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λ
of a wave of frequency f and speed
v is λ=v /f . Since
the speed of a wave inside the prism is smaller
than outside, it does not have the "SAME
WAVELENGTH". What is red outside is not the same
wavelength inside but shifted toward the
blue.
v W
N which is the force which
the table exerts on the ball, vertically up;
since there is no motion in the vertical
direction, these two forces must be equal and
opposite and thus cancel out. (This is Newton's
first law and not Newton's third law which is
often called "action and reaction".) Finally, the important force for us
is the frictional force f f
Now, three things
can happen, depending on the values of v and ω :
If v is
large and ω is small, the ball will
stop slipping before it stops moving forward and from
that point on it will simply keep rolling forward
(not slipping) with constant speed because
f will disappear if there is no
slipping. (In the real world there is still some
friction from other forces and the ball will
eventually stop.)
If v is
small and ω is large, the ball will
stop moving forward before it stops rotating; but if
it is still rotating f
If v
and ω are just right, the ball will
stop moving forward at the instant that it stops
rotating and just stop dead.
Hope that this
answers your question about the "theories of physics"
relevant to your experiment.
g or is in empty space
with an acceleration g ." If your man is
in a box and sees the apple accelerating toward
him with acceleration g , it could be
either that the box is accelerating up with an
acceleration g or that he is in a
gravitational field the gravitational
acceleration of which is g . There is no
way, other than opening the box and looking
around, that he could distinguish between the
two possibilities. The equivalence principle is a
keystone of the general theory of relativity.
F of the air on an object with
area A . As you note, the force depends
on the density of the air and other properties
of the air. But, you need to appreciate that any
calculation of air drag is going to be an
approximation. There is a very handy little
formula which makes it easy to estimate this for
air near sea level pressure: P=F /A =¼v 2
where v is the speed of the air. The
only catch is that this works only if you used
SI units, F in N (newtons), A in m2
(square meters), and v in m/s (meters
per second). But, this is really no big deal,
you just need to be able to go back and forth
between SI and English system. There is an
excellent little program, named
Convert ,
which allows you to do this and it is easy to
use. I will walk you through an example: Suppose
you want to exert a force of 1 lb on an area of
1 ft2 ; convert lb to N and ft2
to m2 :
So you want a pressure of 4.45 N/0.0929 m2 =47.9
N/m2 . Now, our formula says that this
is
47.9=¼v 2 .
Solving, v =√(4x47.9)=13.8 m/s.
Now, maybe you want this in miles per hour, so
convert:
So, you need a wind velocity of 30.9 mph. As you
can see, volume of air has nothing to do with
it, just the area on which the wind is blowing.
QUESTION :
There is an often seen problem of finding the angle at which a particle sliding from the top of a smooth sphere leaves the sphere.
However, I am trying to calculate the time it takes the particle to reach the point of departure.
When I try to so the integration, I get an infinite time... .
How do I resolve the problem?
θ at which a point of mass
m
, starting from rest at θ =0 and sliding
frictionlessy down the surface of a sphere or
cylinder of radius R leaves the surface.
The answer is
θ 0 =cos-1 (2/3)=48.20 =0.841
radians.
In solving this problem one uses energy
conservation and finds the velocity
v =√[2Rg (1-cosθ )].
Now, if we are interested in time we write the
angular velocity
ω =dθ /dt =v /R= √[(2g /R )(1-cosθ )].
Now, integrating,
∫ dt =√(R /(2g ))∫[dθ /√(1-cosθ )]R /(2g ))sin(θ /2)[ln(sin(θ /4))-ln(cos(θ /4))]/√[1 - cos(θ )]+constant
The integral is shown below for R =1 m,
g =9.8
m/s2 .
(I did not do the integral off the top of
my head! I used the wonderful resource Wolfram
Alpha .)
Now, integrating from t =0 to t
and from θ= 0 to 0.841, one gets t= ∞,
just as you found. But, this is not unexpected
because when θ= 0, the mass is in
equilibrium and will remain at rest until you
nudge it. If you calculate the definite integral
from some θ to 0.841, you get the
other curve shown (red). This is instructive but
probably not what you want because each point θ
on the curve is the time it takes before exiting
if it starts with the velocity it would have had
if it fell from the top; for example, for θ= 0.4,
t =0.23 s is the time before leaving the surface
if the mass started there with speed
v =√[2Rg (1-cosθ )]=√[2x1x9.8(1-cos0.4)]=1.2
m/s. An example which more closely would be of
interest to you would be the smallest angle I
plot, θ= 0.005=0.290 , where the time to
launch is about 1.6 s.
To get a little more quantitative, we could do a
small angle approximation, θ« 1.
Then, 1-cosθ≈θ 2 /2,
so
∫ dt≈ √(R /g )∫[dθ /θ ]=√(R /(g ))lnθ,
and the definite integral if you start at θ 0
is t≈ [√(R /g )]ln(0.841/θ 0 ).
So if you start at θ= 0.005, t ≈1.64
s. It is quite remarkable how well the small
angle approximation works as shown by the green
curve in the picture.
Thanks to the questioner for pointing out an
error in my first answer!
This, I believe, answers the question. It might
be interesting to calculate times and angles of
launch as a function of starting angles θ> 0
and the particle initially at rest; I will add
this later if I get around to working it out.
0 <θ 0 ≤900 .
The object starts at rest at θ 0
and leaves the surface at an angle θ x
after a time t . The main
difference between this and the earlier is that
the starting potential energy is mgR cosθ x ;
this changes the velocity to
v =√[2Rg (cosθ 0 θ )]
and
θ x =cos-1 (2cosθ 0 /3).
Everything else proceeds the same as above
except the indefinite integral now has the form
∫t = 2√(R /(2g ))F (θ /2|csc2 (θ 0
/2)/√(cosθ 0 -1constant
Where F
is an elliptic integral of the first kind. Plots
of t and
θ x
are plotted in the graph for 00 <θ 0 ≤900 .
Also, I have added the t graph above
(black curve) so it can be compared to the
calculations done earlier in this question.
1 and
2 ) addressed this issue.
The takeaway from those is that, although one
possible interpretation of GR is the deformed
spacetime, GR is a field theory and therefore it
can also be interpreted as there being forces
associated with the fields. Regarding your final
problem, tides, the reason is quite simple and
is called the "tidal
force "; the gravitational force from the
moon gets smaller as distance increases.
Therefore the force on the side closest to the
moon is bigger than the force on the center
which is bigger than the force on the far side.
All three forces are toward the moon but,
relative to the center of the earth, the
near-side and far-side forces both point away
from the center of the earth.
earlier answer which
investigates rolling of a bus.
*By elastic I mean that if you deform it a small
amount it will tend to go back to its original
shape.
E=mc 2
has nothing to do with your question. It is not
clear what you mean by the "energy of a
gravitational field". It is shown in a
NASA publication that the energy density
of a point mass M at a distance
r is u G =g 2 /(8πG )
where g=MG /r 2 and
G is the universal gravitational
constant. This quantity is definitely positive
and so would be the total energy of this field
in any volume in space. Maybe you are thinking
of the gravitational potential energy of a point
mass, U=-MG /r +MG /r 0
where r 0 is the position
where we choose the potential energy to be zero.
If we choose r 0 =∞, U=-MG /r< 0
everywhere. This is convenient because the total
energy of a particle is E=U+K and if
E <0, the particle is bound (orbit) and
if E >0, it is unbound (scattering).
f is
proportional to the force which presses the two
together and that force is proportional to the
weight W times the cosine of the incline angle
θ ;
the proportionality constant μ is called the
coefficient of kinetic friction and depends on
the nature of the surfaces sliding on each
other, snow and tube in your case. So we have
f=-μW cosθ. (The minus
sign means f acts up the hill.) But
there is also the force of gravity down the hill
which is F=W sinθ , so the net force on the
block will be F+f=W (sinθ -μ cosθ ). Now, this
force (F+f ) causes an acceleration which is
proportional to the force divided by the weight;
the proportionality in this case is 32 if you
are measuring weight in pounds. So we have a =32(F+f )/W= 32(sinθ -μ cosθ ).
Since the weight cancels out, the motion down the hill should be identical
no matter how much weight you have. For example
if μ =0.1 and θ =300 ,
a=32(0.5-0.1x0.87)=13 ft/s2 , about
40% of gravitational acceleration (32 ft/s2 ),
regardless of the weight.
I believe that your data do not show the
expected behavior because snow is difficult to
assign a coefficient of friction to. Snow can be
wet, dry, loose, compacted, etc .; for
example , μ for waxed wood (ski) on wet
snow is about 0.1 but on dry snow about 0.4, a
huge difference. Furthermore adding more weight is
very likely to change the frictional force by
causing more packing or by "plowing" up a little
pile of snow in the front. In order to
successfully make these measurements, you would
have to take extraordinary pains to keep the
conditions of the snow identical for the many
runs. Overall, snow is not going to provide a
really scientific study of friction.
Here is an important lesson to be learned from
all this. You know Galileo's experiment
demonstrating that heavy objects, when dropped,
fall at the same rate as light objects (assuming
that air drag is not important). The reason for
this is just the same as the reason that the
objects of different weights sliding down an
incline should move down at the same rates. So
you can demonstrate to him the overriding
principle by dropping a baseball and a bowling
ball from the same height simultaneously.
"wormholes" , but that is indeed speculative.
2 is pressure, not force. If they are, for example, 100 square inches the thrust would be 7,000,000 lb.
However, your numbers seem far off compared to data I found on
Wikepedia . The Vulcan B.1 had a maximum takeoff weight of 170,000 lb and a thrust from its four engines of 4x11,000=44,000 lb. Hence thrust/weight=0.25.
This plane cannot fly without wings! By the way,
the wings do not provide acceleration, they
provide lift.
12 m. On both ships
time moves perfectly normally. But on the fast
ship the distance to pluto is drastically
reduced by length contraction, L'=L √(1-(v /c )2 )=5.9x1012 √(1-(0.99)2 )=8.3x1011
m. So the fast ship will take, by his clock,
t fast =8.3x1011 /(0.99x3x108 )=2739
s=46 min. For the slow ship, which experiences
almost no length contraction, the time according
to his ship's clock is t slow =5.9x1012 /104 =5.9x108
s=19 yr.
R =3MG /2 from a black hole
of mass M , photons will move in a
circular orbit. If you were this distance from a
black hole and looked in a tangential direction
you would see the back of your head! Elliptical
orbits are not possible because the speed of a
photon has a constant speed.
j 3 r t 3 ,
is called jerk. It is also the time derivative
of acceleration, j a t ,
and aptly named because if acceleration is not
smooth and constant, like when someone is
learning how to drive a standard transmission
car, motion is jerky. The time derivative of
jerk, s j t ,
is called snap. After snap, the 5th and 6th
derivatives are called crackle and pop, a
tribute to Rice Crispies, I guess! If you want
to read a whole article about this question,
check out this
link .
2 serve in E=mc2 other than to make E a ridiculously large quantity of really small units instead of E=m where E is a 'normal' quantity of ridiculously large units?
m and you could figure out
how, the amount of energy you could extract from
that mass is, in fact, very huge. Most processes
in nature do not result in mass just
disappearing and some energy appearing, rather
some small fraction of mass disappears and
energy appears. That is how the sun works: when
you create helium from hydrogen you end up with
about 1% less mass than you started with. One
example of complete annihilation mass is when an
electron meets its antiparticle, the positron,
they both disappear and electrictromagnetic
radiation with energy precisely equal to the sum
of their masses times c 2
appears. Finally, I should note that E=m is an
impossible equation because mass and energy do
not have the same units; for example, the
equation 1 meter=4 kilograms make no sense.
etc . But, in the late 19th
and early 20th centuries when scientists started
examining things going very fast, things being
very small, etc., it was inevitable that very
large or very small numbers would arise in our
theories and calculations. In many cases, the
mathematics emerging from theories could be much
more concisely written by introducing different
systems of units. One which you would like, I
think, is the planck system where many
fundamental constants are chosen to be equal to
1: the speed of light, c =1; rationalized Planck
constant, ℏ =1; permittivity of free space,
ε 0 =1; permiability
of free space, μ 0 =1;
etc . So, you see, E=m is the
famous equation in natural units. If you want to
look into this more deeply, see the Wikepedia
article on
natural units .
There will be a frictional drag force on the
fluid as it slides on the inner surface of
the drum, F drum
on fluid which tends to pull the fluid
surface along with the motion of the drum.
This answers your first question—the
fluid will rotate in the same direction as
the drum but it will take some time to get
all the fluid rotating. Note that this drag
force will be of different magnitude at
different positions along the surface of the
drum but all will act tangentially,
producing some net torque tending to get the
fluid rolling.
Because of Newton's third law, the drum will
experience an equal and opposite force, F fluid
on drum , which would cause the
drum to accelerate down the incline more
slowly than if those forces were not there.
I have also shown the frictional force which
the incline exerts on the drum to keep it
from sliding; this is a static friction
force whereas the forces of the drum-fluid
are kinetic friction. I assume that this
force is large enough that the drum does not
slip down the incline.
Now, to understand how this system behaves for
various situations, I will look at a couple of
extreme situation quantitatively. I will assume
that the mass of the fluid is M and the
mass of the drum is m and that the
radius of each is R . Suppose that the
fluid has an extremely high viscosity, i.e.
it is essentially a solid.
In the case of very low (essentially zero)
friction between the drum and the fluid, the
fluid will never get rolling. You can use
energy conservation and note that only the
drum has any rotational kinetic at the
bottom: m+M )gh =½(m+M )v 2 +½mR 2 (v/R )2
=½v 2 (2m+M ).v 0 =√[(2gh (m+M )/(2m+M )].
(Note that I have used mR 2
as the moment of inertia (I ) of a
hollow cylinder and v /R as the
angular velocity of the drum about its
center of mass.)
In the case of very high (essentially locked
together) friction between the drum and the
fluid, the fluid will immediately get
rolling. In that case,m+M )gh =½(m+M )v 2 +½[mR 2 +½MR 2 ](v/R )2 =½v 2 (2m+ 3M /2).v∞ =√[(2gh (m+M )/(2m+ 3M /2)].
This is smaller than the case where the
fluid does not rotate. (Here I have used
I =½mR 2 for the
rotating fluid.)
If the viscosity is 0, there is no shear force
which can get the any of the fluid inside its
outer surface rotating, so the result will be
the same as for a solid with no friction with
the drum.
Finally let's allow values intermediate between
0 and ∞ for the viscosity. In these cases the
fluid will have different angular velocities
depending how far from the axis it is. If you
wait long enough, have a long enough incline,
eventually the whole fluid will be rotating with
the same angular velocity as the drum, just like
the solid case. But, for a given length incline,
the speed at the bottom will decrease from v 0
to v∞
as the viscosity increases from zero.
K
with a vertical velocity at y =0 in a
uniform gravitaitional field will rise
(neglecting air drag) to a height h=K /(mg ).
You are right and your text is wrong.
etc . contain the oxygen they need to
explode.
site ground rules . "I no longer answer questions which are based on premises like: "...if we could travel faster than the speed of light..." or "...ignoring the fact that nothing can go faster than the speed of light...", "...if I were traveling at the speed of light..., etc." The laws of physics forbid anything with mass from traveling with the speed of light.
However, I think I can address your question if
99.99% of c is the speed of your
traveler. The traveler sees the distance L= 10
ly length-contracted to the distance L'= 10√(1-.99992 )=0.0141
ly so the time of travel would be 0.00141
yr=12.4 hr. The stationary observer would see
the time elapsed to be 10x0.9999=9.999 yr.
faq page .
Huygen's principle —every point on a
wave front acts like a point source of the
waves, so those points on the edges of the beam
emit light with components transverse to the
direction of the beam. Laser light through a
square aperature is shown in the figure. The
central spot is saturated, much brighter than
the fringes.
v : v= √(2gh )
where g =9.8 m/s2 is the
acceleration due to gravity and h is
the height. Note that the speed does not depend
on the mass. For example, drop a bowling ball
from h =10 m; v =√(2x9.8x10)=14
m/s=46 ft/s=31 mph. However, if air drag is
important the object will speed up at first, but
the faster it goes the greater the drag so
eventually it falls with constant speed called
the terminal velocity, v t =√(mg /D )
where m is the mass and D is a
constant which depends on the geometry of the
object, the density of the air, etc .
The drag force due to air drag is proportional
to the square of the velocity of the object and
D is the proportionality constant,
F drag =Dv 2 .
For example, for a crumpled up piece of paper I
estimate that the terminal velocity would be
about v t =1.6 m/s=3.6 mph
and, if dropped from 10 m, it would quickly
accelerate to this speed and fall most of the
rest of the way down with constant velocity.
exchange
interaction . The summary explanation from
Wikepedia is "In simple terms, the electrons, which repel one another, can move
'further apart' by aligning their spins, so the spins of these electrons tend to line up. This difference in energy is called the exchange energy."
Note that, if the outer electrons in adjacent
atoms can get farther apart, they reduce their
energy and will therefore tend to align their
spins. To get the details of the exchange
interaction in ferromagnetism, see the
Wikepedia article .
Now, you might ask, "what about the baseball
which can be made to break right or left?" The
baseball is not thrown so that its spin axis is
parallel to the velocity direction like the
bullet spins, so it can have lateral forces on
it.
Fr =(4000
kg)(9.8 m/s2 )(0.04 m)=1568 N·m=1157
ft·lb. (Note that I have converted the
mass 4000 kg to force in newtons using F=mg .)
I have assumed that the force is in metric tons.
3 Al5 O12 :Ce)
which absorbs some of the light from the LED and
then emits yellow light. I would guess that,
since both yellow and blue are present, the
design of the device could be such that at
certain orientations one would see only blue or
yellow. As the bulbs in your sheet move
slightly, you might be looking at just the right
orientation that you separately see the blue and
yellow. You might try taking a video to see if
you can catch the phenomenon. I am pretty sure
that your "quantum effect in the brain" is not
what is happening!
old
answer .
f , the energies of the photons which
comprise it are E=hf where h
is Planck's conatant.
This is a pretty
mathematical question, so I will answer it in
pieces as I get the time to work it out. To find
the time it takes to fall, you start with
Newton's second law, ma=m [d2 r /dt 2 ]=-GMm /r 2 =dv /dt
and integrate it once to get v =dr /dt =√[2GM (R-r )/(Rr )]
where, for your question, R is the
distance from the center of the earth to the
point from which
it was dropped, r is where it is at the
time t (r ), G is the
universal gravitation constant, and M
is the mass of the earth. (The mass m
of the ship does not matter.) (You can more
easily get v (r ) from energy
conservation.) Next integrate v (r )=dr /dt to get
t (r )=-2.69x105 {0.5tan-1 ((2x -1)/(2√[x (1-x )]-√[x (1-x )]-π /4}
where x=r /R ; this is not a simple
integral to perform, but you can find good
integrators on the web, for example
Wolfram
Alpha . Once you have gotten the indefinite
integral you are not finished, you need to find
the integration constant which will put the ship
at rest at x=r /R =1; that is
where the π /4 comes from. Putting in your
numbers, I find t =4.22x105 s=117 hr=4.89 days. t (x ).
W bird ,
the box W box , and the air
W air ; the total weight I
will denote as W .
Before the bird does
anything is just sitting on the bottom, there is
a force down on the whole system (box, bird, and
air) of magnitude W and the ground
exerts an upward force on the whole system of
magnitude W because of the system is at
rest so has zero net force on it.
Now suppose
the bird hovers in the middle of the box. The
air is exerting an upward force W bird
on the bird because the bird is pushing down on
the air with a downward force of magnitude W bird
(N3). Now the forces on the air are its own
weight down W air , the
force the bird exerts down on it W bird ,
and the force up from the bottom which must
equal W air +W bird
because the air is in equilibrium. But now
the air exerts a force down (N3) on the
bottom of the box equal in magnitude to the
force W air +W bird
the box exerts up on it, so all the forces
on the box are this force plus the weight of
the box W box ; the box is
in equilibrium so the force the ground
exerts on the box is the sum of all weights
W , exactly the same as if the bird
were sitting on the bottom of the box. All
the forces which happen internally in this
system, cancel out. The box still weighs
W in spite of the bird's hovering.
Now the
bird flies up to the top of the box and
exerts an upward force F on the box
which is greater than the total weight of
the system, W , so that he can lift
the system up in the air; in order to do
this he must exert a downward force F+W bird
on the air. So the air must exert a downward
force on the bottom equal to its own weight
plus the force the bird is exerting on it,
magnitude F+ W air +W bird F+W-F=W because the top of the box
exerts a force of magnitude F on
the bird. Therefore, the bird cannot lift
the box plus air plus himself from inside.
The bottom line here is
that any forces which happen inside a system,
forces between individual pieces of a system,
all cancel out because of N3 and can be ignored.
This is a case where the general case is easier
to understand than a specific case. If a system
consists of a large number of objects, the force
between any pair of objects will add to zero
because of N3, so only external forces need be
considered; in the case being discussed above,
the only forces external to the system are its
weight and the force the ground exerts.
c . But when you add the two together
you are left, astonishingly, with light of the
same frequency but which travels at a speed
slower than c . So you see, neither the
density nor size of the molecules is relevant;
the speed you end up with is simply determined
by how the particular molecule responds to
electric and magnetic fields. I think any more
quantitative explanation is probably beyond the
mathematical capabilities of a ninth grader, but
if you want to explore this further you can look
at Wikepedia articles on
refractive index and the
Ewald-Oseen extinction theorem .
a descending rappeller). What is the resultant maximum strength of the system? Is it 40KN (because the tensile force is distributed across two 20KN ropes), or is it 20KN (because there is only one strand at the back of the bollard)? I say 40KN, but I can't explain away the 20KN argument.
