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.
f acting on the
car which stops it is f= μmg where μ
is the coefficient of friction between the tires and the road, m
is the mass of the car, and g =9.8 m/s2 is the
acceleration due to gravity. The work done by the friction if the car
skids a distance s while stopping is W =-μmgs , the
negative sign meaning the work took kinetic energy from the car. The
work is also the change in kinetic energy, W =ΔK =½m 02 -½mv 2 =-½mv 2 =-μmgs.
So, solving for s , s=v 2 /(2μg ).
So, you see, if you double the speed you increase the stopping distance
by a factor of four.
or a wave, it is a particle and a wave. If
you design an experiment to "prove" that light is a wave (particle) you
will "find" that it is a wave (particle). If you have light waves with
some frequency f , that light can equally correctly be thought
of as a collection of particles (photons) which have an energy E=hf
where h is Planck's constant. The oscillating wave depected
inside your "football" is a cartoon representation and really has no
quantitative meaning. Incidentally, you can never have a precise
knowledge of the energy of a photon because of the uncertainty
principle; therefore you cannot have a precise knowledge of the
frequency of the wave. Your "football" is called the envelope of the
wave function and is representative of the uncertainty of the position
of the photon.
—energy
is energy. (Thermodynamics has never been my forté, so I welcome
comments here!)
v t is the maximum speed something
achieves when dropped from a very high height and may be
shown ,
for of a sphere of mass m and radius R , to be
approximately v t ≈√ (9.7m /R 2 )
at sea level. The density of ice is about 103 kg/m3 ,
so we can write m =4x103 πR 3 /3=4.2x103 R 3 .
So, the speed of a hail stone of radius R may be approximated
as
v t =√ (4.1x104 R );
for example a golf-ball (R ≈0.021 m) sized hail stone will
have a speed of about 29 m/s and a mass of about 0.039 kg.
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.
https://www.youtube.com/watch?v=Dt0JyXMjbNs
At 18:48, why the partial differentiation notation can be "taken out" of the bracket?
(∂/∂x){ψ* (∂ψ /∂x )-(∂ψ* /∂x )ψ}={(∂ψ* /∂x )(∂ψ /∂x )+ψ* (∂2 ψ /∂2 x )-(∂ψ* /∂x )(∂ψ /∂x )-(∂2 ψ* /∂2 x )ψ }=ψ* (∂2 ψ /∂2 x )-(∂2 ψ /∂2 x )ψ.
mg T F F-T sin θ =0 and mg-T cosθ =0.
Solving, tanθ =x /y =F /(mg ),
so x=y tanθ =Fy /(mg ). Now, how
can we get F ? There is a very good approximation to the drag
force by air at sea level moving with speed v : F ≈¼Av 2
where A is the area of the object presents to the onrushing air
(and which works only for SI units). Finally, x =Av2 y /(4mg ).
If I approximate g ≈10 m/s2 and A ≈1
m2 and use your numbers, x ≈9 m.
±20%. Also, for
your situation, the general approximation for x as a function
of y is x≈y /7. Also, if the horizontal displacement seems
too large, keep in mind that 20 m/s is a very strong wind, about 45 mph
which is gale force.
each
would think the others clock was slow…" was wrong —they
were both slow as observed by the other. I think you would understand
the whole thing better if you read my earlier answer on the
twin paradox .
d ≈2 cm=0.02 m; for simplicity,
assume the orange stops with constant deceleration. If an orange of mass
m =¼ lb=0.11 kg and speed v hits the wall and
stops in a time t , the force it experiences is F=mv /t
and, because of Newton's third law, this will also be the force the
orange exerts on the window. During the collision, there will be some
average area of contact between the orange and the window; I will
estimate that area to be a circle of radius R=0.01 m so that the area
will be about A=πR 2 ≈3x10-4 m2 .
The pressure will then be P=mv /(At ); converting to SI
units, P =50 psi=3.45x105 N/m2 . For the
assumptions I have made (uniform acceleration) it is easily shown that
t= 2d /v , so P=mv 2 /(2Ad )=3.45x105 =0.11v 2 /(2x3x10-4 x0.02);
solving, v =6.1 m/s=13 mph and t =0.0066 s. Keep in mind
that this is a very rough estimate of the speed, but it does indicate
that it is not surprising if the glass broke since almost anybody can
throw an orange with a speed more than 13 mph.
is the very reason for time dilation and length contraction! So,
the answer is no; the speed of light is the same for all observers.
s =128-4=124
mm=0.124 m and the area of the top is about π (0.03)2 =2.8x10-3
m2 . The force which the press exerted on the can was
F=PA =6x105 x2.8x10-3 =1680 N. This means if
you put a mass 1680/9.8=170 kg it will crush the can. Your idea, I
presume, is to drop a smaller mass from some height (0.5 m) above can.
The way I will attack this problem is to first calculate the work done
by the hydraulic press W 1 and equate it to the work
done by the dropped mass W 2 ; I can then solve for
the mass for any height. I will do it in general so you can estimate the
mass from any height. W 1 =Fs =1680x0.124=208
J; v =√(2gh ) where g =9.8 m/s2
and h is the height (0.5 m in your case); W 2 =½mv 2 -mgs =mgh-mgs =208
J. Solving, m =208/[g (h-s )]. In your case,
h =0.5 m and s =0.124 m, m =56 kg.
Keep in mind that this is just an
estimate. I would be curious to know if it was close. There is an
earlier answer which might
be of interest to you; it might give you an idea of how you might
improve your crusher.
earlier answer on
this topic. The equation you can use to compare elapsed time is T away =T home √[1-(v 2 /c 2 )]
where
T home is the elapsed time for the earth-bound
twin,
T away is the elapsed time for the traveling
twin, v is the speed of the ship, and c is the speed
of light. The easy way to understand this is that the traveling twin
sees the distance to the star shorter by a factor of √[1-(v 2 /c 2 )].
earlier answer
where your question is answered for a speed of 0.8c . Notice
that your speculated "distortion" is experienced by both observers, not
just the traveler. It also depends on the direction of travel —toward
or away from the earth. For v =0.1c , the effect would
be much smaller but qualitatively the same—for the traveler moving
away, both observers find the communication slowed down, for the
traveler moving toward the earth, both observers find the communication
speeded up. Your statement that "…they
can not communicate with fixed objects…" is false; it is just
slowed down or speeded up.
No homework.
REPLY:
to ask the question again in a couple of months to certify that this is not a homework question or does "no homework" amount to no homework "type" questions that could require tutoring or classes?
Well, you seem sincere about wanting to learn the physics, so I am happy
to answer your question. The algebra to get the final solution is very
tedious, but it seems you are most interested in getting the physics
right, so I will set the problem up and if you really need all the final
answers for a particular situation, I will leave that to you. Since you
refer to conservation of energy, I assume the problem of interest is an
elastic collision. The system is shown before and after the collision in
the figure. A point mass m with velocity v M and length L ;
v d from the center of mass of the
rod. After the collision, the center of mass of the rod has a velocity
U u ω
about the center mass. Because the rod is uniform, its moment of
inertia about the center of mass is I=ML 2 /12; the
angular momentum and kinetic energy of a rigid body are L=I ω
and K =½I ω 2
respectively. There are no external forces or torques on this system, so
angular momentum and linear momentum are both conserved. We now have
three equations:
energy
conservation: ½mv 2 =½mu 2 +½MU 2 +ML 2 ω 2 /24
linear
momentum conservation: mv=mu+MU
angular
momentum conservation: mvd=mud +ML 2 ω /12
Take stock of
what we have: three equations and three unknowns—u , U ,
and ω. The physics is done, only algebra remains.
F d =cv 2
where c is a constant (but which depends on the geometry of the
falling object) and v is the speed; this is a force acting in
the opposite direction from the velocity. When you jump from the plane,
you originally have a horizontal velocity but the horizontal component
of the drag will take away the horizontal velocity and you will end up
falling vertically; I will assume the skydiver is falling vertically (or
just jump from a hovering helicopter). In addition to drag, the only
other force acting on the skydiver is his weight W=mg , a force
vertically down. So, writing Newton's second law, ma =cv 2 -mg ,
where a is the acceleration. You start out with an acceleration
g down, but as v increases a gets smaller and
smaller until eventually v =v t = √(W /c );
this is called the terminal velocity. Now the acceleration is zero so
you fall with a constant speed v t . At no time was
the drag bigger than the weight. But, can the drag be greater than the
weight? If you were propelled downward from the helicopter with a speed
greater than the terminal velocity, you would slow down until your
reached the terminal velocity. In that case, the drag is greater than
the weight; however, I cannot imagine this scenario in skydiving. By the
way, you can control the terminal velocity—spread eagle will give
a slower final speed than curled up in a ball.
the
actual 'real' distance traveled ". The distance traveled in one
frame of reference will be different than for another frame. For your
question, if the ball is dropped from a mast of height H , it
will travel a distance H in a time t where t = √(2H /g )
where g =9.8 m/s2 is the acceleration due to gravity.
Now, look at the path of the ball from a dock the boat is passing; if
the boat moves with a speed v , the distance the boat travels
before the ball hits the deck is vt=v √(2H /g ).
The distance between the starting and ending points of its path (blue
line) is √[(2v 2 H /g )+H 2 ].
For example, if H =4 m and v =5 m/s, then t =0.9
s and the distance is 6.03 m. The path followed by the ball in the dock
frame is a parabola (red line), so its length is the distance the ball
has traveled; it is rather complicated mathematically to compute this
distance.
I got curious and calculated the distance the ball actually traveled
in the dock frame. You need to integrate ds = √[(dx )2 +(dy )2 ]=[√(1+(dy /dx )2 )]dx= [√(1+ 4x 2 )]dx
from x =0 to x=vt . The result is s =¼[2vt √(1+4v 2 t 2 )+sinh-1 (2vt )];
for the example above, s =9.19 m.
2 - or time itself must slow down, which is both unlikely.
Your 'complicated mathematical computation' supposedly addresses the problem. But can a mathematical computation correct two conflicting observable facts?
I am thinking of Albert Einstein's skeptical quote, "So far as the theories of mathematics are about reality, they are not certain; so far as they are certain, they are not about reality."
"…the ball must…accelerate
faster than 9.8 m/sec2 …"
Since there is no satisfactory
theory of quantum gravity, I do not want to poo-poo your theory, but
something is causing a repulsive force in the universe and if that field
were quantized there would have to be a quantum of the field, and maybe
a repelon would be a good name for it. I should remind you here that it
is stated on the site that I do not usually do
astronomy/astrophysics/cosmology.
5
N/m2 =10 N/cm2 , about equal to the weight of 1 kg.
That is still a lot, though. The reason this pressure does not crush you
is that we evolved in this atmosphere and inside every cell there is a
pressure on the inside which is equal to atmospheric pressure. On a more
macroscopic level, the pressure in the cavities in your body is about 105 N/m2 .
earlier answer .
X s are the centers of gravity
(COG) for the unladen drone and the load; the load COG is a distance
L below the drone COG. The red X is the COG for
the combination. Suppose the weight of the drone is W drone
and the weight of the load is W load . Then the
distance d by which the new COG is below the COG of the drone
is d=LW load /(W drone +W load d =(7/12)L . Clearly, when L
is increased d is increased. And, contrary to your expectation,
the COG goes farther down when you hang the load from a rope increasing
L . What might be a problem for stability, though, is that if
the drone is not horizontal the rope will still be vertical and so the
COG will no longer be on the center line of the drone.
Would you mind walking me through this one conversion example using Schrodinger Equation.?
I am not a physicist and this Schrodinger Equation formula and values are Greek to me. But if you could set up the equation with this example at 10 nm wavelength, then I can simply cut and paste and copy and the do all the quantum math I want after that.!!!
Thank you for considering my request.!
EX-A)
Bands formed after wave passes thru Slit(s)
Theoretical Photonic wave
single-slit diffraction pattern . You apparently found
a reference for someone who did this using the Schrödinger
equation, but you really do not have to go through all that in order to
get an analytical expression for the intensity pattern on the screen;
all you need to do is to use
classical physical optics , a standard first-year physics
calculation. I presume that the first figure, where you have used some
computer code to get the photon energy of radiation of a particular
wavelength is not a mystery to you; but you do not need this at all for
the intensity, you just need the wavelength λ and the
slit width a . As shown in the figure, you may specify a
point on the screen by specifying either the angle θ
or the distance y ; the results are I=I 0 sin2 [πa sinθ /λ ]/[πa sinθ /λ ]2
or I=I 0 sin2 [πay /(λD )]/[πay /(λD )]2
where D is the distance from the slit to the screen; the
quantity I 0 is some constant which would be
determined by the intensity of the radiation on the slit and the
exposure time. The details are shown on the
Hyperphysics web site. Let's do an example, λ= 5 nm,
D= 10 cm, a= 32 nm; then I /I 0 =sin2 [2y ]/[2y ]2
where y is in cm. The result is plotted in the graph of I /I 0
vs. y .
closed box. A good explanation of this kind (closed box) of
problem can be found in an old
answer . In your question, the answer depends on how the bird takes
off. If she propels herself upward by just using her wings, the reading
of the scale will decrease by the weight of the bird; it's the same as
if you simply lifted her off the perch. If the bird takes off by
jumping, that is pushing off using her legs, the reading will increase
and then, when she is airborne, it will return to the weight of the cage
without the weight of the bird.
m 0 this is the mass referred to in Newtonian
mechanics and this is what the mass seems to be unless the velocities
are comparable to the speed of light, speeds we never see in everyday
life. In general the mass m depends on its speed v ,
m=m 0 / √[1-(v /c )2 ]
where c is the speed of light.
As you heat it up, you cannot know
for sure what will happen in detail; it will depend on how many atoms
you have and the volume in which they are confined. You can be sure that
the first things that happen will be at the atomic level—molecules
dissociate and atoms ionize. But as long as the energies of the atoms
and electrons are not great enough to cause nuclear reactions, the whole
system will be dynamic, constantly changing as atoms capture free
electrons and other atoms get more ionized.
As all this happens, many photons will be generated and absorbed. As
more energy is added, collisions begin to excite and dissociate nuclei
and add more photons to the mix. As more energy is added, exotic and
unstable elementary particles are created and decay. If you now start
taking energy away, can you really imagine that it could possibly go
back to its original state? You would eventually get back to where you
had only neutral atoms, but it is impossible to know what atoms you
would end up with. It would be imperative to know all the details of how
you added and then subtracted the energy, and even then you could not
predict everything that would happen.
recent
answer ; the main difference is that in that example the acceleration
was due speeding up rather than due to
turning. Essentially, you are accelerating because the jet is
turning and no matter what, there must be a net force on you which
provides that acceleration.
3 kg-1 s2 ?
This classical formula is very interesting but very challenging to visualize. The numbers all work but an explanation of the resulting units are difficult to find. Please if anybody is able to visualize the logic behind 9.8 meters a second squared and can share, it would be GREATLY appreciated.
m and M separated by a
distance d is proportional to the product of the masses and
inversely proportional to the square of their separation, F ∝Mm /d 2 ;
this is also true for two spheres if d is the separation of
their centers. To convert this into an equation, we introduce a
proportionality constant G,
F =GMm /d 2 . For a given M ,
m , and d we must measure F to determine G :
G=Fd2 / (Mm ). Now, what are the units of
G ? N⋅m2 /kg2 . This probably answers your
question, but let's finish it off. If M>>m , m will
accelerate toward M with an acceleration a=F /m =GM /d 2 , à
la Newton's second law. Since 1 N=1 kg⋅m/s2 , the units
of a are [(N⋅m2 /kg2 )(kg)/m2 ]=[(kg⋅m/s2 )(m2 /kg2 )(kg)/m2 ]=m/s2 .
ANSWER:
F =dp t where
p mv is the linear
momentum, so if there is no net force the momentum does not change.
Then, if you watch the water coming off one drop at a time you can
calculate the new speed of the ball due to the expulsion of the drop;
therefore the ball does accelerate in that sense but not because the
mass is changing but that the momentum stays the same. As a simple
example, let the mass of the ball plus one drop be M and the
mass of the drop be m; the velocity u V M and the velocity of the ball after the drop has left (now
of mass M-m ) is U p 1 =MV and after
the drop leaves is p 2 =mu +(M-m )U.
Since p 1 =p 2 , the new speed
of the ball is U =(MV-mu )/(M-m ). Notice that
if m is very small compared to M , U ≈V .
ANSWER:
μm wavelengths and is about 90% transmitted through
the glass. There is a precipitous drop in transmission around 2.5 μm
and the glass becomes nearly opaque to wavelengths above 5 μm. The
region 0.75-2.5 is referred to as near infrared (IR), so the glass does
transmit most of that IR, but the IR radiation which is radiated by
objects with temperatures between the freezing and boiling points of
water is on the order of 10-12 μm. You are right, glass lenses are
useless for infrared optics. Lenses have to be made from a material
which transmits IR; common materials which are used are calcium flouride
and germanium. This is part of the reason that IR optical devices are
expensive. This property of glass is also how greenhouses work. Visible
light enters gets absorbed and heats the inside. But as the inside warms
up, it radiates IR but that radiation cannot get back out. The heating
you feel from light coming through a window results from absorption of
visible and near IR which does get through.
F D =½ρv 2 C D A
where ρ is density of interstellar space; which should be about 2 million protons per m3 .
However I am not sure if in relativistic mechanics it is as such:
F D =ρv 2 γ 2
where γ is the lorentz factor (gamma factor).
Both of these equations will give different results.
Furthermore, what is the kinetic energy when matter within the vacuum of space impacts on cross-section area (π500^2) of the ship facing the direction of travel?
The equation is as follows:
KE=(γ−1)p c^2
However the results from these equations are about 26.6 N, 0.0028 N and 6.6e−4 J respectively. The latter is not much! However in many articles on the web and as per certain experts, the KE should have been greater and as explosive as nuclear bombs?! Where am I wrong?
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 bearing.
