R=v 2 sin(2θ )/g
where R =130 ft, v is the speed, θ is the
launch angle, and g=32 ft/s2 is the gravitational
acceleration; the "optimum" launch angle is 450
for maximum range. So v =√(130x32)=64.5
ft/s=44 mph. For a smaller launch angle the speed would be
larger; e.g . at 300 , v =50 mph.
earlier
answer . The issues you raise could mostly be handled by
making a judicious choice of target. Stellar evolution is
reasonably well understood and the life time and age of a
star could be decuced from its properties. For example, if
the star you have in mind is within a million years of
running out of fuel and is a million light years away, it
would foolish to plan a journey there. Regarding distortion,
the earlier answer emphasizes that problem but the most
important distortion is that caused by the high speed of the
vehicle itself; "distortion" due to motion of the target
star itself as well as gravitational lensing could be
adjusted for in flight if you could actually see and lock
onto the star at high speeds. Your last question has no
answer because you need to specify how small a lag you would
call "current"; we are quite close to the sun but light from
it takes about 8 minutes 20 seconds to get to us.
(To give a background on why I asked the question. I have a habit of clicking my pens on my desk so they shoot a few inches into the air. I always try to catch them right at the height of their motion. I started wondering how long I actually had to do that. When I asked some engineering friends they said the pen instantaneously goes from traveling up to falling down. I’m not sure if that’s true but if so it introduces a bit of a crisis. If the pen spends zero time at its maximum height, is it ever actually there? And would I ever be able to actually catch it at is maximum height? Can something be somewhere but spend zero time there?)
To get more technical, what is the probability of finding
the object at a particular point? It is zero. That is
probably what is bothering you. So you could ask for the
probability of catching the pen during the interval 1/1000th
of a second before and after the zero speed time. I will not
get into quantum physics here, but you cannot know precisely
both the speed and location of your pen at any time; but
this truth implies an incredibly small uncertainty,
certainly not important for a macroscopic object like your
pen. One final thought, we do not know whether or not there
exists a shortest possible time or length, sometimes
referred to as the Planck length and the Planck time; if it
exists, the Planck time is expected to be about 5x10-44
s, so perhaps your pen stands still for that time!
faq page .)
2 until they get all the
way to the ground. The air drag is a force on the falling
object upward (opposite the velocity) which gets bigger as
the object falls faster; eventually the drag force becomes
equal in magnitude to the weight which is a force down, so
the object falls with constant velocity after that. That
speed is called the terminal velocity and that depends on
both the mass and the geometry of the object. The speed of
120 mph is the approximate terminal velocity of a falling
person. A bowling ball would have a much larger terminal
velocity and a feather a much smaller one. If the earth had
no atmosphere, anything would continue accelerating until it
hit the ground. If you want to learn more about air drag,
you can find links to many of my older answers on the
faq page . If there
were no air in the hole you want to drill and the earth were
a uniform sphere (which it isn't), the speed would be about
18,000 mph at the center.
etc .
Is a block of iron a "different type" of solid from a block
of tin? Yes. Similarly, a plasma composed of iron ions is a
"different type" of plasma than one composed of tin irons.
v . I will assume that the time which the
collision lasts, Δt , is the same for any situation
I consider. There will be two extremes for a collision—perfectly
elastic where energy and linear momentum are conserved, and perfectly inelastic
where the objects are stuck together after the collision and
energy is not conserved but linear momentum is. The impulse
received by the object originally at rest (your opponent) is
equal to the linear momentum he has after the collision and
is equal to F Δt ; since I specified that Δt
is the same for all collisions, the force experienced is
F=mv' /Δt , where v' is the speed of the struck
object after the collision. First consider a collision
between two equal masses m . For the elastic
collision the incoming mass is at rest after the collision
and the outgoing mass has a speed v ; for the inelastic
collision both masses (stuck together) have a speed ½v .
Therefore the force experienced by your opponent is given by
½mv /Δt ≤F ≤mv /Δt .
Now suppose that your mass M is much larger than your
opponent's mass m , M>>m . In this
case, the inelastic collision will have both objects moving
with approximately a speed of v ; the elastic
collision will have M moving
approximately with the speed v and m
moving approximately with the speed 2v . Therefore
the range of possible values for F experienced by
m is mv /Δt ≤F ≤2mv /Δt ,
which is twice the force for equal masses. So my simplistic
view is that being anchored to the floor is like increasing
your mass allowing you to deliver more impulse to your
opponent as compared with if you are flying toward him which is
more like the collision of equal masses. As I said, it is
much more complicated than this but this might give a little
insight. I would welcome any comment on this treatment.
(Lengthy enough answer for you?)
that direction.
2 )=102,900 N
and the total area is (4 m)x(7 m)=28 m2 . So the
pressure is 3675 N/m2 . Other units used for
pressure which you might prefer are 375 kg/m2 or
76.75 lb/ft2 .
earlier answer (skip down to the
last paragraph ).
p=q d q separated by some distance
d ; imagine these as vibrating
sinesoidally along the direction of the vector d p p here .
Brief Answers to Big Questions ) it is stated that a chemical rocket with a thrust velocity of 3 km/sec, that jettisons 30% of its fuel, attains a velocity of 'about half a km/sec'. Using
Tsiolkovsky's rocket equation I get a value of 1.07 km/s.
Have I made a mistake?
earlier answer , v=u ln[(M+m )/M ]
where M+m is the initial mass, m is the
fuel burned, u is the exhaust speed, and v
is the speed acquired by the rocket. So I find v=3ln(0.7-1 )=1.07
km/s. I do not know how Hawking came up with his answer, but
it might hinge on what he means by "30% of its fuel"; you
and I have interpreted that to mean the exhausted fuel has a
mass equal to 30% of the mass of the rocket before the fuel
was burned, hence M /(M+m )=0.7. Also, maybe
he was thinking about a launch from earth which would
require some of the energy from the rocket to overcome the
gravitational force and the energy loss from air drag. Or
maybe he was guesstimating the effects of efficiency of the
engines. In any case, you did not make a mistake.
0 C, if the air
in the bottle chilled to that temperature, the pressure in
the bottle would be smaller than 15 psi. Using the
Gay-Lussac Law, P /T =const., I find the
pressure would be about 11 psi, a net of about 7 psi. (If
you want to check this, you must use absolute temperature,
not 0 C.)
FAQ page you will
find many links to earlier answers which relate to this.
laminar
which means that if you focus your attention on a small
volume of the fluid it will move along a smooth line and be
followed and preceded by many other small volumes following
the same smooth line. Under other circumstances, the fluid
will become turbulent , moving chaotically. So
something happens which causes the smoke from your incense
stick to change from laminar to turbulent at some point.
This is usually quantified by a dimensionless variable
called the Reynolds' number, Re=ρuL /μ
where ρ is the density of the fluid, u
is its speed, L is a length which describes the
geometry, and μ is the dynamic viscosity of the
fluid. So, Re changes in such a way that the
original laminar flow becomes turbulent after rising a
certain distance from the source of the smoke.
a
or in zero field but having a uniform acceleration a ;
buoyancy would be observed in an accelerating rocket ship in
empty space.)
E i A =½mv i 2 +mgh ,
E f A =½mv f 2
and so v f =√(v i 2 +2gh ).
Exact same calculation for E B .
d =0 in your (simplistic) W=f x
d definition of work. If the saucer is rising with a
constant speed v, then when it has gone some distance d, the
thrust T has done W saucer =Td
and gravity has done work W gravity =-mgd
because T=mg for it to be rising with constant speed. The
net work W on the saucer is still zero, W =W gravity +W saucer =0.
Your last question "…simply converts KE to PE…"
is again using PE unnecessarily. Gravity does work on an
object as it falls because there is a force mg
acting down.
