QUESTION:
Me and my brother are having a debate about running on a treadmill vs running on a track. Not counting mental stuff like learning how to pace i feel that the only real difference is wind resistance. He seems to think inertia comes into play differently for each. I feel like once you are on the treadmill and it starts moving that the treadmill track is your inertial frame of reference so any change in inertia would require just as much energy as it would on an outdoor track. He seems to think that since you don't let the treadmill ever move you actually that you never have to work against inertia of rest while you're accelerating or decelerating on a treadmill. I mean if the treadmill was already moving at 10mph and then you just jumped on it i could see his point but not if you start off standing on the treadmill while its at rest and accelerate along with it.

ANSWER:
In terms of simple introductory physics, you are correct.
However, there are often subtle differences between simple
physics and the real world when applied to very complicated
systems like the human body. Your brother is also wrong
because he is just trying to explain any differences in
terms of simple physics also. In fact, there are many
differences between running on a track and running on a
treadmill due to biomechanics. A good article to read is a
post on the
RunnersConnect blog. Or just google treadmill vs.
track .

QUESTION:
It takes 375 Joules of energy to crack a bone. Moreover, a baseball player hits a ball. The speed of the ball off the bat is 90 mph. The ball the player hit has a weight of 5 oz. If this ball were to hit someone in the head would it crack their skull? What other injuries could be expected?

ANSWER:
This question makes no sense to me. Surely it is harder to crack a femur than a skull.
Besides, you would specify a force needed to break the bone,
not an energy. The kinetic energy of
a 5 oz baseball at 90 mph is about 145 J.

QUESTION:
One of the ways to define horsepower is 550 lb-ft/sec. My understanding is that means 1 hp is required to maintain a velocity of 1 ft/sec straight up with a weight of 550 lbs. However wouldn't it take more than 1 hp to lift a 550 lb weight that is sitting on the floor motionless 1 foot in 1 second? How would you calculate the hp needed to lift 550 lbs from a dead stop to a height of 1 foot in 1 second?

ANSWER:
"Maintain a velocity" means just that —the
object begins, ends, and always has that velocity. So your
worrying about how it got that velocity in the first place
is really irrelevant to this example.

Let's get straight what all these
units are.

A pound (lb) is a
unit of force. A net force on an object accelerates it.
For example, the weight of an object is a force on it
downward and if you exert an equal force upward the
object will move upward (or downward) with a constant
velocity, any velocity.

A foot (ft) is a unit of length.

A foot-pound (ft-lb) is a unit of energy. If one lb is
exerted over a distance of one ft, the energy delivered
is 1 ft-lb.

A second (s) is a unit of time.

1 ft-lb/s is a unit of power and it measures the rate at
which energy is being delivered. Other units of power
are the horsepower (hp) and the Watt (W). 1 hp is 745.7
W.

Your question about how much hp it would to take to move an
object up 1 ft starting and ending at rest is meaningless
because there are an infinite number of ways to do that but
you could not do it with a constant power. For example you
could deliver a lot of power for 0.1 s, ending up with a
speed (at 0.1 s) which would be just right to get the object up
to 1 ft in an additional 0.9 s before falling back down; or
you could do the same thing with a smaller power to 0.2 s,
etc . Regardless of
how you achieve this for a 550 lb object, the total energy
required is 550 ft-lb. So, if it takes 1 s, the
average power is 550 hp.

QUESTION:
How does a credit card reader work? Specifically, when you slide your card in one?

ANSWER:
See an
earlier answer .

QUESTION:
If two dice were floating in space about a centimeter apart, what would eventually happen to them and why?

ANSWER:
They would be attracted to each other and, after a time,
stick together. The details are given in an
earlier answer
to a question very similar to yours except the dice were
separated by 10 cm there. If you want to calculate the time
your dice would take to come together, you can follow the
calculation given there changing the numbers. I find the
time to be surprisingly small.

QUESTION:
This question concerns angular momentum as related to motorcycles. If a motorcycle rests atop a trailer, but sits on a roller system, the bike will stand upright without any supports or tethers if the wheels are rolling (throttle is locked in the
"on" position and wheels are spinning). If the trailer itself travels forward in a straight line, the bike should remain upright. But if the trailer takes a sharp turn, what happens? Does the bike fall, or does it instead do what a rider does to achieve a turn-countersteer and remain upright?

ANSWER:
I have waited a long time to answer this question because I
am bothered by the way the problem is stated. First of all,
if the throttle is locked on, only the rear wheel will be
spinning, so we can discuss the problem by looking only at
the wheel. It is certainly correct that if the truck goes
straight the wheel will continue running upright (assuming
that the center of gravity of the bike is in the vertical
plane passing through the center of gravity of the wheel).
Imagine the bike and rollers to be mounted on a big "lazy
Susan" the base of which is bolted to the truck bed. So, if
a north-bound truck turns to the west, the angular momentum,
experiencing no torque, will remain constant and continue
pointing in the same direction (originally either east or
west). Viewed from inside the truck it will appear that the
whole bike rotated through 90^{0} relative to the
truck.

Now,
if the rollers are attached to the truck bed, when the truck
turns the rollers turn and the wheel, trying to not turn,
will come off the rollers at some point. We first need to
understand the physics relationship between the torque and
the angular momentum of the wheel. The rotational form of
Newton's second law is τ =ΔL /Δt ,
torque equals the time rate of change of the angular
momentum. The first figure shows the wheel, as seen from
above, turning through some small angle which results in a
change of angular momentum from L _{1}
to L _{2} and ΔL =L _{2} -L _{1} .
So the torque which you must apply to make it turn this
way is in the direction of ΔL .

If
you think you can just steer it as if it were not rotating,
you would fail. The second figure shows what would happen if
you try to steer it like your intuition would have you do it
by exerting a force like F in the
figure. The torque points up and so the wheel would not turn
but lean in the opposite direction from the way you would
lean on the bike if you were riding it and making a turn.
You can find some videos showing this by googling
gyroscope in a suitcase video .

If you want the wheel
to turn with the truck, you need to have a torque which
causes that. One way I thought of to achieve this was to
have strings attached from the axles on each side and the
truck bed below. These need to have no tension on them when
the truck is going straight. When the truck turns as
indicated, the tension in the string on the right-side
string will be bigger than the other side and there will
therefore be a net force N on the wheel. This results in a
torque in the horizontal plane which will cause a change in
angular momentum in the direction consistent with the wheel
turning with the truck. Two strings are needed because the
truck might turn either left or right.

QUESTION:
I wonder, how step by step, is parallax method used to find distance of a star?
When you find the parallax angle, then it's easy to find distance using tangent formula or other trigonometric formule.
But how is that parallax angle exactly determined?
In some online articles, like in
Wikipedia
or
Sky and Telescope they show such pictures.
But in this picture, only distance between A to B is known. Distance between A to Eros or B to Eros is unknown. Angle of A and B are unknown. So we can't know angle of p with so little information.
What we see only is a star on the sky do change its position (like you see right side of the picture) slightly (some millimeters) every 6 months.
How do astronomers determine from that motion the parallax angle of the star?

ANSWER:
The angles A and B are not unknown. These are measured by
the direction in which the telescopes at those locations are
pointed. Once you know those angles you know the angle p ;
and if you know p and the distance AB you can also
calculate the distances from A and B to Eros.

QUESTION:
Can you explain the differences between circular motion & rotational
motion?

ANSWER:
Rotation normally refers to an object spinning about some
fixed axis; e.g ., the earth rotates about the axis which passes
through the north and south poles with a period of 24 hours.
Circular motion refers to the motion of an object which
moves in a circle; e.g. , a race car on a circular
track is in circular motion. If the circular motion is with
constant speed it is called uniform circular motion;
e.g. a point on the equator moves in a circle whose
radius is the radius of the earth with constant speed. More
general than circular motion is revolution; e.g. ,
the earth revolves around the sun in an elliptical path once
a year.

QUESTION:
Is there a formula for calculating the side-ways deflection wind has on a lawn bowl(over and above the bias deflection ) running at 12
s, the time a bowl takes from delivery to stop over a 26 m distance over
bowling green grass?

ANSWER:
Once again, doing Ask the Physicist has led me to learn
something new. I never really knew anything about lawn bowls
other than it is done on grass and rolling balls are
involved. For the benefit of others who are ignorant of the
game, let me summarize by describing the ball. (A good
article on the physics of lawn bowls balls can be found
here .) The ball is not a sphere but rather an oblate
spheroid which makes it sort of like a door knob but not so
extremely flattened; but it is slightly more flattened on
one side of the ball than on the other which results in a
center of gravity being displaced to one side of the
equatorial plane as shown in figure (a). This results in a
tendency for the ball to curve left if it is rolling the
angular velocity shown in the figures; this motion is the
"bias" referred to by the questioner which I am to ignore.
When rolling in the x direction (figure (b)), there
is a frictional drag force called, rolling friction
D , which opposes the motion (v )
and eventually brings the rolling to a halt. If there is a
wind, there is a force W due to
the wind which tries to make the ball roll to the right
(figure (a)) but if it does roll, there will also be rolling
friction trying to keep it from rolling. In order for the
wind to have any effect at all, it is clear that we must
have W>D ; if this is not the case, there will only
be static friction in the y direction which will be
equal and opposite to W . A lawn
bowls ball has a mass of about m =1.5 kg and a
radius of about R =6 cm=0.06 m.

To get the equations of
motion for the x and y motions, we first
need expressions for D and W . The rolling
friction may be expressed as D=-μmg where μ
is the coefficient of rolling friction and mg is
the weight of the ball. The force due to the wind may be
approximated as W ≈¼AV ^{2}
where A=πR ^{2} is the cross sectional
area of the ball and V is the speed of the wind;
this approximation is only correct if SI units are used. The
equations of motion in the x -direction are

Here t is the
time and v _{0 } is the speed of the ball at
t =0. If the ball is rolling in the y-direction
because of the wind, the equations of motion are:

It should be noted that
if (¼AV^{2} /m )<μg , these
equations imply that the ball will accelerate opposite the
direction of the wind, obviously not correct; hence the wind
will have no effect on the ball if V <√(4μmg /A ).
In that case, a_{y} =v_{y} =y =0.

So, having found the
general solutions, let us now apply the solutions to the
specific case from the questioner. We are told that when
t =12 s, v_{x} =0 and x =26 m.
With that information you can solve the x -equations
to get v _{0} =4.32 m/s and μ= 0.037,
reasonable values compared to numbers in the
article I read. The area is 3.14x0.06^{2} =0.0113
m^{2} . The first question we should ask is what is
the minimum speed of the wind to have any effect at all:
V _{min} =√(4x0.037x1.5x9.8/0.0113)=13.9
m/s=31 mph=50 km/hr; this is a pretty stiff wind, so the
wind probably has no effect on bowling under normal
conditions. So, just to complete the problem, consider V =15
m/s=34 mph=54 km/hr.

The^{ }
trajectory during the 12 seconds is shown in the graph
below; after 12 seconds the ball will continue accelerating
in the y direction.

So the bottom line is that unless you are playing in a gale-force wind, the wind has no effect on the ball if the wind has no component along the original direction of the ball (which I have called the
x -axis). You can tell if wind makes a difference by simply setting the ball on the ground--unless the wind blows the ball away, you need not worry about its effect. If the wind is blowing in the +x or -x direction, that is a whole different thing, but
the questioner asked for the sideways deflection.

ADDED
THOUGHTS: This question continues to intrigue me and
I have carried my investigation further. The question
originally stipulated "over and above the bias deflection"
so my whole discussion totally ignored the fact that the
ball, owing to its off-center center of mass, will curve. At
the very end of my answer I noted that if the wind is not
perpendicular to the path of the ball, it would be a
different story; indeed for a spherically symmetric ball I
showed that, except for very strong winds, a wind
perpendicular to the path has no effect at all. However, for
an actual lawn bowls ball, the path curves to where a wind
in the y -direction might have a significant
component along the path. I have calculated (graphed below)
the x and y positions of a realistic path
with no wind using equations (10) and (11) of the
article referred to above. To do these I used all the
numbers used above (R , m , A ,
v _{0} , μ ); I used the
moment of inertia for a solid sphere ( I _{0} =I _{cm} +mR ^{2} =(7/5mR ^{2} ))
and chose the COM off-center distance to be d =1 mm.
As you can see, the curving is substantial, carrying the
ball about 4 m from its original direction. You can see that
now a wind of any magnitude can have an effect on the
trajectory. The angle φ which the tangent to
the trajectory is given in the article as φ =(2/p )ln(v _{0} /(v _{0} -μgt ))
where p is a constant also given in the article. As
can be seen, once the trajectory leaves the x-axis the wind
contributes with the component of its force along the
trajectory; this has the effect of reducing the effect of
the frictional force causing the ball to slow down less
rapidly. However, this is now like having a time dependent
force of friction which, I believe, will lead to equations
of motion which will not have an analytical solution but
would have to be solved numerically.

QUESTION:
How can rockets move in space. Because they have nothing to push against won't trust be useless. Like for example the reason the rocket gets out of the atmosphere is because of the thrust pushing against the earth then on the air in the atmosphere and if space has no air how can rocket move in space.

ANSWER:
The reason is that your statement that "the thrust pushing
against the earth then on the air" is the reason the rocket
accelerates is totally wrong. The reason a rocket
accelerates is that hot gas is expelled out the back at a
high velocity. The reason this propels the rocket can be
illustrated by the following example: an astronaut is
floating in empty space and happens to be holding a bowling
ball; she throws the ball and the result is the balls moves
away and she recoils so she is moving also. This is called
conservation of linear momentum —if the ball has
1/20^{th} as much mass as she does, she recoils with
1/20^{th} the speed the ball has. The rocket is not
"pushing against" anything.

QUESTION:
Earth moves around sun at 67000mph. A rocketmoving in same direction as earth at 25000mph isnt it actually moving at 25000 plus 67000 mph?

ANSWER:
There is no such thing as "actually moving". The only way a
stated velocity has any meaning is if it is relative to
something else. If the rocket has a velocity of 25,000 mph
relative to the earth , then its velocity
relative to the sun is 92,000 mph if the velocity of
the earth relative to the sun is 67,000 mph.

QUESTION:
This is something I remember discovering when I was younger. If i would take a small cylindrical object (like a AA battery) , set it on a flat hard surface, then using my fingers I would apply downward pressure on the edge of the battery. This would create backspin on the battery but also shoot the battery across the floor. When the battery's finally stopped moving forward it would spin rapidly on its axis until stopping completely. Could you explain the physics behind movement of the battery?

ANSWER:
Problems like this are standard in intermediate-level
classical mechanics. Round objects can move translationally
on a horizontal surface by rolling without slipping or by
sliding. The two classic extremes of the slipping scenario
are a skid where the object is initially not spinning at all
(e.g . a bowling ball begins by approximately not
rotating but may end up rolling without slipping), or is at
rest but spinning (like "peeling out"); in both cases, the
problem is usually to find the time elapsed or distance
traveled before rolling sets in. In your case, the object is
initially both translating horizontally and rotating with a
backspin. What will happen depends on the initial conditions —the
initial speed and the initial angular velocity; also the
properties of the object will also matter—its mass
m , radius R and shape, and the coefficient of
kinetic friction μ . Three possibilities are that

it will reverse directions and come back still
spinning and slipping, eventually stop slipping and roll (what I think
you are remembering),

it will stop
slipping and continue rolling in the initial direction, or

it will stop
dead.

In the figure
there are three forces on the object: its weight mg ,
the normal force N up from the floor, and the
frictional force f which the floor exerts on the
sliding object; the object begins with velocity v _{0}
and angular velocity ω _{0} . The mass
of the object is m and its radius is R . The friction slows
down both the velocity and the angular velocity. Sliding ceases when
v=-Rω (see footnote*). If v _{0} is very small
and ω _{0} is very large, possibility #1 will happen;
if v _{0} is very large and ω _{0} is
very small, possibility #2 will happen; if v _{0} and ω _{0}
are just right, possibility #3 will happen. I think that this gives you a
good qualitative overview of the physics of the problem.

For those interested in the
quantitative solution, I will give it here for a uniform solid cylinder. For
the translational motion the two equations (taking +x as to the
right, +y up) are -f=ma and N-mg =0; since f=μN=μmg
we can write a=-μg . For rotational motion about the center of
mass (choosing the direction it is initially spinning as positive), only the
frictional force exerts a torque, so -fR =-μmgR=Iα
where I is the moment of inertial about the center of mass and α
is the angular acceleration. For a cylinder I =½mR ^{2} ,
so α=- 2μg /R. So, we can write the
equations for the velocity and angular velocity as functions of time t :
v=v _{0} +at =v _{0} -μgt
and ω=ω _{0} +αt=ω_{0} -2μgt /R.
Now, the time when slipping ceases will be when v=-Rω ;
solving, t =(v _{0} +Rω _{0} )/(3μg ).
Finally, put this into the equation for v to get v (t )=(v _{0} -Rω _{0} )/3.
So if Rω _{0} >v _{0} , possibility #1
will happen; if Rω _{0} <v _{0} ,
possibility #2 will happen; and if Rω _{0} =v _{0} ,
possibility #3 will happen.

*The condition v=-Rω is
a little tricky. The negative sign is because of my choice of the positive
direction for ω which is opposite that which would be the
case for no slipping.

QUESTION:
Do cars need friction to move?

ANSWER:
If they are not moving, they need friction to move. If they
are moving, they need friction to stop. (I have assumed that
you mean that you control things from inside the car.
Somebody outside the car who was anchored to the ground an
held a rope attached to the car, for example, could move the
car. Also, if the car had a rocket engine it could move.)

QUESTION:
Sometimes gravitational force is proportional to r^2 and sometimes to 1/r^2.
What's the reason?

ANSWER:
The gravitational force depends on the way that mass is
distributed. For example, if you have two spherically
symmetric spheres of mass, like the earth and the moon, the
force between them is proportional to 1/r ^{2} .
If you are inside a sphere which has its mass uniformly
distributed inside, the force you would feel is proportional
to r . If you are inside a sphere which has its mass
distributed so that the density varies linearly (e.g. ,
the density at the surface is twice as large as the density
halfway to the center), the force you would feel is
proportional to r ^{2} .

QUESTION:
Can you please explain why putting water into a bottle AFTER putting in baby formula does not give the correct exact amount of water necessary for proper amounts. E.g,the instructions on the formula tin state 1 scoop of formula for 30mls of water

ANSWER:
Because the formula occupies volume also. So when you put
the formula into the bottle and then fill the bottle up to
the 30 ml level, you will have less than 30 ml of water. If
you measure the 30 ml of water before adding it to the
bottle, it should make no difference. (Although, sometimes
if you are mixing something dry with something wet, how
easily the dry dissolves depends on the order of mixing.)

QUESTION:
If a weight of 7.5 kg is falling 0.6 m and land on the head of somebody, I understand that
the energy at impact will be 44.1 J.
But how can I translate this in g as it is the value reported by accelerometers?
In other words, if the person has an helmet with an accelerometer, which value in g the instrument will register?

ANSWER:
When you measure an acceleration you are measuring a force,
not an energy. It is true that a 7.5 kg mass has a kinetic
energy of about 44.1 J if dropped from a height of 0.6 m (mgy =7.5x0.6x9.8),
but that does not tell you the average force exerted by the
mass as it came (accelerated) to a stop. The speed which the
mass has when it hits is v =√(2gy )=3.43
m/s; if it stops in a time t , then the average
force over that time is F=mv /t =25.7 N.
For example, if t =0.1 s, F =257 N. To
convert this to g-force, the weight of the object is
7.5x9.8=71.3 N so the force in gs is F =257/71.3=3.6
gs.

QUESTION:
How would you convince a relative who had not studied physics that the international space station is really a falling body?

ANSWER:
My favorite example is Newton's mountain. The idea is that
you imagine a cannon on top of a very high mountain. You
fire the cannon in a horizontal direction and the cannonball
falls to the ground eventually. The more speed you give to
the cannonball, the farther it goes until it eventually goes
all the way around the world, falling all the time but never
hitting the ground. The illustration here is the original
drawing by Newton which was published in his classic book,
Principia . There is a nice animation you can play
with
here .

QUESTION:
I observed the following:
Placing a photo flashgun very close to a butterfly produced a wisp of what looked like
steam from the butterfly's wings. It seems that the intense burst of light is vaporising some dust on its wings? The creature does not seem to notice and flies
off after three flashes all of which caused the steam or dust to rise?
Was I seeing dust or vaporised dust?

ANSWER:
It is well known that there is "dust" on butterfly and moth
wings. In fact, this is not dust but tiny scales on the
surface of the wing. It comes off rather easily as you have
demonstrated. Although I have seen explanations of similar
results of flashguns which attribute the effects of the
flash to the momentum transfer of the photons in the flash
(for example, the "singing cymbal" demonstration where a
cymbal is made to sound with a flashgun), that has been
shown to be incorrect; what is happening is that there is a
thermal pulse caused by the light pulse which is responsible
for, in your case, dislodging some of the scales.

QUESTION:
What is weight and why there is no weight while falling body (pulling) by earth..?

ANSWER:
Several years ago I had to teach an introductory physics
course using a textbook in which the author defined weight
as what a scale reads . I do not know a single
physicist who likes this definition. Historically and almost
universally weight means the force which the earth (or other
massive object) exerts. When something is dropped it
accelerates downward because the weight is a force pointing
vertically down. Therefore, your question is wrong because
"there is no weight while falling" is incorrect —the
force the earth exerts on you does not disappear when you
are falling. Similarly, astronauts in orbit are not
"weightless" as is usually said because there is still a
force of attraction to the earth.

Nevertheless, it is clear that the astronauts seem to have
zero net force on them. The reason is that they are in "free
fall" when they are in orbit and are therefore accelerating.
There is a legend about Albert Einstein related to this. The
legend is that Einstein saw a painter fall from a ladder and
he said to himself "there is no experiment which he could do
which would allow him to say that he is not at rest in empty
space". This is called the equivalence principle and forms
one of the keystones of the theory of general relativity.

QUESTION:
What is weight and why there is no weight while falling body (pulling) by earth..?

ANSWER:
Several years ago I had to teach an introductory physics
course using a textbook in which the author defined weight
as what a scale reads . I do not know a single
physicist who like this definition. Historically and almost
universally weight means the force which the earth (or other
massive object) exerts. When something is dropped it
accelerates downward because the weight is a force pointing
vertically down. Therefore, your question is wrong because
"there is no weight while falling" is incorrect —the
force the earth exerts on you does not disappear when you
are falling. Similarly, astronauts in orbit are not
"weightless" as is usually said because there is still a
force of attraction to the earth.

Nevertheless, it
is clear that the astronauts seem to have zero net force on them. The reason
is that they are in "free fall" when they are in orbit and are therefore
accelerating. There is a legend about Albert Einstein related to this. The
legend is that Einstein saw a painter fall from a ladder and he said to
himself "there is no experiment which he could do which would allow him to
say that he is not at rest in empty space". This is called the equivalence
principle and forms one of the keystones of the theory of general
relativity.

QUESTION:

Imagine a cart moving on a frictionless surface with a cannon aimed opposite the direction of travel. At 40 km/hr the cannon fires a projectile accelerating the cart-cannon to 50 km/hr. A second projectile (identical to the first) is then fired accelerating the cart to 60 km/hr. The impulse (and thus recoil) against the cart should be the same no matter what the moving velocity of the cart was. After all, the cart is only moving within my frame of reference. If I selected a frame of reference in which the cart was stationary, an impulse acceleration of 10 km/hr would be the same irregardless of whether in another frame of reference the cart was initially moving at either 40 or 50 km/hr. It's true that the second firing of the cannon would be acting on slightly less mass, after the first cannon ball was no longer part of the system—but that would seem a trivial difference. I also understand that the momentum of the total system is the same before and after the cannon is fired.

Does it take more energy to accelerate an object from 40 km/hr to 50 km/hr than it takes the same object to accelerate from 50 km/hr to 60 km/hr ? I know that the Kinetic Energy at 60 km/hr is 60/50 squared or 1.44 times greater than at 50 km/hr so the answer is undoubtedly, yes. What am I missing about Kinetic Energy - does it depend on your frame of reference?

ANSWER:

I have rearranged
your question so we can talk about one question at a
time. Your first question, not really a question,
basically says that, although the momentum of any object
is dependent on your frame of reference, that the change
of momentum if you do something to it (impulse) is not
frame dependent. Why is that? The reason is that
Newton's second law may be written as F =dp /dt
and the rate of change of momentum must be frame
independent because we know that the force is frame
independent.

Now, when you change the momentum you also change the
kinetic energy because either the speed or the mass is
changing. But does everyone see the same change in
kinetic energy? The answer is no. And your example shows
very clearly that the change in kinetic energy does
depend on the frame of reference. A more general
example, calculating the work done by a constant force
as seen in different frames, can be found in an
earlier answer .

Q&A OF THE WEEK,
4/7/2018

QUESTION:
For a snow plow that is very heavy, is there an advantage to having the
connection point of the winch line up high so that there is less weight to
pull or is there no difference?
I can send a picture for clarification.

ANSWER:
Yes send me a picture. You are talking about a winch which is used for what?
Pulling a stuck vehicle? Lifting and lowering the plow? What?

FOLLOWUP QUESTION:
Yes, lowering and lifting a plow. The problem is that the plow is out very far out and that my winch is mounted pretty low on my machine. So instead of the winch simply lifting the plow up, right now it is mostly pulling backwards and then that is making the plow come up.
I attached a picture of what the manufacturer recommends but I haven't had too good of luck with them in the past. The last two pictures are of my machine. You can see how far out the plow is and how low my winch is. Would that pulley and cable system helped at all?

ANSWER:
You may not want to get the full physics explanation here,
so I will first give you a qualitative explanation. The
tension T in the strap is what is
lifting plow and any part of it which is horizontal (T_{H} )
is wasted. Your gut feeling is right, "…it is mostly
pulling backwards…" Anything which you can do to
increase the vertical part (T_{V} ) will
make the lift easier, and moving the winch up is a good way
to do this but it might be easier to have a pulley
higher up which then brings the strap back down to the
winch.

The figure shows all
the forces on the plow assembly: the weight W
which acts at the center of gravity (yellow X), the tension
T in the strap (where I have shown
vertical (T_{V} ) and horizontal pieces (T_{H} )), and the force the
truck exerts on the support which I have represented as its
vertical and horizontal parts, V
and H respectively. (Note that the
various forces are not drawn to scale since T has
to be much larger than W to lift the plow.) Suppose T
is just right that the plow is just about to lift. Then the
sum of all the forces must add to zero, or V+T_{V} -W =0
and H-T_{H} =0. The sum or torques must also
add to zero; summing torques about the point of attachment
to the truck (light blue X), WD+T_{H} s-T_{V} d =0.
Note that the T_{V} is trying to lift the
plow but T_{H} is trying to push it down.
Now, in order to get a final answer for the unknowns (which
are T , V , and H ) we note that
T_{V} =T sinθ and T_{H} =T cosθ
where θ is the angle which the strap to the
winch makes with the horizontal. The final answers I get
are:

T=DW /(d sinθ-s cosθ )

V =W-T sinθ

H=T cosθ

I put in some reasonable numbers just
to get an idea of the answers, W =500 lb, θ =20^{0} ,
D =2m, d =1.8 m, and s =0.1 m. Then T =1920
lb, H =1800 lb, and V =-157 lb. The negative value for V
means that V is down, not up. In this scenario,
the pulling force has to be nearly four times greater than than the weight
being lifted.

Now
we need to look at whether the manufacturer's suggestion will be better than
a straight shot to the winch. Now there are two forces pulling up, the
tension T from the pull point to the winch and the
tension P from the pull point to some anchor
higher up. Of course the magnitudes of these two tensions are the same,
P=T . The picture shows only the pulling forces, the rest are the same
as in the picture above. There are still three unknowns, T , V ,
and H . I will call the angle that P makes
with the horizontal φ . I will not show the details, just give
the final results:

As a numerical example, I will use the
same numbers as above and add φ= 40^{0} . Then T =624
lb, H =1070 lb, and V =-115 lb. It is definitely
advantageous to use the manufacturer's suggestion here which, in my
numerical example, reduced the force the winch needed to exert by a factor
of about 3.

QUESTION:
Is it possible that if one person lives in another planet (say Mars) can we chat with them live? (like on a mobile phone) or is it impossible? also will that person age differently compared to me?

