QUESTION:
Where can I get a graph (or other information) about the increasing "relativistic resistance" to the acceleration of a particle (an electron, hopefully) as its velocity is increased to near-relativistic speeds? If such a graph is not available, then how can I calcuate this increasing force that resists a particle being accelerated when it is already traveling at some relatively high percentage of the speed of light, like maybe 70%?

ANSWER:
It would definitely behoove you to read an
earlier
answer first which has lots of details and discussion of
acceleration. There are two ways to approach this problem:

Suppose
the observer in the inertial frame is doing the pushing and wants to
know how hard to push the particle to achieve a particular
acceleration.
Resistance to acceleration is usually called inertial mass and the
inertial mass m of a particle with rest mass m _{0}
and speed v is m=m _{0} / √[1-(v /c )^{2} ].
The first figure above plots
m /m _{0} as a function of v /c .
So, to achieve an instanteous acceleration of a , a force of
F=ma =m _{0} a /√[1-(v /c )^{2} ];
so for your example of v /c =0.7, you can read off
the graph that you would need to exert a force 1.4 times larger than
you would if the particle were moving slowly. This is probably what
you want if you are interested in electrons accelerating since you
would be accelerating them.

Suppose
the observer was on the particle and the particle was a rocket ship.
You adjust your engines so that the force F which they
exert on the ship causes a constant acceleration of a _{0} =F /m
where m is the rest mass of the ship. What
acceleration a does another observer in an inertial frame
(on earth maybe) see when the rocket has a speed v ?
Starting with the velocity derived in the
earlier
answer , v =(a _{0} t )/√[1+(a _{0} t /c )^{2} ],
we can calculate the acceleration by differentiating with respect to
t , [dv /dt ]/a _{0} =a /a _{0} =(1+(a _{0} t/c )^{2} )^{-3/2} .
Now, reading off the second graph, when v /c =0.7,
a =0.36a _{0} ; the stationary observer will
only see 36% of the acceleration seen on the ship.

Q&A OF THE WEEK
(2/12-18/2017)

QUESTION:
If I am moving 55 MPH East (or West) at the equator how much weight would I gain (or lose) due to the Eötvös Effect. Thank You in advance. I am 73 years old and too dumb to figure this out myself.

ANSWER:
First of all, weight is the force the earth exerts on you so you
never gain or lose weight when you are moving; you might want to say
"apparent weight" which is the force which would be measured by a scale
you were standing on. You experience two real forces, your weight
W down
and the normal force N (a scale, for example) up. One way to solve this
problem is to note that an object with mass m with speed v
moving in a circle of radius R has an acceleration a=v ^{2} /R
which points toward the center of the circle; then apply Newton's second
law, F=ma =mv ^{2} /R=W-N and solve for
N to get your apparent weight of W -mv ^{2} /R ,
smaller than your actual weight. This is the
"Eötvös effect".
But there is another way to approach the problem. Rather
than solving the problem from the outside the earth, we might want to
solve it here on the earth. But Newton's laws are not valid in an
accelerating reference frame (accelerating because it is rotating). You
can force Newton's second law to work, though, by inserting a fictitious
force which I will call E for
Eötvös but it is more
commonly known as the centrifugal force; E =mv ^{2} /R
pointing radially out. Newton's first law now applies, N+E-W =0,
so, again, N=W -mv ^{2} /R=W [1-v^{2} / (gR )]
where g =32 ft/s^{2} . In the figure above you have a velocity
v=v _{Earth} +v _{man} . If you are at
rest, v=v _{Earth} =1040 mph=1525 ft/s and R =3959
mi=2.09x10^{7} ft; so N=W (1-0.00348) and a scale will
read 0.348% smaller than your actual weight. If you move with a speed of
55 mph=81 ft/s in an east direction, v =1525+81=1606 ft/s and
N=W (1-0.00386), 0.386% smaller than your actual weight. If you
move with a speed of 55 mph=81 ft/s in a west direction, v =1525-81=1444
ft/s and N=W (1-0.00312), 0.312% smaller than your actual
weight.

FOLLOWUP QUESTION:
I made a donation, but the way I read your answer you have both directions being SMALLER, if I read it right. I know that West is an increase and East is a decrease, just thought you would like to know.

ANSWER:
Thanks for your support! Very generous, particularly because you
say that I am wrong! I must stand by my calculations, though. It is
certainly not unheard-of that I make an error, but this answer is right.
Since I define weight to be what a scale would read if the earth were
not rotating (or at north or south poles), you will see that the
apparent weight (what the scale reads) increases if you go west and
decreases if you go east, just the same as what you "know"! All apparent
weights are smaller than the actual weight; the only exception is if you
go west with speed of v _{Earth} in which case the
actual and apparent weight will be the same. If you want to compare to
your apparent weight at rest, it is 0.386-0.348=0.038% lighter going
east, 0.348-0.312=0.036% heavier going west.

QUESTION:
How much "work" (physics definition) is actually accomplished in a gym workout?
I'm currently using F x D x reps = actual work done.
The upward lift ("D") x the amount of weight lifted ("F") x the number or repetitions... to get actual work done.
Am I in making some mistake, here?
Thanks, in advance, for your help.