Eltjo Paselhoff on the website
quora.com . Suppose we plot the velocity v as a
function of time t for the upward trip. At t =0
the magnitude of the slope must be larger than g
because the air drag is slowing the ball down as well. As
the ball gets higher, its speed decreases so the air drag
gets smaller so that finally at the top of its trajectory it
is at rest and the magnitude of the slope is equal to g .
Now, the ball falls back down and speeds up because air drag
is smaller than weight; eventually, though, the two forces
become close to equal and the speed becomes constant at the
terminal velocity v T . One more thing you
must understand now is that the area under the velocity
curve is the distance traveled by the ball, so since up and
down go the same distance the green and red areas need to be
the same. The only way this can be is if the time to fall is
larger than the time to rise.
F -F f F
web site which gives the details of the
solution of the physics problem but is still quite readable.
For the casual reader, I will include here an overview.
Newton's second law is an inhomogeneous second-order
differential equation. What makes the equation inhomogeneous
is the presence of the driving force, the signal your
amplifier sends to the speaker; with no driving force, the
solution of the problem would be x t (t )=A t e-γt ωt+φ t )
where ω is the natural frequency the speaker
would vibrate with on its own, γ is a parameter
which describes the damping of the speaker, and A t
and φ t are constants
determined by the initial conditions. If the
driving force is present, F=F 0 cos(ω 0 t +ψ ),
the solution is a linear combination of the what is called
the steady state solution , x s (t )=A s sin(ω 0 t+φ s ),
and the solution for the undriven oscillator x t (t )
which is referred to as the transient solution: x (t )=A t e-γt ωt+φ t )+A s sin(ω 0 t+φ s ).
The constants A t and φ t
are determined by the initial conditions (speed and position
of the speaker) and the constants A s and
φ s are determined by the driving
force (F 0 and φ t ).
Because of the damping term e-γt
The figure
below shows the transient (red), steady-state (blue), and
total (black) solutions separately.
These show the total displacement for different initial
conditions. In both cases the speaker started at rest. The
first shows the starting position at zero, the second shows
the starting position at 0.5. The important thing is that
they both end up, after the transient has died, as the same.
0 . I found a
little
calculator you can play with; a screenshot is shown
below. All you would need to vary are the velocity, the
angle, and the height of the dock. I found that if I chose
the initial speed to be 26 ft/s (about 18 mph) the distance
the dog would go for a 450 jump would be about 21
ft from a zero-height dock. Now, when I do the calculations
for the 24 in=2 ft and 16 in=1.33 ft, I find the best angles
and distances (for 26 ft/s) are (43.50 , 22.9 ft)
and (42.50 , 22.3 ft), respectively. All in all,
there is just about no difference.
Note that I have not included any air drag. I believe the
speed of the dog is low enough to make drag negligible.
D is hinged to a wall. A force F T d from the hinge; we would find
experimentally that if T =F (D /d )
the stick would not rotate. We would conclude that the net
torque about the hinge is now zero and that this torque
changes linearly with the distance. Of course, the torque
exerted by F T τ=FD-Td= 0.
This is not the most general case but I feel you wanted to
be convinced that torque depends on where the force is
applied relative to an axis about which you are computing
torques, which was easy to see.
first
and second times.
In a nutshell, if there is not air moving over the wings
with sufficient speed, there is not sufficient lift on the
airplane for it to leave the ground. I agree that
Mythbusters'
version of this is not the same as I understand your
question and previous questions where whatever the conveyer
is doing, the plane is not moving forward through the air;
the Mythbusters have the conveyer at rest relative to the
plane but moving forward with the same speed as the plane,
so the wheels are not spinning.
I guess it is possible
that a propeller plane could take off from rest since the
props send a wash of air over the wings, but probably not
unless there were a pretty hefty headwind; you do see some
planes, designed for flying into and out of tight places,
which take off in a
very
short distance.
Let's just call a truce on this whole thing. A 747 at rest
for any reason is not going to take off. And practical
problems with the fabrication and utilization of this stupid
treadmill will ensure that it is never constructed or
tested.
L a distance d from where you are viewing
it. You can see that the angle subtended by the moon is 31
arcminutes≈0.520 . The angle is expressed in
radians as θ=L /d. To convert that to degrees,
multiply by 180/π , so θ= (L /d )x57.3
in degrees. For example, your 19 km dreadnought 100 km away
would have θ =10.90 , about 21 times
larger than the moon.
earlier answer which answers
the same question except for the earth-moon and sun-earth
pairs. For your question it is the electrical force which
holds the electron in its orbit. There is an important
difference, though. The electron-proton system, if you look
at in classical electromagnetic theory, should radiate
energy (it is essentially a tiny antenna) and fall into the
proton. This does not happen; the reason is that for such
small systems classical physics does not work and this was
one of the milestones of physics when Niels Bohr proposed
his model
of the hydrogen atom where certain orbits simply did not
radiate. To understand the details of quantum mechanics,
which eventually evolved from the Bohr model, would require
a much longer explanation.
etc . For
a more detailed explanation, see
this link .
Rutherford scattering ) does result in the
angular momentum of the moving charge q about an axis passing
through the fixed charge is conserved. It is not hard to
understand. What is the condition for which angular momentum
is conserved? There must be no torques on q . The
expression for torque is τ r F qr xE r E
r and E r E q and its
angular momentum will remain constant.
Wikepedia : "Gravitational waves carry energy away from their sources and, in the case of orbiting bodies, this is associated with an in-spiral or decrease in orbit. Imagine for example a simple system
of two masses—such as the Earth-Sun system—moving slowly compared to the speed of light in circular orbits. Assume that these two masses orbit each other in a circular orbit in the
x-y plane. To a good approximation, the masses follow simple Keplerian orbits. However, such an orbit represents a changing quadrupole moment. That is, the system will give off gravitational waves.
In theory, the loss of energy through gravitational radiation could eventually drop the Earth into the Sun. However, the total energy of the Earth orbiting the Sun (kinetic energy + gravitational potential energy) is about 1.14x1036 joules of which only 200 watts (joules per second) is lost through gravitational radiation, leading to a decay in the orbit by about 1x10-15 meters per day or roughly the diameter of a proton. At this rate, it would take the Earth approximately 1x1013 times more than the current age of the Universe to spiral onto the Sun. This estimate overlooks the decrease in
r over time, but the majority of the time the bodies are far apart and only radiating slowly, so the difference is unimportant in this example."
So the answer is that the energy comes from the orbiting
earth which causes its orbit to get smaller as it loses
energy; however, as the calculation shows, the loss of
radius of the orbit is trivially small.
c where c is the speed of
light. In order to see the effect which I think interests
you, you need to choose a much larger speed for
the rocket; I will choose v =0.8c , 80% the
speed of light. I will choose my unit of length to be one
light hour and my unit of speed to be light hours per hour,
so c =1, v =0.8, and L =1. First,
look at what an earth-based observer will measure with his
clock. If t is the time when the light
arrives at the rocket, then vt =0.8t =x and
ct=t =(L-x )=(1-x ); solving, x =0.8/1.8=0.444
and t=x /0.8=0.556. Also, note L-x =(1-x )=0.556.
But, what you asked for is how long a clock on the rocket
takes to see the light. The important thing is that the
distance between the earth and the satellite is not 1 light
hour any more, it is length contracted such that the
distance seen for L by the rocket is L'=L √[1-(v /c )2 ]=0.6
rather than 1. Of course the same factor occurs for x' =0.6x =0.267
and the time this happens is t'=x' /0.8=0.333
because even though the distance to the satellite is
shorter, it still zips past the rocket with a speed of 0.8c .
Finally, c' =(L'-x' )/t' =(0.6-0.267)/0.333=1!
So, you see, c really is the speed seen by all
observers.
relative to what . As a concrete
example, suppose that you throw the bottle with a velocity
of 20 mph relative to you (the speed you would throw it if
you were at rest on the ground). But you are are at rest
relative to the car, so you and anybody else in the car
would see it moving forward with a speed of 20 mph. But
someone by the roadside would see it moving forward with a
speed of 70 mph. Somebody in a car going in the same
direction as you with a speed of 70 mph would see the bottle drop
straight down to the ground (no horizontal velocity). If you
threw it backwards (relative to the direction of your car)
with a speed of 50 mph, someone by the roadside would see it
drop straight to the ground (see a
Mythbusters episode).