ANSWER:
Certainly not using your mobile phone; the current cell
phone network depends on earthbound transmitters and
receivers. But communication is certainly possible or else
we would not be able to communicate with the rovers and
probes already there. Trying to carry on a conversation
would be very frustrating, though, because of the transit
times of the signals. Depending on where earth and Mars are
in their orbits, transit time is 3-13 minutes. So, if you
ask a question, you would have to wait 6-26 minutes for an
answer.

QUESTION:
We are a seventh grade science class, and we are discussing friction. We understand that life as we know it would end if there were no friction, but we were wondering what, exactly, would happen to your body without friction.

ANSWER:
It
would have been impossible for your body to form in the
first place. When two surfaces are in contact with each
other they exert two forces on each other. As an example,
think about a block which is sitting at rest on a slight
incline. The part of the force which the incline exerts on
the block which keeps the block from breaking through the
incline and that is called the normal force. The part of the
force which the incline exerts on the block which keeps the
block from sliding down the incline and that is called the
frictional force. With no frictional force, the block would
slide down the incline. Now imagine a single bone cell on
the surface of your leg bone. It has a weight which makes it
want to slide down your leg to the ground; friction between
it and its neighbors with which it is in contact keeps that
from happening. With no friction, all the cells in you body
would slide down to the floor and spread across the floor.

QUESTION:
When operating a forklift and wearing a hip style seat belt, what is the g force the body is subjected to during a head-on collision with a immovable object at 2-3 mph?

ANSWER:
This question has been answered, although for different
numbers, before . Here is the answer
I gave there but substituting your numbers. You should
definitely read that answer before proceeding here. Let's
say that the forklift stops dead and that it plus the pole
deform by 1 cm so that the distance traveled during the
collision time is d ≈1
cm. Since these are very rough calculations, I will let
the mass of the forklift be m _{f} ≈5000
kg and the initial speed be v =3 mph≈1.3
m/s. If the forklift stops uniformly, it will stop in a
time t =2d /v =0.015 s. The
force F experienced (by both the pole and the
forklift) will be F=m _{f} v /t ≈1.4x10^{6}
N=315,000 lb. To get the g-force you divide by the
weight of the forklift, mg =5000x9.8 N=49,000 N; so,
g-force=6.4. That is the force experienced by the forklift
and would be the same for anything which moves the same. The
driver would experience that force where the lap belt was
restraining him but his upper body would experience a
smaller force since it would take longer to stop. As
emphasized in the earlier answer, this is a very rough
approximation.

QUESTION:
What is the ratio between the height H of a mountain and depth h of a
mine,if a pendulum swings with the same period at the top of the mountain
and at the bottom of the mine?

ANSWER:
This must be a homework question where you assume that the density of the
earth is uniform.

QUESTION:
No it's not. I am a student who is trying to crack NEET. While I was solving questions, I saw this one but I couldn't get the explained answer.
The answer I got was 1,but the answer given here was 1/2. I just wanted to know was my answer correct.

ANSWER:
There is no way to solve this problem without an assumption
regarding the density of the earth. The standard assumption
in introductory physics is to assume that it is uniform
(which is
not a good approximation); I will do that. Also, I
will assume that it is a simple pendulum with a small angle
amplitude so that the period is T ≈2 π √(L /g )
where L is the length and g=MG /R ^{2} is the
acceleration due to gravity. At a distance H above
the earth's surface g _{H} =MG /(R+H )^{2} ;
at a distance h below the surface of a
uniform-density earth, g _{h} =MG (R-h )/R ^{3}
(see footnote*). Now, for the periods to be equal it is
necessary that g _{h} =g _{H} ;
a little algebra shows that equivalently (1-(h /R ))=1/(1+(H /R ))^{2} =g _{H(h)} /g .

It is instructive to
look at h /R as a function of H /R :
h /R =1-1/(1+(H /R ))^{2
} shown in the first graph above. When h=H =0
you are at the surface of the earth; when h /R =1,
H /R =∞ and g _{h} =g _{H} =0.

Next, look at the
plots of g _{H(h)} /g as a function of H /R
(black) and of h /R (Red). There
are only two locations where g _{h} =g _{H} ,
at the surface where h /R =H /R =0
corresponding to g _{h} =g _{H} =g
and near h /R =H /R =0.6.
The third plot shows a closer look around 0.6 showing h /R =H /R =0.618
corresponding to g _{h} =g _{H} =0.382g .

This was a pretty
long-winded answer, but the upshot is that you were right
and the answer key was wrong: h /H =1.

* M is the mass
of the earth, R is the radius of the earth, and
G is the universal gravitational constant.

ADDED NOTE: Actually, I did not need to
solve this problem graphically, I could have solved it
analytically. If you assume (guess) that h /R=H /R≡x ,
then the equation to determine x is 1-x =1/(1+x )^{2}
or (1-x )(1+x )^{2} -1=0=(1-x ^{2} )(1+x )-1=
x ^{3} +x ^{2} -x=x (x ^{2} +x -1)=0.
One solution is x =0 (the surface) and the positive
solution to the quadratic equation is x = ½( √(5)-1)=0.618.
I guess I shied away from this because I know I am not very
good at solving cubic equations, but I can handle this one!

QUESTION:
If you could spin a person on the moon how many revolutions per minute would it take to release them to escape the moons gravity?

ANSWER:
The moon's escape velocity is about 2400 m/s. The angular
frequency for an object moving in a circle with radius R
and speed v is ω=v/R . Taking
R ≈1 m,
ω= 2400 Hz=23,000 rpm. The velocity required
for a low altitude orbit is about 1700 m/s, so that would
require about 16,000 rpm.

QUESTION:
With Star Wars being in fashion again with the new movies coming out, I thought to ask this.

So Emperor Palpatine is known for his ability to shoot lightning out of his fingertips quite effectively.
And here's a
GIF of him doing it:
So how much energy, current and charge generally speaking would producing that much electricity to penetrate that much air require from a human being, and how does it relate to the amount of energy a human body actually can produce, and finally assuming a human could produce the electric power needed to do that, how would a real life Palpatine shoot lightning out of his hands with real life physics?

ANSWER:
For the air to break down and create a spark, about 10,000
V/cm is required. It appears that these sparks are about 3 m
long, so about 300x10,000=3x10^{6} Volts=3 Megavolts
of potential difference is required. Of course, this is
fiction so I will not go any farther with this answer.

QUESTION:
How the f*** can the velocity be negative? And why isn't the acceleration 0 when the velocity is 0?

ANSWER:
Gee, don't get so upset! Velocity is a vector —to
specify the velocity you must specify both magnitude (called
speed) and direction. To specify the direction, it is
convenient to define one direction, say north, to be
positive and the other, south, to be negative; so a car
having a speed 50 mph on a north-south road has a velocity
of +50 mph if going north, -50 mph is going south. (Speed is
never negative.) Acceleration is the rate at which the
velocity is changing. Suppose you throw a ball vertically;
at the very top the ball is at rest, albeit only for an
instant, but at rest. Now, if at the top the acceleration
were zero, that would mean that its velocity is not changing
and it would therefore just stay there. It is possible for
the acceleration and velocity to both be zero; a book
sitting at rest on the table would be one example. It is
just not necessary that both be zero.

QUESTION:
Suppose the Sun and the Earth were each given an equal amount of charge of the same sign. Just sufficient to cancel their gravitational attraction. How many times the charge on an electron would that charge be? Is this number a large fraction of the number of charges of either sign in the earth?

ANSWER:
So we want kq ^{2} /R ^{2} =Gm _{e} m _{s} /R ^{2}
or q = √(Gm _{e} m _{s} /k )≈3x10^{17}
C. The number of electrons this would be is q /e =1.9x10^{36} .
If we roughly estimate that the earth has is 50% protons and
50% neutrons, then the approximate number of protons would
be (m _{e} /2)/m _{p} =1.8x10^{51} ,
far larger than 1.9x10^{36} . If you want to check my
arithmetic, look up all the constants I have used.

QUESTION:
Theoretically, if I look through a telescope at a mirror 2 light years away, the light bouncing back would give me a instant view of my location from 4 years ago, right?
With that, if I was in a ship that safely accelerated to the speed of light, say 1 g acceleration, so that I would reach the speed of light after approximately 1 year, traveling 1/2 of a light year, then turning around and decelerating at 1 g for one year to come to a complete stop, then doing the same to get back to my original location, thereby traveling 4 years total, why would I change 4 years in age, but my original location (and final destination) be more than 4 years older?

ANSWER:
Yes, the mirror would give you the view four years earlier.
However, your understanding of the accelerating spaceship is
very wrong. It is a pretty difficult thing to understand, so
I suggest that you read an
earlier answer
as well as the earlier answer it links to. When you have
done that, you should be able to follow my answer to your
question.

First a qualitative discussion. You decide that you want
your spaceship to have a constant acceleration of g
as measured by you, captain of the ship. You therefore
arrange to have your engines always exert a force on the
ship of F=mg where m is the mass of the
ship. So you will always see your ship having the desired
acceleration. But, as I have emphasized in many earlier
answers, acceleration you measure is not the same as
acceleration someone on earth measures. In Newtonian
physics, everyone measures the same acceleration. This is
really a glorified
twin paradox problem and I am not going to calculate how
long the round trip will be for you (but I will calculate
how long it takes in earth time). What I can tell you is that you
will take longer than 4 years to return to earth and the
time elapsed on earth will be longer than your time. Your
desire to accelerate "to
the speed of light" is impossible, as you will see, so
forget that; with your scenario the fastest you will ever go
is about 86% of c .

What I want to focus on is what someone on
earth will measure regarding your motion. This is mostly
worked out in the earlier answers I have referred you to, in
particular v /c =(gt /c )√[1+(gt /c )^{2} ]
and a /g =[1+(gt /c )^{2} ]^{-3/2} ;
these are plotted in the graph above with the red and blue
curves. Things to notice are that after 1 year you have only
acquired a speed of about 0.7c and your
acceleration has dropped to about 0.35g . What was
not derived earlier was your position as a function of time;
integrating the velocity equation, I find gx /c ^{2} =√[1+(gt /c )^{2} ]-1,
shown by the black curve. Things to notice are that it
starts out looking like a parbola as it would if you had a
constant acceleration of g , but eventually becomes
a straight line as you approach the speed of light which you
never reach. Reading off the graph, the time it takes you to
get halfway, 1 ly, is about 1.7 years (as measured from
earth); your speed will be about 0.86c and your acceleration
(again as measured from earth) will be about 0.13g . This will be ¼
of your total trip, 6.8 years. The time you measure will be somewhere
between 4 and 6.8 years.

Incidentally, the axes
may be shown to be numerically equal to years and light
years if you approximate g ≈10 m/s^{2} .

ADDED
COMMENT: I see that I have misread the question. For some reason the
questioner reverses his thrust at ½ ly out so he only goes to
1 ly before turning the ship back to earth. From the graph you can see that
the time to get to this position is about 1.1 years and the speed there is
about 0.75c . Thus the total elapsed time on earth before your
return would be 4.4 years in this scenario. Light would take 2 years to make
this trip. Again, I have not calculated the time on your clock but it would
be between 2 and 4.4 years.

QUESTION:
Why the projectile motion of a body in air is not three dimensional?

ANSWER:
In the real world the motion is three dimensional. In an
elementary physics course, the situation of the body having
no interaction with the air is first considered for
simplicity. In this case the only force on the object is its
own weight which is vertical and therefore the object moves
in a plane defined by the initial velocity and the vertical.
This is a good approximation for many everyday applications
of projectile motion. However, the object will interact with
the air. If the air is still and the object is a point mass,
it will experience a force opposite the direction of its
velocity and therefore also move in a plane; the path will
not be a parabola. If there is a wind whose velocity does
not lie in the plane defined by the initial velocity and
vertical, the projectile will move out of the plane —true
three-dimensional motion. If the object experiences a
greater air force on one side or the other, for example for
a spinning ball, the projectile will curve out of the
original plane.

QUESTION:
For a book pushed horizontally against a vertical wall, we know that the friction force is equal to the weight of the book. We also know that the friction force is equal to the normal force multiplied by the coefficient of friction. So, technically speaking, the harder you push against the book, the greater the normal force. Therefore, the friction should get bigger the harder the you push, but we know that isn't true since friction is equal to the weight of the book, which doesn't change when you press the book harder to the wall. Thus the only way to appease the equation: (mu)(normal force) = (weight of book), with an increase in normal force (caused by an increase in applied force to the book), that would mean the (mu) or coefficient of friction has to change. So my question is, why does the coefficient of friction change when we apply more force to the book on the wall? Its still the same two surfaces!

ANSWER:
You should have stopped after the first sentence which is
correct! You misunderstand static friction. Although it is
true that f= μ _{k} N
for kinetic friction, for static friction the equation
f= μ _{s} N
is not true. The correct formula is an inequality rather
than an equality:
f ≤ μ _{s} N.
For one single force you may write an equation, f _{max} =μ _{s} N
which is the largest frictional force you can
get for a given N . If you do not push your book
with a force equal to or larger than N=mg /μ _{s}
it will fall to the floor. (The Q&A right
below yours is closely related to your question.)

QUESTION:
How two objects of the same material with different masses when placed on a sloped surface the force of gravity overcomes the force of friction at the same angle for both objects, this would bring the question why would they have the same angle if one object has more inertia than the other.

ANSWER:
An object on an inclined plane which makes an angle
θ to the horizontal is at rest and in
equilibrium. There are three forces acting on it, the
friction f , the normal force N , and the
weight W=mg . Then, using Newton's first law, f-W sin θ =0
and N-W cosθ =0. Solving,
f=W sin θ and N=W cosθ.
If the block is just about to start moving f=μN
where μ is the coefficient of static
friction, so μ= (W sin θ /W cosθ )=tanθ .
As you can see, the angle is independent of the weight; this
is because both N and f are proportional to
W , so
W cancels out when you calculate their ratio.

QUESTION:
In what distance will a car skid to a stop on a dry concrete road if its brakes are locked when it is moving at 89 km/h?

ANSWER:
The coefficient of kinetic friction for rubber on dry
concrete is μ ≈0.7. The speed
of the car is v=89 km/hr=24.7 m/s. The change in kinetic
energy must equal the work done, (K _{2} -K _{1} )=-μ mgd =(0-½mv ^{2} );
the mass m cancels out, so -½x24.7^{2} =-0.7x9.8xd
and therefore d =44.5 m.

QUESTION:
can length contraction of the crew of a spaceship reach the point where it affects the biological processes? I am not talking about time dilation effects. Is thinking changed? Can one be squeezed down to the point of injury? At very near the speed of light could the density of the spaceship be so great that it becomes a black hole? What about charge density in an electron? Can fundamental constants like Planck's constant be affected?

ANSWER:
You misunderstand length contraction. In the frame of
reference of the spaceship there is no length contraction.
Lengths on the spaceship as measured by someone at rest are
shorter, but everyone on the spaceship sees everything just
the same as if it were at rest.

QUESTION:
We have been talking in science class about air pressure and vacuums,and how a straw even in perfect vacuum will only suck water up to ~10m, my question is why does the diameter of the straw not effect the height to which the water is pulled as a greater diameter would mean more mass.

ANSWER:
The mass m of the water in the tube is the density
of water (ρ =10^{3}
kg/m^{3} ) times the volume of water (V=Ah );
therefore the weight of the water is mg =ρgAh ≈10^{4} Ah
(taking g ≈10 m/s^{2} ). The pressure at
the bottom of the tube is atmospheric pressure, P _{bottom} ≈10^{5}
N/m^{2} and at the top the pressure is zero;
therefore there is a force F up on the tube which
is F =(P _{bottom} -P _{top} )A ≈10^{5} A
N. But F is holding up the weight, so F=mg or 10^{5} A= 10^{4} Ah
or h ≈10 m. A cancels out!

Q&A OF THE WEEK, 4/14/2018

QUESTION:
hi, so, there is a fairly recent
video going around the internet of nasa ISS astronaut Randy Bresnik spinning a fidget spinner, in space, in a quite low gravity environment and then apparently grabbing hold of the spinner then video cuts and we, presumably some short time later, see the entire body of the astronaut holding the spinner spinning fairly rapidly in space.
Perhaps it is a quite elementary question, but, I wonder if that video might have been faked, wondering if the what must be a relatively small amount of energy in the spinner could cause the entire body of mass of astronaut plus the spinner to spin in the fashion observed in the video. That is to say: wouldn't the energy from the spinner, say, be absorbed by the astronauts body or the muscles in his arm, or some such, upon grabbing the spinner; or, if there was movement, wouldn't it be far far less than what is demonstrated in the video?

ANSWER:
You are right, I think the astronauts are messing with you
here! You have looked at it from the perspective of energy
conservation; however, energy would not be conserved here
and energy would actually be lost, not gained as it appears.
So, your reasoning was sound —where did the
energy come from? What is conserved is the angular momentum.
If the moments of inertia are I _{toy} and
I _{man} and the original angular velocity
of the toy is ω _{1} , then the
initial angular momentum is L _{1} =I _{toy} ω _{1}
and the final angular momentum is L _{2} =(I _{toy} +I_{man} ) ω _{2}
where ω _{2} is the angular velocity
of the man+toy. Conserving angular momentum and solving for
ω _{2} , ω _{2} =[I _{toy} /(I _{toy} +I_{man} )]ω _{1} <<ω _{1} .
Now, I actually found the moment of inertia of a fidget
spinner,
I _{toy} ≈7x10^{-5} kg·m^{2}
and I estimate
I _{man} ≈70 kg·m^{2} .
This means that ω _{2} =ω _{1} /1,000,000!
The astronauts have angular velocity of about 1 revolution
per second and you know perfectly well that the fidget
spinner did not have a speed of a million revolutions per
second.

Finally, if you
are interested, you can show that energy is not conserved. The expression
for kinetic energy is E =½Iω ^{2 } so
E _{1} =½I _{toy} ω _{1} ^{2
} and E _{2} =½[I _{toy} /(I _{toy} +I_{man} )]ω _{1} ^{2} ≠E _{1} .

QUESTION:
I have an idea to help enhance mankind's interplanetary manifest destiny.
The idea hinges on pragmatics and so I want to know if it is worth considering further. Without giving away too much of the idea, and without more ado, here is my question.
Is the escape velocity the same at sea level as it is atop a mountain? The only equations I can find show escape surface velocity. As an added bonus question: If my initial velocity is extremely large but I have no propulsion, how would this scenario affect escape velocity?

ANSWER:
First of all, let's be sure we know what escape velocity
from earth is: it is the minimum speed an object must have
at a distance R from the center of the earth to
totally escape (to infinity) the gravitational field of the
object; the expression is v _{escape} =√(2GM /R )
(which assumes that the mass of the escaping object is very
small compared to the mass of the earth, M ).
That is the initial speed with no additional propulsion.
Obviously you could escape any gravity by simply pushing
with a force greater than the force of gravity you
experience; if you had a means of propulsion which kept you
moving at 1 mph, you could completely escape the earth given
enough time. Therefore your "bonus question" essentially has
no meaning. Your first question is whether the escape
velocity atop a mountain is different from sea level; the
answer is yes because R is bigger atop the
mountain, but since the height of even the highest mountain
is tiny compared to the radius of the earth, the difference
is neglegible.

QUESTION:
Why do raindrops fall down at a pronounced angle with the vertical when seen from the window of a speeding train? Is this angle necessarily the same as that of the path of a water drop sliding down the outside surface of the window?

ANSWER:
For the same reason that stationary objects outside the
train appear to be moving backwards with the same speed as
the speed of the train. For example, if the drops are
falling vertically relative to the ground with velocity
v , the negative of the velocity
(horizontal) of the train u must
be added to v ; the velocity
w seen by an observer in the train
will be this vector sum.

This has nothing to do with how drops
adhering to the window move. If the train were still a drop would drip
straight down the window with some speed which would be much smaller than
the speed of a falling rain drop. If the train is moving the drop would see
a wind of speed u which would push it backwards with some speed considerably
less than the speed of the train. Again, the vector sum of the two
velocities would be seen as the drop velocity by an observer in the train.

QUESTION:
Why is it when you sit on a swing and swing really goes smooth but you get to a certain height and it gets bumpey

ANSWER:
It is pretty complicated to discuss in detail. It has to do
with the "pumping" the swinger performs. Look at the little
animation above. Notice that the chains do not stay in a
straight line. If someone were pushing you and you sat
perfectly straight, I believe that the "bumpey (sic )"
motion would not happen until you were swinging above the
bar. When pumping, you eventually reach a point where the
chain between your hands and the seat goes slack which you
can see in the photograph of the swinging girl if you look
carefully (see my yellow line). If the girl is at the very
top of her path, she is at rest and there is no force from
the chain on the swing itself. So when she starts to fall
back she will be in a sort of free fall until the chain
becomes taught; at that instant there will be a jerk.

DISCLAIMER:
I was unable to find any discussion of this anywhere and I
have invented this answer using what I find to be reasonable
arguments. It might not be the right answer!

QUESTION:
So, you have a train that is not moving and inside the train you have a drone that is hovering. No one is controlling the drone it is just hovering in one place. If the train begins to move, would the drone move with the train or would it run into the back wall of the train?

ANSWER:
As viewed by an observer outside the train, for the drone to
accelerate with the train it would have to experience a
force in the direction of that acceleration. But there is no
such force so the drone would stand still; viewed from
inside the train, the drone would appear to accelerate
toward the back wall of the train.

QUESTION:
With the recent discovery of "seeing" gravitational waves emitted by two merging neutron stars, how does one "see" gravitational waves.

ANSWER:
There is a good description of the detectors
here .
What is particularly exciting is that light from the merging
neutron stars could be seen; earlier observations, of
merging black holes could not. Also, the abundance of heavy
elements like gold is much larger than could be understood
using standard models of stellar evolution; it had been
hypothesized some years ago that perhaps the excess could be
produced in neutron star collisions, and the current
observation seems, at first glance of the spectroscopy of
the light, to support this hypothesis.

QUESTION:
What would a chart of an 8g impact look like for an 8g impact with a 100 ms
duration?
I can send a jpg of what I am told fits the requirement.
But I seriously doubt that it does fit the requirement. But then again I am
barely understanding the test. I see spikes to 20-30g.
I am having a hard time finding anyone that can answer this question.
(I asked for more information.)

FOLLOWUP QUESTION:
These are the charts provide by the AF to me when they tested a pallet that we built.

The requirement is:
Ultimate load. When uniformly loaded to 10,000 pounds, the load being restrained to the pallet by chains, the pallet installed between restraining rails locked to the rails by 2 locks through each rail and engaging 2 lock notches on each side of the pallet, and resting on 4 rows of conveyor as specified (see 3.4.5.1), the loaded pallet shall withstand a dynamic load of 3 times the force of gravity (g's) for a period of 0.1 second. The pallet shall be serviceable after undergoing the test. In addition, the pallet shall withstand a dynamic load of 8 g's for a period not less than 0.1 second. The pallet need not be serviceable after undergoing such a load; however, the pallet shall remain in one piece.

These are the result of those tests.
These charts just do not look right to me in their conclusions.
I am not a math major, but I do understand some stuff and this does not look correct. ie. "Area under the curve." or correct averaging.
If you can set me straight or put me on another path it would be greatly appreciated.

ANSWER:
So, I must admit that I do not understand all the details of
the requirements (locks, chains, notches, etc .)
What I can do, though, is estimate the area under the curve
which seems to be one of your main concerns. I drew in a
rough fit (green) to the data (red) and calculated the area
under the green curves (keeping in mind that the area of the
triangular segment below the axis is negative). As you can
see, the area I got as a rough estimate is 0.37 compared to
the actual area of 0.35. The average would simply be the
area/time=3.5. My guess is that your pallets did not fulfil
the requirements since the precipitous drop before the 0.1 s
had passed would seem to indicate a collapse of the pallet.
Both tests show this behavior, with the 8g test having the
"collapse" occur earlier as would be expected. However,
since I do not understand the details of the test well
enough, I would not take my guess as gospel.

QUESTION:
How would i explain bus or a truck(high CG)flip over, when driving in a curve with high speed, from a inertial reference frame? If centrifugal force is fictitious and bound to non-inertial reference frame, what torque causes truck to flip over observed from an inertial ref. frame?

ANSWER:
I have attached figure from the
earlier answer where I did essentially this problem in
the noninertial frame; this saves me having to draw the
whole picture again. For your problem, the vector labelled
C is zero. Part of what makes your
problem harder than in the accelerating frame is that you
must sum torques about the center of mass (green x). For
equilibrium, the sum of torques must be zero; summing
torques gives 0=H (f _{1} +f _{2} )+LN _{1} -LN _{2} .
(Translational equations are f _{1} +f _{2} =mv ^{2} /R
and N _{1} +N _{1} -W=0.)
Now, when the truck is about to tip over the left wheel is
about to leave the ground so N _{1} =f _{1} =0,
f _{2} =mv ^{2} /R ,
and N _{1} =W ; therefore Hmv ^{2} /R-LW= 0.
So, if v > √[LWR /(Hm )]
the truck will flip over.

QUESTION:
If you jumped vertically upwards, strictly speaking and according to Newton's 3rd Law, what would the Earth do?

ANSWER:
The earth would recoil but with a velocity much less than
you do. Suppose that you have a mass m and initial
speed v and the earth has a mass M and
initial speed V ; then conservation of linear
momentum (essentially Newton's 3^{rd } law) says
mv+MV =0 or V =-(m /M )v ;
the recoil velocity of the earth is unmeasurably smaller
than your velocity.

QUESTION:
Suppose I am in space, in a 100 meter barrel strapped to one end. The barrel and I have mass 100 kg (I'm pretty light). I throw a 100 kg ball down the center of the barrel towards the other end, accelerating it to 1m/s velocity. Presumably, it takes the ball 100 seconds to reach the other end and it sticks because it is covered in velcro. I presume the barrel started in motion in space when I threw the ball from the reaction from the force I applied to the ball and it stopped when the ball hit the other wall. Could I keep moving the barrel by throwing more balls? How is this possible without violating Newtons laws? Surely this is not a case of propellantless thrust?

ANSWER:
These 100 kg balls just materialize from nothing, do they?
So you must have a supply of them which means that the total
mass of the entire barrel and its contents is much bigger
than 100 kg which means you go much less far with each
throw. Eventually you will run out of balls, but you will
have gone some distance. What propelled the barrel? You did.

FOLLOWUP QUESTION:
Thank you so much for you previous answer. My question about throwing balls at the back of a barrel in space was inspired by what I had read about some experts criticizing the "EM Drive" for violating Newton's Laws. Basically the writers quote experts as saying that you can't have forward thrust without throwing mass out the back " no such thing as propellantless thrust". I suppose these experts are simplifying their response for the lay press because qualifying their answer would take up to much column space. In retrospect, I should have known that some net movement was possible as I believe NASA uses large gyroscopes to adjust the pointing of some space telescopes.
This did get me to thinking though. Suppose I attach a laser to one end of the barrel and fire a single photon towards the other end. Presumably, momentum is imparted to the wall attached to the laser and the barrel moves through space as the photon moves to the opposite wall where it is absorbed. Now I wonder why I can't keep firing photons and moving the barrel. Is the hitch in my "propellantless thrust" scheme come from the absorption process? Is the heat generated through absorption dissipated out of the barrel preferentially so as to counteract my momentum drive?

ANSWER:
The
situation is similar to your first question since, as you
apparently know, photons have energy and momentum. But, to
operate your laser you have to supply energy somehow.
Suppose that you use a fusion reactor to generate that
energy. Then as you shoot more and more photons, the whole
ship gets less and less massive. But, energy has to be
conserved (mass is a form of energy, mc ^{2} )
and when the photons get absorbed, the heat generated will
"retrieve" the lost mass. Eventually you will run out of
fuel for your reactor and you will be right back to being at
rest. (I am assuming that the rules of the game forbid
letting the heat at the back radiate into space.)

QUESTION:
Can you calculate the negative g when a 38sq metr vehicle hits sea level atmosphere at 5,000kph?

