Online
Etymology Dictionary has the following entry for mass:
"'lump, quantity, size,' late 14c., from Old French masse
'lump, heap, pile; crowd, large amount; ingot, bar' (11c.), and directly from Latin massa
'kneaded dough, lump, that which adheres together like dough,' probably from Greek maza
'barley cake, lump, mass, ball,' related to massein 'to knead," from PIE root *mag-
'to knead, fashion, fit.' Sense extended in English 1580s to
'a large quantity, amount, or number.' Strict sense in physics is from 1704."
I was unable to
determine who first (1704) used the word for the current
physics useage. It is difficult to determine because
Newton's
Principia was written in Latin and translations
would incorporate current physics usage. I believe, but am
not certain that in the first edition (1687) of
Principia he used the Latin gravitas
(heaviness) for both force and mass and that would be sorted
out in the second edition (1713) and in subsequent
translations. Because of linguistics twists and turns, it is
probably a fool's errand trying to give credit to a single
person, sort of like "…who invented TV?"
1 and the mass of large box be m2 . Now imagine that you are in the box, now the combined mass would be m1 +m2 . Now start applying the force on the floor of the box from inside the box. Does the mass of the body vary or its just the weight that varies?
m 1 +m 2 ,
does not change. The weight of the body, (m 1 +m 2 )g ,
does not change either because the weight is defined as the
force which the earth exerts on the body; if the body is
sitting on a scale, this is what the scale reads. If I push
down on the floor with a force F , the floor pushes
back on my hand with a force F (Newton's third
law); since these are both forces on the body, they cancel
out so the scale reads the same. (I have assumed that the
body is not being accelerated.)
T cr )
there is no distinguishing between vapor and liquid. Above
the critical pressure the water is called a
supercritical fluid , below it is called a
supercritical gas .
Contrary to what "many
people" have been telling you, the
sample retains no memory of how it got where it is on the
phase diagram; if you were exactly at the critical point, I
guess that it might be argued that the fractions of liquid
and vapor depend on how you got there, but you have
specified that the water is supercritical, beyond that
point. What does matter, though, is how you reduce the
pressure. You can either reduce the temperature (keeping the
volume constant), increase the volume (keeping the
temperature constant), or some combination of the two. If
you increase the volume while holding the temperature
constant, as indicated by the orange arrow A, you would
still be supercritical but you would perhaps call it
"supercritical steam" instead of "supercritical water". If
you then cooled it, it would become normal water vapor below
T cr . (I have also included an accurate
phase diagram for water.)
F (in
Newtons) is F ¼Av 2
where A (in m2 ) is the area
presented to the wind and v (in m/s) is the speed.
(This only works if you work in SI units; I will convert the
answer to pounds in the end.) So, taking v =70 mph≈30
m/s and A ≈ ½ m2 ,
F ≈¼x½x302 ≈113
N≈25 lb.
earlier answer .
(Key to understanding this is the notion that the speed of
light one measures is independent of how you are moving.) It
would also help you to understand if you read about the "twin
paradox ".
publication I could find, I figure that the flux of
3 He in the solar wind is less than 3000 atoms/cm2 /s.
That is not a very promising source of an isotope!
here
(it is for electrons instead of photons, but the idea is the
same).
V , so V =75.8
mph.
Wikepedia article
on hydrogen states that "Its monatomic form (H) is the most abundant chemical substance in the Universe, constituting roughly 75% of all baryonic mass."
Most of this resides in the interstellar medium, the vast
regions betweens stars and galaxies. Temperatures greater
than 3000 K break the molecules into atoms.
earlier
question . This answer emphasizes that this idea of
photon exchange is a "cartoon" to help visualize the
qualitative role of virtual photons in a quantized field and
should not be taken too seriously or literally. Your
statements regarding the photons "borrowing energy" is
totally unnecessary. No energy is borrowed from anywhere.
When a virtual photon appears, the energy of the system
suddenly increases, violating energy conservation; it turns
out that, because of the Heisenberg uncertainty principle,
it is perfectly ok for this to happen as long as it doesn't
happen for too long a time.
Earth's
electromagnetic field". If you could extract energy from the
air, surely it would have been done by now. I think there
was something like that in Ayn Rand's novel Atlas
Shrugged . but that was purely science fiction.
F R W d F from the front axel and
a distance d R from the rear axel, it may
be shown that F= ½Wd R /(d R +d F ).
So, if d R =d F ,
F=W /4 or if d R = 2d F ,
F=W /3.
By the way,
F and R
here and
here .
faq page.
etc .) and the stickers were to
notify the recipient that this had been done. But do not
worry—the x-rays are long gone and have done no damage
nor have caused any lingering radiation.
earlier answer . The key is to realize
that the center of gravity of the balance itself is below
the pivot point.
question I answered many years ago.
fl owing c ontinuously, but there were waves at roughly equal distanc es, rolling down pretty rapidly on the surfac e.
The water supply was
c ontinuous on the elevated end, so the emergenc e of the waves was not apparent, but the resulting pattern was spec tac ularly regular.
Why do these waves appear? What is their veloc ity? What is the time interval between waves?
17 ft;
so the number of ET to Vega is about 8x1014 ET so
a round trip is about 16x1014 ET. Avagadro's
number is about 6x1023 , so the number of round
trips is about 6x1023 /16x1014 ≈3.8x108 =380,000,000.
Physics Stackexchange and more or less summarizes why
your question is basically unanswerable: "Spinon, orbiton and holon are quasiparticles, they are not actual constituents of the electron. We know about particles by their interactions, and it appears that in condensed matter physics some phenomena are best described as interactions of new kind of particles which are actually arising from complex collective behavior."
An electron is a particle without constituents. However,
sometimes under special conditions where there are very many
electrons all closely packed, the results of their
interactions which involve a very large number of them can
be well described by inventing new particles. These
particles do not really exist, per se , but assuming
that they do gives accurate descriptions of physical
phenomena. As far as I could find, these collective
behaviors are best observed in one dimensional systems. If
you google holon physics you will find places where you can
read about experiments.
c , you are assuming that each would see
the other as having a speed of 1.8c , right? You are
thinking classically and, indeed, that is not the right way
to add velocities when the speeds are not small compared to
c . See an
earlier answer . The correct way to calculate the
relative speed if one particle has a speed u and the other
is moving in the opposite direction with speed v is
v' =(u+v )/[1+(uv /c 2 )].
So, for example, if u=v =0.9c , v'=1.8c /(1+0.92 )=0.9945c .
This is tricky stuff,
and you have some confusion about electric potential which
is common to almost all students when they first learn it. I
prefer to talk about electric potential, leave the term
"voltage" out of it; if you insist on using voltage, it is
the same a the difference in electric potential between two
points in space. Potential is not the same thing as
potential energy, just the same as electric field is not the
same thing as electric force.
The force felt by a
charge Q in an electric field E
F =QE . x
direction. (This very question was dealt with very
recently except for gravitational
forces and fields.) Now, the difference in potential energy
(PE) between two points, x 1 and x 2 ,
is the negative of the work done by the electric
field on a charge Q moving from x 1
to x 2 : ΔU=U final -U initial =U (x 2 )-U (x 1 )=-QE (x 2 -x 1 ).
Now, suppose we choose a coordinate system such that x 1 =0,
x 2 =x , and U (x 1 )=0;
Then U (x )=-QEx . The minus sign
says that as a positive charge moves in the direction of the
field (+x ), the PE gets smaller (just as in the
case of a gravitational field); but as a negative charge
moves in the direction of the field, its PE becomes larger.
So, left to its own devices a positive charge will "fall" in
the direction of the field but a negative charge will "fall"
opposite the direction of the field. So, one reason one
might want to introduce electric potential is to avoid this
ambiguity. We follow the same procedure to define as was
done to define electric field —divide out the
charge: E =F Q
and, now, Δ V= Δ U /Q.
So, you see, you have to be careful how you use the
terms, force, field, potential, and potential energy. Force
is N, field is N/C, potential energy is J, potential is J/C.
Now I will finally
address your question which, if I understand correctly, is
why the electrons in a conducting wire don't just keep
accelerating because their potential energy should be
decreasing all the time. In fact, the electrons are
accelerating all the time but they are trying to make their
way through a forest of atoms and they do not go very far at
all before they bump into an atom and transfer their kinetic
energy to the atom and have to start all over again. So it's
go-stop, go-stop, go-stop, go-stop, etc . And, what
happens to the electron energy losses from all those tiny
collisions? It goes to heating up the wire. Don't you worry—energy
is conserved!
E.g ., if you have have an object which has
some electric charge Q and you turn on an upward
electric field E , the object will float if EQ=W
where W is the weight of the object. In the last 20
years it has been observed that the stars which we can
observe near the outer edge of the universe are not only
moving away from us (that has been known for nearly 100
years now) but they are actually speeding up. If gravity
were the only force which those stars see, this would be
impossible because gravity is an attractive force; therefore
there must some repulsive force pushing out on those distant
start and its origin has been called dark
energy . Nobody really understands what dark energy is,
but in your fiction you could imagine that somebody has
managed to harness dark energy and if you had a dark energy
power generator on your space ship, you could use it to push
your ship away from a planet. So, dark energy could be
thought of as antigravity.
m
experiences at that point in space is proportional to m .
You could write this as an equation, F =mG
where G
G =F /m. F
is a force measured in Newtons (N) and G G F particular mass will
feel, field tells you the force any mass will feel.
PV =constant, so if V becomes
half as big, the pressure becomes twice as big. But, you
have to allow energy to flow out of the bottle in order for
the temperature to remain constant; one way to do this is to
squeeze very slowly so that the temperature of the gas can
equilibrate with the ambient temperature. On the other hand,
if you reduce the volume very quickly or thermally insulate
the bottle very well, it is called an adiabatic compression.
For this situation it may be shown that PV γ =γ=C P /C V
and C P and C V are
the specific heats at constant pressure and volume,
respectively. For a monatomic gas γ= 5/3 and
air is pretty well described by γ= 7/ 5.
So, let's assume your bottle is filled with air. Then P 1 V 1 7/5 =P 2 V 2 7/5
so P 2 /P 1 =27/5 =2.64.
You can also find the temperature because it may be shown
that TV γ -1T 1 V 1 2/5 =T 2 V 2 2/5
so T 2 /T 1 =22/5 =
Q&A OF THE WEEK,
3/24/2018
FOLLOWUP :
I must mention that I modified the design to make the walkway eight feet longer, so that it is now 16 feet long. Also, I had another empty 55-gallon barrel on hand, and placed it lengthwise under the walkway at the end where the walkway joins the cross of the tee. The salient data for the components are as follows:
T-shaped ramp (total weight 648 lb)
Walkway (463 lb) is 3 feet wide and 16 feet long; a 55-gallon drum is placed under the walkway on the lake end.
Cross of tee (185 lb): 5 feet wide and 3 feet long, covering two 55-gallon drums.
The dock platform weighs 515 lb and is 8'x8'. Four 55-gallon drums support it.
My dock will be located in a tributary of the Potomac River. As such, the water is tidal, varying in depth between 2 and 5 feet. (This is the reason I did not join the ramp and the platform rigidly, as I anticipate that the angle of surface of the ramp relative to that of the platform will fluctuate, given that one end of the ramp rests on the
shore.)
In order to help visualize the situation, I am attaching three JPG images which I created in Sketchup. Please note that these images do not include the latest modification, whereby I lengthened the walkway and placed a barrel under its far end. Also, I recognize that you will probably make certain simplifications in order to expedite the analysis.
That is fine with me; I only wish to obtain a rough idea of the limits of motion of the ramp and dock as I walk upon them.
At high tide the walkway and platform are approximately
horizontal. At low tide the water level is about 2-5 feet
down.
ANSWER:
Some preliminaries:
The volume of each
barrel is 55 gallons=7.35 ft3 ;
the density of
water is 64.2 lb/ft3 ;
each barrel is 45%
submerged, so each barrel provides a buoyant force of
7.35x64.2x0.45=212 lb;
each barrel has a
weight of 21.5 lb;
I will first do the
platform which is easiest if you assume that the load is at
the center.
The weight of the
platform is 515 lb;
the weight of the
four barrels is 21.5x4=86 lb;
the weight of the
load will be denoted as W ;
the buoyant force
of the four barrels will be 4x212=848 lb.
Therefore,
W = 848-515-86=247
lb.
If
the load is not in the center, the total buoyant force will
still have to be 848 lb but the platform will tilt so that
two of the barrels will be submerged more than 45% and the
other two less than 45%. I made an estimate of how much the
platform would tilt if the load were moved over to 1 ft
from one side edge. Without going into details, the heavy side
would go down by about 2.8 inches and the other side would
go up by the same distance; the corresponding tilt would be
about θ =4.50. This would make two
of the barrels 65% submerged, beyond your desired limit.
Next I will look at the
walkway.
the weight of the
tee and two barrels under it as well as the net buoyant
force of those barrels act at 17.5 ft from the shore ;
the weight and
buoyant force of the third barrel act at 14.5 ft from
the shore
the maximum weight
W acts at a distance x from the shore;
there is a force
F exerted up by the shore which we will not
need to know;
I have assumed that
the ramp has no interaction with the platform, since
reference is made to "slack ropes".
All this is shown in my
diagram. If one now sums the torques about the point of
shore contact and sets that sum equal to zero, the product
Wx can be solved for.
0=212x14.5+424X17.5-228x17.5-21.5x14.5-Wx -463x8=2488-Wx .
Wx =2488 ft⋅lb.
So, for x =19 ft, the end of the ramp, W =131
lb. I am guessing that this result does not make you happy!
I am not sure how
rigidly coupled the platform and walkway are ("slack ropes"), but suppose
that they are coupled as if, when horizontal, they were
rigidly attached. So now the summed torque equation is
0=212x14.5+424X17.5-228x17.5-21.5x14.5-Wx -463x8+848x23-601x23=8169-Wx .
So W =8169/x . So, for x =19, W =430
lb and for x =27, W =303 lb. It would seem
that it is important that the coupling be designed such that
the platform can help hold up the walkway. I figure that the
walkway will only go down a maximum of about 150
at low tide; this should
not significantly alter the estimates I made for the horizontal
situation. So you
should allow a fairly rigid coupling like some kind of
hinge.
"We know that two spheres of the same volume with
differing mass will fall at a nearly identical speed …"
Sorry, that statement is wrong. It is true, as you state,
that, since they have the same size and shape, the drag
forces will be the same on both. But, that force will have a
much bigger effect on the less massive sphere because it has
much smaller inertia. Adding identical parachutes will still
result in the drag forces being identical, but the more
massive ball+parachute will still fall faster. (You can
understand this intuitively. Imagine a bowling ball and a
balloon the same size. Drop them from a height of 2 m and
surely the bowling ball will hit the floor first.)
To be a bit more
quantitative, the drag force is of the form f D =Cv 2
where C is a constant determined by geometry and
v is the speed the object if falling. As the object
falls from rest, v gets bigger and bigger.
Eventually f D will equal the weight
W and thereafter it will fall at constant speed because
the two forces add to zero: W =Cv 2 ,
so v t =√(C /W )
where v t is called the terminal
velocity.
0 and W± .
Any charged particle feels the electromagnetic field, any
particle made up of quarks feels the strong field, and the
particles called leptons (the electron being one) feel the
weak field. Interestingly, the fourth fundamental field,
gravity, has not been quantized and so, although a graviton
has been hypothesized, there is no successful theory of
quantum gravity.
ρgd where ρ is the density
of the fluid, g ≈10 m/s2 is the
acceleration due to gravity, and d is the diameter
of your inflated cylindrical balloon. For example, if d =1 cm=0.01
m
the pressure on the back side is 0.1 N/m2 =1.5x10-5
psi larger at the front for compared to 100 N/m2 =0.015
psi for water. In other words, the back side of the
esophagus (I assume that is what this is) must support the
weight of the water creating a larger push by the balloon on
the back than the front. The density of the fluid does make
a significant difference; I have no way of judging whether
the difference is enough to make a difference in the efficacy of
your device.
QUESTION OF THE WEEK 3/17/2018
There are four forces acting on the ladder: the weight
W
F which you apply, the force N f F θ
the force necessary to hold it can be determined by summing
torques about the corner of the wall and floor: Στ =0=FD- ½LW cosθ.
So, the force you must exert to hold up the ladder is
F =½W [(L /D )cosθ ].
Now, this is not so simple because as the ladder gets
higher, cosθ gets smaller but L /D
gets bigger. So, we have to make a further assumption about
how the ladder is lifted. Having done this, my recollection
is that I start off lifting the end straight up over my head
and then raise it by walking toward the wall lifting as I
go. So F h above the ground; then sinθ h /D
and
cosθ= √[1-(h /D )2 ].
So, F =½W (L /D )√[1-(h /D )2 ].
So we should put in some numbers, W =300 lb, L =21',
h =7': F =(3150/D )√[1-(7/D )2 ].
The graph illustrates how F varies: starting at
D =21' the force you must apply increases from a little
less than half the total weight until you get about halfway
(10', just as you perceived qualitatively) where it is about
225 lb. For the rest of the way F decreases rapidly
to zero when the ladder is upright. The angle for the
maximum F is about 350 , also about what
you observed.
F , set its derivative equal to zero and solve for
D . dF /dD =LW (2h 2 -D 2 )/[2D 4 h /D )2 )]=0.
The root of interest to us is D=h √2, so for
h =7', D =9.90'. (The other real roots are
at D=-h √2 and D =± ∞.)
+ 18,000)v
where v is the speed of the two after the collision.
Solving, v ≈9 mph. If the collision lasted
some time t , your head experienced an accleration
of about a =(9 mph)/t . To translate that
into a force F , use Newton's second law, F=ma
where m is the mass of your head. Suppose the
collision lasted about 1/20th of a second, t =0.05
s. Then, if you work that out, the average force your head
felt during the collision was about 80 lb, quite enough to
cause a neck injury, I would think.
A large box of mass
M is moving on a horizontal surface at speed
v 0 . A small box of mass m
sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are
µ s and µ k , respectively. Find an expression for the shortest distance
d min in which the large box can stop without the small box slipping.
A pickup truck with a steel bed is carrying a steel
file cabinet. If the truck's speed is 15 m/s, what is the shortest distance in which it can stop without the
file cabinet sliding?
The
first thing to do is to choose a body upon which to
focus your attention; ignore everything else at this
point. I choose to look at m .
Next, draw a diagram and show all the forces acting on
the body you have chosen. In this problem, there are
three forces: the weight of the box, mg N M
exerts upward on m , unknown for now; and the
frictional force f M exerts on
m in the negative direction, also unknown.
Also, choose a coordinate system and resolve the forces;
in this simple example, all the forces are already
trivially resolved: N x =(mg )x =f y =0,
N y =N , (mg )y =-mg.
f x =-f . The acceleration
a is in the negative x -direction,
a x =-a and therefore f x
is in the negative x -direction. (If you are not
sure of the direction of a force, choose it to be in the
positive direction and if you find that the magnitude of
the force is negative, you drew it in the wrong
direction.)