Assume the bollard is stronger than the maximum system tension, the bollard is frictionless, and ignore the resulting internal tension/compression within the rope as it curves around the bollard. The latter is usually the real-world point of rope failure in a doubled rope system, as more tensile force is taken by the outside of the rope as it curves around the bollard.
i.e. both halves
are held as one so that each holds half the
total load, up to 40 kN is possible. If indeed
the bollard is frictionless, each half of the
doubled rope will hold half the load regardless
of how it is held. If there were friction, one
half could go slack requiring only one rope, up
to 20 kN, to take the load. In the real world,
ensure that the ropes are held together.
YouTube . The one where Brian
Cox of BBC humanuniverse does the gravity experiment at that NASA lab. So
according to Brian Cox Einstein's theory suggested that the bowling ball and
the feather in space are not moving at all and in this clip if there were no
background that's what it would look like. Does this then mean that the
earth is moving towards the objects?
special relativity is that
the laws of physics are the same in all
inertial frames of reference . This is
called the principle of relativity and
says that once you have found one inertial
frame, one in which Newton's first law is true,
any other frame moving with
constant velocity
relative to that one is also an inertial frame.
Another way of stating this is that there is no
special frame which is the universal "at-rest
frame" since no experiment you can do can tell
you which frame is "at rest", all are
indistinguishable. The problem is that the
frame of the ball/feathers is not an inertial
frame; it therefore seems wrong to me to state,
as Cox does, that in fact the ball/feather are
at rest because they are accelerating. The
appropriate way of understanding this is the
equivalence principle which may be stated
in two ways. First you can say that there is no
experiment you can do which can distinguish
whether you are at rest in empty space (or
moving with constant velocity, which is the same
thing because of the principle of relativity) or
you are accelerating due to a gravitational
field (free fall). Alternatively you can say
that there is no experiment you can perform to
distinguish whether you are accelerating in
empty space (by burning your rockets, e.g .)
or whether you are at rest in a gravitational
field. The first of these is most applicable to
the case at hand: if you were inside the bolwing
ball and falling, it would be the same as if you
were in the middle of space and not falling.
Finally, the equivalence principle, along with
the principle of relativity, lead to the
principle of general relativity : the laws
of physics are true in all frames of reference.
You have hit upon the main flaw of Cox's
statement: just taking them at rest would imply
that the earth is accelerating toward them which
is incorrect.
Here is what Cox was probably referring to as
Einstein's happiest thought (although it is
probably apocryphal): Einstein, bored at his
desk in the Swiss Patent Office, was watching a
house painter across the square on a ladder
painting a building. The painter fell. As he
fell Einstein thought "there is no experiment
that painter can perform which will distinguish
whether he is at rest in empty space or falling
in a gravitational field!" You can read a bit
more detail
here .
e.g ., anti-atoms like
antihydrogen with an antiproton with an orbiting
positron; now, we can fairly easily contain
charged antiparticles with electric and magnetic
fields, but an anti-atom has no charge and could
not be easily contained. The difficulty in
making an anti-atom is that the antiparticles we
create usually have very high speeds and the
components of the anti-atom need to be nearly at
rest relative to each other to form. I do not
understand your last question.
Why does do we get heat as a by product in electric machines or during generation of PV panels why cant that heat be used up in production of electricity altogether??.
Can this two different phenemenon influence a object to create electricity at the same time.
recent post . The upper limit for the mass of
an electron neutrino is 1.1 eV where eV is
electron-volt; 1 eV=1.6x10-19 J. Note
that, although physicists usually call an eV a
mass by convention, it is really the rest mass
energy of the particle, i.e . mc 2 .
So the upper limit of the mass energy of an
electron neutrino is 1.6x10-19 /(3x108 )2 =1.8x10-3 6
kg. Now, most of the mass (not counting dark
matter) of the universe is in protons and
neutrons each of which has a mass of about 1
GeV=109 eV, a billion times bigger
than the the neutrino. Electrons have a mass of
500 keV=500,000 eV; we would expect there to be
about as many electrons in the universe as
protons so electrons would only contribute about
.025% of the total mass. There are about
337 neutrinos/cm3 =337x106
m-3 in the universe, so the mass
energy density is about 3.7x104 eV/m3 ;
the average ordinary mass density of the
universe is about
4x10-28 kg/m3 =3.6x10-11
J/m3 =2.25x108 eV/m3 .
So the neutrinos contribute less than about
3.7x104 /2.25x108 =1.6x10-4 =0.016%
of the ordinary mass. Keep in mind that all this
represents order of magnitude calculations; the
numbers should not be taken too seriously but
the relative numbers should be right to better
than one order of magnitude.
escape velocity of
a projectile, the minimum speed at the earth's
surface for which a projectile will never come
back. This will slightly overestimate the speed
necessary to reach the moon for two reasons: the
moon is not infinitely far away and the moon has
its own small gravity. Still, to do it more
exactly will not really be necessary to get to
the heart of this question. The escape velocity
is 1.12x104 m/s, so the necessary
energy of a spear of mass m is E =½m (1.12x104 )2 =6.3x107 m
J. I will take m =0.8 kg which is the
mass of an olympic javelin, so Fd =5x107
J where F is the force necessary and d is the
distance over which the force is applied which I
will take to be 2 m. Hence, the force is about
2.5x107 N=5,600,000 lb=2,800
tons=3,100 metric tons.
earlier answer and find the time the earth
will take to get to the sun is about 2 months. I
see no reason it should explode; it will simply
fall into the sun and be totally incinerated.
And there is no way that it will "go[es] out of
the solar system with masive acceleraration".
V /A has
the same units as an ohm and so 1 Ω=1 kg·m2 /(s3 ·A2 ).
One tesla has units 1 T=1 kg/(s2 ·A)
so one tesla per second is 1 kg/(s3 ·A).
In other volts per ampere is not equilivant to
teslas per second. Since I do not know the
details of your experiment, I cannot help you
further.
W down, the force up from
the floor N f f
v =√(2gh )√(2x9.8x1)=4.4
m/s. You can't know the force though. If you look at the faq page
(search for how much
force on the faq page) you will see that
what matters is either how far the 10 kg went
before stopping or how long it took it to stop.
E.g. , suppose that it stopped in 0.1 s. Then the force
would be approximately F=mg+mv /t =10x9.8+10x4.4/0.1=538 N (about
55 kg-force). So for a tenth of a second it
would be like having a 55 kg mass on the
person's head. The weight is the force the earth
exerts on the mass which is mg =10x9.8=98
N; this never changes.
how much
force on the faq page) you will see that
it is not the force which matters but rather the
force and the distance and/or time over which
that force acts. The velocity you would need
to get to h =100 ft would be more than
v =√(2gh )=√(2x32x100)=80
ft/s=54.5 mph. I say "more than" because a 1 lb
object would occupy enough volume that air drag
would not be negligible. Now, how much energy
would you have to impart to the object? Again
for no air drag considered, and therefore more
than E=Wh =1x100=100
ft·lb. So, suppose you could have a bow
with a draw of 2 ft; the force you would need
would be more than 100 lb. (I have modeled the
bow as a simple spring which would have a spring
constant of 50 lb/ft. See an
earlier
answer if you are interested in the
details.) With the likely air drag plus the drag
of the string being, I presume, trailing behind,
I would not be surprised if you needed as much
as twice the energy I have calculated here or
more. Since you have to send a pound, I would
think one of those air cannons they use to send
tee shirts into the croud at atheletic events
would be a better bet.
p
of a mass m with velocity v is
p=mv and so the
car has a linear momentum of 6 kg·m/s.
Otherwise it is a typographical error and should
have been 6 km/s which would be v .
It would
be simplest if you could make your marble go
straight up; that is not imperative, but it
will be easier to analyze. At the very
least, the marble should leave your
contraption at the same angle relative
relative to the horizontal for every
measurement.
You will also get better results if you can
arrange for the falling weight to stick to
the contraption and always hit at the same
place.
Use a good number of falling weights so you
can get a lot of data.
Plot however you measure what the marble
does (height, distance horizontally, or
whatever) versus the dropped weight.
I will help you after you have data if you need
help.
0 , it will end up
crashing into the earth before it completes one
orbit. The minimum possible orbit would have a
radius equal to the earth's radius and have a
speed given by v =√(gR )
where R= 6.4x106 m is the
radius of the earth and g =9.8 m/s2
is the acceleration due to gravity. Doing the
arithmetic, v =7.92x103 m/s which is
about 17,700 mph.
It is
possible to independently get an approximation
for the speed from the data for the braking
period. To do this I assume that the
acceleration is constant and the coefficient of
friction between rubber and dry asphalt is about
0.8. Without going into detail, I find the speed
of the car to be 76.7 ft/s=52.3 mph. So these
data would certainly not hold up in a court of
law! Possible sources of error are:
The reaction time was perhaps overestimated
by nearly a factor of three; 20/0.26=76.9
ft/s.
I have assumed that the car did not skid
(antilock brakes); possible values of the
coefficient of friction have a rather large
range depending on conditions, but the
smallest likely value is about 0.6 which
would result in a speed about 66.5 ft/s,
still quite a bit larger than 26.7 ft/s.
If the car skidded the coefficient would be
somewhat smaller but not enough to yield a
speed in agreement with 26.7 ft/s.
If the road were wet or even icy and she
skidded, the speed would be quite a bit
smaller.
I looked up a typical human reaction time and
found that the average reaction time is about
0.25 s. If Dana's reaction time was average,
everything would point to a speed of around 50
mph.
Eddington who observed a shift in the
apparent position of a star during a solar
eclipse due to bending of light passing close to
the sun. So, gravity does interact with light
and bends it. The reason light cannot escape a
black hole is pretty straightforward. You can
calculate the escape velocity (the speed
necessary to move away and never return) from
any object as a function of how far away from it
you are. At a distance called the event horizon
from a black hole the escape velocity is equal
to the speed of light and larger for closer
distances. Hence, since nothing can exceed the
speed of light, nothing can escape once inside
the event horizon.
PV /(NRT )=Constant,
where P is pressure, V is
volume, N is some measure of how much
gas you have, T is the temperature, and
R is a constant the value of which
depends on
the units you use to measure the variables.
Boyle's law is constant only if the temperature
and amount of air in the baloon are constants,
which could be approximately true for T
but certainly not N since you are
adding air as you blow the balloon up.
A balloon is made of rubber. For simplicity,
let's think first about a rubber band. As you
start to stretch it, it acts like a spring, F ≈kx
where F is the force that you pull
with (and therefore the force the rubber band
exerts on you) and x is the distance it
is stretched. Then, when you pull it far enough,
it starts having less and less stretch for each
increase of force. Finally it breaks, or if you
pull it past where the stretch is linear and
then let is go back you find that the rubber
band is longer than it was when you started.
This property of not coming back to where you
started is called hysteresis . The
balloon is analogous to the rubber band as it is
blown up. The graph depicts this behavior, but
needs a little explanation. Instead of plotting
length as a function of force, we plot stress as
a function of strain. Strain is essentially the
volume (analogous to length for the rubber band)
but as a ratio to the original volume; stress is
the pressure times the surface area of the
balloon which is a force. So the stress
(pressure) starts out as a straight line, then
maintains almost constant stress (pressure) as
the balloon gets bigger, and finally it gets
harder to blow it up farther so the stress
(pressure) increases rapidly but the strain
(volume) increases more slowly until it pops.
Finally, we look at the graph where the balloon
deflates. Now the stress (pressure) and strain
(volume) both rapidly decrease not following the
inflation curve, come to a point where the
stress (pressure) is nearly costant, then both
finally decrease approximately linearly to zero
stress (pressure) but with a strain (volume)
bigger than before the balloon was inflated.
There is a nice discussion of balloon physics at
this
link . (I borrowed their graphs.)
similar
question recently answered.
v F F C
and tangential F T . The
tangential force causes an acceleration which
decreases the speed and the centripetal force
causes the velocity to turn. In the case of a
circular orbit of radius R , F T =0
and the centripetal force is FC =Mv 2 /R ;
the centrifugal force would be a vector with
magnitude Mv 2 /R
pointing away from the sun but it doesn't really
exist.
entire ice cube, not just
the submerged part of it.
contraction ,
not dilation. Dilation would mean the length
increased.)
W , and the air drag upwards which is
proportional to your speed, D=Cv 2 ,
where C is a constant determined
essentially by the area you present by to the
onrushing air; so, your terminal velocity is
when the weight down equals the drag up or v= √(W /C ).
Before you open your parachute your speed
increases until it is about 120 mph. But that
terminal velocity is much bigger than what it
will be when you open your parachute. So, when
you open the chute you are well past your
terminal velocity (as per your question), so the
increased drag now slows you down until you
reach the terminal velocity, about 20 mph. The
answer to your question is that the air drag
will slow you down until you are traveling at
the terminal velocity.
cannot find a trig identity for "a - sin(a)"
a ) and a -2π /3.
The first graph shows that a ≈2.6.
If you need it more accurately, zoom in as I
have in the second graph; now a ≈2.6053.
After I did that I found a handy
on-line calculator which verified my
calculation and went one place farther to a ≈2.60533.
WolframAlpha is very handy.
c 2 or
c 3 or c 4
or etc . has the units of speed, so it
could not possibly have any reference to a
superluminal speed. The units of speed are, for
example, m/s and c has the units of m3 /s3 .
And, yes, mass may be created from energy; for
example, if a very energetic photon interacts
with a strong electric field (no mass anywhere
there) it is possible that its energy will be
converted into an electron-positron pair (called
pair production ), which has the mass of two
electrons. Well, the mass of the universe had to
come from something and E=mc 2
demonstrates that mass is just another form of
energy.
answered
a question just like yours earlier except for
cars. You should look at that answer because it
provides the best answer I can give you here.
I did find a lot of sites which address the
specific case of bicycles by googling. In
one case , somebody in Holland did wind
tunnel experiments and found some interesting
results, one of which is that the lead biker
also gets a modest boost (about 3%).
h and velocity v : v =√(2gh )
where g =9.8 m/s2 is the
acceleration due to gravity, h is in meters and
v is in m/s; note that it is
independent of mass m which will be in
kg. The easiest way to relate the speed of the
ball to the force it applies is to use energy
methods: the work W done on the ball by the bone
equals the change in energy of the ball W=E final -E initial .
The initial energy is ½mv 2 =mgh
and the final energy is -mgd where
d is the distance the ball moved before it
stopped; the work done is -Fd where
F is the magnitude of the force the bone
exerts on the ball. (Because of Newton's third
law, the force the ball exerts on the bone must
be equal and opposite the force the bone exerts
on the ball.) Put it all together and you get
h=d [(F /(mg ))-1]. For
example, the weight of a bowling ball is about
50 N (m is about 5 kg) and the force
needed to break a femur is about 4000 N. If you
observe d to be about 1 cm, h =79
cm. For a baseball, m =0.145 kg, F /mg =2760,
so h= .01x2759=27.59 m for d =1cm.
But you have to be careful as the mass gets
small because air drag, which I have neglected,
gets important and the object will never get
enough speed to break the bone. E.g. ,
for a ping pong ball with m =2.47x10-3
kg you find that h =1619 m such that
v =180 m/s=403 mph when it hits the bone;
but the terminal velocity for a ping pong ball
is about 8 m/s so that is how fast it would
actually be going when it hit the bone. Unless
somebody has a gun which can launch a 400 mph
ping pong ball at point-blank range, you do not
need to worry about getting a broken bone by
being hit by a ping pong ball. Also you
have to be judicious about your guess for d
because a soft ball, like a volley ball, for
example, will compress when it hits so d
will be larger which makes F smaller.
2 . I understand that because matter is energy, by going faster, you are adding energy and therefore effectively increasing mass. Right? Anyone who has experienced acceleration can basically feel this in action, I think. Or if you’ve ever had the opportunity to compare falling down to a bicycle crash at speed, same thing.
But why is the increase exponential? Why is it not simply additive?
M ) increases like
M at.v =M at.rest /√[1-(v /c )2 ]
where v is the speed and c is
the speed of light; only when v is
quite close to c is there an
appreciable change. What happens at lower speeds
is well-described by classical mechanics: your
kinetic energy ½Mv 2
increases while M stays approximately constant.
So, what does this have to do with E=Mc 2 .
When Einstein developed the theory of special
relativity, one of the most important
restrictions on it was that any results had to
reduce to classical values when v was small
compared to c because at these speeds classical
mechanics is spectacularly successful. It turned
out that when he calculated the kinetic energy
assuming that M was a constant he found out that
it did not reduce to ½Mv 2 ;
but, if he let M depend on velocity,
M at.v =M at.rest /√[1-(v /c )2 ],
it did. The bonus to this was that the energy of
a mass M has energy of Mc 2
even when it is at rest which means that mass is
a form of energy itself and if you figure out
how to convert it you can use it to power your
car, or your lights, or your refrigerator, and
on and on. In fact we began to learn out how to
do this millenia ago: fire is essentially
chemistry. And, chemistry is gasoline, TNT,
coal, natural gas, etc . However
chemistry is extremely inefficient and the
change in mass when you burn gasoline, for
example, is so small that you would be hard put
to measure it. That is why chemists have the
"law of conservation of mass" which is wrong but
works if the change in mass is too small to
measure. Nuclear energy is enormously more
efficient with a noticeable amount of mass being
converted to energy. Since the speed of light is
so large, the energy of even a tiny amount of
mass is huge. If you have a mass of 1 gram and
calculate the energy it has you get E =9x1013
Joules; the biggest power plant in the US is
about 4 GigaWatts=4x109
Joules/second, which means our 1 gram of fuel
could deliver that power for 22,500 seconds=62.5
hours!
m 1 v 1 =m 2 v 2
or v 2 =(127/28000)v 1 =0.0045v 2 .
The range of a projectile is proportional to the
square of its
speed, so the range would be reduced by a factor
of about 0.00452 =2x10-5 . (I have neglected air drag.) Of
course, I know of no way you could make this
happen.
CORRECTION: R=v 2 sin(2θ )/g.
14 m2 .
Suppose your "rock" has a volume of 1 m3 ,
pretty big. Then this volume will be spread out
evenly over the whole area (we assume), so we
have 1 m3 =(3.6x1014 m2 )t
where t is the height which the ocean
level rose; therefore, t =2.8x10-15
m; but a water molecule has a size of about
2.8x10-10 m, about 10,000 times
bigger than t ! I guess your rock would
have to be about 104 m3 in
order to add a single molecule film to the whole
area. So your question really has no meaning.
However, if you added a much bigger rock so that
the rise in sea level was measurable, the
fastest that the information could be
transmitted to raise the surface is the speed of
sound in water. (All this assumes we do not need
to think about things like tides, storms,
etc .)
faq page.
Wikepedia .
What you really want is the force
you have to apply to pull the tire; the weight
of the tire is the weight of the tire. First,
the simple case which is if you are pulling
horizontally. The force you need to apply to
move it with constant speed is given by F=μN
where μ is the coefficient of friction for
rubber on rubber and N is the force which the
tire and floor are pushed together, the weight
in this case. The coefficient is about 1.2, so
the force you must exert is about 210 lb. If you
are pulling at some angle θ , it
is a bit trickier because you are both pulling
forward which moves the tire but also pulling
upward which causes N to get smaller.
Without going into details, the result is F=μW /[cosθ +μ sinθ ].
I have made a graph which shows this result for
your case. As you can see, the closer to
horizontal you can pull, the better workout you
will get. Also note that if you pull at around
50° you will exert the least force
this answer on Quora, it says a particle cant be in two (or more) places at once. how can this go with what we read on the internet?
for example, there is this study which based on the conception of the particles being in places (parallel universes)“When the team fired a photon at the cavity, the atom’s dual personality caused two things to happen at once. In one of the 'parallel universes' of its superposition, the one in which the atom was resonant with the cavity, the photon did not enter.
and many many science articles on the internet for the ordinary readers (not specialized) says it can, and it is. some related informations about Superposition and Wave-Function i found.
nothing
to do with parallel universes. That said, I have
to say that the answer on Quora is excellent.
One could actually say that saying that a
particle is anywhere at all is meaningless
unless you have made a measurement of the
position and then you only know where it was at
the time you observed it. It is meaningless
because all you can specify is the probability
of its being inside any volume at any given
time. I have answered this question several
times in the past and you can see those via the
faq page .
etc . The
electron is therefore spread out all over the
place but you wouldn't say that it is "located"
in many places; rather, there are many places
where you might find it to be if you went
looking for it.
drag ) is proportional to the
square of the velocity, F∝v 2
or, F=Kv 2 where K
is a proportionality constant and depends on the
geometry of the object and the density of the
fluid. The actual
drag equation is F =½ρC D Av 2 ,
where ρ is the density of the fluid,
C D is the
drag coefficient which depends on the shape
of the object, and A is the area which
the object presents to the onrushing fluid. So,
for a parachute, since A is much bigger
when open than when packed, the drag is much
larger and therefore you survive if you jump out
of an airplane with one. There is an
approximation to the drag equation for motion
through air at atmospheric pressure which is
convient for estimating drag for those special
cases, F ≈¼Av 2 ;
this approximation does not work unless you do
your calculation in SI units, m2 for A , m/s for v , and N for F .
Here is an example: a skydiver and his equipment
have a weight of 1000. She accelerates to some
speed v where she experiences an upward drag of
F =¼Av 2 =½v 2
(I have estimated A =2 m2 ); but
her weight is another force on her, so the net
force on her is F net =1000-½v 2
and when v gets big enough she falls
with constant speed because the net force is
zero. This is called the terminal velocity and,
for this case, v =√2000=45 m/s=101 mph and she
will likely die. If she had opened her parachute
and it had an area of 25 m2 , her terminal
velocity would have been about v =√160=12.7
m/s=28 mph and she would have survived.
static electricity it was discovered that
the forces objects exert on each other can be
either attractive or repulsive; this led to the
conclusion that there is not only one property
called charge but charge must come in two
different "flavors". Asking what charge "is made
of" is like asking what "blue is made of"; blue
is just a property some objects have.
L close =Mv close R close .
At the point in the orbit farthest from the sun
the angular momentum is L far =Mv far R far .