ANSWER:
By "upward lift" I assume you mean the distance lifted. So,
lifting a weight F over a distance D you would do
W=FD units of work on the weight. For example, the weight of a 2 kg
mass is about 19.6 N and the work to lift it 1 m is 19.6 J. But, and
here is the catch, you use more energy than 19.6 J to lift that weight
because your body is not a simple machine like a lever or a pulley. To
understand why, see the faq page. In a
nutshell, the reason is that to just hold up a 2 kg mass, not move it up
at all, requires input of energy —you get tired trying to
hold up a weight at arm's length, right? And, what about lowering the
weight back down? The work done on the weight is negative which implies
that energy is being put back into you but know that it also takes
energy for you to lower the weight at a constant speed. A biological
system is considerably more complex than systems we talk about in
elementary physics classes. I think that it is of little use to try to
analyze a workout in terms of elementary physics.

QUESTION:
Speed of light again becomes 3×10^{8} m/s when it emerges out in air from denser material without the loss of energy. Why?

ANSWER:
Just because it speeds up does not mean that it gains energy.
For light, the energy is determined by the frequency, not the speed.
When the light enters a dense medium its speed v decreases but
its frequency f stays the same. Since v = λf ,
where λ is the wavelength, the wavelength decreases.
Another way to look at it is to think of the light as a swarm of
photons. The energy of a photon is hf where h is
Planck's constant, so the energy of a photon depends only on frequency.

QUESTION:
If I'm driving and hit the gas and turn left would the angle
between the velocity vector and acceleration vector be less than, greater
than, or equal to 90 degrees. I would think it's greater.

ANSWER:
You would think wrong! The velocity vector
v points
straight ahead. The tangential acceleration
a _{t} points
parallel to the velocity vector because you are speeding up. The
centripetal acceleration vector a _{c}
points toward the center of the circle you are turning. As you can see
in the figure, the total acceleration vector
a makes an acute (less than 90^{0} )
angle with v .

QUESTION:
If the earth is curved how is it you can get a laser to hit a target at same height at sea level more then 8 km away?
How is it that it's bent around the earth?

ANSWER:
First of all, light is not bent around the earth; it travels in
a perfectly straight line and therefore, because the earth is curved,
there is a maximum distance away for a target at the same altitude. What
that distance is depends on the altitude of the laser. You say that the
laser is exactly at sea level by which I presume you mean the surface of
the earth; at this altitude you could not hit any target also at sea
level. In the figure I have drawn the earth, radius R , a point
a distance h above the earth's surface (laser location), and
another point a distance h above the earth's surface (target
location). The distance between them is 2d . Focus your
attention on one of the triangles with hypotenuse (R+h ).
From the Pythagorean theorem, d =√[(R+h )^{2} -R ^{2} ]=√[2Rh +h ^{2} ];
if h<<R , d ≈√(2Rh ). For example, if
h =10 km, about the height a commercial jet flies, 2d ≈714 km is the most distant target at the same altitude which you could hit.

QUESTION:
if gravity can hold back all the seas and heavy objects to earth how can a fly or leafs move threw the air does gravity not have same force on everything I don't get how it can hold back everything but same time let small things move so easy ?

ANSWER:
For starters, the force of gravity (often called the weight) is
proportional to the mass of the object; the weight is ten times bigger
for a 10 kg object than for a 1 kg object. Second, Isaac Newton taught
us more than 300 years ago that to understand how an object moves (or
doesn't) you need to consider all forces on the object. For example, for
a leaf being blown upward, the force of the air up on the leaf is
greater than the force of gravity pulling down.

QUESTION:
I have an old fashion balance scale, center fulcrum and two dishes on either side of equal weight. If I place two weights on either side and the weights are nearly the same, the heavier side dips slightly, if the difference between the weights are large, the heavier side dips much more. I do not understand why this is so. Logic says that if there is any difference at all, the heavier side should continue to drop until it reaches a barrier to the fall no matter what the difference in the weight.
I asked a physics instructor and he did not have the answer either.

ANSWER:
The center of mass
⊗ of the scale
itself must be below the fulcrum (suspension point). Then, as shown
above, if the beam is off horizontal for the empty scale or equal
weights in each pan, there will be a restoring torque to force a balance
only for the horizontal beam.

FOLLOWUP QUESTION:
I don't think the answer given expained the phenomena described in the question. It only explained why when the weights are equal and there is a force on one side that it will bounce up again and continue to rock back and forth until it eventually evens out again. The problem I posed was a different issue, as to how the scale behaves when different weights are put on the heavier side.

ANSWER:
It does answer your question, but just not explicitly because I
did not include examples of unequal weights. So let me do that now.
Given the explanation in the original answer, I can model the scale as a
massless T with all the scale mass M located a
distance d below the pivot at the bottom of the T .
As shown in the figure, when one side is loaded with a mass m
the scale rotates to an angle θ with the
horizontal and reaches equalibrium. The sum of the torques about the
fulcrum is mgL cosθ -Mgd sinθ =0,
so θ =tan^{-1} [mL /(Md )]. As
m gets larger, θ gets larger. For small angles, θ ≈mL /(Md )
in radians.

QUESTION:
While i am reading Einstein book 'evolution of physics', i have encountered a very confusing reasoning process which i can not understand it.
The topic is about how we can deduce that the gravitational mass and inertial mass are equal just by knowing that the acceleration due to gravity for all objects at the same height is constant.
And here is the text i can not understand how he deduced that:
"Now the earth attracts
a stone with the force of gravity and knows nothing about its inertial mass. The
"calling" force of the earth depends on the gravitational mass. The "answering"
motion of the stone depends on the inertial mass. Since the "answering" motion is always the same —all bodies dropped from the same height fall in the same way—it
must be deduced that gravitational mass and inertial mass are equal.
More pedantically a physicist formulates the same conclusion: the
acceleration of a falling body increases in proportion to its
gravitational mass and decreases in proportion to its inertial mass.
Since all falling bodies have the same constant acceleration, the two
masses must be equal."

