wooden telephone pole, which is much heavier, would float? I'm sure the shape and the material the object is made from plays a role but I would just like to know the science behind it.

ANSWER:
The science behind this is over 2000 years old.
Archimedes was a Greek mathemetician who
discovered what we call today Archimedes'
Principle. It says that there is an upward force
(called the buoyant force) on an object in a
fluid which is equal to the weight of the fluid
which it displaces. It turns out that, applying
this principle, anything with a density smaller
than water floats and anything with a density
greater than water sinks. Wood is less dense
than whatever bowling balls are made of. You
might then ask why a steel battleship can float.
It is because it is hollow and therefore the
total density is much lower than the density of
steel; a hollowed out bowling ball would float.

QUESTION:
I read that metals conduct heat well because the electrons of metallic elements have free electrons in its outer shell that move quickly when heated. These quickly moving electrons transfer their kinetic energy to other electrons which vibrate quickly and collide with other electrons and transfer the heat energy and so and so forth.
I also read that metals conduct electricity well because the electrons of the outer shell are free to move (as is the definition of electricity - the movement of charged particles).
So it would reason to stand that heating a metal should generate electricity or at the least facilitate the production of electricity as both heat conductance and electricity involves the movement of electrons in terms of metals. To my surprise after a google search, I learned that heating a metal does not generate electricity, and furthermore hampers (not facilitate) electric conductance.
Please dear physicist help me to understand why hot metals are poor conductors of electrons.

ANSWER: In a conductor, there are "free
electrons" which essentially behave as an
electron gas. There is normally no net current
because, just as there is no wind in still air,
the electrons move in random directions and the
net flow is zero. If you apply a voltage across
a conductor, these moving electrons, on average,
flow from the negative to the positive
terminals. Since they are now experiencing a
force due to the electric field, they
accelerate; but, obviously, they do not really
speed up the whole time that they move from one
end to the other of the conductor but, on
average, move with some constant speed called
the drift velocity. The reason is that they are
not really free; rather, they accelerate for
some short distance and then collide with one of
the atoms in the solid which scatters them, then
accelerate again, then collide again, etc .
Now, all the atoms in the solid are constantly
vibrating as if attached to tiny springs, and
the hotter the metal gets, the larger the
amplitudes of their vibrations becomes. This
means that they present bigger targets for the
electrons to hit and so collisions happen more
often which reduces the drift velocity and
therefore the conductivity.

QUESTION:
I was confused about the strict physics definition of a wave. Considering it is a traveling disturbance that carries energy, would wind on a field, an audience created "wave", and dominoes falling be waves? I thought the audience would not be a wave, and I was unsure about the dominoes because there is no wavelength or period.

ANSWER:
The "strict physics definition of a wave" is
that it is something which satisfies the wave
equation. The solutions of a one-dimensional
wave equation are always of the form f (x-vt )
where f is any function; for
example where x is the position, t
is the time, and v is the speed at
which the disturbance travels through the medium
which supports it. For your examples: dominos
falling and people in a stadium sitting and
standing are both waves, but wind is not. Wind
is an example of the entire medium moving but
that would not be described by a wave equation.
The stadium wave would not be a wave if the fans
all all just ran around together, like a
hurricane of people.

QUESTION:
In my understanding of time dilation if one has a clock in constant motion relative to me I will measure the time ticking off on that clock as slower. However from the other clocks view it is just as valid to say that I AM in motion so an observer traveling along with that clock would measure my clock as moving slower. How is that reconciled? Especially if at some time in the future both clocks are brought together at rest with each other to compare clocks? Am I missing the part that acceleration plays here?

ANSWER: There is no need to reconcile
that each observer measures the other's clock to
run slower unless you bring the two of them back
together into the same reference frame.
Acceleration has nothing directly to do with it.
So this is the "twin paradox" which we know is
not really paradoxical at all. Go to the faq
page and find "twin paradox ".

QUESTION:
We have been taught that the big bang created all heavenly bodies as we understand them. I'll disagree with that. Were the exo-planets we have discovered thus far created by the same creation event that brought forth us, Mars and Venus or was it something else that lead to the Gliese's and the Kepler's?

ANSWER:
Good grief, nobody claims that the big bang
created everything as it is right now. The big
bang created a huge amount of energy (from
where, nobody knows) and physical laws. From
that point the universe evolved over time
eventually atoms, stars, galaxies, planets,
etc . appeared. To see a timeline of the
universe, click
here .