ANSWER:
I presume that you want to know the acceleration of the
vehicle due to the air drag. There is an approximate
expression (only valid for SI units) for the drag for an
area A and speed v , F= - ¼Av ^{2} .
The problem is that I do not really know how accurate this
is at such a high speed as 5000 km/hr≈1400 m/s. So,
view this as a very rough calculation. F ≈-2x10^{7}
N. Now, to get the acceleration you need to know the mass
m (in kg). Using Newton's second law, F=ma ,
a ≈2x10^{7} /m m/s^{2} .
To express this in g s, divide by 9.8 m/s^{2} .
For example, if the mass of the vehicle is 10,000 kg, a ≈2000
m/s^{2} ≈200 g s.

FOLLOWUP QUESTION:
What would be the approximate negative g on a rocket having a 7sqr metre frontal area when exiting from an evacuated tube doing 5,000kph at sea level?
This is a further question related to impact of air on a vehicle.

ANSWER:
The only thing you changed is the area A . Since the
drag force is proportional to A, all estimates given above
need only be multiplied by 7/38=0.18.

QUESTION:
My question is using the flight time equation of (1/2)x g(flight time/2)^2 or the more simplified (g x (flight time)^2)/8 i have the answer and know it is m but i was just wondering what the origins of the equation is, such as what is it called or what is the rule.
Or if it is derived and what from. this is just for myself to have a deeper understanding of what i have done and worked out. for me the the equations was (1/2) x 9.81(0.0557/2)^2=0.38m or 38cm. i know how to use the equation just want to know where it comes from.

ANSWER:
The equations of motion for the vertical component of
projectile motion are v=v _{0} -gt
and y=y _{0} +v _{0} t - ½gt ^{2} ;
note that this is only for the vertical component of the
motion, the horizontal motion does not interest us for your
question. Here v is (the vertical component of) the
speed at time t , v _{0} is (the
vertical component of) the speed at time t =0, y
is the altitude at time t , and y _{0}
is the altitude at time t =0 (which I will choose to
be 0). You are interested in the height to which the
projectile rises, and at that position v =0;
therefore v _{0} =gt and so y=h =(gt )t- ½gt ^{2} =½gt ^{2} .
But my time here is only half the "flight time" of the
projectile, so h =½g (t _{flight} /2)^{2} .
You do not need to memorize "the flight time equation".
Everything you need to know is contained in the equations of
motion which you should memorize.

QUESTION:
A Newton's Cradle is some times used to demonstrate the Law of Conservation of Momentum but, I noticed anomaly. Specifically, whenever the steel balls collide, I hear a clicking sound and that requires energy. This tells me some of the kinetic energy of a moving ball transforms into thermal and accoustic. How do physicists explain that momentum is conserved when a ball leaving a collision is moving slightly slower? If there is less kinetic energy but no change in the total energy, it seems logical to think a certain quantity of motion has vanished after any collision. Both momentum and kinetic energy use the same variables. I also understand that when I pull one ball back, after a collision one ball emerges. If two balls are pulled back and released, two balls pop out.

VIDEO

ANSWER:
You are right, energy is not conserved in the collisions.
And, you can tell that it is not conserved because the
machine eventually stops. But momentum is always conserved
in an isolated system (no external forces on the colliding
bodies). I commend you for thinking about this which seems
paradoxical. To illustrate, let's choose the simplest
situation, the collision of two balls of equal mass m ,
one at rest the other with speed v , in one
dimension. As you know, if the collision is perfectly
elastic the first ball ends up at rest, the second moving
with speed v , so the momentum is conserved because it is
mv before and after the collision. Next you say that if
the second ball emerges with speed u=v-δ that
momentum has been lost. But you have assumed that the first
ball is at rest, but it cannot be if momentum is conserved:
mv=mu'+m (v-δ ) and so the first ball
has a speed u'= δ after the collision.
In the
video I have inserted above, look closely at the first
demo, with one ball striking four and one popping out; you
will clearly see that the next to last ball is not at rest
after the collision(s).

QUESTION:
If I take two solenoids with dimensions of 6 inches in length, and 2 inches in radius, each producing .5 Tesla and I combined the opposite poles so they attract each other, will the combined solenoids be a 1 Tesla solenoid?

QUESTION:
If I take two solenoids with dimensions of 6 inches in length, and 2 inches in radius, each producing .5 Tesla and I combined the opposite poles so they attract each other, will the combined solenoids be a 1 Tesla solenoid?

ANSWER:
No. The argument goes like this: Suppose you have an
infinitely long solenoid; it has a perfectly uniform field
0.5 T aligned with the axis of the solenoid throughout its
whole volume. Now cut out two 6" sections of that solenoid;
each will still have a nearly uniform field 0.5 T near its
center but the field will become weaker as you approach the
ends. Now, put those two together. It is just the same as if
you had cut out a single 12" section to start with which
would have an approximately uniform field of 0.5 T except
near its ends.

Here
is a history of questions and answers processed by "Ask the Physicist!".
If you like my answer, please consider making a
donation to help support this service.

If there is a link to a previously
answered question, be patient. Since the files containing the older
answers are rather large, it takes some time (maybe as much as 15
seconds or so) to find the appropriate bookmark.

QUESTION:
As I understand it, a helium balloon can float upward because of the difference in pressure between the helium in the balloon and the atmosphere. Suppose the balloon had a strong hard surface and the inside of the balloon was a partial vacuum, Would the balloon float upward or at least weigh less? In this case there would also be a difference in pressures like helium.

ANSWER:
Your understanding is incorrect. The reason anything floats
up in the air is that the atmospheric pressure up on the
bottom of the balloon is greater than the atmospheric
pressure down on the top resulting in a net upward force
F . If F>W where W is the weight of
the balloon plus its contents, there is a net force up
causing the balloon to accelerate upwards; this force is
called the buoyant force and also explains why things float
in water. The reason you use helium is that it is lighter
than air. Then the answer to vacuum question is clear:
taking away the helium but keeping the balloon the same size
would make it lighter and therefore the net upward force
would be larger.

QUESTION:
What would be the difference between a bowling ball size meteorite hitting the ground at an average meteor speed compared to a meteor that has been propelled past lightspeed? I'm just wondering how much more damage that would do and if we would even see it coming. Sorry just a random thought.

ANSWER:
First, it is impossible for anything to be "propelled
past lightspeed"; no material object can go faster than (or
as fast as) the speed of light. I will take the meteor mass
to be 300 kg and its "average speed" to be 5x10^{4}
m/s. Then the energy would be E = ½mv ^{2} =4x10^{11}
J. For the super fast meteor, let the speed be 99% the speed
of light. Then the energy would be E=mc ^{2} / √(1-.99^{2} )=2x10^{20}
J. For comparison, the energy of the Nagasaki atomic bomb
was about 10^{14} J.

QUESTION:
If I threw a ball straight up into the air and it stayed there for 6hrs would it drop back to excactly the same spot?

ANSWER:
Any object thrown vertically will not fall back exactly to
where it started. To understand why, see a
recent answer . Your question
has a slightly different twist: when this type of problem is
calculated it is usually assumed that g , the
acceleration due to gravity, is constant; if your ball stays
in the air for six hours, that will not be the case.
Nevertheless, the ball would always have a Coriolis force on
it so it would be deflected westward by some amount. If you
had said something like the ball goes up 1000 m, you could
have done a pretty good calculation since 1000 m is very
small compared to the radius of the earth and it would be a
reasonable approximation to say that g =9.8 m/s^{2}
the whole time (which would be about 28 s). It can be shown
that, for a latitude of λ , on the way up the
Coriolis deflection is [(4ω /3) √(8h ^{3} /g )]cosλ
west and on the way down it is
[(ω /3) √(8h ^{3} /g )]cosλ
east, so the net deflection is
x=ω √(8h ^{3} /g )cosλ
west; the angular velocity of the earth is
ω= 7.27x10^{-5} s^{-1} , so
x =2.1 m west at the equator (λ= 0).

Finally we should note that at the poles (cosλ =0)
there is no deflection due to the earth's rotation, but since
the earth revolves around the sun you are still in a frame where
there would be a Coriolis force but it would be much smaller; it
would be smaller yet because the angular velocity of the motion
around the sun is almost parallel (about 23^{0 } away) to
the vertical at the poles. The deflection for the example above
(h =1000 m)would be about 2 mm compared to 2 m; the
direction would be opposite the direction the earth is moving in
its orbit.

All of the above ignores
air drag.

QUESTION:
Is the graviitational force on Earth the same everywhere or is the force different near the south or North Pole?

ANSWER:
If the earth were a perfect sphere with a spherically
symmetric mass distribution, the gravitational force would
be everywhere the same. Since it is not, there are small
variations as you move around the surface. Even if it were,
it would appear to vary as you moved from the poles
to the equator because of the earth's rotation. A scale
would read your weight (gravitational force on you)
correctly at the poles but would, because of the centrifugal
force (a "fictitious" force), read slightly less at the
equator.

QUESTION:
I can get a neon lamp to ignite (light up) at less than ignition voltage two ways. One is at a voltage a little lower than ignition voltage to shine a UV lamp on the neon bulb which causes it
to light. This can be explained by the photoelectric effect.
The other way, in the same situation as above, is to generate a spark near the neon lamp. One way is to use a piezo electric barbecue igniter which generates about 15KV. This produces the same effect as the UV light - the neon lamp comes on at a lower voltage than its ignition voltage. There is also a YouTube video of a fluorescent that will no longer come on starting when a spark is generated near it. I know a spark produces RF radiation but it is said to not have enough energy to produce a photo electric effect. I have be unable to find any explanation of what is happening here.

ANSWER:
Well, you can be sure that the spark does not generate only
RF because you can see it! Seems to me that if visible light
is being produced by the spark, there is probably also UV
radiation being produced.

QUESTION:
If time is relative does that mean space would be too/ or to put it another way is space actually fixed or can it be and has it been streatched or bent like time in the famous high altitude clock experiments??

ANSWER:
Yes, space is also relative to frame of reference. The most
obvious example is length contraction where lengths along
the direction of relative velocity are shortened by the
factor √(1-(v /c )^{2} ) where v is the
velocity and c is the speed of light. The most
elegant way to think of this is to not think of space and
time as being two things, rather think of space-time.

QUESTION:
Why does a satellite shot straight up from the equator deflect to the west?

ANSWER:
There are two ways to look at this.

If you view it from outside the earth,
the satellite will continue moving directly away from the
center of the earth. But, viewing the earth from above the
north pole, the earth rotates counterclockwise, west to e ast.
So the earth is actually rotating under it which gives it
the appearance of deflecting to the west.

The second is a little fancier and harder to understand. If
you view Newton's laws from a rotating coordinate system
(the rotating earth with you on it) you find that they are
wrong; these "laws" of physics only work in inertial frames
of reference. However you can force Newton's laws to be
correct if you invent just the right fictitious forces. I
will not go into all the complexity here, but one of the
fictitious forces is called the Coriolis force and can be
written as F =-2m ω xv
. Since the vector ω
points along the earth's axis and out through
the north pole and the velocity vector v
points radially outward, the negative of their
cross product points west. Hence, it will appear that the
satellite is pushed west when viewed from the ground.

QUESTION:
I am a safety coordinator in a warehouse operation. I am attempting to impress the need for safety with my co-workers. To that end, I am trying to answer the following:
A 10,636 lb. forklift (4825 kg) traveling at a speed of 8 mph (13 kph) strikes a fixed object (metal pole). How much force is transferred during this collision?
.
Secondly, same forklift, same traveling speed strikes a 150 lbs. (68 kgs) person. How much force is transferred to the person in this collision?

ANSWER:
There is no accurate way to calculate the forces. They depend on the details of the collisions.
In particular, how long do the actual collisions last? Also, how do the forklift and the person move after the collision has occurred (assuming the metal pole stays stationary)?
I could make some rough estimates to get an idea:

For the first problem, let's say that the forklift stops
dead and that it plus the pole deform by 1 cm so that the
distance traveled during the collision time is d ≈1
cm. Since these are very rough calculations, I will let the
mass of the forklift be m _{f} ≈5000 kg
and the initial speed be v =8 mph≈3.6 m/s. If
the forklift stops uniformly, it will stop in a time t =2d /v =0.0056
s. The force F experienced (by both the pole and
the forklift) will be F=m _{f} v /t ≈3.2x10^{6}
N=720,000 lb.

For the man-forklift collision, assume the forklift keeps
right on moving with the same speed but with the man stuck
to the front. Suppose that the man's body compresses 3 cm
during the collision, so the time of collision will be
t=0.06/3.6=0.017 s. The force now is F=m _{m} v /t =68x3.6/0.017=1.44x10^{4}
N=3200 lb.

QUESTION:

do the proton and the neutron have exactly the same mass?

how do masses of the proton and neutron compared to the mass of the electron?
w

which particles make the greatest contribution to the mass of an atom?

which particles make the greatest contribution to the chemical properties of an atom?

ANSWER:
I usually do not answer multiple questions, but these are pretty
much one question about properties of atoms.

The neutron is about 0.14% heavier than the proton.

The electron is about 1837 times lighter than a proton.

All stable atoms but ^{1} H have neutrons as having
to largest fraction of the weight. All stable nuclei except
hydrogen have more or an equal number of neutrons than
protons.

The electrons, which orbit the nucleus comprised of protons
and neutrons, determine the chemical properties of the atom

QUESTION:
CAN A STRUCK GOLF BALL SLICE, IF HIT IN A VACCUM, LIKE THE MOON ?

ANSWER:
Because a slice is curving due to a spinning ball moving
through air, it will not occur in a vacuum. For a more
detailed discussion of golf on the moon, see an
earlier answer .

QUESTION:
After searching through so many questions on here relating to the speed of light (each one being fobbed off by saying 'it just can't get faster') my question is why ?
If a craft was able to reach light speed and had a torch pointing forwards, the light emitted from that torch will be travelling at twice light speed (craft speed (of the source of light) and the actual light projected from the source) ? Don't say it just can't.

ANSWER:
Well, all I can say is that you must not have done a very
good search of my site because both your questions have been
answered multiple times. The first thing you should always
do in such a situation is go to the site's
FAQ page if there
is one—there is one on AskThePhysicist.com! For the
question regarding why the speed of light is the highest
possible, see
this
link . Your question/statement that the speed of light
would be the speed of light plus the speed of the source is
simply wrong. To understand why, see
this
link .

QUESTION:
In your book From Newton to Einstein , page 2-15, you show the resulting solution to variable gravity effect on increasing mass.
Is there a web site where the step by step solution (rather than the end solution) can be found?

ANSWER:
In the
original Q&A there is a
link to a more detailed derivation although I do not
think it will make this simpler for you. Rewrite the first
equation as

(g /c )dt =√(1-β ^{2} )·d[β/ √(1-β ^{2} )]=dβ /(1-β ^{2} ).

There was a little tricky differential calculus here to get
the the last step:

d[β/ √(1-β ^{2} )]=(1-β ^{2} )^{-1/2} d β +β (-½)(-2β )(1-β ^{2} )^{-3/2} dβ= dβ /(1-β ^{2} )^{3/2} .

If, like me, your integration skills are rusty, I suggest
the Mathematica on-line integral
calculator which gets you to the solution for gt /c ;
don't forget that the difference of logs is the log of the
ratio. Then, to get the final answer, exponentiate 2gt /c
to get rid of the natural log and algebraically solve for β .
Of course, the starting point is recalling m=m _{0} / √(1-β ^{2} ).

QUESTION:
I am interested in the temporal phase relationship of electric field
excitation and magnetic field excitation for a circularly polarized electromagnetic wave.
I was taught that linearly polarized waves had temporal fields in sync, while circularly polarized waves had an orthogonally delayed temporal phase difference.
An Electrical Engineer with 50 odd years in antenna design tells me that this is a common misunderstanding of circularly polarized waves - it is his opinion is that ultimately ALL EMWs have temporally synchronised pahase relationships, even circularly polarized EMW!

ANSWER:
I believe your friend is correct. An electromagnetic wave,
regardless of its polarization has a magnetic field which at
every point and time is proportional to the electric field.
Therefore they are in phase. Quoting the article in
Wikepedia on circular polarization, "Since this is an electromagnetic wave each electric field vector has a corresponding, but not illustrated, magnetic field vector that is at a right angle to the electric field vector and proportional in magnitude to it. As a result, the magnetic field vectors would trace out a second helix if displayed."
The electric field, which has a constant magnitude, is
illustrated in the animation; the helix traced out by the
magnetic field would be π /2 out of phase
with the electric field which may be the source of
confusion.

QUESTION:
Another question relating to gravity is what does 9.8 meters per second squared actually mean or at least the "squared" part of it? When I square 9.8 I get 96.04 then when I square the resulting answer the number escalates to over 9,223 and falling objects clearly don't reach 9,223 m/s in just 2 seconds!

ANSWER:
Oh my, this is so fundamental to understanding physics. One
of the first things you learn in an introductory physics
class is kinematics, the specification of position, rate of
change of position (velocity), and rate of change of
velocity (acceleration). Position is specified by a length,
so many meters (m) from somewhere. If the position is
changing at some constant rate, called velocity, is the
change in position divided by the elapsed time (m/s). If the
velocity is changing at some constant rate, called uniform
acceleration, is the change in velocity divided by the
elapsed time ((m/s)/s). So, if a car goes from 0 to 60 in 6
seconds, its acceleration is (60 mi/hr)/6 s)=10 mi/hr/s. If
you drop a stone and measure its speed after it has fallen
one second, you will find that the velocity has speed up to
9.8 m/s; therefore its acceleration is (9.8 m/s)/(1 s)=9.8
m/s/s=9.8 m/s^{2} . Students usually find the s^{2}
as strange —what in the world is a square
second? So, m/s^{2} is just shorthand for m/s/s.
That makes sense, right: 1/4/4=1/4^{2} =1/16.

QUESTION:
How does lightning conductor works?

ANSWER:
See an
earlier answer .

QUESTION:
If a person ran around in a circle at the speed of light, would he cause something like a cyclone?

ANSWER:
No material object can travel the speed of light or faster.
But let's imagine someone who could run a million miles per
hour (way smaller than the speed of light). In order to
cause "something like a cyclone", she would have to be
running in air. But way before she even got to top speed she
would burn up due to the air friction.

QUESTION:
what is the terminal velocity of a soccer ball sized piece of hail?

ANSWER:
I don't think there is hail that large! But, the calculation
is straightforward. I usually estimate the air drag of
something of cross section A having a speed v
to be ¼Av ^{2} =0.038v ^{2} .
When this is equal to the weight of the object, mg =(4πR ^{3/} 3) ρg =402
N, the object will have the terminal velocity; so v =√(402/0.038)=103
m/s=230 mph. I have used R =0.22 m, ρ= 920
kg/m^{3} , and g =9.8 m/s^{2} .

QUESTION:
If you were traveling at a speed of 60mph, what would happen if you ran into a 180mph wall of wind.

ANSWER:
Assuming that you did not change the accelerator setting,
the speed of the air hitting your car would change from 60
mph to 240 mph. Since the drag force is proportional to the
square of the speed, the drag force on the car would become
(240/60)^{2} =16 times larger. This would be like
brakes and probably eventually stop the car. Depending on
the details of the car, it would possibly eventually reverse
the direction of the car.

QUESTION:
I need to know when a pingpong ball is under 10' of water and when ball is released how much work is the ball capable?

ANSWER:
This is a peculiar question. To find out the work you can
get out of the ball you need to calculate the work you have
to do to get it there. Imagine pushing the ball down very
slowly so that you can neglect the drag forces; also neglect
the work you need to do to initially submerge it since that
will certainly be neglegible compared to the work the rest
of the way down. The data for the ball are a mass of m =2.7x10^{-3}
kg and a radius of R =0.02 m; the volume is V =4πR ^{3/} 3=3.35x10^{-5}
m^{3} . The forces are its weight down mg =9.8x2.7x10^{-3} =0.0265 N
and the buoyant force B up which is the weight of
the displaced water (the density of water is 1000 kg/m^{3} ),
B=ρgV =1000x9.8x3.35x10^{-5} =0.328
N; the net force on the submerged ball is therefore F =0.328-0.0265=0.301
N up. So, since s =10 ft=3.05 m, the work done to move
place it down there is W =Fs =0.301x3.05=0.918
J=2.55x10^{-7} kW ·hr=0.677 ft·lb.
Keep in mind that if simply released the ball will quickly
reach terminal velocity and move up with a constant speed,
most of the work being against the frictional drag force.

QUESTION:
A mathematician while he was a kid was punished by teacher to add 1 to 100. But he gave the answer in less than 10 minutes? How is it possible with out modern devices in 1700 NC

ANSWER:
It is actually pretty simple: 1+99=100, 2+98=100, …
49+51=100, so that is 4900. But we still have 50 and 100, so
the answer is 5050. (This is not physics, though!)

QUESTION:
Recently in my university, we had to do an experiment on beta spectroscopy. They gave us a sample of radioactive strontium-90 and told us that it decays beta particles at energies of 0.546MeV and 2.274 MeV and below. After finishing the experiment and submitting the report, the lab instructors told me that i had understood the concept wrongly. I had thought that at the energies of 0.546MeV and 2.274 MeV i would find a high number of electrons emitted. They told me that this was not the case and that at these energies you would expect to find no electrons emitted. They told me that electrons were not emitted at this energies because the neutons would absorb all the energy and leave the electrons in the sample thus no electrons would be detected. However, my understanding is that at these supposed max energies shouldn't the eletrons be emitted and leave the atom behind, allowing us to detect it? Why does the electron not leave the atom when it has high energies? Also during the experiment if there is supposed to be no electrons detected why did i detect a lot of electrons at these energies that is orders of magnitudes higher than that of the background radiation?

ANSWER:
Let's first talk about what happens in β ^{-}
decay: a nucleus with too many neutrons to be stable changes
a neutron to a proton (which stays in the nucleus), an
electron, and a neutrino. The energy released by this decay
is therefore shared between the electron and the neutrino
(assuming that the nucleus, being enormously heavier than
the other two particles, will have almost no recoil energy.
The electron, as you demonstrate in your lab assignment, is
pretty easy to observe. The neutrino, however, is extremely
difficult
to detect—the sun produces copious amounts of them and
nearly all pass completely through the earth without
interacting at all. When β ^{-} decay
was first discovered, it was a huge mystery because the
electrons, the only thing observed, were not emitted with a
single energy but with a broad spectrum of energies. A
typical electron spectrum is shown in the graph above;
in that graph you see a maximum near 0.1 MeV electron energy
but, in this example, the total energy should be about 1.15
MeV. There seemed to be only two explanations: either energy
is not conserved or else there is a third particle which was
not observed and carries off some of the energy. Everyone
believed the second explanation and the existence of
neutrinos was totally accepted by all physicists. They were
so hard to observe that it was 26 years after Wolfgang Pauli
proposed them (1930) and they were observed by Cowan and
Reines (1956).

In your experiment, there
were two decays going on simultaneously: ^{90} Sr
decaying to ^{90} Y and ^{90} Y decaying to ^{
90} Zr. Both have a maximum but they are far smaller than
the total energies of the decays. Nearly nothing after the first
sentence starting with "They told me…" makes little or no
sense but that is probably because you did not understand what
they were trying to tell you.

QUESTION:
The centre of mass of a uniform ring lies at its geometrical centre i.e. outside the body but as we generally define centre of mass that if we apply a force at that unique point the whole body will move in a same way of whole of the mass is assumed to be concentrated there. So my question is how can we apply a force to a point not on body and still see the force's effect??

ANSWER:
If the force can be represented as a field, put the ring in
a uniform field. For example, near the surface of the earth
there is a nearly uniform gravitational force. If you drop
the ring it will not rotate regardless of how you orient it
at the time you drop it. The effect is exactly the same as
if all the weight were acting at the center. Another way
would be to put a very thin membrane across the ring whose
mass was negligibly small compared to the mass of the ring;
then apply a force at the center of the ring.

QUESTION:
We know that electric field lines do not intersect because they shows two direction but it is a vector quantity so we can find the resultant of the two direction but this not happen.
Why?

ANSWER:
You may observe the electric field at any point in space and
map the field. You will never find that that field crosses
itself. Of course, if you have more than one source of field
and you look at each separately, those fields will cross
each other just about everywhere. However, the net field is
the vector sum of all the fields everywhere. The electric
field obeys superposition.

QUESTION:
We know that if a body rotates about an axis that is not the axis of symmetry then the angular momentum about a point on the axis does not point along the axis. If we rotate such body about such axis at a constant angular speed then a contradiction arises: since angular acceleration is zero, the torque about that axis has to be zero but since the angular mometum vector is changing (since the axis is not the axis of symmetry) a net torque is required. Where am I getting wrong? Please help.

ANSWER:
Your question is a little ambiguous since you do not clearly
specify which axis you are talking about after the first
sentence of your question; if you just say axis, I do not
know if you mean the symmetry axis or the rotation axis. In
the laboratory frame, though, the angular momentum points
along the rotation axis and does not change. As you state,
there are no torques on the system about the rotation axis
so the angular momentum is not expected to change. But, you
are probably looking at the problem from the body frame and
see that the angular momentum is changing in that frame; but
the body frame is not an inertial frame and therefore
Newton's laws are not applicable (in particular, torque is
not necessarily equal to the rate of change of angular
momentum). That is probably what you are getting wrong.

QUESTION:
If I take my foot off the gas pedal on a flat road will my car eventually stop or do I have to hit the breaks? I tried to test it myself but I felt unsafe towards the end each time.

ANSWER:
That
depends on the car. Some cars, when in gear, will move
forward when idling unless brakes are applied. Certainly a
car with manual transmission will not stop unless it is in
neutral but will probably eventually stall if in a high gear
at sufficiently low speeds; if it stalls, it will certainly
stop soon after.

QUESTION:
can a body have a charge of 1.8×10^-19 C?

ANSWER:
The electric charge on an object is due to the excess or
deficiency of electrons. The charge of an electron is 1.6x10^{-19}
C, so the net charge must be an integral multiple of this.
So the answer to your question is no.

QUESTION:
What is the velocity of the electron when it orbits around nucleus?
I mean the distance taken by electron on the perimeter of the orbit divided by time not its angular speed

ANSWER:
See
this video for a derivation. For a hydrogen atom in its
ground state, v =2.16x10^{6} m/s.

QUESTION:
How many Suns would it take, laid side by side, to reach a star 9.3 light-years away?

ANSWER:
The diameter of the sun is about 865,000 miles which is
about 1.47x10^{-7} ly. Therefore 9.3 ly is about
9.3/1.47x10^{-7} =6.33x10^{7} sun diameters
(63.3 million).

QUESTION:
If the space inside an atom is totally empty, therefore a vacuum, why doesn't the atom collapse?

ANSWER:
If
the space inside the solar system is totally empty,
therefore a vacuum, why doesn't the solar system collapse?
Because there is a force, gravity, pointing from the object
to the center of the sun which keeps all the moving objects
in the solar system moving in elliptical orbits. The same is
true of the Bohr model of the atom except the force is the
electrostatic force rather than gravity. You should be
aware, however, that the inside of an atom is not a vacuum.
The electrons are not really in simple orbits as the Bohr
model would have it, but they have wave functions
(distributions of the probabilities of their being at any
place in space) which extend all the way into the nucleus of
the atom.

QUESTION:
I've been thinking about this one, and cannot find a good way to reason it out. if you are in an elevator that is going down and you jump in there while its going down would your head hit the elevator's ceiling??

ANSWER:
If the elevator is moving with constant speed, it is exactly
as if it were standing still —if you give
yourself an adequately large velocity you will hit the
ceiling. If d is the distance from your head to the
ceiling at the instant that your feet leave the floor, then
you will hit the ceiling if the speed v you launch
with is greater than √(2gd ) where g
is the acceleration due to gravity. If the elevator has an
acceleration a , the critical speed will be √(2(g-a )d ).
For example, if a=g you are in free fall and even
the tiniest velocity will cause you to hit the ceiling
(assuming that the elevator does not get to the ground
before you get to the ceiling).

QUESTION:
I understand that the motion of a light source doesn’t add to or subtract from the speed, or change the direction, of the light that it emits. So, if I was a passenger on a spaceship, and I shone a flashlight from one side of the ship to the other, perpendicular to the direction of travel, the light beam wouldn’t strike the wall exactly where the flashlight was pointed, but rather a little towards the rear, as the opposite wall would have moved forward some small distance in the time it took the photons to cross the ship. And if the ship was traveling at, say, half the speed of light, the light beam would leave the flashlight at a noticeable angle. Is this correct?