ANSWER:
Newton's universal law of gravity says that the force is
proportional to the gravitational mass of the object, F=k _{1} m _{g}
where k _{1} is a constant. Newton's second law of
motion says that the acceleration is inversely proportional to the
inertial mass and directly proportional to the applied force, F=k _{2} m _{i} a ;
in SI units, k _{2} =1.
Therefore, a =k _{1} m _{g} /m _{i} .
(This is Einstein's statement "…the
acceleration of a falling body increases in proportion to its
gravitational mass and decreases in proportion to its inertial mass .")
Now, since it is an experimental fact that a is the same regardless of
the mass, m _{g} /m _{i} is a constant.
If we choose to measure both m _{g} and m _{i}
in kilograms, and since k _{1} expresses the strength of
the gravitational field, m _{g} /m _{i} =1.
(For the field near the earth's surface, k _{1} =M _{earth} G /R _{earth
} where G is the universal gravitational constant, M _{earth}
is the gravitational mass of the earth, and R _{earth}
is the radius of the earth.)

QUESTION:
well, i saw a theory that if you rotate the ISS, which weighs 450 tons at 10 rotations per second, it would generate its own gravity. would it be possible to spin something of proportion size at a proportional speed to generate artifical gravity like the ISS?

ANSWER:
You should read earlier answers regarding "artificial gravity"
in rotating space stations. The idea is that if you are in an
accelerating frame of reference you feel like there is a force on you;
for example, when you drive fast around a curve you feel like you are
being pulled toward the outside of the curve. It is not a "theory", it
is elementary physics. The mathematics you need is that your
acceleration if you are going in a circle of radius R with
speed v your acceleration is a=v ^{2} /R .
For a given R , if you choose v such that a=g =9.8
m/s^{2} where g is the acceleration due to gravity, you
will feel like you are at the surface of the earth. In your case, v
can be determined from the frequency 10 rotations/second: each rotation
has a distance of 2πR so v =20πR /1. This
would imply that the acceleration would be a =9.8=(20πR )^{2} /R =400π ^{2} R ;
solving, R =.0025 m=2.5 mm. This is obviously nonsense, so you
must have remembered the the frequency incorrectly. Looking at the
figure, the main cabins appear to have a diameter of about 5 m, R ≈2.5
m, so if they spun about their central axis the required speed would be
about v ≈√(9.8x2.5)=5 m/s and so the
frequency would be f =5/(5π )=0.32
rotations per second=19 revolutions/minute. But this would not work at
all for the ISS because an astronaut's head would feel almost no
"gravity" because it would be very close to the axis of rotation; if she
were to lay on the floor/wall/ceiling it would be close to what it would
be like on earth. The ISS is just too small.

QUESTION:
I've been told by many that the fastest thing is vacuumed light. since light travels at different wave lengths does some light travel than other. in other words does a ultraviolet rays travel the same linear speed from A to B the same as say an x-ray. If it does than would that mean we don't have a solid speed limit of light. if they do not do we just use the longest light wave to measure the max speed, OOORRRR do we throw light like a laser... but wouldn't that still have a wave of some form.

ANSWER:
Every-day language usually interprets "light" as that which we
see with our eyes. When a physicist says "the speed of light", she means
"the speed of electromagnetic radiation". All electromagnetic radiation
travels with the same speed in a vacuum.

QUESTION:
I made a statement to somebody that a plane hitting a building was the same as if the building hit the plane at exactly the same speed,the plane now stationary. The results would be the same. In other words, if a man with large hands slapped my hand at 50 mph, it would be same as me slapping his hand at 50 mph.....its interchangable.....the other person said, no, the mass of the building and hand would have different results...

ANSWER:
Either you or your friend could be right depending on what you
mean by "different results". Let me try to set up a simple example to
demonstrate why.

Imagine we have a 2 lb ball of putty moving
with a speed of 5 mph striking and sticking to a 18 lb bowling ball
at rest; the time it takes to collide is 0.1 s. After the collision,
the two move together with a speed of v _{1} . To
find v _{1} , use momentum conservation: 2x5=(18+2)v _{1} ,
v _{1} =0.5 mph.

Next, imagine we have a 18 lb bowling ball
moving with a speed of 5 mph striking and sticking to a 2 lb ball of
putty at rest; the time it takes to collide is 0.1 s. After the
collision, the two move together with a speed of v _{2} .
To find v _{2} , use momentum conservation:
18x5=(18+2)v _{2} , v _{2} =4.5 mph.

So, you see
that the two scenarios have different speeds after the colliision. But,
suppose that you were the putty ball. During the collision you feel a
force and the force is what is going to hurt you. Do you get hurt as
badly, not as badly, or equally as badly during the collision? What
determines the force you feel is the acceleration you experience during
the collision, how quickly your velocity changes, which is your final
velocity minus your initial velocity divided by the time of the
collision.

For the
putty ball moving initially, (v _{final} -v _{initial} )/t =(0.5-5)/0.1=-45
mph/s.

For the
bowling ball moving initially, (v _{final} -v _{initial} )/t =(0-4.5)/0.1=-45
mph/s.

You could go
through through the exact same process to find that the bowling ball
experienced exactly the same force regardless of who moved initially. A
physicist would say that you were right, but the ambiguity of your
statement means that the other guy could split hairs. As far as physics
is concerned, the only thing which matters is the relative
velocities of the two before the collision. If the putty ball were
moving 105 mph and the bowling ball were moving 100 mph in the same
direction, the result of the collision which matters (the force) would
be the same.