-29 g/cm3 ,
so it isn't even close.
https://www.bbc.com/news/science-environment-47873592
Why this cannot be done by one single telescope and how the group of telescope functions in order to work (I cannot think why it needed so many to be focused in a single point). Also the earth is rotating so most of them are not aligned with the point of focus.
Second question is why we only have an image? if we can take a photo, we can surely take more photos and create a video. In that way we can see the moving plasma or whatever is there.
-3 m wavelength; this
wavelength is expected to originate only from the accretion
disk-7 m. Increasing the
size of a telescope gives it better resolution, so the size
S for a millimeter telescope comparable to the best
optical telescopes would be S =10-3 (10/5x10-7 )=20,000
m=10 km; but, looking at an object so far away (55 million
light years) would be impossible for visible light but,
since the earth has a diameter of about 13,000 km, we can
get very much better resolution with a large millimeter
telescope array by having the individual telescopes have
separations comparable to the size of the earth. Although the size is very
huge, the total area of all the radio telescopes is
extremely
small, so trying to make an image would be like blacking out
99.999% of the mirror in an optical telescope—you
would still get an image but you would have to wait a very
long time. So the data from all the telescopes in the array
were gathered over many hours during a 10 day observation
period. Then, all the data had to be put together using a
computer to reconstruct the image; the data analysis took
two years to complete, a very difficult problem.
Finally, you should ask what the dark circular region in the
center is—it is not the black hole itself which has
zero size. The outer edge is what is called the photon
radius, the distance from the black hole where a photon will
orbit. Somewhere inside of the photon radius is the event
horizon, the sphere inside of which nothing, including light
can escape.
A warning: as I clearly state on the site, I do not usually
answer complex questions in
astronomy/astrophysics/cosmology, so take what I say with a
grain of salt!
L was quantized, L=n ℏ where ℏ
is the rationalized Planck constant. This was actually
wrong, but it gave the right answer for the spectrum of
light observed from a hydrogen atom and that was a great
leap forward; but purely empirical. Improvements
and
extensions of this model were made, but they were all
equally empirical. Still, this first empirical model set
everyone on the right track to achieve a more fundamental
model. During this time, it was beginning to be understood
that trying to imagine a model based on classical ideas,
ideas like an electron orbiting in a circular orbit, were
invalid at such tiny scales—that realization ultimately gave
rise to the development of quantum mechanics. You can no
longer even think of an electron as a little point charge
when it is bound in an atom. Rather, you solve an equation,
known as the Schrödinger equation, which gives you a
mathematical function, called the wave function, which tells
you the likelihood of finding the electron at any particular
place in space. So the best way to think of an electron in
an atom is akin to picturing a cloud over which the electron
is smeared. The shape of this cloud is dependent on four
quantum numbers which quantize the energy, the angular
momentum, the projections of the angular momentum and the
spin angular momentum on some axis. Some examples are shown
in the figure. To visualize the "clouds", imagine rotating
each around a horizontal axis through the center; so,
e.g. , the upper right cloud is donut-shaped.
ground rules , you would have seen that "I no longer answer questions which are based on premises like:
'...if we could travel faster than the speed of light...' or
'...ignoring the fact that nothing can go faster than the speed of light...',
'...if I were traveling at the speed of light...', etc."
Suppose that you were traveling at 99.99% the speed of
light. The light which you would see if at rest would be
visible, but in your moving frame the wavelength would be
Doppler-shifted to a longer wavelength, into the region of
far infrared which your eye cannot detect.
241 Am. In addition to
the four energies of alpha particles there are gamma rays of
energies up to about 100 keV. I cannot explain why water is
the least absorptive than any other material you tested.
here , the
second is linked to from that answer.
F =m a
In rotational physics, moment of inertia I plays
the role of mass and angular acceleration α F r τ r x F τ Iα .
I do not know much about engineering notation semantics but
if what you say is correct, that torque means that which is
zero if the object is not rotating at all (or with a
constant angular velocity), then that is what would be
called net torque in physics which seems to make
more sense to me. According to Wikepedia, mechanical
engineers (and also British physicists) refer to (physics)
torque as moment of force , not torque moment as you
state. Similarly, the definition by mathematicians is that
torque is equal to the rate of change of angular momentum
which is exactly the same as τ Iα
and so torque, again contrary to your
statement, is the same as in physics. So, I read the
Wikepedia article differently from you and, to me, there is
no difference in anybody's definition of what I would call
torque other than semantics. And I read nothing there which
implies that the word torque means the sum of all moments of
force to an engineer. So my final answer is that everybody's
definitions of the physical quantities are the same, only
the words used to label them differ.
Why doesn't the moon fall to the earth?
the volume of a sphere of radius R is 4πR 3 /3;
the cross sectional area A of a sphere of
radius R is πR 2 ;;
the air drag force on an object having speed v
is proportional to Av 2 ; and
the terminal velocity U of an object of mass
m is
proportional to √(m /A ).
I
will use subscripts 1 and 2 to denote the original and final
drops, respectively; so m 2 =2m 1 and
R 2 /R 1 =21/3 .
Therefore, A 2 /A 1 =22/3 .
Finally, U 2 /U 1 =[(m 2 /m 1 )(A 1 /A 2 )]1/2 =[2⋅ 2-2/3 ]1/2 =21/6 .
So, finally, U 2 =10⋅ 21/6 =11.22
cm/s.
R with the current
I distributed uniformly throughout the wire. The
figure shows the field due to the wire, the vector
B having a constant magnitude at a
distance r from the axis of the cylinder defined by
the wire and always tangential to the circle of radius r
around the wire. Hence field lines, as shown are circles,
the direction defined by the right-hand rule as shown. The
magnitude of the field is determined by Ampere's law,
∫B ⋅ ds μ 0 I enc
where the integration
is all the way around the path defined by the circle and
I enc is the amount of current which passes
through the area defined by that closed path. Given the
cylindrical symmetry of this special case, the integral is
simple because B is constant and B s
∫B ⋅ ds πrB=μ 0 I
so Br>R =μ 0 I /(2πr )
for r>R . This is only correct outside the wire
itself because the integration path to determine B
inside the wire encloses less current. Inside, I enc =I (r 2 /R 2 )
so Br<R =μ 0 Ir /(2πR 2 )
for r<R. Note that when r=R both results
give the same value of B , B =μ 0 I /(2πR ).
So, the field increases linearly with r inside the
wire and decreases like 1/r outside.
The questioner also
wanted to express this in terms of the velocity
v Q=-e and having a density of
n
electrons/m3 . The fact that the electrons have
negative charge means that, in the figure above, the
direction of v t the
electrons go a distance ΔL and so sweeps out a
volume ΔV which contains a total charge ΔQ =-ne ΔV =-ne ΔL (πR 2 I =ΔQ /Δt=-ne (ΔL /Δt )(πR
2 )=-nevπR 2
v in a cylindrical beam of radius R . The second part of my answer showed you how to relate
I in the first part of the answer to the charge, speed, density of charges, and
R . So the quantity you seek results from setting
I=nQvπR 2 into
B=μ 0 I /(2πR ) (assuming that "adjacent" means right on the edge of the moving charge distribution).
In the first case the two electrons are very close together and are moving to the right at say 50 m/s, what would be the
"B" field be 50 mm directly above, i.e. in the positive y axis above these electrons?
In the second case, same two electrons traveling to the right at 50 m/s but this time they are separated. Call the first electron,
"e1"and the second "e2".