Since the angular momentum is conserved, Mv far R far =Mv close R close
or v close /v far R far /R close .
Therefore v close >v far L is always the same.
suggestions that perhaps the 4-neutron
system or a 2-neutron system might be bound,
but neither has been experimentally observed unambiguously in
spite of many attempts. Also, a
neutron star is predominantly a bound
neutron matter, but the binding is due to
gravity, not the strong interaction.
-15 m)
which is much more than the wavelength of
visible light (on the order of 5x10-7
m). It is therefore impossible to "see" a
nucleus by imaging with visible light like you
might do with a bacterium. In that sense, it
makes no sense at all to try to talk about its
color. If a nucleus radiates, the wavelength is
never in the visible region, x-rays being the
longest wavelength you will detect. An
individual neutron or proton is smaller than a
nucleus so the same arguments apply to them. The
bottom line is that none of these things have
color.
t =√(750/16)=6.8 s
and the speed would be v =32x6.85=219
ft/s=149 mph. So, if your mass is m , your kinetic
energy is ½mv 2 =2.4x104 m
ft lb. Suppose there are 20 steps per
floor, so each step is a height 15/20=3/4 ft. So
the amount of work you do going up one step is
(3/4)32m =24m ft lb. So the
work you do going up is 24x1000m =2.4x104 m
ft lb. So, you see that the energy you put in by
climbing was exactly equal to the energy you got
out by falling back. There is no such thing as
conservation of kinetic energy. You do work to
increase your energy (called potential energy in
this case), so you end up with this potential
energy (and no kinetic energy) at the top. If
you jump off, that potential energy is turned
into kinetic energy.
E i =E f
where i denotes initial and f denotes
final. But, there is no such law as conservation
of mass because Einstein showed that mass is
just a kind of energy. So the mass of a system
need not be the same before and after. Chemists
still use a "law of mass conservation" because
the mass changes in chemical reactions are too
tiny to measure; still, as we know in physics,
the mass does change in chemical reactions. Now,
you know the famous equation E=mc 2 ,
so how about
m i c 2 +E i =m f c 2 +E f
where m i
and m f are the amount of
mass initially and finally and now E
means all energy
except mass energy.
does have angular momentum before the
collision. Consider the bullet as a point mass;
a point mass has a moment of inertia I bullet =mr 2 where
m is the mass and r is the distance to the axis
about which you are computing the angular
momentum, in this case the hinge of the beam.
The angular momentum of the bullet at the
instant of impact is L=I bullet ω =(mr 2 )(v /r )=mrv .
Because the hinge has no friction, there are no
external torques on it and therefore angular
momentum will be conserved. The beam has a
moment of inertia I beam =ML 2 /3
where M is its mass and L its
length. After the collision the beam plus bullet
have an angular velocity ω' .
Conserving angular momentum, mrv= (ML 2 /3+mr 2 )ω'
which can be solved for ω' .
What about
linear momentum? It is not conserved because the
hinge exerts a force on the system, the absense
of which is the condition for linear momentum
conservation. If this were all happening in
empty space, both angular and linear momentum
would be conserved; afterwards the beam+bullet
would rotate about their center of mass and the
center of mass would move in the same direction
the bullet came in with a speed v'=mv /(m +M ).
The bottom line is that linear momentum is
conserved if there are no external forces on the
system and angular momentum is conserved if
there are no external torques on the system
3 and a
mass of 1 g=10-3 kg. Its density is
10-3 kg/m3 . Now, put into
that box an object whose mass is 1000 kg and
whose volume is 10-3 m3 ;
the density of this object is 106
kg/m3 , but the density of the box
plus object is 1000.001/1 kg/m3 , 1000
times smaller. So, now let's do the same thing
for a single atom (which can be approximately
scaled to a macroscopic object). I will take
your number for the nuclear density, 1017
kg/m3 ; I am sure that you know that
the mass of an electron is about 2000 times
smaller than that of a proton or neutron, so the
nucleus mass is roughly 4000 times greater than
the mass of all the electrons; therefore I will
approximate the mass of the electrons to be
zero. The ratio of the radius of the atom to the
radius of the nucleus is about 105
and therefore the ratio of their volumes is
about 1015 . Therefore the density of
the atom will be on the order of 100 kg/m3 ;
this number is comparable to typical macroscopic
densities. This is just a rough approximation
but the average
density of an atom is simply its mass divided by
its volume, and that is what matters when you
determine the density of some object made up of
a great many atoms, not what the density of some
little part of it is.
answered this question.
photoelectric effect to understand where the
notion that the energy of an electromagnetic
wave was proportional to the square of its
amplitude was found to fail. The solution was
simply that the EM field is quantized which
means that the radiation has a minimum amount of
energy, called a photon, in which it can exist.
In essence this means that if you have a
particular wave and wish to reduce its energy,
you cannot do it smoothly but only in steps of
the energy of one photon at a time. This had
never been observed in a laboratory because the
energy of a single photon is so tiny compared to
the energy of a macroscopic wave. So, if the
wave interacts in a macroscopic way, for example
a radio wave interacting with a radio antenna,
you find that the energy is indeed dependent
only on its amplitude. On the other hand, when
you try to make that wave interact with an
individual electron, the electron sees and
interacts with individual photons of energy
hf as you note in your original question.
How can we reconcile these two facts? Simply
that if the radiation interacts as a wave,
amplitude matters, but if the radiation
interacts as a photon, frequency matters. If you
have a blue wave and a red wave with equal
amplitudes, they carry the same energy; but the
blue photons have a higher energy than the red
photons, so there are more red photons in the
red wave than there are blue photons in the blue
wave.
So, you asked whether the energy of light
is "…dependent only on amplitude or [on] amplitude
and
frequency…"; the answer is neither, it
depends on amplitude
or frequency!
F=μN , is not a
physical law, it is just an approximate method
for getting the frictional force if an object is
sliding and based on experimental measurements.
N is just how hard the
two (sliding and stationary) surfaces are pressed together
and μ depends only on what the two
surfaces are. Note that the area of contact is
nowhere in this equation, so that is not
relevent for your question. Secondly, the number
you calculate is not a torque, it is the force
you would have to exert to just pull the whole
hose. The torque is determined by how long the
handle you crank with is. I envision a crank of
length L attached to the axle of the
drum on which you exert a force F at
its end and perpendicular to its length; I
imagine a ratchet which will lock the crank to
the drum when turning counterclockwise (taking
in hose) and slipping when the crank is
repositioned by moving it clockwise. The
resulting torque about the axle of the drum is
FL. Meanwhile the hose is exerting a
force T , as shown, on the drum and its
torque is -TR . (This is negative
because it is trying to spin the drum clockwise
whereas the torque due to F is trying
to turn it counterclockwise and we have called
that torque positive.) If the hose is at rest or
moving with constant speed, the sum of the
torques is zero, so FL-TR =0 or F=T (R /L ).
In your case, T =2843 lb, so if, for
example, L =5R , F =569
lb. The resulting torque would be 17,058 ft lb.
Practical notes:
The
more hose you bring in, the smaller the
friction will be;
as hose
piles up on the drum, R gets
bigger; and
static
friction is bigger than kinetic friction, so
it will be harder to get it moving than to
keep it moving.
recently answered a question along these
lines in which I found that "warping of
spacetime" is just one possible interpretation
of the nontrivial mathematics which is the
framework of general relativity; this
visualization is attractive because it allows
persons like yourself to comprehend the
implications of the theory. However, the
mathematics of the theory may also be
interpreted as a field theory, and fields can be
quantized. The
essay linked to is short and not too
technical; reading it may put you in a state let
you might be able to get a good night's sleep!
0 F).
If there were no atmosphere, this spectrum would
look like black body radiation, the dashed line
labeled 320 K. Note that its peak is near 15 μm.
But there is atmosphere and it absorbs a lot of
the radiation resulting in large discrepancies
between observed and predicted spectra. Most
notable is CO2 , the prime culprit in global
warning, which reradiates this infrared (heat)
back to the earth. Also ozone (O3 ), methane
(CH4 ), and water vapor absorb a lot of enery. I
do not understand what the argument is; I do not
know what is meant by
I do know that data clearly show that the earth
radiates at all wavelengths, some of which do
not fully escape the earth because of
interactions with the atmosphere; wavelengths
far from the maximum black body intensity for a
given temperature will be small enough intensity
to be negligible.
V where V is the
speed of the two of you right after the
collision, so V =1.23 m/s. Now, the rate
of change of your momentum was equal to the
force on you in Newtons, so (1.23x78-0)/0.2=480
N=108 lb. You see that the time of collision is
crucial, for example if you used 0.1 seconds the
average force during the collision would have
been 216 lb. For 0.01 seconds, the force would
be over a ton.
R =kA ΔT /d
where R is the rate
of energy loss (typically Watts in SI units), k
is the thermal conductivity of the material, ΔT
is the temperature difference, A is the
area, and d is the thickness. For
glass, k is about 1 W/(K⋅m). For
your case, A =18 ft2 =1.67 m2
and d =2 in=0.0508 m. Therefore
R =32.9ΔT. For example,
if ΔT =(20-0) 0 C=20
0 C=20 K (about (68-32) 0 F),
R =656 W=0.622 BTU/hr.
faq page and look for
twin paradox .
precisely a simple variation of the twin paradox.
Below I have worked out the answer to your specific question (using 99% of light speed for
"close to the speed of light").
This follows exactly the method of my earlier
answer on the
twin paradox
but using your numbers, although I have not
redrawn the graph for your numbers because that
would be impossibly tedious.
My favorite way to understand the twin paradox is to suppose that Superman and someone on earth, each using his own clock, sends 100 light pulses per year to the other. Each receives all the pulses from the other, but Superman sends fewer since, because of length contraction, he has less far to travel in his frame than
the earthbound person observes; Superman travels to a star
and back which is 5 light years away with a speed of 99% the speed of light. Since Superman, because of length contraction, sees a distance to the star of only 5√(1-0.992 )=0.71 light years, he sends out
71 light pulses on the way out and 71 on the way back, and all are received by the earthbound observer, so both agree that he has aged
1.42 years. The earthbound someone sends out
990 pulses and Superman receives them all, so both agree that he has aged
9.9 years.
I have done this in the frame in which both the
earth and the star are at rest. If you want, you
can do it in the frame where Superman is at
rest, but the result would be the same because
the distance between the earth and the star
would still be contracted to 0.71 light years
for Superman.
a=F /m . If you compare two
objects with the same mass, the one with the
largest force applied wins. Similarly, if you
compare two objects encountering the same force,
the one with the smallest inertia applied wins.
Maybe your game is designed such that boats are
always much heavier than land vehicles.
The Physics of Star Trek
by Lawrence Krauss, which has a lengthy
discussion of teleportation.
answer to a question similar to yours but
with considerably more detail
earlier answer .
E /B=c
where c is the speed of light. Since
this is a constant, your conclusion that the
E and B fields are not in phase
must be wrong because two waves out of phase
have different ratios at different places. Where
you went wrong is thinking of B as just being
created by some current; in fact, another origin
for a B field is a time-varying
electric field. Similarly, a time-varying
magnetic field causes an electric field. In some
sense, the E and B fields are
keeping themselves going, sort of a perpetual
feed-back loop.
The second
part of your question involves the
photoelectric effect . What happens when you
send electromagnetic wave into matter cannot be
described by thinking about waves. Understanding
that it could only by acknowledging that
sometimes these waves fo not behave like waves
at all, but must be thought of as particles
called photons. Do not waste your time trying to
understand the photoelectric effect in terms of
electromagnetic waves.
W D D down on the air, but since
D W
earlier answer . For another question similar
to yours, see another
earlier answer .
ground rule forbidding "questions based on
unphysical assumptions". Anyhow, I can talk a
bit about uniformly charged spheres and
demonstrate how unphysical this question really
is. The total energy it takes to assemble a
uniformly charged sphere of radius R to
a total charge Q is E =3kQ 2 /(5R )
where k is Coulomb's conatant and is
about 9x109 N m2 /C2 .
Now, the mass of an electron is about 9x10-31
kg and its charge is about -1.6x10-19
C. So I will call 1 gram=10-3 kg
"considerable" and will therefore need about 1027
electrons; their total charge will be Q =-1.6x108
C and Q 2 =2.6x1016
C2 . Suppose we jam them into a sphere
of radius 1mm=10-3 m, plenty big to
"be visible"; then the energy to assemble these
electrons would be E =1.4x1028
J. If "suddenly" means 1 s, this is a power
requirement of 1.4x1028 W. The total
power consumption of the earth is 44x1012 ,
miniscule compared to that the create this
electron ball. Furthermore, there is no force to
bind this ball and it will blow itself apart
with the energy you put into it; you would not
want to be around it!
earlier answer . So what we are doing is an
academic exercise not really very close to
reality. Still, it is an interesting exercise.
Above the earth the
prediction of the pressure as a function of
altitude is fairly standard. I urge you to look
over that derivation since I will follow the
same procedure underground. The result there is
P=P A exp(-mgh /(kT ))
where h is the altitude, m is the mass
of a molecule of air, g is the
acceleration due to gravity, k is
Boltzmann's constant, P A is atmospheric
pressure, and T is the absolute
temperature. This derivation assumes that g
is constant over the range of h that contains
most of our atmosphere; that is not a bad
approximation since h is much smaller
than the radius of the earth. It is important to
note that there are really two unknowns here,
P and T . This model is usually
used assuming that T is constant; since
T actually decreases with h , if you measure
P and
then solve for T , you will find that
the measured pressure is larger than the
predicted pressure.
So, to predict the pressure in your hole I will
follow the derivation given in the
hyperphysics
site and start with ΔP=-ρg (r )Δr
where ρ is the density of the air, r
is the distance from the center of the earth,
and g (r )
is the acceleration due to gravity at r .
For a uniform sphere it is easy to show that
g (r )=gr /R
where R is the radius of the earth and
g =9.8 m/s2 . Now use the
ideal gas formula to write ρ=mP /(kT ) so
that the differential equation for P is
dP /dr =[mPg /(ktR )]r ;
integrating, I find that P =P A exp[-(C (1-(r 2 /R 2 ))]
where C=mgR /(2kT ). Choosing
T =300 K, I find C≈ 350
so P /P A ≈ exp[-350(1-(r 2 /R 2 ))].
I have plotted, below, the predicted
pressure both inside (red) and outside (blue for
constant g , black for g (r )=g (R /r )2 )
the earth; note that the
graph covering 0<r /R <2 is
logarithmic because the pressure is almost zero
very rapidly. Both inside and outside the
pressure falls very quickly away from the
surface, faster for inside because of the
quadratic rather than linear term in the
exponential function. The other graph shows the
curves 1% above and below the earth's surface;
inside the pressure is nearly zero, outside 80%
below atmospheric pressure.
The center of the earth has a temperature of
about 6000 K which would result in a much larger
pressure of about 10-8 P A ,
but still very small. Just as the case of the
prediction of P above the earth's
surface has difficulties related to the
temperature profile, so does the prediction of
P below the surface.
There was an error in my original post in
the calculation for r>R which has been
corrected in the graphs. I also did the
analogous calculations for r>R where
the variation of g was included.
mg D D
is proportional to the square of the speed v 2 .
So, if you drop something it starts with very
little drag and the weight is the main force
which causes the object to accelerate downward;
eventually v gets big enough so that
the drag is equal to and opposite the weight so
the object then falls with a constant speed,
called the terminal velocity, for the rest of
the fall. There is a well-known expression for
the
drag force which works well in most
every-day situations, D =½ρv 2 C D A.
You can learn more about it at the link given
above, but for our purposes here, for an object
which presents an area A to the onrushing air at
sea level, a fairly good approximation is D =¼Av 2 .
(Warning: this equation only works if you work
in SI units.) So, we can estimate the terminal
velocity, v T =√(2mg /A ).
Let's estimate the mass of a parachutist plus
unopened parachute as 100 kg and her area about
1 m2 ; then you could estimate v T
for a sky diver plus unopened parachute as about
45 m/s=100 mph. Now, when the parachute is
opened maybe the area increases to 20 m2 ;
now v T is about 10 m/s=22
mph. These are just rough approximations to
emphasize that the area is key. Actual numbers
will depend a lot on the design of the
parachute.
0 and the
change in horizon would be trivially small. I
assume that you are looking for something coming
in from far away, that is it is ok to have blind
spots near the earth as long as anything far
from the earth can be
seen.
Imagine one gun at each pole; they see anything
above a plane tangent to the earth at the pole.
Imagine one gun at the equator, another at the
equator on the opposite side of the planet; they
can see anything above their tangent planes.
Finally station guns 900 longitude from the
other two on the equator. You have made an
imaginary cube around the planet using only six
gunners into which no one can enter undetected.
You are screwed, though, if somebody gets
inside. It just occurred to me that you could
put the planet in a tetrahedron with only
four gunners.
concepts and should be not thought of as
formulas. I will try to specify these
conceptually, not mathematically.
An object which experiences zero net force
will remain at rest or moving with a
constant speed in a straight line.
The acceleration (change of its velocity) of
an object is proportional to the net force
applied to it and inversely proportional to
its mass.
If one object exerts a force on a second
object, the second object exerts an equal and
equal force and opposite on the first.
D moving with
speed v in a fluid with density ρ
and speed
v R m/s, D =½ρ (v +v R )2 C D A ;
C D is called the drag
coefficient and depends only on the shape of the
barge and A is the area presented to
the oncoming water. I have assumed the barge is
moving upstream and this force, of course, is
downstream. Now, what you want to know is what
is the increase in drag when you change the
draft. The only thing changed by the the draft
is A and A is proportional to the draft. So, if
we take a ratio D 9.5 /D 9 =9.5/9=1.06;
so the drag increases by about 6% if the two
draft situations have the same speed. If they
have different speeds, then you would have to
calculate that, D 9.5 /D 9 =1.06[(v 9.5 +4)/(v 9 +4)]2 .
For example if some force (tow boat) pulls the
barge and the speed is 5 mph for the 9.5 ft
draft and 7 mph for the 9 ft draft, D 9.5 /D 9 =1.06[(5+4)/(7+4)]2 =0.71;
there is about 29% less drag for the 9.5 ft
draft because it is going slower. I can't get
any more quantitative without more information
about the barge and the way it is being towed.
v 0 at the origin. Then we have four equations
describing the subsequent motion:
vx =v 0 cosθ
vy =v 0 sinθ -gt
x =v 0 t cosθ
y=v 0 t sinθ- ½gt 2
What do you know? t , θ , and g .
What do you not know?
vx , vy ,
x , y , and v 0 .
Since you have more unknowns than equations, you
cannot solve the problem. The answer to your
question is no.
W F F is greater than W . Finally, we
invoke Newton's third law which says that if you
exert a force on the pot, the pot exerts an
equal and opposite force on your foot and so you
experience, for a brief time, a force larger
than 10 lb.
m =2
oz=0.055 kg and his length is L =5
in=0.13 m. The initial speed is v 0 =64 mph=29
m/s. Estimating his width as 1 cm=0.01 m, his
area is roughly A =0.01x0.13=1.3x10-3
m2 . So you can estimate the force on him as
F =-¼Av 2 =ma =m (dv /dt )
or (dv /dt )=-v 2 x6x10-3 .
Integrating this differential equation, 6x10-3 t= (1/v )-(1/29)
or t =167(1/v -.035) seconds. Oh
dear, I started this answer expecting to
reassure you that the little fellow would be
fine. But, suppose that we wanted him to have a
speed of 2 m/s to be ok when he hit. Then the
time to slow to that speed would have been about
78 s and he would have hit long before that. So,
I calculated his terminal velocity, the speed he
would attain if he had simply been dropped from
some large height. That would be v terminal =√(2mg /A )=28.8
m/s, just about the speed he already had when he
became airborne! If there had been no air drag
at all and you dropped him from about 3 m, he
would have hit with a speed of 7 m/s=15 mph and
it would only have taken him about 0.4 s. In
this time, there is no hope of his slowing down
appreciably. I am afraid that the gecko met an
unhappy end.
I really have no idea what you are trying to get
at when you talk about the motion of the air. It
is important for you to understand that there is
no net motion of the air itself is any
direction. Focus on any air molecule and it
simply moves back and forth parallel to the
direction the wave is moving there. The wave
moves in some direction, the air does not.
Considerable water mass is stored in Earths polar ice-caps. If the ice-caps melted, and the stored water released, would this raise the sea-level more in equatorial regions? If so, would this in turn slow the earths rotation and thus reduce the total days in say a century?
1 = 0.01 Ohm that's let's say twice as tall as me (doesn't really matter in my question, it has to
strike the rod first) and lightning strikes the rod. Let's assume that human body in this situation has resistance of R2 = 1,000,000 Ohm. If the lightning strikes the rod it will go through parallel resistor (me + rod) hence the total R will be 0.00999 Ohm. If we calculate total current it will be 10,000,000,100 A. 10,000,000,000 A will go through the rod, 100 A will go through my body. I know I simplified it a lot, I'm not even a decent at that, but the question is - there's a saying that electricity takes path of the lowest resistance - but why would I prefer to hold super conductive rod than to hold anything less conductive, if the current through my body will remain unchanged? I total = I1 through R1 + I2 through R2 , right? So that would mean lower R1 means lower I2 , but it doesn't, because with lower R1 , Rtotal is lower hence I is higher, so I2 remains unchanged. I believe it has to do something with power, I'm just not sure.
R net =R 1 R 2 /(R 1 +R 2 ).
If R 1 =0 then R net =0,
all the current will flow through R 1 .
No rod will have zero resistance, but as R 1 ⇒0,
I 2 ⇒0.
a or
you are in a gravitational field where the
gravitational field causes an acceleration a
at the point where you are performing the
measurement. Usually, like your rotating
cylindrical room, we think of it as having an
angular velocity ω and a radius
R and if
we adjust these so that g=v 2 /R =(Rω )2 /R =Rω 2 ,
then it will be just like being on the earth.