QUESTION:
The atmosphere is heavy. If the weight of a column of air above your desk is about the same weight as the bus you rode to school in, why doesn’t air pressure crush your desk?

ANSWER:
Because the atmospheric pressure also acts up under your desk.

QUESTION:
I have always wondered how much energy do you do with if you let a kettle at 1800 W be running for two minutes? What is the approximate cost for this?
this is not a homework question.. just a question i wonder =)

ANSWER:
A
Watt is one Joule per second, 1 W=1 J/s. Energy consumed by an 1800 W
kettle in 2 min is 1800x120=216,000 J. But, we are more used to
measuring electrical energy in kilowatt hours, (1 kW ·hr)(1000
W/1 kW)(3600 s/1 hr)=360,000 J. So the energy used by the kettle is
(216,000 J)/(360,000 J/kW·hr)=0.6 kW·hr. A kW·hr
costs on the order of 5¢-15¢, so the cost would be between 3¢
and 9¢.

QUESTION:
Why do unstable elements always give off alpha particles with 2 protons and 2 neutrons. Essentially a Helium nuclei. Why not a hydrogen nuclei or a heavier nucleus

ANSWER:
Alpha-decay is only prevalent in very heavy unstable elements. Most
unstable nuclei decay by beta decay, the ejection of an electron or
positron along with a neutrino. The reason that alpha-decay happens is
that the alpha particle is an extremely tightly bound particle and
therefore there is a fairly high probability that it will spontaneously
form inside a very heavy nucleus where there are a lot of neutrons and
protons to contribute. For a more extensive discussion, see an
earlier answer .

QUESTION:
Every day my wife reads the latest fake news about planet 9 and sits weeping in fear the whole time . As a scientist ,surely you can tell me one thing that just shuts this whole fiasco down . Please. Tell me that undeniable fact that will convince her that we are in fact safe from a collision or near miss from this nonexistent space oddity and it 's cohorts . It s affecting her health . And I love her . So it' i m feeling her pain as well .

ANSWER:
First of all, "Planet 9" is a serious astronomical topic. Minor
irregularities in the orbits of some of the distant planets suggest the
presence of another planet farther out.
Serious efforts are underway to try to observe it. But its
anticipated period is more than 10,000 years, so if it is far away now,
it is not likely to be a problem for us for far longer than the lives of
any of us. All reputable astronomers have declared that (if it actually
exists) it is absolutely no danger to earth. Google "planet 9" to get
lots of good information. But, even if there were a planet much closer,
like the "fake news" "planet x", there is an amazingly low likelihood of
its colliding with earth. Most lay folk like your wife have no
comprehension of the vastness of space; the probability of two
particular objects in the solar system colliding is, for all practical
purposes, zero. It is often posited that a "rogue planet" passing
through the asteroid belt would "shake loose" a "storm" of asteroids
toward the earth. I recently
answered a question
addressing this possibility which you might find useful.

QUESTION:
Can tension ever be negative?

ANSWER:
This
is an ambiguous question. If you mean can the magnitude of the tension
force exerted by a string be negative, the answer is no; a string can
only pull, it can never push. But, if you mean can the tension ever have
a component which is negative, the answer is yes; it simply depends on
how you have chosen your coordinate system. But, if you have drawn the
tension vector T such that the string is
pulling on something and you solve the problem and the magnitude T <0, you have made
a mistake somewher.

Q&A OF THE
WEEK (1/21-28/2017)

QUESTION:
My question is about the maximum tension experienced by a bow string. I'm
specifically concerned with a traditional or recurve bow NOT a compound bow
with pullies. I want to know the max tension compared to the draw weight so
I have an idea how strong to make my strings. So here's the scenario, what's
the maximum tension in the string for a recurve with a 70 lbs draw weight
and a physical weight of 25 oz? I'm assuming the maximum tension is when
it's at brace (not full draw), right after the arrow leaves. I think this
because not only does the string have to oppose the restoring force of the
bow limbs but it also has to stop the momentum of the limbs that isn't
transferred to the arrow.

ANSWER:
To compute the tension I would need to know the geometry of the bow. I can
tell you that the tension will be at the maximum draw for a simple bow, not
where the arrow leaves the string.

REPLY:
The bow is 64 inches long and the string is about 4-5 inches shorter than the bow. It Is braced at 6 inches and has tips that are 3 inches recurved behind the handle. Its draw length is 28 inches and has an elipitcal/circular tiller shape at full draw.

ANSWER:
(An incorrect answer was posted earlier. This is a reposting, correct
now, I hope!)
When researching the physics of archery I discovered that this can be a
very complicated problem requiring very sophisticated numerical
calculations on computers if you want precision descriptions of all the
details. You, however, require only a rough calculation for estimating
the strength of the string. I can do that and it is more appropriate for
the spirit of this site —to solve problems with simple physics
concepts. This problem requires facility with trigonometry,
understanding of Hooke's law, and application of Newton's first law. The
simple model I will use was one used before the advent of computers; the
bow is modeled as two straight rods (purple) the ends of which move on a
circle as the string (red) is drawn. With this simple model, most of the
details of your bow are not necessary. When the string is braced
(undrawn) there is a certain tension in the string and this tension will
increase as the bow is drawn. So, the maximum will be at the maximum
draw. The figure shows, roughly to scale, the situation. Using simple
trigonometry (law of cosines), I find β =59.6^{0} .
The point where the draw weight W is being applied must be in
equilibrium, W -2T cosβ =0; solving, T =69.2
lb. I think that your concern about the string having to "stop the
momentum of the limbs" is misplaced because bows tend to be quite
elastic so that nearly all the energy imparted to the bow by drawing it
is imparted to the arrow and the limbs will end nearly at rest. Not so
if you draw and release without an arrow, though; what I have read is
that in that situation you are more likely to break your bow than the
string. I am working on a general solution which I will later add here
but thought I would post the part of the solution which answers your
question regarding the tension in the string.

