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 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 .
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.
MR 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.
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).
twin paradox . How fast a clock runs and how fast it
appears to actually runs are two different things; see
earlier answer .
C r ),
W R ),
N
f
F
Note that I have
resolved F
H V
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
If you are
interested in the entire bike, not just the rear wheel,
the force C
C on the rear sprocket but a force
-C
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.
F
N-V-W =0. But in the horizontal direction there will
be an acceleration, ma=H-f . Only f F F f
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.
*I have drawn
the forces on the ball, which are
the ball's
weight W
the normal force
N
the frictional
force f
the force
F exerted on the ball.
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.
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/s2 )=0.016 lb·s2 /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.
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, 1.68x105 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.
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.
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.
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.
recently answered this question in some detail.
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.
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.)
https://youtu.be/hs4iv7IHvGg
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.
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 =6.4x106
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 m3 , 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.
18 m3 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 5.15x1018 kg; so you can calculate
the volume if the whole atmosphere were at sea level
pressure where the density is 1.225 kg/m3 ,
5.15x1018 /1.225=4.2x1018 m3 .
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, 6.4x106 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!
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 m3 kg-1 s-2
is the universal gravitational constant and c =3x108
m/s is the speed of light. For example, let the location
be on the surface of the earth, M =6x1024
kg, R =6.4x106 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 RS , the time
dilation equation becomes t'=t √[1-RS /R ]
so time stops when R=RS . If R= 2RS ,
t' =0.7t .
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.
The physics is 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.
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 F F F + F
A·B =AB cosθ AB
where θ AB
is the angle between the vectors
A and
B A and B
are the magnitudes of the vectors. An example of
a scalar product is the work done by a force
F d W =F·d · i.e .
A·B =B·A .
The vector product or cross product is defined
as
A×B =u AB sinθ AB
where u
A and B
A·B =-B·A .
An example of a vector product is torque
τ F r
τ =r×F .
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√(1-0.992 )=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.
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).
m =30
grams=0.03 kg. I took the acceleration to be
about g =10 m/s2 . 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.
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.
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:
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.
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.
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 arrive—the 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.
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.
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.
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 =104 /(3x108 )2 =10-13
kg=0.1 nanograms! This is approximately the mass
of a single yeast cell.
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.
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.
"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 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 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 Wx 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.
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.
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 )=(2x1030 /6x1024 )(3.8x108 /1.5x1011 )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.
-11 /102 =6.67x10-8
N=1.5x10-8 lb.
v 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.
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?
R =6.4x106 m
(radius of the earth)
h =4x105
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/s2 =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 -vR+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.
h ~200
km, after leaving the ISS, but I could find no reference
to its altitude in the video.
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.
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.
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.
twin paradox ".