But that is wrong because the field close to the
surface of the earth is very nearly uniform—g
is very nearly the same at both your head and
feet. That is not the case in the rotating room
because the acceleration depends on R
and therefore your head "weighs more" when you
are standing up than when you are lying down. If
you could simulate a gravitational field which
increased linearly with R as you moved
away from some straight line (analogous to the
rotation axis of your room), there would be no
way you could determine whether you were in that
field or in your rotating room. So for the
rotating space station to be a good simulator of
gravity on earth, it would need a very large
radius. If you want to read much longer
discussions of rotating gravity simulators,
there are several
earlier answers to questions similar to
yours.
a or
you are in a gravitational field where the
gravitational field causes an acceleration a
at the point where you are performing the
measurement. Usually, like your rotating
cylindrical room, we think of it as having an
angular velocity ω and a radius
R and if
we adjust these so that g=v 2 /R =(Rω )2 /R =Rω 2 ,
then it will be just like being on the earth.
But that is wrong because the field close to the
surface of the earth is very nearly uniform—g
is very nearly the same at both your head and
feet. That is not the case in the rotating room
because the acceleration depends on R
and therefore your head "weighs more" when you
are standing up than when you are lying down. If
you could simulate a gravitational field which
increased linearly with R as you moved
away from some straight line (analogous to the
rotation axis of your room), there would be no
way you could determine whether you were in that
field or in your rotating room. So for the
rotating space station to be a good simulator of
gravity on earth, it would need a very large
radius. If you want to read much longer
discussions of rotating gravity simulators,
there are several
earlier answers to questions similar to
yours.
a or
you are in a gravitational field where the
gravitational field causes an acceleration a
at the point where you are performing the
measurement. Usually, like your rotating
cylindrical room, we think of it as having an
angular velocity ω and a radius
R and if
we adjust these so that g=v 2 /R =(Rω )2 /R =Rω 2 ,
then it will be just like being on the earth.
But that is wrong because the field close to the
surface of the earth is very nearly uniform—g
is very nearly the same at both your head and
feet. That is not the case in the rotating room
because the acceleration depends on R
and therefore your head "weighs more" when you
are standing up than when you are lying down. If
you could simulate a gravitational field which
increased linearly with R as you moved
away from some straight line (analogous to the
rotation axis of your room), there would be no
way you could determine whether you were in that
field or in your rotating room. So for the
rotating space station to be a good simulator of
gravity on earth, it would need a very large
radius. If you want to read much longer
discussions of rotating gravity simulators,
there are several
earlier answers to questions similar to
yours.
0 (100,000/250,000,000)=0.140 .
T required is T =(300,000,000
events)x(40 seconds/event)=1.2x1010
seconds. That is the answer, but maybe we should
put it into different units of time, years,
let's say: T =(1.2x1010
seconds)x(1 hour/3600 seconds)x(1 day/24
hours)x(1 year/365.25 days)=380 years.
W
net force which slows her down if she is
going up and speeds her up if she is going down.
The other woman has her feet in contact with the
box and is either just landing of just about to
go up. There are now two forces on her, her
weight W N net force
must point up, so N>W as shown. If she
is just about to go up, she is speeding up and
so her acceleration also needs to be upward
because she is moving upward; in this case the
net force must also point up, so
N>W as shown. Newton's third law (N3) now
states that, if the box exerts an upward force
of magnitude N on you, you exert a
downward force of magnitude N on the
box.
Here is what you said correctly: you said that
the force with which you would have to push down
on the box would be bigger than your weight.
Here is what you said incorrectly: you said that
"gravity would hit back with more
than my bodyweight". You were apparently
thinking that W N W N
If the woman
is just standing on the box, N W
c plays an
important role in quantum mechanics, atomic
physics would be different, perhaps making the
formation of atoms impossible given the other
physical constants. And, since E=mc 2
would still be true, stars, if any existed at
all, would burn out approximately 108
times sooner; for example, the sun will have a
total life time of about 10 billion years but
would only have lasted about 100 years with your
value of c . I believe that for atoms,
nuclei, elementary particle physics, etc . to
work at all, you would have to adjust all the
other physical constants to also be different.
After writing the above I googled around a bit
and discovered that it is a very complex
situation. A thread on
Physics Forums is enlightening. This should
give you some food for thought!
summary from the CDC of which units are
useful for which situations.
https://www.popularmechanics.com/science/a25581/science-behind-train-tracks-wheels/
https://www.quora.com/How-do-trains-turn
https://www.quora.com/When-a-train-makes-a-turn-isnt-its-outer-wheel-covering-more-distance-than-the-inner-one-How-come-the-train-doesnt-come-off-the-tracks
link . It is incorrect to say that matter is
mostly empty space. Although true that most of
the mass occupies a very small volume, the
intervening space is occupied by a "cloud" of
negative charge which the electrons occupy.
Bricks are stronger than glass because they are
"proper" solids, i.e . the atoms arrange
in regular crystaline patterns which tend to be
strong, whereas in glass they do not.
Macroscopic properties of matter have nothing to
do with the nuclear force, they are determined
by atomic forces among atoms in the material.
-3 N due to the
moon, and 5.9x10-3 due to the sun. So
I will estimate
the effect on the total weight for four
situations, new moon (moon between the earth and
the sun) midday and midnight, and full moon
(earth between the sun and the moon) midday and
midnight.
New moon day: 686-2.4x10-3 -5.9x10-3
New moon night: 686+2.4x10-3 +5.9x10-3
Full moon day: 686+2.4x10-3 -5.9x10-3
Full moon night: 686-2.4x10-3 +5.9x10-3
So
you are "lightest" with a new moon during the
day and "heaviest" with a new moon at night.
earlier answer also addressing the
possibility of a very small fusion reactor.
Regarding getting the fuel from moisture from
the air, any practical fusion reactor cannot
operate on single-proton hydrogen, 1 H,
but requires deuterium, 2 H and/or
tritium,3 H. There is very little
deuterium in plain water and virtually no
tritium.
a=F /m =100/40=2.5
m/s2 . There are two equations you
need to work with for uniform acceleration if you
start at x =0 from rest, x =½at 2
and v=at . In your problem you know that
v =40 m/s so you can solve for t ,
t =40/2.5=16 s. You can also find the
distance traveled, x =2.5x162 /2=320
m.
That an open vessel is filled with water and fitted with a submersed pump;
That the water in the vessel is under normal atmospheric pressure, and at all times, remains so;
That the vessel is immersed in, and is level with, a surrounding pool of water;
That the surrounding pool of water is infinite in size;
That the surrounding pool of water has a pressure of 2 atmospheres;
That friction losses are ignored;
That the pump operates at 100% efficiency.
I need to know how many joules of energy would be required to pump 1 litre of water, from the vessel, horizontally, into the surrounding pool of water.
a between
the two fluid containers which is at a depth
d below the surface of each; the volume of
moved water is La , imagined as the
volume shown in the diagram. The pressure at
depth d is P A +ρgd
inside the hole and 2P A +ρgd
outside, so the pressure difference is ΔP=P A
and the force necessary to push water through
the hole is a ΔP=aP A .
The work to expel the fluid is then W =aLP A =VP A .
Taking P A =105 N/m2
and V =1 l=10-3 m3 ,
W =100 J. It becomes a more difficult
problem if the volume of the low-pressure water
is not very large; in that case, as the level
fell the pressure difference would become
larger.
v and find yourself heading straight for a brick wall that intersects the line of your path at 90°. Assuming that the coefficients of friction for stopping and for turning are the same, are your chances of avoiding a crash better if you continue straight ahead while braking or if you simply turn along a circular path at a constant speed?
μ
and the car has a mass m , weight mg , and the frictional
force is at its maximum, f=μN=μmg for both
crash-avoidance options.
The straight-ahead option starts
with kinetic energy K =½mv 2 and has work done on it
W=-fd=-μmgd before it stops with
no energy, where d is the distance to stop; therefore, ½mv 2 -μmgd= 0
or d=v 2 /(2μg ).
The turning option moves in a circle of
radius R and the centripetal force
is μmg=mv 2 /R .
Therefore, R=v 2 /(μg ).
Since d=R/ 2, straight ahead is a much
better option!
Assuming you
go up with approximately constant speed, this is
a Newton's first law problem—all the forces on
the wheelchair plus patient must add up to zero.
There are four forces on the wheelchair plus
patient—the force F
W of the patient straight
down, the normal force N f
θ =tan-1(1/6)=9.460 , so the
component of the weight down the plane is W x =-240sin(9.460 )=-39.4
lb; the component of the weight normal to the
plane is W y =-240cos(9.460 )=-237
lb. The most problematic force is the friction;
the most important thing is that f is
proportional to N , that is f=μN
where μ is called the coefficient of
friction. What this constant is depends on
things like the quality of the bearings in the
wheels, the nature of the surface you are
rolling on (a rug would have a bigger μ than a
hard surface, for example), how inflated the
tires are, etc . Wheelchairs are
presumably designed to minimize friction so I
will make a guess that μ ≈0.1. Now, the
physics:
y
forces add to zero, so N =237 lb;
therefore, f =-0.1x237=-23.7 lb;
x
forces add to zero so
if
there were no friction, F =39.4 lb.
For going
down the ramp, the force diagram is just the
same except now f points up the ramp rather than
down—friction now works with you, not against
you. N and W are the same but
now F =39.4-23.7=15.7 lb. I noted that
you said you are "pulling her down the ramp" but
with my choice of coefficient of friction I find
you are actually "holding her back" from
accelerating down the ramp. If you need to pull
her down the ramp, then my estimate of μ
is too small which would mean you have to push
more than F =2x39.4=78.8 lb when going
up.
F
e.g . 63.1, 15.7, 78.8 in
various situations). F you must apply on the
wheel chair to move it up or down the incline with a
constant speed. Newton's third law says that if you exert a
force on the wheel chair, the wheel chair exerts an equal
and opposite force on you. This is a force on your hands.
There is no way I can deduce what this implies about your
spine, your shoulders, your legs, etc .; the body is
a very complicated machine and how you push might be
different from how somebody else might push, so there is no
simple answer. [For example, consider the simpler example of
lifting of a box. If you ask me for the force you need to
lift it, I could tell you that you must exert an upward
force equal to the weight of the box. You have probably
heard "lift with your legs, not with your back". So for two
diferent people doing the exact same thing but in different
ways, one might end up with an injured back and the other
not.] Regarding the 78.8 lb force I
derived, you can probably simply ignore it because it is
only for the special case of very high friction in the
wheelchair which I think unlikely. You can easily test
whether the friction is very high by carefully letting go of
the wheelchair at rest: if it begins begins moving down the
ramp, the friction is pretty low, but if it just sits there,
the friction is large. High friction would mean that you
have to exert a force up the ramp on the chair to move it up
the ramp of at least
78.8 lb (pointing up the ramp). The condition for high
friction would be μ ≥39.4/237=0.17. Keep in mind that I am only guessing about
the coefficient of friction μ ; if you do the
test above and find the chair will roll down if you do not
hold it, at least you will know that μ <0.17.
I have removed the part about a force
G in my original answer because it is too confusing.
14 C
which is commonly used for carbon dating of
archeological items. Now, the decay is a
statistical process and since this nucleus is
bound you might have to wait for a while for the
decay to happen. But, somehow the nucleus knows
that if it can turn one of its protons into a
neutron, making 14 N, 0.156 million
electron volts of energy will be released;
depending on the detailed structured of this
nucleus, you may have to wait a long while. In
this case you will have to wait 5730 years for
half the 14 C nuclei to beta decay to
14 N. 14 N is stable against
beta decay which would be to 14 O,
which means that if you want it to happen you
would have to add energy. Incidentally, there is
another type of beta decay where a neutron turns
into a proton, a positron (a positively charged
electron), and a neutrino. So, if a nucleus has
too many protons it will beta minus decay, too
many neutrons it will beta positive decay. A
neutron, unlike a proton, will also beta decay
when outside a nucleus.
v in the opposite direction (the negative
sign denotes velocity opposite the ball). Linear momentum is
given as mass times velocity, so the ball has momentum
5x20=100 kg m/s and the astronaut has a momentum 100v .
But the ball plus astronaut had zero momentum before she
threw the ball and they must have zero momentum afterwards,
so 100-100v =0 or v =1 m/s. That's all a
rocket is: you throw stuff out and recoil in the opposite
direction. It has nothing to do with platforms repelling or
electrons in the ejected gasses. (Incidentally, conservation
of linear momentum is easy to derive just using Newton's
second and third laws. It is only true if there are no other
forces acting. But the recoil idea is true of a rocket going
up in gravity, it is just that a given amount of fuel
ejected in space will have a larger acceleration than when
leaving earth.)
D on an object with a cross-sectional area A presented to the oncoming fluid
and moving with speed
v through a
fluid with density ρ , is given by
D =½ρv 2 C D A
where C D is the drag
coefficient, a constant which depends only on
the object's geometry and orientation relative
to the fluid flow. In our case, ρ =103
kg/m3 and v =4 mph=1.8 m/s. Now we come
to your weight; it is clear that it is made of
lead which has a density of 1.13x104
kg/m3 . The problem is that the drag
coefficient for this shape is not known and you
really do not know what angle it would assume
relative to the direction of the line, so you
would either have to do extensive measurements
or make simplifying approximations; since we
just want a rough approximation, so I am going
to assume that the weight is a sphere of radius
R , C D =0.47 and A=πR 2 .
(You may be amused by
"consider a
spherical cow ".) R for a solid
sphere of mass m may be
calculated using the density of lead. Putting this all
together, I find that D =9.2m 2/3 .
So that takes care of the fluid dynamics part of
the problem, the part where approximations are
made. The rest is pretty simple first-semester
Newtonian physics. The weight W is in
equilibrium so W=T sinθ and
D=T cosθ . For your
desired situation, sinθ= 25/50=0.5,
so θ =300 . Doing all
the arithmetic and algebra here I find that
m =0.067 kg=2.4 oz.
Finally, I emphasize again that this is a rough
calculation. My advice would be to go out in
your boat with a protractor, drop the weight,
and go 4 mph trying varous
weights. Since you want the sinθ= ½,
get the angle to near 300 . Choose the
weight which is closest. I am guessing it will
be greater than 2.4 oz because I think the drag
coefficients for your geometry will be smaller
than for a sphere. If you do this, I would be
interested in what you find.
d=λL /x , where λ is the wavelength of the laser (around 632 nm) and
L is the distance between the hair and the “screen.” (This formula is derived from
d=λ /sinθ .) The confusion comes from measuring
x . According to some sources, we should measure the distance between the center of the central bright band and the 1st order maximum (the center of the first bright area on either side). Other sources say to measure from the center to a minimum, either the center or an edge of one of the dark bands!
Here’s my question: Which is it (to a maximum or a minimum), and why? What measurement do we use?
mλ=d sinθ
locates maxima; those authors believe the pattern is the same
as a double slit with spacing d . Other references say that mλ=d sinθ
locates minima; those authors believe the pattern is the same
as a single slit with width d . The first of these
is flat-out wrong. There is a well-known rule, Babinet's principle,
which states that the diffraction pattern from some
obstruction is the same as the diffraction pattern from an
aperture of the same shape. The only difference is that the
central area of the pattern is much brighter since the part
of the incident beam which would have been stopped by
missing the slit carry right on through. Approximating sinθ∼x /L , d= mλL /x m
where x m is the location of the mth
minimum relative to the central maximum. I have never done
this experiment, but it occurs to me that it would be
difficult to get a good estimate of the center spot because
it is so bright and broad. Rather, I would try to get the
distance x m,m+1 =x m+1 -x m , the distance between two
adjacent
minima, and write d=λL /x m+1,m .
a special case of the ideal gas equation, PV /(NT )= constant, where N is some measure
of the amount of gas in a container of volume V , pressure
P ,
and temperature T ; N can be anything you like—number
of molecules, mass, number of moles, etc .—because,
as you will see, only ratios matter in my analysis. Boyle's
law assumes that N and T are constant
which is not the case here. In these calculations it is
important that T be absolute temperature (Kelvins)
and P be the absolute pressure which is the gauge
pressure plus atmospheric pressure which is 14.7 psi.
I am assuming that there is a valve or other
mechanism between the two tanks which keeps the
pressure at 40 psi in the beer keg as the beer
is dispensed.
I think it is instructive to first do the case
where only 64 oz of beer is dispensed and
both vessels are at 40 psi at the end. First, I
need to convert all volumes to in3 .
64 oz=116 in3 and 448 oz=809 in3 .
In the compressed air tank the initial absolute pressure
is P , the final absolute pressure is 40+14.7=54.7
psi, the initial amount of gas is N ,
and the final amount is N-n where n
is the amount of gas filling the dispenser after
all the beer has been dispensed. Therefore, 54.7/(N-n )=P /N .
Rewriting, 54.7/P =1-(n /N ).
Now the ratio of the volumes is 116/26=n /(N-n )
because when all the beer is out, this ratio is
the same as the ratio of the amounts of gas in
each volume; don't forget that the pressures are
now equal. Solving I find that 1-(n /N )=0.183
and so P≡P 1 =299 psi (gauge pressure
is 284 psi).
Suppose now that we want to have enough pressure
so that two 64 oz tanks of beer can be
dispensed. The previous paragraph now represents
the second of the two dispensings, so the first
dispensing would finish with P 1 =299
psi in the tank and start with an unknown
pressure P 2 . The calculation
proceeds the same as for the calculation of
P 1 except that the ratio n /(N-n )
and the ratio for the volumes are no longer
equal because the pressures are not the same.
Following the same procedure as for P 1 ,
we now have 299/P 2 =1-(n /N ).
I will now denote the pressure and volume in the
beer tank as p =54.7 psi and v =116
in3 ; similarly, P 2
and V =26 in3 for
the compressed air tank. Therefore from the
ideal gas equation, P 2 V /(pv )=(N-n )/n.
So, 1-(n /N )=1-pv /(P 2 V+pv )=P 2 V /(P 2 V+pv ).
So, finally, 299/P 2 =26P 2 /(26P 2 + 6350)=P 2 /(P 2 + 244),
a quadratic equation for P 2 :
P 2 2 -299P 2 -73,000=0.
Solving, P 2 =458 psi (gauge
pressure 444 psi).
The above procedure can be generalized as
P j-1 /P j =1-(n /N )=P j /(Pj + 244)
P j 2 -P j-1 P j -244P j-1 =0
Pj =[Pj-1 +√(Pj-1 2 +976P j-1 )]/2
So,
P 3 =[458+√(4582 +976x458)]/2=634
psi (620 psi gauge)
P 4 =[634+√(6342 +976x634)]/2=822
psi (807 psi gauge)
P 5 =[822+√(8222 +976x822)]/2=1019
psi (1004 psi gauge)
P 6 =[1019+√(10192 +976x1019)]/2=1220
psi (1205 psi gauge)
P 7 =[1220+√(12202 +976x1220)]/2=1428
psi (1414 psi gauge)
3 keg
ala the calculation for P 1 above. I
was never able to convince myself that this
would be equivalent, so I trudged through the
step-by-step solution (which wasn't so bad after
I figured out P 2 ). When I
had finished I did the one-step calculation:
809/26=n /(N-n ). Solving I find that 1-(n /N )=0.0321
and so P 7 =1702 psi which is
not correct.
How fast would Lady Valarian be traveling to go 17 light yrs in nine minutes?
ANSWER: L =9x106 light minutes in 17 light
years. The speed v must be adequate to shrink, via
length contraction, L to L' =9 light minutes. So,
L'=L √[1-(v /c )2 ].
I find v =√(1-10-6 )c ≈(1-0.0000005)c =0.9999995c .
donation to help support this service.
If there is a link to a previously
answered question, be patient. Since the files containing the older
answers are rather large, it takes some time (maybe as much as 15
seconds or so) to find the appropriate bookmark.
F of attraction between two spheres (masses M
and m ) is inversely proportional to the distance
r between their centers, F=MmG /r 2
where G is the universal gravitational constant. In
the case you are asking, M is the mass of the earth
and m is the mass you are measuring F with. If
g denotes the acceleration of m , then g (r )=F /m=MG /r 2 ;
g (r ) indicates that the value of g
depends on what r is. Note that if r=R
where R is the radius of the earth, g (R )=9.8
m/s2 . So this explains why g decreases
for r>R— as r gets bigger, g
gets smaller. But shouldn't that now imply that as
r gets smaller, g gets bigger? The answer is
no because of something called Gauss's law which says that
only the mass inside where you are contributes to the force,
not the whole mass of the earth. If you assume that the
density of the earth has a constant value of ρ=M /(4πR 3 /3),
it can be shown that
g (r )=4πρGr which decreases
linearly to zero at the center of the earth.
It is interesting to note, though, that it is a poor
approximation to assume the earth has uniform density. In
fact the core of the earth is much denser than the mantle
and crust. See an
earlier answer for more detail on this.
video
demonstrating this on YouTube.
earlier answer . This also
discusses electromagnetic theory and quantum theory as
applied to the H atom. By the way, simple models like the
Bohr model can be useful in getting an understanding of the
crux of something; even if they are "outdated" it does not
necessarily mean that everything about them is wrong.
ways
to get around this but they are pretty tricky. If you want a
constant torque, why not just have a weight falling weight
as in grandfather clocks? Torque and force are not the same
thing. It would maybe help me answer your questions if you
gave me some idea of what you are trying to do.
t 2 and 82=vt where
v is the speed she left the road and t is the time to hit. I find that
t =0.61 s and v =134 ft/s=91 mph. Other guesses would be drops of 7 ft, 8 ft, and 9 ft giving speeds of 85 mph, 79 mph, and 74 mph,
respectively.
Hope this helps. There is no additional information which I can add.
R and with a speed v ,
the acceleration points toward the center of the earth and
has a magnitude a=v 2 /R ; this is
called the centripetal acceleration. Of course there has to
be a force which provides the acceleration and that is
simply the weight of the satellite. If you are close to the
earth's surface (compared to the radius R E
of the earth) the weight is mg and so
F=ma or mg =mv 2 /R E ;
so the speed you must have to orbit close to the surface of
the earth is v =√(gR E ).