ANSWER:
If the spaceship were moving with constant velocity, the
light beam would go straight across the ship. As viewed from
outside, the light would have a component of its velocity
along the direction of motion equal to the speed of the
ship, so it would still hit exactly opposite of your
position which also moves with speed v). The speed of the
light would still be c. If the width of the ship is W, you
would see the light take a time W /c but
the outside observer would see it take a time (W /c )/ √[1-(v /c )^{2} ].
If the spaceship were accelerating you would see the light
hit somewhere backward (forward) of where you aimed it
depending if you were speeding up (slowing down).

FOLLOWUP QUESTION:
Your answer confirmed what I thought was the case for three-dimensional objects propelled mechanically through space. For example, an arrow shot from a bow across the spaceship would hit the bullseye or miss, according to whether the ship's velocity was constant or accelerating/decelerating.
However, I thought electromagnetic radiation, such as light, was a different animal. What led to my original question was reading that the speed of light is a constant which can't be exceeded, so I assumed it couldn't take on additional velocity from the motion of its frame of reference. When you say "the light would have a component of its velocity along the direction of motion equal to the speed of the ship" it sounds like you're saying the light is leaving the flashlight at speed c, and then the ship's speed of, say, 10 km/sec is being added to make the effective speed c + 10 km/sec (viewed from outside) which I thought was impossible.
My mental image of light radiating is of energy being "conducted" (analogous to electricity conducted by a copper wire) at a constant speed through an intrinsically motionless medium (space), not of a physical object being propelled.

ANSWER:
First a minor point. The arrow, speed u across the
ship which moves with speed v , will have a speed u'= √(u ^{2} +v ^{2} ),
not u+v as you suggest. However, as you correctly
surmise, this cannot be true for light since, in your
example, you could not have the light with speed √(c ^{2} +10^{2} )
as seen by an outside observer. The speed must still be
c . So, if the component along the direction of the ship
is c_{x} =v , you must have c_{y} ^{2} +c_{x} ^{2} =c ^{2} =c_{y} ^{2} +v ^{2}
or c_{y} =c √[1-(v /c )^{2} ].
Regarding what you would see in the ship, the principle of
relativity states that the laws of physics are the same in
all inertial frames of reference; if your ship is at rest,
we could agree that anything aimed straight acoss will go
straight across. In other words, there is no such thing as
absolute rest —any inertial frame with constant
velocity relative to another inertial frame would have the
same laws of physics.

QUESTION:
How did the eclipse of the sun move from east to west across the earth.

ANSWER:
The shadow the moon casts on the earth (which is the
eclipse) moves west to east, not east to west. A clear
explanation was provided by
NASA : "Because the Moon moves to the east in its orbit at about 3,400 km/hour. Earth rotates to the east at 1,670 km/hr at the equator, so the lunar shadow moves to the east at 3,400 – 1,670 = 1,730 km/hr near the equator. You cannot keep up with the shadow of the eclipse unless you traveled at Mach 1.5."

QUESTION:
I teach middle school science, however I've spent more time advancing my understanding in several of the other science disciplines I teach so my ability to work through this problem is above my current skill set. However I would really like to be able to walk my students through this scenario, in part because I'm not good with subtlety. I would like to know what other variables I need to measure/know in order to calculate the force a hypothetical student's head would experience at the moment it hit the floor after they tipped their chair back too far (despite numerous warnings not to).
I imagine I'll need the Mass of the hypothetical student + the mass of the chair. The Time from when the hypothetical student's center of gravity passes over the point of rotation for the chair, the Height of the hypothetical student's head is above the floor. With Gravity being constant, I "should" have all the necessary information to calculate this (no chance at all will I let my students try to measure the force in real life) but I still have the niggling sense that I'm missing some variable.

ANSWER:
Let's first do the simplest estimate: imagine dropping the
head from about h =1 m above the floor. The speed
the head would hit the floor could be found from energy
conservation, ½mv ^{2} =mgh
or v =√(2gh )=√(2x9.8x1)=4.4
m/s. Now, you either need to know the distance the head
stops in or the time it takes to stop. There is not much
give in either the head or the floor, so let's just guess
that the distance it takes to stop is 3 mm. If you do the
kinematics, the acceleration of something moving 4.4 m/s and
stopping in 3 mm is about 3000 m/s^{2} , about 300
gs! If you take the mass of the head to be about 4.5 kg, the
force is about F=ma =13,500 N=3000 lb. The time over
which this force acts is very short, about 4.4/3000=0.0015
s.

To be fancier, you
would have to do the rotational problem. The student plus
the chair have a certain moment of inertia I and
when they rotate about the rear chair legs they acquire a
rotational velocity ω when the center of mass
has fallen a distance h . In that case ½Iω ^{2} =mgh
and so ω =√(2mgh /I );
now you can get the speed the head hits the ground by
writing v=Lω where L is the distance
of the head from the rear legs of the chair. But to
calculate the moment of inertia of the student/chair would
be very complicated and, in the end, the speed of the head
would not be all that different from the simple calculation
above. And you still have to make approximations over how
far it would travel to stop. Just use the simple head
dropping calculation to convince the students that the order
of magnitude of the force would be a few thousand pounds.

QUESTION:
A swing is suspended by nylon ropes and connected to a limb that is 30 feet in the air. The problem is that the kids cannot build any momentum in order to swing. They pull and kick their legs but the swing will not swing. Is there a way to fix this swing?

ANSWER:
30 ft is a very long pendulum. The period T of a
pendulum (the time it takes for one swing back to where it
started) is approximately T ≈2π √(L /g )
where L is the length and g =32 ft/s^{2}
is the acceleration due to gravity. In your case, L =30
ft, T ≈6 s. Now, if you watch a kid swing a much
shorter swing, you will see that they "pump" once per cycle;
this is called a driven oscillator and, driving with the
same frequency as the natural frequency of the swing is
called resonance and each pump will increase the amplitude.
No doubt the kids, being used to a much shorter swing, are
pumping with a period much shorter than 6 s, far off
resonance, therefore not having the effect of increasing the
amplitude. I suggest that you start them with a good push
and tell them to pump each time they reach the bottom of the
arc and moving forward.

QUESTION:
When a man going from pointA to pointB which is at 1km and return via same path according to physics the work done is nothing but man still feel tired why?

ANSWER:
First, what force moves the man along the path? It is the frictional
force between the ground and his feet. Since friction is not
a conservative force, the work done over a closed path is
not zero. But walking is more complicated than that ,
way too complicated to discuss here. But, it is pretty easy
to understand why we get tired—any time the muscles
exert a force they are doing work whether or not they are
actually moving something (see
earlier
answers ).

QUESTION:
With respect to special relativity, is it correct (or not) to say that time dilation & length contraction are "real" because Nature has them in place to protect the upper speed limit "c"? Is there a better way to say this?

ANSWER:
First of all, what does it mean for something to be "real"?
I would argue that, for example, length contraction is real
because experimental measurements confirm its reality.
Second, I find anthropomorphizing "Nature" to be fatuous;
"Nature" does not "place" or "protect". Finally, you have it
exactly backwards: time dilation and length contraction are
the result of the fact that the speed of light is a
universal constant, not vice versa . I would also
point out that these would also be true if the speed of
light were different from what it is; as illustrated by
George Gamow's book
Mr.
Thompkins in Wonderland , they would be much more
evident if the speed of light were very small.

QUESTION:
I am working on a science fiction book and have a bombardment of a planet happening. I would like to get an idea of the force of impact of 2 kinetic kill vehicles
1 is a 10 pound Depleted Uranium ball.
2 is a 100 pound Depleted Uranium dart.
They are traveling at 1/10th light speed.
They impact an Earth like planet.
I only need a ballpark value. If either or both are some how destroyed interacting with the atmosphere or cause some other effect, that would be good to know too.

ANSWER:
I am not sure what the material being depleted uranium has
to do with anything. Also, "force of impact" cannot be
calculated unless you know the details of the collision, in
particular how long the collision lasted. You could estimate
the kinetic energy classically for your projectiles because
the speed is much less than the speed of light. 1 lb is
about 4.5 kg, so K ≈½mv ^{2} =½x4.5x(3x10^{7} )^{2} =2x10^{15}
J=2000 TJ (terajoules); the Hiroshima bomb had an energy of
about 63 TJ. If the collision lasted 10 s, this would
correspond to a power of 200 TW (terawatts). To give you a
feeling for the magnitude of this power, the total output of
all power sources on earth is about 15 TW. The 100 lb
projectiles would have 10 times the energy and power as the
10 lb projectiles. So those things would have quite a punch.
Here is the problem that most sci-fi writers never think
about, though. Whoever is firing these projectiles has to
give them this energy—where are you going to get 200
TW in the middle of empty space?

Now, the second part of your question. These things
have a speed (3x10^{4} km/s) which is much faster
than the fastest meteor (72 km/s) and you know what happens
to them —they burn up and break up. The recent
(2013)
Chelyabinsk meteor
exploded at an altitude of about 20 mi and most of its
energy (about 1500 TJ) was absorbed by the atmosphere. It
had a much smaller speed than your projectiles (about 20
km/s compared to about 30,000 km/s) but a much larger mass
(about 10^{7} kg compared to 4.5 kg), so the
energies were quite comparable (1500 TJ compared to 2000
TJ). So I would guess that your projectiles would not do
much damage to the objects on the surface of the planet.

QUESTION:
i'm traveling at the speed of sound and a gunshot is fired at the exact moment I am passing the gun, what is the resulting sound that I hear and for how long?

ANSWER:
The figure shows the plane at three times:

just after the gun has been shot;

at the time when the sound reaches
where the plane had been at time 1;

at a time twice as long as 2.

Also
shown are the corresponding spherical wave fronts of the sound
from the shot. As you can see, the wave fronts never catch up
with the plane. You will never hear the gun.

QUESTION:
I am a nurse at a long term care facility. My back hurts every time I get finished pushing a medication cart for the 8 hour shift. My question is...If the medication cart weighs 220 pounds when assembled, how much weight am I pushing given the fact it is on wheels?

ANSWER:
You may assume that the wheels almost remove the frictional force of moving forward; in other words, it takes very little force to keep it moving once it is up to speed. The times you need to exert a significant force on the cart would be when it speeds up or slows down. The more quickly you bring it up to speed or bring it to rest, the larger force you need to exert. So plan ahead and speed it up or slow it down gradually. Here is an example: if you sped the cart up to a speed of about 6 ft/s in 1 s, you would need to exert a force of about 40 lb, whereas if you took 2 s, the needed force would be about 20 lb.
Also, when you turn a corner, go slowly since it takes a
force to turn the cart also and the faster you take the
corner, the greater the required force.

QUESTION:
I have a small (15' diameter) swimming pool.
I have a round leaf net, about 17' in diameter, which I suspend over the pool as follows:
I took 5 10' lenghts of 1" PVC electrical conduit, and joined them end-end to form a hoop/circle ca. 16' in diameter.
The round leaf net has a drawstring, so I placed the 17' diameter leaf net on the ground, then placed the 16' diameter on top of the net.
Then I folded the excess leaf net up above the conduit hoop and tightned the drawstring.
So far so good -- net below hoop, edges wrap around hoop, drawstring holds the net on the hoop.
I suspend this hoop-supported net from a single point in the middle of the circle formed by the hoop/net.
From this central point, I have 10 "spokes" of cord (parachute cord) going to 10 points along the hoop.
The pieces of cord forming these "spokes" were measured carefully to all be the same length.
Here's my question: When I suspend this hoop/net from this central point and the 10 lengths of cord (spokes), the hoop/net does not want to stay in a plane.
Instead, 2 opposite sides (e.g. at 12 and 6 on a clock face) lift

up, and the other opposite sides (e.g. at 3 and 9 in the clock face) drop down.
I know in German, when a bycicle wheel gets stressed too much and deforms, this is referred to as an "eight" -- I suppose because from the edge it may resemble an eight.
I suspect there is something similar going on with my hoop/net, but don't really understand enough to know for sure.
As a followup question, I would be grateful for any thoughts as to how I might get this hoop/net to remain in a plane. Many people suggest just tightening up (shortening) some of the "spokes" but I have tried this and it does not help.

ANSWER:
The surface defined by your deformed loop is called a saddle
point. Because the pvc is pretty flexible (you were able to
easily bend it into a hoop), unless the weight is
distributed very symmetrically, it is quite possible for
this kind of warping to take place. To minimize this uneven
distribution of weight, your spokes should be connected to
the points where you have used straight through connectors
to connect the pvc pipes and the points opposite as shown in
blue in my drawing. But I suspect that there will still be
an asymmetric distribution of mass and the circle will
deform again. If so, then you will need to add crossbars as
shown in red; this should keep the hoop pretty resistant to
deformation. You will probably only need 3 spokes now and if
it does not hang so the plane of the hoop is horizontal, you
can just adjust the lengths. It would also be good to have a
cord coming straight up from the center, maybe another piece
of pvc to which the spokes could be attached or just another
cord. You could get some idea of the balance by just hanging
it from the center.

QUESTION:
Even though radio waves can't raise electrons to higher energy levels due to their extremely low energies do they still interact with orbiting electrons in some way? Just curious, how exactly do radio waves and microwaves interact with orbiting electrons within atoms when their energies are far too low to excite electrons to higher energy levels?

ANSWER:
You know that a radio wave will not travel unaffected
through a large mountain, so something is interacting with
it. The electromagnetic wave is electric and magnetic
fields, and electric and magnetic fields interact with
electric charges. The wave sees mainly electrons. In a
conductor, the conduction electrons are essentially
free-floating and respond to the fields approximately like a
free electron would. In a nonconductor, there is still an
interaction with the crystal as a whole, but again via the
atomic electrons. These interactions result in the
properties of refraction and attenuation of the incident
wave. The details of how you model these interactions are
beyond the purposes of this site.

QUESTION:
I was fishing and the water was rough. Waves crashing all over. Then I look over and see a perfect about 20M circle of calm water while the whole lake is thrashing about. It lasted for about 5mins. The water was about 20feet deep and there was not large rock anything different where the circle appeared. The calm circle did not appear to be like a vortex or whatever of water. It was calm and not rotating. Can you explain the physics behind this or at least give me a guess of what you think caused this. (It almost looked like an object that was not visible to the naked eye caused it but I know thats not true)

ANSWER:
I was just about to reply that I have no idea but just
googled calm water in a storm. It turns out that
there is an old sailor's trick to calm water in choppy
conditions: just pour a little oil on the surface. There
seems also to be
physics understanding of this phenomenon. For a good
overview, see the discussion on
Physics Stack Exchange . Perhaps there was some oil
in your calm area.

QUESTION:
If a bullet of mass m moving at v strikes a stationary rod of mass
M at its center and passes through exiting with u , there is simple conservation of momentum if the rod is free to move/slide:
m (v-u )=MU . What happens if the bullet strikes and passes through the rod at its end (same
v and u ), causing the rod to move forward while at the same time start to spin?

ANSWER:
Your question is essentially the same as an
earlier question except that in
that question the collision was stipulated to be elastic. In
your case, there is no conservation of energy equation and
so there are only two equations —conservation of
linear momentum and angular momentum. But, there are only
two unknowns, U and ω , because you
stipulate u to be known. (Also, d=L /2 for
your question.)

ADDED
THOUGHTS:
I thought it might be interesting to look at the energy
change for these collisions. It is straightforward algebra to
show that U=m (v-u )/M and ω= 6U /L .
The energy before the collision is E _{1} =½mv ^{2} .
Referring to the earlier answer, the energy after the collision
is ½mu ^{2} +½MU ^{2} +½I ω ^{2} .
For the simple case of hitting the center (ω= 0) I
find E _{2} =½m (u ^{2} +(m /M )(v-u )^{2} ),
and for the case of hitting the end I find E _{2} ' =½m (u ^{2} +4(m /M )(v-u )^{2} ) .
Since E _{2} ' >E _{2} ,
less energy will be lost in the case where the collision happens
off center (assuming u and v are the same in
each case). For example, if m=M /4 and u =½v ,
I find ΔE =-(11/ 32)mv ^{2}
and ΔE' =-¼mv ^{2} . In both
cases, energy is lost in the collision.

QUESTION:
If the farther out in space we look, we are looking back through time how is it we do not see ourselves?

ANSWER:
You are looking at how things which are far away looked when they
were at an earlier time. But none of those things are at the
position where you were in the past so you would not see
yourself.

QUESTION:
How could I measure the viscosity of pizza sauce (and other materials)
using "at home" equipment? I want to determine the viscosity of a sauce,
then take the pizza sauce and place it on a turntable whose speed can be
controlled and see what speed is required to make the sauce flow from the
center to the edge.

Then, I want to alter the sauce and make it more and/or less watery
(changing its viscosity) and measure that new sauces viscosity.

Then, I'll
take the new sauce and re-measure the turntable speed necessary to make the
new sauce flow from the center to the edge.

Last, I want to replicate the tests enough times so I can create an equation
that would allow me know what turntable speed would be necessary to
correctly flow the sauce based upon the viscosity of the substance.

ANSWER:
Why not just experiment with sauces until you find the right thickness to
achieve what you want? Measuring the viscosity (not an easy task) is just an
unnecessary step in the process.

FOLLOWUP QUESTION:
The reason is that I want to be able to alter the sauce, test the viscosity, then know how much to alter the speed of the rotating table consistently.
Knowing, for example, a 1% increase in viscosity requires a 5% increase in rotation speed allows me to continuously alter the sauces and know with certainty the speed the table must rotate.

ANSWER:
This must be a science fair project or something because it
will certainly not be of any help in making pizzas in the
real world. In a pizza you want to spread the sauce
uniformly over the whole area, right? What causes the sauce
to move out on the rotating turntable? There is a
(fictitious) force F , the centrifugal force, which
pushes the sauce out, F=mv ^{2} /R
for a mass m with speed v when at a
distance R from the center. But, the speed depends
on the distance from the center, v=R ω
where ω is the angular velocity; so F=mRω ^{2
} and an ounce of sauce experiences a bigger force as it
moves out. In other words the sauce will tend to all be
pushed out the the rim of the pizza regardless of its
viscosity. If the viscosity of the sauce too large, the
centrifugal force might be too small to move the sauce at
all, so there would be the tendency for the sauce to stay in
the center. In any case, I cannot imagine that it is
possible to use rotation to get a uniform spread of sauce.

If you still want to pursue
this, I found the description of a straightforward
experiment to measure viscosity μ . You get a
tall cylindrical container and fill to a depth d with
the fluid (density ρ _{f} ). Drop a sphere
(density ρ _{s} radius R ) into the
fluid and measure the time t it takes to reach the
bottom; then the velocity of the falling sphere was v=d /t .
The viscosity is then μ= (4R ^{2} g (
ρ _{s} -ρ _{f} ))/(9v ).
It will be a little tricky since the sauce is not transparent.
Also, it is important that the sauce be homogeneous, no chuncks
in it.

QUESTION:
Lately I've been dealing with the Work, Force and Energy thing. I can't deny the fact that understanding some of the equations included in most of the textbooks has been one hell of a challenge to me. I'm currently trying the understand something here: we know that the work done (W) is equal to the change of kinetic energy. Is it also the same when it comes to potential energy? Does it amount to the done work as well? I never see the work of the gravity force included in the equations. I think I know the answer but I just want to make sure. There's something more bugging though--when we write the conservation of energy equations for a certain setting, do we include the work there or is it already included? For example, if there's a body on the top of an inclined plane and we take friction into account, should the equation be "potential energy at the top + kinetic energy at the top = potential energy at the bottom (0=0) + kinetic energy at the bottom + the work done by the friction"? Getting an answer would mean a lot to me as an engineer-to-be. :)

ANSWER:
This question verges on a violation of a "concise,
well-focused" question as required by site groundrules. I
will do a couple of examples to try to clarify work-energy
for you. The "guiding principle" which is always true is
that the the total work W _{total} done by
external forces on an object of mass m is equal to
the change in kinetic energy, ΔK =½mv _{2} ^{2} -½mv _{1} ^{2} .
The picture is a mass m on an incline θ
being pushed by a constant force F up the incline
and moving a distance s ; for simplicity, I choose
m
to be at rest when you start pushing. Other forces on m
are its own weight W=mg , vertically down, and a
possible frictional force f pointing down the
plane. For the time being I will assume f =0. Only
the component of W along the plane W_{s} =-mg sinθ
will do work. The net work w (sorry if it is confusing to
have big W and little w ) is w=Fs-W_{s} s=Fs-mgs sinθ= ΔK= ½mv^{2}
and so v =√[2s (F -mg sinθ )/m ].

Rather than derive the
gravitational potential energy, U=mgy , I will
assume that you already know that. I will now do the problem
over but using potential energy. w=Fs=E _{2} -E _{1} =(K _{2} -K _{1} )+(U _{2} -U _{1} )=½mv^{2} +mgh= ½mv^{2} +mgs sinθ=Fs.
Solving this, you find exactly the same answer: v =√[2s (F -mg sinθ )/m ].
So here is how I look at it: potential energy is a very
clever bookkeeping device to keep track of the work done by
a force which is always present like gravity. It just makes
life a lot easier to not always have to calculate the work
done by some force that you know is always there. So, here
is the so-called work-energy theorem: the change in total
energy, kinetic energy plus any potential energy you have
included, is equal to the work done by all external
forces, E _{2} -E _{1} =W _{ext} ;
here an external force is any force for which you have not
introduced a potential energy function. I like to rearrange
the work-energy theorem as E _{2} =E _{1} +W _{ext} —what
you end up is what you started with plus what you added (or
subtracted if W _{ext} <0).

Finally, let's include the frictional force, f=μW_{N} =μmg cosθ.
(μ is the coefficient of kinetic friction.) The work which the friction does is negative because it
is opposite the direction of s , w _{f} =-μmgs cosθ.
Therefore, W _{ext} =Fs -μmgs cosθ=½mv^{2} +mgs sinθ ;
solving, v =√[2s (F -mg (sinθ+μ cosθ ))/m ].
You might wonder why we did not introduce a potential energy
function for f. The reason is that there are two kinds of
forces in nature, conservative forces and nonconservative
forces and the latter kind cannot have a meaningful
potential energy function; friction is a nonconservative
force. But this answer has rambled on long enough and that
is a topic for another day!

ADDED
NOTE:
I see that I did not answer your question ("…if
there's a body on the top of an inclined plane and we take
friction into account, should the equation be "potential
energy at the top + kinetic energy at the top = potential
energy at the bottom (==0) + kinetic energy at the bottom +
the work done by the friction"?" ) explicitly.
You have it wrong. It should be that the total energy at the
bottom equals the total energy at the top plus the work done
by friction; don't forget that, as in my example going up
the plane, the work done by the friction is negative. Some
books like to write it your way replacing the friction part
by the work done against
the friction ; I actually do not like that one bit.

QUESTION:
Why does dust move behind fast moving cars?

ANSWER:
Because there is a pocket of turbulent air
behind the car which moves with
the vehicle.

QUESTION:
If you were on the moon and used an electric motor - would the armature spark as it does here in an atmosphere like ours?

ANSWER:
It is not possible for a visible spark to form in a vacuum,
so the answer is no.

QUESTION:
As a physicist do you think it's theoretically possible for energetic photons of x-rays and gamma rays (keV-MeV) to be converted to visible light photons (1.77-3.1 eV) with the compton scattering effect despite the HUGE energy difference?

ANSWER:
It is pretty easy to figure this out. The change in
wavelength in compton scattering is λ'-λ =(h /mc )(1-cosθ )
where h is Planck's constant, m is the
mass of an electron, c is the speed of light, and
θ is the angle of the scattered photon; (h /mc )=2.4x10^{-12}
m=2.4x10^{-3} nm and is called the Compton
wavelength of the electron. The largest shift is for the
largest angle, θ =180^{0} where cos180^{0} =-1
so the largest possible shift is (Δλ )_{max} =4.8x10^{-3}
nm. Visible light has wavelengths around 1000 nm, soft
x-rays (the closest to visible light) around 1 nm: so that
would require (Δλ )_{max} ≈1000
nm>>4.8x10^{-3} nm. So the answer to your question
is no.

QUESTION:
You can't pressurize a liquid, correct? So why is it when you turn your hose off at the faucet bib, the water still is under pressure when you squeeze the sprayer trigger? It still shoots our with force.

ANSWER:
Certainly you can pressurize a liquid. Most liquids are almost imcompressible, so increasing pressure does not significantly decrease volume.

FOLLOWUP QUESTION:
BUT, the pressure is from a giant water tank, correct. So , when you shut off the faucet, you've isolated the water tank. Therefore, the pressure should be gone.

ANSWER:
Why do you think that isolating the water makes the pressure
go away? The water deep in your giant tank has to support
the weight of all the water above it, so the pressure gets
bigger as you go deeper. And the pressure at the top is not
zero because atmospheric pressure is pushing down on the
surface. I suggest you put on a bathing suit and swim down
to the bottom of the tank and then decide whether there is
pressure in the water down there.

QUESTION:
I am writing a novel and am trying to understand how my interstellar spacecraft will behave. It is going to Iota Persei. 34.38 ly away. I was thinking it would travel at .6c (somewhat arbitrary but...). If I accelerated the space craft at a consistant 9.8m/s, it would take ~212.44 days at a distance of ~.1744 ly. Presumably, it would take about the same time and distance to slow the craft as it approached Iota Persei, roughly speaking.
While I've no idea of my craft's mass, it would be large, like 40,000 people large, and be propelled by an anti-matter drive. Would the energy output of the antimatter drive need to increase due to the increased mass as speed increased? Is it feasible to suggest that an antimatter drive would be able to produce enough energy to accomplish the above scenario?

ANSWER:
I have answered variations on this question several times in the past.
The most useful for you to read and understand compares
acceleration of an object as measured from both a rest frame
and from the frame of the object; I recommend you read both
this and all the links in
that answer . [The derivations of
the equations below can be found in the earlier answers.]
From the perspective of your space ship the velocity v
as a function of time t is v =(a _{0} t )/√[1+(a _{0} t /c )^{2} ] where
a _{0} is the acceleration as measured by the ship (g in your case). In order to achieve this acceleration, you must exert a force
F=ma _{0} where m is the mass of the ship.
The velocity is plotted in red in the graph; reading from
the graph, the time when v /c =0.6 is when a _{0} t /c ≈0.8=9.8t /3x10^{8} . Solving, t =2.45x10^{7} s=0.78
yr=284 days. The position as a function of t is x =(mc ^{2} /F )(√[1+(Ft /(mc ))^{2} ]-1)=(c ^{2} /g )(√[1+(gt /c )^{2} ]-1);
for gt /c =0.8, x =2.58x10^{15}
m=0.273 ly. At this point you turn off your engines and
coast at v /c =0.6. The distance you need to
go to get halfway to the star is 16.9 ly and it will take
you 16.9/0.6=28.2 years. The second half of the journey will
take the same time, so the time for the whole trip is T =2(28.2+0.78)=58
years.

Your answers (212
days, 0.17 ly) are different from mine (284 days, 0.273 ly)
because you have assumed that the acceleration, viewed from
the earth, is the same as the acceleration (9.8 m/s^{2} ) seen from the ship. The
black curve in the graph shows the ratio of the acceleration
seen from the earth compared to the acceleration seen from
the ship; reading from the graph, at 0.6c the ratio
is about 0.5, 4.9 m/s^{2} . Be sure to read the
discussions of acceleration in the
earlier answers if you want to understand this issue
with acceleration in special relativity.

Since there is no such
thing as an antimatter drive engine, I can make only a rough
assessment of how feasible this is. The force you need to
apply during the 0.78 year accelerations is F=mg . What is m?
The space station has a mass of about 400,000 kg and you
have 40,000 passengers. Let's say each passenger needs one
space station of mass to support his/her needs, so the
required force would be 1.6x10^{9} x9.8=1.57x10^{10}
N. The energy you have to supply to the ship to v /c =0.6
I figure to be about 36x10^{24} J and the amount of
mass converted with 100% efficiency into energy would be
about 4x10^{8} kg and you would need that much again
to stop it. So your ship of mass 1.6x10^{9} kg would
have to carry 8x10^{8} kg of fuel, one third of the total
weight! Obviously, I have not included this mass change in
my estimates of time. And, of course, there is the issue of
how you are going to store 4x10^{8} kg of
antimatter. The whole scenario certainly does not look
feasible to me!