QUESTION:
I'm designing a special suction tube for dental practitioners (I'm an orthodontist) and have a basic question I feel I should know. I want to maintain as MUCH air flow through the tip of the suction tube(end of the tube by the patient's mouth) as possible. The lumen of the suction at the dental unit has a cross sectional area of 75mm2. This tube sucks the air from the patient's mouth into the dental unit. If I use a tube that has a much smaller lumen EXCEPT at the patient's mouth where the cross sectional area would be 75mm2, will I lose air flow? Another way of asking this is: am I required to keep the cross sectional area of the ENTIRE tube at 75mm2 in order to maintain the same flow rate at the tip of the suction (patient's mouth) that I have at the dental unit?

ANSWER: I learned something new: in
physics, lumen is a measure of luminous flux.
Biolgists, though, use it as a measurement of
inner space of a tubular structure; apparently
it is just the cross-sectional area of the tube
since the questioner specifies the area of the
opening. If you were dealing with an ideal
fluid, incompressible, no wall drag, no
viscosity, you could argue that the flow would
be constant, the speed in the narrower part of
the tube would be larger, but the volume per
unit time flow would be the same. But air is not
incompressible and energy loss effects are not
negligible. In order for you to maintain the
same volume flow rate you would need a more
powerful pump which is pulling the air. Think of
trying to drink through a pinched straw–you
have to suck harder to get the same amount of
drink.

QUESTION:
I have a question for you. So from what I've learnt, Einstein's Theory of Relativity says that an object with mass will bend light towards it by way of its gravity and the source and direction of the light will therefore appear differently to an observer. Besides that, Einstein also says that E=mc^2 and thus m=E/C^2 so something like light, with measurable energy, would have a measurable mass. My question is, just like that object with mass pulls light towards it with its gravity, can light also pull that object towards it with its gravity? Similarly to how an object falling to Earth pulls on the Earth itself? Or does light not have gravity in the same sense as objects with actual mass on account that its mass is relativistic and not actual, classical-physics mass?

ANSWER: For the first part of your
question, see an
earlier answer . (You could also have found
this on my
faq page.) The answer to your second
question is that the theory of general
relativity finds that any energy density warps
spacetime, that is causes a gravitational field;
a photon has energy and therefore creates
gravity; keep in mind how small this is when
compared to any macroscopic mass.

QUESTION:
This is not a homework question, I am actually a teacher and saw this question in a textbook and would like to know if this is a valid question to ask a third-grader. I may be wrong but I do not believe that the pound and ounce are units of measurement for mass. I have tried looking it up online and did not find anything useful.
Which two units of measurement are used to measure mass?
Gram
Pounds
Ounce
Kilogram

ANSWER:
Kilograms (kg) and grams (g) (1 kg=1000 g) are
units for mass. Pounds (lb) and ounces (oz) (1
lb=16 oz) are units for force. This is very
confusing because of the ways weight is measured
in different countries. In countries where
Imperial units are used, weight (the force which
the earth pulls down on something) is measured
in pounds (which is correct) and we think that
this must also be a measure of mass (the amount
of stuff there is). In countries where SI units
are used weight is measured in kilograms (which
is incorrect) and we think that this must also
be a measure of force. I will leave it up to you
to decide whether third-graders can understand
this subtle difference!

QUESTION:
I know that in a solid the particles are still in motion, but I was wandering if their is any point in the state of matter where the particles are completely motionless? (something like before solid)

ANSWER: The Heisenberg uncertainty
principle stipulates that you cannot know to
absolute precision both the position x and
momentum p (mass times velocity). Stated
mathematically, Δx Δp ≤ℏ
where ℏ is the rationalized Planck constant
(a very tiny number and Δ denotes the uncertainy of the quantity.
So if the velocity (momentum) uncertainty is
zero (particle perfectly at rest), the
uncertainty of the position of the particle is
infinite—the particle could be anywhere in
the universe.

QUESTION:
how can the calculations for the curvature of space-time under general relativity be correct without the mass-energy associated with dark matter taken into account? I am just a retired attorney trying to educate my self about our physical world.

ANSWER:
Keep in mind that I specify on the site that I
do not do astronomy/astrophysics/cosmology, so
you may wish to take my answer with a grain of
salt. When you refer to the "curvature of
space-time" you are actually talking about the
gravitational field. And, you are certainly
right, dark matter, if it actually exists, would
affect the field. I believe that it is actually
the other way around: by studying the
gravitational field you can infer something
about the distribution of dark matter. The most
well-known such inference has to do with
rotating galaxies and how the angular velocities
of the stars varies with distance from the
center. If the only gravity in our Milky Way
galaxy were from stars and gas which we know are
there, the galaxy would not be able to hold
together, stars at larger radii would have
speeds far too large. The Wikepedia article on
dark matter halos would be of interest to
you.