Apply Newton's laws; the box is in equilibrium in the
y -direction (Newton's first law) and has an
acceleration in the x -direction (Newton's
second law): ΣF y =0=N-mg
so N=mg , and ΣF x =ma=f
so a=f/m .
Now, since the box is not slipping, what we know about
f is that f≤μ s N =μ s mg .
Therefore, a≤μs g so the
largest the acceleration can be is a=μs g.
Now apply the kinematic equations of motion which in
this case would be v=v 0 -μs gt
and d min =v 0 t -½μs gt 2 where
t is the time for the box to stop, i.e .
for v =0. Finally, t=v 0 /(μs g )
so d min =½v 0 2 /(μs g ).
M or μ k .
It is very instructive, while we are at it, to look at
the bottom box now. Suppose the coefficient of kinetic
friction between M and the ground is μ k .
Now, what are the forces on M ? Clearly its own weight,
Mg F
N' up from the ground. But M
is also experiencing forces from its contact with m .
Newton's third law tells you that the forces which m
exerts on M are equal and opposite to the forces
M exerts on m ; therefore, there are forces -f μ s mg ) and a force
-N mg ) on
M . So now, going through the same procedure as above, ΣF y =0=N'-Mg-mg
so N' =(M+m )g and, ΣF x =Ma=-Mμs g=μs mg-μ k (M+m )g
so μ k =μ s .
It was a little surprising to me that the kinetic friction
coefficeint between ground and M is the same as the
static friction coefficient between M and m
for maximum acceleration. It is worth going through this to
illustrate Newton's third law.
Here is a little
warning. It is tempting to say that the weight of m
is a force down on M . But this is wrong because
that is a force on m , not on M . If you had
that mistake in this problem, you would have ended up with
the right answer. However, there will be many problems (e.g. ,
blocks on inclines) where making that mistake will cause the
final answers to be incorrect.
link .
Stern-Gerlach experiment with all the "color box" stuff
in the reference. I told him that I could explain the
experiment with just classical E&M and a little QM. If he
cares, I have attached
the appendix from my second book,
Atoms and Photons and Quanta, Oh My!
8
m/s), your clocks will, for all intents and purposes, still
be synchronized and it will take 70 minutes for the light
from his waving to reach you; you will not be surprised to
see his clock reading 70 minutes earlier than yours when you
see the waving. On the other hand, if he is traveling very
rapidly, say 80% the speed of light, he will get there,
according to your clock, 70/0.8=87.5 minutes. But his clock
will not read 87.5 minutes because he sees (length
contraction) the distance to be shorter, 702 )=42
light-minutes, so you will see him wave at 87.5+70=147.5
minutes on your clock; you will be surprised, however, if
your telescope is good enough, to see his clock reading 42
minutes.
ω radians/second. Since it will be rotating about the hinges, you need to know its moment of inertial
I about that axis; if you model the door as a uniform rectangle,
I=ML 2 /3=27x12 /3=9 kg m2 . If it had an angular velocity of
ω =1 rev/s=2π s-1 then its rotation kinetic energy is
K =½Iω 2 =178 J.
ω =v /R= (1.1 m/s)/(1 m)=1.1 s-1 and
therefore,
K =½Iω 2 =½x(9 kg m2 )x(1.1
s-1 )2 =5.45 J. The equation you want is
K =½Iω 2 =½(ML 2 /3)(v /R )2
=M (vL /R )2 /6 where
L is the distance from the hinge to the outer edge of
the door and R is the distance from the hinge to
the point where v is measured. If L=R ,
then K =Mv 2 /6. (Keep in mind
that this (I=ML 2 /3) assumes that the
door's mass is uniformly distributed in the radial direction
which will certainly not be exactly true. It would be most
accurate if you could get a good measurement of the moment
of inertia about its hinges.)
K =½Mv 2 .
To illustrate the first
consideration, I will model the catapult as a simple spring
pushing the mass m (a sphere of aluminum foil) up an
incline. The easiest way to solve this is to use
conservation of energy. The spring is compressed by an
amount s and so has a potential energy of ½ks 2
where k is the spring constant; the ball is at rest
and at a position I define to be y =0, so it has no
energy. At the instant of launch, the ball has a kinetic
energy of ½mv 2 and a potential
energy of mgh . Equating the initial and final
energies and solving for v , I find v =√[(ks 2 /m )-2gh ].
So, you can see, the bigger the mass m , the smaller
the speed at launch. If you ignore all friction, it is clear
that the lighter ball will go farther. However, it is quite
likely that the air drag (friction due to the air) will play
a role.
To illustrate the
second consideration, look at the ball at the instant of
launch. There are two forces on it, its own weight
mg straight down and the air drag
f which points in the direction
opposite the velocity. Now, the magnitude may be
approximated as f ≈-¼Av 2
where A=πR 2 is the area presented to
the onrushing wind and R is the radius of the ball.
So the air drag causes an acceleration opposite the
direction of v which is a =-¼πR2 v 2 /m ;
the negative sign means the ball is slowing down because of
air drag. If the balls had the same velocity, the one with
smaller mass would lose speed faster and therefore go less
far.
So, the first
consideration gives a higher speed to the lighter ball but
the second takes speed away faster from the lighter ball. My
hunch is that the air drag will be more important and the
more massive ball will go the farthest; my reason is that
the catapult force mechanism ("spring" in my simplified
model) also will have to accelerate the bucket and arm, so
the difference in the velocities leaving the machine will be
quite small. I am quite sure that the heavier ball will go
farther.
8 m/s) with the acceleration g ? how
far will it travel during this interval and what fraction of a light year is
the answer to the previous question?
Reddit .
cheatatmathhomework ! And if you can
read you see that AskThePhysicist.com does not do homework
so I should just trash this question. However, this question
was submitted to AskReddit a year ago so the present
questioner is presumably interested in the answer since it
is marked UNSOLVED . But,
the problem is not unsolved because the one comment points
the reader to a site where the one-dimensional, constant
acceleration kinematics is discussed; this is the
appropriate way to help someone do their homework, not
working it out to the final answer. The solution to the
problem is a simple application of these equations. So here
is my answer:
You have not accurately conveyed the problem. Most important, the final speed is not 3x108 m/s but rather 1% of that number. But that is a small enough fraction of the speed of light that you can do the problem classically, not relativistically
(which would be much more difficult). The problem is not unsolved since the single answer points you to the appropriate equations to use, Specifically,
x ≈½gt 2 and v≈gt . Here,
g =9.8 m/s2 , v =3x106
m/s. Do the algebra to solve for the two unknowns, convert
t from seconds to days and convert x from
meters to light years.
I worked out the answers and
found that the problem, as stated, does not have the correct
answer for the distance traveled. The correct answer is
4.6x1011 m=4.9x10-5 ly.
v of the space ship relative to
someone on the earth. The time t' according to her
clock elapsed when she returns to earth after a time t
elapsed on earth is t'=t v /c )2 ]
where c =3x108 m/s is the speed of light.
You mentioned a space station. The ISS has a speed of about
4.76 mi/s=7660 m/s, so if she is up there for 80% of 250 y,
200 y as measured on earth, her clock will read, upon
returning to earth,
t'= 200 √(1-(7.66x103 /3x108 )2 )=199.9999999348
y, 0.652x10-7 y=2.06 s younger than she would
have been if not traveling. On the other hand, if she were
on a spaceship going half the speed of light,
t'= 200 √0.75= 173
y, 27 years younger.
E=m 0 c 2 /v /c )2 ]
where v is the speed of the particle and m 0
is the mass it has when it is at rest. So, the faster you
get it going, the more energy it will have. In your "scheme"
you start out with a particle-antiparticle pair at rest, say
an electron and a positron which have an energy of 2m 0 c 2
where m 0 is the rest mass of an
electron. The energy they create is a pair of photons with
no mass but they still have between them a total energy of 2m 0 c 2
and, in fact each will have energy m 0 c 2
because they must come off in opposite directions with equal
energy to conserve momentum and energy. So getting them back
together will be a trick which could probably achieved with
mirrors. But, I know no way that these two photons could be
made to coalesce into an electron/positron pair but if that
could be achieved, they would still have the total energy of
2m 0 c 2 and be at
rest, so you have gained nothing. Why is it that you think
this would be dangerous? The total rest energy which gets
converted to photons is really tiny and while the photons
are of X-ray energies, there are only two of them and one
X-ray photon is not the least bit dangerous.
18 s≈3x1011 yr. So you really need to say how
long you want it to take to give the ship this energy, that
is, what power do you need to supply? For example, there are
about 3x107 in a year, so if you want to achieve
crusing speed in a year, the power required would be (5x1018
J)/(3x107 s)=1.67x1011 W=167 GW. The
largest non-hydroelectric power plant in the world, the
Kashiwazaki-Kariwa Nuclear Power Plant
in Japan, generates about 8 GW, so the notion of this ship
achieving the speed you want in a reasonable time is, for
all intents and purposes, impossible.
g force it will never
feel like a 10g force regardless of how fast you
are going or how long the force is applied. There is one
simple example I can give where you are going at a constant
speed (not velocity) but experiencing a force.
Shown in the figure is a popular carnival ride where you
(the guy in the striped shirt) is moving in a circle of
radius R with a speed v because the whole
thing is spinning like a wheel. You feel like there is a
force F pushing you outward; this is called the
centrifugal force but what is really happening is that the
wheel is pushing you toward the center (centripetal force).
The magnitude of your acceleration is a=v 2 /R .
If you want to express the force in g s, the force
is a /g where g =9.8 m/s2
is the acceleration due to gravity. Suppose R =4 m.
Then if a =3.7x9.8, v =g . Similarly,
if a =10x9.8, v = √(10x9.8x4)=19.8
m/s and you feel a force of 10g . (You can express
v instead as revolutions per second, rps, by
f=v /(2 πR ) so f 3.7 =0.48
rps and f 10 =0.79 rps.)
My question is—how would a radio signal behave if this ship pointed a transmitter at the earth when it started this journey and broadcast a constant radio signal as it grew closer and closer to light speed traveling further away?
c is not
really that big for such a long distance. If you go at that
speed, the distance (as seen by the ship) would be 13.9x109 √(1-0.99992 )=2x108
ly. Since the speed of the ship is 0.9999 ly/yr, the time
necessary would be 2x108 /0.9999≈2x108
yr, not 30. I have seen this kind of estimate before and the
situation being described was probably that the ship has a
constant acceleration (as seen by the ship) equal to the
acceleration of gravity (about 10 m/s2 )
essentially forever; so the speed (as seen from earth) gets
to be much larger than 0.9999c . You can see the
calculation of the speed of accelerating objects as seen
from an inertial frame in an
earlier answer . If you want a really detailed discussion
of getting to the edge of the universe in a few decades, see
this post by
Dave Goldberg ; his number is more like 99.9999999999999999998% of
c !
But your question does
not depend on the details of how far you went and how long
it took you to get there. The radio signals being received
on earth will be Doppler shifted to lower frequency, the
shift getting larger as the speed gets larger. This is the
same idea as the Doppler shift of a siren going away from
you getting changed to a lower pitch. The expression for the
shift in frequency is f observed =f source √[(1-(v /c ))/(1+(v /c ))]=√[(1-0.9999/(1+0.9999)]=0.0007f source ,
a much lower frequency.
v . You look at a rear tire of the car in
front of you and what you see is the tire not moving either
toward or away from you but spinning with a rotational
velocity where each point on the circumference is moving
with a velocity v
Newton's Cradle .
dQ /dt =(kA /s )(T high -T )=3.17x103 (290-T )
where k =0.47
W/(m·K) is the thermal conductivity of
high density polyethylene, s =0.25"=0.00635
m is the thickness of the barrier, A =43.15 m2
is the area of this tank, T is the temperature of
the water as it warms, and T high =290 K.
dQ /dt is the rate at which energy (heat)
enters the tank.
The equation involving
the rate of increase of the temperature of water to which
heat is being added at a rate dQ /dt is
dQ /dt =mC (dT /dt )=8.51x107 (dT /dt )
where m =20.5x103
kg is the mass of the water and C =4.19x103
J/(kg·K) is the specific heat of water.
Combining the two
equations, I find [dT /(290-T )]=3.73x10-5 dt
. Integrating and solving for T I find T =290-8·exp(-3.73x10-5 t )=290-8·exp(-0.134t );
the first expression is for t in seconds, the
second for t in hours, and the temperature is in K.
The final result is shown in the graph converting the Kelvin
temperatures to 0 C. Note that the temperature
never (theoretically) reaches 170 C, but a day to
a day and a half would probably be fine for your purposes.
(By the way, the volume which I calculated and used was 20.5
m3 , a tad larger than the volume you quoted, 18.9
m3 .)
x ) is standard alternative notation for ex x ) is the natural log,
ln(x ) in more common notation, and x =eln(x ) .
So, if you have an equation of the form ln(x )=y
and you want to solve for x , x =ey T .
You also need to remember the properties of logarithms,
viz ., ln(a )-ln(b )=ln(a /b )
and ln(a )=-ln(1/a ).
sic )
the object
way less
than the energy consumed by the fans.
f changes with θ .
You are probably thinking that f= μN=μmg cosθ ,
but this is static friction and the correct expression is
that f ≤μN , so in this problem
f is whatever it has to be in order to satisfy Newton's
laws, as long as it does not exceed μN , in which
case the wheel will not roll without slipping. This problem
has three unknowns, so you must generate three equations:
-f+mg sinθ=ma
N-mg cosθ= 0
fR=Iα=Ia /R=mRa
To simplify the algebra I have assumed that the wheel is a
hoop of radius R , so the moment of inertia is
I=mR 2 . So, from equations 1 and 3,
f=ma=m (g sinθ-ma )
or a =g sinθ
and f =½mg sinθ. You
can see that f increases with θ. The
third unknown is easily found, N=mg cosθ.
You can also calculate
the largest angle for which the hoop can roll without
slipping since the maximum static friction you can get is
f max =μN=μmgcos θ max =½mg sinθ max .
Solving, θ max -1 (2μ ).
For example, if μ= 0.5, θ max =450 .
here
and
here on the faq page) you will understand this better.
Often the mass of a moving object is talked of as increasing
with speed because mass is a measure of how much inertia
(resistance to acceleration) an object has and it certainly
gets harder and harder to accelerate an object as it gets
closer and closer to the speed of light. But, questioner #2,
there is no such thing as "real mass" because if you think
of mass as inertia, it is something which we say is not an
invariant quantity, it depends on the motion of the
particle. And, questioner #1, the only thing that matters as
to whether a star becomes a black hole is the mass it has in
its own rest frame , in other words its rest mass.
d before stopping would
be about 25,000/d lb if d is measured in
inches; Newton's third law says that the head would
experience an equal and opposite force. So the average force
for the head moving 3 inches would be about 8,000 lb.
Information I could find indicated that 1000 lb should
be adequate to break a neck but that the actual force would
depend on circumstances —the individual, where
the force was applied and its direction, and whether the
neck was initially bent or not. I think that 8,000 would do
the trick for almost all situations.
F=Mg where M is the
mass of the earth; who is exerting that force? Your analysis
assumes (I presume) that you will also see the earth
accelerating away from you with a=g but, you will
see the acceleration getting smaller and smaller as time
goes on. The graph above shows the velocity and acceleration
which you would observe. This comes from an
earlier answer which, along with links to even earlier
answers, fully describes how to handle acceleration in
special relativity. In this graph, a 0 =g .
Using the result from the earlier answer, I find that the
speed you would measure after 6000 years is given by v /c =0.999999974,
very close to but not larger than c ; the folks on
the "earth" would all the time see a constant acceleration
of g . If you are actually going to discuss this
with a flat earther, ask her where the force providing the
acceleration is going to come from? God?
R 8
m and mass of about M≈ 2x1030 kg
and no atmosphere. In that case the gravitational
acceleration would be g≈GM /R 2 ≈272
m/s2 . Then t ≈ √(2h /g )≈3.4
s; I used h =1 mi≈1600 m and G =6.67x10-11
N ·m2 /kg2 .
earlier answer .
g is usually used to denote the
magnitude of the acceleration of gravity at the surface of
the earth. Since the magnitude of a vector is a
positive-definite number, g is then positive. If we
call the vertical direction the y-axis, then the y- component
of the acceleration of an object in free fall is ay =-g
if +y is up or ay =g if +y is
down.
Q&A OF THE WEEK,
4/21/2018
A rod (blue) of length 2m and mass of 1kg can freely pivot about it's center (orange dot) which is connected to a frictionless track (purple) running in the Y direction. The rod is positioned at 45 degrees to the track, and a ball of mass 1kg travelling at 15 m/sec strikes the rod at 0.5m from the rod's center. Collision is perfectly elastic, friction is zero. How can I determine the final velocity of the rod along the track, and what is it's rotational speed? And more importantly, I need to know what effect the rod angle has on these two answers. Then I can solve and plot a chart for all angles from 0 to 90 and compare with my Inventor results. I'm really having trouble with the moment of inertia part, being that it's not struck in the center, or the end.
y direction (not conserved in the
x
direction) demands that the speed of the center of mass (COM) of the rod
along the rail would be 15 m/s meaning it would not be rotating. And, the
angular momentum relative to the COM of the incoming ball is equal and
opposite that of the outgoing ball, so the rod would have to be rotating to
conserve angular momentum. So I thought to simply redraw the picture but
with the final ball velocity having unknown components (2 unknowns). Two
other unknowns are the final speed of the COM and the angular velocity of
the rod about the COM. However, I only can see three equations: conservation
of energy, linear momentum (only y ), and angular momentum. I am still
pondering the question, but there is too little information or I am missing
something.
m 1 =m 2 =m
as soon as I have written the most general solutions to
simplify the algebra after setting up the problem. As noted
above, the problem begins with three conservation equations:
conservation of
linear momentum in the y -direction:m 1 v 1 =m 1 v 2y +m 2 u 2
conservation of
energy:m 1 v 1 2 =½m 1 v 2 2 +½m 2 u 2 2 +½Iω 2
conservation of
angular momentum: m 1 v 1 S sinθ =Iω+m 1 v 2y S sinθ-m 1 v 2x S cosθ
The moment of inertia
of a thin rod about its COM of mass m and length
L is I=mL 2 /12. Therefore, if
m 1 =m 2 =m ,
the three conservation equations are:
v 1 =v 2y +u 2
v 1 2 =v 2 2 +u 2 2 +L 2 ω 2 /12
v 1 S sinθ =L 2 ω /12+v 2y S sinθ-v 2x S cosθ
v 2x ,
v 2y , u 2 , and ω.
To generate a fourth equation, consider the impulse
J delivered by the rod to the ball:
J = Δp =mv 2 -v 1 )or
J x =mv 2x and J y =mv 2y -mv 1 .
But, the impulse must be normal to the surface of the rod,
so tanθ=|J y /J x |=(v 1 -v 2y )/v 2x .
The fourth equation is
v 2x =(v 1 -v 2y )cotθ .