Finally, note that all
the distances and times are as measured by the earth bound
observer. In other words, the passengers will not be 58
years older when they arrive at their destination. Because
of length contraction, they will have a shorter distance to
travel. Ignoring the accelerating periods, I can estimate
the time of the trip by just assuming they travel 0.6c the
whole trip. In that case, T' ≈T √(1-.6^{2} )=58x0.8=46.4
yr.

QUESTION:
If energy cannot be destroyed or made, only transferred from one form to another, how does the level of matter and energy in the universe stay at a constant, if black holes can consume both matter and radiation in huge quantities?

ANSWER:
If a black hole consumes an amount of energy E , its mass
increases by E /c ^{2} . If it consumes an amount
of mass M , its mass increases by M .

QUESTION:
I have 1,000 kilowatt hours of energy. Could you help me understand how much weight I would need to lift an object up 100 feet in the air to use that 1,000 kWh of energy up completely?
This is not homework. I have tried researching online, but every answer I have seen ends up being different than the last. Trying to understand for a meeting I have with some friends at work.

ANSWER:
I will work in SI units because a Watt is a Joule/second (and because
scientists prefer SI units): 100 ft=30.5 m. The energy to lift a mass
M to a height h =30.5 m is E=Mgh =299M
where g =9.8 m/s^{2} is the acceleration due to gravity;
the answer will have units of Joules if M is in kg. Now the
energy available is 10^{3} kW ⋅hr(3600 s/hr)(1000 W/1
kW)=3.6x10^{9} J. So, 299M =3.6x10^{9} so M =1.204x10^{7}
kg=2.65x10^{7} lb.

QUESTION:
If a fly is in a car that is travelling, say, 50 mph... and the fly is not resting in the car (i.e. he is flying, not attached to any surface), does that mean he has to fly at 50 mph as well? Or would he be pressed to the back window of the car?

ANSWER:
The fly flies relative to the air in the car. But the air in the car is
at rest relative to the car. Therefore the fly flies exactly the same as
she would if the car were at rest.

QUESTION:
How is this possible. My brother while playing baseball hit the ball in such a way that when it broke the glass window in the living room it left a perfect circle. What are the chances of that happening?

QUESTION:
If an asteroid was in space but not moving, and you were within reach of it, and you pushed on the asteroid to get to the spaceship, would you lose force because the asteroid isn't moving and you were acting upon the asteroid? (The asteroid wasn't pushing against you) Wouldn't the asteroid go faster than you?

ANSWER:
You have misunderstood this problem on many levels. First, there is
really no such thing as moving or not moving in an absolute sense. If
you are on some other asteroid and see this one moving with a speed of
500 mph it is not at rest; but if you get in a space ship and go 500 mph
to just keep up with the asteroid, it is at rest. If you are in a frame
where the asteroid is at rest, its lack of motion has nothing to do with
how it would react to a force you might exert on it. And, Newton's third
law says that if you exert a force on the asteroid, it always
exerts an equal and opposite force on you. And, whether you or the
asteroid goes faster after the push depends only on your relative
masses. An asteroid is most likely much more massive than you and
therefore you will be moving faster after the push. For example, if the
asteroid is 1000 times more massive than yours, its speed after the push
will be 1000 times smaller than yours.

QUESTION:
What will be the work done by a person who is moving forward and carrying the bucket in his hand?

ANSWER:
The person does zero work on the bucket
assuming that the bucket moves horizontally with constant speed.
However, the person does do work; see an
earlier answer.

QUESTION:
If you could please explain the experiments that supposedly confirmed quantum entanglement and
"spooky action at a distance" I would appreciate it.
The experiments I read about all involved a pair of entangled particles, that are then moved
"far apart". Since they are entangled, one has an "up" spin and the other a
"down" spin. However, we don't know which is which until they are separated and one is observed.
The it is found that by observing one particle to have, let's say, the "up" spin, miraculously the other particle
"far away"has the down spin, with the information between them being transmitted instantaneously even though they are far apart.
What I don't get is, how is this different than having two cards; and ace and a king. Both face down. You don't know which is which until you observe them. You place them
"far apart". Then you flip one over, and see it is the ace. Instantaneously, the other card becomes the king.
This does not seem like "spooky action at a distance". You observed one of two possible particle states, and therefore, the other particle must have the other state. Obviously, I am missing something, but can never find the explanation.

ANSWER:
If entanglement were as you currently understand it, one up and one
down, this would not be entanglement at all and there would be nothing
mysterious about the second one being, instantaneously, the opposite of
the first once you had revealed it. But that is not how it works because
quantum mechanics, which is the operative physics in the realm of
particles, describes the state of a particle. Let us first start by
looking at a single particle which has two possible states which I will
denote as |+> for up and |-> for down. A particle does not have to be
one or the other, it can be a mixture of the two, for example 25% |+>
and 75% |->; as long as the two percentages add up to 100%, this is a
possible state. Now, when you observe the particle by measuring whether
it is up or down, you will find it in one or the other (three times more
likely to find it in |->) and now the measurement has put it into a
state which is purely up or down; this is called "collapsing the wave
function". In a typical entanglement experiment the two particles are
initially in a state 50% |+> and 50% |->, half up and half down and
therefore the total spin of the two must be zero. When you observe one
to be up, the other is instantaneously observed to be down no matter how
far apart the two are at the time of observation —that is
the spookiness! The trick is to devise an experiment which clearly
demonstrates that this is what is happening, not what you described as
one being up and the other down from the beginning. It is pretty
technical and tricky to understand this, but Roger Pennrose in his book
The
Emperor's New Mind has a pretty good explanation.

QUESTION:
If I were to coil 500' of 1/2" OD rope in a single layer, how large
would the circle be?

ANSWER:
No homework.

FOLLOWUP QUESTION:
I am a 49 yo woman making a rug. Even my kids are too old for that kind of homework.

ANSWER:
If you lay the rope out straight, the total area will be 500x12x ½=3000
in^{2} . If you coil it up, the area will be the same and the
area of a circle is πR ^{2} . Therefore the radius of
your rug would be R =√(3000/π )=30.9 in.

QUESTION:
Frankly I wasn't sure if this was a crank question or deeply profound, I'll let you be the judge.
I've been an electronics hobbyist all my 50 years, and I've used resistors, capacitors, and less often inductors. I started to wonder how many different types of passive electronic components were possible. Were there more than just these? I searched for and found something called a memristor which apparently IS a brand new kind of component that behaves differently than any of those. Wikipedia has a
diagram .
I'm curious, there are components on all the outside edges, what about the diagonals, what do these represent? They have different equations so they MUST behave differently, what ARE these components?

ANSWER:
As in many situations where you are trying to count something, it becomes problematic how to define what you are counting. What about diodes? Are Zener diodes different from diodes in your count? You should really think of this kind of question as qualitative and not quantitative; the answer is not of any importance.
Your questions about the device itself would be better directed to an
electrical engineer. The diagram you refer to is not a circuit, it is
merely a representation of how these four devices relate to each other
and to the quantities voltage, current, charge, and flux. The outer
edges show how devices manipulate the quantities, e.g . changing
the current through a resistor changes the voltage across it (dv=R di ).
The diagonal lines indicate relations between the corner quantities,
e.g . current is the motion of charges (dq=i dt ).

QUESTION:
What happens to the atmosphere inside a cupping vessel when a flame is introduced and causes a partial vacuum (negative pressure) - which allows the therapeutic vessel to adhere to the skin surface? What are the physics that explain this happening?

ANSWER:
Cupping vessels have been in use for thousands of years. Ancient Greek
and Roman physicians used them to assist in
blood letting.
Chinese medicine uses them to
treat a
variety of maladies; cupping therapy is generally considered
pseudoscience by modern medicine. The idea is to provide suction on the
surface of the skin and is achieved by first heating the cup and air
inside and then placing it on the skin. As the air inside cools, the
pressure decreases. The physics of this pressure decrease can be
understood by examining the ideal gas equation which relates pressure
P , temperature T , volume V , and amount of gas
N : PV=NRT where R is a constant which depends
on the units you use. So, as you can see, keeping the volume and amount
of gas constant, if the temperature decreases the pressure must
decrease. The fact that P ∝T is sometimes
called
Gay-Lussac's law .

A similar thing happens when canning food in glass jars. The
canning is done with the contents very hot and a lid which
is slightly domed is affixed. As the contents cool, the
pressure decreases causing the dome to pop inward toward the
contents, signaling that a good seal has been achieved.

QUESTION:
I am a hot air balloon pilot.
I am trying to develop a mathematical formula, for calculating where my Scoring baggy will land, when I participate in a Ballooning event.
I am given a location in a Pilots briefing, as to where the Scoring X's are located. I attempt to fly to that particular location, drop my Scoring Baggy on the X, and earn points depending on how close I am to the center of the X.
(Each Leg of the X is approximately 100' to 300' long, depending on the Ballooning event.)

The Scoring Baggy (6 ounces) has a constant weight.

My Balloon's Speed, as I approach the Target, is a variable. the Scoring Baggy will share that speed once I release it, Correct ?

And, the height I
release the Scoring Baggy from, at the time I release it, is also a variable, correct ?

My question is, is there a Mathematical formula that will calculate, how many feet from where I release the Scoring Baggy, it will land ? And how long it will take ?

ANSWER:
This is not a simple question. Although it would be simple if air drag were neglected, I suspect that it is not negligible for the bag weighing only 6 oz. I will do the calculation without air drag here.
I will not include the details, just the final results. The time t
in seconds it takes for the bag to hit the ground is t = √(2h /g )=¼√h
where h is the height in feet from which you drop it and
g =32 ft/s^{2} is the acceleration due to gravity. For
example if you drop it from h =144 ft, t =3 s. The
distance x in feet it will travel horizontally in this time is
x=vt where v_{x} is the horizontal speed of the balloon in ft/s when
you drop it; so if v_{x} = 20 ft/s (about 13.6 mph) and you drop it
from 144 ft, x =60 ft. The equation for x can be written
as x =¼v_{x} √h which is handy if you
do not care about the time.
Note that the weight of the bag does not come into this at
all.

Warning:
this is pretty mathematical and probably not a computation
you would want to do in the heat of a competition! For my
own interest, I want to estimate how much error is
introduced by neglecting air drag. It is much more
complicated if you include air drag. For this case, I need
to do the calculation in SI units rather than English units.
The reason I need to use SI units is that I will estimate
the air drag force as F _{d} ≈¼(v_{y} ^{2} /A )
where v_{y} is the vertical velocity of the
bag and A is the area it presents to the onrushing
air; this estimate is correct only for SI units because it
has things like the density of the air at sea level built
into it. Now, it is my understanding that the speed of a hot
air balloon moving horizontally is the same as the speed of
the wind; in other words, from the perspective of a person
riding in the balloon, he is in perfectly still air. This
greatly simplifies the problem because the bag will drop
straight down as seen from the balloon, i.e. it
should strike the ground directly below the balloon. So, the
bag sees two forces, its own weight mg down and air
drag up ; Newton's second law becomes ma=m (dv /dt )=-mg +¼(v_{y} ^{2} /A ).
This is a first-order differential equation has a solution
v_{y} =-[√(g /c )]tanh([√(gc )]t )
where c=A /(4m ). This solution is also a
differential equation since v_{y} =dy /dt
where y is the distance above the ground; solving
this differential equation, y =h -(1/c )ln(cosh([√(gc )]t ).
A graph of this function compared to the case above for
dropping from 144 ft (note that 144 ft=44 m) is shown. (I
approximated the area of the bag to be 0.01 m^{2} ≈16
in^{2} =4x4 inches.) The time for the bag to hit the
ground is now 3.33 s rather than 3 s, approximately 10%
longer meaning that you should drop it when you are 66 ft
from the target rather than 60 ft. If you are lower the
correction is smaller, if you are higher the correction is
larger. Given the circumstances under which you must act, I
would expect the inclusion of air drag to be unnecessary
unless you are at a very high altitude and that you should
just drop it when you are about x =¼v_{x} √h
from the target.

ADDED
COMMENT:
It occurs to me that I have assumed that the wind speed and
direction are the same at all altitudes. This will not be
true in the real world and I am told that taking advantage
of this is how hot air balloons can get some control over
direction of travel. Obviously, trying to do a calculation
including this would be impossible other than for a
particular set of wind velocities as a function of altitude.

QUESTION:
The ideal gas law says that pv=nrt. If I have a balloon filled with an
inert gas and put heat into the balloon while holding the volume constant
the pressure will increase on the left side of the equation and the
temperature will increase on the right side of the equation. If I remove the
heat source and allow the balloon to expand the volume will increase and due
to the gas laws the pressure will decrease by the inverse of the volume
keeping the value for pv constant. However, on the right side of the
equation with a constant number of molecules the temperature will decrease.
How can pv remain equal to nrt with the temperature decreasing and pv
remaining a constant value?

ANSWER:
What makes you think that PV will remain constant? This is
called an adiabatic expansion, a process where no heat enters or leaves
the system. What remains constant is PV ^{γ
} where γ is a constant which depends on the gas. For
example, for a monotonic gas γ= 5/3 and for a diatomic gas
γ= 7/5. Once you know what the new P and V
are you can get the new T : T=PV /NR .

QUESTION:
Work done by ship's engine = KE of ship + Work done to overcome frictional forces. Work done by ship's engine - work done to overcome frictional forces = KE of ship Eventually when the ship travels at constant speed, and all work done by engine is used to overcome frictional forces, then mathematically,
Work done by engine - work done to overcome frictional forces = 0
But KE of the ship is not zero.
So where is the source of the ship's KE coming from?

ANSWER:
While the ship is accelerating, the engine is both overcoming drag and accelerating the ship. The drag is dependent on the speed, gets larger as the speed gets larger. Eventually, the drag will become equal the force the engine applies so the net force will be zero so the ship will move at constant speed. But I do not understand your question since the work done by engine was clearly adding kinetic energy to the ship during the acceleration time. Your error was to assume that the work to overcome friction was always the same which is clearly not the case.

QUESTION:
If the Sun is comprised of largely unignited fuel, and enough so that it can last for billions of years, how can it appear to be a fiery ball that is entirely aflame, even to the core? Why is the Sun not a hot, dark, ball?

ANSWER:
It is a basic physics fact that the amount of radiation from an object
with temperature T is proportional to T ^{4} ;
also, the wavelength λ of the most intense radiation is
inversely proportional to T . So, it is not possible for an
object to be both hot and dark.

QUESTION:
I just wanted to know if a ball that is thrown downward would have an acceleration of 9.8m/s^{2} or it would depend upon how much force the ball is thrown with.
Is the acceleration of a stone thrown upward the same as that of a stone
thrown downward?

ANSWER:
The acceleration of any object depends, according to Newton's second
law, only on the sum of all forces acting on it. Ignoring air drag, an
object going either upward or downward has only one force on it, its own
weight, so its acceleration is g =9.8 m/s^{2} downward;
this results in the object speeding up if moving downward and slowing
down if moving upward but both have the same downward acceleration of
magnitude g . During the time you are throwing it down (up)
there is a larger (smaller) acceleration because of the force your hand
is applying, but that disappears the instant the object leaves your
hand. If you take air drag into account, there is an additional force
which always points in the direction opposite the velocity.

QUESTION:
Is inertia a form of energy? If not then why does the blades of a fan move even after the supply of energy is stopped because energy cannot be created nor be destroyed?

ANSWER:
When a fan is spinning at full speed, the only reason it needs power
from the motor is that friction is constantly taking energy away —friction
in the motor and bearings and air drag. With no friction, once you got
up to speed you could turn the motor off. In the real world when you
turn the motor off the fan slowly stops as friction takes its kinetic
energy away and turns it into heat; always a "balance" as you expect.

QUESTION:
If light was trapped in a dome of mirrors and the light source was cut off by another mirror and there were no holes to have the light leak out would the light go on forever.

ANSWER:
Not in the real world. There is no such thing as a perfect mirror. See
an earlier answer.

QUESTION:
I understand that adding energy to an object increases the motion of its molecules, and that the movement of molecules creates heat. But what I don't understand is WHY the movement of molecules creates heat. Why does an increase (or decrease) in molecular motion change temperature?

COMMENT:
First of all, you need to learn the vocabulary used in thermodynamics —see
the faq page .
You will then see that your question should have been written as:

I understand that adding energy to an object increases the
kinetic energy of its molecules, and that the
kinetic energy of molecules determines the
temperature . But what I don't understand is WHY the
kinetic energy of molecules determines temperature . Why does an increase (or decrease) in molecular
kinetic energy change temperature?

ANSWER:
The simplest answer is rather arcane: Absolute temperature
is defined to be proportional to the average
kinetic energy per molecule. In what sense, then, can we
understand why changing the average kinentic energy of the
molecules causes you to feel different. It is easiest to
talk about an ideal gas for which T =2<KE >/(3k )
where <KE > is the average kinetic energy per
molecule and k=1.38x10^{-23} J/K is Boltzmann's
constant. (Most everyday gases are pretty well described as
ideal gases.) Molecules are continually striking your body
and transferring momentum as they bounce off. This causes
the molecules in your skin to gain energy, i.e. get
hotter. Now, your skin is hotter than the interior part of
your body so heat starts flowing, causing the inside part of
your body to become hotter. Meanwhile, your body is doing
what it does to keep your body temperature at 96.8^{0} F.
As the temperature of the air gets hotter, your body has
more and more trouble cancelling out the flow of heat from
the air and eventually fails and you get heat stroke or at
least start to feel sick.

QUESTION:
So while I was a little kid due to numerous headaches I had, I was scanned in an MRI machine. I was feeling a little anxious when the machine was put on and my mom came to me to ease my anxiety. She had her wallet with her and pretty much all her cards went dead. What exactly causes magnetism to destroy payment/membership cards?

ANSWER:
The
magnetic strip is just like magnetic recording tape. There is a layer of
very fine particles which are magnetizable. Data is written on the tape
by using an electromagnet called the recording head; when the magnet is
turned on the particles become magnetized. So the data is written in
stripes in a code, sort of like the UBS labels used to scan products at
the cash register. In a magnetic strip, the card moves by a tiny coil in
which a current is caused to flow when the magnetic stripe goes past it.
Since magnetic fields are used to create the magnetized particles,
magnetic fields can be used to destroy them. Even a relatively weak
field, if present for a long enough time, can mess up the data on a
magnetic strip. An MRI machine has a huge field and it would easily
demagnetize the strip.

QUESTION:
If you spun a mirror-smooth metal ball at just about the maximum possible
RPM... would the angles of light reflection change any due to the speed (as
opposed to from the distortion of shape from centrifugal forces)? If so, would it be a visible difference, or something you'd detect with a
precision instrument?

ANSWER:
There would be an observable effect but not due changes in direction of
reflected rays. Wavelengths of light reflected from parts of the sphere
moving toward (away) from the observer because of the rotation would be
shifted to shorter (longer) wavelengths; this is called a blue (red)
shift. This effect would be very small for speeds which would be
achievable.

QUESTION:
If a fishbowl (full of water and a fish) was propelled forward (any direction really) at 2g.
Would the fish feel the effects of the 2g acceleration in the water in the same way that a person feels this effect in air and therefore cause the fish to be pushed against the side of the bowl if it did not swim to remain in the centre of the body of water or would the fish be held in stasis in the body of water (Water neutralising the effect of the acceleration)?

ANSWER:
I have answered variations of this question before. You should read an
earlier answer to see the complications and how
you can understand your question. In that answer I addressed two
possibilities applicable to your fish —if the fish has a
density greater than that of water, it will move backward until it hits
the back side of the bowl and if it has a smaller density it will move
forward until it hits the front side of the bowl. But your question is
different because the fish has the same density as water.
Therefore it will remain in the center of the bowl if it started there.
But the fish is accelerating, so he has to feel a net force; the water
behind him will exert a force which is larger than the water in front of
him does. So, the effect of the acceleration is not "neutralised", he
feels it. It is analogous to your sitting in an accelerating car and the
back of the seat exerts a forward force on you even though you remain
seated in the seat.

QUESTION:
As I'm sure you know, LIGO has detected 3 black hole mergers in the last few months. This means that there have been and likely are a huge number of black holes in the universe. In parallel, the math says that 95% of the mass of the universe is not visible to us. Could it be that there are millions or billions of black holes floating around a few light-years from the nearest thing and that they would be virtually invisible - we could only detect them if they happened to line up with further distant light sources such that we might detect gravitational lensing. Could they account for the 95%? I know you don't do astrophysics but thought that the answer might be so simple that you might be happy to address it.

ANSWER:
Keep in mind that I have no real expertise in cosmology. Two things, I think, will shoot down your idea. The 95% number you quote is like 20% dark matter and 75% dark energy. The dark energy component cannot be thought of as
"mass-like" because it makes itself known in the accelerating expansion of the universe, so it would appear to be some kind of repulsive force. The dark matter which everyone is trying to observe cannot be in black holes because to cause the things it is thought to cause (like the anomalous rotation of spiral galaxies) it needs to be spread out over huge volumes of space which, of course, black holes are not. My take on dark matter (absolutely not mainstream) is that perhaps
dark matter is the luminiferous æther of the 21st century, i.e . it is not some invisible
"stuff" but rather its purported effects are indicative that we do not really understand gravity as well as we think we do. Certainly general relativity is an incomplete theory since no one has successfully quantized it. You may know, but dark energy, at least the evidence for it, has an interesting history. When Einstein formulated general relativity, the expansion of the universe was not known and the universe was thought to be static; but with just gravity this is not possibleâ€”it will expand forever or expand until it collapses, depending on the initial conditions. Einstein therefore empirically inserted something he called the cosmolgical constant. Later when Hubble discovered that the universe is expanding, Einstein removed the cosmological constant and dubbed it his
"biggest blunder"; now it is coming back into play, but still just empirically.
There are doubtless millions or billions of black holes as you speculate, because all stars greater than a few solar masses end up that way. It is known that nearly all galaxies have a supermassive black hole at their centers. I would guess these came from early-universe stars coalescing. And early-universe stars were huge, thousands of times more massive than the sun because only hydrogen was available to make early stars. Still, I have learned that the size of the universe is just about impossible to comprehend and at its relatively young age still, I would guess these remnants of earlier stars are still a pretty small fraction of the mass of the entire universe.

QUESTION:
how can a photon have a frequency since moving at the speed of light would prohibit any internal vibration since time has stopped inside the photon. Textbooks show a tiny " football " shaped photon with an oscillating wave inside the " football
" photon ?

ANSWER:
You are right, a photon does not have a frequency but does have an
energy which is directly related to a frequency. Electromagnetic
radiation is an example of particle-wave duality. Light is not a
particle or a wave, it is a particle and a wave. If
you design an experiment to "prove" that light is a wave (particle) you
will "find" that it is a wave (particle). If you have light waves with
some frequency f , that light can equally correctly be thought
of as a collection of particles (photons) which have an energy E=hf
where h is Planck's constant. The oscillating wave depected
inside your "football" is a cartoon representation and really has no
quantitative meaning. Incidentally, you can never have a precise
knowledge of the energy of a photon because of the uncertainty
principle; therefore you cannot have a precise knowledge of the
frequency of the wave. Your "football" is called the envelope of the
wave function and is representative of the uncertainty of the position
of the photon.

QUESTION:
If one were calculating the design load for an hvac, would a cooler
operating/"burning" 13 watt LED bulb be expected to add less demand on the
system than a 13 watt halogen bulb which operates at a higher temperature
and gives off less light? For the hypothetical, please assume the bulbs are
installed in identical windowless closets. Does the additional light from
the LED bulb convert to heat when it is absorbed by the closet walls? Would
the temperatures of each closet be the same after a considerable period, or
would the closet with the halogen bulb be warmer?

ANSWER:
To answer this question, you have to make some approximations. Before
turning on the lights, the walls and room air are in equilibrium with
each other, at the same temperature. I will assume that no heat escapes
from the wall/room system and that the walls are ideal blackbodies. So
the walls radiate a blackbody spectrum continuously and absorb energy
from the air at the same rate. Now turn on the lamps, each radiating 13
Joules of energy per second, but it is not blackbody radiation because
there are different amounts of heat (infrared) and visible radiation for
the lamps. Nevertheless, it is my suspicion that eventually the two
rooms will both come to where the walls and air are at the same
temperature but are both increasing because of the constant input of 13
W of power. When that happens, I would think both would have the same
temperature. This is an idealized situation; in the real world, heat
would also flow out of the closet into the environment but, if
everything is identical, both rooms should eventually be the same —energy
is energy. (Thermodynamics has never been my forté, so I welcome
comments here!)

QUESTION:
I am an Electrician and my Fitter friends have been arguing about the answer to a seemingly simple question.
If I have a solar panel and it is rated to withstand a 1kg steel ball from a height of 1 metre, what size hail stone would it take to break it?

ANSWER:
This is a kind of tricky question because you have to estimate how fast
the hail stone will be going when it hits and that speed will depend on
the size of the hail stone. I will assume that the speed
that it hits is the terminal velocity of a stone of a given mass; the
terminal velocity v _{t} is the maximum speed something
achieves when dropped from a very high height and may be
shown ,
for of a sphere of mass m and radius R , to be
approximately v _{t} ≈√ (9.7m /R ^{2} )
at sea level. The density of ice is about 10^{3} kg/m^{3} ,
so we can write m =4x10^{3} πR ^{3} /3=4.2x10^{3} R ^{3} .
So, the speed of a hail stone of radius R may be approximated
as
v _{t} =√ (4.1x10^{4} R );
for example a golf-ball (R ≈0.021 m) sized hail stone will
have a speed of about 29 m/s and a mass of about 0.039 kg.

Now, we need to make an estimate of the average force the steel ball
exerts on the panel when it is stopping. The force is the linear
momentum mv of the ball divided by the time the collision
lasts; I will assume that any hard ball will stop in about the same
time, so what matters is the momentum. It is easy to show that the speed
of the steel ball when it hits is v =4.5 m/s so mv =4.5
kg·m/s; so we need to find the size of a hail stone with a
momentum of 4.5 kg·m/s. But, we know the speed and mass from
above, so 4.5=(4.2x10^{3} R ^{3} )√ (4.1x10^{4} R )=√ (7.2x10^{11} R ^{7} );
solving, R =0.031 m, m =0.13 kg, v _{t} =36
m/s. So my best estimate would say that golf-ball sized hail would not
break your panel but hail larger than a golf ball by 50% or more would.
Keep in mind that all these calculations are estimates, not precise
calculations.

QUESTION:
referring to this link:
https://www.youtube.com/watch?v=Dt0JyXMjbNs
At 18:48, why the partial differentiation notation can be "taken out" of the bracket?

ANSWER:
The equality of interest here is shown in the figure taken from the
youtube video. It works simply because of the product rule for
differentiation:

(∂/∂x){ψ* (∂ψ /∂x )-(∂ψ* /∂x )ψ}={(∂ψ* /∂x )(∂ψ /∂x )+ψ* (∂^{2} ψ /∂^{2} x )-(∂ψ* /∂x )(∂ψ /∂x )-(∂^{2} ψ* /∂^{2} x )ψ }=ψ* (∂^{2} ψ /∂^{2} x )-(∂^{2} ψ /∂^{2} x )ψ.

QUESTION:
I would like to perform a calculation of a man descending a tower using cords and subjected to the action of the winds. I have been trying to find some equations but it is somewhat difficult due to the drag force. My main objective is to calculate the maximum horizontal distance
x that the man could reach due to the wind action against the technician descending the tower using cords.
Tower height: 78 m (please consider up tower as a zero reference).
wind speed: 20 m/s
Mass of man: 70 kg
Man descending with constant speed and slowly.
So, please what is the maximum distance when the man is at 66 m from the top of the tower ?

ANSWER:
As shown in the figure, there are three forces on the man, his weight
mg , the tension in the cords T ,
and the force of the wind F . The equations of
equilibrium are F-T sin θ =0 and mg-T cosθ =0.
Solving, tanθ =x /y =F /(mg ),
so x=y tanθ =Fy /(mg ). Now, how
can we get F ? There is a very good approximation to the drag
force by air at sea level moving with speed v : F ≈¼Av ^{2}
where A is the area of the object presents to the onrushing air
(and which works only for SI units). Finally, x =Av^{2} y /(4mg ).
If I approximate g ≈10 m/s^{2} and A ≈1
m^{2} and use your numbers, x ≈9 m.