The first electron e1 is located directly 1 mm below the point of interest where I want to calculate the combined
"B" field. The second electron e2 is located directly 98mm below e1 or 99 mm below the point of interest where I want to determine the combined
"B" field. In effect from the first case I have moved e1 closer to
"B" by 49 mm and e2 farther away by 49 mm.
I was using the formula B=μ 0 NqV /(4πR 2 ) where
μ 0 is permeability of free space, N is the number of charges, q is charge on an electron and V is the velocity of the electrons, R is the vertical distance between the charges and the point of interest for
"B" the strength of the magnetic field.
B R B V B=μ 0 Nq V R πR 3 );
note that the magnitude of V R VR sinθ where θ is the angle between
V and R θ =900 , your equation is correct. Be
careful, though, because the direction matters when you are
adding vectors from two sources; the case you are looking at
has the point the same side of both electrons so the
magnitude is the sum of the magnitudes, but if you look at a
point between the electrons, the magnitude will be the
magnitude of the difference. The form of Ampere's law which
I used in the earlier answer is simply not applicable in
this situation because making reference to a current I only
makes sense in magnetostatics. Finally, I should note that
all this is only approximately correct when you are in a
nonrelativistic regime which you certainly are for V =50 m/s.
twin paradox (I have attached here the figure from that
post for your convenience) which shows how the traveling twin "sees"
how the earth clock runs by having the earth-bound twin send
him one message per year; this is different from your
situation where essentially a daily rather than an annual
message is sent, but the ideas are identical. When he
arrives home, the traveling twin has "seen" the same number
of years (days) on earth pass as the earth-bound twin has
but, as you can see, a different number of years (days) have
passed on his on-board clocks. "Watching" some clock in
another frame of reference is not in any way relevant to
time in your frame; the traveling twin would count his
birthdays using his clock, not earth's clock. Note also that
the traveling twin "sees" the earth spinning very slowly on
the trip out but spinning very rapidly on the trip home.
mv 2 ,
so it is natural to think that the maximum possible kinetic
energy would be ½mc 2 . But
Newtonian physics is not valid at speeds close to c ,
and in relativity the kinetic energy is ½mv 2 /√[1-(v 2 /c 2 )].
To learn more, see an
earlier answer
to the same question; you might also look at the answer
following that one.
earlier answer . Spin is nothing more than a
specification of the angular momentum of a system; e.g. ,
the spin quantum number of an electron, proton, neutron, and
other so-called fermions is s =½. The actual
angular momentum is given by S =ℏ√(s (s +1))
where ℏ is the rationalized Planck's constant. Usually, we think of spin
as the intrinsic angular momentum of an elementary particle
or entire quantum system like an atom; particles may also
have orbital angular momentum, for example an electron
orbiting around a nucleus in an atom. But each angular
momentum will be quantized (have a specific quantum number)
and the total angular momentum J of any angular
momentum number j will be J =ℏ√(j (j +1)).
Orbital angular momentum quantum numbers are integers rather
than half integers like electrons; so, an electron in a
"p-orbit" in a Bohr atom has an orbital quantum number 1 and
a spin quantum number ½. As to where the idea came from,
there were several things which suggested it. The experiment
which showed what the spin of an electron is was the
Stern-Gerlach experiment , but there were also hints from
spectroscopy and, notably, that there were twice as many
states in quantum systems (like atoms) as expected from the
Pauli exclusion principle .
spinning top does. The earth's axis is inclined by
about 23.40 relative to the normal of the plane
of its orbit but that axis precesses around the
perpendicular axis about once every 26,000 years. This
precession is caused by torques exerted on the earth by the
sun and the moon. Currently the axis is pointing almost
exactly to the the north star, Polaris; however, in a few
thousand years there will be some new star taking that role.
The last ice age was about 2.4 million years ago and lasted
much longer than 26,000 years, so the answer to your
question is that it pointed many directions during the ice
age. You are asking about the geographic north pole which is
essentially constant in its location (although the land
masses move over the surface because of plate techtonics).
The gravitational north pole, though, is forever wandering
around and has actually jumped to the southern hemisphere
around Antactica and back many times.
book , PHYSICS IS … The
Physicist Explores Attributes of Physics , has a chapter
titled Physics Is Beautiful , devoted to beauty and
elegance. However, I have never thought of it as art. A
sunrise over the Sangre de Christo mountains is beautiful
but you would certainly not call it art. I found that it is
impossible to find an unambiguous or not vague definition of
art; art is qualitative which sets it apart from science
which is quantitative. And, consider this: There are many
paintings of the of Chartre Cathedral, four of which I have
copied here, and all are indisputedly art; there is no
"correct" painting of this subject. The theory of
electromagnetism, on the other hand, is beautiful and the
only one of its kind; you might think up another
electromagnetic theory, but it would be wrong even though it
might be considered very imaginative, creative, even
beautiful. There is no well-defined test which specifies
which painting is the best; there is a test as to which is
the theory of electromagnetism —it must
quantitatively describe what really happens in nature and if
it fails to do so, it is false. I have always thought of
string theory as not a theory (and not beautiful) because it
makes no attempt to quantitatively interface with the real
world and is therefore not falsifiable.
F=- μs N
(negative because the force is opposite the velocity) where
μs is the coefficient of static
friction between rubber and dry asphalt which is about
0.5-0.8, and N is the normal force of the truck on
the road which, on level ground, is the weight; I will
choose μs = 0.65, halfway between the
extremes. At this point I will switch to SI units which
scientests prefer: weight N =7,200 lb=32,027 N, mass
M =32,027/9.8=3,268 kg, and speed v =45 mph=20.1
m/s. Now, using Newton's second law, F=Ma =- μs N ;
so acceleration a =-6.37 m/s2 . Now, you
can compute the time to stop: v =0=v 0 +at =20.1-6.37t,
so t=3.16 s. Finally, the distance traveled is d =v 0 t +½at 2 =20.1x3.16-½x6.37x3.162 =31.7
m=104 ft. It is interesting that this answer is independent
of how heavy the truck is. The reason is that the
acceleration is proportional to the weight/mass which is
just 9.8 m/s2 .
-5 s and the
acceleration is about 2.3x107 m/s2 .
The corresponding force to stop the beanie baby is
F=ma=.0128x2.3x107 ≈3x105 N≈67,000
lb. I think that if the beany baby hit you anywhere except
possibly your limbs, you would be killed. Keep in mind that
this is a very rough calculation, but certainly in the right
ballpark.
V= 4πr 3 /3 is
the volume of a sphere and if it is increasing with a constant rate dV /dt=C ,
then C =4πr 2 dr /dt .
Solving, dr /dt=C /(4πr 2 ) which is velocity at which the
surface is expanding.
So the acceleration of the surface is (d2 r /dt 2 )=-[2C /(4πr 2 )]dr /dt=- 2C 2 / [(4π )2 r 5 )] or, combining all constants,
kr -5 . I just don't see if the radius of the
sphere is expanding at a rate of Cr -2 , how it is accelerating at kr -5 ?
k is k =- 2C 2 / (4π )2 .
I think that the problem you are having is that you have
carried the minus sign into your constant. But that minus
sign is important because it tells you (since everything
else in k is constant) that the radius of the surface is
increasing, but the rate at which it is increasing is
decreasing as r changes. So, if the volume of
a sphere is increasing with a constant rate, its radius is
increasing like 1/r 2 , but the rate at
which the speed of the surface is decreasing goes
like 1/r 5 .
recent answer
is very similar to yours.
here . For example, for your current weight, a 5K race at
3.1 mph (which is about equal to 5 km/hr, so it takes one
hour for you to complete), I find "You burned 1151 calories
on your 5 kilometer run. At that pace, your calorie burn
rate is 230 calories per mile and 1151 calories per hour."
Sorry, I could not find any information on temperature and
humidity.
v Mv 2 .
where M is the mass of the universe except for your
mass. And just a minute ago that energy was not there.