I first solved these
four equations for θ= 450 , L =2
m, S =0.5 m, and v 1 =15 m/s, the
conditions specified by the questioner; note that, because
it is a quadratic equation we are solving, we get two
solutions:
v 2y =(15,
8.33) m/s
v 2x =u 2 =(0,
6.67) m/s
ω =(0,
14.15) s-1 =(0, 2.25) rev/s.
The meaning of the
first (trivial) solution is that there was no interaction
between the rod and the ball (you missed!). Another
situation for which we can check the solution intuitively is θ= 900 ,
for which v 2x must be zero:
v 2y =(15,
4.1) m/s
v 2x =(0,
0) m/s
u 2 =(0,
10.9) m/s
ω =(0,
16.35) s-1 =(0, 2.6) rev/s.
Finally, the general
solutions are:
v 2y =v 1 (A -1)/(A +1)
where A =cot2 θ +1+[12S 2 /(L 2 sin2 θ )]
v 2x =(v 1 -v 2y )cotθ= 2v 1 cotθ /(A +1)
u 2 =(v 1 -v 2y )=2v 1 /(A +1)
ω =[12(v 1 -v 2y )S /(L 2 sinθ )]=[24v 1 S /((L 2 sinθ )(A +1))].
The plots requested by
the writer are shown to the right.
v 2
is to specify its magnitude, v 2 , and the
angle φ it makes with the +x
axis:
Physics Forums for helping me realize how to get the
desperately sought fourth equation!
m and speed v has a kinetic energy
classically of K =mv 2 ;
but if you are running with speed v along with it,
it has zero kinetic energy. In special relativity, the
kinetic energy relative to you is
K=m 0 c 2 [(1/ √(1-v 2 /c 2 )-1]
where v is its speed relative to you; if you start
moving, the kinetic energy of the object changes relative to
you.
q 1 and q 2 separated by a
distance r in a medium with
permittivity ε= ε r ε 0
is F =q 1 q 2 /(4πεr 2 ).
The answer would depend on what kind of oil it is.
K=m 0 c 2 [(1/v 2 /c 2 )-1]
where c =3x108 m/s is the speed of light. For
v /c =½
and
m 0 =1
kg, K =1.39x1016 J. K for a 5 kg would
just be five times as large.
Etc .,
etc .
RunnersConnect blog. Or just google treadmill vs.
track .
Let's get straight what all these
units are.
A pound (lb) is a
unit of force. A net force on an object accelerates it.
For example, the weight of an object is a force on it
downward and if you exert an equal force upward the
object will move upward (or downward) with a constant
velocity, any velocity.
A foot (ft) is a unit of length.
A foot-pound (ft-lb) is a unit of energy. If one lb is
exerted over a distance of one ft, the energy delivered
is 1 ft-lb.
A second (s) is a unit of time.
1 ft-lb/s is a unit of power and it measures the rate at
which energy is being delivered. Other units of power
are the horsepower (hp) and the Watt (W). 1 hp is 745.7
W.
Your question about how much hp it would to take to move an
object up 1 ft starting and ending at rest is meaningless
because there are an infinite number of ways to do that but
you could not do it with a constant power. For example you
could deliver a lot of power for 0.1 s, ending up with a
speed (at 0.1 s) which would be just right to get the object up
to 1 ft in an additional 0.9 s before falling back down; or
you could do the same thing with a smaller power to 0.2 s,
etc . Regardless of
how you achieve this for a 550 lb object, the total energy
required is 550 ft-lb. So, if it takes 1 s, the
average power is 550 hp.
earlier answer .
earlier answer
to a question very similar to yours except the dice were
separated by 10 cm there. If you want to calculate the time
your dice would take to come together, you can follow the
calculation given there changing the numbers. I find the
time to be surprisingly small.
0 relative to the
truck.
Now,
if the rollers are attached to the truck bed, when the truck
turns the rollers turn and the wheel, trying to not turn,
will come off the rollers at some point. We first need to
understand the physics relationship between the torque and
the angular momentum of the wheel. The rotational form of
Newton's second law is τ L t ,
torque equals the time rate of change of the angular
momentum. The first figure shows the wheel, as seen from
above, turning through some small angle which results in a
change of angular momentum from L 1
to L 2 and ΔL =L 2 -L 1 .
So the torque which you must apply to make it turn this
way is in the direction of ΔL .
F
gyroscope in a suitcase video .
If you want the wheel
to turn with the truck, you need to have a torque which
causes that. One way I thought of to achieve this was to
have strings attached from the axles on each side and the
truck bed below. These need to have no tension on them when
the truck is going straight. When the truck turns as
indicated, the tension in the string on the right-side
string will be bigger than the other side and there will
therefore be a net force N
Wikipedia
or
Sky and Telescope they show such pictures.
But in this picture, only distance between A to B is known. Distance between A to Eros or B to Eros is unknown. Angle of A and B are unknown. So we can't know angle of p with so little information.
What we see only is a star on the sky do change its position (like you see right side of the picture) slightly (some millimeters) every 6 months.
How do astronomers determine from that motion the parallax angle of the star?
p ;
and if you know p and the distance AB you can also
calculate the distances from A and B to Eros.
e.g ., the earth rotates about the axis which passes
through the north and south poles with a period of 24 hours.
Circular motion refers to the motion of an object which
moves in a circle; e.g. , a race car on a circular
track is in circular motion. If the circular motion is with
constant speed it is called uniform circular motion;
e.g. a point on the equator moves in a circle whose
radius is the radius of the earth with constant speed. More
general than circular motion is revolution; e.g. ,
the earth revolves around the sun in an elliptical path once
a year.
here .) The ball is not a sphere but rather an oblate
spheroid which makes it sort of like a door knob but not so
extremely flattened; but it is slightly more flattened on
one side of the ball than on the other which results in a
center of gravity being displaced to one side of the
equatorial plane as shown in figure (a). This results in a
tendency for the ball to curve left if it is rolling the
angular velocity shown in the figures; this motion is the
"bias" referred to by the questioner which I am to ignore.
When rolling in the x direction (figure (b)), there
is a frictional drag force called, rolling friction
D , which opposes the motion (v )
and eventually brings the rolling to a halt. If there is a
wind, there is a force W W>D ; if this is not the case, there will only
be static friction in the y direction which will be
equal and opposite to W m =1.5 kg and a
radius of about R =6 cm=0.06 m.
To get the equations of
motion for the x and y motions, we first
need expressions for D and W . The rolling
friction may be expressed as D=-μmg where μ
is the coefficient of rolling friction and mg is
the weight of the ball. The force due to the wind may be
approximated as W ≈¼AV 2
where A=πR 2 is the cross sectional
area of the ball and V is the speed of the wind;
this approximation is only correct if SI units are used. The
equations of motion in the x -direction are
D=-μmg=max
⇒ ax =-μg
vx =v 0 -μgt
x=v 0 t- ½μgt 2 .
Here t is the
time and v 0 is the speed of the ball at
t =0. If the ball is rolling in the y-direction
because of the wind, the equations of motion are:
It should be noted that
if (¼AV2 /m )<μg , these
equations imply that the ball will accelerate opposite the
direction of the wind, obviously not correct; hence the wind
will have no effect on the ball if V <√(4μmg /A ).
In that case, ay =vy =y =0.
So, having found the
general solutions, let us now apply the solutions to the
specific case from the questioner. We are told that when
t =12 s, vx =0 and x =26 m.
With that information you can solve the x -equations
to get v 0 =4.32 m/s and μ= 0.037,
reasonable values compared to numbers in the
article I read. The area is 3.14x0.062 =0.0113
m2 . The first question we should ask is what is
the minimum speed of the wind to have any effect at all:
V min =√(4x0.037x1.5x9.8/0.0113)=13.9
m/s=31 mph=50 km/hr; this is a pretty stiff wind, so the
wind probably has no effect on bowling under normal
conditions. So, just to complete the problem, consider V =15
m/s=34 mph=54 km/hr.
The
trajectory during the 12 seconds is shown in the graph
below; after 12 seconds the ball will continue accelerating
in the y direction.
So the bottom line is that unless you are playing in a gale-force wind, the wind has no effect on the ball if the wind has no component along the original direction of the ball (which I have called the
x -axis). You can tell if wind makes a difference by simply setting the ball on the ground—unless the wind blows the ball away, you need not worry about its effect. If the wind is blowing in the +x or -x direction, that is a whole different thing, but
the questioner asked for the sideways deflection.
This question continues
to intrigue me and I have carried my investigation further.
The question originally stipulated "over and above the bias
deflection" so my whole discussion totally ignored the fact
that the ball, owing to its off-center center of mass, will
curve. At the very end of my answer I noted that if the wind
is not perpendicular to the path of the ball, it would
be a different story; indeed for a spherically symmetric ball I showed that, except for very strong winds,
a wind perpendicular to the path has no effect at all.
However, for an actual lawn bowls ball, the path curves to
where a wind in the y -direction might have a
significant component along the path. I have calculated
(graphed below) the
x and y positions of a realistic path with no
wind using equations (10) and (11) of the
article referred to above. To do these I used all the
numbers used above (R , m , A ,
v 0 ,
μ ); I used the moment of inertia
for a solid sphere ( I 0 =I cm +mR 2 =(7/5mR 2 ))
and chose the COM off-center distance to be d =1 mm. As you can see, the curving is
substantial, carrying the ball about 4 m from its original
direction. You can see that now a wind of any magnitude can
have an effect on the trajectory. The angle φ
which the tangent to the trajectory is given in the article
as φ =(2/p )ln(v 0 /(v 0 -μgt ))
where p is a constant also given in the article. As
can be seen, once the trajectory leaves the x-axis the wind
contributes with the component of its force along the
trajectory; this has the effect of reducing the effect of
the frictional force causing the ball to slow down less
rapidly. However, this is now like having a time dependent
force of friction which, I believe, will lead to equations
of motion which will not have an analytical solution but
would have to be solved numerically.
th as much mass as she does, she recoils with
1/20th the speed the ball has. The rocket is not
"pushing against" anything.
relative to the earth , then its velocity
relative to the sun is 92,000 mph if the velocity of
the earth relative to the sun is 67,000 mph.
e.g . a bowling ball begins by approximately not
rotating but may end up rolling without slipping), or is at
rest but spinning (like "peeling out"); in both cases, the
problem is usually to find the time elapsed or distance
traveled before rolling sets in. In your case, the object is
initially both translating horizontally and rotating with a
backspin. What will happen depends on the initial conditions
m , radius R and shape, and the coefficient of
kinetic friction μ . Three possibilities are that
it will reverse directions and come
back still spinning and slipping, eventually stop
slipping and roll (what I think you are remembering),
it will stop slipping and continue rolling in the
initial direction, or
it will stop dead.
In the figure there are three forces on the object: its
weight mg N f v 0 and angular
velocity ω 0 . The
mass of the object is m and its radius is R .
The friction slows down both the velocity and the angular
velocity. Sliding ceases when v=-Rω (see
footnote*). If v 0 is very small and ω 0
is very large, possibility #1 will happen; if v 0
is very large and ω 0 is very
small, possibility #2 will happen; if v 0
and ω 0 are just right, possibility
#3 will happen. I think that this gives you a good
qualitative overview of the physics of the problem.
For those interested in
the quantitative solution, I will give it here for a uniform
solid cylinder. For the translational motion the two
equations (taking +x as to the right, +y
up) are -f=ma and N-mg =0; since f=μN=μmg
we can write a=-μg . For rotational motion about
the center of mass (choosing the direction it is initially
spinning as positive), only the frictional force exerts a
torque, so -fR =-μmgR=Iα where
I is the moment of inertia about the center of mass
and α is the angular acceleration. For a
cylinder I =½mR 2 , so α=- 2μg /R.
So, we can write the equations for the velocity and
angular velocity as functions of time t : v=v 0 +at =v 0 -μgt
and ω=ω 0 +αt=ω0 -2μgt /R.
Now, the time when slipping ceases will be when
v=-Rω ; solving, t =(v 0 +Rω 0 )/(3μg ).
Finally, put this into the equation for v to get
v (t )=(v 0 -Rω 0 )/3.
So if Rω 0 >v 0 ,
possibility #1 will happen; if Rω 0 <v 0 ,
possibility #2 will happen; and if Rω 0 =v 0 ,
possibility #3 will happen.
*The condition v=-Rω
is a little tricky. The negative sign is because of my
choice of the positive direction for ω which
is opposite that which would be the case for no slipping.
r 2 .
If you are inside a sphere which has its mass uniformly
distributed inside, the force you would feel is proportional
to r . If you are inside a sphere which has its mass
distributed so that the density varies linearly (e.g. ,
the density at the surface is twice as large as the density
halfway to the center), the force you would feel is
proportional to r 2 .
the energy at impact will be 44.1 J.
But how can I translate this in g as it is the value reported by accelerometers?
In other words, if the person has an helmet with an accelerometer, which value in g the instrument will register?
mgy =7.5x0.6x9.8),
but that does not tell you the average force exerted by the
mass as it came (accelerated) to a stop. The speed which the
mass has when it hits is v =√(2gy )=3.43
m/s; if it stops in a time t , then the average
force over that time is F=mv /t =25.7 N.
For example, if t =0.1 s, F =257 N. To
convert this to g-force, the weight of the object is
7.5x9.8=71.3 N so the force in gs is F =257/71.3=3.6
gs.
Principia . There is a nice animation you can play
with
here .
what a scale reads . I do not know a single
physicist who likes this definition. Historically and almost
universally weight means the force which the earth (or other
massive object) exerts. When something is dropped it
accelerates downward because the weight is a force pointing
vertically down. Therefore, your question is wrong because
"there is no weight while falling" is incorrect
Nevertheless, it is clear that the astronauts seem to have
zero net force on them. The reason is that they are in "free
fall" when they are in orbit and are therefore accelerating.
There is a legend about Albert Einstein related to this. The
legend is that Einstein saw a painter fall from a ladder and
he said to himself "there is no experiment which he could do
which would allow him to say that he is not at rest in empty
space". This is called the equivalence principle and forms
one of the keystones of the theory of general relativity.
Imagine a cart moving on a frictionless surface with a cannon aimed opposite the direction of travel. At 40 km/hr the cannon fires a projectile accelerating the cart-cannon to 50 km/hr. A second projectile (identical to the first) is then fired accelerating the cart to 60 km/hr. The impulse (and thus recoil) against the cart should be the same no matter what the moving velocity of the cart was. After all, the cart is only moving within my frame of reference. If I selected a frame of reference in which the cart was stationary, an impulse acceleration of 10 km/hr would be the same irregardless of whether in another frame of reference the cart was initially moving at either 40 or 50 km/hr. It's true that the second firing of the cannon would be acting on slightly less mass, after the first cannon ball was no longer part of the system—but that would seem a trivial difference. I also understand that the momentum of the total system is the same before and after the cannon is fired.
Does it take more energy to accelerate an object from 40 km/hr to 50 km/hr than it takes the same object to accelerate from 50 km/hr to 60 km/hr ? I know that the Kinetic Energy at 60 km/hr is 60/50 squared or 1.44 times greater than at 50 km/hr so the answer is undoubtedly, yes. What am I missing about Kinetic Energy - does it depend on your frame of reference?
I have rearranged
your question so we can talk about one question at a
time. Your first question, not really a question,
basically says that, although the momentum of any object
is dependent on your frame of reference, that the change
of momentum if you do something to it (impulse) is not
frame dependent. Why is that? The reason is that
Newton's second law may be written as F p t
and the rate of change of momentum must be frame
independent because we know that the force is frame
independent.
Now, when you change the momentum you also change the
kinetic energy because either the speed or the mass is
changing. But does everyone see the same change in
kinetic energy? The answer is no. And your example shows
very clearly that the change in kinetic energy does
depend on the frame of reference. A more general
example, calculating the work done by a constant force
as seen in different frames, can be found in an
earlier answer .
Q&A OF THE WEEK,
4/7/2018
T TH )
is wasted. Your gut feeling is right, "…it is mostly
pulling backwards…" Anything which you can do to
increase the vertical part (TV ) will
make the lift easier, and moving the winch up is a good way
to do this but it might be easier to have a pulley
higher up which then brings the strap back down to the
winch.
The figure shows all
the forces on the plow assembly: the weight W T TV ) and horizontal pieces (TH )), and the force the
truck exerts on the support which I have represented as its
vertical and horizontal parts, V H T has
to be much larger than W to lift the plow.) Suppose T V+TV -W =0
and H-TH =0. The sum or torques must also
add to zero; summing torques about the point of attachment
to the truck (light blue X), WD+TH s-TV d =0.
Note that the TV is trying to lift the
plow but TH is trying to push it down.
Now, in order to get a final answer for the unknowns (which
are T , V , and H ) we note that
TV =T sinθ and TH =T cosθ
where θ is the angle which the strap to the
winch makes with the horizontal. The final answers I get
are:
T=DW /(d sinθ-s cosθ )
V =W-T sinθ
H=T cosθ
I put in some
reasonable numbers just to get an idea of the answers, W =500
lb, θ =200 , D =2m, d =1.8
m, and s =0.1 m. Then T =1920 lb, H =1800
lb, and V =-157 lb. The negative value for V
means that V
T
P from the pull point to some anchor
higher up. Of course the magnitudes of these two tensions
are the same, P=T . The picture shows only the
pulling forces, the rest are the same as in the picture
above. There are still three unknowns, T , V ,
and H . I will call the angle that P φ . I will not show
the details, just give the final results:
As a numerical example,
I will use the same numbers as above and add φ= 400 .
Then T =624 lb, H =1070 lb, and V =-115
lb. It is definitely advantageous to use the manufacturer's
suggestion here which, in my numerical example, reduced the
force the winch needed to exert by a factor of about 3.
before . Here is the answer
I gave there but substituting your numbers. You should
definitely read that answer before proceeding here. Let's
say that the forklift stops dead and that it plus the pole
deform by 1 cm so that the distance traveled during the
collision time is d ≈1
cm. Since these are very rough calculations, I will let
the mass of the forklift be m f ≈5000
kg and the initial speed be v =3 mph≈1.3
m/s. If the forklift stops uniformly, it will stop in a
time t =2d /v =0.015 s. The
force F experienced (by both the pole and the
forklift) will be F=m f v /t ≈1.4x106
N=315,000 lb. To get the g-force you divide by the
weight of the forklift, mg =5000x9.8 N=49,000 N; so,
g-force=6.4. That is the force experienced by the forklift
and would be the same for anything which moves the same. The
driver would experience that force where the lap belt was
restraining him but his upper body would experience a
smaller force since it would take longer to stop. As
emphasized in the earlier answer, this is a very rough
approximation.
not a good approximation); I will do that. Also, I
will assume that it is a simple pendulum with a small angle
amplitude so that the period is T π √(L /g )
where L is the length and g=MG /R 2 is the
acceleration due to gravity. At a distance H above
the earth's surface g H =MG /(R+H )2 ;
at a distance h below the surface of a
uniform-density earth, g h =MG (R-h )/R 3
(see footnote*). Now, for the periods to be equal it is
necessary that g h =g H ;
a little algebra shows that equivalently (1-(h /R ))=1/(1+(H /R ))2 =g H(h) /g .