GENERALIZED
SOLUTION:
To get a better understanding of this problem it is worthwhile to
find an analytical solution for the tension as a function of draw
distance. My research showed me that a traditional or recurve bow
behaves, to an excellent approximation, like a simple spring
(Hooke's law), the draw weight being proportional to the draw distance,
i.e . W≈kx where x is the distance the
string is drawn and k is the spring constant. In this case, since W =70 lb when x =28
in, k =2.5 lb/in. Using the law of cosines, cosβ =(L ^{2} +(x+d )^{2} -R ^{2} )/(2L (x+d )).
Again, the point where the force W is applied
is in equilibrium so W -2T cosβ =0 or T (x )=kx /(2cosβ ).
Now, note that in the limit where x → 0, β → 90^{0}
and cosβ → x /L . Therefore T (0)=kL /2.
Using your numbers, T (0)=37.5 lb and the angle and tension for
all points are plotted below. At full draw, β= 56^{0}
and T =64 lb. It was interesting to me that in order for the
calculated values to be correct at zero draw, very precise relative values of
R and L
had to be used because otherwise the expression for cosβ would not be 0 exactly when x =0. The value was
R =30.59411708 inches for L =30.

QUESTION:
If a charged particle oscillates, it produces a propagating electromagentic wave. What happens when the motion of the charged particle is not oscillatory, but Brownian? Is the emitted radiation much weaker?

ANSWER:
It is acceleration which causes radiation. An oscillating charge is always accelerating except at the instants it passes through its equilibrium position so continuously radiating. Brownian motion is a series of very brief accelerations followed by much longer periods of constant velocity and therefore
the radiation is a series of pulses. There is no way you can predict
which is weaker by only knowing the acceleration. It depends on the
magnitude of the charges involved, the magnitudes of the accelerations,
and the frequencies of the accelerations.

QUESTION:
My question is related to photons. We have coherent light IE laser emission which over distance spread out much more slowly. Our sun emits incoherent light, which is the same for all observant stars. My query is why do we still see the stars as pinpoints of light no matter if they are near or very distant.

ANSWER:
The reason all the stars look like pinpoints is that they are very far
away, not because they have tightly collimated beams like a laser; if
the sun were very far away it would look that way too. If the light you
see from a star were a tightly collimated beam, it would be pointed
directly at you and if you stepped to the side it would disappear; you
would see almost no stars at night because nearly all of them would be
pointing elsewhere.

QUESTION:
when i kick the ball ,i exert a force on it and it exert the same force on me on the opposite direction according to 3rd law of newton ,bet what is the effect of the ball on me .the effect of the force that i exert on the ball make it move ,but my leg do not move .

ANSWER:
Do you feel it when you kick the ball? Of course you do and what you are
feeling is the force the ball exerts on your foot. But your foot moves in
response to all the forces on it, not just that one. During the kick
your leg is exerting a force on your foot much larger than the ball's
force on you. Your foot will be moving a little bit more slowly after
the kick than if you had not kicked the ball.

QUESTION:
With the understanding that "gravity" is a fictitious force created or experienced by Earths acceleration what I don't understand is what is the source of this acceleration? How is it that the Earth is continuously accelerating us upwards to produce weight?

ANSWER:
Your understanding is wrong. Gravity is not a fictitious force caused by
acceleration. Maybe it would help you to read some of my earlier answers
regarding gravity and general relativity listed on the
faq page .

QUESTION:
This is odd, but My family has just moved into a huge house with little outdoor space. We live in a climate that is cold in the winter, and I want my children to get some exercise on a daily basis. We own a trampoline, and have space for it indoors on the Second floor of our house. The ceilings are 12 feet high, so there would be no problem with the kids hitting their heads on the ceiling. My question is whether or not the house would stand up to the force generated by the trampoline. The walls of the house are made of concrete (you can't nail into it.) I am assuming the floors are quite solid as well, as they must support the weight of the house. They are concrete as well.
My Youngest child is quite large (6 ft, 260 lbs)--he is only twelve. We need the activity.

ANSWER:
First, a disclaimer: I can give you an idea of how much force the floor
will experience. I cannot predict whether this will cause your floor to
fail because I have no information about your floor other than that it
might be concrete. I have watched some videos and it seems that the
jumper never goes as high as h =2 m and the trampoline never
goes down as far as s =1 m. So I will just do my calculations
with those to get an upper limit on what force might be expected. Your
son's mass is about m =120 kg. An object falling from h =2
m will hit the trampoline with a speed of about v = √(2gh )≈√(2x10x2)=6.3
m/s. I will treat the trampoline as a simple spring so that I can write ½mv ^{2} =½ks ^{2} -mgs ^{
} where k is the spring constant. Putting in m ,
v , and s and solving for k I find k =7200
N/m; since the force exerted by a spring is F=ks ,
the largest force the trampoline exerts on your son is about 7200 N=1600 lb;
Newton's third law tells you that this is also the force your son exerts
down on the trampoline. Therefore, the trampoline exerts a force down on
the floor of 1600+W where W is the weight of the
trampoline. This is a little more than the weight of a grand piano.Keep
in mind that this is the greatest force and just for an instant; the
average force over the collision time would be half this. This is a
little more than the weight of a grand piano.