Obviously, the total energy of that universe cannot be the
same in all frames. But, if you stay in one frame and keep
track of the total energy of the universe in your frame, it
will not change.
g 2 . The acceleration due to gravitation does
vary at various places on the earth. The animation shows you
how it varies. The variation is small, no larger than 50
milligals; a gal is 1 cm/s2 =0.01 m/s2 ,
so 50 milligal=0.005 m/s2 . So, compared to 9.8
m/s2 , the variations are only on the order of 0.05%. There
are several reasons for gravitational variations, notably:
altitude (the farther you are from
the center of the earth, the weaker the gravity);
local geology (large volumes of
more dense earth will increase the gravity);
the earth
is not a perfect sphere.
If you want a
lot more detail, see the
Wikepedia article on the earth's gravity.
N =0.001, N would be the number of
seconds it would take for the intensity of the light to drop
to 0.1% of its original value. Using
Wolfram Alpha , I
find N
f in water is a
quadratic
function of the velocity v , f= ½ρw CD Av 2 , where
ρw = 103
kg/m3 is the density of water, A
is the area presented to the onrushing water as the
object falls, and CD is a constant which depends on the
geometry of the object, called the drag coefficient. For a sphere of radius R
the drag coefficient is given by CD =0.47 . The upward force f is
only one force on the object as it sinks; it also has a
weight down W=mg and a buoyant force up B=Vρw g
where V =4πR 3 /3
is the volume of the sphere. If the sphere is
solid and has a density ρ=m /V , we can
write B=mg (ρw /ρ ).
Also note that we can write R as R =3 m /(4πρ )].
Finally, we can write the total force on a sphere: F =[mg (ρw -ρ )/ρ ]+½ρw CD Av 2 .
Note that when a speed is reached where F =0, the
sphere will fall with a constant velocity called the
terminal velocity, vt =√{[mg (ρ-ρw )/ρ ]/[½ρw CD A ]} .
So, let's do a specific example to be able to answer your
question for a particular situation. Suppose we have two 1
kg balls of solid iron for which ρ= 7.87x103
kg/m3 . Then the radius of each ball will be R =0.0312
m. Doing the arithmetic, F =-8.56+0.72v 2 .
So each of two balls of mass 1 kg, tied together will
accelerate until their speeds would be vt =F =-17.1+1.14v 2 .
So one ball of mass 2 kg will accelerate until
vt =√(17.1/1.14)=3.87 m/s.
You could do more examples to come to the conclusion that
the answer to your question is that, in general, two spheres
tied together will not sink at the same rate as one sphere
with their combined mass. I am sure, by carefully tailoring
the shapes of the objects, you could force the rates be the
same, but that is not you would find in general.
f can be approximated as f ¼Av 2
where A is the area presented to the onrushing air
and v is the speed. (You will note that Av 2
is not the same dimensionally as a force; this approximation
only works if you work in SI units, A in m2 ,
v in m/s and f in N.) So the net force
F on each ball (which I assume to be solid) is F =¼Av 2 -mg
where m is its mass; and, by Newtons second
law, the acceleration a is a =(¼Av 2 /m )-g ;
and there is a velocity U , called the terminal
velocity, at which the drag equals the weight so the
acceleration is zero and U= 2√(mg /A ).
Iron and aluminum have different densities, but it is pretty
straightforward to compute the radii of eachA for each. I find A Fe =0.014
m2 and A Al =0.0064 m2 .
The accelerations are a Fe =-(9.8-3.5x10-4 v 2 )
and a Al =-(9.8-1.6x10-3 v 2 ).
At any given speed the magnitude of the acceleration for the
iron ball is larger than for the aluminum ball. The terminal
velocities would be U Fe =170 m/s and
U Al =78 m/s. The iron ball always has a
larger speed than the aluminum ball and will hit the ground
first.
It is also enlightening to find the distance traveled as a
function of time, s= (U 2 /g )ln(cosh(gt /U )).
This may beyond your math ability (ln is a natural logarithm
and cosh is a hyperbolic cosine), but the graph shows the
behavior of each ball and clearly the iron ball is falling
faster. Note that for the first few seconds (6 or 7 s) the
balls pretty much keep pace with each other and after about
15 s both have nearly reached their respecitve terminal
velocities because the curves are nearly linear.
When this kind of problem is
presented as an exercise in elementary physics, it is
usually assumed that the density of the earth is uniform
throughout—a cubic meter of earth from the surface
has the same mass as a cubic meter at the center. In
this case, it is easy to show that the acceleration due
to gravity, g (r ), is linear: g (r )=9.8(r /R )
m/s2 where r is the distance from
the center and R is the radius of the earth.
Since the acceleration is always toward the center, you
speed up until you reach the center (not decelerate as
you suggest) and then slow down until you reach the
other end, neglecting any frictional forces. More
realistically there will be at least air drag which will
get bigger as you speed up until you eventually reach a
constant speed; then at the center, you start slowing
down until you stop somewhere before the other end. Then
you speed up back to the center getting there with a
smaller speed than the last time you were at the center,
etc. eventually stopping at the center as
k =9.8/R
N/m.
But, the earth is not a uniform
sphere. The core is much denser than the mantle and
therefore the acceleration is not a nice linear
function. The figure shows that the acceleration is
almost constant until about one third of the way to the
center. The above qualitative description of your motion
remains the same except for the similarity to the simple
spring.
mg )
decreases linearly to zero at the center. In the realistic
model, the weight remains constant for a while, then
increases slightly, then falls almost linearly to zero at
the center.
First, though, let me
address your reference to the ball being "heavier" or
"lighter" depending on the spin on the ball. This is totally
incorrect as the ball always has the same weight, a force
which points vertically downward. The way that the ball
moves results from the forces which the ball experiences
after it leaves your racquet; there is the weight straight
down, the force the ground exerts on it when it hits, and
the forces which the air exerts on it. The figure above
shows a ball with top spin; the top of the ball is spinning
in the same direction as the ball is moving to the right.
(If you are confused, the diagram is shown in the frame of
the ball, so the air is shown like a wind blowing in the
opposite direction as the ball is moving.) Here the ball is
experiencing the force of its weight straight down (not
shown), a drag force to the left slowing it down (not
shown), and an aerodynamic force called the magnus force
F F
For
completeness I would like to talk about what happens when
the ball hits the ground because these are strategically
important in tennis. Shown in the second figure are the
forces (in red) which the ground exerts on the ball in top
and back spin cases. For clarity, I have not shown the
forces due to the weight or the air. In each case there is a
normal force N
f which slows down the spin
in both cases. In the the top spin case, the friction both
slows down the spin and speeds up the ball so it comes up
faster and lower than a nonspinning ball would; in the the
back spin case, the friction both slows down the spin and
slows down the ball so it comes up slower and higher than a
nonspinning ball would. If the spin were large enough, the
back spin ball could actually bounce straight up or even
backwards!
hf , which the photon
had is given to the electron; the electron ends up with a
kinetic energy equal to hf minus the
binding energy it had in the atom. The rest mass of the
electron remains unchanged.
17 kg/m3 , so your volume would be
about 102 /2x1017 =5x10-16 m3 .
So, if you are a cube, your dimensions would be 3 5x10-16 =8x10-6
m ≈10 microns. A flea has a size of about 1
mm=1000 microns, one hundred times bigger than the
compressed you. And, yes, you would still have a mass of 100
kg.
link .
R , the R ω ,
etc . all distract from the crux of the problem, in
my opinion). I have redrawn the problem showing some axially
symmetric object of radius R , mass m rolling down
an incline of angle θ without
slipping. There are only three forces acting on the object,
the frictional force f N ,
mg N
f a F ma
and Στ Iα );
here I is the moment of inertial about the axis
about which torque is calculated and α
F ma
I find N=mg cosθ and ma =mg sinθ
-f. The first equation nails N , one of our
unknowns, the second is one equation for the two remaining
unknowns, so we are not done. So the only place to get a
second equation is the rotational form of Newton's second
law; I think this is the trickiest part of this problem. I find many students
have an inclination to choose the symmetry axis to sum
torques; they have been told, when they learned statics of
non point objects, that you can choose any axis you like,
but this is not a statics problem and there is a very
important constraint—Newton's laws are not valid
in an accelerating system , and the center of the
rolling object is accelerating. (See correction below!) So Στ Iα
yields Iα=Ia /R=mgR sinθ ,
or a= (mR 2 /I )g sinθ ;
knowing a , then f =m (g sinθ -a )
But, what is I ? It isn't what you look up in a book
because usually what is listed is the moment of inertia
about an axis passing through the center of mass (Icm ).