The value of the acceleration g gets smaller as you
get farther away from the surface.
Rosetta , was sent to a comet and orbited it;
the escape velocity from the comet is only about 1 m/s so
the probe orbitted with a smaller speed than this.
N W mv 2 /r=W-N
or N=W -mv 2 /r . So, when
v gets big enough N =0; so the ring could
be taken away and this would be the right speed for a "real"
orbit at this r . And the ring would not work
for speeds larger than the orbital speed, the object would
leave the ring; any smaller v
would result in the speed remaining constant as the mass
rolled around.
L i ,
a weight W i , and a center of gravity
r i . Since the panels are uniform and have
weight density of 2.08 lb/ft, r i =½L i
and W i =2.08L i . I
will also specify the number of panels with the length L i
to be N i . The center of gravity of a
stack of panels of length L i is r i = ½N i W i L i /(N i W i ).
Now, suppose there are two different lengths stacked. Then
the position of the center of gravity is
r CG =(½N 1 W 1 L 1 +½N 2 W 2 L 2 )/(N 1 W 1 +N 2 W 2 )
=½x2.08x(N 1 L 1 2 +N 2 L 2 2 )/(2.08x(N 1 L 1 +N 2 L 2 ))
=½(N 1 L 1 2 +N 2 L 2 2 )/(N 1 L 1 +N 2 L 2 ).
So, if you have j different lengths, the final answer is
r CG =½(N 1 L 1 2 +N 2 L 2 2 +...N j L j 2 )/(N 1 L 1 +N 2 L 2 +...N j L j ).
The more compact way to write this is
r CG =½ΣN i L i 2 /ΣN i L i .
Here Σ denotes summing from i=1 to i=j.
Note that the denominator is the total weight of
the whole stack divided by 2.08 and that the
weight per foot is not needed to find the
location of the center of gravity.
groundrules forbid unphysical questions, and an infinite
number of windmills is obviously unphysical. However, let's
investigate whether a reasonable number of windmills could
have a significant impact on the weather. At the end of 2018
the total global wind power was about 600 GW, that's 0.6x1012
W; the total world power production is about 44x1012
W; an average hurricane has about 1.5x1012 W; the
Nagasaki atomic bomb was about 1013 W (1014
J and I guessed the time of the explosion was 10 s). It has
been
estimated that if we put wind farms covering all land
and all off-shore water, 80x1012 W would be
generated. That same
article estimated that you would need to have enough
wind mills to generate more than 250x1012 W to
have a noticeable effect on weather, about four times the
number feasible.
Bay of Fundy during tidal changes; it is harder for the
windmills because it is not easy to estimate the energy of a
"typical" storm, so the following calculation is a very
approximate one. A typical windmill generates about 3x106
W and guess that a typical strong storm is 10 times smaller
than a hurricane, 1.5x1011 W; so the percentage
by which the energy of the storm is reduced is 0.02%, so
5000 windmills would use up most of the wind energy.
r about the axis with
some frequency f (f times around every
second) and, if r is in ft, the speed it travels is
v =2πrf ft/s. For example if you are
doing 2 revolutions per second and r =2 ft, v =25
ft/s=17 mph. But, since the speed depends on r ,
every point moves with a different speed, the top being the
fastest. I have shown also the weight w and
centrifugal force c F with which the rope exerts on you; w
is the same for any piece of the rope the same size but
c depends on r , the largest being at the very
top in the figure and also at the very bottom as. When the
rope is straight up as shown, the forces F are
their smallest because c and w are in
opposite directions; when the rope is straight down, the
forces F are their largest because c and
w are in the same direction. But to actually
calculate the force is a formidable problem because we need
to add all the pieces which are all different (this adding
is integration in calculus if you have ever studied it).
Just to give you some idea of the answer you sought I will
assume that the rope has a negligible weight and the whole 3
lb is located in the middle of the rope and is a distance of
3 ft from the axis. You want v =10 mph=14.7 ft/s,
and c=mv 2 /r =6.75 lb. (Here
m is the mass which is 3/32 lb/(ft/s2 ), but
you don't need to worry about that.) So, at the top, the
force to hold it is 6.75-3=3.75 lb and at the bottom it is
6.75+3=9.75 lb. This is what you are calling the "equivalent
weight", I guess.
P =290/0.5=580 ft lb/s=1.055 horsepower=786 W.
t later. The moving
paper moves with velocity v perpendicular to the
not-moving edge and the intersection point moves with
velocity u along the not-moving edge. The distance
moved is v Δt and the distance
between the earlier and later edges of the moving paper is s =v Δt cosθ .
During this time interval, the intersection point has moved
a distance u Δt and s =u Δt sinθ.
Therefore,v Δt cosθ=u Δtsinθ
or u=v /tanθ. Suppose we want
u =2c and cause the paper to move with a
speed of v=c /2; then θ= tan-1 (v /u )=tan-1 (0.25)=140 .
So, indeed, the point of intersection can move faster than
the speed of light. How can this be? Isn't c the
universal "speed limit"? It is the limit for transmitting
information or for any object with mass; this intersection
neither has mass nor transmits any information. There is a
similar example where something can have a speed greater
than the speed of light without violating special
relativity. Imagine pointing a laser at the moon and
sweeping the spot on the moon across the surface by rotating
the laser; because of the long arm, it would be pretty easy
to get the spot going faster than c but, just as in
your example, the spot has no mass and conveys no
information.
discussion there.
ad absurdum .
v+c , still illuminating the
front of the ship. But that is not what he is doing. Instead
he is saying that there will be Cerenkov radiation (from the
flashlight, I guess) which is like the shock wave of a
supersonic aircraft. This is in the form of a cone behind
and that is the light supposedly illuminating the back. I
find this a very far fetched example and would advise you to
simply ignore it!
here .
W 1 ,
the weight of the rider; W 2 ,
the weight of the driver; W 3 ,
the weight of the motorcycle; D f etc .;
N 1 , the force up on the
rear wheel by the road; N F N 1 +N 2 =W 1 +W 2 +W 3
and F-f-D =0. The most important thing to appreciate
is that f is proportional to the normal force, f=C (N 1 +N 2 )
where C is some constant. So when the rider falls
off, W 1 disappears and so N 1
and N 2 get smaller and so f gets
smaller. F, however, does not depend on N 1
because it is static friction; it is determined by how much
power is being delivered by the engine so it does not
change. I will denote the new frictional force as
f' and note that it is smaller than
f F-f'-D =Ma where M
is the mass of the rider plus bike, M 2 +M 3 =(W 2 +W 3 )/g .
where g is the acceleration due to gravity.
Finally, the bike has an acceleration a =[(F-f'-D )/(W 2 +W 3 )]g .
m at sea level is
given by F 0 =mMG /R 2 =W
where R is the radius of the earth, M is
the mass of the earth, and G is the universal
gravitational constant; if you go up to an altitude h ,
the weight gets smaller, Fh =mMG /(R+h )2 .
Therefore, if you want to hover at an altitude h you must
exert an upward force of Fh = [R 2 /(R+h )2 ]W
where W is the weight at sea level. Even several
hundred miles is small compared to R which is about 4000
miles, so we can make an estimate by doing a binomial
expansion: R 2 /(R+h )2 =(1+(h /R ))-2 ≈1-(2h /R )+…
So suppose that h=R /20; then Fh ≈0.9W .
(The exact calculation would be 0.907W .) You asked
for the force "in terms of g-factors"; that would be the
factor of W , [R 2 /(R+h )2 ].
1% of that exerted by the radiation. By the way, light
having momentum is not a "theory", it is a well-measured
fact.
H and the number of times she
jumped during the semester N . The total time, in
seconds, that
child spent in the air would be T =0.072N √H
if H is measured in inches and T =0.045N √H
if H is measured in centimeters. For example, if
the child typically jumped 8 inches and jumped 200 times,
the total air time is T =0.072x200x√8=40.7
seconds. I would only worry that an elementary school
student might not be familiar with a square root. (You only
asked for an equation; I can also post the derivation of the
equations I posted here if you are interested.
recent answer . Think of the
passer as your pirate ship and the receiver as the target on
the ground. If the passer (ship) in the first figure is not
running (moving), the ball will go to the location of the
receiver (target) who will receive (which will be hit.) it
if not running. If the passer (ship) is running (moving), he
(it) will give his (its) velocity to the ball (cannonball),
in addition to the velocity he (it) imparts by throwing
(shooting); if the receiver (ship) is not running (moving),
he (it) will not catch it (be hit) because it will go to the
position he (the target) would have been if he (it) were
running (moving) too.
bike
plus rider needs to be halfway between the two axles
for the wheels to have equal friction. The left-to-right
location would have nothing to do with having equal
friction, so that can be ignored for your purposes; however,
since the the legs are moving up and down and forwards and
backwards, not sideways, the left-to-right centering never
changes anyway. The first figure shows one pedal up and one
down; the second figure shows level pedals. In the first
case, the right knee is quite forward and the left leg is
straight down; in the second case the right knee is less
forward than in the first case and the left knee is slightly
less forward than the right but substantially forward from
the first case. To my mind it is a tossup as to which has
the center of gravity more forward; and besides, where is
the center of gravity of the bike rider system? Furthermore,
in my opinion, any difference between the two cases will
make a trivial difference regarding whether one wheel or the
other is more likely to lose traction. In addition, if you
have an extreme lean or you are on terrain which is sloping
upward in the direction you are turning, you are less likely
to catch your pedal on the ground if it is all the way up.
Do whatever you are most comfortable with; this issue does
not merit "a great debate"!
is rate of change of velocity, then why a body is said to have a constant
acceleration due to gravity when a body is at rest at a plane horizontal surface?
g plays an
important role for the body which, if its mass is m ,
has a weight W given by W=mg .
E to each, E =½mv 2 ,
where m is the mass and v is the speed.
Therefore, ½m t v t 2 =½m b v b 2
or v b =v t √(m t /m b ).
Putting in the masses of the tennis ball and the baseball,
m t =0.058 kg and m b =0.145
kg, I find v b =58 mph.
W =3000
lb=13,345 N, 1.5 ft=0.467 m, 8.6 ft=2.62 m, 25 ft=7.62 m, and
6 ft=1.83 m. The answer depends on which way the wind blows.
So, I will make the calculation for the wind blowing
directly on the largest surface whose area will be A =7.62x0.467=3.56
m2 . There is a pretty good approximation for
calculating the drag force F on an object of cross
section A in a wind with speed v , F =¼Av 2 ,
so v =2√(F /3.56). (N.B. ,
this works only in SI units.) The forces on your mass are
shown in the figure. When I do the analysis I find F =3576
N=804 lb, and v =63.4 m/s=142 mph. Please keep in
mind that this is an approximation as are all calculations
of air drag.
A ,
thickness t , and temperatures T in
inside and T out outside. The rate of
heat flowing through the material depends on four things:
what the material is, what A is, what t is,
and the temperature difference ΔT =|T out -T in |.
(This assumes that conduction through the container is the
most important means of having heat flow, not convection or
thermal radiation.) So take two examples: inside cold water
at 10 C, outside room temperature at 200 C,
ΔT =190 C; and inside hot soup at 900 C,
outside room temperature at 200 C, ΔT =700 C.
So the soup would cool down faster than the cold water would
heat up. Why would evaporation have anything to do with it?
via one of the springs attached to it, on
its neighbor which then propogates the push to the next
neighbor, etc . Clearly, the speed of this sound,
displacements of molecules, will depend on the spring
constant of the springs; but the spring constants are an
approximate measure of the appropriate modulus. If we think
of the spacing of the molecules as being roughly constant
regardless of the composition of the solid, the mass of a
molecule is going to be proportional to the density. But,
the mass of a particle on a spring determines the speed of
the motion of the particle—the greater the mass, the more
slowly the particle moves when attached to a spring. By this
reasoning, the speed of sound depends on both modulus and
density.
is a battery. As long as it
can supply the current needed at the right voltage (power),
no other battery need be added.
y=x tanθ -(½g /(v 0 cosθ ))x 2
where the acceleration due to gravity is g =9.8 m/s2 ,
θ is the launch angle, and v 0
is the launch speed. In fact, since speeds are relatively
modest, air drag will probably be negligible. If you include
air drag, each bounce segment will be a different shape
because the drag depends on the speed of the ball.
F W F'= F -W F' v . Your coworker is right
that the force which provides the acceleration,
F' F W
F has a magnitude twice the weight of
the adult, approximately 1640 N (360 lb), so F' =820
N. If this force acts over 0.5 m for the adult, he launches
with a speed of about 3.2 m/s, while the force acting over
0.25 m launches super child with a speed of about 7.3 m/s.
With these initial speeds, the adult goes up about 0.5 m while
the child goes up about 2.7 m. Now, we must calculate the
height the adult will rise on the moon exerting the same
force over the same 0.5 m; but now the launching force is
F'=F-mg /6=1640-137=1503 N. I find the launching
speed to be about 4.3 m/s and the height to be about 5.5 m.
So the adult on the moon goes about 5.5/2.7=2 times as
high as child with equal leg strength does on earth, and
about 11 time higher than he could jump on earth; the reason
for this large factor of 11 is twofold—first, the fact
that the weight is much smaller means that the given leg
strength will be much more effective in accelerating during
the launch; second, once launched the weaker gravity cause
him to slow down more slowly. So your coworker's
guess about how the compared jumps would be is pretty far
off, a factor of two in terms of height achieved. However,
he got it qualitatively correct that a smaller person with
equally strong legs could jump higher on earth; it just
doesn't work out quantitatively when comparing to the jump
on the moon.
But, what is the real difference? If you look at the videos
they look like they are taken in slow motion, but they are
not. Because the gravity is much smaller, it will take much
more time to get up to and back down from a given height;
even if the height achieved by the super child on earth were
exactly the same as for the adult on the moon, the two would
appear very different because the child would be back to
ground level much faster than the adult would be.
gedanken , a
thought experiment; the cone is at the point of unstable
equilibrium and you would not be able to actually balance a
cone on its tip in the laboratory. Any force which has no
vertical component and has a line of action through the
center of mass will cause horizontal translational motion
but no rotational motion. In fact, any force passing through
the center of mass will cause no rotational motion. The more
general case is shown in the diagram. The effect of the
horizontal component (F x F y )
is to reduce the normal force (N mg Fy >mg , the cone would have an
acceleration in the y
direction as well, fly off the surface. The components of
the acceleration would be ax =Fx /m and ay = (Fy /m )-g.
In no case would the cone have any rotational motion.
ion propulsion in this link by NASA.
Mythbusters episode debunking the scene from that Sandra Bullock movie, but I'm interested in the theoretical point at which it could happen.
The diagram above shows the forces on the bus if it is turning
right (as seen by the driver) around a circle of radius R
(center to your left), traveling with speed v ; the
coefficient of static friction between the tires and the
road is μ . The bus is just about to skid and so
the frictional forces are at their limits, μN . The driver's side tires have a normal
force N 1 and a frictional force f 1 =μN 1 ;
similarly, the tires on the other side have a normal force
N 2 and a frictional force f 2 =μN 2 .
The weight of the bus, W=mg , acts at the center of mass of
the bus; I will assume that the center of mass is halfway
between the left and right wheels. The bus is in equilibrium
in the vertical direction, so N 1 +N 2 =mg .
The bus has a centripetal acceleration to your left a=v 2 /R
so f 1 +f 2 =mv 2 /R.
The greatest that f=f 1 +f 2 =can
be is f=μ (N 1 +N 2 )=μmg=mv2 /R.
Therefore the smallest radius that the bus can have if
its speed is v is R min,slide =v 2 /(μg ).
In your case, v =50 mph=22 m/s, g =9.8 m/s2 ,
and μ≈ 0.8 for rubber on dry asphalt, R min,slide ≈ 62
m=203 ft.
But you are interested
in the radius when the bus will start to tip over; this
happens when N 2 =f 2 =0,
N 1 =mg , f 1 =mhv 2 /R . Now we need to calculate
the torque about some axis and set it equal to zero, but
that axis must pass through the center of mass because the
bus is accelerating. Now, you need to know the height above
ground h of the center of mass and the distance 2d ,
between the left and right wheels. Doing that, the torque equation is (mhv 2 /R )-mgd =0
or R max,start =hv 2 /gd.
This is the radius at which the bus will just start to
tip. For smaller radii the inner wheels will
come
off the ground but the bus will not tip over until you
reach some critical smaller radius. So let's just guess that
approximate values of h and d are
equal at
about 2 m each: R max,start =v 2 /g =49
m=161 ft. This choice of h and d result in
the bus never tipping because it would slide before it
started to tip; to reach this point, f =μmg=mv 2 /R ,
μ= 222 /(9.8x49)=1.01.
Consider the bus when it has tipped through some angle
θ relative to the horizontal as shown in the second
figure. Now the moment arms to compute torques are D=d cosθ-h sinθ
for N=mg and H =h cosθ +d sinθ
for f=mv 2 /R . (I
must admit that it took me a while of floundering with my
trig to figure these out!) So now the torque equation is
mgD =Hmv 2 /R or, R=Hv 2 /(gD ).
In the graph below you can see that, for d=h , the radius gets
smaller and smaller as the tip angle gets larger; all these
calculations assume that the bus does not skid, but unless there is
something else besides static friction to provide the
centripetal acceleration it will likely skid before it tips. Note that if the bus tips until the center of mass is directly above
the outer wheels, where D =0 and θ= 450
for d=h , it is on the verge of
fully tipping over. The bus
will roll over when the angle it makes with the ground is greater
than tan-1 (d /h ).
If you manage to get the bus on two wheels and at the angle
just before it tips over you can go essentially straight
(making small steering corrections), not turning as the
video
linked to above. The bottom line is that a bus is extremely
unlikely to roll over in a turn because it will likely skid
before it rolls. I think my choices of d and h are very
reasonable and probably, because so much of the weight is in
the frame, suspension, axles, transmission, and engine of the bus, the center of gravity is
likely to be lower than 2 m making a roll even less likely.
It would be a different story for a train on tracks where
the "friction" is almost limitless because the train is
constrained by the track from moving sideways ("skidding")
at all.
does stop but only for zero time. But an engineer will have to take into account that a real pendulum might stick for a brief time, maybe a microsecond, at the extrema. But I cannot give you an answer to your question because the amount of stick time will depend entirely on the details of the construction of the pendulum.
(I believe that the main reason for sticking would be the
difference between static and kinetic coefficients of
friction.)
h , plays more roles than just
setting the smallest value of something. Most famous of this
role is that, if you have electromagnetic waves with a
frequency f , the smallest amount of energy, called
a photon, which
such electromagnetic waves can have is hf . But,
this idea of minimum amount possible does not extend to
everything, certainly not to something like the number of
digits in any irrational number like π . The idea
that something has a minimum value usually has to do with
what we call "fields"; when the electromagnetic field is
quantized, the photon is the "messenger" which conveys the
field.
More interesting, I believe, is the role played by Planck's
constant in the structure of systems of particles like atoms
or atomic nuclei. The great hypothesis by Niels Bohr that the
hydrogen atom can only exist in particular energies owes its
inception to the idea of quantization. He assumed that the
electron in a hydrogen atom moved in circular orbits but not
in any old orbit but only in those for which the angular
momentum was quantized, L=nh /(2π ) where
n =1, 2, 3, 4…; the angular momentum is the
product of the electron mass, the orbit radius, and the
speed of the electron. This led to quantized energy levels;
this model excellently explained everything known about the
hydrogen atom at that time. See the Wikepedia article on the
Bohr
model. (The Bohr model, while easy to visualize, is a
very simplistic and incorrect compared to the correct modern
quantization of the atom.)
If you think about it,
energy quantization is an astounding surprise. It is like
saying that a pendulum may have an amplitude of 50
or 70 but not 60 . But nature does hand
us surprises sometimes.
But energy quantization does not apply to everything. If a
system of particles is not bound together it can have any
amount of energy.
What
is not known but speculated about is whether space and time are
quantized. Is there a shortest possible distance? Is there a
shortest possible time? Nobody knows but speculations called the
Planck
length and the
Planck time
are extraordinarily small, far smaller than current measuring
capabilities.
Finally, look
at the following Q&A for another example of the role of Planck's
constant.
x
and linear momentum p , and the product of their
uncertainties cannot be smaller than the rationalized Planck
constant, h /(2π ). So, the more
accurately you measure either, the less accurately you can
know the other. Another pair are energy E and time
t . So, for example, if a radioactive system has a
half life of Δt , the most accurately you can
measure the energy is ΔE >[h /(2π )]/Δt .
F AB on object B,
object B exerts an equal and opposite force F BA =-F AB
on object A. F AB tends
to accelerate object B and F BA
tends to accelerate object A.
F AB on object B,
object B exerts an equal and opposite force F BA =-F AB
on object A. F AB tends
to accelerate object B and F BA
tends to accelerate object A.
These are two equations for two unknowns, v and
t , and can be solved using simple algebra. The
solutions are t =0.563 s and v =7.33 ft/s=5
mph.
v . There is no difference in the
"damage".
θ subtended by 300θ= 300/5000=0.06
radians=3.440 . Looking at the diagram, cosθ=R /(R+d )=[1+(d /R )]-1 ≈1-(d /R );
therefore d≈R (1-cosθ )=9 miles.