It is instructive to
look at h /R as a function of H /R :
h /R =1-1/(1+(H /R ))2
shown in the first graph above. When h=H =0
you are at the surface of the earth; when h /R =1,
H /R =∞ and g h =g H =0.
Next, look at the
plots of g H(h) /g as a function of H /R
(black) and of h /R (Red). There
are only two locations where g h =g H ,
at the surface where h /R =H /R =0
corresponding to g h =g H =g
and near h /R =H /R =0.6.
The third plot shows a closer look around 0.6 showing h /R =H /R =0.618
corresponding to g h =g H =0.382g .
This was a pretty
long-winded answer, but the upshot is that you were right
and the answer key was wrong: h /H =1.
*M is the mass
of the earth, R is the radius of the earth, and
G is the universal gravitational constant.
Actually, I did not need to
solve this problem graphically, I could have solved it
analytically. If you assume (guess) that h /R=H /R≡x ,
then the equation to determine x is 1-x =1/(1+x )2
or (1-x )(1+x )2 -1=0=(1-x 2 )(1+x )-1=
x 3 +x 2 -x=x (x 2 +x -1)=0.
One solution is x =0 (the surface) and the positive
solution to the quadratic equation is x = √(5)-1)=0.618.
I guess I shied away from this because I know I am not very
good at solving cubic equations, but I can handle this one!
R
and speed v is ω=v/R . Taking
R ≈1 m,
ω= 2400 Hz=23,000 rpm. The velocity required
for a low altitude orbit is about 1700 m/s, so that would
require about 16,000 rpm.
GIF of him doing it:
So how much energy, current and charge generally speaking would producing that much electricity to penetrate that much air require from a human being, and how does it relate to the amount of energy a human body actually can produce, and finally assuming a human could produce the electric power needed to do that, how would a real life Palpatine shoot lightning out of his hands with real life physics?
6 Volts=3 Megavolts
of potential difference is required. Of course, this is
fiction so I will not go any farther with this answer.
kq 2 /R 2 =Gm e m s /R 2
or q =(Gm e m s /k )≈3x1017
C. The number of electrons this would be is q /e =1.9x1036 .
If we roughly estimate that the earth has is 50% protons and
50% neutrons, then the approximate number of protons would
be (m e /2)/m p =1.8x1051 ,
far larger than 1.9x1036 . If you want to check my
arithmetic, look up all the constants I have used.
earlier answer
as well as the earlier answer it links to. When you have
done that, you should be able to follow my answer to your
question.
First a qualitative discussion. You decide that you want
your spaceship to have a constant acceleration of g
as measured by you, captain of the ship. You therefore
arrange to have your engines always exert a force on the
ship of F=mg where m is the mass of the
ship. So you will always see your ship having the desired
acceleration. But, as I have emphasized in many earlier
answers, acceleration you measure is not the same as
acceleration someone on earth measures. In Newtonian
physics, everyone measures the same acceleration. This is
really a glorified
twin paradox problem and I am not going to calculate how
long the round trip will be for you (but I will calculate
how long it takes in earth time). What I can tell you is that you
will take longer than 4 years to return to earth and the
time elapsed on earth will be longer than your time. Your
desire to accelerate "to
the speed of light" is impossible, as you will see, so
forget that; with your scenario the fastest you will ever go
is about 86% of c .
What I want to focus on is what someone on
earth will measure regarding your motion. This is mostly
worked out in the earlier answers I have referred you to, in
particular v /c =(gt /c )√[1+(gt /c )2 ]
and a /g =[1+(gt /c )2 ]-3/2 ;
these are plotted in the graph above with the red and blue
curves. Things to notice are that after 1 year you have only
acquired a speed of about 0.7c and your
acceleration has dropped to about 0.35g . What was
not derived earlier was your position as a function of time;
integrating the velocity equation, I find gx /c 2 =√[1+(gt /c )2 ]-1,
shown by the black curve. Things to notice are that it
starts out looking like a parbola as it would if you had a
constant acceleration of g , but eventually becomes
a straight line as you approach the speed of light which you
never reach. Reading off the graph, the time it takes you to
get halfway, 1 ly, is about 1.7 years (as measured from
earth); your speed will be about 0.86c and your acceleration
(again as measured from earth) will be about 0.13g . This will be ¼
of your total trip, 6.8 years. The time you measure will be somewhere
between 4 and 6.8 years.
Incidentally, the axes
may be shown to be numerically equal to years and light
years if you approximate g ≈10 m/s2 .
I see that I have misread
the question. For some reason the questioner reverses his thrust at c . Thus the total elapsed time on earth
before your return would be 4.4 years in this scenario.
Light would take 2 years to make this trip. Again, I have
not calculated the time on your clock but it would be
between 2 and 4.4 years.
f= μ k N
for kinetic friction, for static friction the equation
f= μ s N
is not true. The correct formula is an inequality rather
than an equality:
f ≤ μ s N.
For one single force you may write an equation, f max =μ s N
which is the largest frictional force you can
get for a given N . If you do not push your book
with a force equal to or larger than N=mg /μ s
it will fall to the floor. (The Q&A right
below yours is closely related to your question.)
θ to the horizontal is at rest and in
equilibrium. There are three forces acting on it, the
friction f , the normal force N , and the
weight W=mg . Then, using Newton's first law, f-W sin θ =0
and N-W cosθ =0. Solving,
f=W sin θ and N=W cosθ.
If the block is just about to start moving f=μN
where μ is the coefficient of static
friction, so μ= (W sin θ /W cosθ )=tanθ .
As you can see, the angle is independent of the weight; this
is because both N and f are proportional to
W , so
W cancels out when you calculate their ratio.
μ ≈0.7. The speed
of the car is v=89 km/hr=24.7 m/s. The change in kinetic
energy must equal the work done, (K 2 -K 1 )=-μ mgd =(0-½mv 2 );
the mass m cancels out, so -½x24.72 =-0.7x9.8xd
and therefore d =44.5 m.
m of the water in the tube is the density
of water (ρ =103
kg/m3 ) times the volume of water (V=Ah );
therefore the weight of the water is mg =ρgAh ≈104 Ah
(taking g ≈10 m/s2 ). The pressure at
the bottom of the tube is atmospheric pressure, P bottom ≈105
N/m2 and at the top the pressure is zero;
therefore there is a force F up on the tube which
is F =(P bottom -P top )A ≈105 A
N. But F is holding up the weight, so F=mg or 105 A= 104 Ah
or h ≈10 m. A cancels out!
Q&A OF THE
WEEK, 4/14/2018
video going around the internet of nasa ISS astronaut Randy Bresnik spinning a fidget spinner, in space, in a quite low gravity environment and then apparently grabbing hold of the spinner then video cuts and we, presumably some short time later, see the entire body of the astronaut holding the spinner spinning fairly rapidly in space.
Perhaps it is a quite elementary question, but, I wonder if that video might have been faked, wondering if the what must be a relatively small amount of energy in the spinner could cause the entire body of mass of astronaut plus the spinner to spin in the fashion observed in the video. That is to say: wouldn't the energy from the spinner, say, be absorbed by the astronauts body or the muscles in his arm, or some such, upon grabbing the spinner; or, if there was movement, wouldn't it be far far less than what is demonstrated in the video?
I toy and
I man and the original angular velocity
of the toy is ω 1 , then the
initial angular momentum is L 1 =I toy ω 1
and the final angular momentum is L 2 =(I toy +Iman ) ω 2
where ω 2 is the angular velocity
of the man+toy. Conserving angular momentum and solving for
ω 2 , ω 2 =[I toy /(I toy +Iman )]ω 1 ω 1 .
Now, I actually found the moment of inertia of a fidget
spinner,
I toy ≈7x10-5 kg·m2
and I estimate
I man ≈70 kg·m2 .
This means that ω 2 =ω 1 /1,000,000!
The astronauts have angular velocity of about 1 revolution
per second and you know perfectly well that the fidget
spinner did not have a speed of a million revolutions per
second.
Finally, if you are interested, you can show that energy is
not conserved. The expression for kinetic energy is E =½Iω 2
so E 1 =½I toy ω 1 2
and E 2 =½[I toy /(I toy +Iman )]ω 1 2 ≠E 1 .
R from the center of the earth to
totally escape (to infinity) the gravitational field of the
object; the expression is v escape =√(2GM /R )
(which assumes that the mass of the escaping object is very
small compared to the mass of the earth, M ).
That is the initial speed with no additional propulsion.
Obviously you could escape any gravity by simply pushing
with a force greater than the force of gravity you
experience; if you had a means of propulsion which kept you
moving at 1 mph, you could completely escape the earth given
enough time. Therefore your "bonus question" essentially has
no meaning. Your first question is whether the escape
velocity atop a mountain is different from sea level; the
answer is yes because R is bigger atop the
mountain, but since the height of even the highest mountain
is tiny compared to the radius of the earth, the difference
is neglegible.
v , the negative of the velocity
(horizontal) of the train u v w
This has nothing to do
with how drops adhering to the window move. If the train
were still a drop would drip straight down the window with
some speed which would be much smaller than the speed of a
falling rain drop. If the train is moving the drop would see
a wind of speed u which would push it backwards with some
speed considerably less than the speed of the train. Again,
the vector sum of the two velocities would be seen as the
drop velocity by an observer in the train.
sic )"
motion would not happen until you were swinging above the
bar. When pumping, you eventually reach a point where the
chain between your hands and the seat goes slack which you
can see in the photograph of the swinging girl if you look
carefully (see my yellow line). If the girl is at the very
top of her path, she is at rest and there is no force from
the chain on the swing itself. So when she starts to fall
back she will be in a sort of free fall until the chain
becomes taught; at that instant there will be a jerk.
here .
What is particularly exciting is that light from the merging
neutron stars could be seen; earlier observations, of
merging black holes could not. Also, the abundance of heavy
elements like gold is much larger than could be understood
using standard models of stellar evolution; it had been
hypothesized some years ago that perhaps the excess could be
produced in neutron star collisions, and the current
observation seems, at first glance of the spectroscopy of
the light, to support this hypothesis.
These are the result of those tests.
These charts just do not look right to me in their conclusions.
I am not a math major, but I do understand some stuff and this does not look correct. ie. "Area under the curve." or correct averaging.
If you can set me straight or put me on another path it would be greatly appreciated.
etc .)
What I can do, though, is estimate the area under the curve
which seems to be one of your main concerns. I drew in a
rough fit (green) to the data (red) and calculated the area
under the green curves (keeping in mind that the area of the
triangular segment below the axis is negative). As you can
see, the area I got as a rough estimate is 0.37 compared to
the actual area of 0.35. The average would simply be the
area/time=3.5. My guess is that your pallets did not fulfil
the requirements since the precipitous drop before the 0.1 s
had passed would seem to indicate a collapse of the pallet.
Both tests show this behavior, with the 8g test having the
"collapse" occur earlier as would be expected. However,
since I do not understand the details of the test well
enough, I would not take my guess as gospel.
earlier answer
where I did essentially this problem in the noninertial
frame; this saves me having to draw the whole picture again.
For your problem, the vector labelled C
H (f 1 +f 2 )+LN 1 -LN 2 .
(Translational equations are f 1 +f 2 =mv 2 /R
and N 1 +N 1 -W=0.)
Now, when the truck is about to tip over the left wheel is
about to leave the ground so N 1 =f 1 =0,
f 2 =mv 2 /R ,
and N 1 =W ; therefore Hmv 2 /R-LW= 0.
So, if v > √[LWR /(Hm )]
the truck will flip over.
m and initial
speed v and the earth has a mass M and
initial speed V ; then conservation of linear
momentum (essentially Newton's 3rd law) says
mv+MV =0 or V =-(m /M )v ;
the recoil velocity of the earth is unmeasurably smaller
than your velocity.
mc 2 )
and when the photons get absorbed, the heat generated will
"retrieve" the lost mass. Eventually you will run out of
fuel for your reactor and you will be right back to being at
rest. (I am assuming that the rules of the game forbid
letting the heat at the back radiate into space.)
v , F= - ¼Av 2 .
The problem is that I do not really know how accurate this
is at such a high speed as 5000 km/hr≈1400 m/s. So,
view this as a very rough calculation. F ≈-2x107
N. Now, to get the acceleration you need to know the mass
m (in kg). Using Newton's second law, F=ma ,
a ≈2x107 /m m/s2 .
To express this in g s, divide by 9.8 m/s2 .
For example, if the mass of the vehicle is 10,000 kg, a ≈2000
m/s2 ≈200 g s.
A . Since the
drag force is proportional to A, all estimates given above
need only be multiplied by 7/38=0.18.
v=v 0 -gt
and y=y 0 +v 0 t - ½gt 2 ;
note that this is only for the vertical component of the
motion, the horizontal motion does not interest us for your
question. Here v is (the vertical component of) the
speed at time t , v 0 is (the
vertical component of) the speed at time t =0, y
is the altitude at time t , and y 0
is the altitude at time t =0 (which I will choose to
be 0). You are interested in the height to which the
projectile rises, and at that position v =0;
therefore v 0 =gt and so y=h =(gt )t- ½gt 2 =½gt 2 .
But my time here is only half the "flight time" of the
projectile, so h =½g (t flight /2)2 .
You do not need to memorize "the flight time equation".
Everything you need to know is contained in the equations of
motion which you should memorize.
VIDEO
m ,
one at rest the other with speed v , in one
dimension. As you know, if the collision is perfectly
elastic the first ball ends up at rest, the second moving
with speed v , so the momentum is conserved because it is
mv before and after the collision. Next you say that if
the second ball emerges with speed u=v-δ that
momentum has been lost. But you have assumed that the first
ball is at rest, but it cannot be if momentum is conserved:
mv=mu'+m (v-δ ) and so the first ball
has a speed u'= δ after the collision.
In the
video I have inserted above, look closely at the first
demo, with one ball striking four and one popping out; you
will clearly see that the next to last ball is not at rest
after the collision(s).
F . If F>W where W is the weight of
the balloon plus its contents, there is a net force up
causing the balloon to accelerate upwards; this force is
called the buoyant force and also explains why things float
in water. The reason you use helium is that it is lighter
than air. Then the answer to vacuum question is clear:
taking away the helium but keeping the balloon the same size
would make it lighter and therefore the net upward force
would be larger.
propelled
past lightspeed"; no material object can go faster than (or
as fast as) the speed of light. I will take the meteor mass
to be 300 kg and its "average speed" to be 5x104
m/s. Then the energy would be E = ½mv 2 =4x1011
J. For the super fast meteor, let the speed be 99% the speed
of light. Then the energy would be E=mc 2 / √(1-.992 )=2x1020
J. For comparison, the energy of the Nagasaki atomic bomb
was about 1014 J.
recent answer . Your question
has a slightly different twist: when this type of problem is
calculated it is usually assumed that g , the
acceleration due to gravity, is constant; if your ball stays
in the air for six hours, that will not be the case.
Nevertheless, the ball would always have a Coriolis force on
it so it would be deflected westward by some amount. If you
had said something like the ball goes up 1000 m, you could
have done a pretty good calculation since 1000 m is very
small compared to the radius of the earth and it would be a
reasonable approximation to say that g =9.8 m/s2
the whole time (which would be about 28 s). It can be shown
that, for a latitude of λ , on the way up the
Coriolis deflection is [(4ω /3) √(8h 3 /g )]cosλ
west and on the way down it is
[(ω /3) √(8h 3 /g )]cosλ
east, so the net deflection is
x=ω √(8h 3 /g )cosλ
west; the angular velocity of the earth is
ω= 7.27x10-5 s-1 , so
x =2.1 m west at the equator (λ= 0).
Finally we should note that at the poles (cosλ =0)
there is no deflection due to the earth's rotation, but
since the earth revolves around the sun you are still in a
frame where there would be a Coriolis force but it would be
much smaller; it would be smaller yet because the angular
velocity of the motion around the sun is almost parallel
(about 230 away) to the vertical at the poles.
The deflection for the example above (h =1000
m)would be about 2 mm compared to 2 m; the direction would
be opposite the direction the earth is moving in its orbit.
All of the above
ignores air drag.
appear to vary as you moved from the poles
to the equator because of the earth's rotation. A scale
would read your weight (gravitational force on you)
correctly at the poles but would, because of the centrifugal
force (a "fictitious" force), read slightly less at the
equator.
√(1-(v /c )2 ) where v is the
velocity and c is the speed of light. The most
elegant way to think of this is to not think of space and
time as being two things, rather think of space-time.
If you view it from outside the
earth, the satellite will continue moving directly away
from the center of the earth.
The second is a little fancier and harder to understand.
If you view Newton's laws from a rotating coordinate
system (the rotating earth with you on it) you find that
they are wrong; these "laws" of physics only work in
inertial frames of reference. However you can force
Newton's laws to be correct if you invent just the right
fictitious forces. I will not go into all the complexity
here, but one of the fictitious forces is called the
Coriolis force and can be written as F m ω v
ω
v points radially outward, the
negative of their cross product points west. Hence, it
will appear that the satellite is pushed west when
viewed from the ground.
For the first problem, let's say that the forklift stops
dead and that it plus the pole deform by 1 cm so that
the distance traveled during the collision time is d ≈1
cm. Since these are very rough calculations, I will let
the mass of the forklift be m f ≈5000
kg and the initial speed be v =8 mph≈3.6
m/s. If the forklift stops uniformly, it will stop in a
time t =2d /v =0.0056 s. The
force F experienced (by both the pole and the
forklift) will be F=m f v /t ≈3.2x106
N=720,000 lb.
For the man-forklift collision, assume the forklift
keeps right on moving with the same speed but with the
man stuck to the front. Suppose that the man's body
compresses 3 cm during the collision, so the time of
collision will be t=0.06/3.6=0.017 s. The force now is
F=m m v /t =68x3.6/0.017=1.44x104
N=3200 lb.
do the proton and the neutron have exactly the same mass?
how do masses of the proton and neutron compared to the mass of the electron?
w
which particles make the greatest contribution to the mass of an atom?
which particles make the greatest contribution to the chemical properties of an atom?
I usually do not answer multiple questions, but these are pretty
much one question about properties of atoms.
The neutron is about 0.14% heavier than the proton.
The electron is about 1837 times lighter than a proton.
All stable atoms but 1 H have neutrons as
having to largest fraction of the weight. All stable
nuclei except hydrogen have more or an equal number of
neutrons than protons.
The electrons, which orbit the nucleus comprised of
protons and neutrons, determine the chemical properties
of the atom
earlier answer .
FAQ page if there is one—there
is one on AskThePhysicist.com! For the question regarding
why the speed of light is the highest possible, see
this link . Your
question/statement that the speed of light would be the
speed of light plus the speed of the source is simply wrong.
To understand why, see
this link .
From Newton to Einstein
original Q&A there is a
link to a more
detailed derivation although I do not think it will make
this simpler for you. Rewrite the first equation as
(g /c )dt =√(1-β 2 )·d[β/ √(1-β 2 )]=dβ /(1-β 2 ).