ADDED
COMMENTS:
I should have emphasized that air drag calculations are only
rough calculations, probably accurate to maybe ±20%.
Also, for your situation, the general approximation for
x as a function of y is x≈y /7. Also, if the
horizontal displacement seems too large, keep in mind that
20 m/s is a very strong wind, about 45 mph which is gale
force.

QUESTION:
why a pen in a cup of water appears to bend away from the normal since the refraction of light entering a higher-index medium bends towards the normal?

ANSWER:
Because the light which allows you to see the submerged part of the pen
is going from water to air, entering a lower-index medium.

QUESTION:
I understand why, if the speed of light is the constant from any viewpoint, the variable between two observers has to be their clocks but I don't understand why one is the reference clock and the other is the clock that, in the example below, is the one that is slower. On a train going a meaningful percentage of the speed of light past an observer on earth where both an observer on a train and the observer on the earth can see a light source on the train, both will measure the speed of light at the same rate so their clocks have to be calculating at different rates. My question is, how does one determine which observer is called stationary and which observer is moving?
As a corollary question, if a rocket ship is moving away from the earth what is the difference between the rocket moving away and the earth moving away from the rocket?

ANSWER:
Actually, any inertial frame may be chosen as the rest frame. A's clock runs slow as B measures it and vice versa.

FOLLOWUP QUESTION:
Thank you for your reply. In that example neither person would actually age at a different rate though each would think the others clock was slow?

ANSWER:
You are thinking classically, but there is no such thing as absolute
time. The rate at which any clock which moves relative to you (that's
why they call it relativity) runs more slowly than yours, period. It is
not a case of those clocks "seeming" or "appearing" to run slowly, they
do run slowly. Your statement "…each
would think the others clock was slow…" was wrong —they
were both slow as observed by the other. I think you would understand
the whole thing better if you read my earlier answer on the
twin paradox .

QUESTION:
If wired glass takes 50 lbs of pressure to break and you throw a 1/4 pound
orange at the window from 35 feet away how fast would you have to throw the
orange to break the window?

ANSWER:
There is no way to calculate this with the information given.

FOLLOWUP QUESTION:
Alright. Well the orange is ripened so it is firm. Like fresh from the grocery store naval orange. I am not sure how to quantify that. The orange is the size of a baseball as well. So presuming that the orange is roughly the size and shape of a standard baseball but not quite as soft, we could use a baseball instead of an orange. And i did mean 50 psi not lbs. Would that help take out some of the confounding variables?

ANSWER:
I can only do a very rough estimate. And I certainly cannot replace the
orange with a baseball since the orange is much softer, but I will use
the size of the baseball, about 7.5 cm=0.075 m in diameter. So, I will
assume that the orange compresses by an amount about a quarter of its
diameter, say d ≈2 cm=0.02 m; for simplicity,
assume the orange stops with constant deceleration. If an orange of mass
m =¼ lb=0.11 kg and speed v hits the wall and
stops in a time t , the force it experiences is F=mv /t
and, because of Newton's third law, this will also be the force the
orange exerts on the window. During the collision, there will be some
average area of contact between the orange and the window; I will
estimate that area to be a circle of radius R=0.01 m so that the area
will be about A=πR ^{2} ≈3x10^{-4 } m^{2} .
The pressure will then be P=mv /(At ); converting to SI
units, P =50 psi=3.45x10^{5} N/m^{2} . For the
assumptions I have made (uniform acceleration) it is easily shown that
t= 2d /v , so P=mv ^{2} /(2Ad )=3.45x10^{5} =0.11v ^{2} /(2x3x10^{-4} x0.02);
solving, v =6.1 m/s=13 mph and t =0.0066 s. Keep in mind
that this is a very rough estimate of the speed, but it does indicate
that it is not surprising if the glass broke since almost anybody can
throw an orange with a speed more than 13 mph.

QUESTION:
If i was travelling towards a light pulse in opposite direction to its velocity wouldnt length contraction and time dilation make its velocity even bigger than c, for my length measuring rods will become shorter so i will measure more distance travelled by the light and my clocks will run slower so more time for light to travel ? I get it when the distance between the pulse and observer is increasing but not the other way.

ANSWER:
The very fact that the speed of light is the same for all observers
is the very reason for time dilation and length contraction! So,
the answer is no; the speed of light is the same for all observers.

QUESTION:
Hi. I'm creating an aluminum can crusher and I need to know the weight required in order to crush the can from a set height knowing the amount of Pa required to crush the can.
I'm crushing a regular sized coke (375ml) can and need to get it to about 4mm long once crushed. On a pneumatic crusher I only needed 600000 Pa or 6 Bar to achieve this result. However I am trying to create another method of crushing the can using a weight (unknown) that drops 0.5m on top of the can, theoretically crushing it to the required length.
My question is: what weight in Kg is required to crush a can when dropped 0.5m to reach a crushed can length of 4mm, or a conversion of 6 bar? And maybe the equation used to work it out?

ANSWER:
I looked up the geometry of the can: the height is about 128 mm and the
radius is about 30 mm, so the crush distance is s =128-4=124
mm=0.124 m and the area of the top is about π (0.03)^{2} =2.8x10^{-3}
m^{2} . The force which the press exerted on the can was
F=PA =6x10^{5} x2.8x10^{-3} =1680 N. This means if
you put a mass 1680/9.8=170 kg it will crush the can. Your idea, I
presume, is to drop a smaller mass from some height (0.5 m) above can.
The way I will attack this problem is to first calculate the work done
by the hydraulic press W _{1} and equate it to the work
done by the dropped mass W _{2} ; I can then solve for
the mass for any height. I will do it in general so you can estimate the
mass from any height. W _{1} =Fs =1680x0.124=208
J; v =√(2gh ) where g =9.8 m/s^{2}
and h is the height (0.5 m in your case); W _{2} =½mv ^{2} -mgs =mgh-mgs =208
J. Solving, m =208/[g (h-s )]. In your case,
h =0.5 m and s =0.124 m, m =56 kg.
Keep in mind that this is just an
estimate. I would be curious to know if it was close. There is an
earlier
answer which might be of interest to you; it might give you an idea
of how you might improve your crusher.

QUESTION:
I am a high school math teacher and have a student who asked a question I wasn't prepared to try to answer:
Looking at the twins paradox in relation to special relativity, someone who left their twin behind on Earth while travelling near the speed of light would return to a much older twin while seemingly to have not aged as much themselves.
Is there an equation or way to calculate this relationship in terms of how much one person would age in relation to another given a specific velocity in relation to each other?

ANSWER:
To understand the twin paradox intuitively, I recommend your student
read my earlier answer on
this topic. The equation you can use to compare elapsed time is T _{away} =T _{home} √[1-(v ^{2} /c ^{2} )]
where
T _{home} is the elapsed time for the earth-bound
twin,
T _{away} is the elapsed time for the traveling
twin, v is the speed of the ship, and c is the speed
of light. The easy way to understand this is that the traveling twin
sees the distance to the star shorter by a factor of √[1-(v ^{2} /c ^{2} )].

QUESTION:
I'm reading the book "Aurora" and a question occurred to me: If a spaceship is traveling at 0.1C and is in communication with Earth (via radio waves), and C is relative to the observer, wouldn't the communications be distorted or "out of phase" between a "fixed" point (Earth) and the ship?
Compare that to "Ender's Game", where as people are traveling close to C, they can not communicate with fixed objects.

ANSWER:
I do not know what you mean by "…C is relative to the observer."
The speed of light is the same for all observers. To answer your
question, you should read an
earlier answer where your question is answered for a speed of 0.8c .
Notice that your speculated "distortion" is experienced by both
observers, not just the traveler. It also depends on the direction of
travel —toward or away from the earth. For v =0.1c ,
the effect would be much smaller but qualitatively the same—for
the traveler moving away, both observers find the communication slowed
down, for the traveler moving toward the earth, both observers find the
communication speeded up. Your statement that "…they
can not communicate with fixed objects…" is false; it is just
slowed down or speeded up.

QUESTION:
If a ball impacts a rod off center from the rods center of mass how does
the rotational momentum and energy gained by the rod affect the amount of
gained translational momentum and energy? For example, if the ball impacted
at the rods center of mass the equations for conservation of momentum and
kinetic energy could be used to determine the velocities of the objects
after the collision, but with energy and momentum being transferred to the
rotation of the rod would the ball loss more or less velocity or would the
energy transferred remain the same? Would the rotational energy and momentum
for the objects after the collision be solved for separately from the
translational transfer?

ANSWER:
No homework.

REPLY:
Thank you for the response. I'm unsure if this could be a homework question for someone else, but I've posted in forums with no response or had the post deleted. This is an attempt to gain an understanding of basic physics concepts and I was given this site as a possibility to find answers from a local university. Would it be possible for me
to ask the question again in a couple of months to certify that this is not a homework question or does "no homework" amount to no homework "type" questions that could require tutoring or classes?

ANSWER:
Well, you seem sincere about wanting to learn the physics, so I am happy
to answer your question. The algebra to get the final solution is very
tedious, but it seems you are most interested in getting the physics
right, so I will set the problem up and if you really need all the final
answers for a particular situation, I will leave that to you. Since you
refer to conservation of energy, I assume the problem of interest is an
elastic collision. The system is shown before and after the collision in
the figure. A point mass m with velocity v
approaches a uniform thin rod of mass M and length L ;
v is normal to the rod and the collision
occurs at a point a distance d from the center of mass of the
rod. After the collision, the center of mass of the rod has a velocity
U , the point mass has a velocity u ,
and the rod has an angular velocity ω
about the center mass. Because the rod is uniform, its moment of
inertia about the center of mass is I=ML ^{2} /12; the
angular momentum and kinetic energy of a rigid body are L=I ω
and K =½I ω ^{2}
respectively. There are no external forces or torques on this system, so
angular momentum and linear momentum are both conserved. We now have
three equations:

energy conservation: ½mv ^{2} =½mu ^{2} +½MU ^{2} +ML ^{2} ω ^{2} /24

linear momentum conservation: mv=mu+MU

angular momentum conservation: mvd=mud +ML ^{2} ω /12

Take stock of what we have: three equations and three
unknowns—u , U , and ω.
The physics is done, only algebra remains.

QUESTION:
I would like clarification on force
experienced by a skydiver with air
resistance factored in (not talking
about during opening shock of a
parachute)
Someone at my work stated that you will
experience a number of negative G's
while simply skydiving and with only
air resistance causing the negative G
Force.
Does this seem to be accurate?
My understanding is that air resistance could not cause more than 1G of negative force or else you would technically be going upwards due to gravity only exerting 1 G of force downward while skydiving.

ANSWER:
The air drag can be well approximated as F _{d} =cv ^{2}
where c is a constant (but which depends on the geometry of the
falling object) and v is the speed; this is a force acting in
the opposite direction from the velocity. When you jump from the plane,
you originally have a horizontal velocity but the horizontal component
of the drag will take away the horizontal velocity and you will end up
falling vertically; I will assume the skydiver is falling vertically (or
just jump from a hovering helicopter). In addition to drag, the only
other force acting on the skydiver is his weight W=mg , a force
vertically down. So, writing Newton's second law, ma =cv ^{2} -mg ,
where a is the acceleration. You start out with an acceleration
g down, but as v increases a gets smaller and
smaller until eventually v =v _{t} = √(W /c );
this is called the terminal velocity. Now the acceleration is zero so
you fall with a constant speed v _{t} . At no time was
the drag bigger than the weight. But, can the drag be greater than the
weight? If you were propelled downward from the helicopter with a speed
greater than the terminal velocity, you would slow down until your
reached the terminal velocity. In that case, the drag is greater than
the weight; however, I cannot imagine this scenario in skydiving. By the
way, you can control the terminal velocity—spread eagle will give
a slower final speed than curled up in a ball.

QUESTION:
A ball dropped from the top of the mast of a moving ship falls straight down parallel to the mast. But with the movement of the ship, it also falls down on an inclined LONGER path. In other words, it travels over two different distances at the same time. Which is the actual 'real' distance traveled?

ANSWER:
There is no such thing as "the
actual 'real' distance traveled ". The distance traveled in one
frame of reference will be different than for another frame. For your
question, if the ball is dropped from a mast of height H , it
will travel a distance H in a time t where t = √(2H /g )
where g =9.8 m/s^{2} is the acceleration due to gravity.
Now, look at the path of the ball from a dock the boat is passing; if
the boat moves with a speed v , the distance the boat travels
before the ball hits the deck is vt=v √(2H /g ).
The distance between the starting and ending points of its path (blue
line) is √[(2v ^{2} H /g )+H ^{2} ].
For example, if H =4 m and v =5 m/s, then t =0.9
s and the distance is 6.03 m. The path followed by the ball in the dock
frame is a parabola (red line), so its length is the distance the ball
has traveled; it is rather complicated mathematically to compute this
distance.

ADDED
THOUGHT:
I got curious and calculated the distance the ball actually traveled
in the dock frame. You need to integrate ds = √[(dx )^{2} +(dy )^{2} ]=[√(1+(dy /dx )^{2} )]dx= [√(1+ 4x ^{2} )]dx
from x =0 to x=vt . The result is
s =¼[2vt √(1+4v ^{2} t ^{2} )+sinh^{-1} (2vt )];
for the example above, s =9.19 m.

FOLLOWUP QUESTION:
Still confused, I appreciate your reply. The ball in the first frame of reference hits the deck at the same time as in the second, where it travels the longer parabolic path or distance. These are two observable facts.
To travel the longer distance in the same time in the second frame of reference, the ball must either accelerate faster than 9.8 m/sec^{2} - or time itself must slow down, which is both unlikely.
Your "complicated mathematical computation" supposedly addresses the problem. But can a mathematical computation correct two conflicting observable facts?
I am thinking of Albert Einstein's skeptical quote, "So far as the theories of mathematics are about reality, they are not certain; so far as they are certain, they are not about reality."

ANSWER:
The photograph above is a time-lapse photo of two balls falling. The red
ball is simply dropped. The white ball is given a horizontal velocity.
The only force on each ball is its own weight which points vertically
down and is responsible, of course, for the downward acceleration of the
balls. Since there is no force in the horizontal direction, the white
ball's velocity in the horizontal direction remains constant. The thing
to note is that both balls, having equal vertical accelerations, fall at
exactly equal distances in equal times. This is exactly what is
happening for your question. Your error was the false assumption that
"…the ball must…accelerate
faster than 9.8 m/sec^{2} …"

QUESTION:
I question if there is an antimatter equivalent to the graviton which for now I will call a repelon. The repelon would be a massless, repelling force carrying particle. The antigravity force of antimatter would act the same but opposite of gravity.
If that were the case, then at the creation of matter and antimatter an equal amount of gravity and antigravity would exist. Following subsequent loss of most of the antimatter, the repelon wave would still be prorogating through the universe, as indicated by the detection of gravity waves. If so, could this energy wave account for some or all of the dark energy?
If there is a repelling force associated with antimatter, could the interaction of the gravity and antigravity forces also contribute to the loss of homogeneity in the soup at the big bang?

ANSWER:
A graviton (hypothetical particle which has never been observed and
there is no good theoretical description of) would be a boson and bosons
do not have antiparticles.
Since there is no satisfactory
theory of quantum gravity, I do not want to poo-poo your theory, but
something is causing a repulsive force in the universe and if that field
were quantized there would have to be a quantum of the field, and maybe
a repelon would be a good name for it. I should remind you here that it
is stated on the site that I do not usually do
astronomy/astrophysics/cosmology.

QUESTION:
Why does the sun does not heat up the air as it heat up Earth and Water?

ANSWER:
The air does heat up, but because it is a gas most of the radiation
which strikes it passes through without leaving much energy to heat it.

QUESTION:
when we stand on the surface of earth, every square centimetre of our body experiences atmospheric pressure that is equal to mass of 100 kg. This pressure is sufficient to grind our body, but nothing happens to us. Why?

QUESTION:
As I understand the situation (perhaps incorrectly), if you were to throw a charged baseball, the baseball would radiate owing to the moving charge. The radiation would transport energy and momentum from the charged object into the field. Robbed of energy, it would eventually stop. Here, stop means that it would revert to a radial electric field and display no magnetic field at all. Is that true? If it were true, our physics seems to be very off. This is probably a beginner question, but it is not for some homework. What's the truth here?

ANSWER:
An accelerating charge radiates, one moving with constant velocity does
not. If there is no gravity and no air drag, the ball will not radiate
and not lose energy. But, here is something really interesting: if you
are in a frame which is accelerating relative to the ball, the ball will
radiate in that frame. See an
earlier answer .

QUESTION:
I have attached some images to show the question. In one, a heavy
package is affixed to the drone firmly.
In the other it hangs by a rope that is affixed on the same level as the engines.
The weight of the drone is 5 lb and the weight of the load is 7 lb. Where has the center of gravity moved?

ANSWER:
In the diagram the yellow X s are the centers of gravity
(COG) for the unladen drone and the load; the load COG is a distance
L below the drone COG. The red X is the COG for
the combination. Suppose the weight of the drone is W _{drone}
and the weight of the load is W _{load} . Then the
distance d by which the new COG is below the COG of the drone
is d=LW _{load} /(W _{drone} +W _{load} ).
In your example, d =(7/12)L . Clearly, when L
is increased d is increased. And, contrary to your expectation,
the COG goes farther down when you hang the load from a rope increasing
L . What might be a problem for stability, though, is that if
the drone is not horizontal the rope will still be vertical and so the
COG will no longer be on the center line of the drone.

QUESTION:
I practice medicine in Houston Texas. I am trying to workout a math formula that you are probably very familiar with. It
is called the Schrodinger Equation.
Below is one calculation based on Time dependent Theory.
This would be a calculation for a single PHOTON PARTICLE, whereas this equation is for a wave of Photons?
At any rate, Its all greek to me!!!
I was hoping you help me in calculating the "pattern " of x rays (ionized Photon particles) that pass thru a slit and it calculated projected bands density, shape and size of the bands. (See Ex A below for what I am trying to calculate)
Could you tell me which formula I would most likely use for the best predictable result?
So, If took the Photonic energy wave (lets say...a wave length at 10 nm this is the conversion:
As you can See, I could use Joules, Volts, Hertz ,etc...

Would you mind walking me through this one conversion example using Schrodinger Equation.?
I am not a physicist and this Schrodinger Equation formula and values are Greek to me. But if you could set up the equation with this example at 10 nm wavelength, then I can simply cut and paste and copy and the do all the quantum math I want after that.!!!
Thank you for considering my request.!

EX-A)
Bands formed after wave passes thru Slit(s)
Theoretical Photonic wave

ANSWER:
Something tells me that you are making this way harder than
you need to. It looks to me like all you are really
interested in is the pattern, called a single-slit
diffraction pattern . You apparently found a reference
for someone who did this using the Schrödinger
equation, but you really do not have to go through all that
in order to get an analytical expression for the intensity
pattern on the screen; all you need to do is to use
classical physical optics , a standard first-year physics
calculation. I presume that the first figure, where you have
used some computer code to get the photon energy of
radiation of a particular wavelength is not a mystery to
you; but you do not need this at all for the intensity, you
just need the wavelength λ and the slit
width a . As shown in the figure, you may
specify a point on the screen by specifying either the angle
θ or the distance y ; the results are
I=I _{0} sin^{2} [πa sinθ /λ ]/[πa sinθ /λ ]^{2}
or I=I _{0} sin^{2} [πay /(λD )]/[πay /(λD )]^{2}
where D is the distance from the slit to the
screen; the quantity I _{0} is some constant
which would be determined by the intensity of the radiation
on the slit and the exposure time. The details are shown on
the
Hyperphysics web site. Let's do an example, λ= 5
nm, D= 10 cm, a= 32 nm; then I /I _{0} =sin^{2} [2y ]/[2y ]^{2}
where y is in cm. The result is plotted in the
graph of I /I _{0 } vs. y .

QUESTION:
A bird is just started to fly in a open cage in spring balance system when it just started to fly no change in reading why, the reading should increase, if not, then why?

ANSWER:
This question is different from the on I usually get which is a bird in
a closed box. A good explanation of this kind (closed box) of
problem can be found in an
old answer .
In your question, the answer depends on how the bird takes off. If she
propels herself upward by just using her wings, the reading of the scale
will decrease by the weight of the bird; it's the same as if you simply
lifted her off the perch. If the bird takes off by jumping, that is
pushing off using her legs, the reading will increase and then, when she
is airborne, it will return to the weight of the cage without the weight
of the bird.

QUESTION:
The way I understand accelerating to the speed of light, is that mass increases, so that at the speed of light we would have infinite mass. What happens if we slow to an absolute stop â€“ we stop walking, the earth stops spinning, the earth stops rotating around the sun, the sun stops rotating around the galaxy and the galaxy stops moving through the universe â€“ Would our mass drop to zero and would we cease to exist? Or in other words, do we exist because we are moving?

ANSWER:
In relativity, you can always imagine yourself in a frame in which the
object you are observing is at rest. The mass of something has meaning
only relative to some frame of reference. If the object is at rest in
some frame, it does not have zero mass, it has what is called the rest
mass, m _{0} this is the mass referred to in Newtonian
mechanics and this is what the mass seems to be unless the velocities
are comparable to the speed of light, speeds we never see in everyday
life. In general the mass m depends on its speed v ,
m=m _{0} / √[1-(v /c )^{2} ]
where c is the speed of light.

QUESTION:
If you were to heat up a substance and heat it up to the point where is protons and neutron split into quarks and cool it down,will you get the same elements the substance consisted of or will you get the elements created during the big bang(hydrogen,helium,lithium and beryllium)?

ANSWER:
For starters, you cannot ever have a free quark.
As you heat it up, you cannot know
for sure what will happen in detail; it will depend on how many atoms
you have and the volume in which they are confined. You can be sure that
the first things that happen will be at the atomic level—molecules
dissociate and atoms ionize. But as long as the energies of the atoms
and electrons are not great enough to cause nuclear reactions, the whole
system will be dynamic, constantly changing as atoms capture free
electrons and other atoms get more ionized.
As all this happens, many photons will be generated and absorbed. As
more energy is added, collisions begin to excite and dissociate nuclei
and add more photons to the mix. As more energy is added, exotic and
unstable elementary particles are created and decay. If you now start
taking energy away, can you really imagine that it could possibly go
back to its original state? You would eventually get back to where you
had only neutral atoms, but it is impossible to know what atoms you
would end up with. It would be imperative to know all the details of how
you added and then subtracted the energy, and even then you could not
predict everything that would happen.

QUESTION:
If you are encased in water in a jet and go into a high g turn would you feel any g force?

ANSWER:
I think you will find the answer to this in a recent
answer ; the main difference is that in that example the acceleration
was due speeding up
rather than due to
turning. Essentially, you are accelerating because the jet is
turning and no matter what, there must be a net force on you which
provides that acceleration.

QUESTION:
A steel ball is released just below the surface of thick oil in a
cylinder. How would you describe, qualitatively, the motion of the ball?

ANSWER:
There are three forces on the ball. Its weight down, a buoyant force up, and the frictional drag up. The weight and buoyant force are constant. The drag depends on the speed of the ball, probably
approximately proportional the speed, but certainly increasing with
speed. The ball starts at rest and starts falling down. Therefore, the drag begins increasing which means that the net force down is decreasing, so the magnitude of the acceleration is decreasing. As the speed continues to increase (but at a decreasing rate), eventually the sum of the buoyant and drag forces becomes equal to the weight and the ball will fall with constant speed.

QUESTION:
We know that when a magnet is taken near to iron it gets attracted.
It is because of induction of magnetic field of opposite nature in iron.
Is there no any phenomena that field of some nature is induced in iron so that the iron gets repelled?

ANSWER:
There is no other field which will have the opposite effect on iron.
However, there are some materials where they are repelled by a magnet.
These materials are called diamagnetic and the magnitude of the effect,
compared the attraction for iron, is very small. Some elements which are
diamagnetic are bismuth, copper, and lead. Actually, all atoms are
diamagnetic but in most cases they are smaller than ferromagnetic or
paramagnetic forces.

QUESTION:
I have to understand the classic formula for finding acceleration due to gravity's force.
G x (m x M)/d Squared = m x (9.8 m/s/s)
dividing the mass of the object out . . .
G x M/d Squared = 9.8 m/s/s
Finding acceleration from the mass x the universal constant and the inverse square of the distance to the center is so incredible and useful but still perplexing. My main question is:
How are the units of Meters per second every second; m/s/s obtained when you are dividing Kilograms of mass? maybe the units of big "G" drives the conversion to m/s/s through the units m^{3} kg^{-1} s^{2} ?
This classical formula is very interesting but very challenging to visualize. The numbers all work but an explanation of the resulting units are difficult to find. Please if anybody is able to visualize the logic behind 9.8 meters a second squared and can share, it would be GREATLY appreciated.

ANSWER:
Let's start from the beginning: Newton's law of universal gravitation
states that the force between two point masses m and M separated by a
distance d is proportional to the product of the masses and
inversely proportional to the square of their separation, F ∝Mm /d ^{2} ;
this is also true for two spheres if d is the separation of
their centers. To convert this into an equation, we introduce a
proportionality constant G,
F =GMm /d ^{2} . For a given M ,
m , and d we must measure F to determine G :
G=Fd^{2} / (Mm ). Now, what are the units of
G ? N⋅m^{2} /kg^{2} . This probably answers your
question, but let's finish it off. If M>>m , m will
accelerate toward M with an acceleration a=F /m =GM /d ^{2} , à
la Newton's second law. Since 1 N=1 kg⋅m/s^{2} , the units
of a are [(N⋅m^{2} /kg^{2} )(kg)/m^{2} ]=[(kg⋅m/s^{2} )(m^{2} /kg^{2} )(kg)/m^{2} ]=m/s^{2} .

QUESTION:
A photon of blue light has more energy than a photon of red light. Correct?
Yet, when Herschel divided the sun's light into its colors with a prism and measured the resulting temperature produced by each color in thermometers, he found that red light increased a thermometer's temperature more than blue light.
Blue light has more energy; so, why doesn't blue light make a thermometer warmer than red light? Also, apparently the thermometers were painted black. If that makes a difference.

ANSWER:
The answer is rather prosaic. As shown in the figure, since the red
light is refracted less than the blue light, the region spanned by the
thermometer at the blue end of the spectrum contains a smaller fraction
of the total light than at the red end.

QUESTION:
I take my chocolate Lab to the river to play. The ball-flinger (Chuck-it) imparts backspin to the ball, and water flied off. F=MA - the ball is losing mass; is it accelerating?

ANSWER:
The mass is not disappearing, so the total mass of the system is not
changing. You should actually think of Newton's second law as
F =dp /dt where
p =mv is the linear
momentum, so if there is no net force the momentum does not change.
Then, if you watch the water coming off one drop at a time you can
calculate the new speed of the ball due to the expulsion of the drop;
therefore the ball does accelerate in that sense but not because the
mass is changing but that the momentum stays the same. As a simple
example, let the mass of the ball plus one drop be M and the
mass of the drop be m; the velocity u of the
drop is in the same direction as the original velocity V
of M and the velocity of the ball after the drop has left (now
of mass M-m ) is U . The momentum
before the drop leaves is p _{1} =MV and after
the drop leaves is p _{2} =mu +(M-m )U.
Since p _{1} =p _{2} , the new speed
of the ball is U =(MV-mu )/(M-m ). Notice that
if m is very small compared to M , U ≈V .

QUESTION:
I've been watching Bill Nye's new show on Netflix, and, while I love Bill Nye, I think he just said something that's totally incorrect.
He said "glass blocks infrared light." Sure, glass will hold in a gas or fluid, but he said that it actually blocks infrared light.
That seems impossible to me because it would mean that infrared cameras couldn't have glass lenses, and also because sunlight through a window feels warm on my face.
Does glass block infrared light?

ANSWER:
Actually, what he said is completely true. The graph shows the
transmission of radiation as a function of wavelength through several
thicknesses of a particular type of glass. Visible light has about
0.4-0.75 μm wavelengths and is about 90% transmitted through
the glass. There is a precipitous drop in transmission around 2.5 μm
and the glass becomes nearly opaque to wavelengths above 5 μm. The
region 0.75-2.5 is referred to as near infrared (IR), so the glass does
transmit most of that IR, but the IR radiation which is radiated by
objects with temperatures between the freezing and boiling points of
water is on the order of 10-12 μm. You are right, glass lenses are
useless for infrared optics. Lenses have to be made from a material
which transmits IR; common materials which are used are calcium flouride
and germanium. This is part of the reason that IR optical devices are
expensive. This property of glass is also how greenhouses work. Visible
light enters gets absorbed and heats the inside. But as the inside warms
up, it radiates IR but that radiation cannot get back out. The heating
you feel from light coming through a window results from absorption of
visible and near IR which does get through.