Now, you are probably thinking that all the sound waves move
in straight lines, that the sound waves coming from the
detonation are radiating out as shown in my second figure
thereby missing the earth's surface except very close to the
source. But, since sound is composed of waves, it is subject
to diffraction. Huygen's principle says that the wave a
short time later is determined by treating each point on the
wave front as a new point source. Finally, I have drawn one
of the wave fronts which is off the earth in the earlier
picture and how Huygen's principle predicts the wave a short
time later. Note that the wave where it originally ended is
now bending and lengthening; this is diffraction and allows
the sound to follow the contour of the earth. This is the
same principle where when you speak to someone on the other
side of a sound-absorbing wall with an opening in it (but
she is not visible to you), you can be heard just fine;
sound diffracts around corners.
mg f down which the upper
cube exerts on it in the direction of the flow, and the
force f up which the lower cube exerts on it opposite the
direction of the flow. Notice that all cubes exert pushing
forces on their neighbors, not pulling, so "sucked" is never
what is going on. Now, the cubes are flowing downstream with
constant velocity, so the sum of all forces in that
direction must add to zero (note that only the downstream
component of the weight is mg θ
pushes the water downstream): f up -f down -mg sinθ =0.
So, f up =f down +mg sinθ.
So, you see, gravity and the water upstream push the
water downstream while the water downstream tries to hold
the water back. There is a greater force upstream from the
water itself than there is a force downstream. Without
gravity or with no slope to the riverbed the water would not
flow at all.
Here is a nice explanation of how magnets are rated.
Here is a real-world example.
mg , that component is mg sinθ
where θ is the angle of the ramp, 100
in your case. The force pointing up the ramp which
must be applied is therefore mg sinθ.
Note that the speed is irrelevant.
earlier answer . I am sure that astronomers are perfectly
aware that masses will alter the position of the images they
observe relative to the objects they are observing are not
perfect. However, the position recorded is simply the
position in our sky at this time .
Positions of objects can change simply because they are
moving.
link .
Kind regards
lifts the airplane. Regarding
"scientific/reliable" and "uncredited/verified", I am not
writing an encyclopedia here where footnotes and references
are required. I am presenting myself as someone who has a
degree of expertise in physics who can answer questions
about the subject; it is your responsibility, since there
can be lots of BS on the internet, to convince yourself that
I do indeed have some physics credibility. To that end I
post information about myself, on the page
"The
Physicist" where you can get the information you need to
judge my qualifications. Incidentally, the reason I chose
the video which I did was because the Bernoulli aspect of
wing lift is presented as the whole story in a great many
introductory physics textbooks and only when I was taking
flying lessons years ago did I learn that "angle of attack"
is much more important in actually creating lift; I like the
way this video presented both of these important concepts.
t over which the
fish is accelerating and the change in linear momentum Δp =m Δv ; F =Δp /Δt .
In your case, the mass is m =1 lb=0.45 kg and the speed is
v =23
mph=10 m/s so the change in momentum is 4.5 kg⋅m/s if
the fish starts at rest and accelerates to 23 mph or
vice versa . I
will assume that the line does not break and then compare
forces with the pound-test of the line; then I will compare
with a four pound test leader. If your
rod had zero flex, was perfectly rigid, the fish would be
unable to move at all unless the line broke, so assuming
that he could exert a force of four pounds or more which he
probably could, you would lose that fish. If he hit the fly
going 23 mph and stopped in 0.1 seconds, the average force
would be 4.1/0.1=41 N=9.2 lb—bye bye fish! But, that is unrealistic because
what really happens is that the rod bends and what that does
is lengthen the time the strike lasts; that means that the
force on the line will be smaller. Suppose that it takes 2 s for
the fish to get up to his maximum speed while hooked; during
that time the average force exerted will be 4.5/2=2.25
N=0.51 lb, and you have caught that fish. Now you understand
why fishing rods are designed to bend.
3 .
The closest metal to this is germanium with a density of
5.32 g/cm3 ; but germanium is brittle and would
not be pliable. Lead has a density of 11.34 g/cm3
and silver 10.49 g/cm3 , both about twice as dense
as your sample. So it must be something else. To sort it out
would require a chemical analysis, I guess.
R=v 2 sin(2θ )/g
where R =130 ft, v is the speed, θ is the
launch angle, and g=32 ft/s2 is the gravitational
acceleration; the "optimum" launch angle is 450
for maximum range. So v =√(130x32)=64.5
ft/s=44 mph. For a smaller launch angle the speed would be
larger; e.g . at 300 , v =50 mph.
earlier
answer . The issues you raise could mostly be handled by
making a judicious choice of target. Stellar evolution is
reasonably well understood and the life time and age of a
star could be decuced from its properties. For example, if
the star you have in mind is within a million years of
running out of fuel and is a million light years away, it
would foolish to plan a journey there. Regarding distortion,
the earlier answer emphasizes that problem but the most
important distortion is that caused by the high speed of the
vehicle itself; "distortion" due to motion of the target
star itself as well as gravitational lensing could be
adjusted for in flight if you could actually see and lock
onto the star at high speeds. Your last question has no
answer because you need to specify how small a lag you would
call "current"; we are quite close to the sun but light from
it takes about 8 minutes 20 seconds to get to us.
(To give a background on why I asked the question. I have a habit of clicking my pens on my desk so they shoot a few inches into the air. I always try to catch them right at the height of their motion. I started wondering how long I actually had to do that. When I asked some engineering friends they said the pen instantaneously goes from traveling up to falling down. I’m not sure if that’s true but if so it introduces a bit of a crisis. If the pen spends zero time at its maximum height, is it ever actually there? And would I ever be able to actually catch it at is maximum height? Can something be somewhere but spend zero time there?)
To get more technical, what is the probability of finding
the object at a particular point? It is zero. That is
probably what is bothering you. So you could ask for the
probability of catching the pen during the interval 1/1000th
of a second before and after the zero speed time. I will not
get into quantum physics here, but you cannot know precisely
both the speed and location of your pen at any time; but
this truth implies an incredibly small uncertainty,
certainly not important for a macroscopic object like your
pen. One final thought, we do not know whether or not there
exists a shortest possible time or length, sometimes
referred to as the Planck length and the Planck time; if it
exists, the Planck time is expected to be about 5x10-44
s, so perhaps your pen stands still for that time!
faq page .)
2 until they get all the
way to the ground. The air drag is a force on the falling
object upward (opposite the velocity) which gets bigger as
the object falls faster; eventually the drag force becomes
equal in magnitude to the weight which is a force down, so
the object falls with constant velocity after that. That
speed is called the terminal velocity and that depends on
both the mass and the geometry of the object. The speed of
120 mph is the approximate terminal velocity of a falling
person. A bowling ball would have a much larger terminal
velocity and a feather a much smaller one. If the earth had
no atmosphere, anything would continue accelerating until it
hit the ground. If you want to learn more about air drag,
you can find links to many of my older answers on the
faq
page . If there were no air in the hole you want to drill
and the earth were a uniform sphere (which it isn't), the
speed would be about 18,000 mph at the center.
etc .
Is a block of iron a "different type" of solid from a block
of tin? Yes. Similarly, a plasma composed of iron ions is a
"different type" of plasma than one composed of tin irons.
v . I will assume that the time which the
collision lasts, Δt , is the same for any situation
I consider. There will be two extremes for a collision—perfectly
elastic where energy and linear momentum are conserved, and perfectly inelastic
where the objects are stuck together after the collision and
energy is not conserved but linear momentum is. The impulse
received by the object originally at rest (your opponent) is
equal to the linear momentum he has after the collision and
is equal to F Δt ; since I specified that Δt
is the same for all collisions, the force experienced is
F=mv' /Δt , where v' is the speed of the struck
object after the collision. First consider a collision
between two equal masses m . For the elastic
collision the incoming mass is at rest after the collision
and the outgoing mass has a speed v ; for the inelastic
collision both masses (stuck together) have a speed ½v .
Therefore the force experienced by your opponent is given by
½mv /Δt ≤F ≤mv /Δt .
Now suppose that your mass M is much larger than your
opponent's mass m , M>>m . In this
case, the inelastic collision will have both objects moving
with approximately a speed of v ; the elastic
collision will have M moving
approximately with the speed v and m
moving approximately with the speed 2v . Therefore
the range of possible values for F experienced by
m is mv /Δt ≤F ≤2mv /Δt ,
which is twice the force for equal masses. So my simplistic
view is that being anchored to the floor is like increasing
your mass allowing you to deliver more impulse to your
opponent as compared with if you are flying toward him which is
more like the collision of equal masses. As I said, it is
much more complicated than this but this might give a little
insight. I would welcome any comment on this treatment.
(Lengthy enough answer for you?)
that direction.
2 )=102,900 N
and the total area is (4 m)x(7 m)=28 m2 . So the
pressure is 3675 N/m2 . Other units used for
pressure which you might prefer are 375 kg/m2 or
76.75 lb/ft2 .
earlier answer (skip down to the
last paragraph ).
p =q d q separated by some distance
d ; imagine these as vibrating
sinesoidally along the direction of the vector d p p here .
Brief Answers to Big Questions ) it is stated that a chemical rocket with a thrust velocity of 3 km/sec, that jettisons 30% of its fuel, attains a velocity of 'about half a km/sec'. Using
Tsiolkovsky's rocket equation I get a value of 1.07 km/s.
Have I made a mistake?
earlier answer , v=u ln[(M+m )/M ]
where M+m is the initial mass, m is the
fuel burned, u is the exhaust speed, and v
is the speed acquired by the rocket. So I find v=3ln(0.7-1 )=1.07
km/s. I do not know how Hawking came up with his answer, but
it might hinge on what he means by "30% of its fuel"; you
and I have interpreted that to mean the exhausted fuel has a
mass equal to 30% of the mass of the rocket before the fuel
was burned, hence M /(M+m )=0.7. Also, maybe
he was thinking about a launch from earth which would
require some of the energy from the rocket to overcome the
gravitational force and the energy loss from air drag. Or
maybe he was guesstimating the effects of efficiency of the
engines. In any case, you did not make a mistake.
0 C, if the air
in the bottle chilled to that temperature, the pressure in
the bottle would be smaller than 15 psi. Using the
Gay-Lussac Law, P /T =const., I find the
pressure would be about 11 psi, a net of about 7 psi. (If
you want to check this, you must use absolute temperature,
not 0 C.)
FAQ
page you will find many links to earlier answers which
relate to this.
laminar
which means that if you focus your attention on a small
volume of the fluid it will move along a smooth line and be
followed and preceded by many other small volumes following
the same smooth line. Under other circumstances, the fluid
will become turbulent , moving chaotically. So
something happens which causes the smoke from your incense
stick to change from laminar to turbulent at some point.
This is usually quantified by a dimensionless variable
called the Reynolds' number, Re=ρuL /μ
where ρ is the density of the fluid, u
is its speed, L is a length which describes the
geometry, and μ is the dynamic viscosity of the
fluid. So, Re changes in such a way that the
original laminar flow becomes turbulent after rising a
certain distance from the source of the smoke.
a
or in zero field but having a uniform acceleration a ;
buoyancy would be observed in an accelerating rocket ship in
empty space.)
E i A =½mv i 2 +mgh ,
E f A =½mv f 2
and so v f =√(v i 2 +2gh ).
Exact same calculation for E B .
d =0 in your (simplistic) W=f x
d definition of work. If the saucer is rising with a
constant speed v, then when it has gone some distance d, the
thrust T has done W saucer =Td
and gravity has done work W gravity =-mgd
because T=mg for it to be rising with constant speed. The
net work W on the saucer is still zero, W =W gravity +W saucer =0.
Your last question "…simply converts KE to PE…"
is again using PE unnecessarily. Gravity does work on an
object as it falls because there is a force mg
acting down.
This is the most popular way to explain the answer without
resorting to very opaque mathematics, but I find it just a bit
unsatisfying even though it is certainly correct. I found a more
convincing argument put forward by Eltjo
Paselhoff on the website
quora.com . Suppose we plot the velocity v as a
function of time t for the upward trip. At t =0
the magnitude of the slope must be larger than g
because the air drag is slowing the ball down as well. As the
ball gets higher, its speed decreases so the air drag gets
smaller so that finally at the top of its trajectory it is at
rest and the magnitude of the slope is equal to g . Now,
the ball falls back down and speeds up because air drag is
smaller than weight; eventually, though, the two forces become
close to equal and the speed becomes constant at the terminal
velocity v T . One more thing you must
understand now is that the area under the velocity curve is the
distance traveled by the ball, so since up and down go the same
distance the green and red areas need to be the same. The only
way this can be is if the time to fall is larger than the time
to rise.
F - F f F
web site which gives the details of the
solution of the physics problem but is still quite readable.
For the casual reader, I will include here an overview.
Newton's second law is an inhomogeneous second-order
differential equation. What makes the equation inhomogeneous
is the presence of the driving force, the signal your
amplifier sends to the speaker; with no driving force, the
solution of the problem would be x t (t )=A t e-γt ωt+φ t )
where ω is the natural frequency the speaker
would vibrate with on its own, γ is a parameter
which describes the damping of the speaker, and A t
and φ t are constants
determined by the initial conditions. If the
driving force is present, F=F 0 cos(ω 0 t +ψ ),
the solution is a linear combination of the what is called
the steady state solution , x s (t )=A s sin(ω 0 t+φ s ),
and the solution for the undriven oscillator x t (t )
which is referred to as the transient solution: x t (t )=A t e-γt ωt+φ t )+A s sin(ω 0 t+φ s ).
The constants A t and φ t
are determined by the initial conditions (speed and position
of the speaker) and the constants A s and
φ s are determined by the driving
force (F 0 and φ t ).
Because of the damping term e-γt
Below are shown a couple of examples:
These show the total displacement for different initial
conditions. In both cases the speaker started at rest. The first
shows the starting position at zero, the second shows the
starting position at 0.5. The important thing is that they both
end up, after the transient has died, as the same.
0 . I found a
little
calculator you can play with; a screenshot is shown
below. All you would need to vary are the velocity, the
angle, and the height of the dock. I found that if I chose
the initial speed to be 26 ft/s (about 18 mph) the distance
the dog would go for a 450 jump would be about 21
ft from a zero-height dock. Now, when I do the calculations
for the 24 in=2 ft and 16 in=1.33 ft, I find the best angles
and distances (for 26 ft/s) are (43.50 , 22.9 ft)
and (42.50 , 22.3 ft), respectively. All in all,
there is just about no difference.
Note that I have not included any air drag. I believe the speed
of the dog is low enough to make drag negligible.
D is hinged to a wall. A force F T d from the hinge; we would find
experimentally that if T =F (D /d )
the stick would not rotate. We would conclude that the net
torque about the hinge is now zero and that this torque
changes linearly with the distance. Of course, the torque
exerted by F T τ=FD-Td= 0.
This is not the most general case but I feel you wanted to
be convinced that torque depends on where the force is
applied relative to an axis about which you are computing
torques, which was easy to see.
first and
second times. In a nutshell, if there is not air moving
over the wings with sufficient speed, there is not
sufficient lift on the airplane for it to leave the ground.
I agree that Mythbusters'
version of this is not the same as I understand your
question and previous questions where whatever the conveyer
is doing, the plane is not moving forward through the air;
the Mythbusters have the conveyer at rest relative to the
plane but moving forward with the same speed as the plane,
so the wheels are not spinning.
I guess it is possible
that a propeller plane could take off from rest since the props
send a wash of air over the wings, but probably not unless there
were a pretty hefty headwind; you do see some planes, designed
for flying into and out of tight places, which take off in a
very short
distance.
Let's just call a truce on this whole thing. A 747 at rest for
any reason is not going to take off. And practical problems with
the fabrication and utilization of this stupid treadmill will
ensure that it is never constructed or tested.
L a distance d from where you are viewing
it. You can see that the angle subtended by the moon is 31
arcminutes≈0.520 . The angle is expressed in
radians as θ=L /d. To convert that to degrees,
multiply by 180/π , so θ= (L /d )x57.3
in degrees. For example, your 19 km dreadnought 100 km away
would have θ =10.90 , about 21 times
larger than the moon.
earlier answer which answers
the same question except for the earth-moon and sun-earth
pairs. For your question it is the electrical force which
holds the electron in its orbit. There is an important
difference, though. The electron-proton system, if you look
at in classical electromagnetic theory, should radiate
energy (it is essentially a tiny antenna) and fall into the
proton. This does not happen; the reason is that for such
small systems classical physics does not work and this was
one of the milestones of physics when Niels Bohr proposed
his model
of the hydrogen atom where certain orbits simply did not
radiate. To understand the details of quantum mechanics,
which eventually evolved from the Bohr model, would require
a much longer explanation.
etc . For
a more detailed explanation, see
this link .
Rutherford scattering ) does result in the
angular momentum of the moving charge q about an axis passing
through the fixed charge is conserved. It is not hard to
understand. What is the condition for which angular momentum
is conserved? There must be no torques on q . The
expression for torque is τ r F qr xE r E
r and E r E q and its
angular momentum will remain constant.
Wikepedia : "Gravitational waves carry energy away from their sources and, in the case of orbiting bodies, this is associated with an in-spiral or decrease in orbit. Imagine for example a simple system
of two masses—such as the Earth–Sun system—moving slowly compared to the speed of light in circular orbits. Assume that these two masses orbit each other in a circular orbit in the
x–y plane. To a good approximation, the masses follow simple Keplerian orbits. However, such an orbit represents a changing quadrupole moment. That is, the system will give off gravitational waves.
In theory, the loss of energy through gravitational radiation could eventually drop the Earth into the Sun. However, the total energy of the Earth orbiting the Sun (kinetic energy + gravitational potential energy) is about 1.14×1036 joules of which only 200 watts (joules per second) is lost through gravitational radiation, leading to a decay in the orbit by about 1×10−15 meters per day or roughly the diameter of a proton. At this rate, it would take the Earth approximately 1×1013 times more than the current age of the Universe to spiral onto the Sun. This estimate overlooks the decrease in
r over time, but the majority of the time the bodies are far apart and only radiating slowly, so the difference is unimportant in this example."
So the answer is that the energy comes from the orbiting
earth which causes its orbit to get smaller as it loses
energy; however, as the calculation shows, the loss of
radius of the orbit is trivially small.
c where c is the speed of
light. In order to see the effect which I think interests
you, you need to choose a much larger much larger speed for
the rocket; I will choose v =0.8c , 80% the
speed of light. I will choose my unit of length to be one
light hour and my unit of speed to be light hours per hour,
so c =1, v =0.8, and L =1. First,
look at what an earth-based observer will measure with his
clock. If t is the time when the light
arrives at the rocket, then vt =0.8t =x and
ct=t =(L-x )=(1-x ); solving, x =0.8/1.8=0.444
and t=x /0.8=0.556. Also, note L-x =(1-x )=0.556.
But, what you asked for is how long a clock on the rocket
takes to see the light. The important thing is that the
distance between the earth and the satellite is not 1 light
hour any more, it is length contracted such that the
distance seen for L by the rocket is L'=L √[1-(v /c )2 ]=0.6
rather than 1. Of course the same factor occurs for x' =0.6x =0.267
and the time this happens is t'=x' /0.8=0.333
because even though the distance to the satellite is
shorter, it still zips past the rocket with a speed of 0.8c .
Finally, c' =(L'-x' )/t' =(0.6-0.267)/0.333=1!
So, you see, c really is the speed seen by all
observers.
relative to what . As a concrete
example, suppose that you throw the bottle with a velocity
of 20 mph relative to you (the speed you would throw it if
you were at rest on the ground). But you are are at rest
relative to the car, so you and anybody else in the car
would see it moving forward with a speed of 20 mph. But
someone by the roadside would see it moving forward with a
speed of 70 mph. Somebody in a car going in the same
direction as you with a speed of 70 mph would see the bottle drop
straight down to the ground (no horizontal velocity). If you
threw it backwards (relative to the direction of your car)
with a speed of 50 mph, someone by the roadside would see it
drop straight to the ground (see a
Mythbusters episode).
-29 g/cm3 ,
so it isn't even close.
https://www.bbc.com/news/science-environment-47873592
Why this cannot be done by one single telescope and how the group of telescope functions in order to work (I cannot think why it needed so many to be focused in a single point). Also the earth is rotating so most of them are not aligned with the point of focus.
Second question is why we only have an image? if we can take a photo, we can surely take more photos and create a video. In that way we can see the moving plasma or whatever is there.
-3 m wavelength; this
wavelength is expected to originate only from the accretion
disk-7 m. Increasing the
size of a telescope gives it better resolution, so the size
S for a millimeter telescope comparable to the best
optical telescopes would be S =10-3 (10/5x10-7 )=20,000
m=10 km; but, looking at an object so far away (55 million
light years) would be impossible for visible light, but
since the earth has a diameter of about 13,000 km, we can
get very much better resolution with a large millimeter
telescope array by having the individual telescopes have
separations comparable to the size of the earth. Although the size is very
huge, the total area of all the radio telescopes is
extremely
small, so trying to make an image would be like blacking out
99.999% of the mirror in an optical telescope—you
would still get an image but you would have to wait a very
long time. So the data from all the telescopes in the array
were gathered over many hours during a 10 day observation
period. Then, all the data had to be put together using a
computer to reconstruct the image; the data analysis took
two years to complete, a very difficult problem.
Finally, you should ask what the dark circular region in the
center is—it is not the black hole itself which has
zero size. The outer edge is what is called the photon
radius, the distance from the black hole where a photon will
orbit. Somewhere inside of the photon radius is the event
horizon, the sphere inside of which nothing, including light
can escape.
A warning: as I clearly state on the site, I do not usually
answer complex questions in
astronomy/astrophysics/cosmology, so take what I say with a
grain of salt!