QUESTION:
I recently watched with great interest a PBS program which described in layman's terms how Uranium 238 transforms into the different chained elements, to include U235. It also explained the basics of the chain reaction caused by splitting U235 using E=MC^2 as the basis for energy release. This is where I was a little unclear.
The split was described as one U325 nucleus splitting into two separate nuclei with some individual particles released (can't remember if they are protons or neutrons) Those particles then collide with other U235 atoms in proximity triggering subsequent splits and particle releases as part of a chain reaction.
My question is that the mass doesn't appear to be transforming into energy (E=MC^2). Rather it appears that it is simply splitting and being cast off, so what causes the energy release? This assumes that the number of particles in the remaining two nuclei + the particles independently released still equal 235. There was a general reference in the program to how the Strong Force reduces the size of the resultant smaller nuclei, but it didn't say if matter within each was converted to energy or if the number of particles are additionally reduced through such a conversion. Thanks for any clarification you can provide.

ANSWER:
Suppose that you weigh one ^{235} U and one neutron. Now, when
you add the neutron to the ^{235} U it fissions and, after all is said
and done you have two lighter atoms and a few neutrons; if you weigh all
these byproducts, you will find that approximately 0.1% of the original
weight is missing. Where did it go? Most of it went into kinetic energy
of the byproducts, that is they are all moving faster. Kinetic energy of
atoms is essentially what thermal energy is —the reactor (or
bomb) has gotten hotter. For a bomb it all gets enormously hotter
resulting in the explosion. For a reactor, the rate of fissioning is
controlled and the heat is extracted to drive turbines to create
electricity. More detail can be found in an
earlier answer .

FOLLOWUP QUESTION:
I'm assuming (correct me if I'm wrong) that such motion was some percent of the speed of light which would account for the transformation of about
0.1% of its matter to energy.

ANSWER:
First things first—there was an error in my original answer, now
corrected throughout: the amount of mass converted to energy is
about 0.1%; nuclear fusion is about 1%. No, the motion of the atoms is
nowhere near the speed of light. It is simply classical ½mv ^{2}
type of kinetic energy. The "transformation" is simply that—mass
energy transformed into kinetic energy. To understand, see a
recent answer which explains why a bound system has
less mass than if it is pulled apart and the mass measured.
It turns out
that heavy nuclei like uranium are less tightly bound than nuclei with
roughly half their mass; therefore when they split the products are less
massive. That is why fission works as an energy source.

QUESTION:
If a capacitor is made of oppositely charged plates, why do they look like cylinders inside computers, remote control cars and other electronics

ANSWER:
The easiest form of capacitor to understand and analyze in an
introductory physics class is the parallel plate capacitor. But any two
conductors insulated from each other is a capacitor. One possible
capacitor is a wire along the axis of a hollow cylinder, but that is not
what the common capacitor you are referring to is. Rather, it is a parallel
plate capacitor! The analysis of the capacitance of two parallel plates
shows that the capacitance is proportional to the area of the plates and
inversely proportional to the distance between them. So, take a two
ribbons of foil as long as a football field to make the area big and
separate them by sandwiching a ribbon of mylar between them to make the
separation small; then just roll it up so you can fit it in your device!

QUESTION:
Suppose a block is moving with constant velocity towards right on a frictionless surface and during its motion another block of slightly smaller mass lands on top of it from a negligible height.
I argue that the lower block will eventually start moving to the left and upper block will end up moving towards right provided that there is friction between the blocks but not between lower block and ground . My friends can't accept my reasoning. Am I wrong? Please help!

ANSWER:
I hate to tell you, but you are wrong. This is actually a simple
momentum conservation problem. Call the masses of the upper and lower
blocks m and M , respectively. Before they come
together the momentum is Mv where v is the incoming
speed of M . When the masses come in contact they will slide on
each other but, because there is friction, they will eventually stop
sliding and both will move with a velocity u ; the linear momentum will
now be (M+m )u . Conserving momentum, u =[M /(M+m )]v .
They both end up going with speed u and move to the right.

You can also determine the time it takes for the sliding to stop. m
will feel a frictional force to the right of magnitude f= μmg
and M
will feel a frictional force to the left of magnitude f= μmg
(Newton's third law). So, choosing +x to the
right, the acceleration of m is a =μg and
the acceleration of M is A =-μg (m /M ).
The velocities as a function of t are v _{m} =μgt
and v _{M} =v -μgt (m /M );
we are interested in the time when v _{m} =v _{M} ,
so solving for t , t=Mv /[μg (M+m )].
If you substitute this back into v _{m} or v _{M} ,
you will find the same value we found for u above: v _{m} =v _{M} =u =[M /(M+m )]v .

FOLLOWUP QUESTION:
If the block of mass M is not too long, i.e., the total distance that the upper block can slide is less that the distance it could move in your calculated time " t" , wouldn't the two blocks get separated?

ANSWER:
Well, of course the block has to be big enough, otherwise m
will drop down on the frictionless surface. It would be a good exercize
for a student to calculate how far the block would slide for some
μ . And then, if this is greater than the size of the bottom
block, how fast will each be moving after separating.