QUESTION:
Assuming a relativistic rocket travelling at 0.95 times the speed of light (c), what would be the drag force on the cross-section area (π500^2) of the ship facing the direction of travel assuming here that drag coefficient is 0.25. The equation for force drag in classical mechanics is:
F _{D} =½ρv ^{2} C _{D} A
where ρ is density of interstellar space; which should be about 2 million protons per m^{3} .
However I am not sure if in relativistic mechanics it is as such:
F _{D} =ρv ^{2} γ ^{2}
where γ is the lorentz factor (gamma factor).
Both of these equations will give different results.
Furthermore, what is the kinetic energy when matter within the vacuum of space impacts on cross-section area (π500^2) of the ship facing the direction of travel?
The equation is as follows:
KE=(γ−1)p c^2
However the results from these equations are about 26.6 N, 0.0028 N and 6.6e−4 J respectively. The latter is not much! However in many articles on the web and as per certain experts, the KE should have been greater and as explosive as nuclear bombs?! Where am I wrong?

ANSWER:
Calculations of air drag are always an approximation. And the equations you are trying to use are certainly not going to be valid under such extreme situations—huge speed and tiny density. Furthermore, when these protons collide with the ship, they will very likely penetrate into the metal unlike the conditions for which equations like you are using are applicable where the gas molecules just bounce off the object. This will cause radiation damage to the ship's hull. Furthermore, at such high speeds interaction with photons will become important, in particular the cosmic microwave background, causing additional drag.
Your relativistic drag equation (which I do not where it came from)
cannot be dimensionally correct.)

QUESTION:
I am trying to figure out the actual force on a RV or Tinyhouse when traveling against a head wind. From what I recall it is not as simple as taking the vehicle speed plus head wind. For example, lets assume you are towing a tiny house at 60 mph against a head wind of 40 mph - one would think the total wind speed/force on the tiny house to be 100 mph.
However, I recall reading that the wind force increases as the speeds of both the trailer (tiny house in this case) and the head wind increase. E.g. 60 mph + 40 mph = 105 mph force but 60 mph + 50 mph head wind could have a force of 120+ mph, not 110.
I could not find info online for this. Is this true and if so is there a formula to work this out?
In the end I am trying to figure out if tiny houses should be built for hurricane force winds. If in fact, the above is correct and towing a tiny house in a head wind can be likened to a tiny house sustaining hurricane winds.

ANSWER:
Calculating air drag forces is an approximate exercise at best. For the
speeds you are talking about, the approximation that the force is
proportional to the square of the speed of the air is probably accurate
enough for your purposes, F ∝v ^{2} .
There is no such thing as "mph force". There is no difference, as far as
air drag is concerned, between having a ground speed (GS) of 60 mph and
a head wind (HW) of 40 mph; and being at rest with a HW of 100 mph; and
having a GS of 100 mph in still air. I think what you are getting at is
that the force does not increase linearly with air speed; so if the
speed increases from 100 mph to 110 mph (an increase by a factor of
1.1), the force increases by a factor of (110/100)^{2} =1.21.
That said, I can now give you a way to estimate the force (very
approximately) on the tiny house. Let A be the area which the
house presents to the wind; then F≈ ¼Av ^{2} ,
which only works for SI units (meters, kilograms, seconds). So, for
example, if v =100 mph=44.7 m/s and A =(4 m)^{2} =16
m^{2} , F ≈¼x44.7^{2} x16≈8000
N≈1800 lb. This would correspond to a pressure of about 0.07 PSI.
(A complicating factor is the presence of the towing vehicle. To some
degree, the tiny house would be shielded from the oncoming wind by the
vehicle. I see no way to estimate this effect because it would be
dependent on the vehicle.)

QUESTION:
What is the weight of impact (kgs) when a 0.5kg handsaw falls 11 floors at a speed on impact of 57mph?

ANSWER:
There is no such thing as "weight of impact". If you mean what is the
force which the saw exerts on whatever it lands on, you need to know the
time it takes it to stop, t . In that case, the average force
F during that time, it is
given by F =mv /t . We
need to convert mph to m/s, 57 mph=25.5 m/s. For example, if t =0.1
s, F =0.5x25.5/0.1=127.5 N;
the mass of a 127.5 weight is 127.5 N/9.8 m/s^{2} =13 kg.

QUESTION:
Please explain me that why if a thin layer of water is spilled on a rough surface like plastered floor and we place our finger in it then why water move away from the point of contact of our finger on that surface and it appear to be dried around.

ANSWER:
This
is probably akin to the similar phenomenon of a foot pressing down on
wet sand and the area close to the foot is visibly dried. I found an
explanation on
Physics Forums which seems to be correct:
"The phenomenon being described is called 'Dilatancy' and was discovered by Reynolds about 100 years ago. It works only when you have well compacted sand that contains just enough water to cover all the individual grains. When you stand on the sand you create a stress / force which causes the sand to move. In order for the sand to move / flow individual grains have to be able to move past one another. Imagine a bunch of oranges stacked as you might see them at a grocers. The first layer has them all tightly arranged and then the second layer sits down into the gaps between the oranges on the first and third layer. Now imagine trying to move one of the oranges in the second layer. In order to move it the oranges on the first and third layer mut move down and up respectively to enable the orange to move. Effectively the volume of the pile of oranges or grains of sand increases with bigger gaps in between. So when you put your foot down on the sand it shoves sand out the way but in doing so the volume in between grains has to increase temporarily to allow the grains to move relative to one another. Consequently all the fluid at the surface is sucked by surface tension into the extra gaps made by the rearrangement of the sand. Since there is no longer any fluid at the surface the grains of sand are now dry."
If you go to the original Physics Forums question, ignore all the
early answers which are wrong.

QUESTION:
In magnetism, how can the magnetic field be used to trap charges inside a magnetic bottle? as this requires deceleration of the moving charge at either ends of the bottle and changing its direction of movement, this will need a work done by the magnetic force although the force will always be perpendicular to the velocity vector of the moving charge from Biot-Savart Law.
Please, if there is an image for clarification of directions of magnetic force and velocity vectors for more explanation.

ANSWER:
In the diagram, note that the forces on the charge are always toward the
center of the bottle. A full explanation can be found at
Physics Stack Exchange .

QUESTION:
If the law of conservation of mass states that matter can't be created or destroyed, then how is it possible in the deep future, after black holes die etc., the universe can be left with literal nothingness?

ANSWER:
There is no such thing as the law of conservation of mass. Mass
can be created or destroyed. In chemistry, conservation of mass is assumed in chemical reactions because the mass changes are so small as to be almost unobservable.

QUESTION:
Some people like to think about what would happen if the universe's constants were changed (see fine-tuning argument for the existence of God). But what if the laws of nature were themselves altered instead? What if, instead of E = mc^2, we had E = mc^3 or E = mc^4? Or what if the law of conservation of mass-energy or the equations governing electromagnetism were altered? Or what if instead of F = ma, we had F = m/a or F = 2ma?

ANSWER:
Your first example, changing the power to which c is raised in
the famous equation E=mc ^{2} , is impossible because for
any power other than 2 the equation is not dimensionally correct —e.g .,
mc ^{3} does not have the units of energy; it would be
like saying "the speed of my car is 55 pounds/foot". Your second example
has two errors. F=m /a could be written as a=m /F
which would mean that the harder you push on something, the less it
would speed up, in violation of what happens. F =2ma would be
perfectly ok but would imply that you have defined what you mean by
force differently. Newton's second law is actually that the acceleration
is proportional to the applied force and inversely proportional to the
mass, a∝F/m or a=kF/m , where
k
is a proportionality constant. Assuming you have already defined what
mass (kg), length (m), and time (s) are, your choice of k
determines what you mean by force. The standard choice is k =1
which means that the unit of force (which we call 1 Newton) is that
force which causes a 1 kg object to have an acceleration of 1 m/s^{2} .
If you were to choose k =½ (your F= 2ma ), one
unit of force would be that force which caused a 1 kg object to have an
acceleration of ½ m/s^{2} .

You need to keep in mind that laws of physics are simply
expressions of how the universe works. For example, Coulomb's law simply states that the force between two
charges is inversely proportional to the square of the
distance between them, F∝ 1/r ^{2}
because this is what we find doing the most accurate
experiments that we can. But, we can not really be sure that
the "true" law is not F∝ 1/r ^{2.000000001}
until we are able to perform an experiment which rules it
out.

QUESTION:
I have two magnets, North poles facing each other (repulsive) Magnet A moves in the positive X direction, Magnet B moves in the negative X direction so that total momentum of system is zero.
The magnets come to a stop do to the repulsion of the same N poles. When they come to a stop, I "pin" them in place so they cant move. Total momentum was zero to begin with, both magnets stopped, total momentum is still zero. The kinetic energy of both magnets rushing towards each other, now stopped, the energy is now in the field between the magnets, if I was to weigh this system, with a sensitive enough scale, I would find the total mass of the system to be: Magnet A + Magnet B + Field so far so good?
Ok, I "unpin" the magnets, they fly apart with equal but opposite momentum, their kinetic energy coming from the energy stored in the field as mass. Is it wrong to say that the energy in the field is transferred directly to the magnets? or, is it more correct to say that "virtual photons" mediated the transfer of energy from the field to the magnets?

ANSWER:
This is a very difficult question even though it looks simple. First, I
think it is a mistake to say that the field has mass. The field has an
energy density U=B ^{2} /(2 μ _{0} )+E ^{2} (ε _{0} /2);
yes, there is an electric field because the magnetic field is changing
in your example. As the two magnets approach each other the fields
change and therefore the energy content of the fields changes also. To
avoid the electrodynamics implied by your question (time varying fields)
and the resulting radiation (carrying energy away) which would occur, I
think we can get to the crux of your question by starting the two
magnets at rest and bringing them together by doing a certain amount of
work W and having them end at rest. If you measure the mass of
the whole system before moving the magnets to be M , its mass
after moving will be M+W /c ^{2} . Again,
thinking of the field as having mass is not the right way to think about
it, you need to just think about the mass of the system. For example, if
a nucleus has N neutrons and Z protons, its mass is less than Nm _{n} +Zm _{p} ;
you would not want to say that the field holding the nucleus together
had negative mass, would you? Your second question, I think you can see,
is not simple either. Because the magnets are moving, there will be an
induced electric field. Because the magnets are accelerated, there will
be electromagnetic radiation carrying energy out of the system. I think
it is best to be thinking always about the whole system and not where
the energy "resides".

QUESTION:
I need to calculate the effort to hold up ONE end of an 80 pound beam to a 40 degree angle. What is the formula?

ANSWER:
I am not sure what you mean by "…the effort to hold up…" I
will assume you mean the force F (see figure)
you need to exert to hold the beam at rest. In general, it depends on
where the center of gravity of the beam is, how you exert the force, and
how long the beam is. The smallest force you would need to exert is a
force straight up, so I will assume that is the case. If you analyze the
equilibrium problem for the beam in the figure, you find that F=Wd /L .
Note that it does not depend on the angle at all. For your case, if the
beam is uniform, i.e. d=L /2, then F=W /2=40 lb.

FOLLOWUP QUESTION:
Re: latest answer. The force required to hold up the end of the beam must involve a cosine of the angle.

ANSWER:
Details: Sum of vertical forces: ∑F _{y} =0=F+N-W ;
sum of torques about point touching the ground: ∑τ =0=Wd cosθ -FL cosθ=Wd-FL.
Solutions: F=Wd /L and N=W (1-(d /L )).
This results from choosing to have F be vertical. I could have
chosen to have F perpendicular to the beam in which case the
answer would depend on θ. (I now see that my
statement in the original answer that a vertical force would be the smallest is
wrong, although my answer for a vertical F was correct and
independent of θ .) The new torque equation would be ∑τ =0=Wd cosθ -FL
so F =Wd cosθ /L and the 80
lb uniform beam at 40^{0} would require a force of F =½(80)(0.77)=30.6
lb. It would require no force to hold the beam vertical since cos90^{0} =0.

QUESTION:
If gravity has the same effect as acceleration (according to the theory of relativity ) gravity curves light, does acceleration also cause a curve ?
And if it why do light particles move in a straight line ?

ANSWER:
The operative principle here is the equivalence principle: there is no
experiment you can do to distinguish between being in a uniform
gravitational field with associated acceleration g and
having an acceleration g in zero gravitational field. Suppose
that you are in an accelerating elevator in empty space which has a
small hole in the side. Someone shines a beam of light into the hole.
That beam will follow a parabolic path as seen by you inside your
elevator. Therefore, light will also be bent by a gravitational field.
There are two answers to your last question. First, if you watch the
photons from your Euclidian frame of reference, the photons do not
follow a straight line in that space. On the other hand, what mass does
is warp the space around it so a photon will follow the shortest path
between two points which you would call a straight line in that space
which is, itself, curved.

QUESTION:
So, I am learning about the basics of waves in class and I was bored so I started messing around with some equations.
λ=h/p p=mv, in the case of light, p=mc, c=λf E=hf c/f = λ c/f = h/p c/f = h/mc h=E/f c/f = E/mcf c=E/mc
E=mc^2
Is this a correct derivation of E=mc^{2} ??

ANSWER:
Well, "messing around" with equations you do not really understand
seldom leads to good results! First, E=mc ^{2} does not
apply to light (photons) because light has no mass and you would
therefore you would conclude that light carries no energy which would be
nonsense; for more detail, see the
faq page . Your
major error, though, is right at the start writing p=mv which
cannot be true for a photon since it has momentum but does not have
mass. In fact, this is not even true for particles with mass since, in
relativity, the momentum is given by p=mv / √[1-(v ^{2} /c ^{2} )].
Incidentally, the energy of a photon is E=hf and the momentum
is p=h /c .

QUESTION:
a light photon moving at the speed of light, can it have spin? and if so can the spin be added to its speed giving it a faster than light speed breaking the laws of physics? I have always wondered this but never seem to find out if they have spin....

ANSWER:
Every photon has a spin of 1; this means that it has an intrinsic spin
angular momentum of L = √[1(1+1)] ℏ.
So, I am guessing that you are visualizing the photon as a
little spinning ball and think that you can conclude (see my
figure) that the "equator" on the near side of the ball is
moving forward with a speed of c+v . However, spin
in quantum mechanics cannot be visualized using such a
simple classical model, even though we often do think of
spin this way to get a qualitative feel for spin.
For example, if you visualize an electron as a spinning
uniform sphere with a reasonable radius, you find that the
surface must have a speed greater than the speed of light.

QUESTION:
Say I fill an airtight barrel with water and have a valve at the bottom and a feeder hose at the top.
If this barrel is uphill and I have the feeder hose down lower say in a pond will the draining of the barrel through the lower valve create enough vacuum to pull the water uphill creating a siphon?

ANSWER:
First of all, I would call what you are proposing a pump, not a siphon.
You are trying to "suck" water uphill using a vacuum. The first thing
that comes to mind is that there will be a limit on how high the hill is
above the water level below. Even if you have a perfect vacuum, the
highest you can lift water this way is 10.3 m=33.9 ft. But you start off
with a hose full of air, so you will never get a vacuum, so you will be
limited further in the height to which you can pull the water from
below. For example, if the volume of the air in the hose were 1/10 of
the volume of the barrel, you could only lift the water 9.3 m. Or, if
the volume of the air in the hose were equal to the volume of the
barrel, you could only lift the water 5.1 m. So there is no simple
answer to your question, but this is probably not a very workable way to
lift water.

QUESTION:
If a person were in a closed container filled with water and the container was accelerated at high speeds, would the person in the container feel the g-forces the same as if they were not in the container?

ANSWER:
You would move backward until you hit the back wall and then the back
wall would exert a force forward on you. If there were no water, the
back wall would exert a force F _{1} =ma _{
} on you where m is your mass. If there were water, the
water would exert a force F _{2} toward the back on you
and the wall would exert a force F _{3} forward on you.
So now, F _{3} -F _{2} =ma so the
force on you from the wall would be bigger (F _{2} +ma )
than if there were no water. Plus the water would be trying to
crush you.

QUESTION
ABOUT THIS ANSWER:
You recently answered a question regarding the effect of acceleration on a person in a closed container of water. You suggested the result would be movement to the back of the container and an increase in the force experienced by the person.
Buoyancy is dependent on relative densities, so a person will float with the same percentage immersed regardless of the local gravity/acceleration. This implies that the victim would, at least at moderate levels of acceleration, be forced to the front of the container.
Initially I thought that if the container was not full, it would be quite a comfortable experience since the accelerating force would be evenly distributed. However, humans are not of uniform density, so the persons chest with its air-filled lungs would be forced to the front and his bony, less buoyant extremities would be dragged the the back. Unfortunately, the questioner filled the container completely so the person would be pressed uncomfortably against the front wall.
At higher acceleration the body would be compressed sufficiently that he would become denser than the water, only then would he move to the back of the container to be further crushed by the water column.

ANSWER:
You are right, the motion depends on the density of the astronaut
relative to the density of the water. If the ship is in empty space (no
gravity) and not accelerating, there would be no buoyant force in any
direction and the astronaut would float either at rest or at constant
velocity until he hit something. The equivalence principle says that
there is no experiment you can do to distinguish between being in a
uniform gravitational field with associated acceleration g and
having an acceleration g in zero gravitational field. If the
ship had an acceleration a forward it would be the same as being in a
gravitational field pointing backward in the ship. Therefore, there
would be a buoyant force which would cause objects with smaller density
than water to move forward ("float") and objects with larger density
than water to move backward ("sink"); you correctly point that out and
in my original answer I was assuming an astronaut whose overall density
is larger (quite possible if he were wearing a heavy space suit, for
example). In the back wall case, my original answer was correct. In the
front wall case, the water would be pushing you forward and the wall
backward, so F _{2} -F _{3} =ma
or F _{2} =F _{3} +ma ; now the
water pushes on you with a force greater than the wall pushes back.

QUESTION:
Most physicist say that an antimatter engine is impossible because there is no way to store the antimatter but if you were able to get raw positrons and electrons and store them in seperate electrically charged tanks (each with the same charge as the particle they are holding). would you be able to sucessfully store the matter and antimatter and decide when to open it and close it. So basically is the atraction between matter and antimatter stronger than the electroweak force?

ANSWER:
If you have a hollow conducting tank and charge it up with electrical
charge, the field inside is zero, so your plan to store charge inside
will definitely not work; the only force felt by the charges inside will
be the forces between those charges which will have the effect of
pushing them all out to the tank. All the positrons in the positron tank
would annihilate with electrons in the tank; all the electrons in the
electron tank would end up on the outer surface of the tank. And, by the
way, the only force between an electron and a positron is electrostatic.

QUESTION:
In the part of electromotive force, when we connect a wire to the terminals of the source of EMF like for example a battery, the potential difference between the two terminals create an electric field inside the conductor.
Haw fast this field been established, is it simultaneous just at the time of closing the circuit or it takes time even if it is extremely small amount of time?

ANSWER:
In a vacuum, an electric field propogates at the speed of light. In the wire
it will be a bit slower, but still close to light speed. The electric
field causes conduction electrons to move, hence creating the current.

QUESTION:
I have been studying atoms and strong and the electroweak force for about a week and i came across something that caught my attention. When a radioactive isotope undergoes beta decay one of its neutrons ejects an electron and an antineutrino and leaves behind a proton. my question is if it leaves behind a proton and jettisions an electron is there any chance that the nucleus with one extra proton than electrons in orbit around the nucleus could catch the loose electron?

ANSWER:
Why not just ask if you can combine an electron and a proton to make a
neutron? The answer is yes, and it is called inverse beta decay .
If a proton absorbs an electron, an electron neutrino will be ejected:
p ^{+} +e ^{-} →n ^{0} +ν _{e} .
This most commonly happens when an atomic electron is captured into the
nucleus, a process called electron capture. This process also results in
the emission of an x-ray because the hole in the K-shell is filled by a
higher-orbital electron. It is also the main way
neutron
stars are formed.

Q&A
OF THE WEEK,

3/19-25/2017

QUESTION:
Having a bit of a debate about whether this tennis ball would've landed in with a tennis player and we have a $100 bet on it. The ball machine fed the ball from the other side of the court at the baseline the player that hit the ball is a top ranked junior player...
the ball hit the ball machine edge 5 inches off the ground at the top of the wheel base and the player claims that it would've landed on the line if it had not hit the ball machine of which the picture demonstrates the point of impact is 5 inches above ground at the back edge of the line and there are no external elements such as wind as we are playing indoor.
the ball was traveling at approximately 30 mph at time of impact. We have attached an image for reference. We appreciate any clarity you could provide :-)
[Note that the angle relative to the horizontal is specified to be 60^{0} -70^{0}
in the photograph attached by the questioner.]

ANSWER:
My first reaction was to say that, of course, it would not hit the line.
That was because, as physicists often do, I was thinking of the ball as
a point and ignoring its size. You can see from the figure that there
could easily be a combination of h and v _{0}
where the ball would strike the line had the obstruction not been there.
As best as I could tell, some part of the ball must touch the line so if
we calculate where the bottom-most point of the ball strikes the ground
(y =0) the ball will be in if x >0 in my coordinate
system. My guess is that if we simply assumed that the bottom point went
in a straight line in the direction of its initial velocity we would get
the right answer; but the ball is actually a projectile and moves in a
parabolic path so, since this is a $100 bet, I better do it right! The
equations of motion are

x (t )=x _{0} +v _{0x} t

y (t )=y _{0} +v _{0y} t- ½gt ^{2}

where t is the time it takes to hit the ground and g =9.8
m/s^{2} is the acceleration due to gravity. Being a scientist, I
prefer to work in SI units, so I will do that. OK, let's summarize
everything we know. I will assume θ =70^{0} since
that gives the best chance of hitting.

x _{0} =R =2.7"=0.0686 m

y _{0} =h-R =5"-2.7"=2.3"=0.0584 m

v _{0x} =-v _{0} cosθ=- 30cos70^{0}
mph=-4.583 m/s

v _{0y} =-v _{0} sinθ=- 12.59
m/s

y (t )=0.

So now the task is to put these into the equations above, solve the
y equation for t and put that value of t into the
x equation to find x . I find that t =0.00463 s
and x =0.047 m=1.9". Because x is positive, the ball
will hit the line. Going through the same procedure for θ =60^{0} ,
I find t =0.00502 s and x =1.4", again hitting the line.
When the bet is settled, don't forget to
reward The
Physicist !

ADDED
THOUGHT:
The curvature of the parabolic path is, as I had speculated,
miniscule. For θ =70^{0} above
I found x =0.0474 m and if you just assume the ball
went in a straight line to the ground you find that x =0.0473
m.

FOLLOWUP
QUESTION: For clarity sake I am sending you
two more pictures from the exact set up of which we have not
moved. Previously you only were sent the Sideview so I have
included below sent a top view showing that the ball machine
is actually out and the court view from the approximate
angle that the ball was struck showing the approximate
ballpark at the ball was hit from approximately contact was
made 3 feet off of the ground. I still can't wrap my head
around if an apparatus is actually out and a ball strikes it
from 5 inches above the court going in the opposite
direction that it can still be in unless air resistance is
involved of which there are no air elements. Just want to
make sure that you still think the ball Woodland in from
more data provided before I pay $100?

ANSWER:
The top view is helpful in
verifying what I had assumed in the original answer â€”the
surface which is hit is aligned with the outer edge of the
line. The best way that I can convince you that it is
possible that my first answer is correct is to show two
figures, one showing a trajectory of the ball which lands it
in bounds and one out:

Each figure shows the ball at the
instant of impact with the machine and the instant of impact
with the floor had the machine not been there. The dashed
red line shows the trajectory of the center of the ball and
the solid red line shows the trajectory of the bottom of the
ball. The diameter of a tennis ball is 5.4". The ball with
the steeper trajectory (which would include your 60 ^{0} -70 ^{0} trajectory)
is clearly in bounds by standard tennis rules.

FOLLOWUP QUESTION:
We did not know how to define the arc so we just said 60 70 as an ignorant guesstimate but we are ok with the actual real representation of the arc that we sent you in the last image...we just want to understand based on the ACTUAL real flight path arc that the ball really took when struck and its possibility of landing on the line of which the image with the actual art provided is a more accurate visual representation... and we are not sure of how to define the angle of approach based on that visual.
Our original explanation of the ball angle degree of approach to the baseline could have been lost in translation somewhat we are not certain... that's why we contacted you and that's why I sent a VISUAL representation because we really aren't sure of how to define that...we reckon the more information we provide you the more accurately u can clarify our understanding.
In saying that.... what approximate degree is the arc approaching the baseline from the image we provided this morning... assuming the ball was struck at about 3 feet off the ground traveling approximately 3 4 feet over the net? The distance from the net to the baseline is 39 feet and the player was approx 3 feet behind the baseline at time of contact so that would be approx 42 feet from the net at the time the contact was made. Does this change whether or not the ball lands on the line or is it irrelevant?

ANSWER:
This Q&A is turning into quite a tome! With the information you sent, I
can calculate a pretty good estimate of the trajectory of the ball
ignoring air drag. It turns out that your guess of 60^{0} -70^{0}
for the trajectory angle was way off. The algebra is tedious, so I will
just give the final results. To check that I made no algebra errors, I
plotted the trajectory, shown in the figure. I have worked in meters.
The ball is hit at x =0 from a height of 1 m, passes 1 m above
the 1 m high net at x =13 m, and then hits the machine at x =25
m just above the ground (y =0 m). You can see that my result
describes the path you specified quite well. Do not be deceived by the
picture, though, because the scales of the two axes are very different,
only 2.5 m for vertical motion compared to horizontal motion of 25 m, a
factor of 10. The inset shows the trajectory as it would look to the
eye. As you can see, the angle is much smaller than you estimated. The
analytical solution is that θ =15.6^{0 } and t =1.1 s; your estimate of the initial speed was pretty
good, 23.5 m/s=52.6 mph but the final speed (although it is not
important) is 23.9 m/s=53.5 mph not 30 mph.

Now I am compelled to recalculate with the best possible numbers whether
the ball hits in bounds or not. However, given all that I learned above,
I can make it brief: there will be a critical angle
θ _{c} =tan^{-1} (y _{0} /x _{0} )=tan^{-1} (2.3/2.7)=40.4^{0} ;
any angle smaller than θ _{c } will be out of
bounds. For 15.6^{0} , x =2.7-(2.3/tan15.6)=-5.5", 5.5
inches out of bounds.

Finally, just for completeness, I would like to estimate the effect of
air drag. Using the approximation in an
earlier answer I find that t =1.13
s (0.03 s longer) and v =22.1 m/s (1.8 m/s slower). This would
correspond to the angle being a bit bigger, about 17^{0} , but not nearly
enough to cause the ball to drop in bounds.

Q&A OF THE WEEK,

3/12-18/2017

QUESTION:
How much lateral force is needed to damage bearings in the hub
motors of a skateboard when carving? So...if you roll in a straight
line, on a skateboard, you exert a radial load on the eight bearings
contained in the four wheels. But skating is more fun when you ride in
wavy lines - carving! So the bearings start getting a lateral or axial
load. How "hard" would the carve have to be (let's assume a rider of 100kg) to
break the weakest of the bearings? The
rotor is supported by two bearings, one that has an radial maximum force of 3.5
kN and the other of 7 kN. So the maximum axial forces would half those.

(The questioner and I had several
exchanges. The bearings are cylindrical and he was only guessing that
the axial force was half the radial force based on data for similar
bearings. I have edited his several emails to get the gist of things in
the question above.)