There was a little tricky differential calculus here to get
the the last step:
d[β/ √(1-β 2 )]=(1-β 2 )-1/2 d β +β (-½)(-2β )(1-β 2 )-3/2 dβ= dβ /(1-β 2 )3/2 .
If, like me, your integration skills are rusty, I suggest
the Mathematica on-line integral
calculator which gets you to the solution for gt /c ;
don't forget that the difference of logs is the log of the
ratio. Then, to get the final answer, exponentiate 2gt /c
to get rid of the natural log and algebraically solve for
β . Of course, the starting point is recalling
m=m 0 / √(1-β 2 ).
Wikepedia on circular polarization, "Since this is an electromagnetic wave each electric field vector has a corresponding, but not illustrated, magnetic field vector that is at a right angle to the electric field vector and proportional in magnitude to it. As a result, the magnetic field vectors would trace out a second helix if displayed."
The electric field, which has a constant magnitude, is
illustrated in the animation; the helix traced out by the
magnetic field would be π /2 out of phase
with the electric field which may be the source of
confusion.
2 . Students usually find the s2
as strange —what in the world is a square
second? So, m/s2 is just shorthand for m/s/s.
That makes sense, right: 1/4/4=1/42 =1/16.
earlier answer .
A having a speed v
to be ¼Av 2 =0.038v 2 .
When this is equal to the weight of the object, mg =(4πR 3/ 3) ρg =402
N, the object will have the terminal velocity; so v =√(402/0.038)=103
m/s=230 mph. I have used R =0.22 m, ρ= 920
kg/m3 , and g =9.8 m/s2 .
2 =16 times larger. This would be like
brakes and probably eventually stop the car. Depending on
the details of the car, it would possibly eventually reverse
the direction of the car.
m =2.7x10-3
kg and a radius of R =0.02 m; the volume is V =4πR 3/ 3=3.35x10-5
m3 . The forces are its weight down mg =9.8x2.7x10-3 =0.0265 N
and the buoyant force B up which is the weight of
the displaced water (the density of water is 1000 kg/m3 ),
B=ρgV =1000x9.8x3.35x10-5 =0.328
N; the net force on the submerged ball is therefore F =0.328-0.0265=0.301
N up. So, since s =10 ft=3.05 m, the work done to move
place it down there is W =Fs =0.301x3.05=0.918
J=2.55x10-7 kW ·hr=0.677 ft·lb.
Keep in mind that if simply released the ball will quickly
reach terminal velocity and move up with a constant speed,
most of the work being against the frictional drag force.
…
49+51=100, so that is 4900. But we still have 50 and 100, so
the answer is 5050. (This is not physics, though!)
β -
decay: a nucleus with too many neutrons to be stable changes
a neutron to a proton (which stays in the nucleus), an
electron, and a neutrino. The energy released by this decay
is therefore shared between the electron and the neutrino
(assuming that the nucleus, being enormously heavier than
the other two particles, will have almost no recoil energy.
The electron, as you demonstrate in your lab assignment, is
pretty easy to observe. The neutrino, however, is extremely
β - decay
was first discovered, it was a huge mystery because the
electrons, the only thing observed, were not emitted with a
single energy but with a broad spectrum of energies. A
typical electron spectrum is shown in the graph to the left;
in that graph you see a maximum near 0.1 MeV electron energy
but, in this example, the total energy should be about 1.15
MeV. There seemed to be only two explanations: either energy
is not conserved or else there is a third particle which was
not observed and carries off some of the energy. Everyone
believed the second explanation and the existence of
neutrinos was totally accepted by all physicists. They were
so hard to observe that it was 26 years after Wolfgang Pauli
proposed them (1930) and they were observed by Cowan and
Reines (1956).
In your experiment,
there were two decays going on simultaneously: 90 Sr
decaying to 90 Y and 90 Y decaying to
90 Zr. Both have a maximum but they are far
smaller than the total energies of the decays. Nearly
nothing after the first sentence starting with "They told me…"
makes little or no sense but that is probably because you
did not understand what they were trying to tell you.
-19
C, so the net charge must be an integral multiple of this.
So the answer to your question is no.
this video for a derivation. For a hydrogen atom in its
ground state, v =2.16x106 m/s.
-7 ly. Therefore 9.3 ly is about
9.3/1.47x10-7 =6.33x107 sun diameters
(63.3 million).
—if you give
yourself an adequately large velocity you will hit the
ceiling. If d is the distance from your head to the
ceiling at the instant that your feet leave the floor, then
you will hit the ceiling if the speed v you launch
with is greater than √(2gd ) where g
is the acceleration due to gravity. If the elevator has an
acceleration a , the critical speed will be √(2(g-a )d ).
For example, if a=g you are in free fall and even
the tiniest velocity will cause you to hit the ceiling
(assuming that the elevator does not get to the ground
before you get to the ceiling).
W /c but
the outside observer would see it take a time (W /c )/ √[1-(v /c )2 ].
If the spaceship were accelerating you would see the light
hit somewhere backward (forward) of where you aimed it
depending if you were speeding up (slowing down).
u across the
ship which moves with speed v , will have a speed u'= √(u 2 +v 2 ),
not u+v as you suggest. However, as you correctly
surmise, this cannot be true for light since, in your
example, you could not have the light with speed √(c 2 +102 )
as seen by an outside observer. The speed must still be
c . So, if the component along the direction of the ship
is cx =v , you must have cy 2 +cx 2 =c 2 =cy 2 +v 2
or cy =c √[1-(v /c )2 ].
Regarding what you would see in the ship, the principle of
relativity states that the laws of physics are the same in
all inertial frames of reference; if your ship is at rest,
we could agree that anything aimed straight acoss will go
straight across. In other words, there is no such thing as
absolute rest —any inertial frame with constant
velocity relative to another inertial frame would have the
same laws of physics.
NASA : "Because the Moon moves to the east in its orbit at about 3,400 km/hour. Earth rotates to the east at 1,670 km/hr at the equator, so the lunar shadow moves to the east at 3,400-1,670 = 1,730 km/hr near the equator. You cannot keep up with the shadow of the eclipse unless you traveled at Mach 1.5."
h =1 m above the floor. The speed
the head would hit the floor could be found from energy
conservation, ½mv 2 =mgh
or v =√(2gh )=√(2x9.8x1)=4.4
m/s. Now, you either need to know the distance the head
stops in or the time it takes to stop. There is not much
give in either the head or the floor, so let's just guess
that the distance it takes to stop is 3 mm. If you do the
kinematics, the acceleration of something moving 4.4 m/s and
stopping in 3 mm is about 3000 m/s2 , about 300
gs! If you take the mass of the head to be about 4.5 kg, the
force is about F=ma =13,500 N=3000 lb. The time over
which this force acts is very short, about 4.4/3000=0.0015
s.
To be fancier, you
would have to do the rotational problem. The student plus
the chair have a certain moment of inertia I and
when they rotate about the rear chair legs they acquire a
rotational velocity ω when the center of mass
has fallen a distance h . In that case ½Iω 2 =mgh
and so ω =√(2mgh /I );
now you can get the speed the head hits the ground by
writing v=Lω where L is the distance
of the head from the rear legs of the chair. But to
calculate the moment of inertia of the student/chair would
be very complicated and, in the end, the speed of the head
would not be all that different from the simple calculation
above. And you still have to make approximations over how
far it would travel to stop. Just use the simple head
dropping calculation to convince the students that the order
of magnitude of the force would be a few thousand pounds.
T of a
pendulum (the time it takes for one swing back to where it
started) is approximately T ≈2π √(L /g )
where L is the length and g =32 ft/s2
is the acceleration due to gravity. In your case, L =30
ft, T ≈6 s. Now, if you watch a kid swing a much
shorter swing, you will see that they "pump" once per cycle;
this is called a driven oscillator and, driving with the
same frequency as the natural frequency of the swing is
called resonance and each pump will increase the amplitude.
No doubt the kids, being used to a much shorter swing, are
pumping with a period much shorter than 6 s, far off
resonance, therefore not having the effect of increasing the
amplitude. I suggest that you start them with a good push
and tell them to pump each time they reach the bottom of the
arc and moving forward.
,
way too complicated to discuss here. But, it is pretty easy
to understand why we get tired—any time the muscles
exert a force they are doing work whether or not they are
actually moving something (see
earlier
answers ).
result of the fact that the speed of light is a
universal constant, not vice versa . I would also
point out that these would also be true if the speed of
light were different from what it is; as illustrated by
George Gamow's book
Mr.
Thompkins in Wonderland , they would be much more
evident if the speed of light were very small.
K ≈½mv 2 =½x4.5x(3x107 )2 =2x1015
J=2000 TJ (terajoules); the Hiroshima bomb had an energy of
about 63 TJ. If the collision lasted 10 s, this would
correspond to a power of 200 TW (terawatts). To give you a
feeling for the magnitude of this power, the total output of
all power sources on earth is about 15 TW. The 100 lb
projectiles would have 10 times the energy and power as the
10 lb projectiles. So those things would have quite a punch.
Here is the problem that most sci-fi writers never think
about, though. Whoever is firing these projectiles has to
give them this energy—where are you going to get 200
TW in the middle of empty space?
Now, the second part of your question. These things
have a speed (3x104 km/s) which is much faster
than the fastest meteor (72 km/s) and you know what happens
to them —they burn up and break up. The recent
(2013)
Chelyabinsk meteor
exploded at an altitude of about 20 mi and most of its
energy (about 1500 TJ) was absorbed by the atmosphere. It
had a much smaller speed than your projectiles (about 20
km/s compared to about 30,000 km/s) but a much larger mass
(about 107 kg compared to 4.5 kg), so the
energies were quite comparable (1500 TJ compared to 2000
TJ). So I would guess that your projectiles would not do
much damage to the objects on the surface of the planet.
just after the gun has been shot;
at the time when the sound reaches
where the plane had been at time 1;
at a time twice as long as 2.
Also shown are the corresponding spherical wave fronts of
the sound from the shot. As you can see, the wave fronts
never catch up with the plane. You will never hear the gun.
calm water in a storm. It turns out that
there is an old sailor's trick to calm water in choppy
conditions: just pour a little oil on the surface. There
seems also to be
physics understanding of this phenomenon. For a good
overview, see the discussion on
Physics Stack Exchange . Perhaps there was some oil
in your calm area.
m moving at v strikes a stationary rod of mass
M at its center and passes through exiting with u , there is simple conservation of momentum if the rod is free to move/slide:
m (v-u )=MU . What happens if the bullet strikes and passes through the rod at its end (same
v and u ), causing the rod to move forward while at the same time start to spin?
earlier question except that in
that question the collision was stipulated to be elastic. In
your case, there is no conservation of energy equation and
so there are only two equations —conservation of
linear momentum and angular momentum. But, there are only
two unknowns, U and ω , because you
stipulate u to be known. (Also, d=L /2 for
your question.)
U=m (v-u )/M and ω= 6U /L .
The energy before the collision is E 1 =½mv 2 .
Referring to the earlier answer, the energy after the
collision is ½mu 2 +½MU 2 +½I ω 2 .
For the simple case of hitting the center (ω= 0)
I find E 2 =½m (u 2 +(m /M )(v-u )2 ),
and for the case of hitting the end I find E 2 ' =½m (u 2 +4(m /M )(v-u )2 )E 2 ' >E 2 ,
less energy will be lost in the case where the collision
happens off center (assuming u and v are
the same in each case). For example, if m=M /4 and
u =½v , I find ΔE =-(11/ 32)mv 2
and ΔE' =-¼mv 2 . In
both cases, energy is lost in the collision.
Then, I want to alter the sauce and make it more and/or less watery
(changing its viscosity) and measure that new sauces viscosity.
Then, I'll
take the new sauce and re-measure the turntable speed necessary to make the
new sauce flow from the center to the edge.
Last, I want to replicate the tests enough times so I can create an equation
that would allow me know what turntable speed would be necessary to
correctly flow the sauce based upon the viscosity of the substance.
F , the centrifugal force, which
pushes the sauce out, F=mv 2 /R
for a mass m with speed v when at a
distance R from the center. But, the speed depends
on the distance from the center, v=R ω
where ω is the angular velocity; so F=mRω 2
and an ounce of sauce experiences a bigger force as it
moves out. In other words the sauce will tend to all be
pushed out the the rim of the pizza regardless of its
viscosity. If the viscosity of the sauce too large, the
centrifugal force might be too small to move the sauce at
all, so there would be the tendency for the sauce to stay in
the center. In any case, I cannot imagine that it is
possible to use rotation to get a uniform spread of sauce.
If you still want to
pursue this, I found the description of a straightforward
experiment to measure viscosity μ . You get a
tall cylindrical container and fill to a depth d
with the fluid (density ρ f ). Drop a
sphere (density ρ s radius R )
into the fluid and measure the time t it takes to
reach the bottom; then the velocity of the falling sphere
was v=d /t . The viscosity is then μ= (4R 2 g (
ρ s -ρ f ))/(9v ).
It will be a little tricky since the sauce is not
transparent. Also, it is important that the sauce be
homogeneous, no chuncks in it.
W total done by
all forces on an object of mass m is equal to
the change in kinetic energy, ΔK =½mv 2 2 -½mv 1 2 .
The picture is a mass m on an incline θ
being pushed by a constant force F up the incline
and moving a distance s ; for simplicity, I choose
m
to be at rest when you start pushing. Other forces on m
are its own weight W=mg , vertically down, and a
possible frictional force f pointing down the
plane. For the time being I will assume f =0. Only
the component of W along the plane Ws =-mg sinθ
will do work. The net work w (sorry if it is confusing to
have big W and little w ) is w=Fs-Ws s=Fs-mgs sinθ= ΔK= ½mv2
and so v =√[2s (F -mg sinθ )/m ].
Rather than derive the
gravitational potential energy, U=mgy , I will
assume that you already know that. I will now do the problem
over but using potential energy. w=Fs=E 2 -E 1 =(K 2 -K 1 )+(U 2 -U 1 )=½mv2 +mgh= ½mv2 +mgs sinθ=Fs.
Solving this, you find exactly the same answer: v =√[2s (F -mg sinθ )/m ].
So here is how I look at it: potential energy is a very
clever bookkeeping device to keep track of the work done by
a force which is always present like gravity. It just makes
life a lot easier to not always have to calculate the work
done by some force that you know is always there. So, here
is the so-called work-energy theorem: the change in total
energy, kinetic energy plus any potential energy you have
included, is equal to the work done by all external
forces, E 2 -E 1 =W ext ;
here an external force is any force for which you have not
introduced a potential energy function. I like to rearrange
the work-energy theorem as E 2 =E 1 +W ext —what
you end up is what you started with plus what you added (or
subtracted if W ext <0).
Finally, let's include the frictional force, f=μWN =μmg cosθ.
(μ
is the coefficient of kinetic friction.) The work which the friction does is negative because it
is opposite the direction of s , w f =-μmgs cosθ.
Therefore, W ext =Fs -μmgs cosθ=½mv2 +mgs sinθ ;
solving, v =√[2s (F -mg (sinθ+μ cosθ ))/m ].
You might wonder why we did not introduce a potential energy
function for f . The reason is that there are two kinds of
forces in nature, conservative forces and nonconservative
forces and the latter kind cannot have a meaningful
potential energy function; friction is a nonconservative
force. But this answer has rambled on long enough and that
is a topic for another day!
"…if
there's a body on the top of an inclined plane and we take
friction into account, should the equation be "potential
energy at the top + kinetic energy at the top = potential
energy at the bottom (==0) + kinetic energy at the bottom +
the work done by the friction"?" ) the work done against
the friction ; I actually do not like that one bit.
λ'-λ =(h /mc )(1-cosθ )
where h is Planck's constant, m is the
mass of an electron, c is the speed of light, and
θ is the angle of the scattered photon; (h /mc )=2.4x10-12
m=2.4x10-3 nm and is called the Compton
wavelength of the electron. The largest shift is for the
largest angle, θ =1800 where cos1800 =-1
so the largest possible shift is (Δλ )max =4.8x10-3
nm. Visible light has wavelengths around 1000 nm, soft
x-rays (the closest to visible light) around 1 nm: so that
would require (Δλ )max ≈1000
nm>>4.8x10-3 nm. So the answer to your question
is no.
this and all the links in
that answer . [The derivations of
the equations below can be found in the earlier answers.]
From the perspective of your space ship the velocity v
as a function of time t is v =(a 0 t )/√[1+(a 0 t /c )2 ] where
a 0 is the acceleration as measured by the ship (g in your case). In order to achieve this acceleration, you must exert a force
F=ma 0 where m is the mass of the ship.
The velocity is plotted in red in the graph; reading from
the graph, the time when v /c =0.6 is when a 0 t /c ≈0.8=9.8t /3x108 . Solving, t =2.45x107 s=0.78
yr=284 days. The position as a function of t is x =(mc 2 /F )(√[1+(Ft /(mc ))2 ]-1)=(c 2 /g )(√[1+(gt /c )2 ]-1);
for gt /c =0.8, x =2.58x1015
m=0.273 ly. At this point you turn off your engines and
coast at v /c =0.6. The distance you need to
go to get halfway to the star is 16.9 ly and it will take
you 16.9/0.6=28.2 years. The second half of the journey will
take the same time, so the time for the whole trip is T =2(28.2+0.78)=58
years.
Your answers (212
days, 0.17 ly) are different from mine (284 days, 0.273 ly)
because you have assumed that the acceleration, viewed from
the earth, is the same as the acceleration (9.8 m/s2 ) seen from the ship. The
black curve in the graph shows the ratio of the acceleration
seen from the earth compared to the acceleration seen from
the ship; reading from the graph, at 0.6c the ratio
is about 0.5, 4.9 m/s2 . Be sure to read the
discussions of acceleration in the
earlier answers if you want to understand this issue
with acceleration in special relativity.
Since there is no such
thing as an antimatter drive engine, I can make only a rough
assessment of how feasible this is. The force you need to
apply during the 0.78 year accelerations is F=mg. What is m?
The space station has a mass of about 400,000 kg and you
have 40,000 passengers. Let's say each passenger needs one
space station of mass to support his/her needs, so the
required force would be 1.6x109 x9.8=1.57x1010
N. The energy you have to supply to the ship to v /c =0.6
I figure to be about 36x1024 J and the amount of
mass converted with 100% efficiency into energy would be
about 4x108 kg and you would need that much again
to stop it. So your ship of mass 1.6x109 kg would
have to carry 8x108 kg of fuel, one third of the total
weight! Obviously, I have not included this mass change in
my estimates of time. And, of course, there is the issue of
how you are going to store 4x108 kg of
antimatter. The whole scenario certainly does not look
feasible to me!
Finally, note that all
the distances and times are as measured by the earth bound
observer. In other words, the passengers will not be 58
years older when they arrive at their destination. Because
of length contraction, they will have a shorter distance to
travel. Ignoring the accelerating periods, I can estimate
the time of the trip by just assuming they travel 0.6c the
whole trip. In that case, T' ≈T √(1-.62 )=58x0.8=46.4
yr.