QUESTION:
What would it take for a falling body to travel 25' horizontally from a 350' height?
I'm sorry, I didn't explain well. This is an actual situation. There is a guard booth at the base of the bridge, 25' away. If someone was to jump off the bridge at a height of approx 300+ feet, would it be possible that they would be able to strike the booth?

ANSWER:
One possibilty is if there is a steady wind blowing in the right
direction. You should know that calculation of air drag, the force of
the wind in this case, is always a rough estimate, not something you can
predict with precision. I prefer to work in metric units, but I will
switch back to ft and mph at the end. A rough estimate of the wind force
is F ≈¼Aw ^{2 } where A
is the area presented to the wind and w is the speed of the
wind (this approximation only works for SI units). I will assume that
the horizontal speed acquired by the jumper is small compared to the
wind speed so that w is a constant during the fall. So,
approximating A ≈1 m^{2} , the
acceleration horizontally is a _{x} =F /m =w ^{2} /(4m )
where m is the mass of the jumper. The time to fall 350 ft is
about t =4.7 s and the horizontal distance is then x =½a _{x} t ^{2} =w ^{2} t ^{2} /(8m ).
Now, x =25 ft=7.6 m and I will take m =150 lb=68 kg.
Solving for w I find w =13.7 mps=31 mph. A steady wind
of 31 mph could cause the jumper to move 25 ft horizontally.

You have not told me where the the booth is. If it is under the bridge,
then the jumper could not propel himself in that direction. But if the
booth is 25 ft out from under the bridge, the jumper could jump out as
well as drop down. The speed v _{x} he would have to
give himself horizontally can be easily calculated: v _{x} =x /t= 25/4.7=5.3
ft/s=3.6 mph.

QUESTION:
In electromagnetism we compute the intensity of a wave by taking the square of its amplitude. Why do we not do exactly the same thing with quantum mechanical waves?

ANSWER:
Actually, you could say that is exactly what we do. You just have to ask
what intensity means for the wave function. In electromagnetism,
intensity is just the energy density flux, the power per unit area,
measured in W/m^{2} . The square of the wave function is the
probability density and so is a measure of the likelihood of finding the
particle in one small volume in space. If you add up the square of the
wave function at all points in space (integrate), you must get the
answer of 1 because the probability of finding the particle somewhere in
space must be 1 for this interpretation to make sense; this is called
normalization.

QUESTION:
My dad told me about your website, very interesting reading. My question deals with molecules. When a molecule emits a photon, the mass of the molecule decreases to account for the energy in the photon. So, the mass of the molecule as a whole decreases, but this mass does not come from the "parts" of the molecule. In other words, the mass of the constituent electrons does not decrease, the mass of the protons does not decrease, so the energy must come from the electric field between the electrons and protons.
But the electric field has energy, not mass. Now mass is a form of energy, but I don't think you can say that the field has mass? But yet, it is said that the mass of the molecule decreases. The electric field contributes to the mass of the molecule, but yet it is incorrect to say that the field has mass?

ANSWER:
Actually, this is not as complicated as you are trying to make it. It
all boils down to the fact that mass is a form of energy and must be factored into
any energy conservation that occurs in an isolated system. You say that
the masses of the protons and electrons do not change, but that is not
right. Look at the simplest case, a hydrogen atom. If you measure the
mass of this atom it will be less than if you measure the masses of a
free electron and a free proton. Here is how you can see that: if you
pull the electron away from the proton, that is you ionize the atom, do
you have to do any work? Of course you do because the electron and
proton are bound together. So, you have added energy to the system (p+e)
and that energy shows up as mass. In a system as complicated as a
molecule you cannot say which particle or particles changed their
masses, but you can say for sure that the total mass of the molecule
changed by exactly the energy of the emitted photon divided by c ^{2} .

QUESTION:
This is my 2nd question related to the issue of time dilation - this one being related to the issue of motion [the other being based on gravity]. Since time dilation occurs for all moving objects, and considering further that the Earth has been revolving around the sun at 30 km per second for the last 4 billion years -- and further that our solar system is moving at roughly 45K mph through space, can't it be said that, compared to other objects in the universe, time dilation has occurred to a significant degree for our planet over those 4 billions years? And that really every object in the universe likewise has its own unique time dilation associated with it? Can't it also be said that every consolidated arrangement of matter in the universe is moving along at different "rates of time?" Wouldn't, over the course of several billion years, these "pockets" of different time spans become more and more "incompatible" with each other?

ANSWER:
The two speeds you quote are about the same (45,000 mph ≈2x10^{4}
m/s and 30x10^{3} km/s=3x10^{4} m/s). So let's just
choose the larger one and see how much time dilation there is. Relative
to the sun, an elapsed time T =4x10^{9} y would
correspond to T'=ΥT wwhere Υ= 1/√(1-(v /c )^{2} )=1/√(1-(3x10^{4} /3x10^{8} )^{2} )≈(1+0.5x10^{-8} ).
Therefore T'≈ (T +20), a difference of 20 years. This
may sound like a pretty long time to you, but relative to 4 billion it
is less than 10^{-6} %. That is not "significant" to my mind.
You are right that every object has its own clock which, relative to
other clocks, is not necessarily the same; every object also has its own
meter stick, not necessarily the same as other meter sticks in the
universe. The important thing is that you always must talk about
velocity with respect to what.

QUESTION:
The movie "Interstellar" did a nice job of explaining how time dilation works in a massive gravity field. My question relates to how we on Earth measure the age of the universe to be 14.3B years. If I could make that measurement on the planet orbiting the "Interstellar" singularity [and since, theoretically, if I could view life on the singularity planet from Earth, it would all look to be in super slow motion], what would my measurement of the universe's age be from the my perspective on that planet? If time moves more slowly on the singularity planet, wouldn't my estimate of the universe's age be much less?