Here the axis is a distance R from the center of
mass, so using the parallel axis theorem, I=Icm +mR 2 .
I will do one example, a solid sphere where Icm =2mR 2 /5,
so I =7mR 2 /5, so a =5g sinθ /7
and f =2mg sinθ /7. If you sum torques
about the center of mass, you get the wrong answer.
OK, I slipped up here, albeit
in a fairly minor way. While it is true that Newton's laws
are not correct in an accelerating frame of reference, there
is one exception: the sum of the torques about an axis
passing through the center of mass is indeed Icm α .
A little detail I had forgotten about classical mechanics.
Let me work that out for my example: Icm α= 2maR /5=fR ,
so, using the result above, f =2ma /5=m (g sinθ-a )
or a =5g sinθ /7. Then it
easily follows that f =2mg sinθ /7.
But you should try to sum torques about an axis halfway
between the point of contact and the center of mass—you
will not get the right answer.
Nevertheless, it is important that your students be careful
in choosing the axis. I have edited the original answer
somewhat.
Most Massive Black Holes . This is the most
up-to-date information I could find.
article I found, recent research indicates that the
outer core rotates with the same speed as the mantle (once a
day) but the inner core's rotational speed is slightly
faster by about 0.3-0.5 degrees per year.
CD
for the three shapes, determined experimentally, are shown
in the figure. The drag force is given by FD = ρv 2 CD A
where ρ is the density of the air, v
is the speed of descent in still air, and A is the
area. Therefore the drag force is least on the circular
parachute so it reaches the ground first. (A proviso is that
the values of the drag coefficient depend on a quantity
called the Reynolds number which depends on the velocity.
However, the values given are for low Reynolds numbers which
are suitable for parachutes.)
paper on special relativity, On the
Electrodynamics of Moving Bodies , states two main
problems in electromagnetic theory at the end of the 19th
century:
The first
paragraph states, essentially, that Maxwell's equations
will not have the same form in a frame of reference
moving with constant velocity relative to a frame where
they are the laws governing electromagnetism if you
transform them using Galilean transformations.*
The
beginning of the second paragraph acknowledges "…the unsuccessful attempts to discover
any motion of the earth relatively to the 'light medium' [luminiferous æther,
e.g. the
Michaelson-Morley experiment]…"
Einstein had a
strongly-held philosophical belief that, for something to be
a true law of physics, it must be the same in all reference
frames—regardless of where you are or how you are
moving, the laws of physics for you are the same as for
anybody else in the universe; this is the principle of
relativity . Since he believed Maxwell's equations to be
laws of physics, he concluded that Galilean relativity was
not correct and this led to his postulate that the speed of
light was independent of the motion of the source or the
observer. To my own mind, I believe that the constancy of
the speed of light is not a necessary postulate, rather a
consequence
of the principle of relativity applied to Maxwell's
equations.
*A Galilean transformation is, for
example, observing a car moving 60 mph east relative to the
ground to be moving 100 mph east if you are in a car moving
40 mph west relative to the ground.
M moving with velocity V MV after the
collision or any time during the collision. What will not be
conserved is the kinetic energy MV 2 .
recent answer , you mention the Cosmic Microwave Background and say that it "provides a frame of reference for the whole universe".
It was my understanding that one of the underpinnings of the basic theory of relativity is that there CANNOT be such a universal frame of reference (UFOR). I assume you're not claiming Relativity has been disproven (well, shown to be a special case of something else), so can you explain the difference between the frame of reference meant for relativity and the CMB as you described it?
v v
0 F) of fall would result in the
outside being burnt and the inside being raw. It might be
fun to do a couple of very rough calculations just to get an
order of magnitude feeling for what you are interested in.
I will approximate the turkey to have the properties of
water, in particular that its specific heat is C =4.186
J/gram⋅0 C; so the energy which will raise
the temperature of a M =6000 gm turkey is E=MC ΔT =6000x4.186x56=1.41x106
J. (700 F→1700 F is 210 C→770 C,
hence ΔT =560 C.)
The change in potential energy in falling from a height
h is V=Mgh =6x9.8h= 59h , so if the turkey falls
with constant velocity the potential energy lost goes into
heat. If the heat is 1.41x106 J, the height is
h =1.41x106 /59=24,000 m. The atmospheric
pressure at this altitude is only about 3% what it is at sea
level; so it is clear that, if your turkey is given any
velocity at this altitude, it will not continue moving with
constant velocity. (Whereas, if the pressure were
atmospheric and you gave the turkey a speed equal to the
terminal velocity it would continue moving with that speed
as it fell.) If you dropped it from very far away
(essentially infinitely far), it would arrive at the upper
atmosphere going nearly the escape velocity, 11,200 m/s. At
this speed, which is faster than the reentry speed of the
shuttle which needed special ceramic tiles to shield against
burning up, the turkey would be burned to a crisp, probably
would not reach ground before totally burnt. Although there
would be some speed of entry to the atmosphere where the
turkey would absorb 1.41x106
J, it would almost certainly not be "perfectly cooked".
3 He
(one neutron, two protons), and als3 He now fuse, throwing
off two protons in the process and one alpha particle (4 He,
two protons and two neutrons). The two thrown-off protons
are now free to continue the process, so this chain, called
the proton-proton cycle, is self-sustaining until the star
runs out of hydrogen which ends the life of the star.
For more earth-bound hydrogen fusion, like fusion reactors
or hydrogen bombs, single-step reactions are sought. The
most frequently used is the deuterium-tritium reaction,
involving the fusion of one deuteron and one triton (3 H,
one proton and two neutrons) shown in the other figure.
t
elapsed in the gravitational field of a nonrotating
spherically symmetric mass M is t=t 0 / δ )
where δ= 2GM /(Rc 2 )=2aR /c 2 ;
here, R is the distance from the center, t 0
is the time elapsed when R =∞, G
is the universal gravitational constant, M is the
mass of the sphere, and c is the speed of light,
and a is the local acceleration due to gravity. If a=g
and R =6.4x106 m (earth radius), it
is easy to calculate that δ= 1.39x10-9 .
Now, t g t ½g =[√(1- ½δ )]/[√(1- δ )];
using the binomial expansion
(1 +δ )x ≈1+xδ+ …
if δ<< 1, we finally get t ½g ≈tg (1+¼δ )=tg (1+3.5x10-10 ),
quite a bit different from t ½g =2tg !
old
answer and the links it contains. I do not know what
"accelerate wildly" means in this context but, since you
mentioned "bumped", mechanical bumping probably caused
unexpected behavior of the radiometer.
groundrule for "single, concise, well-focused questions". I can tell you that what is learned in elementary physics courses that airplanes get lift from the Bernouli effect of lower pressure due to higher velocity of air on the top of a wing is a relatively small part of how an airplane is able to fly. As you seem to refer to in your question, the main thing is that air leaves the wing with a downward component
of its velocity; this means that the wing exerted a force down on that air which means, via Newton's third law, that the air exerted a force up on the airplane. To quote Wolfgang Langewiesche, author of
Stick and Rudder ,
the private pilot's bible on the art of flying, "The main fact of all heavier-than-air flight is this:
the wing keeps the airplane up by pushing the air down ."
e.g. , another galaxy or the
center of our local cluster of galaxies. However, the
discovery of the cosmic microwave background (CMB) has now
given us a reference frame which seems to define the
reference frame of the whole universe. Very careful
measurements and data analyses from the satellite COBE have
shown that our galaxy moves with a speed of about 580 km/s
relative to the CMB.