ANSWER:
The way I see the problem is shown to the right. The forces on the
skater plus skateboard are his weight mg vertically down, the
normal forces on the inner (N _{2} ) and outer (N _{1} )
wheels, and the corresponding frictional forces f _{1}
and f _{2} . Whatever the rated axial (along the wheel
axis) maximum force is, that is what we want to use as the to find the
limiting conditions for the "carve"; the determining factors will be the
mass m of the skater plus board, the speed v he is
going, and the radius R of the path. For the lean, d
and h are the distances, respectively, of the center of mass
horizontally and vertically from the front wheels; note tanθ =d /h .
The wheel base is s . The normal forces and frictional forces
shown each represent the forces on two wheels. Note that the inner
wheels carry the most force.
The easiest way to do this problem, since it is an accelerating system,
is to introduce a fititious centrifugal force C =mv ^{2} /R
pointing outward. Newton's equations are N _{1} +N _{2} -mg =0
(vertical forces), f _{1} +f _{2} -C =0
(horizontal forces), and N _{1} s+mgd-Ch =0
(torque about inner wheels). Now, from the torque equation, there will
be a maximum speed you can go for a given m and R
before the outer wheels leave the ground. At this time N _{1} =0=(Ch-mgd )/s
or v ^{2} /(gR )=d /h =tanθ.
Also, f _{1} =0, so f _{2} =C=mv ^{2} /R .
Now, the inner two wheels each are experiencing the axial force of F =½mv ^{2} /R .
Just to do an example, let v =10 mph=4.47 m/s, R =5 m,
and m =100 kg; then F =0.2 kN. If your guess that F _{max} ≈1.75
kN is correct, you should be ok. For this example, θ =tan^{-1} [v ^{2} /(gR )]=22^{0} . If you execute the turn with
a smaller lean angle, all four wheels will share part of the load. To do
the general solution where both wheels have N ≠0 would
not be two difficult but would be quite a bit messier algebraically and
probably not all that useful.

FOLLOWUP QUESTION:
Thanks for the explanation on your website. It wasn't really what I was looking for because it doesn't really give the answer something layman can understand.
Also, you compare your estimated lateral force of 1.96 kN
(0.2 kN, see below) to 1.75 kN, although they would be the breaking limit for only one bearing.
You assumed that just the 2 inner wheels receive the load, so isn't that a total of 4 bearings... shouldn't the breaking limit be 4*1.75?

ANSWER:
I am glad you asked this question because it got me looking more
carefully at what I had done late last night and I found an extraneous
factor of g had crept into the final stage of my calculation, ( " …so f _{2} =C=mg v ^{2} /R …")
so my final answer was too big by a factor of 9.8 meaning that the
lateral force per wheel is F= 0.2 kN; this has been corrected in the
original answer. So you should have no problem after all. I was in the
process of writing an email trying to "verbalize" my answer somewhat to
make it more accessible when I discovered this. Here is what I wrote:

You asked me "How 'hard' would the carve have to be (let's assume a rider of 100kg) to break the weakest of the bearings?" That is what I gave you. I assumed that the weaker of two bearings in a wheel can break even if the stronger does not.
(I assume they are coaxial, one inside the other.) Sorry if the explanation was too technical, but that is as basic as I can get to convey the details. A few comments to try to clarify:

What I call
C , the centrifugal force, determines how "hard" the carve is.
C=mv ^{2} /R . For the example I did, C =100x4.47^{2} /5=400 N.

When carving, the harder you carve the greater the load will be carried by the inner wheels on each axel. Eventually, the outer wheels will just lift off the ground which necessarily results in the inner wheels taking all the load, both lateral and radial.
Since you asked me for how to calculate the maximum lateral force
any wheel would experience for a given m , v , and
R , that is what I gave you.

I do not
understand your last question but I do know that if two wheels
experience a force of 0.4 N then each experiences a force of 0.2 N
(assuming they equally share the load).

ADDED
THOUGHT: I guess I have not really fully answered
your question
"How 'hard' would the carve have to
be…to break the weakest of the bearings?" I suggest
that the appropriate equation would be 1.75x10^{3} =½mv ^{2} /R.
Here are a couple of examples:

What is the fastest a 100 kg rider could go in a curve
of radius 10 m? v =√(2x10x1.75x10^{3} /100)=18.7
m/s=67 km/hr=42 mph.

What is the tightest curve a 100 kg rider could turn at
30 mph=48 km/hr=13.4 m/s? R =½x13.4^{2} x100/1.75x10^{3} =5.1
m.

These, of course, assume 1.75 kN is the strength of the
weakest bearing.

QUESTION:
In a black hole what becomes of the system's angular momentum?

ANSWER:
Black holes have angular momentum. Any object which has angular
momentum, will add its angular momentum to the black hole's when it is
captured, conserving angular momentum.

QUESTION:
If the earth is spinning at 1040mph to the EAST, why do trips from LA to NY, take the same amount of time in either direction, give or take 30min?
If the plane is traveling at 500mph, wouldn't flight times be significantly different from LA to NY vs NY to LA?

ANSWER:
The atmosphere rotates with the earth, to first approximation. The
airplane flies relative to the air and therefore takes the same time in
either direction because it is flying in still air. In reality, at the
altitudes where commercial airliners fly, there is a strong air current
called the jet stream which moves, relative to the ground, in a west to
east direction. Therefore the trip from NY to LA is usually longer.

QUESTION:
Since a perfect black body would absorb all electromagnetic radiation would it mean that it would also have a perceived temperature of absolute zero and wouldn't it also mean that eventually the temperature of whatever the material the black body itself is made up of would eventually reach the planck temp since it's always absorbing radiation and never releasing it?

ANSWER:
A black body is also a radiator. If it is in a radiation field it will
absorb all radiation striking it and, as it absorbs this energy, it will
increase in temperature. As it gets hotter it will increase its own
radiation. Eventually it will come into thermal equilibrium with its
environment, absorbing and radiating at the same rates.

QUESTION:
Good Day ! Thank you for being there and dedicating the time to answer our questions. I have many... Can I start with the "Inverse Square Law" as it pertains to light. E = I/r2 , Using 100 as "S", S/4pi2 = I ( intensity at surface of sphere ) as the source strength, the farthest we would be able to "see" would not even get us to PC. it is 4.2 light years away, 5,878,499,817 x 4.2 = 24,689,699,231.40 miles... Assuming 100% at "S". Reaching the surface of the earth, using a nominal, 8,000 mile diameter, that works out to be about 1 (one) photon for every 1,205. square miles.... And, that is purported to be our closest star. How can this be ?

ANSWER:
I generally do not try to find errors in questioners' calculations, I
just do the calculation myself. The sun is a pretty average star and its
photon
rate is about Φ ≈10^{45} photons per
second. Using R =4.2 ly=4x10^{16} m, I =Φ /(4πR ^{2} )= 10^{45} /( 64π x10^{32} )≈ 5x10^{10}
photons/s/m^{2} . That would be the photon intensity 4.2 ly from
a sun-like star.

QUESTION:
If you would like to send a space probe to an asteroid that is 7 AU away, how many years would the journey take using the minimum energy direct trajectory?

ANSWER:
You seem to think that energy must be expended to keep it moving. In
fact, once you give it a velocity greater than the escape velocity, it
will coast the whole way. The minimum velocity you must give it is just
slightly less than the escape velocity, but for the purposes of this
answer, let's just say that we give it the escape velocity which is
about 10^{4 } m/s. I find that the speed at r =7 AU would
be about 28 m/s and the time to get there would be about 1.5x10^{12}
s≈48,000 years. So the energy required would be that amount needed
to give your probe the escape velocity from earth's surface. Keep in
mind that this estimate ignores everything except the earth. It would be
a better question to ask, for example, how much energy would be needed
for the probe to arrive in 10 years.

QUESTION:
I'm hoping this is not considered off the wall but our moon does not have an atmosphere. And our moon and earth are spinning (quite fast) and also our solar system is spinning while traveling through space. so my question would be how those men weren't blown clear off the moon when they landed on the moon?

ANSWER:
Where did you get the idea that the moon is "spinning quite fast"? The
moon spins once on its axis approximately every 28 days so that it
always keeps the same side pointed toward the earth. The earth, of
course, rotates once every 24 hours. You can easily calculate the
centrifugal force on something on the equator of the moon or earth, that
force which I presume you are assuming "would blow [us] clear off". On
earth that force is less than ½% or your weight; on the
moon that force is 2.5x10^{-16} % of your moon weight! So, you
see, in no way do you have to worry about being "blown clear off"!

QUESTION:
I am a Middle school science teacher and we do an unit on cell phones as part of our NGSS "waves" unit.
The performance task is to build a device that blocks a cell phone signal. The kids quickly figure out (from the unit lessons and the internet) that aluminum foil is the best solution.
My Question(s): What exactly is happening with the aluminum foil? everything I read says the waves are being absorbed, are they creating a Faraday cage, or just creating a barrier that blocks the waves?
I have had students create "Faraday Cages" (with copper wire, following internet plans) and it did not work. What exactly is a Faraday Cage and how does it work?
My teaching partner noticed a correlation between when the phone is touching the box, and when it is insulated from the box (with a cell phone case). Is there something to this?
This is the second year of this project and I am in awe at the opportunities for learning for myself and the kids, thanks for your time.

ANSWER:
Do your students know what an electromagnetic wave is composed of? There
are oscillating electric and magnetic fields as shown in the figure. The
important thing is that there are electric fields and electric fields
cannot penetrate into a hollow conductor (the Faraday cage). To
understand how the Faraday cage works, look at the animation. When an
electric field is applied the electrons experience a force opposite the
field and migrate to one side of the cage while the other side has a net
positive charge because of the missing electrons. Amazingly, the
electrons arrange themselves in just the right way that their electric
field is exactly opposite what the external electric field inside would
be, giving a net field of zero inside. The waves are not so much
absorbed as they are cancelled out. A Faraday cage does not have
to be completely closed if you want to keep radio waves out as long as
the size of the holes is very small compared to the wavelength of the
radiation; it could be made of chicken wire, for example, as long as the
wavelength is much greater than a couple of inches. From what I can find
out the frequencies are in the not too far from f =1 GHz=10^{9}
s^{-1} ; since the wavelength of a wave is the speed divided by
the frequency, λ=c /f =3x10^{8} /10^{9} =0.3
m. So, I would have thought that a cage you made would have gaps much
less than 30 cm. Were all the wires in good contact with all others?
Often copper wires have a varnish-like coating on them to serve as
insulation. I don't know what you mean by "correlation", so I cannot
answer that part of your question. (By the way, the aluminum foil is
a Farady box if it completely surrounds the phone.)

QUESTION:
How much heat would it take to heat 1 gallon of water to 600 deg F in a pressurized system, from 70 deg F to 600 F in 1 hour. Not counting the ss vessel.
Also since the water is not allowed to change states are the calculations just the Sensible heat cals or are there special calculations needed.
This is part of a R&D Application.

ANSWER:
For my approximation to be fairly accurate, the water must
remain liquid at a constant volume. I will work in SI units
so 70^{0} F=21^{0} C and 600^{0} F=315^{0} C;
I will convert back to Imperial units for the final answer.
I looked up
data for the specific heat of water which turns out to
have a significant temperature dependence as shown in the
figure (black). I did a quadratic fit (green) to these data
and integrated over the temperature range to get E =1356
kJ/kg. The mass of a gallon of water is about 3.8 kg so the
total heat is Q =1356x3.8=5.2x10^{3} kJ=1.44
kW⋅hr=1240 kilocalories. Keep in mind that the pressure
will be very large at 600^{0} F, about 1800 psi.

QUESTION:
I have been reading a lot lately about the prospects of nuclear fusion reactors; especially regarding electrical power plants. My question is this: if the hot plasma from the reaction is shielded from the equipment to protect it from the heat by a magnetic bottle, how then do you access the heat to make electrical power via steam for turbine generators or direct conversion from heat to electricity.

ANSWER:
The purpose of the magnetic bottle is to keep the hot plasma isolated
from the physical containment; just touching the containment vessel
would present two problems —it would cause instabilities in
the plasma and it would damage the vessel. But the plasma is hot and
anything hot will radiate heat energy; in other words, the magnet bottle
contains the plasma, not the heat. This radiant heat would rapidly heat
up the vessel which would heat up some coolant mechanism (imagine having
water tubes coiled around the outside of the vessel) which would carry
away the heat to drive the turbines.

QUESTION:
We could induce artificial gravity through centripetal acceleration. For example a ring-like structure in a spaceship could rotate about 1.34 rpm if the radius to the centre is 500 meters. This will give 1 g at the edges of the ring.
However we can also induce artificial gravity in the spaceship through its propulsive power and hence a constant acceleration at 1 g of the spaceship is required.
But what if the spaceship is accelerated at more than 1 g in order to achieve 75% the speed of light, even though the spaceship consists of the 500m radius ring-like structure that rotates at 1.34 rpm. What type of artificial gravity will the occupants feel? Will the excessive acceleration cancel out the centripetal effects in the ring? Is there an equation or formula that combine both linear acceleration of an object while the objects is also rotating?

ANSWER:
You seem to think that you need an acceleration greater than
g to get to 0.75c , but any acceleration will do. But,
since you seem interested in a greater than g acceleration, I
will choose a _{0} =1.5g. So now a person of
mass m in the ship sees two fictitious forces, mg
which is radially out and 1.5mg which is
toward the rear of the ship. A person in the ship would experience a net
force of about 1.8mg pointing out and back. This would not be a
comfortable situation. If you are just going to accelerate
constantly, I would recommend not messing with the rotation
at all, set

your
acceleration to g , and make all the "floors" in the ship
perpendicular to the acceleration. If you are interested in how long it
would take to reach 0.75c with a _{0} =g ,
you can read off the graph (which I took from an
earlier answer ) that gt /c =1.8,
so t =5.5x10^{7}
s=1.7 yr.

QUESTION:
A worker was pushing a pallet forklift and when he put the metal against the metal at the bottom of a glass door - the glass shattered to pieces. He just barely touched the metal on the bottom of the door. What could have happened to make it shatter like that?

ANSWER:
Glass sometimes breaks under circumstances where it does not seem that
enough force has been applied. As you probably know, glass is formed in
very high temperatures and allowed to cool. Sometimes the cooling does
not occur uniformly over the whole volume of the glass and the finally
cooled object can have places where there are very large internal
stresses. Just a small force at such a location can then cause the glass
to break. I cannot verify that this was the case for your situation
because it is a little hard to imagine a forklift having "just barely
touched" anything!

QUESTION:
Is it true that the electric energy of one electron that is moved through a potential difference of 1 volt is called electron volt?

ANSWER:
You are on the right track, but your terminology is a little shaky (it
is not clear what "electric energy" means). There are two ways you might
want to define the electron volt (eV). First, if an electron at rest is
allowed to accelerate across a potential difference of 1 V, it acquires
a kinetic energy of 1 eV. Second, if you push an electron across a
potential difference of 1 V (opposite the direction it wants to go), you
will do 1 eV of work. 1 eV=1.6x10^{-19} J. You need to realize
that if the potential difference is greater than a few hundred volts,
the kinetic energy will not be exactly ½mv ^{2}
because of relativistic effects.

QUESTION:
why are sparks more likely to occur between two charged particles closer together rather than far.

ANSWER:
For air to become a conductor, there must be a sufficiently strong
electric field to ionize the air molecules. There must therefore be a
potential difference (voltage) between the two electrodes. The air has a
dielectric breakdown strength of about 30 kV/cm which means that 30,000
V are needed for a gap of 1 cm but only 3000 V is needed for a gap of
1mm.

QUESTION:
If I drove my car home with the boot lid open, why would the air resistance be higher than usual?

ANSWER:
It probably would be higher, but not necessarily. If it were higher, it
would be due to the fact that the area presented to the onrushing air
was larger and the drag is proportional to the area. On the other hand,
aerodynamics can be very nonintuitive. The spoiler on some cars is
designed to break up the smooth flow of air over the car which actually
results in lower drag at high speeds. The dimples on golf balls and the
hairs on tennis balls have the same purpose, to break up smooth air
flow.

Here is an anectdote which illustrates that your intuition is not always right regarding drag. Some years ago somebody called in to
Car Talk on NPR and asked about these nets you can buy to replace the tailgate in a pickup truck to reduce air drag. Makes sense, right? The tailgate is like a wall in the wind and to get rid of it will reduce drag and increase your mileage. Click and Clack said that they thought these things were a great idea for reducing drag and increasing fuel efficiency. During the intervening week before the next show an engineer from GM called in and told them that removing the tailgate in fact greatly increases the overall drag on the truck. The reason is that the tailgate traps a bubble of air which rides along with the truck and the headwind slips over it. There was a lot of crow-eating that week at Car Talk Plaza!

QUESTION:
Does temperature affect a magnet's magnetism?

ANSWER:
Magnetism is a vast field and the temperature dependence depends on the
material. I will just address a simplified description of a simple
ferromagnet, a common permanent magnet. The basic way that these work is
that electrons interact with their neighbors in nearby atoms such that
their magnetic moments align with each other giving the bulk material a
net magnetization. The most important temperature dependence is that,
for any ferromagnetic material, there is a critical temperature called
the Curie temperature where the magnetization disappears. So, if you
want to destroy a permanent magnet, heat it up above the Curie
temperature; unless you cool it in an exteranal magnetic field, it will
not be a very good magnet when cooled. The behavior below the Curie
temperature is likely to be complicated, but there will be generally a
decrease in magnetization as it heats up.

QUESTION:
Photons don't have mass, but they have momentum. So this means that if a photon hit a mirror, for example, it pushes the mirror a little bit forward. Thus, the photon transferred momentum to the mirror. So this would mean that the photon lost a fraction of its speed or its mass decreased because p=m*v! But how could this be if photons are massless and only can travel at the speed of light?

ANSWER:
The linear momentum of a photon is not p=mv (since it could not
have any momentum if m =0), but rather p=hf /c
where h is Planck's constant, c is the speed of light,
and f is the frequency the corresponding light wave. In
everyday life, the light which reflects from a mirror is the same color
of light that went in; this means that the mirror did not recoil at all,
essentially has infinite mass. If you take into account the recoil of
the mirror, some of the photon's momentum would be transferred to the
mirror which would mean the photon would have to lose some momentum. But
h and c are both constants of nature, so f
would have to decrease; that would mean that the actual color of the
photon would be slightly shifted toward the red end of the spectrum. You
could never do this experiment, the shift is too tiny. But, if you
"reflect" a photon from an electron (shoot gamma rays or x-rays at a
solid which has many electrons in it) you can easily measure the change
in wavelength of the radiation; this is called
Compton
scattering and was one of the pivotal experiments in the development
of quantum physics.

QUESTION:
is the terminal velocity of a matchstick falling from 1000 metres the same as a 3000lb car falling from the same height.

ANSWER:
The terminal velocity is the largest speed something will acquire if it
could fall forever. So it has nothing to do with how far it falls; a
feather may acquire 99.999% of its terminal velocity after falling one
inch but a bowling ball might have to fall a mile before it reached
99.999% of its terminal velocity. This problem you can do just using
your common sense (although I will do a rough estimate below). Something
as light as a matchstick will not fall far at all before it is moving
with nearly its terminal velocity; if you had a terminal velocity as low
as a match's, you could jump off the Empire State Building and not get
killed. The car would go much faster.

This may be all that you want, but for the interested I will
roughly calculate some numbers. The terminal velocity v
of something with mass m which presents an area
A to the onrushing air can be roughly approximated as
v =2√(mg /A ) where g ≈10
m/s^{2} is the acceleration due to gravity. The
match I estimate to have m ≈0.1 grams=10^{-4}
kg and area A ≈(5 cm)x(1 mm)=5x10^{-5}
m^{2} ; then v ≈9 m/s. The car I
estimate to have m ≈3000 lb≈1400 kg and
area A ≈(2 m)x(3 m)=6 m^{2} ; then
v ≈97 m/s.

QUESTION:
I was playing with my nephew and his toy cars and came to notice that when pushed, after a short distance, the cars begin to veer off course. Some simple "testing" determined this to be a (seemingly) random effect. This was not really new to me, but curious nephew eyes then began to ask me "why this is so". I assume it to be a mix of many effects arising due to the (lacking) production quality of the cars, but I didn't find a really satisfying answer. Perhaps you could give a "best guess" as to which effect is the main contributor?

ANSWER:
Suppose there is some small force which pushes the car left or right at
a rate of 0.4 ft /s/s and there is some larger force which causes
it to slow down at the rate of 3 ft/s/s; now push the car so it has an
initial speed 10 ft/s. The path followed by the car until it stops will
resemble the path shown in the figure. That is my best guess!

QUESTION:
In space like sci fi dogfights (gundam or macross) is it possible to accelerate from a stationary (from the pilots perspective like a ship) object straight forward (thinking in 4 axis) then if accelerate directly down from your original vector then accelerate again in a new vector does the orignal speed change or do only the amount of gravitational energy change? Also what would it do to the person inside the craft? If your already doing 500mph and you change directions like I said down at another 500mph then again at 500mph does it add up or is it like launching from a craft already doing 500mph where the energy felt going foward is 0 from their perspective? Like when a astronaut inside the ISS goes out for a space walk? Do they feel the pressures when they wear those jetpacks (sorry for childish name) and change directions? Again what type of math would I study to learn the answer to these types of questions? and hopefully solve them some day because space has space for everyone and I want people to one day go to the stars ALIVE!

ANSWER:
I
am afraid that your question is quite muddled, but I think it shows that
you do not really understand what acceleration is. Average acceleration
is defined to be the change in your velocity (the difference between the
final and initial velocities)
divided by the time to affect the change. If I understand your example
you are originally moving with speed v _{initial} =500
mph in one direction and later are moving with speed v _{final} =500
mph in a direction perpendicular to your original direction of travel;
it is important that you realize that velocity is a vector so that it
changes if the direction changes but the speed stays the same. The
difference between the two velocities is shown in the figure, Δv =v _{final} -v _{initial} .
So the average acceleration is a =Δv /Δt
where Δt is the time to make the maneuver. The average
force F you will experience is in the
direction of a and proportional to your mass
m , F =ma . For
example, suppose your weight is 160 lb (then your mass is m =160/32=5 lb⋅s^{2} /ft), Δv= 500√2=707
mph=1037 ft/s, and Δt= 1 s. The average force you will
feel is F =5x1037/1=5185 lb! You do not want to change your velocity too
quickly!

QUESTION:
I'm just curious: is there any geometric structure that could be built-up infinitely without collapsing under it's own gravity? A tetrahedral mesh, like diamond maybe?

ANSWER:
Let's refrain from talking about building up "infinitely" since
there is not an infinite amount of energy in the entire universe. The
gravitational field g of a point mass m
is g =-mG 1 _{r} /r ^{2}
where G is the universal gravitational constant, 1 _{r}
is a unit vector pointing radially out and r is the distance
away from the point. To get the gravitational field for an object not a
point you need to divide it up into infinitesmal pieces, each having a
field dg =-dmG 1 _{r'} /r' ^{2}
where r' is the vector from dm to the place where you
wish to calculate the field; you then integrate over the entire object,
usually not a trivial exercise. This is one of the main reasons that
Newton had to invent the calculus, so that he could prove that he could
treat the sun and the planets (spheres) as point masses. I understand
that the difficulty of proving that the field of a spherically symmetric
mass distribution outside the object is identical to the field there if
all the mass were in a point at the center caused a delay of something
like 20 years in the publication of his theory of gravity. I will only
talk about spherically symmetric masses here because, the point mass
field being spherically symmetric, all large bodies will tend toward a
spherical shape. This is why the stars and planets are spheres; see an
earlier answer on cylindrical masses. Calculation inside the object
is trickier: a uniform sphere of radius R and mass M
has zero field at the center which increases linearly until the surface
where it has magnitude g=MG /R ^{2} ; a hollow
sphere of radius R and mass M has zero field
everywhere inside and then jumps discontinuously to g=MG /R ^{2}
at the surface. Since we know that a star with a large enough mass will
collapse to a neutron star and/or a black hole, nothing will stop that
from happening, certainly no "mesh" of any kind. On the other hand, if
you want to create a hollow sphere with a thin outer shell, I believe
that there would be no limit to how large you could make it without its
collapsing. Suppose you have a hollow sphere of radius R _{1}
and and mass M _{1} and it does not collapse; the field
pushing in at the surface is g _{1} =GM _{1} /R _{1} ^{2} .
Now suppose you have a hollow sphere of radius R _{2}
and and mass M _{2} =2M _{1} ; it is easy
to show that if you keep the surface density of the shell σ=M /(4πR ^{2} )
constant then R _{2} =(√2)R _{1}
and so g _{2} =GM _{2} /R _{2} ^{2} =2GM _{1} /(2R _{1} ^{2} )=g _{1} .
The field remains the same and so the field per unit area actually gets
smaller as the shell gets larger, making it even less prone to collapse.

QUESTION:
What's the physics of orbiting? Why don't satellites lose their velocity over time and fall straight to the earth's surface as they're orbiting (falling) around the planet?

ANSWER:
Orbital mechanics is just applications of Newton's laws of
motion: the motion is determined by all the forces on an object. Start
with the simplest assumptions: there are only to bodies and the mass of
one body is hugely greater than the other and the only force is gravity.
This is what is called the Kepler problem. See an
earlier answer for the several possibilities of orbits but the
important thing is that a stable orbit is the solution to Newton's
equations and it simply continues forever until some other force changes
it. Next you take into account the masses of both objects and find that
the solutions are essentially the same except that the two orbit around
their center of mass and the reduced mass μ=Mm /(M+m )
replaces the mass M in your equations. Now you can add other
non-ideal forces to the problem. For example, although earth satellites
are above most of the earth's atmosphere, they are not above all of it
and the air drag with the very thin atmosphere gradually takes energy
away from the satellite and its orbit eventually decays to where it hits
the ground. The farther out the satellite is, the slower this decay is.
Other things can cause the orbit to change also. For example, the moon
causes tides on the earth (in both the oceans and the solid earth) which
causes some of the energy lost by the earth to be gained by the moon so
it is gradually moving farther away (by like about 4 cm a year). And of
course any third body perturbs the nature of the orbit. For example, the
planet Neptune was discovered in 1846 because the observed orbit of
Uranus was not exactly what its Kepler orbit should have been because of
interaction with another body, Neptune; calculations predicted the
necessary orbit of the unknown planet and it was subsequently observed
where it was predicted to be. Imagine doing such a complicated
calculation long before computers.

QUESTION:
Would a stove-top pot heat liquid faster if the bottom surface was concave instead of flat? My theory is the increased surface area on the liquid in the pot would cause it to heat faster, and heat would travel 'up' the concave 'cone' area so it would not lose any heat (or proportianally not a lot) compared with if that area was flat and closer to the flame.

ANSWER:
Such a pot would definitely not be best for an electric range.
Here you want the element to be in contact with the metal pot bottom so
that conduction is the main way of getting heat into the pot; with your
pot convection would be the mechanism and result is smaller heat flow.
For a gas range, it is possible that some advantage could result as you
speculate, but I doubt it. The hottest point of a flame is in the
visible part of the flame that you see, and you want this to be close to
the pot.

QUESTION:
If traveling on an airplane from North Pole to Brazil, the pilot cannot fly on a straight line or he will miss his target; the pilot must actually fly on curved path to reach his desired destination. What is the reason for this?

ANSWER:
This is probably a homework question, forbidden here, but I am
going to answer it because of its ambiguity. The problem with this
question is that it does not define what is meant by a curved path. Of
course, since we live on the surface of a sphere, no path between any
two points on the surface of the sphere is a straight line in the three
dimensional space from which we might view that path. Only by digging a
tunnel between the two points is the path a straight line. But if you
travel through this straight tunnel and your path is observed from
outside the earth you follow a curved path because the earth is
rotating. How do we usually define a straight line? It is the shortest
path between two points. So if you observe things from our point of view
on earth and define a straight line on this two-dimensional surface to
be the shortest distance between two points, you can connect any two
points on earth by a straight line. If the pilot flies due south on the
longitude 43.2^{0} W, he will end up in Rio de Janeiro. The
pilot does not have to keep steering his plane to account for the
earth's rotation (which is what is hinted at here, I think); he flies
relative to the air and the air rotates with the earth. If this is a
homework question, it is a pretty stupid one.

Q&A OF THE WEEK,
2/18-25/2017

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.

ADDED
NOTE:
It is important to note that the acceleration observed depends
on the inertial frame you are in. In way#2 above, two
different inertial frames, say with v /c =0.7
and 0.9, will observe the ship having different
accelerations. This is why acceleration, and therefore
also force, in special relativity do not play an
important role as they do in Newtonian physics where all
inertial observers see the same acceleration. This is
discussed in the
earlier
answer .

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 3x10^{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≈7 14 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 doesnâ€™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 â€”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:
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 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 .

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.