ANSWER:
As I say on the site, I do not usually answer questions on
astronomy/astrophysics/cosmology. Maybe I can make a stab at this. The
microwave background radiation which pervades the universe is generally
considered the best source for information about the big bang and
measurements are probably the best determinations of the age of the
universe. In your high-gravity position, you would see the same
microwave radiation I do. Likely, you would have to make corrections to
your observations due to the intense gravity, but you would still
conclude the same age as I did.

QUESTION:
hi, can and are earthquakes be caused by celestial alignments ie planets?

ANSWER:
Let's take a simple example. As seen from earth, Mars and Jupiter are
aligned. I estimated the force on a 1 kg object which is sitting, let's
say, on the San Andreas fault: F =3x10^{-11} N; the
weight of that 1 kg object is about 10 N. I would say that putting a 1
kg object on the ground is a great deal more likely to cause an
earthquake than those planets, wouldn't you?

QUESTION:
So there's a powerline outside my bedroom window, and I thought, huh. Turns out I'm sleeping with my head in a 6mG AC magnetic field (according to two meters). Help me use physics to stop caring.
How do I estimate/calculate which puts more force on the charged particles (calcium, potassium, sodium) in my brain: a) An aqueous solution at 98.6 degrees Fahrenheit or b) a magnetic field acting on charged particles moving at some estimated speed in said aqueous solution.
My hope here is that the force of (b) is like an order of magnitude or two, or something, below the "noise floor" of (a) and then I can stop caring forever.

ANSWER:
How about this: the earth's magnetic field is about 0.6 G, two orders of
magnitude bigger than the field due to the power line, and you are
exposed to it 24 hours a day. It is also possible that there is some
other source of field closer by than the power line which, though a much
smaller current, would produce a much bigger field. For example, if
there were a wire in the wall carrying a typical household current of 1
A, the field 2 m away would be 1 mG. There is no good scientific
evidence that any magnetic fields you are likely to encounter have any
effect, good or bad, on the functioning of your body.

QUESTION:
I do not want a theoretical answer, but has any experimentalist ever put a very sensitive weight balance below a vacuum chamber before and after vacuating it? Does it get lighter or... heavier? I do not have a sensitive balance nor a vacuum chamber.
The reason I ask is that it would say something about the density, or absence of density, of the vacuum.
If I understand pressure correctly, the scale would read a smaller weight value, due to less gas being in the column of air directly above it, but there might also be new physics there, if it is not the case. I simply do not know.

ANSWER:
You do not want a "theoretical answer" but you clearly do not understand
the physics so I am obliged to give you one anyway. Let us assume the
simplest possible "weight balance" so that I do not have to worry that
it might operate differently in a vacuum. Envision just a simple string
with a tiny butcher's scale which will measure the tension in the string
and then hang an unknown weight of mass M and volume V
from the string. Besides the string, there are two forces on the object
being weighed, its weight Mg and the buoyant force B= ρVg
where ρ is the density of air (about 1 kg/m^{3} at
atmospheric pressure) and g =9.8 m/s^{2} is the
acceleration due to gravity. The scale will read W=Mg-B , an
incorrect measure of the weight. Putting the whole device in a vacuum
will change B to zero because the air is gone, so W=Mg ,
the correct weight. To get an idea of how important this is, consider
weighing a solid block of iron whose mass is 1 kg. The density of iron
is ρ _{iron} =7870 kg=M /V , so the
volume of the block is V =1/7870=1.27x10^{-4} m^{3} .
So, the true weight is W =9.8 N and the measured weight in air
is (9.8-1.27x10^{-4} ) N=9.7999 N, an error of 0.0013%. However,
there are certainly examples where the effect of buoyancy would be very
important. For example, consider an air-filled balloon. I did a rough
calculation and estimated that the volume of an inflated balloon is
about 5x10^{-3} m^{3} so it contains about 5x10^{-3
} kg of air; the mass of an uninflated balloon is about 5 gm=5x10^{-3}
kg, so the total weight of the inflated balloon is about 9.8x10^{-2}
N. But if you weighed it in air you would only measure half that amount.
Your question, has anybody ever actually observed this, is a no brainer:
since the existence of a buoyant force has been known and understood for
well over 2000 years (Archimedes' principle), anyone wanting to make an
extremely accurate measurement of a mass would either correct for it or
eliminate it.

QUESTION:
Several years ago, I was caught in a massive windstorm in a skyscraper. I was on the 54th floor (approx. 756 feet from street level, full building height is 909 feet) , pulling cable, and I stopped for a break. I left a cable pulling string hanging from the ceiling (48 inches free hanging length) in the office, with a 1/4 lb weight attached, and when the storm hit, the weight began swinging like a pendulum. The arc was 16 inches (eyeballing it), and traversing the length of the arc took about 1 second. How can I calculate how far (full arc) the skyscraper was moving by observing what the pendulum in the building was doing?

ANSWER:
A 48" pendulum has a period of about 2.2 s, the time to swing over the
arc and back. Since you were estimating, the pendulum was swinging with
about the period it would if the building were not moving at all. I
would conclude that either the pendulum got swinging somehow and the
building was not perceptibly moving or that the period of the building's
motion was about the same. If the building was swinging with a period
significantly different from 2 s, the pendulum would be swinging with
that same period; that is called a driven oscillator.