E , its mass
increases by E /c 2 . If it consumes an amount
of mass M , its mass increases by M .
M to a height h =30.5 m is E=Mgh =299M
where g =9.8 m/s2 is the acceleration due to gravity;
the answer will have units of Joules if M is in kg. Now the
energy available is 103 kW ⋅hr(3600 s/hr)(1000 W/1
kW)=3.6x109 J. So, 299M =3.6x109 so M =1.204x107
kg=2.65x107 lb.
always
exerts an equal and opposite force on you. And, whether you or the
asteroid goes faster after the push depends only on your relative
masses. An asteroid is most likely much more massive than you and
therefore you will be moving faster after the push. For example, if the
asteroid is 1000 times more massive than yours, its speed after the push
will be 1000 times smaller than yours.
on the bucket
assuming that the bucket moves horizontally with constant speed.
However, the person does do work; see an
earlier answer.
—that is
the spookiness! The trick is to devise an experiment which clearly
demonstrates that this is what is happening, not what you described as
one being up and the other down from the beginning. It is pretty
technical and tricky to understand this, but Roger Pennrose in his book
The
Emperor's New Mind
½=3000
in2 . If you coil it up, the area will be the same and the
area of a circle is πR 2 . Therefore the radius of
your rug would be R =√(3000/π )=30.9 in.
diagram .
I'm curious, there are components on all the outside edges, what about the diagonals, what do these represent? They have different equations so they MUST behave differently, what ARE these components?
e.g . changing
the current through a resistor changes the voltage across it (dv=R di ).
The diagonal lines indicate relations between the corner quantities,
e.g . current is the motion of charges (dq=i dt ).
blood letting.
Chinese medicine uses them to
treat a
variety of maladies; cupping therapy is generally considered
pseudoscience by modern medicine. The idea is to provide suction on the
surface of the skin and is achieved by first heating the cup and air
inside and then placing it on the skin. As the air inside cools, the
pressure decreases. The physics of this pressure decrease can be
understood by examining the ideal gas equation which relates pressure
P , temperature T , volume V , and amount of gas
N : PV=NRT where R is a constant which depends
on the units you use. So, as you can see, keeping the volume and amount
of gas constant, if the temperature decreases the pressure must
decrease. The fact that P ∝T is sometimes
called
Gay-Lussac's law .
A similar thing happens when canning food in glass jars. The
canning is done with the contents very hot and a lid which
is slightly domed is affixed. As the contents cool, the
pressure decreases causing the dome to pop inward toward the
contents, signaling that a good seal has been achieved.
The Scoring Baggy (6 ounces) has a constant weight.
My Balloon's Speed, as I approach the Target, is a variable. the Scoring Baggy will share that speed once I release it, Correct ?
And, the height I
release the Scoring Baggy from, at the time I release it, is also a variable, correct ?
My question is, is there a Mathematical formula that will calculate, how many feet from where I release the Scoring Baggy, it will land ? And how long it will take ?
t
in seconds it takes for the bag to hit the ground is t = √(2h /g )=¼√h
where h is the height in feet from which you drop it and
g =32 ft/s2 is the acceleration due to gravity. For
example if you drop it from h =144 ft, t =3 s. The
distance x in feet it will travel horizontally in this time is
x=vt where vx is the horizontal speed of the balloon in ft/s when
you drop it; so if vx = 20 ft/s (about 13.6 mph) and you drop it
from 144 ft, x =60 ft. The equation for x can be written
as x =¼vx √h which is handy if you
do not care about the time.
Note that the weight of the bag does not come into this at
all.
F d ≈¼(vy 2 /A )
where vy is the vertical velocity of the
bag and A is the area it presents to the onrushing
air; this estimate is correct only for SI units because it
has things like the density of the air at sea level built
into it. Now, it is my understanding that the speed of a hot
air balloon moving horizontally is the same as the speed of
the wind; in other words, from the perspective of a person
riding in the balloon, he is in perfectly still air. This
greatly simplifies the problem because the bag will drop
straight down as seen from the balloon, i.e. it
should strike the ground directly below the balloon. So, the
bag sees two forces, its own weight mg down and air
drag up; Newton's second law becomes may =m (dvy /dt )=-mg +¼(vy 2 /A ).
This is a first-order differential equation has a solution
vy =-[√(g /c )]tanh([√(gc )]t )
where c=A /(4m ). This solution is also a
differential equation since vy =dy /dt
where y is the distance above the ground; solving
this differential equation, y =h -(1/c )ln(cosh([√(gc )]t ).
A graph of this function compared to the case above for
dropping from 144 ft (note that 144 ft=44 m) is shown. (I
approximated the area of the bag to be 0.01 m2 ≈16
in2 =4x4 inches.) The time for the bag to hit the
ground is now 3.33 s rather than 3 s, approximately 10%
longer meaning that you should drop it when you are 66 ft
from the target rather than 60 ft. If you are lower the
correction is smaller, if you are higher the correction is
larger. Given the circumstances under which you must act, I
would expect the inclusion of air drag to be unnecessary
unless you are at a very high altitude and that you should
just drop it when you are about x =¼vx √h
from the target.
PV will remain constant? This is
called an adiabatic expansion, a process where no heat enters or leaves
the system. What remains constant is PV γ
γ is a constant which depends on the gas. For
example, for a monotonic gas γ= 5/3 and for a diatomic gas
γ= 7/5. Once you know what the new P and V
are you can get the new T : T=PV /NR .
T is proportional to T 4 ;
also, the wavelength λ of the most intense radiation is
inversely proportional to T . So, it is not possible for an
object to be both hot and dark.
2 or it would depend upon how much force the ball is thrown with.
Is the acceleration of a stone thrown upward the same as that of a stone
thrown downward?
g =9.8 m/s2 downward;
this results in the object speeding up if moving downward and slowing
down if moving upward but both have the same downward acceleration of
magnitude g . During the time you are throwing it down (up)
there is a larger (smaller) acceleration because of the force your hand
is applying, but that disappears the instant the object leaves your
hand. If you take air drag into account, there is an additional force
which always points in the direction opposite the velocity.
—friction
in the motor and bearings and air drag. With no friction, once you got
up to speed you could turn the motor off. In the real world when you
turn the motor off the fan slowly stops as friction takes its kinetic
energy away and turns it into heat; always a "balance" as you expect.
earlier answer.
—see
the faq page . You will then see
that your question should have been written as:
I understand that adding energy to an object increases the
kinetic energy of its molecules, and that the
kinetic energy of molecules determines the
temperature . But what I don't understand is WHY the
kinetic energy of molecules determines temperature . Why does an increase (or decrease) in molecular
kinetic energy change temperature?
defined to be proportional to the average
kinetic energy per molecule. In what sense, then, can we
understand why changing the average kinentic energy of the
molecules causes you to feel different. It is easiest to
talk about an ideal gas for which T =2<KE >/(3k )
where <KE > is the average kinetic energy per
molecule and k=1.38x10-23 J/K is Boltzmann's
constant. (Most everyday gases are pretty well described as
ideal gases.) Molecules are continually striking your body
and transferring momentum as they bounce off. This causes
the molecules in your skin to gain energy, i.e. get
hotter. Now, your skin is hotter than the interior part of
your body so heat starts flowing, causing the inside part of
your body to become hotter. Meanwhile, your body is doing
what it does to keep your body temperature at 96.80 F.
As the temperature of the air gets hotter, your body has
more and more trouble cancelling out the flow of heat from
the air and eventually fails and you get heat stroke or at
least start to feel sick.
earlier answer to see the complications and how
you can understand your question. In that answer I addressed two
possibilities applicable to your fish —if the fish has a
density greater than that of water, it will move backward until it hits
the back side of the bowl and if it has a smaller density it will move
forward until it hits the front side of the bowl. But your question is
different because the fish has the same density as water.
Therefore it will remain in the center of the bowl if it started there.
But the fish is accelerating, so he has to feel a net force; the water
behind him will exert a force which is larger than the water in front of
him does. So, the effect of the acceleration is not "neutralised", he
feels it. It is analogous to your sitting in an accelerating car and the
back of the seat exerts a forward force on you even though you remain
seated in the seat.
i.e . it is not some invisible
"stuff" but rather its purported effects are indicative that we do not really understand gravity as well as we think we do. Certainly general relativity is an incomplete theory since no one has successfully quantized it. You may know, but dark energy, at least the evidence for it, has an interesting history. When Einstein formulated general relativity, the expansion of the universe was not known and the universe was thought to be static; but with just gravity this is not possible—it will expand forever or expand until it collapses, depending on the initial conditions. Einstein therefore empirically inserted something he called the cosmolgical constant. Later when Hubble discovered that the universe is expanding, Einstein removed the cosmological constant and dubbed it his
"biggest blunder"; now it is coming back into play, but still just empirically.
There are doubtless millions or billions of black holes as you speculate, because all stars greater than a few solar masses end up that way. It is known that nearly all galaxies have a supermassive black hole at their centers. I would guess these came from early-universe stars coalescing. And early-universe stars were huge, thousands of times more massive than the sun because only hydrogen was available to make early stars. Still, I have learned that the size of the universe is just about impossible to comprehend and at its relatively young age still, I would guess these remnants of earlier stars are still a pretty small fraction of the mass of the entire universe.
Now, we need to make an estimate of the average force the steel ball
exerts on the panel when it is stopping. The force is the linear
momentum mv of the ball divided by the time the collision
lasts; I will assume that any hard ball will stop in about the same
time, so what matters is the momentum. It is easy to show that the speed
of the steel ball when it hits is v =4.5 m/s so mv =4.5
kg·m/s; so we need to find the size of a hail stone with a
momentum of 4.5 kg·m/s. But, we know the speed and mass from
above, so 4.5=(4.2x103 R 3 )√ (4.1x104 R )=√ (7.2x1011 R 7 );
solving, R =0.031 m, m =0.13 kg, v t =36
m/s. So my best estimate would say that golf-ball sized hail would not
break your panel but hail larger than a golf ball by 50% or more would.
Keep in mind that all these calculations are estimates, not precise
calculations.
ANSWER:
ANSWER:
F ∝v 2 .
There is no such thing as "mph force". There is no difference, as far as
air drag is concerned, between having a ground speed (GS) of 60 mph and
a head wind (HW) of 40 mph; and being at rest with a HW of 100 mph; and
having a GS of 100 mph in still air. I think what you are getting at is
that the force does not increase linearly with air speed; so if the
speed increases from 100 mph to 110 mph (an increase by a factor of
1.1), the force increases by a factor of (110/100)2 =1.21.
That said, I can now give you a way to estimate the force (very
approximately) on the tiny house. Let A be the area which the
house presents to the wind; then F≈ ¼Av 2 ,
which only works for SI units (meters, kilograms, seconds). So, for
example, if v =100 mph=44.7 m/s and A =(4 m)2 =16
m2 , F ≈¼x44.72 x16≈8000
N≈1800 lb. This would correspond to a pressure of about 0.07 PSI.
(A complicating factor is the presence of the towing vehicle. To some
degree, the tiny house would be shielded from the oncoming wind by the
vehicle. I see no way to estimate this effect because it would be
dependent on the vehicle.)
ANSWER:
t . In that case, the average force
F F =mvt . We
need to convert mph to m/s, 57 mph=25.5 m/s. For example, if t =0.1
s, F 2 =13 kg.
ANSWER:
Physics Forums which seems to be correct:
"The phenomenon being described is called 'Dilatancy' and was discovered by Reynolds about 100 years ago. It works only when you have well compacted sand that contains just enough water to cover all the individual grains. When you stand on the sand you create a stress / force which causes the sand to move. In order for the sand to move / flow individual grains have to be able to move past one another. Imagine a bunch of oranges stacked as you might see them at a grocers. The first layer has them all tightly arranged and then the second layer sits down into the gaps between the oranges on the first and third layer. Now imagine trying to move one of the oranges in the second layer. In order to move it the oranges on the first and third layer mut move down and up respectively to enable the orange to move. Effectively the volume of the pile of oranges or grains of sand increases with bigger gaps in between. So when you put your foot down on the sand it shoves sand out the way but in doing so the volume in between grains has to increase temporarily to allow the grains to move relative to one another. Consequently all the fluid at the surface is sucked by surface tension into the extra gaps made by the rearrangement of the sand. Since there is no longer any fluid at the surface the grains of sand are now dry."
If you go to the original Physics Forums question, ignore all the
early answers which are wrong.
ANSWER:
Physics Stack Exchange .
ANSWER:
can be created or destroyed. In chemistry, conservation of mass is assumed in chemical reactions because the mass changes are so small as to be almost unobservable.
ANSWER:
c is raised in
the famous equation E=mc 2 , is impossible because for
any power other than 2 the equation is not dimensionally correct —e.g .,
mc 3 does not have the units of energy; it would be
like saying "the speed of my car is 55 pounds/foot". Your second example
has two errors. F=m /a could be written as a=m /F
which would mean that the harder you push on something, the less it
would speed up, in violation of what happens. F =2ma would be
perfectly ok but would imply that you have defined what you mean by
force differently. Newton's second law is actually that the acceleration
is proportional to the applied force and inversely proportional to the
mass, a∝F/m or a=kF/m , where
k
is a proportionality constant. Assuming you have already defined what
mass (kg), length (m), and time (s) are, your choice of k
determines what you mean by force. The standard choice is k =1
which means that the unit of force (which we call 1 Newton) is that
force which causes a 1 kg object to have an acceleration of 1 m/s2 .
If you were to choose k =½ (your F= 2ma ), one
unit of force would be that force which caused a 1 kg object to have an
acceleration of ½ m/s2 .
You need to keep in mind that laws of physics are simply expressions of
how the universe works. For example, Coulomb's law simply states
that the force between two charges is inversely proportional to the
square of the distance between them, F∝ 1/r 2
because this is what we find doing the most accurate experiments that we
can. But, we can not really be sure that the "true" law is not F∝ 1/r 2.000000001
until we are able to perform an experiment which rules it out.
ANSWER:
U=B 2 /(2 μ 0 )+E 2 (ε 0 /2);
yes, there is an electric field because the magnetic field is changing
in your example. As the two magnets approach each other the fields
change and therefore the energy content of the fields changes also. To
avoid the electrodynamics implied by your question (time varying fields)
and the resulting radiation (carrying energy away) which would occur, I
think we can get to the crux of your question by starting the two
magnets at rest and bringing them together by doing a certain amount of
work W and having them end at rest. If you measure the mass of
the whole system before moving the magnets to be M , its mass
after moving will be M+W /c 2 . Again,
thinking of the field as having mass is not the right way to think about
it, you need to just think about the mass of the system. For example, if
a nucleus has N neutrons and Z protons, its mass is less than Nm n +Zm p ;
you would not want to say that the field holding the nucleus together
had negative mass, would you? Your second question, I think you can see,
is not simple either. Because the magnets are moving, there will be an
induced electric field. Because the magnets are accelerated, there will
be electromagnetic radiation carrying energy out of the system. I think
it is best to be thinking always about the whole system and not where
the energy "resides".
ANSWER:
F (this statement is
incorrect, see below) , so I will assume that is the case. If you analyze the
equilibrium problem for the beam in the figure, you find that F=Wd /L .
Note that it does not depend on the angle at all. For your case, if the
beam is uniform, i.e. d=L /2, then F=W /2=40 lb.
ANSWER:
∑F y =0=F+N-W ;
sum of torques about point touching the ground: ∑τ =0=Wd cosθ -FL cosθ=Wd-FL.
Solutions: F=Wd /L and N=W (1-(d /L )).
This results from choosing to have F be vertical. I could have
chosen to have F perpendicular to the beam in which case the
answer would depend on θ. (I now see that my
statement in the original answer that a vertical force would be the smallest is
wrong, although my answer for a vertical F was correct and
independent of θ .) The new torque equation would be ∑τ =0=Wd cosθ -FL
so F =Wd cosθ /L and the 80
lb uniform beam at 400 would require a force of F =½(80)(0.77)=30.6
lb. It would require no force to hold the beam vertical since cos900 =0.
ANSWER:
g and
having an acceleration g in zero gravitational field. Suppose
that you are in an accelerating elevator in empty space which has a
small hole in the side. Someone shines a beam of light into the hole.
That beam will follow a parabolic path as seen by you inside your
elevator. Therefore, light will also be bent by a gravitational field.
There are two answers to your last question. First, if you watch the
photons from your Euclidian frame of reference, the photons do not
follow a straight line in that space. On the other hand, what mass does
is warp the space around it so a photon will follow the shortest path
between two points which you would call a straight line in that space
which is, itself, curved.
2 ??
ANSWER:
E=mc 2 does not
apply to light (photons) because light has no mass and you would
therefore you would conclude that light carries no energy which would be
nonsense; for more detail, see the
faq page . Your
major error, though, is right at the start writing p=mv which
cannot be true for a photon since it has momentum but does not have
mass. In fact, this is not even true for particles with mass since, in
relativity, the momentum is given by p=mv / √[1-(v 2 /c 2 )].
Incidentally, the energy of a photon is E=hf and the momentum
is p=h /c .
ANSWER:
L = √[1(1+1)] ℏ.
So, I am guessing that you are visualizing the photon as a little
spinning ball and think that you can conclude (see my figure) that the
"equator" on the near side of the ball is moving forward with a speed of
c+v . However, spin in quantum mechanics cannot be visualized using
such a simple classical model, even though we often do think of spin
this way to get a qualitative feel for spin. For example, if
you visualize an electron as a spinning uniform sphere with a reasonable
radius, you find that the surface must have a speed greater than the
speed of light.
ANSWER:
ANSWER:
F 1 =ma
on you where m is your mass. If there were water, the
water would exert a force F 2 toward the back on you
and the wall would exert a force F 3 forward on you.
So now, F 3 -F 2 =ma so the
force on you from the wall would be bigger (F 2 +ma )
than if there were no water. Plus the water would be trying to
crush you.
ANSWER:
g and
having an acceleration g in zero gravitational field. If the
ship had an acceleration a forward it would be the same as being in a
gravitational field pointing backward in the ship. Therefore, there
would be a buoyant force which would cause objects with smaller density
than water to move forward ("float") and objects with larger density
than water to move backward ("sink"); you correctly point that out and
in my original answer I was assuming an astronaut whose overall density
is larger (quite possible if he were wearing a heavy space suit, for
example). In the back wall case, my original answer was correct. In the
front wall case, the water would be pushing you forward and the wall
backward, so F 2 -F 3 =ma
or F 2 =F 3 +ma ; now the
water pushes on you with a force greater than the wall pushes back.
ANSWER:
ANSWER:
ANSWER:
inverse beta decay .
If a proton absorbs an electron, an electron neutrino will be ejected:
p + +e - →n 0 +ν e .
This most commonly happens when an atomic electron is captured into the
nucleus, a process called electron capture. This process also results
in the emission of an x-ray because the hole in the K-shell is filled by
a higher-orbital electron. It is also the main way
neutron
stars are formed.