QUESTION:
If a toy car hits a wooden block and pushes it along the track, do the wooden block and the car have the same level of energy or does the energy of the car decrease and the energy of the block increase?

ANSWER: You have described a perfectly
inelastic collision, one in which the two
colliding bodies stick together after the
collision. I am assuming that the block is at
rest before the collision. So linear momentum is
conserved and energy is not conserved. I will
denote the velocity of the car before the
collision to be V , the velocity of the
car/block after the collision to be U ,
the mass of the car to be M , and the
mass of the block to be m . Conservation
of linear momentum gives MV =(M+m )U
or U=V [M /(M+m )]. Note
that U<V , so the car has lost energy
and the block has gained energy.

If you are interested you can calculate the
change in energy (ΔE=E _{after} -E _{before} )
for each:

ΔE _{car} = ½M (U ^{2} -V ^{2} )=½MV ^{2} [
(M ^{2} /(M+m )^{2} )-1]

ΔE _{block} = ½MU ^{2} =½MV ^{2} [M ^{2} /(M+m )^{2} ]

For example, if M=m ,
ΔE _{car} =- (3/4)[½MV ^{2} ]
and ΔE _{block} =(1/4)[½MV ^{2} ].
The car lost three fourths of its initial energy
and the blocked gained one fourth of the initial
energy. In this case , half of the total
energy was lost. In this case , half the
speed was also lost.

QUESTION:
The question is about Maximum gradient a vehicle can climb.
The maximum friction force that can be transferred to the road cannot be more than
μmg cosθ . As far as I know, μ cannot be greater than 1 (irrespective of contact patch size) for a Rubber tyre on concrete road. At 45 degrees, the drag due to gradient (mg sinθ ) is equal to the Normal force (mg cosθ ). Any more than 45 degrees, and the drag force increases and Normal force decreases, thus making the vehicle's tyres slip due to loss of traction.
Now, I see videos of RC cars on
youtube
climbing gradients well over 54 degrees.
Could you explain this? Is the μ value greater than 1?

ANSWER: Your analysis is fine if the
vehicle is a point mass. But, the video you
refer to has the failure to climb a steep
incline due to the vehicle tipping over
backwards rather than slipping. I will assume the
truck shown above is at rest because it is
easier to visualize and exactly the same as if
it were moving up with constant speed. The whole
weight mg may be considered to act at the center
of mass of the truck (white cross) which is
located a distance h above the ground,
a distance d _{1} behind the
front axle, and a distance d _{2}
in front of the rear axel, as shown. There are
three equations, sum of the forces along the
ramp equals zero, sum of the forces
perpendicular to the ramp equals zero, and sum
of the torques about the point where the rear
wheel touches the road equals zero:

f _{1} +f _{2} -mg sinθ =0

N _{1} +N _{2} -mg cosθ =0

(d _{1} +d _{2} )N _{2} +hmg sinθ-d _{2} mg cosθ= 0

Now, we are interested in when
the truck will start rotating backward about the rear
axle; when it is just about to do that, the normal
force N 2 =0. So, the third equation
tells us that tanθ=d _{2} /h ;
therefore the maximum angle before tipping is θ _{tip} =arctan(d _{2} /h ).
For example, if d _{2} =1.5 m and h =1
m, θ _{tip} =56.3^{0} . What
determines tipping is just where the center of mass is;
you want it as far forward and as close to the ground as
you can.

But, if you lose traction, slip,
before you rotate, you do not need to worry about
rotating. As you correctly surmised above, the truck will
start slipping when θ _{slip} =arctan(μ )
where μ is the coefficient of static. I agree
with you that μ= 1 is a reasonable estimate
for rubber on dry concrete. However, the video you sent
is not done on a concrete ramp, rather the surface
appeared to be carpet. Since the tire tread can press
into the carpet pile, it is not surprising that μ >1
for those videos.

QUESTION:
How does a single speaker unit play an audio clip consisting of sound frequency of different magnitude? It's the same speaker and has only one 'vibrating membrane' to induce the sound in the air to propagate. How does this single membrane generate sound waves of different frequency and different amplitude simultaneously?

ANSWER:
Sound waves are linear, i.e. if several are
acting on a point in space, the net action is
simply the sum of all the soundwaves. As an
example, I have plotted three waves representing
the individual sounds, each with different
frequencies and different amplitudes, at some
point in space as a function of time. The net
sound at this point in space is simply the sum
of the three, shown by the black curve. Your
speaker need only be driven to vibrate like
this; this is pretty easy to do, usually with a
magnet attached to the speaker cone and inside a
coil, the current in which varies like the black
curve.

If you cannot find what you are looking for, it was
probably archived. You may find the most recent archived
file

here .