L was quantized, L=n ℏ where ℏ
is the rationalized Planck constant. This was actually
wrong, but it gave the right answer for the spectrum of
light observed from a hydrogen atom and that was a great
leap forward; but purely empirical. Improvements
and
extensions of this model were made, but they were all
equally empirical. Still, this first empirical model set
everyone on the right track to achieve a more fundamental
model. During this time, it was beginning to be understood
that trying to imagine a model based on classical ideas,
ideas like an electron orbiting in a circular orbit, were
invalid at such tiny scales—that realization ultimately gave
rise to the development of quantum mechanics. You can no
longer even think of an electron as a little point charge
when it is bound in an atom. Rather, you solve an equation,
known as the Schrodinger equation, which gives you a
mathematical function, called the wave function, which tells
you the likelihood of finding the electron at any particular
place in space. So the best way to think of an electron in
an atom is akin to picturing a cloud over which the electron
is smeared. The shape of this cloud is dependent on four
quantum numbers which quantize the energy, the angular
momentum, the projections of the angular momentum and the
spin angular momentum on some axis. Some examples are shown
in the figure. To visualize the "clouds", imagine rotating
each around a horizontal axis through the center; so, e.g. ,
the upper right cloud is donut-shaped.
ground rules , you would have seen that "I no longer answer questions which are based on premises like:
'...if we could travel faster than the speed of light...' or
'...ignoring the fact that nothing can go faster than the speed of light...',
'...if I were traveling at the speed of light...', etc."
Suppose that you were traveling at 99.99% the speed of
light. The light which you would see if at rest would be
visible, but in your moving frame the wavelength would be
Doppler-shifted to a longer wavelength, into the region of
far infrared which your eye cannot detect.
241 Am. In addition to
the four energies of alpha particles there are gamma rays of
energies up to about 100 keV. I cannot explain why water is
the least absorptive than any other material you tested.
here , the
second is linked to from that answer.
F =m a
In rotational physics, moment of inertia I plays
the role of mass and angular acceleration α F r τ r x F τ Iα .
I do not know much about engineering notation semantics but
if what you say is correct, that torque means that which is
zero if the object is not rotating at all (or with a
constant angular velocity), then that is what would be
called net torque in physics which seems to make
more sense to me. According to Wikepedia, mechanical
engineers (and also British physicists) refer to (physics)
torque as moment of force , not torque moment as you
state. Similarly, the definition by mathematicians is that
torque is equal to the rate of change of angular momentum
which is exactly the same as τ Iα
and so torque, again contrary to your
statement, is the same as in physics. So, I read the
Wikepedia article differently from you and, to me, there is
no difference in anybody's definition of what I would call
torque other than semantics. And I read nothing there which
implies that the word torque means the sum of all moments of
force to an engineer. So my final answer is that everybody's
definitions of the physical quantities are the same, only
the words used to label them differ.
Why doesn't the moon fall to the earth?
the volume of a sphere of radius R is 4πR 3 /3;
the cross sectional area A of a sphere of
radius R is πR 2 ;
the air drag force on an object having speed v
is proportional to Av 2 ; and
the terminal velocity U of an object of mass
m is
proportional to √(m /A ).
I
will use subscripts 1 and 2 to denote the original and final
drops, respectively; so m 2 =2m 1 and
R 2 /R 1 =21/3 .
Therefore, A 2 /A 1 =22/3 .
Finally, U 2 /U 1 =[(m 2 /m 1 )(A 1 /A 2 )]1/2 =[2⋅ 2-2/3 ]1/2 =21/6 .
So, finally, U 2 =10⋅ 21/6 =11.22
cm/s.
R with the current
I distributed uniformly throughout the wire. The
figure shows the field due to the wire, the vector
B having a constant magnitude at a
distance r from the axis of the cylinder defined by
the wire and always tangential to the circle of radius r
around the wire. Hence field lines, as shown are circles,
the direction defined by the right-hand rule as shown. The
magnitude of the field is determined by Ampere's law,
∫B ⋅ ds μ 0 I enc
where the integration
is all the way around the path defined by the circle and
I enc is the amount of current which passes
through the area defined by that closed path. Given the
cylindrical symmetry of this special case, the integral is
simple because B is constant and B s
∫B ⋅ ds πrB=μ 0 I
so Br>R =μ 0 I /(2πr )
for r>R . This is only correct outside the wire
itself because the integration path to determine B
inside the wire encloses less current. Inside, I enc =I (r 2 /R 2 )
so Br<R =μ 0 Ir /(2πR 2 )
for r<R. Note that when r=R both results
give the same value of B , B =μ 0 I /(2πR ).
So, the field increases linearly with r inside the
wire and decreases like 1/r outside.
The questioner also
wanted to express this in terms of the velocity
v Q=-e and having a density of
n
electrons/m3 . The fact that the electrons have
negative charge means that, in the figure above, the
direction of v t the
electrons go a distance ΔL and so sweeps out a
volume ΔV which contains a total charge ΔQ =-ne ΔV =-ne ΔL (πR 2 I =ΔQ /Δt=-ne (ΔL /Δt )(πR 2 )=-nevπR 2
v in a cylindrical beam of radius R . The second part of my answer showed you how to relate
I in the first part of the answer to the charge, speed, density of charges, and
R . So the quantity you seek results from setting
I=nQvπR 2 into
B=μ 0 I /(2πR ) (assuming that "adjacent" means right on the edge of the moving charge distribution).
In the first case the two electrons are very close together and are moving to the right at say 50 m/s, what would be the
"B" field be 50 mm directly above, i.e. in the positive y axis above these electrons?
In the second case, same two electrons traveling to the right at 50 m/s but this time they are separated. Call the first electron,
"e1"and the second "e2".
The first electron e1 is located directly 1 mm below the point of interest where I want to calculate the combined
"B" field. The second electron e2 is located directly 98mm below e1 or 99 mm below the point of interest where I want to determine the combined
"B" field. In effect from the first case I have moved e1 closer to
"B" by 49 mm and e2 farther away by 49 mm.
I was using the formula B=μ 0 NqV /(4πR 2 ) where
μ 0 is permeability of free space, N is the number of charges, q is charge on an electron and V is the velocity of the electrons, R is the vertical distance between the charges and the point of interest for
"B" the strength of the magnetic field.
B R B V B=μ 0 Nq V R πR 3 );
note that the magnitude of V R VR sinθ where θ is the angle between
V and R θ =900 , your equation is correct. Be
careful, though, because the direction matters when you are
adding vectors from two sources; the case you are looking at
has the point the same side of both electrons so the
magnitude is the sum of the magnitudes, but if you look at a
point between the electrons, the magnitude will be the
magnitude of the difference. The form of Ampere's law which
I used in the earlier answer is simply not applicable in
this situation because making reference to a current I only
makes sense in magnetostatics. Finally, I should note that
all this is only approximately correct when you are in a
nonrelativistic regime which you certainly are for V =50 m/s.
twin
paradox (I have attached here the figure from that post for
your convenience) which shows how the traveling twin "sees" how the
earth clock runs by having the earth-bound twin send him one
message per year; this is different from your situation where
essentially a daily rather than an annual message is sent, but
the ideas are identical. When he arrives home, the traveling
twin has "seen" the same number of years (days) on earth pass as
the earth-bound twin has but, as you can see, a different number
of years (days) have passed on his on-board clocks. "Watching"
some clock in another frame of reference is not in any way
relevant to time in your frame; the traveling twin would count
his birthdays using his clock, not earth's clock. Note also that
the traveling twin "sees" the earth spinning very slowly on the
trip out but spinning very rapidly on the trip home.
mv 2 ,
so it is natural to think that the maximum possible kinetic
energy would be ½mc 2 . But
Newtonian physics is not valid at speeds close to c ,
and in relativity the kinetic energy is ½mv 2 /√[1-(v 2 /c 2 )].
To learn more, see an
earlier answer to the same question; you might also look
at the answer following that one.
earlier answer . Spin is nothing more than a
specification of the angular momentum of a system; e.g. ,
the spin quantum number of an electron, proton, neutron, and
other so-called fermions is s =½. The actual
angular momentum is given by S =ℏ√(s (s +1))
where ℏ is the rationalized Planck's constant. Usually, we think of spin
as the intrinsic angular momentum of an elementary particle
or entire quantum system like an atom; particles may also
have orbital angular momentum, for example an electron
orbiting around a nucleus in an atom. But each angular
momentum will be quantized (have a specific quantum number)
and the total angular momentum J of any angular
momentum number j will be J =ℏ√(j (j +1)).
Orbital angular momentum quantum numbers are integers rather
than half integers like electrons; so, an electron in a
"p-orbit" in a Bohr atom has an orbital quantum number 1 and
a spin quantum number ½. As to where the idea came from,
there were several things which suggested it. The experiment
which showed that spin of an electron was the
Stern-Gerlach experiment , but there were also hints from
spectroscopy and, notably, that there were twice as many
states in quantum systems (like atoms) as expected from the
Pauli exclusion principle .
0 relative to the normal of the plane
of its orbit but that axis precesses around the
perpendicular axis about once every 26,000 years. This
precession is caused by torques exerted on the earth by the
sun and the moon. Currently the axis is pointing almost
exactly to the the north star, Polaris; however, in a few
thousand years there will be some new star taking that role.
The last ice age was about 2.4 million years ago and lasted
much longer than 26,000 years, so the answer to your
question is that it pointed many directions during the ice
age. You are asking about the geographic north pole which is
essentially constant in its location (although the land
masses move over the surface because of plate techtonics).
The gravitational north pole, though, is forever wandering
around and has actually jumped to the southern hemisphere
around Antactica and back many times.
My third
book , PHYSICS IS … The
Physicist Explores Attributes of Physics , has a chapter
titled Physics Is Beautiful , devoted to beauty and
elegance. However, I have never thought of it as art. A
sunrise over the Sangre de Christo mountains is beautiful
but you would certainly not call it art. I found that it is
impossible to find an unambiguous or not vague definition of
art; art is qualitative which sets it apart from science
which is quantitative. And, consider this: There are many
paintings of the of Chartre Cathedral, four of which I have
copied here, and all are indisputedly art; there is no
"correct" painting of this subject. The theory of
electromagnetism, on the other hand, is beautiful and the
only one of its kind; you might think up another
electromagnetic theory, but it would be wrong even though it
might be considered very imaginative, creative, even
beautiful. There is no well-defined test which specifies
which painting is the best; there is a test as to which is
the theory of electromagnetism —it must
quantitatively describe what really happens in nature and if
it fails to do so, it is false. I have always thought of
string theory as not a theory (and not beautiful) because it
makes no attempt to quantitatively interface with the real
world and is therefore not falsifiable.
F=- μs N
(negative because the force is opposite the velocity. where
μs is the coefficient of static
friction between rubber and dry asphalt which is about
0.5-0.8, and N is the normal force of the truck on
the road which, on level ground, is the weight; I will
choose μs = 0.65, halfway between the
extremes. At this point I will switch to SI units which
scientests prefer: weight N =7,200 lb=32,027 N, mass
M =32,027/9.8=3,268 kg, and speed v =45 mph=20.1
m/s. Now, using Newton's second law, F=Ma =- μs N ;
so acceleration a =-6.37 m/s2 . Now, you
can compute the time to stop: v =0=v 0 +at =20.1-6.37t,
so t=3.16 s. Finally, the distance traveled is d =v 0 t +½at 2 =20.1x3.16-½x6.37x3.162 =31.7
m=104 ft. It is interesting that this answer is independent
of how heavy the truck is. The reason is that the
acceleration is proportional to the weight/mass which is
just 9.8 m/s2 .
-5 s and the
acceleration is about 2.3x107 m/s2 .
The corresponding force to stop the beanie baby is
F=ma=.0128x2.3x107 ≈3x105 N≈67,000
lb. I think that if the beany baby hit you anywhere except
possibly your limbs, you would be killed. Keep in mind that
this is a very rough calculation, but certainly in the right
ballpark.
V= 4πr 3 /3 is
the volume of a sphere and if it is increasing with a constant rate dV /dt=C ,
then C =4πr 2 dr /dt .
Solving, dr /dt=C /(4πr 2 ) which is velocity at which the
surface is expanding.
So the acceleration of the surface is (d2 r /dt 2 )=-[2C /(4πr 2 )]dr /dt=- 2C 2 / [(4π )2 r 5 )] or, combining all constants,
kr -5 . I just don't see if the radius of the
sphere is expanding at a rate of Cr -2 , how it is accelerating at kr -5 ?
k is k =- 2C 2 / (4π )2 .
I think that the problem you are having is that you have
carried the minus sign into your constant. But that minus
sign is important because it tells you (since everything
else in k is constant) that the radius of the surface is
increasing, but the rate at which it is increasing is
decreasing as r changes. So, if the volume of
a sphere is increasing with a constant rate, its radius is
increasing like 1/r 2 , but the rate at
which the speed of the surface is decreasing goes
like 1/r 5 .
recent answer
is very similar to yours.
here . For example, for your current weight, a 5K race at
3.1 mph (which is about equal to 5 km/hr, so it takes one
hour for you to complete), I find "You burned 1151 calories
on your 5 kilometer run. At that pace, your calorie burn
rate is 230 calories per mile and 1151 calories per hour."
Sorry, I could not find any information on temperature and
humidity.
v Mv 2 .
where M is the mass of the universe except for your
mass. And just a minute ago that energy was not there.
Obviously, the total energy of that universe cannot be the
same in all frames. But, if you stay in one frame and keep
track of the total energy of the universe in your frame, it
will not change.
g 2 . The acceleration due to gravitation does
vary at various places on the earth. The animation shows you
how it varies. The variation is small, no larger than 50
milligals; a gal is 1 cm/s2 =0.01 m/s2 ,
so 50 milligal=0.005 m/s2 . So, compared to 9.8
m/s2 , the variations are only on the order of 0.05%. There
are several reasons for gravitational variations, notably:
altitude (the farther you are from the
center of the earth, the weaker the gravity);
local geology (large volumes of more
dense earth will increase the gravity);
the earth is
not a perfect sphere.
If you want a lot
more detail, see the
Wikepedia article on the earth's gravity.
N =0.001, N would be the number of
seconds it would take for the intensity of the light to drop
to 0.1% of its original value. Using
Wolfram Alpha , I
find N
f in water is a
quadratic
function of the velocity v , f= ½ρw CD Av 2 , where
ρw = 103
kg/m3 is the density of water, A
is the area presented to the onrushing water as the
object falls, and CD is a constant which depends on the
geometry of the object, called the drag coefficient. For a sphere of radius R
the drag coefficient is given by CD =0.47 . The upward force f is
only one force on the object as it sinks; it also has a
weight down W=mg and a buoyant force up B=Vρw g
where V =4πR 3 /3
is the volume of the sphere. If the sphere is
solid and has a density ρ=m /V , we can
write B=mg (ρw /ρ ).
Also note that we can write R as R =3 m /(4πρ )].
Finally, we can write the total force on a sphere: F =[mg (ρw -ρ )/ρ ]+½ρw CD Av 2 .
Note that when a speed is reached where F =0, the
sphere will fall with a constant velocity called the
terminal velocity, vt =√{[mg (ρ-ρw )/ρ ]/[½ρw CD A ]} .
So, let's do a specific example to be able to answer your
question for a particular situation. Suppose we have two 1
kg balls of solid iron for which ρ= 7.87x103
kg/m3 . Then the radius of each ball will be R =0.0312
m. Doing the arithmetic, F =-8.56+0.72v 2 .
So each of two balls of mass 1 kg, tied together will
accelerate until their speeds would be vt =F =-17.1+1.14v 2 .
So one ball of mass 2 kg will accelerate until
vt =√(17.1/1.14)=3.87 m/s.
You could do more examples to come to the conclusion that
the answer to your question is that, in general, two spheres
tied together will not sink at the same rate as one sphere
with their combined mass. I am sure, by carefully tailoring
the shapes of the objects, you could force the rates be the
same, but that is not you would find in general.
f can be approximated as f ¼Av 2
where A is the area presented to the onrushing air
and v is the speed. (You will note that Av 2
is not the same dimensionally as a force; this approximation
only works if you work in SI units, A in m2 ,
v in m/s and f in N.) So the net force
F on each ball (which I assume to be solid) is F =¼Av 2 -mg
where m is its mass; and, by Newtons second
law, the acceleration a is a =(¼Av 2 /m )-g ;
and there is a velocity U , called the terminal
velocity, at which the drag equals the weight so the
acceleration is zero and U= 2√(mg /A ).
Iron and aluminum have different densities, but it is pretty
straightforward to compute the radii of each
ball and from that compute A for each. I find A Fe =0.014
m2 and A Al =0.0064 m2 .
The accelerations are a Fe =-(9.8-3.5x10-4 v 2 )
and a Al =-(9.8-1.6x10-3 v 2 ).
At any given speed the magnitude of the acceleration for the
iron ball is larger than for the aluminum ball. The terminal
velocities would be U Fe =170 m/s and
U Al =78 m/s. The iron ball always has a
larger speed than the aluminum ball and will hit the ground
first.
It is also enlightening to find the distance traveled as a
function of time, s= (U 2 /g )ln(cosh(gt /U )).
This may beyond your math ability (ln is a natural logarithm
and cosh is a hyperbolic cosine), but the graph shows the
behavior of each ball and clearly the iron ball is falling
faster. Note that for the first few seconds (6 or 7 s) the
balls pretty much keep pace with each other and after about
15 s both have nearly reached their respecitve terminal
velocities because the curves are nearly linear.
When this kind of problem is
presented as an exercise in elementary physics, it is
usually assumed that the density of the earth is uniform
throughout—a cubic meter of earth from the surface has
the same mass as a cubic meter at the center. In this case,
it is easy to show that the acceleration due to gravity,
g (r ), is linear: g (r )=9.8(r /R )
m/s2 where r is the distance from the
center and R is the radius of the earth. Since the
acceleration is always toward the center, you speed up until
you reach the center (not decelerate as you suggest) and
then slow down until you reach the other end, neglecting any
frictional forces. More realistically there will be at least
air drag which will get bigger as you speed up until you
eventually reach a constant speed; then at the center, you
start slowing down until you stop somewhere before the other
end. Then you speed up back to the center getting there with
a smaller speed than the last time you were at the center,
etc. eventually stopping at the center as
k =9.8/R
N/m.
But, the earth is not a uniform
sphere. The core is much denser than the mantle and
therefore the acceleration is not a nice linear function.
The figure shows that the acceleration is almost constant
until about one third of the way to the center. The above
qualitative description of your motion remains the same
except for the similarity to the simple spring.
mg )
decreases linearly to zero at the center. In the realistic
model, the weight remains constant for a while, then increases
slightly, then falls almost linearly to zero at the center.
First, though, let me
address your reference to the ball being "heavier" or
"lighter" depending on the spin on the ball. This is totally
incorrect as the ball always has the same weight, a force
which points vertically downward. The way that the ball
moves results from the forces which the ball experiences
after it leaves your racquet; there is the weight straight
down, the force the ground exerts on it when it hits, and
the forces which the air exerts on it. The figure above
shows a ball with top spin; the top of the ball is spinning
in the same direction as the ball is moving to the right.
(If you are confused, the diagram is shown in the frame of
the ball, so the air is shown like a wind blowing in the
opposite direction as the ball is moving.) Here the ball is
experiencing the force of its weight straight down (not
shown), a drag force to the left slowing it down (not
shown), and an aerodynamic force called the magnus force
F F
For
completeness I would like to talk about what happens when
the ball hits the ground because these are strategically
important in tennis. Shown in the second figure are the
forces (in red) which the ground exerts on the ball in top
and back spin cases. For clarity, I have not shown the
forces due to the weight or the air. In each case there is a
normal force N
f which slows down the spin
in both cases. In the the top spin case, the friction both
slows down the spin and speeds up the ball so it comes up
faster and lower than a nonspinning ball would; in the the
back spin case, the friction both slows down the spin and
slows down the ball so it comes up slower and higher than a
nonspinning ball would. If the spin were large enough, the
back spin ball could actually bounce straight up or even
backwards!
hf , which the photon
had is given to the electron; the electron ends up with a
kinetic energy equal to hf minus the
binding energy it had in the atom. The rest mass of the
electron remains unchanged.
17 kg/m3 , so your volume would be
about 102 /2x1017 =5x10-16 m3 .
So, if you are a cube, your dimensions would be 3 5x10-16 =8x10-6
m ≈10 microns. A flea has a size of about 1
mm=1000 microns, one hundred times bigger than the
compressed you. And, yes, you would still have a mass of 100
kg.
link .
R , the R ω ,
etc . all distract from the crux of the problem, in
my opinion). I have redrawn the problem showing some axially
symmetric object of radius R , mass m rolling down
an incline of angle θ without
slipping. There are only three forces acting on the object,
the frictional force f N ,
mg N
f a F ma
and Στ Iα );
here I is the moment of inertial about the axis
about which torque is calculated and α
F ma
I find N=mg cosθ and ma =mg sinθ
-f. The first equation nails N , one of our
unknowns, the second is one equation for the two remaining
unknowns, so we are not done. So the only place to get a
second equation is the rotational form of Newton's second
law; I think this is the trickiest part of this problem. I find many students
have an inclination to choose the symmetry axis to sum
torques; they have been told, when they learned statics of
non point objects, that you can choose any axis you like,
but this is not a statics problem and there is a very
important constraint—Newton's laws are not valid
in an accelerating system , and the center of the
rolling object is accelerating. (See correction below!) So Στ Iα
yields Iα=Ia /R=mgR sinθ ,
or a= (mR 2 /I )g sinθ ;
knowing a , then f =m (g sinθ -a )
But, what is I ? It isn't what you look up in a book
because usually what is listed is the moment of inertia
about an axis passing through the center of mass (Icm ).
Here the axis is a distance R from the center of
mass, so using the parallel axis theorem, I=Icm +mR 2 .
I will do one example, a solid sphere where Icm =2mR 2 /5,
so I =7mR 2 /5, so a =5g sinθ /7
and f =2mg sinθ /7. If you sum torques
about the center of mass, you get the wrong answer.
OK, I slipped up here, albeit in a fairly minor way. While
it is true that Newton's laws are not correct in an
accelerating frame of reference, there is one exception: the
sum of the torques about an axis passing through the center
of mass is indeed Icm α .
A little detail I had forgotten about classical mechanics.
Let me work that out for my example: Icm α= 2maR /5=fR ,
so, using the result above, f =2ma /5=m (g sinθ-a )
or a =5g sinθ /7. Then it
easily follows that f =2mg sinθ /7.
But you should try to sum torques about an axis halfway
between the point of contact and the center of mass—you
will not get the right answer.
Nevertheless, it is important that your students be careful
in choosing the axis. I have edited the original answer
somewhat.
Most Massive Black Holes . This is the most
up-to-date information I could find.
article I found, recent research indicates that the
outer core rotates with the same speed as the mantle (once a
day) but the inner core's rotational speed is slightly
faster by about 0.3-0.5 degrees per year.
CD
for the three shapes, determined experimentally, are shown
in the figure. The drag force is given by FD = ρv 2 CD A
where ρ is the density of the air, v
is the speed of descent in still air, and A is the
area. Therefore the drag force is least on the circular
parachute so it reaches the ground first. (A proviso is that
the values of the drag coefficient depend on a quantity
called the Reynolds number which depends on the velocity.
However, the values given are for low Reynolds numbers which
are suitable for parachutes.)
paper on special relativity, On the
Electrodynamics of Moving Bodies , states two main
problems in electromagnetic theory at the end of the 19th
century:
The first
paragraph states, essentially, that Maxwell's equations
will not have the same form in a frame of reference
moving with constant velocity relative to a frame where
they are the laws governing electromagnetism if you
transform them using Galilean transformations.*
The
beginning of the second paragraph acknowledges "…the unsuccessful attempts to discover
any motion of the earth relatively to the 'light medium' [luminiferous æther,
e.g. the
Michaelson-Morley experiment]…"
Einstein had a
strongly-held philosophical belief that, for something to be a
true law of physics, it must be the same in all reference frames—regardless
of where you are or how you are moving, the laws of physics for
you are the same as for anybody else in the universe; this is
the principle of relativity . Since he believed
Maxwell's equations to be laws of physics, he concluded that
Galilean relativity was not correct and this led to his
postulate that the speed of light was independent of the motion
of the source or the observer. To my own mind, I believe that
the constancy of the speed of light is not a necessary
postulate, rather a
consequence of the principle of relativity applied to
Maxwell's equations.
*A Galilean transformation is, for
example, observing a car moving 60 mph east relative to the
ground to be moving 100 mph east if you are in a car moving 40
mph west relative to the ground.
M moving with velocity V MV after the
collision or any time during the collision. What will not be
conserved is the kinetic energy MV 2 .
recent answer , you mention the Cosmic Microwave Background and say that it "provides a frame of reference for the whole universe".
It was my understanding that one of the underpinnings of the basic theory of relativity is that there CANNOT be such a universal frame of reference (UFOR). I assume you're not claiming Relativity has been disproven (well, shown to be a special case of something else), so can you explain the difference between the frame of reference meant for relativity and the CMB as you described it?
v v
0 F) of fall would result in the
outside being burnt and the inside being raw. It might be
fun to do a couple of very rough calculations just to get an
order of magnitude feeling for what you are interested in.
I will approximate the turkey to have the properties of
water, in particular that its specific heat is C =4.186
J/gram⋅0 C; so the energy which will raise
the temperature of a M =6000 gm turkey is E=MC ΔT =6000x4.186x56=1.41x106
J. (700 F→1700 F is 210 C→770 C,
hence ΔT =560 C.)
The change in potential energy in falling from a height
h is V=Mgh =6x9.8h= 59h , so if the turkey falls
with constant velocity the potential energy lost goes into
heat. If the heat is 1.41x106 J, the height is
h =1.41x106 /59=24,000 m. The atmospheric
pressure at this altitude is only about 3% what it is at sea
level; so it is clear that, if your turkey is given any
velocity at this altitude, it will not continue moving with
constant velocity. (Whereas, if the pressure were
atmospheric and you gave the turkey a speed equal to the
terminal velocity it would continue moving with that speed
as it fell.) If you dropped it from very far away
(essentially infinitely far), it would arrive at the upper
atmosphere going nearly the escape velocity, 11,200 m/s. At
this speed, which is faster than the reentry speed of the
shuttle which needed special ceramic tiles to shield against
burning up, the turkey would be burned to a crisp, probably
would not reach ground before totally burnt. Although there
would be some speed of entry to the atmosphere where the
turkey would absorb 1.41x106
J, it would almost certainly not be "perfectly cooked".
light isotope of helium, He (one neutron, two protons), and also a gamma
ray. Two of these 3 He now fuse, throwing off two
protons in the process and one alpha particle (4 He,
two protons and two neutrons). The two thrown-off protons
are now free to continue the process, so this chain, called
the proton-proton cycle, is self-sustaining until the star
runs out of hydrogen which ends the life of the star.
For more earth-bound hydrogen fusion, like fusion reactors
or hydrogen bombs, single-step reactions are sought. The
most frequently used is the deuterium-tritium reaction,
involving the fusion of one deuteron and one triton (3 H,
one proton and two neutrons) shown in the other figure.
Most certainly not! The expression for the time t
elapsed in the gravitational field of a nonrotating
spherically symmetric mass M is t=t 0 / √(1- δ )
where δ= 2GM /(Rc 2 )=2aR /c 2 ;
here, R is the distance from the center, t 0
is the time elapsed when R =∞, G
is the universal gravitational constant, M is the
mass of the sphere, and c is the speed of light,
and a is the local acceleration due to gravity. If a=g
and R =6.4x106 m (earth radius), it
is easy to calculate that δ= 1.39x10-9 .
Now, t g t ½g =[√(1- ½δ )]/[√(1- δ )];
using the binomial expansion
(1 +δ )x ≈1+xδ+ …
if δ<< 1, we finally get t ½g ≈tg (1+¼δ )=tg (1+3.5x10-10 ),
quite a bit different from t ½g =2tg !
old
answer and the links it contains. I do not know what
"accelerate wildly" means in this context but, since you
mentioned "bumped", mechanical bumping probably caused
unexpected behavior of the radiometer.
groundrule for "single, concise, well-focused questions". I can tell you that what is learned in elementary physics courses that airplanes get lift from the Bernouli effect of lower pressure due to higher velocity of air on the top of a wing is a relatively small part of how an airplane is able to fly. As you seem to refer to in your question, the main thing is that air leaves the wing with a downward component
of its velocity; this means that the wing exerted a force down on that air which means, via Newton's third law, that the air exerted a force up on the airplane. To quote Wolfgang Langewiesche, author of
Stick and Rudder ,
the private pilot's bible on the art of flying, "The main fact of all heavier-than-air flight is this:
the wing keeps the airplane up by pushing the air down ."
e.g. , another galaxy or the
center of our local cluster of galaxies. However, the
discovery of the cosmic microwave background (CMB) has now
given us a reference frame which seems to define the
reference frame of the whole universe. Very careful
measurements and data analyses from the satellite COBE have
shown that our galaxy moves with a speed of about 580 km/s
relative to the CMB.
So, what can we do to answer your question which is,
essentially, how far does the earth move in 57 yr=1.8x109
s? (Your 5 minute question is essentially impossible to
answer since it depends on the time of year. Over years the
orbital motion averages out to zero.) Relative to the center
of the galaxy, the distance is D =220x1.8x109
km= 4x1011 km=0.013 pc. The center of the galaxy
in that time will move 580x1.8x109 km=1012
km=0.032 pc. But I do not know the relative directions of
these two velocities, the sum of which would be the average
velocity of the solar system over these 57 years; this speed
would be anything between 360 and 800 km/s. However, in my
researching this question I found that the speed of sun
relative to the CMB is about 370 km/s, so the final answer
is that the earth moved about 370x1.8x1011 =6.7x1011
km=0.022 pc. It is interesting to me that, at this time, the
sun and the center of the galaxy are moving in nearly
opposite directions (370 km/s compared to 360 km/s); since
the time for the sun to go all the way around the galaxy is
about 225 million years, in about 110 million years our
speed relative to the CMB will be about 800 km/s.
Keep in mind that I am not an
astronomer/astrophysicist/cosmologist! I have gathered these
data as best I can.
2 )=1.93368x10-5
PSI=1.31579x10-6 atm.
As the conversation progressed, we considered the implications and began to wonder if it was possible to shield against cosmic radiation? To the best of my knowledge, I explained that no substance I know of can block cosmic radiation and that the Earth's magnetic field would have little effect either. The only effects I know of are when they impact atoms in the atmosphere or the earth. If I understand this, if it doesn't hit something fairly dense, it just keeps going, understandable given the space between atoms.
I have read about the gravitational lensing effects on light by massive objects like neutron stars and galaxies that bend light around them. I have also read about the possibility of warping space as a method space travel by changing the shape of space in front of and behind a craft so that it "falls" in a particular direction.
While we were discussing these effects, he looked at me and said something to the effect, "Maybe we could bend space and make them go around?". To me, this was a pretty profound leap in my book. So I decided to ask physicist if he could be right.
Please forgive the length of this post before asking our question. I wanted to provide sufficient context before asking. So, here it is. Is it possible to make cosmic rays go around something, either by gravitational or microwave warping of space?
B.T.W. I really really enjoy your web site. It is awesome!
But, I can speak to your main question and comment on some
of your statements.
Cosmic
radiation refers to a broad range of radiations but
mostly protons and light nuclei; but also it could refer to
neutrinos, electrons, positrons, antiprotons, gamma rays,
x-rays, and others. What is not the case, though, is that "no
substance…block cosmic radiation"; just about any
cosmic radiation can be blocked by a dense solid or liquid.
Neutrinos are the exception and most pass all the way
through the earth without any interactions. You may know
many neutrino experiments are performed in deep abandoned
mines to block out all other radiation. The atmosphere also
is very effective since a single energetic proton will
strike an atom and many other particles, all with less
energy than the original proton, will form a "shower" of
particles. And magnetic fields are not very effective
deflecting the very highest energy charged particles but are
known to be an effective shield for lower energy particles.
Finally, let's address
the question of warping of space as a way to deflect the
rays. The only way to warp space that I know of is to have
an enormous mass nearby. The earth already warps the space
around it which would cause a particle passing close by, but
not on a collision course with earth, to be deflected a tiny
bit (viz. gravity). But putting a star sized object near
enough the earth to "shield" us would obviously be a
catastrophic thing to do! The whole solar system would be
disrupted. I have no idea what "microwave warping of space"
means.
I commend your son for his hypothesizing and you for your
tutoring and encouragement of his curiosity.
and electric fields
earlier answer to understand the answer
here. The graph here is from that earlier answer and shows
(in red) v /c as a function of a 0 t /c
where a 0 is the acceleration measured in
the frame of the object which is accelerating and t
is the time since the object was at rest. Reading off the
graph, you can see that v /c =0.9 when
a 0 t /c= 2. Now, t =30
days=2.6x106 s, so a 0 t /c=a0 (2.6x106 /3x108 )=8.64x10-3 a 0 =2;
solving, a 0 =232 m/s2 =23.6 Gs. I would not
suggest having anybody aboard this object!
m is walking up a
hill with incline θ with a constant
speed of v . His potential energy is E=mgy
where y is his height above the bottom of the hill
and g= 9.8 m/s2 is the acceleration due
to gravity. The rate at which his energy is changing is dE /dt=P =mg (dy /dt )=mgvy =mgv sinθ ;
so this rate (called power) depends only on the vertical,
not the horizontal component of his velocity. Now, you have
made things difficult for me because you have not specified
the angle, but rather the inclination which is actually the
tangent of the angle (rise/run) times 100%. I will call this
inclination the grade G =100tanθ. For
example, let's take a man with mass 100 kg walking with a
speed of 2 m/s up the 10% and 20% inclines you specify. The
angles would be θ 10% = tan-1 (10/100)=5.710
and θ 20% = tan-1 (20/100)=11.310 .
So the power values are P 10% = 100x9.8x2xsin(5.71)=195.0
Watts and P 20% = 100x9.8x2xsin(11.31)= 384.4
Watts. The ratio of these two is 1.97, close to, but not
equal to, 2.0; for small angles the sine and tangent are
approximately linear functions but as the angle (hence
inclination or grade) increases this is no longer a good
approximation. The graph shows the power for grades up to
200% (about 63.40 ); as you can see, the power
increases approximately linearly only up to about a grade of
about 30% which is about 170 .
When I say "trajectory"
I refer to the path from collision point to the ground. The
horizontal distance gone from the point of collision to the
point of landing depends on the angle at which the cat
started; if launched at 450 relative to the
horizontal he will go much farther than if launched at 200 .
Again, unless you witnessed the accident, there is no way to
know the angle.
2 )=0.31;
so if you went to a star 1 light year from earth, it would
only take you 0.31/0.95=0.33 years to get there but the
earth-bound clock would say it took you 1/0.95=1.05 years.
earlier
answer . In the case where v /c =0.95, the graph shows
how the angle at which you see something depends on where it
would be if you were at rest; for example, a star actually
at θ =1200 (300
behind you) would be seen as being about θ' ≈300
in front of you!
R =2GM /c 2 ,
would be about 1 cm. R is the radius within which
nothing can escape. So, if you could get the earth down to
that size, it would then collapse the rest of the way to a
black hole, I presume.
m approaches the end of a thin uniform rod of mass
M and length 2L with velocity v L from where m
exerts the force. Linear momentum and angular momentum are
both conserved but, as we shall see, energy is not
conserved. After the collision the center of mass has a
velocity u ; conserving linear momentum, (M+m )u=mv
or u=v [m /(M+m )]. Before the
collision m brings in an angular momentum mvL .
After the collision, the rod plus mass has angular momentum
(I +mL 2 )ω
where I=M (2L )2 /12=ML 2 /3
is the moment of inertia of a thin rod about its center of
mass. Conserving angular momentum I find ω [m +(M /3)]L 2 =mvL
or
ω=mv /{[m+ (M /3)]L }.
Just for fun I put in some numbers to illustrate what would
happen in a specific situation: m =1 kg, M =3
kg, L =1 m, v =4 m/s. With these I find:
u =1 m/s, ω =2 radians/s=19.1 rpm. You can
also calculate the energy before and after the collision:
E before = ½mv 2 =8
J, E after =½(M+m )u 2 +½([m +(M /3)]L 2 ω 2 =6
J. So, energy was not conserved in this collision, 2 Joules
were lost.
x=mL /(M+m )
toward where m is stuck. It is this center
of mass which moves forward with speed u , and u
is still v [m /(M+m )]. The rotation
is about this center of mass but now the moment of
inertia is different; the net (rod plus mass) moment of
inertia is ML 2 /3+Mx 2 +m (L-x )2 ,
so the corrected value of the angular velocity is
ω=mvL /[ML 2 /3+Mx 2 +m (L-x )2 ].
In the example I did, x =1/4 m, so ω= 16/7 radians/s=21.8 rpm and E after =46/7=6.57
J (1.43 J lost).
earlier answer and the earlier answers referred to in
that answer. The graph here shows the results derived or
cited in that earlier answer. The three graphs all show what
is observed of your motion as seen by the twin on earth. The
red curve is your velocity v relative to c
as seen by the earth-bound twin, v /c =(gt c )/√[1+(gt c 2 ];
the blue curve is your acceleration a g , a /g =[1+(gt c 2 -3/2 ;
the black curve is your position expressed (approximately)
in light years, gx /c 2 =√[1+(gt /c )2 ]-1.
Now, since I have set this up as seen from earth, let me
specify the time you travel as the time of your trip out as
observed from earth: earth-bound twin observes you to arrive
at you destination in two years. Then, as you can read off
the graph, gx /c 2 =√[1+(2)2 ]-1=1.24
ly; the return trip will be the same length of time, so the
earth-bound twin's clock will have advanced 4 years. Light
would take 2.48 yr to travel that distance, so your clock
will have advanced a time somewhere between 2.48 and 4
years. Your maximum speed would have been about v /c =(2)/√[1+(2)2 ]=0.89.
We can get a rough estimate of your clock reading by
approximating your average relative speed as 0.89/2=0.445;
so you would have seen, on average, the distance contracted
by √[1-(0.445)2 ]=0.9 and your time would
have been 0.9x4=3.6 years.
g appears to be
somewhat smaller than it really is, but the amount is very
tiny. This effect depends on your latitude and is zero at
the poles and greatest at the equator. Also because of this
effect, the acceleration is not directly toward the center
of the earth except at the poles and equator.
earlier answer
illuminating.
Kibble balance which is used. Also see
Wikipedia for more detail
I stated that it was incorrect based on the fact that
"made up of" usually means putting two or more things together (aka addition). Net radiated energy is the difference between emitted energy and absorbed energy. I have yet to see my exam paper but I believe that the mark I lost was due to this question on basis that “made up of” means that it is a result of a mathematical relation between the two values and not necessarily addition (in this case subtraction). What is the correct answer in this case and why?
angular
speed of the object in the smaller orbit is twice that
of the object in the larger orbit even though their linear
speeds are the same. You cannot do this with planets or
moons, though, because the speed of a satellite in a
circular orbit depends on the radius.
essay about general relativity, curvature of
spacetime, gravity. Many nonscientists (and scientists as
well) are familiar with and accept as gospel that general
relativity demonstrates that the distortion of spacetime by
the presence of mass is what gravity is; the "cartoon" which
illustrates this is the "bowling ball on a trampoline
example in which a mass causes the distortion of the
two-dimensional space defined by the trampoline surface. The
implication is that gravity is just geometry. On the other
hand, many theorists are struggling to find a successful
theory of quantum gravity and to do that one must treat
gravity as a force. The take-away from this essay, though,
is that curved spacetime is not a unique representation of
what the mathematics which are the framework of the very
successful theory of general relativity mean. The author,
Rodney Brooks, notes that "…you can view gravity as a force field that, like the other force fields in QFT
[quantum field theory], exists in three-dimensional space and evolves in time according to the field equations."
He also goes so far as to say "…shame on those who try to foist and force the four-dimensional concept onto the public as essential to the understanding of relativity theory."
a and a mass M . The forces on
the car are its own weight W W=Mg )
acting at the center of mass a distance d from the
real axle; the normal and frictional forces on the rear
wheels, N 1 and
f 1 ; and the normal and
frictional forces on the front wheels, N 2
and f 2 ; the radius of
the wheels is R and the distance between front and
rear axels is D . I will ignore the rotational
motion of the wheels, that is, I will approximate their
masses as zero. If the front wheels are not lifting up from
the ground, the sum of torques about the rear axel is (f 1 -f 2 )R +N 2 D -Mgd= 0;
Newton's first law for the horizontal and vertical
directions are (f 1 -f 2 )=Ma
and N 1 +N 2 Mg =0.
From these equations it is easy to show that N 2 =M (gd-aR )/D .
This is interesting because it shows that N 2 =0
if a=gd /R ; for larger accelerations, the
front wheels will lift off the ground. (Note that as N 2
approaches zero, f 2 also approaches
zero.) I have ignored the suspension of a real car, so this
lift will not be abrupt; it will happen gradually as the
front springs decompress and the rear springs compress; so
overall the nose goes up but the rear end goes down.
If
it is a front-wheel drive, the directions of the frictional
forces reverse. The corresponding torque equation (this time
about the front axle) is
(f 2 -f 1 )R-N 1 D+Mg (D-d )= 0;
Newton's first law for the horizontal and vertical
directions are (f 2 -f 1 )=Ma
and N 1 +N 2 Mg =0.
From these equations it is easy to show that N 1 =M (aR+g (D-d ))/D .
As a increases in magnitude, N 1
gets bigger and bigger until N 1 =Mg
which necessarily means that N 2 becomes
zero. It is easy to show that this happens when a=gd /R ,,
exactly the same result as for the rear wheel drive.
a ,
so most likely the wheels will slip well before N 2 =0
when f 2 must be zero. For very large
accelerations, like required for drag racers, rear-wheel
drive is imperative.
i.e. all
objects near the surface of the earth have an acceleration
of 9.8 m/s2 . The acceleration (which is what
describes the orbit) is F /m but the force
is MmG /r 2 , so the acceleration
is MG /r 2 .
t . Its
momentum drops from mv (where m is the
mass and v is the speed when it hits) to zero
because an impulse I was delivered to it by the
ground. The impulse may be written as I=F avg Δt .
where F avg is the average
force. Therefore you can find the average acceleration a avg
during the collision, but not the acceleration as a function
of time: a avg =F avg /m =I /(m Δt )=v /Δt.
So you cannot get detailed kinematics about the
collision without measuring how the speed decreases with
time.
here .
F= ¼A (u-v )2
where u is the speed of the wind, v is the
speed of the ball, and A is the cross sectional
area of the ball. Applying Newton's second law, ¼A (u-v )2 =ma =m dv /dt
. This differential equation may be shown to have the
solution t=cv /[u (u-v )] where
c =4m /A and t is the time it takes the
ball to reach the speed v if the wind speed is
u . I find that c =137 s2 /m. Note
that after a very long time the speed will be constant at
the wind speed; we are only interested in the time it takes
to get to 35.8 m/s (80 mph). I have plotted t for
three values of u , 70, 80, and 90 m/s.
Now comes the hard part. We are interested in finding the
value of u for which v will be 35.8 m/s when the ball has
gone a distance x =18.4 m. This means that we need
to write the equation for t as
t=c {dx /dt }/[u (u- {dx /dt })]
and solve that differential equation for t as a
function of x . That is actually doable, but the
solution is so messy that I looked for a simpler way to do it.
This took some trial and error, but I got a pretty good
estimate, I believe. Look at the black line in the graph; it
is not a straight line, but it is pretty close, so I am going
to approximate the acceleration as being constant over this
time interval of 1 s, a2 . Now, an object with constant acceleration
starting from rest goes a distance d = ½at 2
in a time t , so if u =90 m/s=201 mph the ball will have a
speed of 35.8 m/s when d =35.8/2=17.9 m=59 ft. As
far as I am concerned, I have adequately solved this problem—it
takes a wind of approximately 200 mph to accelerate a
baseball from rest to the speed of 80 mph in a distance of
about 60 ft.
groundrules state
that I will not answer questions of this sort. Your first
question is equivalent to the question "why can nothing with
mass travel as fast or faster than light?" which is included
in the FAQ page.