Q&A OF THE WEEK, 3/19-25/2017
0 -700
in the photograph attached by the questioner.]
ANSWER:
h and v 0
where the ball would strike the line had the obstruction not been there.
As best as I could tell, some part of the ball must touch the line so if
we calculate where the bottom-most point of the ball strikes the ground
(y =0) the ball will be in if x >0 in my coordinate
system. My guess is that if we simply assumed that the bottom point went
in a straight line in the direction of its initial velocity we would get
the right answer; but the ball is actually a projectile and moves in a
parabolic path so, since this is a $100 bet, I better do it right! The
equations of motion are
x (t )=x 0 +v 0x t
y (t )=y 0 +v 0y t- ½gt 2
where t is the time it takes to hit the ground and g =9.8
m/s2 is the acceleration due to gravity. Being a scientist, I
prefer to work in SI units, so I will do that. OK, let's summarize
everything we know. I will assume θ =700 since
that gives the best chance of hitting.
x 0 =R =2.7"=0.0686 m
y 0 =h-R =5"-2.7"=2.3"=0.0584 m
v 0x =-v 0 cosθ=- 30cos700
mph=-4.583 m/s
v 0y =-v 0 sinθ=- 12.59
m/s
y (t )=0.
So now the task is to put these into the equations above, solve the
y equation for t and put that value of t into the
x equation to find x . I find that t =0.00463 s
and x =0.047 m=1.9". Because x is positive, the ball
will hit the line. Going through the same procedure for θ =600 ,
I find t =0.00502 s and x =1.4", again hitting the line.
When the bet is settled, don't forget to
reward The
Physicist !
ADDED
THOUGHT:
θ =700 above I found x =0.0474
m and if you just assume the ball went in a straight line to the ground
you find that x =0.0473 m.
ANSWER:
—the surface which is hit is aligned with the outer
edge of the line. The best way that I can convince you that it is
possible that my first answer is correct is to show two figures, one
showing a trajectory of the ball which lands it in bounds and one out:
Each figure shows the ball at the instant of impact with the machine and
the instant of impact with the floor had the machine not been there. The
dashed red line shows the trajectory of the center of the ball and the
solid red line shows the trajectory of the bottom of the ball. The
diameter of a tennis ball is 5.4". The ball with the steeper trajectory
(which would include your 600 -700 trajectory) is
clearly in bounds by standard tennis rules.
ANSWER:
0 -700
for the trajectory angle was way off. The algebra is tedious, so I will
just give the final results. To check that I made no algebra errors, I
plotted the trajectory, shown in the figure. I have worked in meters.
The ball is hit at x =0 from a height of 1 m, passes 1 m above
the 1 m high net at x =13 m, and then hits the machine at x =25
m just above the ground (y =0 m). You can see that my result
describes the path you specified quite well. Do not be deceived by the
picture, though, because the scales of the two axes are very different,
only 2.5 m for vertical motion compared to horizontal motion of 25 m, a
factor of 10. The inset shows the trajectory as it would look to the
eye. As you can see, the angle is much smaller than you estimated. The
analytical solution is that θ =15.60 and t =1.1 s; your estimate of the initial speed was pretty
good, 23.5 m/s=52.6 mph but the final speed (although it is not
important) is 23.9 m/s=53.5 mph not 30 mph.
θ c =tan-1 (y 0 /x 0 )=tan-1 (2.3/2.7)=40.40 ;
any angle smaller than θ c will be out of
bounds. For 15.60 , x =2.7-(2.3/tan15.6)=-5.5", 5.5
inches out of bounds.
Finally, just for completeness, I would like to estimate the effect of
air drag. Using the approximation in an
earlier answer I find that t =1.13
s (0.03 s longer) and v =22.1 m/s (1.8 m/s slower). This would
correspond to the angle being a bit bigger, about 170 , but not nearly
enough to cause the ball to drop in bounds.
Q&A OF THE WEEK, 3/12-18/2017
(The questioner and I had several
exchanges. The bearings are cylindrical and he was only guessing that
the axial force was half the radial force based on data for similar
bearings. I have edited his several emails to get the gist of things in
the question above.)
ANSWER:
mg vertically down, the
normal forces on the inner (N 2 ) and outer (N 1 )
wheels, and the corresponding frictional forces f 1
and f 2 . Whatever the rated axial (along the wheel
axis) maximum force is, that is what we want to use as the to find the
limiting conditions for the "carve"; the determining factors will be the
mass m of the skater plus board, the speed v he is
going, and the radius R of the path. For the lean, d
and h are the distances, respectively, of the center of mass
horizontally and vertically from the front wheels; note tanθ =d /h .
The wheel base is s . The normal forces and frictional forces
shown each represent the forces on two wheels. Note that the inner
wheels carry the most force.
The easiest way to do this problem, since it is an accelerating system,
is to introduce a fititious centrifugal force C =mv 2 /R
pointing outward. Newton's equations are N 1 +N 2 -mg =0
(vertical forces), f 1 +f 2 -C =0
(horizontal forces), and N 1 s+mgd-Ch =0
(torque about inner wheels). Now, from the torque equation, there will
be a maximum speed you can go for a given m and R
before the outer wheels leave the ground. At this time N 1 =0=(Ch-mgd )/s
or v 2 /(gR )=d /h =tanθ.
Also, f 1 =0, so f 2 =C=mv 2 /R .
Now, the inner two wheels each are experiencing the axial force of F =½mv 2 /R .
Just to do an example, let v =10 mph=4.47 m/s, R =5 m,
and m =100 kg; then F =0.2 kN. If your guess that F max ≈1.75
kN is correct, you should be ok. For this example, θ =tan-1 [v 2 /(gR )]=220 . If you execute the turn with
a smaller lean angle, all four wheels will share part of the load. To do
the general solution where both wheels have N ≠0 would
not be two difficult but would be quite a bit messier algebraically and
probably not all that useful.
(0.2 kN, see below) to 1.75 kN, although they would be the breaking limit for only one bearing.
You assumed that just the 2 inner wheels receive the load, so isn't that a total of 4 bearings... shouldn't the breaking limit be 4*1.75?
ANSWER:
g had crept into the final stage of my calculation, (" …so f 2 =C=mg v 2 /R …")
so my final answer was too big by a factor of 9.8 meaning that the
lateral force per wheel is F= 0.2 kN; this has been corrected in the
original answer. So you should have no problem after all. I was in the
process of writing an email trying to "verbalize" my answer somewhat to
make it more accessible when I discovered this. Here is what I wrote:
You asked me “How ‘hard’ would the carve have to be (let's assume a rider of 100kg) to break the weakest of the bearings?” That is what I gave you. I assumed that the weaker of two bearings in a wheel can break even if the stronger does not.
(I assume they are coaxial, one inside the other.) Sorry if the explanation was too technical, but that is as basic as I can get to convey the details. A few comments to try to clarify:
What I call
C , the centrifugal force, determines how "hard" the carve is.
C=mv 2 /R . For the example I did, C =100x4.472 /5=400 N.
When carving, the harder you carve the greater the load will be carried by the inner wheels on each axel. Eventually, the outer wheels will just lift off the ground which necessarily results in the inner wheels taking all the load, both lateral and radial.
Since you asked me for how to calculate the maximum lateral force
any wheel would experience for a given m , v , and
R , that is what I gave you.
I do not
understand your last question but I do know that if two wheels
experience a force of 0.4 N then each experiences a force of 0.2 N
(assuming they equally share the load).
ADDED
THOUGHT: "How 'hard' would the carve have to
be…to break the weakest of the bearings?" I suggest that the
appropriate equation would be 1.75x103 =½mv 2 /R.
Here are a couple of examples:
What is
the fastest a 100 kg rider could go in a curve of radius 10 m? v =√(2x10x1.75x103 /100)=18.7
m/s=67 km/hr=42 mph.
What is
the tightest curve a 100 kg rider could turn at 30 mph=48 km/hr=13.4
m/s? R =½x13.42 x100/1.75x103 =5.1
m.
These, of
course, assume 1.75 kN is the strength of the weakest bering.
ANSWER:
ANSWER:
ANSWER:
ANSWER:
rate is about Φ ≈1045 photons per
second. Using R =4.2 ly=4x1016 m, I =Φ /(4πR 2 )= 1045 /( 64π x1032 )≈ 5x1010
photons/s/m2 . That would be the photon intensity 4.2 ly from
a sun-like star.
ANSWER:
4 m/s. I find that the speed at r =7 AU would
be about 28 m/s and the time to get there would be about 1.5x1012
s≈48,000 years. So the energy required would be that amount needed
to give your probe the escape velocity from earth's surface. Keep in
mind that this estimate ignores everything except the earth. It would be
a better question to ask, for example, how much energy would be needed
for the probe to arrive in 10 years.
ANSWER:
½% or your weight; on the
moon that force is 2.5x10-16 % of your moon weight! So, you
see, in no way do you have to worry about being "blown clear off"!
ANSWER:
f =1 GHz=109
s-1 ; since the wavelength of a wave is the speed divided by
the frequency, λ=c /f =3x108 /109 =0.3
m. So, I would have thought that a cage you made would have gaps much
less than 30 cm. Were all the wires in good contact with all others?
Often copper wires have a varnish-like coating on them to serve as
insulation. I don't know what you mean by "correlation", so I cannot
answer that part of your question. (By the way, the aluminum foil is
a Farady box if it completely surrounds the phone.)
ANSWER:
0 F=210 C
and 6000 F=3150 C; I will convert back to Imperial
units for the final answer. I looked up
data for the specific heat of water which turns out to have a
significant temperature dependence as shown in the figure (black). I did
a quadratic fit (green) to these data and integrated over the
temperature range to get E =1356 kJ/kg. The mass of a gallon of
water is about 3.8 kg so the total heat is Q =1356x3.8=5.2x103
kJ=1.44 kW⋅hr=1240 kilocalories. Keep in mind that the pressure
will be very large at 6000 F, about 1800 psi.
ANSWER:
—it would cause instabilities in
the plasma and it would damage the vessel. But the plasma is hot and
anything hot will radiate heat energy; in other words, the magnet bottle
contains the plasma, not the heat. This radiant heat would rapidly heat
up the vessel which would heat up some coolant mechanism (imagine having
water tubes coiled around the outside of the vessel) which would carry
away the heat to drive the turbines.
ANSWER:
You seem to think that you need an acceleration greater than
g to get to 0.75c , but any acceleration will do. But,
since you seem interested in a greater than g acceleration, I
will choose a 0 =1.5g. So now a person of
mass m in the ship sees two fictitious forces, mg mg mg g , and make all the "floors" in the ship
perpendicular to the acceleration. If you are interested in how long it
would take to reach 0.75c with a 0 =g ,
you can read off the graph (which I took from an
earlier answer ) that gt /c =1.8,
so t =5.5x107
s=1.7 yr.
ANSWER:
ANSWER:
-19 J. You need to realize
that if the potential difference is greater than a few hundred volts,
the kinetic energy will not be exactly ½mv 2
because of relativistic effects.
ANSWER:
ANSWER:
Car Talk on NPR and asked about these nets you can buy to replace the tailgate in a pickup truck to reduce air drag. Makes sense, right? The tailgate is like a wall in the wind and to get rid of it will reduce drag and increase your mileage. Click and Clack said that they thought these things were a great idea for reducing drag and increasing fuel efficiency. During the intervening week before the next show an engineer from GM called in and told them that removing the tailgate in fact greatly increases the overall drag on the truck. The reason is that the tailgate traps a bubble of air which rides along with the truck and the headwind slips over it. There was a lot of crow-eating that week at Car Talk Plaza!
ANSWER:
ANSWER:
p=mv (since it could not
have any momentum if m =0), but rather p=hf /c
where h is Planck's constant, c is the speed of light,
and f is the frequency the corresponding light wave. In
everyday life, the light which reflects from a mirror is the same color
of light that went in; this means that the mirror did not recoil at all,
essentially has infinite mass. If you take into account the recoil of
the mirror, some of the photon's momentum would be transferred to the
mirror which would mean the photon would have to lose some momentum. But
h and c are both constants of nature, so f
would have to decrease; that would mean that the actual color of the
photon would be slightly shifted toward the red end of the spectrum. You
could never do this experiment, the shift is too tiny. But, if you
"reflect" a photon from an electron (shoot gamma rays or x-rays at a
solid which has many electrons in it) you can easily measure the change
in wavelength of the radiation; this is called
Compton
scattering and was one of the pivotal experiments in the development
of quantum physics.
ANSWER:
This may be all that you want, but for the interested I will roughly
calculate some numbers. The terminal velocity v of something
with mass m which presents an area A to the onrushing
air can be roughly approximated as v =2√(mg /A )
where g ≈10 m/s2 is the acceleration due to
gravity. The match I estimate to have m ≈0.1 grams=10-4
kg and area A ≈(5 cm)x(1 mm)=5x10-5 m2 ;
then v ≈9 m/s. The car I estimate to have m ≈3000
lb≈1400 kg and area A ≈(2 m)x(3 m)=6 m2 ;
then v ≈97 m/s.
ANSWER:
/s/s and there is some larger force which causes
it to slow down at the rate of 3 ft/s/s; now push the car so it has an
initial speed 10 ft/s. The path followed by the car until it stops will
resemble the path shown in the figure. That is my best guess!
ANSWER:
v initial =500
mph in one direction and later are moving with speed v final =500
mph in a direction perpendicular to your original direction of travel;
it is important that you realize that velocity is a vector so that it
changes if the direction changes but the speed stays the same. The
difference between the two velocities is shown in the figure, Δv v final -v initial .
So the average acceleration is a v t
where Δt is the time to make the maneuver. The average
force F a m , F =ma m =160/32=5 lb⋅s2 /ft), Δv= 500√2=707
mph=1037 ft/s, and Δt= 1 s. The average force you will
feel is F =5x1037/1=5185 lb! You do not want to change your velocity too
quickly!
ANSWER:
g m
is g mG 1 r /r 2
where G is the universal gravitational constant, 1 r
is a unit vector pointing radially out and r is the distance
away from the point. To get the gravitational field for an object not a
point you need to divide it up into infinitesmal pieces, each having a
field dg mG 1 r' /r' 2
where r' is the vector from dm to the place where you
wish to calculate the field; you then integrate over the entire object,
usually not a trivial exercise. This is one of the main reasons that
Newton had to invent the calculus, so that he could prove that he could
treat the sun and the planets (spheres) as point masses. I understand
that the difficulty of proving that the field of a spherically symmetric
mass distribution outside the object is identical to the field there if
all the mass were in a point at the center caused a delay of something
like 20 years in the publication of his theory of gravity. I will only
talk about spherically symmetric masses here because, the point mass
field being spherically symmetric, all large bodies will tend toward a
spherical shape. This is why the stars and planets are spheres; see an
earlier answer on
cylindrical masses. Calculation inside the object is trickier: a uniform
sphere of radius R and mass M has zero field at the
center which increases linearly until the surface where it has magnitude
g=MG /R 2 ; a hollow sphere of radius R
and mass M has zero field everywhere inside and then jumps
discontinuously to g=MG /R 2 at the surface.
Since we know that a star with a large enough mass will collapse to a
neutron star and/or a black hole, nothing will stop that from happening,
certainly no "mesh" of any kind. On the other hand, if you want to
create a hollow sphere with a thin outer shell, I believe that there
would be no limit to how large you could make it without its collapsing.
Suppose you have a hollow sphere of radius R 1 and
and mass M 1 and it does not collapse; the field
pushing in at the surface is g 1 =GM 1 /R 1 2 .
Now suppose you have a hollow sphere of radius R 2
and and mass M 2 =2M 1 ; it is easy
to show that if you keep the surface density of the shell σ=M /(4πR 2 )
constant then R 2 =(√2)R 1
and so g 2 =GM 2 /R 2 2 =2GM 1 /(2R 1 2 )=g 1 .
The field remains the same and so the field per unit area actually gets
smaller as the shell gets larger, making it even less prone to collapse.
ANSWER:
earlier answer for the several possibilities of orbits but the
important thing is that a stable orbit is the solution to Newton's
equations and it simply continues forever until some other force changes
it. Next you take into account the masses of both objects and find that
the solutions are essentially the same except that the two orbit around
their center of mass and the reduced mass μ=Mm /(M+m )
replaces the mass M in your equations. Now you can add other
non-ideal forces to the problem. For example, although earth satellites
are above most of the earth's atmosphere, they are not above all of it
and the air drag with the very thin atmosphere gradually takes energy
away from the satellite and its orbit eventually decays to where it hits
the ground. The farther out the satellite is, the slower this decay is.
Other things can cause the orbit to change also. For example, the moon
causes tides on the earth (in both the oceans and the solid earth) which
causes some of the energy lost by the earth to be gained by the moon so
it is gradually moving farther away (by like about 4 cm a year). And of
course any third body perturbs the nature of the orbit. For example, the
planet Neptune was discovered in 1846 because the observed orbit of
Uranus was not exactly what its Kepler orbit should have been because of
interaction with another body, Neptune; calculations predicted the
necessary orbit of the unknown planet and it was subsequently observed
where it was predicted to be. Imagine doing such a complicated
calculation long before computers.
ANSWER:
ANSWER:
0 W, he will end up in Rio de Janeiro. The
pilot does not have to keep steering his plane to account for the
earth's rotation (which is what is hinted at here, I think); he flies
relative to the air and the air rotates with the earth. If this is a
homework question, it is a pretty stupid one.
ANSWER:
earlier
answer first which has lots of details and discussion of
acceleration. There are two ways to approach this problem:
Suppose
the observer in the inertial frame is doing the pushing and wants to
know how hard to push the particle to achieve a particular
acceleration.
Resistance to acceleration is usually called inertial mass and the
inertial mass m of a particle with rest mass m 0
and speed v is m=m 0 / √[1-(v /c )2 ].
The first figure above plots
m /m 0 as a function of v /c .
So, to achieve an instanteous acceleration of a , a force of
F=ma =m 0 a /√[1-(v /c )2 ];
so for your example of v /c =0.7, you can read off
the graph that you would need to exert a force 1.4 times larger than
you would if the particle were moving slowly. This is probably what
you want if you are interested in electrons accelerating since you
would be accelerating them.
Suppose
the observer was on the particle and the particle was a rocket ship.
You adjust your engines so that the force F which they
exert on the ship causes a constant acceleration of a 0 =F /m
where m is the rest mass of the ship. What
acceleration a does another observer in an inertial frame
(on earth maybe) see when the rocket has a speed v ?
Starting with the velocity derived in the
earlier
answer , v =(a 0 t )/√[1+(a 0 t /c )2 ],
we can calculate the acceleration by differentiating with respect to
t , [dv /dt ]/a 0 =a /a 0 =(1+(a 0 t/c )2 )-3/2 .
Now, reading off the second graph, when v /c =0.7,
a =0.36a 0 ; the stationary observer will
only see 36% of the acceleration seen on the ship.