So, what can we do to answer your question which is,
essentially, how far does the earth move in 57 yr=1.8x109
s? (Your 5 minute question is essentially impossible to
answer since it depends on the time of year. Over years the
orbital motion averages out to zero.) Relative to the center
of the galaxy, the distance is D =220x1.8x109
km= 4x1011 km=0.013 pc. The center of the galaxy
in that time will move 580x1.8x109 km=1012
km=0.032 pc. But I do not know the relative directions of
these two velocities, the sum of which would be the average
velocity of the solar system over these 57 years; this speed
would be anything between 360 and 800 km/s. However, in my
researching this question I found that the speed of sun
relative to the CMB is about 370 km/s, so the final answer
is that the earth moved about 370x1.8x1011 =6.7x1011
km=0.022 pc. It is interesting to me that, at this time, the
sun and the center of the galaxy are moving in nearly
opposite directions (370 km/s compared to 360 km/s); since
the time for the sun to go all the way around the galaxy is
about 225 million years, in about 110 million years our
speed relative to the CMB will be about 800 km/s.
Keep in mind that I am not an
astronomer/astrophysicist/cosmologist! I have gathered these
data as best I can.
2 )=1.93368x10-5
PSI=1.31579x10-6 atm.
As the conversation progressed, we considered the implications and began to wonder if it was possible to shield against cosmic radiation? To the best of my knowledge, I explained that no substance I know of can block cosmic radiation and that the Earth's magnetic field would have little effect either. The only effects I know of are when they impact atoms in the atmosphere or the earth. If I understand this, if it doesn't hit something fairly dense, it just keeps going, understandable given the space between atoms.
I have read about the gravitational lensing effects on light by massive objects like neutron stars and galaxies that bend light around them. I have also read about the possibility of warping space as a method space travel by changing the shape of space in front of and behind a craft so that it "falls" in a particular direction.
While we were discussing these effects, he looked at me and said something to the effect, "Maybe we could bend space and make them go around?". To me, this was a pretty profound leap in my book. So I decided to ask physicist if he could be right.
Please forgive the length of this post before asking our question. I wanted to provide sufficient context before asking. So, here it is. Is it possible to make cosmic rays go around something, either by gravitational or microwave warping of space?
B.T.W. I really really enjoy your web site. It is awesome!
But, I can speak to your main question and comment on some
of your statements.
Cosmic
radiation refers to a broad range of radiations but
mostly protons and light nuclei; but also it could refer to
neutrinos, electrons, positrons, antiprotons, gamma rays,
x-rays, and others. What is not the case, though, is that "no
substance…block cosmic radiation"; just about any
cosmic radiation can be blocked by a dense solid or liquid.
Neutrinos are the exception and most pass all the way
through the earth without any interactions. You may know
many neutrino experiments are performed in deep abandoned
mines to block out all other radiation. The atmosphere also
is very effective since a single energetic proton will
strike an atom and many other particles, all with less
energy than the original proton, will form a "shower" of
particles. And magnetic fields are not very effective
deflecting the very highest energy charged particles but are
known to be an effective shield for lower energy particles.
Finally, let's address
the question of warping of space as a way to deflect the
rays. The only way to warp space that I know of is to have
an enormous mass nearby. The earth already warps the space
around it which would cause a particle passing close by, but
not on a collision course with earth, to be deflected a tiny
bit (viz. gravity). But putting a star sized object near
enough the earth to "shield" us would obviously be a
catastrophic thing to do! The whole solar system would be
disrupted. I have no idea what "microwave warping of space"
means.
I commend your son for his hypothesizing and you for your
tutoring and encouragement of his curiosity.
and electric fields
earlier answer
to understand the answer
here. The graph here is from that earlier answer and shows
(in red) v /c as a function of a 0 t /c
where a 0 is the acceleration measured in
the frame of the object which is accelerating and t
is the time since the object was at rest. Reading off the
graph, you can see that v /c =0.9 when
a 0 t /c= 2. Now, t =30
days=2.6x106 s, so a 0 t /c=a0 (2.6x106 /3x108 )=8.64x10-3 a 0 =2;
solving, a 0 =232 m/s2 =23.6 Gs. I would not
suggest having anybody aboard this object!
m is walking up a
hill with incline θ with a constant
speed of v . His potential energy is E=mgy
where y is his height above the bottom of the hill
and g= 9.8 m/s2 is the acceleration due
to gravity. The rate at which his energy is changing is dE /dt=P =mg (dy /dt )=mgv y =mgvθ ;
so this rate (called power) depends only on the vertical,
not the horizontal component of his velocity. Now, you have
made things difficult for me because you have not specified
the angle, but rather the inclination which is actually the
tangent of the angle (rise/run) times 100%. I will call this
inclination the grade G =100tanθ. For
example, let's take a man with mass 100 kg walking with a
speed of 2 m/s up the 10% and 20% inclines you specify. The
angles would be θ 10% = tan-1 (10/100)=5.710
and θ 20% = tan-1 (20/100)=11.310 .
So the power values are P 10% = 100x9.8x2xsin(5.71)=195.0
Watts and P 20% = 100x9.8x2xsin(11.31)= 384.4
Watts. The ratio of these two is 1.97, close to, but not
equal to, 2.0; for small angles the sine and tangent are
approximately linear functions but as the angle (hence
inclination or grade) increases this is no longer a good
approximation. The graph shows the power for grades up to
200% (about 63.40 ); as you can see, the power
increases approximately linearly only up to about a grade of
about 30% which is about 170 .
When I say "trajectory"
I refer to the path from collision point to the ground. The
horizontal distance gone from the point of collision to the
point of landing depends on the angle at which the cat
started; if launched at 450 relative to the
horizontal he will go much farther than if launched at 200 .
Again, unless you witnessed the accident, there is no way to
know the angle.
2 )=0.31;
so if you went to a star 1 light year from earth, it would
only take you 0.31/0.95=0.33 years to get there but the
earth-bound clock would say it took you 1/0.95=1.05 years.
Also where the stars appear to be relative to where they
would be if you were not moving would be different; see an
earlier
answer . In the case where v /c =0.95, the graph shows
how the angle at which you see something depends on where it
would be if you were at rest; for example, a star actually
at θ =1200 (300
behind you) would be seen as being about θ' ≈300
in front of you!
R =2GM /c 2 ,
would be about 1 cm. R is the radius within which
nothing can escape. So, if you could get the earth down to
that size, it would then collapse the rest of the way to a
black hole, I presume.
m approaches the end of a thin uniform rod of mass
M and length 2L with velocity v L from where m
exerts the force. Linear momentum and angular momentum are
both conserved but, as we shall see, energy is not
conserved. After the collision the center of mass has a
velocity u ; conserving linear momentum, (M+m )u=mv
or u=v [m /(M+m )]. Before the
collision m brings in an angular momentum mvL .
After the collision, the rod plus mass has angular momentum
(I +mL 2 )ω
where I=M (2L )2 /12=ML 2 /3
is the moment of inertia of a thin rod about its center of
mass. Conserving angular momentum I find ω [m +(M /3)]L 2 =mvL
or
ω=mv /{[m+ (M /3)]L }.
Just for fun I put in some numbers to illustrate what would
happen in a specific situation: m =1 kg, M =3
kg, L =1 m, v =4 m/s. With these I find:
u =1 m/s, ω =2 raE before = ½mv 2 =8
J, E after =½(M+m )u 2 +½([m +(M /3)]L 2 ω 2 =6
J. So, energy was not conserved in this collision, 2 Joules
were lost.
x=mL /(M+m )
toward where m is stuck. It is this center
of mass which moves forward with speed u , and u
is still v [m /(M+m )]. The rotation
is about this center of mass but now the moment of
inertia is different; the net (rod plus mass) moment of
inertia is ML 2 /3+Mx 2 +m (L-x )2 ,
so the corrected value of the angular velocity is
ω=mvL /[ML 2 /3+Mx 2 +m (L-x )2 ].
In the example I did, x =1/4 m, so ω= 16/7 radians/s=21.8 rpm and E after =46/7=6.57
J (1.43 J lost).
earlier answer
and the earlier answers referred to in that answer. The
graph here shows the results derived or cited in that
earlier answer. The three graphs all show what is observed
of your motion as seen by the twin on earth. The red curve
is your velocity v relative to c as seen
by the earth-bound twin, v /c =(gt c )/√[1+(gt c 2 ];
the blue curve is your acceleration a g , a /g =[1+(gt c 2 -3/2 ;
the black curve is your position expressed (approximately)
in light years, gx /c 2 =√[1+(gt /c )2 ]-1.
Now, since I have set this up as seen from earth, let me
specify the time you travel as the time of your trip out as
observed from earth: earth-bound twin observes you to arrive
at you destination in two years. Then, as you can read off
the graph, gx /c 2 =√[1+(2)2 ]-1=1.24
ly; the return trip will be the same length of time, so the
earth-bound twin's clock will have advanced 4 years. Light
would take 2.48 yr to travel that distance, so your clock
will have advanced a time somewhere between 2.48 and 4
years. Your maximum speed would have been about v /c =(2)/√[1+(2)2 ]=0.89.
We can get a rough estimate of your clock reading by
approximating your average relative speed as 0.89/2=0.445;
so you would have seen, on average, the distance contracted
by √[1-(0.445)2 ]=0.9 and your time would
have been 0.9x4=3.6 years.
g appears to be
somewhat smaller than it really is, but the amount is very
tiny. This effect depends on your latitude and is zero at
the poles and greatest at the equator. Also because of this
effect, the acceleration is not directly toward the center
of the earth except at the poles and equator.
earlier answer
illuminating.
Kibble balance which is used. Also see
Wikipedia for more detail
I stated that it was incorrect based on the fact that
"made up of" usually means putting two or more things together (aka addition). Net radiated energy is the difference between emitted energy and absorbed energy. I have yet to see my exam paper but I believe that the mark I lost was due to this question on basis that “made up of” means that it is a result of a mathematical relation between the two values and not necessarily addition (in this case subtraction). What is the correct answer in this case and why?
angular
speed of the object in the smaller orbit is twice that
of the object in the larger orbit even though their linear
speeds are the same. You cannot do this with planets or
moons, though, because the speed of a satellite in a
circular orbit depends on the radius.
essay about general relativity, curvature of
spacetime, gravity. Many nonscientists (and scientists as
well) are familiar with and accept as gospel that general
relativity demonstrates that the distortion of spacetime by
the presence of mass is what gravity is; the "cartoon" which
illustrates this is the "bowling ball on a trampoline"
example in which a mass causes the distortion of the
two-dimensional space defined by the trampoline surface. The
implication is that gravity is just geometry. On the other
hand, many theorists are struggling to find a successful
theory of quantum gravity and to do that one must treat
gravity as a force. The take-away from this essay, though,
is that curved spacetime is not a unique representation of
what the mathematics which are the framework of the very
successful theory of general relativity mean. The author,
Rodney Brooks, notes that "…you can view gravity as a force field that, like the other force fields in QFT
[quantum field theory], exists in three-dimensional space and evolves in time according to the field equations."
He also goes so far as to say "…shame on those who try to foist and force the four-dimensional concept onto the public as essential to the understanding of relativity theory."
a and a mass M . The forces on
the car are its own weight W W=Mg )
acting at the center of mass a distance d from the
real axle; the normal and frictional forces on the rear
wheels, N 1 and
f 1 ; and the normal and
frictional forces on the front wheels, N 2
and f 2 ; the radius of
the wheels is R and the distance between front and
rear axels is D . I will ignore the rotational
motion of the wheels, that is, I will approximate their
masses as zero. If the front wheels are not lifting up from
the ground, the sum of torques about the rear axel is (f 1 -f 2 )R +N 2 D -Mgd= 0;
Newton's first law for the horizontal and vertical
directions are (f 1 -f 2 )=Ma
and N 1 +N 2 Mg =0.
From these equations it is easy to show that N 2 =M (gd-aR )/D .
This is interesting because it shows that N 2 =0
if a=gd /R ; for larger accelerations, the
front wheels will lift off the ground. (Note that as N 2
approaches zero, f 2 also approaches
zero.) I have ignored the suspension of a real car, so this
lift will not be abrupt; it will happen gradually as the
front springs decompress and the rear springs compress; so
overall the nose goes up but the rear end goes down.
If
it is a front-wheel drive, the directions of the frictional
forces reverse. The corresponding torque equation (this time
about the front axle) is
(f 2 -f 1 )R-N 1 D+Mg (D-d )= 0;
Newton's first law for the horizontal and vertical
directions are (f 2 -f 1 )=Ma
and N 1 +N 2 Mg =0.
From these equations it is easy to show that N 1 =M (aR+g (D-d ))/D .
As a increases in magnitude, N 1
gets bigger and bigger until N 1 =Mg
which necessarily means that N 2 becomes
zero. It is easy to show that this happens when a=gd /R ,
exactly the same result as for the rear wheel drive.
a ,
so most likely the wheels will slip well before N 2 =0
when f 2 must be zero. For very large
accelerations, like required for drag racers, rear-wheel
drive is imperative.
i.e. all
objects near the surface of the earth have an acceleration
of 9.8 m/s2 . The acceleration (which is what
describes the orbit) is F /m but the force
is MmG /r 2 , so the acceleration
is MG /r 2 .
t . Its
momentum drops from mv (where m is the
mass and v is the speed when it hits) to zero
because an impulse I was delivered to it by the
ground. The impulse may be written as I=F avg Δt .
where F avg is the average
force. Therefore you can find the average acceleration a avg
during the collision, but not the acceleration as a function
of time: a avg =F avg /m =I /(m Δt )=v /Δt.
So you cannot get detailed kinematics about the
collision without measuring how the speed decreases with
time.
here .
F= ¼A (u-v )2
where u is the speed of the wind, v is the
speed of the ball, and A is the cross sectional
area of the ball. Applying Newton's second law, ¼A (u-v )2 =ma =m dv /dt
. This differential equation may be shown to have the
solution t=cv /[u (u-v )] where
c =4m /A and t is the time it takes the
ball to reach the speed v if the wind speed is
u . I find that c =137 s2 /m. Note
that after a very long time the speed will be constant at
the wind speed; we are only interested in the time it takes
to get to 35.8 m/s (80 mph). I have plotted t for
three values of u , 70, 80, and 90 m/s.
Now comes the hard part. We are interested in finding the
value of u for which v will be 35.8 m/s when the ball has
gone a distance x =18.4 m. This means that we need
to write the equation for t as
t=c {dx /dt }/[u (u- {dx /dt })]
and solve that differential equation for t as a
function of x . That is actually doable, but the
solution is so messy that I looked for a simpler way to do it.
This took some trial and error, but I got a pretty good
estimate, I believe. Look at the black line in the graph; it
is not a straight line, but it is pretty close, so I am going
to approximate the acceleration as being constant over this
time interval of 1 s, a2 . Now, an object with constant acceleration
starting from rest goes a distance d = ½at 2
in a time t , so if u =90 m/s=201 mph the ball will have a
speed of 35.8 m/s when d =35.8/2=17.9 m=59 ft. As
far as I am concerned, I have adequately solved this problem—it
takes a wind of approximately 200 mph to accelerate a
baseball from rest to the speed of 80 mph in a distance of
about 60 ft.
groundrules state
that I will not answer questions of this sort. Your first
question is equivalent to the question "why can nothing with
mass travel as fast or faster than light?" which is included
in the FAQ page.