ADDED
NOTE:
v /c =0.7 and 0.9, will observe the
ship having different accelerations. This is why acceleration, and
therefore also force, in special relativity do not play an important
role as they do in Newtonian physics where all inertial observers see
the same acceleration. This is discussed in the
earlier
answer .
ANSWER:
W N m with speed v
moving in a circle of radius R has an acceleration a=v 2 /R
which points toward the center of the circle; then apply Newton's second
law, F=ma =mv 2 /R=W-N and solve for
N to get your apparent weight of W -mv 2 /R ,
smaller than your actual weight. This is the
"Eötvös effect".
But there is another way to approach the problem. Rather
than solving the problem from the outside the earth, we might want to
solve it here on the earth. But Newton's laws are not valid in an
accelerating reference frame (accelerating because it is rotating). You
can force Newton's second law to work, though, by inserting a fictitious
force which I will call E Eötvös but it is more
commonly known as the centrifugal force; E =mv 2 /R
pointing radially out. Newton's first law now applies, N+E-W =0,
so, again, N=W -mv 2 /R=W [1-v2 / (gR )]
where g =32 ft/s2 . In the figure above you have a velocity
v=v Earth +v man . If you are at
rest, v=v Earth =1040 mph=1525 ft/s and R =3959
mi=2.09x107 ft; so N=W (1-0.00348) and a scale will
read 0.348% smaller than your actual weight. If you move with a speed of
55 mph=81 ft/s in an east direction, v =1525+81=1606 ft/s and
N=W (1-0.00386), 0.386% smaller than your actual weight. If you
move with a speed of 55 mph=81 ft/s in a west direction, v =1525-81=1444
ft/s and N=W (1-0.00312), 0.312% smaller than your actual
weight.
ANSWER:
v Earth in which case the
actual and apparent weight will be the same. If you want to compare to
your apparent weight at rest, it is 0.386-0.348=0.038% lighter going
east, 0.348-0.312=0.036% heavier going west.
ANSWER:
F over a distance D you would do
W=FD units of work on the weight. For example, the weight of a 2 kg
mass is about 19.6 N and the work to lift it 1 m is 19.6 J. But, and
here is the catch, you use more energy than 19.6 J to lift that weight
because your body is not a simple machine like a lever or a pulley. To
understand why, see the faq page. In a
nutshell, the reason is that to just hold up a 2 kg mass, not move it up
at all, requires input of energy —you get tired trying to
hold up a weight at arm's length, right? And, what about lowering the
weight back down? The work done on the weight is negative which implies
that energy is being put back into you but know that it also takes
energy for you to lower the weight at a constant speed. A biological
system is considerably more complex than systems we talk about in
elementary physics classes. I think that it is of little use to try to
analyze a workout in terms of elementary physics.
8 m/s when it emerges out in air from denser material without the loss of energy. Why?
ANSWER:
v decreases but
its frequency f stays the same. Since v = λf ,
where λ is the wavelength, the wavelength decreases.
Another way to look at it is to think of the light as a swarm of
photons. The energy of a photon is hf where h is
Planck's constant, so the energy of a photon depends only on frequency.
ANSWER:
v
a t points
parallel to the velocity vector because you are speeding up. The
centripetal acceleration vector a c
a makes an acute (less than 900 )
angle with v
ANSWER:
R , a point
a distance h above the earth's surface (laser location), and
another point a distance h above the earth's surface (target
location). The distance between them is 2d . Focus your
attention on one of the triangles with hypotenuse (R+h ).
From the Pythagorean theorem, d =√[(R+h )2 -R 2 ]=√[2Rh +h 2 ];
if h<<R , d ≈√(2Rh ). For example, if
h =10 km, about the height a commercial jet flies, 2d ≈714 km is the most distant target at the same altitude which you could hit.
ANSWER:
ANSWER:
⊗
ANSWER:
T with all the scale mass M located a
distance d below the pivot at the bottom of the T .
As shown in the figure, when one side is loaded with a mass m
the scale rotates to an angle θ with the
horizontal and reaches equalibrium. The sum of the torques about the
fulcrum is mgL cosθ -Mgd sinθ =0,
so θ =tan-1 [mL /(Md )]. As
m gets larger, θ gets larger. For small angles, θ ≈mL /(Md )
in radians.
ANSWER:
F=k 1 m g
where k 1 is a constant. Newton's second law of
motion says that the acceleration is inversely proportional to the
inertial mass and directly proportional to the applied force, F=k 2 m i a ;
in SI units, k 2 =1.
Therefore, a =k 1 m g /m i .
(This is Einstein's statement "…the
acceleration of a falling body increases in proportion to its
gravitational mass and decreases in proportion to its inertial mass .")
Now, since it is an experimental fact that a is the same regardless of
the mass, m g /m i is a constant.
If we choose to measure both m g and m i
in kilograms, and since k 1 expresses the strength of
the gravitational field, m g /m i =1.
(For the field near the earth's surface, k 1 =M earth G /R earth
where G is the universal gravitational constant, M earth
is the gravitational mass of the earth, and R earth
is the radius of the earth.)
well, i saw a theory that if you rotate the ISS, which weighs 450 tons at 10 rotations per second, it would generate its own gravity. would it be possible to spin something of proportion size at a proportional speed to generate artifical gravity like the ISS?
ANSWER:
earlier answers regarding "artificial gravity"
in rotating space stations. The idea is that if you are in an
accelerating frame of reference you feel like there is a force on you;
for example, when you drive fast around a curve you feel like you are
being pulled toward the outside of the curve. It is not a "theory", it
is elementary physics. The mathematics you need is that your
acceleration if you are going in a circle of radius R with
speed v your acceleration is a=v 2 /R .
For a given R , if you choose v such that a=g =9.8
m/s2 where g is the acceleration due to gravity, you
will feel like you are at the surface of the earth. In your case, v
can be determined from the frequency 10 rotations/second: each rotation
has a distance of 2πR so v =20πR /1. This
would imply that the acceleration would be a =9.8=(20πR )2 /R =400π 2 R ;
solving, R =.0025 m=2.5 mm. This is obviously nonsense, so you
must have remembered the the frequency incorrectly. Looking at the
figure, the main cabins appear to have a diameter of about 5 m, R ≈2.5
m, so if they spun about their central axis the required speed would be
about v ≈√(9.8x2.5)=5 m/s and so the
frequency would be f =5/(5π )=0.32
rotations per second=19 revolutions/minute. But this would not work at
all for the ISS because an astronaut's head would feel almost no
"gravity" because it would be very close to the axis of rotation; if she
were to lay on the floor/wall/ceiling it would be close to what it would
be like on earth. The ISS is just too small.
ANSWER:
ANSWER:
Imagine we have a 2 lb ball of putty moving
with a speed of 5 mph striking and sticking to a 18 lb bowling ball
at rest; the time it takes to collide is 0.1 s. After the collision,
the two move together with a speed of v 1 . To
find v 1 , use momentum conservation: 2x5=(18+2)v 1 ,
v 1 =0.5 mph.
Next, imagine we have a 18 lb bowling ball
moving with a speed of 5 mph striking and sticking to a 2 lb ball of
putty at rest; the time it takes to collide is 0.1 s. After the
collision, the two move together with a speed of v 2 .
To find v 2 , use momentum conservation:
18x5=(18+2)v 2 , v 2 =4.5 mph.
So, you see
that the two scenarios have different speeds after the colliision. But,
suppose that you were the putty ball. During the collision you feel a
force and the force is what is going to hurt you. Do you get hurt as
badly, not as badly, or equally as badly during the collision? What
determines the force you feel is the acceleration you experience during
the collision, how quickly your velocity changes, which is your final
velocity minus your initial velocity divided by the time of the
collision.
For the
putty ball moving initially, (v final -v initial )/t =(0.5-5)/0.1=-45
mph/s.
For the
bowling ball moving initially, (v final -v initial )/t =(0-4.5)/0.1=-45
mph/s.
You could go
through through the exact same process to find that the bowling ball
experienced exactly the same force regardless of who moved initially. A
physicist would say that you were right, but the ambiguity of your
statement means that the other guy could split hairs. As far as physics
is concerned, the only thing which matters is the relative
velocities of the two before the collision. If the putty ball were
moving 105 mph and the bowling ball were moving 100 mph in the same
direction, the result of the collision which matters (the force) would
be the same.
ANSWER:
ANSWER: ·hr)(1000
W/1 kW)(3600 s/1 hr)=360,000 J. So the energy used by the kettle is
(216,000 J)/(360,000 J/kW·hr)=0.6 kW·hr. A kW·hr
costs on the order of 5¢-15¢, so the cost would be between 3¢
and 9¢.
ANSWER: earlier answer .
ANSWER:
Serious efforts are underway to try to observe it. But its
anticipated period is more than 10,000 years, so if it is far away now,
it is not likely to be a problem for us for far longer than the lives of
any of us. All reputable astronomers have declared that (if it actually
exists) it is absolutely no danger to earth. Google "planet 9" to get
lots of good information. But, even if there were a planet much closer,
like the "fake news" "planet x", there is an amazingly low likelihood of
its colliding with earth. Most lay folk like your wife have no
comprehension of the vastness of space; the probability of two
particular objects in the solar system colliding is, for all practical
purposes, zero. It is often posited that a "rogue planet" passing
through the asteroid belt would "shake loose" a "storm" of asteroids
toward the earth. I recently
answered a question
addressing this possibility which you might find useful.
ANSWER: T T <0, you have made
a mistake somewher.
Q&A OF THE
WEEK (1/21-28/2017)
ANSWER:
ANSWER: —to solve problems with simple physics
concepts. This problem requires facility with trigonometry,
understanding of Hooke's law, and application of Newton's first law. The
simple model I will use was one used before the advent of computers; the
bow is modeled as two straight rods (purple) the ends of which move on a
circle as the string (red) is drawn. With this simple model, most of the
details of your bow are not necessary. When the string is braced
(undrawn) there is a certain tension in the string and this tension will
increase as the bow is drawn. So, the maximum will be at the maximum
draw. The figure shows, roughly to scale, the situation. Using simple
trigonometry (law of cosines), I find β =59.60 .
The point where the draw weight W is being applied must be in
equilibrium, W -2T cosβ =0; solving, T =69.2
lb. I think that your concern about the string having to "stop the
momentum of the limbs" is misplaced because bows tend to be quite
elastic so that nearly all the energy imparted to the bow by drawing it
is imparted to the arrow and the limbs will end nearly at rest. Not so
if you draw and release without an arrow, though; what I have read is
that in that situation you are more likely to break your bow than the
string. I am working on a general solution which I will later add here
but thought I would post the part of the solution which answers your
question regarding the tension in the string.
excellent approximation, like a simple spring
(Hooke's law), the draw weight being proportional to the draw distance,
i.e . W≈kx where x is the distance the
string is drawn and k is the spring constant. In this case, since W =70 lb when x =28
in, k =2.5 lb/in. Using the law of cosines, cosβ =(L 2 +(x+d )2 -R 2 )/(2L (x+d )).
Again, the point where the force W W -2T cosβ =0 or T (x )=kx /(2cosβ ).
Now, note that in the limit where x → 0, β → 900
and cosβ → x /L . Therefore T (0)=kL /2.
Using your numbers, T (0)=37.5 lb and the angle and tension for
all points are plotted below. At full draw, β= 560
and T =64 lb. It was interesting to me that in order for the
calculated values to be correct at zero draw, very precise relative values of
R and L
had to be used because otherwise the expression for cosβ would not be 0 exactly when x =0. The value was
R =30.59411708 inches for L =30.
ANSWER:
ANSWER:
ANSWER:
ANSWER: faq page .
ANSWER: h =2 m and the trampoline never
goes down as far as s =1 m. So I will just do my calculations
with those to get an upper limit on what force might be expected. Your
son's mass is about m =120 kg. An object falling from h =2
m will hit the trampoline with a speed of about v = √(2gh )≈√(2x10x2)=6.3
m/s. I will treat the trampoline as a simple spring so that I can write ½mv 2 =½ks 2 -mgs
where k is the spring constant. Putting in m ,
v , and s and solving for k I find k =7200
N/m; since the force exerted by a spring is F=ks ,
the largest force the trampoline exerts on your son is about 7200 N=1600 lb;
Newton's third law tells you that this is also the force your son exerts
down on the trampoline. Therefore, the trampoline exerts a force down on
the floor of 1600+W where W is the weight of the
trampoline. This is a little more than the weight of a grand piano.Keep
in mind that this is the greatest force and just for an instant; the
average force over the collision time would be half this. This is a
little more than the weight of a grand piano.
ANSWER: 235 U and one neutron. Now, when
you add the neutron to the 235 U it fissions and, after all is said
and done you have two lighter atoms and a few neutrons; if you weigh all
these byproducts, you will find that approximately 0.1% of the original
weight is missing. Where did it go? Most of it went into kinetic energy
of the byproducts, that is they are all moving faster. Kinetic energy of
atoms is essentially what thermal energy is —the reactor (or
bomb) has gotten hotter. For a bomb it all gets enormously hotter
resulting in the explosion. For a reactor, the rate of fissioning is
controlled and the heat is extracted to drive turbines to create
electricity. More detail can be found in an
earlier answer .
ANSWER: ½mv 2
type of kinetic energy. The "transformation" is simply that—mass
energy transformed into kinetic energy. To understand, see a
recent answer which explains why a bound system has
less mass than if it is pulled apart and the mass measured.
It turns out
that heavy nuclei like uranium are less tightly bound than nuclei with
roughly half their mass; therefore when they split the products are less
massive. That is why fission works as an energy source.
ANSWER:
ANSWER: m and M , respectively. Before they come
together the momentum is Mv where v is the incoming
speed of M . When the masses come in contact they will slide on
each other but, because there is friction, they will eventually stop
sliding and both will move with a velocity u ; the linear momentum will
now be (M+m )u . Conserving momentum, u =[M /(M+m )]v .
They both end up going with speed u and move to the right.
You can also determine the time it takes for the sliding to stop. m
will feel a frictional force to the right of magnitude f= μmg
and M
will feel a frictional force to the left of magnitude f= μmg
(Newton's third law). So, choosing +x to the
right, the acceleration of m is a =μg and
the acceleration of M is A =-μg (m /M ).
The velocities as a function of t are v m =μgt
and v M =v -μgt (m /M );
we are interested in the time when v m =v M ,
so solving for t , t=Mv /[μg (M+m )].
If you substitute this back into v m or v M ,
you will find the same value we found for u above: v m =v M =u =[M /(M+m )]v .
ANSWER: m
will drop down on the frictionless surface. It would be a good exercize
for a student to calculate how far the block would slide for some
μ . And then, if this is greater than the size of the bottom
block, how fast will each be moving after separating.
ANSWER: F ≈¼Aw 2 where A
is the area presented to the wind and w is the speed of the
wind (this approximation only works for SI units). I will assume that
the horizontal speed acquired by the jumper is small compared to the
wind speed so that w is a constant during the fall. So,
approximating A ≈1 m2 , the
acceleration horizontally is a x =F /m =w 2 /(4m )
where m is the mass of the jumper. The time to fall 350 ft is
about t =4.7 s and the horizontal distance is then x =½a x t 2 =w 2 t 2 m ).
Now, x =25 ft=7.6 m and I will take m =150 lb=68 kg.
Solving for w I find w =13.7 mps=31 mph. A steady wind
of 31 mph could cause the jumper to move 25 ft horizontally.
You have not told me where the the booth is. If it is under the bridge,
then the jumper could not propel himself in that direction. But if the
booth is 25 ft out from under the bridge, the jumper could jump out as
well as drop down. The speed v x he would have to
give himself horizontally can be easily calculated: v x =x /t= 25/4.7=5.3
ft/s=3.6 mph.
ANSWER: 2 . The square of the wave function is the
probability density and so is a measure of the likelihood of finding the
particle in one small volume in space. If you add up the square of the
wave function at all points in space (integrate), you must get the
answer of 1 because the probability of finding the particle somewhere in
space must be 1 for this interpretation to make sense; this is called
normalization.
ANSWER: is a form of energy and must be factored into
any energy conservation that occurs in an isolated system. You say that
the masses of the protons and electrons do not change, but that is not
right. Look at the simplest case, a hydrogen atom. If you measure the
mass of this atom it will be less than if you measure the masses of a
free electron and a free proton. Here is how you can see that: if you
pull the electron away from the proton, that is you ionize the atom, do
you have to do any work? Of course you do because the electron and
proton are bound together. So, you have added energy to the system (p+e)
and that energy shows up as mass. In a system as complicated as a
molecule you cannot say which particle or particles changed their
masses, but you can say for sure that the total mass of the molecule
changed by exactly the energy of the emitted photon divided by c 2 .
ANSWER: ≈2x104
m/s and 30x103 km/s=3x104 m/s). So let's just
choose the larger one and see how much time dilation there is. Relative
to the sun, an elapsed time T =4x109 y would
correspond to T'=ΥT wwhere Υ= 1/√(1-(v /c )2 )=1/√(1-(3x104 /3x108 )2 )≈(1+0.5x10-8 ).
Therefore T'≈ (T +20), a difference of 20 years. This
may sound like a pretty long time to you, but relative to 4 billion it
is less than 10-6 %. That is not "significant" to my mind.
You are right that every object has its own clock which, relative to
other clocks, is not necessarily the same; every object also has its own
meter stick, not necessarily the same as other meter sticks in the
universe. The important thing is that you always must talk about
velocity with respect to what.
ANSWER:
ANSWER: F =3x10-11 N; the
weight of that 1 kg object is about 10 N. I would say that putting a 1
kg object on the ground is a great deal more likely to cause an
earthquake than those planets, wouldn't you?
ANSWER:
ANSWER: M and volume V
from the string. Besides the string, there are two forces on the object
being weighed, its weight Mg and the buoyant force B= ρVg
where ρ is the density of air (about 1 kg/m3 at
atmospheric pressure) and g =9.8 m/s2 is the
acceleration due to gravity. The scale will read W=Mg-B , an
incorrect measure of the weight. Putting the whole device in a vacuum
will change B to zero because the air is gone, so W=Mg ,
the correct weight. To get an idea of how important this is, consider
weighing a solid block of iron whose mass is 1 kg. The density of iron
is ρ iron =7870 kg=M /V , so the
volume of the block is V =1/7870=1.27x10-4 m3 .
So, the true weight is W =9.8 N and the measured weight in air
is (9.8-1.27x10-4 ) N=9.7999 N, an error of 0.0013%. However,
there are certainly examples where the effect of buoyancy would be very
important. For example, consider an air-filled balloon. I did a rough
calculation and estimated that the volume of an inflated balloon is
about 5x10-3 m3 so it contains about 5x10-3
kg of air; the mass of an uninflated balloon is about 5 gm=5x10-3
kg, so the total weight of the inflated balloon is about 9.8x10-2
N. But if you weighed it in air you would only measure half that amount.
Your question, has anybody ever actually observed this, is a no brainer:
since the existence of a buoyant force has been known and understood for
well over 2000 years (Archimedes' principle), anyone wanting to make an
extremely accurate measurement of a mass would either correct for it or
eliminate it.
ANSWER:
