QUESTION:
Let's say you have three 20-foot putts. Same stimp meter (pace on the green). Let's say in all cases, the green is flat after the hole. I.e. don't worry about things like the ball running away or anything.

Putt 1 travels flat and then shortly before the hole, it rises 6 inches.

Putt 2 has an rise of 6 inches right after you hit it, and then travels flat to the hole the rest of the way.

Putt 3 rises 6 inches gently from the ball to the hole at a constant angle.

Do you hit all three putts the same initial speed?
Follow up question : What if you have the same 20 foot putt but halfway to the hole the ground rises 8 inches, and then drops 2 inches and then flattens out the rest of the way to the hole. . .same speed?

ANSWER:
Refer to the figure.
If there were no friction, it would make no difference how the rise
occurred, only the amount of rise. In that case v _{hole} = √(v _{putt} ^{2} -2gh ).
However, there is friction which will slow the ball further and the
frictional force on a slope (f=μmg cosθ ) will
be different than on level ground (f=μmg ); here μ
is the coefficient of friction (essentially, I believe, what is
measured by the "stimp meter"), m is the mass of the ball, and
g is the acceleration due to gravity. Since the frictional
force is smaller on the slope, you might think that less energy is lost
to friction on the slope. But, what matters is not the force but its
product with the distance s over which it acts, specifically
the work done by friction is W _{f} =-fs where
the negative sign indicates that the friction takes energy away from the
ball. When the ball is moving forward along along the segment L _{2} ,
the distance it travels is s=L _{2} /cosθ
and so the work done is W _{f} =fs=(μmg cosθ )(L _{2} /cosθ )=μmgL _{2} ,
exactly the same as if the slope was not there. The total work done by
friction, regardless of the path, is W _{f} =μmgL.
The final velocity is then
v _{hole} = √[v _{putt} ^{2} -2g (h+μL )].

QUESTION:
Hello I have been wondering why basketball shoes have tread. I get why shoes for outdoor activities do but wouldn't a basketball shoe work with no tread like race car tires.

ANSWER:
If the floor is always perfectly dry, there is no reason. However,
because of players sweating there is always a chance of hitting a wet
spot. Treads on a road tire channel water away to prevent hydroplaning
and I assume that would be the case for shoe treads as well. Also,
increasing surface area does not increase traction because the
frictional force for dry friction depends only how hard the shoe is
being pressed to the floor (your weight) and the coefficient of friction
between wood and rubber. Dragsters have "bald" tires because they are
made of a material designed to melt when the heat up, essentially gluing
the tire to the road for super friction.

QUESTION:
I'm coordinating a piece of action for a tvc (tv ad), which requires me to build a rig to raise an actor 3 m off the ground, as if he is being taken away by aliens! It's been a long time since school for me, however I feel that there would be a formula that would assist me in calculating how much counterweight is required on a pulley system with 2:1 MA to raise an 80kg actor 3m in 1.5 seconds?
Are you able to clarify the principals to be used in calculating this problem?
This is definitely not homework!

ANSWER:
The actor will experience an accelleration a such that he moves a
distance y in a time t : y= ½at^{2}
, so for this case a =2y /t ^{2} =2x3/(1.5)=2.7
m/s^{2} . Choosing the man plus pulley above him as the body
(assuming the mass of the pulley is negligible), Newton's second law is
2T-mg =ma or T =½m (g+a )=½x80x(9.8+2.7)=500
N where g =9.8 m/s^{2} is the acceleration due to
gravity. Finally sum forces for the weight Mg which has the
same magnitude of acceleration as the actor: Mg-T=Ma , so
M=T /(g-a )=500/(9.8-2.7)=70.4 kg. You may want to keep in
mind that the speed of the actor at the end of the 1.5 s is v=at =2.7x1.5=4.05
m/s; if the weight is stopped after the 1.5 s, he will continue moving
upwards until he has risen another h =½v ^{2} /g =0.84
m and then fall back.

QUESTION:
I've been trying to figure out how to weigh my boat on its trailer, and it seemed to me that putting a scale under each of the two wheels and the tongue and adding the three weights together might not yield a correct result, but your
table weight answer , makes me think maybe it could be.

ANSWER:
Assuming that you do not have three scales which could be all engaged at
once, lifting one wheel at a time will introduce errors. Refer to the figure above
(noting that I have made the boat on the trailer invisible!): I have
notated 2S as the distance between the wheels, L the
distance from the axle to the tongue support, q the distance
from the ground to the center of mass (COM
© ) of the boat plus trailer, d
the distance of COM forward of the axle, and W the weight of
the boat plus trailer. The normal forces N _{i}
are the forces up on each of the three points of contact; if an ideal
scale of zero thickness were placed under a wheel, the magnitude of the
appropriate normal force is what it would read and N _{1} +N _{2} +N _{3} =W. I have also assumed that the
COM is centered laterally. The geometry and algebra are a bit tedious,
so I give only final answers. In each case I imagine lifting one point
of contact by the angle θ and then measuring the normal
force which I will call N' _{i} . N' _{3}
is the easiest since it does not depend on θ , N' _{3} =Wd /L .
The other two are, by symmetry, the same: N' _{1} =N' _{2} = ½W [1-(d /L )-(½q /S )tanθ ].
If you now compute the sum, N' _{1} +N' _{2} +N' _{3} ≡W'=W [1-(q /S )tanθ ].
So your measurement of the weight will be wrong. As an example, suppose
that S =3 ft and q =5 ft and you lift a wheel h =4
in=1/3 ft off the ground; then tanθ =(1/3)/(2x3)=1/18 and
the measured weight will be W (1-(5/3)/18)=0.91W ,
nearly a 10% error. The culprit here is that the center of gravity is so
far off the ground. To get a more accurate measurement, block the
opposite wheel so that it is also h above the ground (the
tongue support does not need to be blocked); you only have to do this
once since N _{1} =N _{2 } and N _{3}
does not depend on its height.

FOLLOWUP QUESTION:
Also, in your table weight example , wouldn't elevating one end of the table with a 2x4 and putting a scale beneath it shift (rotate) the table's weight (CoG) somewhat toward its other end and lessen the measured weight of the weighed end, and error be particularly noticeable if the span of the table is short and the scale is several inches high? If I could remember my trig I could pose this question more intelligently, but the extreme, if hypothetical, case would be if the table end were elevated all the way to vertical, where it would weigh nothing.

ANSWER:
Congratulations, you have caught one of The Physicist's rare
errors! I must admit that I was thinking small angle here and that is
the only case where it is a good approximation to say that elevating one
end of the table does not change the normal force. And, again, I did not
really appreciate the importance of the fact that the COM is not
colinear with the other two forces until I started puzzling over your
boat trailer question. You can see an example of a case where this is
not an issue in an earlier question about a
boat dock . The correct
answer to the table weight
example would be W' =W [1-(q /S )tanθ ].
(I have changed the notation and assumptions from original problem just
to make the solution more concise. In particular, I have assumed that
the COM is a distance q above the floor and equidistant from
the ends and sides; I have neglected the weight w of the 2x4,
and ½W=W _{1} =W _{2} ; S
is the length of the table.) So, if q =3 ft, S =10 ft,
and h =1/3 ft (COM above floor), W'=W [1-(3/10)(1/3)/10]=0.91W .

QUESTION:
If an evacuated tube encircled the earth at sea level, at what speed
would an object within it have to travel to attain free fall, i.e.
orbital velocity? I m a 66 yr old retired mailman, just musing.

ANSWER:
The radius of the earth is about R =6.37x10^{6} m. An
object which moves in a circle of that radius with a speed v
requires a force pointing toward the center of the circle of F=mv ^{2} /R
where m is its mass. But your object would have its weight
W=mg as the only force acting on it where g =9.8 m/s^{2}
is the acceleration due to gravity. Therefore, mg =mv ^{2} /R ;
solving for v , v = √(gR )=√(9.8x6.37x10^{6} )=7901
m/s=17,674 mph.

Here is an even simpler way to figure it out. An object with speed v
moving in a circle of radius R has an acceleration of v ^{2} /R ,
the direction of which is toward the center of the circle. But an object
near the earth's surface is g , toward the center of the earth.
Therefore,
v = √(gR ).

QUESTION:
how fast would a 6000 lb SUV have to be moving in a head-on collision with
a 3200 lb Sedan to stop the sedan's forward motion and move it backward 75
feet?
I realize there are a lot of variables. The suv was estimated travelling at 40mph. The sedan at 20mph. Presumably the suv driver had applied the breaks just before impact. But the cars dod not stick together. The steel bumper of the suv was bent into a vee shape and the eitire front of the sedan crumpled to the engine block, which pressed in on the firewall but did not penetrate. The road is asphalt. Was slightly wet from a misting rain. Suv came to rest 8 or so feet from sedan. Sedan driver was not able to apply breaks, but transmission remained in gear.
I was hoping for a formula or an estimate, not a precise measurement.

ANSWER:
Keep in mind that this will be a very rough
calculation. I assumed that that the initial speed of the sedan was
u =20 mph and the initial speed v of the SUV was unknown. I
assumed that both cars slide after the collision. The coefficient of
kinetic friction μ for rubber on wet asphalt can be anything
from 0.25 to 0.75 so I did the calculation for both extremes. I first
estimated the speed V immediately after the collision for each
car using their masses M , the distance they slid S : ½MV ^{2} -μMgS= 0,
or V = √(2μgS ).
Next I applied momentum conservation to the collision. M_{SUV} v-M _{sedan} u=M _{SUV} V _{SUV} +M _{sedan} V _{sedan}
and solved for v v= [M _{SUV} V _{SUV} +M _{sedan} (V _{sedan} +u )]/M _{SUV} .
The range of speeds for the SUV before the collision is 42-63 mph. A
word of warning: if you want to do calculations of your own, do not try
to work in units of mph, lb, etc, change to SI units (meter, kilogram,
second) and then change your final answer back to mph.

QUESTION:
I would like to know the formula to calculate the final speed of a
mass sliding down a steep slope and hitting a less steep slope to the bottom.
Obviously I cannot calculate the speed on each slope separately and
add them. Here are the numbers.

First slope : 1.721 Meter high, 35 degrees, friction coefficient
0.14.

Second slope : 2.536 Meter high, 25 degrees, friction
coefficient
0.14.

Obviously the final speed will not be the total of both speeds, so how
do I add the speed of the first slope to the second slope to obtain
final velocity?

ANSWER:
Well, I started to work this out and found that the length of the two
runs are exactly 3 m and 6 m. I therefore surmise that this is
homework and this site is not for homework help.

FOLLOWUP QUESTION:
Hahaha, thank you for answering. At 51, I am surely not going back to school :) In fact the lenght I gave you are not precisely the one I will use for the slides. I want to understand the way to make the right calculations because the first incline has to be steeper to reduce the overall length of the slide. I am a conceptor mostly in transportation but this project is for a waterpark.

ANSWER:
It is easiest to do this using energy concepts. Total energy of a mass
m moving with speed v and at a height h above
some arbitrarily chosen h =0 level is E = ½mv ^{2} +mgh ;
g =9.8 m/s^{2} is the acceleration due to gravity. For a
given system the energy is conserved (the same everywhere) as long as
there is no agent adding or subtracting energy from the system. If there
were no friction in your slides, energy would be conserved. so, if you
chose h =0 at the bottom and started from the top with v =0,
the initial energy would be E _{1} =mgh and the
final energy would be E _{2} =½mv ^{2} ;
setting them equal and solving for v , v =√(2gh );
note that the mass does not matter. In your case, v =√(2x9.8x(1.721+2.536))=9.13
m/s.

Alas, there is friction and the
frictional force f for a mass m sliding on an incline
θ is f=μmg cosθ. The amount
which a force changes the energy if it acts over a distance s
is called the work done and the work done by friction is W _{f} =-μmgs ·cosθ ;
the minus sign indicates that friction takes energy away from the
system. Now, instead of writing E _{2} =E _{1} ,
we write E _{2} =E _{1} +W —the
energy you end up with equals the energy you started with plus what you
added (negative for the case of friction).

Now we can address your specific
case. ½m v^{2} =m gh-μm gs_{1} cosθ _{1} -μm gs _{2} cosθ _{2} =9.8x[(1.721+2.536)-0.14( 3cos35^{0} +6cos25^{0} )]=31.6
m^{2} /s^{2} . Solving for the velocity, v =7.95
m/s.

The formula you seek if you are
doing a slide with two different slopes is v =√{2g [h-μ (s _{1} cosθ _{1} +s _{2} cosθ _{2} )]}.

QUESTION:
How strong is 0.01 newton meters of
torque? I want to build a motorized
camera slider that will pull a Canon 5d mk 2 up a 45 degree slope. It weighs
1.5 kg. How much torque do I need?
T his
is the motor I am thinking of buying:
Rated power 0.01 W Rated speed:20 rpm Rated torque:0.01 N·m

ANSWER:
I will work it out in general and then we will see if this motor can do
the job. There are some things you have not told me, in particular the
speed v you want the camera to move on the slope and the nature
of how the camera moves on the surface (in particular coefficient of
friction μ ). But the general solution will have those
in there and you can calculate with them. You will want to attach a
spool of some sort the the drive shaft of the motor to act as a reel to
pull up the string attached to the camera, say its radius is R .
The angular velocity of this motor is ω =20 rpmx(2π
rad/1 rev)x(1 min/60 s)=2.1 s^{-1} . The weight of the camera is
mg =(1.5 kg)x(9.8 m/s^{2} )=14.7 N. The angle of the
incline is stipulated to be 45^{0} . The force necessary to pull
the camera up the ramp with constant speed may be shown to be F=mg (1+μ )/√2;
since the torque τ is the
known quantity, FR=τ = mg R (1+μ )/√2 .
In order for the camera to move at speed v , the radius of the
spool should be R=v / ω.
If the speed of the camera is v and the force is F , the
power being generated in the motor is P=Fv =mgv (1+μ )/√2.
Let's do an example. Suppose you want the camera to move with a speed of
v =1 cm/s=0.01 m/s and the friction is negligible, μ ≈0.
Then the spool has a radius R =0.01/2.1=0.0048 m=0.48 cm; the
required torque would be
τ =0.0048x14.7/√2=0.05
N·m; the required power would be P =14.7x0.01/√2=0.1
W. I am afraid that your proposed motor is inadequate.

QUESTION:
If I am in a car at a stop. My vehicle can accelerate from 0-60 in
8.5 seconds. I am going up a hill at a 20% grade. How fast would I
be going at
400 feet? (Given that I am fully accelerating.) This isn't homework (I am 48 years old!).
I am trying to prove that I couldn't have been going as fast as the police officer thought I was, based on his years of experience.

ANSWER:
This is a good example of how simple textbook physics sometimes fails
to explain real-world situations. At first I calculated an average force needed to accelerate
the car on level ground to 60 mph in 8.5 s and then applied that force
to the car on a 20^{0} incline; that force was not even big
enough to get the car started! In fact a car is geared and the
assumption that the car's acceleration is uniform is incorrect and there
are therefore too many variables to be able to solve this problem.

Q&A OF THE
WEEK (1/29-2/4/2017)

QUESTION:
I was asked by my son if you get more speed on a flat (fixed slope) skate ramp or a curved ramp (1/4 pipe) of equal height (12ft). I think the flat ramp would result in greater speed as there is no change in energy force and the fixed straight ramp is constant, the 1/4 pipe is vertical and needs to change direction to horizontal resulting in directional change and loss of energy

ANSWER:
Your reference to "change in energy
force" has no meaning in physics. If friction is neglected, the only
thing which matters is the vertical distance fallen and so the speed at
the bottom would be identically the same regardless of the shape of the
ramp. Refer to the figures on the left. Without friction, the speed at
the bottom is v = √(2gh ) where h is the
height. Including friction complicates things.

The ramp is easiest to do: the three forces on
the skater are his weight mg , the normal
force N , and the frictional force
f . The magnitude of the frictional force
is f= μN where μ is the
coefficient of friction. Newton's equations are ΣF _{y} =N-mg cosφ =0
and ΣF _{x} =mg sinφ -μN =ma ;
the solutions are N=mg cosφ and a=g (sinφ -μ cosφ ). Now,
notice that the acceleration depends on the angle of the incline
which means that the speed at the bottom will also depend on the
angle, the steeper the faster. Therefore it is not really fair to
compare the flat ramp with the quarter pipe because the quarter pipe
will always have its horizontal distance equal to its vertical fall.
So the only fair comparison is for the flat ramp to have φ= 45^{0} .
In that case, a=g (1-μ )/√2. I calculate that
the speed at the bottom is v _{flat_ramp} =√(2gh (1-μ )).

For the quarter pipe, the situation is much harder to calculate
because the normal force gets larger as the skater falls thereby
increasing the friction. Newton's equations are ΣF _{x} =mg cosθ -μN =ma
and ΣF _{y} =N-mg sinθ =mv ^{2} /R
where a is the tangential acceleration and R
is the radius of the pipe. I have no idea how to solve these
because they are coupled, second-order, nonlinear differential
equations. I am guessing that there is not a closed-form
analytical solution and the equations would have to be solved
numerically. For this case, though, we are interested in cases
where the friction is low, μ<< 1. We can calculate the
velocity v as a function of θ for the
no-friction case, μ =0 and then use that v to
approximate the normal force as a function of theta and use that
to find the energy lost to friction: v (θ )≈√(2gR sinθ ),
N ≈3mg sinθ , f ≈3μmg sinθ.
Using these, I find v _{¼pipe} ≈√(2gh (1-3μ)).

So, to do a
numerical example, R=h =12 ft, μ= 0.1, g =32
ft/s^{2} , v _{flat_ramp} =26.3 ft/s, v _{¼pipe} ≈23.2
ft/s. (v =27.7 ft/s for zero friction.)

ADDED
COMMENT:
The person who suggested approximating the velocity as the zero-friction
velocity for the ¼ pipe performed numerical solutions for the
differential equations.

His post : "Out of curiosity, I simulated the setup. With μ =0.001 and
m=R=g =1 I got E =0.99701, in agreement with your result.
μ =0.01 leads to
E =0.9704. Even with μ =0.1, the approximation is not bad:
E =0.732. The mass stops at the center for μ >0.60. Tested with 100, 200 and 500 steps: The critical value is somewhere between 0.603 and 0.605. There is no obvious mathematical constant in that range."

Interpreting, his E is
the calculated value of the kinetic energy divided by mgh. For
this quantity, my approximation was [½mv ^{2} /(mgh )]=(1-3μ ).
Comparing the computed values with the approximated values: 0.99701≈0.997
for μ= 0.001, 0.9704≈0.97 for μ= 0.01, and
0.732≈0.7 for μ= 0.1. Thanks to
mfb for
his interest and calculations!

QUESTION:
I have just discovered your fascinating website and my attention was caught by the question of the jumbo
jet on the conveyor belt. I believe that your answer is incorrect.
If an aircraft was like a car, relying on the wheels to push the vehicle forward it would indeed, never gain any airspeed. However, the wheels are not driven, the accelerating
force is applied by the engines throwing tonnes of gas rearwards at high velocity… so long as the brakes are off it doesn't matter whether the conveyor belt moves backwards, forwards or not at all. The aircraft will accelerate and reach take-off speed.

ANSWER:
I see your point entirely.
Although we envision the wheels spinning and the conveyor moving
backwards, there is another possible scenario. The original question
states that the "…conveyor belt is designed to exactly match the
speed of the wheels, moving in the opposite direction." To my mind, this
can mean that the wheels are not moving and the conveyor is not moving.
The figure shows the forces on the plane in blue and the forces on the
conveyor in red. The plane is being pushed forward by the thrust
T . Suppose that the motor running the conveyor is
pulling with a force F =-T and
the conveyor does not move. The conveyor will exert a frictional force
on the plane of f =- T
and, because of Newton's third law, the plane will exert a
force on the conveyor belt of -f = T .
Forces on both the plane and the conveyor are in equilibrium and nothing
moves. It is just the same as if the brakes were locked.

QUESTION:
Say there was no cross-wind and you threw the ball directly up to the same height as a height-altitude aeroplane, when the ball falls back down will it fall into your hand again? Will it not be displaced?

ANSWER:
If you neglect air drag and the altitude is low enough that g
(acceleration due to gravity) can be considered constant, it will land
west of the point where you threw it by a distance d =(4 ωv ^{3} cosθ )/(3g ^{2} )
where v is the initial speed, ω is the angular
velocity of the earth, and θ is the latitude where you
threw it.

QUESTION:
A person pushing hard a huge rock but the rock does not move is the work is done or not if yes then give reason

ANSWER:
See an earlier answer .

QUESTION:
How fast does a 26000 pound truck have to be moving in order to slide 100 ft. on wet roads?

ANSWER:
It makes no difference what the weight is. The initial speed if the
distance traveled while sliding is s is given by v = √(2μgs )
where g =32 ft/s is the acceleration due to gravity and μ is
the coefficient of kinetic friction. Rubber on wet asphalt has a range
of possible μ =0.25-0.75 with a corresponding range of speeds
v =27-47 mph. Rubber on wet concrete has a range of possible μ =0.45-0.75
with a corresponding range of speeds v =37-47 mph.

QUESTION:
I was told that when a car is moving on a circular banked road. If the car reaches a very high velocity then it may move up the slope of the banked road! Is this possible and if so how?

ANSWER:
The
figure shows the car on the track. Three forces on it are the weight
W , the normal force N from the road, and the
frictional force f
from the road. Imagine that it is at rest so that all the forces are in
equilibrium: ΣF _{y} =-W+N cosθ -f sinθ =0
and ΣF _{x} =N sinθ -f cosθ =0.
The solutions are f=W sinθ and N=W cosθ. Now, if the car has some speed v , the car is still in
equilibrium in the y direction, but now has a centripetal acceleration
in the +x -direction of a=v ^{2} /R where R is
the radius of the track. So now, ΣF _{y} =-W+N cosθ -f sinθ =0
and ΣF _{x} =N sinθ -f cosθ =mv ^{2} /R.
Solving these equations, f=m (g sinθ-v ^{2} cosθ /R ).
Now, notice that if v _{0} =√(gR tanθ ), f =0; at
this speed the car can go around the track even if it is perfectly
slippery. For speeds greater than v _{0} , f will be
negative which means it must point down the slope. As the speed
increases beyond v _{0} , the magnitude of the frictional
force will get bigger and bigger down the incline; but, there is a
maximum value which f can be, f _{max} =μN
where μ is the coefficient of static friction. The
corresponding velocity may be
shown to be v _{max} =√[gR (sinθ +μ cosθ )/(cosθ -μ sinθ )].

Not
every angle of incline or every μ will result in the car
slipping, however. Whenever (cosθ -μ sinθ )]=0,
v _{max} =∞ and the car will not slip no matter how
fast it goes. For any given μ there will be a minimum angle θ _{min} = tan^{-1} (1/μ )
beyond which slipping will never occur; for example, for μ= 0.5,
θ _{min} = tan^{-1} (2)=63.4^{0} .
The graph shows the behavior of v _{max} as a function
of angle for several values of μ.

If you want a way to understand
this qualitatively and you have worked with fictitious forces in
accelerated frames, you can look in the frame of the car and there will
be a centrifugal force pointing in the -x direction; this force
will have a magnitude of mv ^{2} /R and, if it is greater than N sinθ -f cosθ=N (sinθ -μ cosθ )
the car will move up the incline.

QUESTION:
There's an e-skate (electric skateboard) that I'm looking at that says it has 7 Nm of constant torque
and the radius of the wheel is 40 mm=0.04 m. How many seconds would it take to bring me (85 kg) to its top speed of 40 km/h? Their 4WD version says it has double torque, so 14 Nm. How long would that take to get to top speed? The e-skate is Mellow Drive. Other brands have lots of initial torque, but it drops as speed increases. Is this possible to calculate in the same way?

ANSWER:
The torque on the drive wheels will result in a frictional force between
the wheels and the road which drives the board forward. I am assuming
that that is the net torque to the two wheels, not the torque to each
wheel. So if the torque is 7
Nm, the force accelerating you is F =7/0.04=175 N. Ignoring all
friction, the
resulting acceleration would be a =175/85=2.1
m/s^{2} . If the final velocity is v =40 km/hr=11.1 m/s,
the time is t=v /a =5.4 s. Now, is friction really
negligible? First consider the frictional force f from the
wheels which could be approximated as f=μmg where μ
is a coefficeint kinetic friction and mg =85x9.8=833 N. A
reasonable approximation for μ would be μ ≈0.1;
with this μ the unpowered board would stop in about about 20
m if it had a speed of 2 m/s. The frictional force would then be about
83 N, not at all negligible. Redoing the calculations above for a net
force of 175-83=92 N, I find t =10.3 s. Finally we should talk
about air drag. The drag force D may be approximated (only in
SI units) as D≈ ¼Av ^{2} where
A is the cross-sectional area you present to the wind, let's say
about A ≈1 m^{2} ; so the biggest D gets
is about 30 N. Since most of the time spent is at lower speeds where
D is much smaller, I will not calculate the the time including
D (a much harder calculation because the force is not constant)
since all this is only a rough caclulation anyway; just think of 10 s as
a lower limit. You can, though, estimate the maximum speed if the torque
really did remain constant by finding the speed for which D =92
N: v _{max} ≈√(4x92/1)≈19 m/s; this is
called the terminal velocity. For
constant torque, doubling it would result in a net force (again
neglecting D but including f ) of 267 N and a time of
3.5 s. If you know how the
torque varies with speed, you could do a similar calculation but it
would be trickier because the force and therefore the acceleration would
change with time.

ADDED
COMMENT:
Just for fun I added the air drag D for the two-wheel drive
case. I found that 11.1 m/s is reached in 11.5 s, only a very small
increase as I anticipated. The graph of v vs. t is
shown; you can see that the curve is nearly linear up to 11.1 m/s and
approaches 19.2 m/s at large times. The solutions are t =17.5·tanh^{-1} (v /19.2)
and v =19.2·tanh(t /19.2). Don't take any of this
too seriously because of the numerous approximations I have made; I
would guess there would be about a 30-40% uncertainty —probably
5-15 s would give the experimental time. If you get one of these, let me
know how long it actually takes.

FOLLOWUP QUESTION:
Someone said the mechanical power needed to ride at 40 kmh was 1900 KW. Is that true?

ANSWER:
The power P delivered by a torque τ at
angular velocity ω is given by P=τω .
In your case τ =7 N·m and ω=v /r= 11.1/0.04=278
s^{-1} , so P =1943 W, not kW.

QUESTION:
Unlike a submarine traveling in a horizontal line at 5 fathoms compared to 50 fathoms (the pressure in the vertical direction would be the same....) But in a vertical line I imagine from 50 fathoms it would rise slower in the first 5 fathoms than it would in its final 5 fathoms.... I.e in its first 5 fathoms of accent it had the external pressure of 50 fathoms descending (dropping to 49 fathoms pressure then 48 etc) compared to 5 fathoms pressure descending to 0 fathoms pressure as it ascends.
Is this correct....

ANSWER:
As long as the water is considered incompressible (which it is for all
intents and purposes), the net force on the submarine due to water
pressure (called the buoyant force) is the same at every depth. Even
though the pressure in the water is enormously bigger at great depths,
the pressure difference between bottom and top of the submarine is the
same. Therefore, if drag forces are neglected there will be a net force
up on the submarine which is constant and equal to the buoyant force
minus the weight of the submarine. However, the drag force is not
negligible and is approximately proportional to the speed. So as the
rising submarine speeds up the drag force down becomes bigger and bigger
until it is eventually equal to the buoyant force minus the weight;
hence the net force is zero and from that depth up to the surface the
speed is constant. You might also be interested in a related
earlier question about
submarines.

QUESTION:
I am a modelmaker. I currently work for an architectural practice in London, UK.
Sometimes I find it hard to explain to architects why there isn't a certain amount of detail to a model.
For example, if i make a 1:500 scale model of a building. The architect will ask why he/she can't see the door handles and I have to explain that it is too fine a detail for the scale.
I would like a better way of explaining it, but can't figure out out to work it out. Basically, I would like to say that the equivalent of a 1:500 scale model is like looking at a building X miles/meters away and that even someone with perfect eyesight would have trouble seeing such small details. How would I work out how far you would have to be away from an object to "see" it at a different scale?

ANSWER:
I found the following useful
description of how small an object the human eye can see:

"What is limited is the eye's resolution: how close two objects can become before they blur into one. At absolute best, humans can resolve two lines about 0.01 degrees apart: a 0.026mm gap, 15cm from your face. In practice, objects 0.04mm wide (the width of a fine human hair) are just distinguishable by good eyes, objects 0.02mm wide are not."

Suppose you had a door knob 10 cm in diameter. Scaling it down by 500
would make the size 0.02 cm=0.2 mm, easily seen by the standards above
but still very small, maybe the size of a period on this page. So maybe
the easiest thing to please your clients would be just to put a dot
there; I can understand how that might violate your aesthetics, though.
To answer your question, what would be the distance R at which
the angle subtended by 10 cm is the same as the angle subtended by 0.02
cm at a distance 15 cm from your eye? The angle, in radians, is
size/distance, so 0.02/15=10/R or R =7500 cm=75 m. To not be able to
distinguish the door knob at all, 0.002 cm, the distance would be 750 m
which would correspond to a scaling of 1:5000.

QUESTION:
Hello, I'm looking for a physics expert to help me with a problem. We were learning Simple Harmonic Motion in our class today and I was struck with curiosity when my professor showed us a ball rolling without slipping on a turntable. The motion that the ball took surprised me because I never imagined that it would move in that fashion. I then started to wonder if it was possible to describe the ball motion in terms of (x,y) for a function of time. I talked to my professor about this problem and it had him stumped. Can you please help me out with this problem and explain in detail how you got your solution.
Here’s the full question I’m proposing and all the conditions:
A spherical ball of mass "m", moment of inertia "I" about any axis through its center, and radius "a", rolls without slipping and without dissipation on a horizontal turntable (of radius "r") describe the balls motion in terms of (x,y) for a function of time.

**The turntable is rotating about the vertical z-axis at a constant unspecified angular velocity.

**Radius and mass of the turntable and the ball are unspecified.

ANSWER:
I was unfamiliar with this problem, but it is interesting. Although the
path of the ball depends on the initial conditions, apparently if you
just set it on the turntable it will
move in a circle
which is at rest in the laboratory frame. I will tell you at the outset
that this is a quite advanced classical mechanics problem and is well
beyond the purposes of this site. However, it was interesting enough to
me that I did a little research to try to get a feeling for how it can
be approached. I will not provide details, but they can be found in a
1979 article in American Journal of
Physics . I do not know your level, but if you are a high school
student, the math is probably beyond your grasp, all the vector
calculus. As you can see from the article, trying to write the solution
in Cartesian coordinates, as you have specified, is not the way to go.
Using the vectors r and ω
as the author does implies using cylindrical polar coordinates (r ,
θ , z ).

QUESTION:
Ok so I am in AP Physics 1 and we are learning about Circular Motion. When
I was doing a problem I noticed a problem in which a car is set at a
constant speed and it asks what the free body diagram looks like at the top
of the hill. As I'm doing multiple questions I think about what the free body
diagram would look like just after passing the top of the hill so not at the
top but not fully down the hill. I tried to do the trick where you set it to
an x and y plane but there is no force inwards from the Normal Force and
then of course there's the Weight Force.... If a more skilled Physcisist could
help me it would help so much in furthering my understanding in physics.

COMMENT:
This student and I had a spirited debate about what constitutes homework
or tutoring help, forbidden on Ask the Physicist . His argument
was "…the
question I brought to your attention does not exsist…" because he
could not find it in a book somewhere and therefore not classifiable as
homework; that, of course, is nonsense because this is a very standard
question. Nevertheless, since he is so avid to learn, I am going to
answer it!

ANSWER:
The question is kind of ambiguous because it refers to
constant speed; but if the car is to maintain constant speed down the
hill, there will have to be some drag friction. If a constant speed
v is maintained, the free body diagram is shown on the left. I will
choose a coordinate system with +y pointing along the direction
opposite the normal force and +x tangent to the circle and
pointing downhill. The car is inequilibrium in the x direction and has
an acceleration of magnitude v ^{2} /R in the
positive y direction. Newton's laws are ΣF _{x} =0=mg sinθ-f
and ΣF_{y} =mv ^{2} /R=mg cosθ-N.
The solutions are simply obtained: f =mg sinθ
and N =m [g cosθ-v^{2} /R ].
An interesting result is that when θ =cos^{-1} (v^{2} /(gR )),
N =0 and the car will leave the road.

Another possible case is that there is no friction and the car has speed
v
at the top of the hill; the car will therefore speed up as it goes down
the hill. Using energy conservation it is easy to show that its speed
when at angle
θ is v' ^{2} =v ^{2} +2gR (1-cosθ )
and so the centripetal acceleration is a_{y} =v ^{2} /R +2g (1-cosθ ).
Notice that there will also be a tangential acceleration, a_{x} ,
an unknown at this stage. Now the equations to solve are ΣF _{x} =ma_{x} =mg sinθ
and ΣF_{y} =m (v^{2} /R+ 2g (1-cosθ ))=mg cosθ-N.
The solutions are a_{x} =g sinθ
and N =m (v^{2} /R+g (2- 3cosθ )).
Again, there will be an angle for which N =0 which is where the car would
leave the road, the determination of which I will leave to my AP
student.

QUESTION:
I am trying to explain to a friend why it is not possible to hold a child in their arms during a rapid deceleration (car accident).
Given a child of 8 kgs in the arms of a parent, a rapid deceleration from 60 kms/hr to zero, what is the resultant force (weight, mass? ) that the 8kgs becomes in the parents arms ? I know that it is not possible to hold a child in such a situation but I'm looking for some simple way to show my friend why.

ANSWER:
60 km/hr is about 17 m/s. Suppose the car stopped in a tenth of a
second, 0.1 s and you were exerting the necessary force to keep the
child pressed to your chest. The child, like everything else attached to
something in the car would experience an acceleration of a =7=17/0.1=170
m/s^{2} . The force necessary to hold the child is ma =8x170=1360
N. This corresponds to a weight 17 times greater than the weight of the
child. Although it might be argued that you might be able to exert such
a force for a tenth of a second, everything happens so fast in an
accident that the child would be ripped from your arms before you had a
chance to react.

QUESTION:
I read somewhere that when you impart energy into something like winding a clock or stretching a rubber band it gains mass. I know it is very minute and almost impossible to measure. But mass and energy being the same if I put mass into an object it gains the mass and therefore gains energetic potential. So if I add energy to it, why wouldn't it gain mass and thereby a little weight?

ANSWER:
Yes, adding energy to will increase the mass. Since weight is the force
of gravity and that force is proportional to the force, you also
increase the weight. Let's get a feeling for how big that mass increase
is. Suppose that we have a spring and it takes a force of 2 N=0.45 lb to
compress it 1 cm=0.01 m. Then, the spring constant is k=f /x =2/0.01=200
N/m. The work you did to compress it was W = ½kx ^{2} =½x200x0.01^{2} =0.01
J which is the energy which you put into the spring. So now the amount
of mass which would correspond to that amount of energy is m=W /c ^{2} =0.01/(3x10^{8} )^{2} =1.1x10^{-19}
kg. The mass of a dust particle is on the order of 10^{-8} kg,
so the mass change is about one ninety billionth of the mass of a
particle of dust. I think we can agree that this would be impossible
(not almost impossible) to observe.

QUESTION:
If I fill my car's gas tank and then drive a couple of hundred miles the cars gas gauge will tell me the tank is empty. I know that energy can't be created, destroyed or used up. So what happens to the energy that was stored in the gasoline?

ANSWER:
Part of that energy went into the kinetic energy of your car. Another
part went to heat both as a result of the combustion as a result of
friction encountered by the car as it moved along. Another part went to
sound produced by your car. Finally, if you ended your trip at an
altitude higher than where you started, some went into potential energy.

QUESTION:
Ordinary car tires have tread. (Racing cars don't, but that's another problem.) Tread reduces surface area in contact with the road, but surface area is not a factor affecting friction. Adding a complex shape to the exterior of a tire increases the manufacturing cost. Nobody likes spending money unnecessarily. What's the point behind this design choice? Why do ordinary car tires have tread?

ANSWER:
The main reason is that on a wet road the tread expels some of the water
lessening the likelihood of hydroplaning. Tread can also help increase
traction on soft surfaces like mud or snow. For a lengthy discussion of
the question of surface area, see an
earlier answer .

QUESTION:
If you took two 1000 mile long metal bars, laid one horizontally on the
ground and stood the other vertically, would they weight the same and if
not, why?

ANSWER:
By weight we mean the force of attraction between the earth and the
object. Therefore, they would not weigh the same because the vertical
bar has most of its mass farther from the center of the earth than the
horizontal rod does. The weight of a point mass on the surface of the
earth is W _{1} =mMG /R ^{2}
where M is the mass of the earth, R is the radius of
the earth, G is the universal gravitational constant, and m
is the mass of the point mass. This is usually written as W _{1} =mg.
But, if the point mass is a distance H above the surface,
its weight is smaller, W _{2} =mMG /(R+H )^{2} =mg /(1+ (H /R ))^{2} .
If you know integral calculus it is not hard to show that the weight of
a vertical uniform bar of length L and mass m is W _{3} =mg /(1+ (L /R )).
In your example L =1000 mi is not small compared to R ≈4000
mi, so
W _{3} =mg /(1.25)=0.8W _{1} .
But, because L is so large, your horizontal bar does not have a
weight of mg either unless it bends to conform with the curved
surface of the earth. But, you can see from the figure that the
distances from the center of the earth are much smaller than for the
vertical bar, so it will surely have a larger weight, just not quite as
big as mg .

ADDED
THOUGHT: Actually, even if the bar conformed to the
surface of the earth, it would weigh less than mg because the
components along the length of the bar of the forces on each piece of
the rod would all cancel out. I think I will not calculate the exact
answer for the horizontal bar, just say that is slightly smaller than
mg .

QUESTION:
We learned about the electromagnetic spectrum in school. I asked a question my teacher could not answer. I understand the history of the electromagnetic spectrum, and how it was not discovered all at once, but rather piece by piece over three centuries. So it goes gamma ray, X-ray, ultraviolet, visible, infrared, microwave, and radio wave. My question is are the lines between each type of wavelength well defined? Or is it possible that there are still, yet to be discovered, waves between any of the already listed wavelengths. Also, is possible that there are wavelengths beyond gamma rays or beyond radio waves. Is the electromagnetic spectrum complete or open ended?

ANSWER:
The lines are not well-defined; depending whose definitions you use you,
will get overlaps at the "boundaries". The names are just qualitative
appelations and have no physics meaning. The electromagnetic spectrum is
continuous from zero to infinite wavelength. Of course, you will never
encounter a wave with zero wavelength because that would correspond to
infinite energy and there is not that much energy in the entire
universe! Also a wave with infinite wavelength would correspond to zero
energy which would mean it was nonexistent. Any electromagnetic wave
with a wavelength shorter (longer) than the most (least) energetic gamma
ray (radio wave) ever observed would still be called a gamma ray (radio
wave).

QUESTION:
Say you have a conveyor belt with constant speed v. I put a bottle on the belt suddenly.
How do I know that it would not tip and whether there is a critical speed for the conveyor belt so tipping would never happen?

ANSWER:
I believe that it depends only on the geometry of the bottle and on the
coefficient of kinetic friction μ between the bottle
and the belt. In the figure, H is the position of the center of
mass ⊗ of the
bottle above the base and R is the radius of the base. When the
bottle touches the belt there will be a frictional force f
in the same direction as the velocity of the belt which will cause an
acceleration of the bottle to bring it up to the belt speed. But, as you
note, the bottle might tip over because of the torques it experiences
due to the forces on it. Because the bottle is trying to tip, the normal
force on the bottle should be expressed as two different forces
N _{1} and N _{2}
acting on the leading and trailing edges of the base. Since the bottle
is accelerating, it is imperative to calculate the torques about the
center of mass of the bottle. The bottle is in equilibrium in the
vertical direction so N _{1} +N _{2} =mg .
The frictional force is f =μ (N _{1} +N _{2} )=μmg.
The torques will sum to zero if the bottle is not tipping, 0=fH+RN _{1} -RN _{2} .
Now, if the bottle is just about to tip over, N _{1} =0
and so the torque equation becomes 0=fH -RN _{2} =μmgH-Rmg
or μ _{max} =R /H ; this is the
maximum that the coefficient of friction could without the bottle
tipping. The result does not depend on either m or v .
For example, if R =3 cm and H =6 cm, then μ ≤0.5.

ADDED
NOTE:
The normal force could have been handled differently by having a single
force N acting a distance d from the leading edge of the base of the
bottle. Then when d =2R the bottle would be about to
tip. You would get the same result as above.

QUESTION:
How can I deduce for uncertainty in measurement of mass by Heisenberg principle or there can not be any uncertainty in mass ?

ANSWER:
Mass is energy, E=mc ^{2} . The uncertainty principle is ΔE Δt ≈h /2π
where h/ 2π= 6.6x10^{-16} eV≈10^{-34}
J·s. Since c =3x10^{8} has no uncertainy, Δm ≈[h /(2πc ^{2} )]/Δt ≈10^{-51} /Δt
kg. Δt is the time over which you observe the mass. For a
stable particle or a stable nuclus, e.g ., Δm= 0 because Δt= ∞.
A neutron has a half life of 880 s, so the uncertainty in its mass is
approximately Δm ≈10^{-51} /880≈10^{-54}
kg.

QUESTION:
A strong magnetic field is applied to a proton at rest then what will
occur?

ANSWER:
If
the field is uniform, the proton will feel no force but will experience
a torque tending to align its magnetic moment with the field. If the
field is nonuniform, the proton will also experience a force the
direction of which will be determined by the alignment of the magnetic
moment.

QUESTION:
the solid Earth density model equation?

ANSWER:
I do not know what your are referring to. First, the earth is not solid
throughout, its core is molten. The density is sufficiently complicated
that there is no single equation which will describe it. For more
detail, see an earlier answer .

QUESTION:
Even after reading the mathematics behind quantum numbers, wave functions and their detailed algebra, I still can't figure out from where does SPIN ANGULAR MOMENTUM? What do you mean by it is intrinsic, when there is nothing that spins??

ANSWER:
See an earlier answer .

QUESTION:
Why do heavy nuclei have an excess of neutrons but lighter have almost equal no or protons and neutrons

ANSWER:
The nuclear force is complicated, so the answer to your question is also
very complicated. You can find an easy-to-understand qualitative
explanation in an earlier answer .

QUESTION:
I am standing on a scale on carriage that is going down the slope without friction. The slope angle is
theta. What will the scale show in two cases:

The carriage is shaped in a way that the scale is horizontal with respect to ground (that is is a special edge shaped carriage)

The carriage is regular so the the scale is not paralleled to the ground.

I am trying to understand how a special wedge or insoles in ski show would effect weight felt by a person. In case 1 I think the answer should be mg(1-(sin(theta)^2)) but I am not sure.
In case 2. I have no idea. I think scales can misfunction if the slope is too strong.

ANSWER:
Before doing either case, imagine that we choose the man and the
carriage as the body to look at. It is the standard introductory physics
problem of something sliding down a frictionless incline. The principle
result is that the acceleration of the man and the carriage are the same
and have a magnitude a=g sin θ down the
slope.

The figure
on the left
shows the forces on the man standing on the horizontal scale. There
is his own weight, mg and the scale exerts
a normal force N vertically up; the scale
will read N . The contact between the man and the scale
cannot be frictionless or else the man would not accelerate along
with the carriage, so there is also a frictional force f .
Newton's equations for the man are ma=m g sinθ=-N sin θ+f cosθ+mg sin θ
and 0=N cosθ -f sinθ-mg sinθ .
Solving, I find N=mg cos^{2} θ and
f=mg sinθ cosθ .

The figure
on the right shows the man on the scale parallel to the incline. Now
no friction is needed and the scale, again, will read N but
N will be different. Newton's equations are 0=-mg cosθ+N
and ma=mg sinθ , just the same as the classic
object on the frictionless incline. So the scale reads N=mg cosθ.

The graph
compares the two cases.

QUESTION:
I was wondering has every single event throughout the time of the universe been inevitable? Correct me if I am wrong, but if the universe has a variety of laws that define what can and can not occur, then can it not be said that since every single event after the Big Bang has been determined by every single law in the universe? If this is correct can it be said that the end of the universe be determined since its beginning? Is the future already set to occur in a certain manner?

ANSWER:
Here is the problem. The actual laws of physics do not necessarily
predict what will happen, rather many laws predict the probabilities
that various things can happen. Therefore I would say that the universe
is not deterministic. (Einstein, however, believed that the universe is
deterministic.)

QUESTION:
if a lamp was placed in space where no gravity effected it...the light turned on would the light then propel the lamp ?

ANSWER:
Yes,
as long as the light did not shine isotropically (equally bright in all
directions). However, the propelling force would be very small for a
simple lamp. See a recent answer to get an
idea of how small the force exerted by light is.

QUESTION:
Imagine another universe that is completely empty except for a large solid sphere at rest. Suddenly, an identical sphere pops into existence 1 trillion kilometres away. what will happen?

ANSWER:
Your question essentially asks at what speed a gravitational field
propogates. It is believed that the speed of gravity is equal to the
speed of light (see faq page). One trillion km is about 0.11 light
years=40 light days. So, assuming that the first sphere has existed for
longer than 40 days, the second sphere will know of the existence of the
first at the instant of its creation. The first, however, will not know
of the existence of the second until 40 days after it was created.

Q&A OF THE WEEK
2/26-3/4/2017

QUESTION:
according to the formula of variation of g above the surface of earth
g' = g(1+2h/R) and the radius of earth is 6400 km .If we put R/2=3200 in the previous formula,did it mean that after 3200 km space start ?

ANSWER:
First, let's see where your "formula" comes from. The acceleration due
to gravity g may be written as g=MG /R ^{2}
where M is the mass of the earth, G is the
gravitational constant, and R is the radius of the earth. For
some height h above the earth, g'=MG /(R+h )^{2} .
Therefore g' /g =R ^{2} /(R+h )^{2} =(1+(h /R ))^{-2} .
This is an exact expression and you can see that g' never
becomes exactly zero but continues decreasing forever as h gets
larger. (Note that your "formula" cannot be correct because g'
gets bigger as h increases.) However, for many applications you
want to know what g' is if h /R is very small;
one can do a binomial expansion of g' /g , g' /g= (1+(h /R ))^{-2} ≈1+(-2)(h/R)+…=1-2h /R .
So your formula was almost right, just need to change the + to -. But
this is only an approximation and will fail when h gets too
big. The graph above compares the exact (black) and approximate
solutions (red); as you can see, the approximation only works up to
about 0.1R . The fact that the approximation goes to zero at
h=R /2 has no meaning.

QUESTION:
Is a board suspended between two points and evenly weighted across it's
length (by books, for instance), less likely to bend if the ends are clamped
or nailed so they don't move? This is in contrast to if the ends simply
rest on end supports.

ANSWER:
Yes. Keep in mind, though, that the board will be exerting a horizontal force
on the support. Be sure the end supports are strongly attached to
the walls.

FOLLOWUP QUESTION:
If the end points are dado slots (the boards are shelves for media storage), and the shelves and dados are cut so that the shelves fit the dado slots very precisely on all three sides, then horizontal travel and twisting could be eliminated, which would obviate the need to attach to the wall. Correct?

ANSWER:
For the board to bend, the ends are pulled toward the center and lifted
up (see figure). I would anchor the ends.

FOLLOWUP QUESTION:
One more question though, and a more challenging one I hope.
If one wanted to screw a square block of wood to a wall, where would the optimal places be to put the screws? Assume that the block of wood is 12" on a side, two edges are parallel to the floor and only two screws are used. The wood will be weight-bearing and the weight will be evenly distributed along the top edge of the block. Also, assume that the thickness of the wood is small relative to the other dimensions and that the screws can find equal purchase in the wall in back of the block of wood being fastened.
My gut tells me that (a) the screws both should be put along a horizontal line parallel to the floor, and, (b) it doesn't matter where that line is, high or low, since the weight is evenly distributed. What I can't intuit, is if it's better to have the screws be 3" or 4" or some other distance from a vertical side, as long as they are symmetrically placed.
In the general case, if N randomly placed screws are used, what is the math that tells me what percentage of the weight each screw bears according to its geometric placement?

ANSWER:
I really do not think it is worth worrying about. The reason is that it,
paradoxically, is likely not the screws which keep the block from
falling, it is the static friction between the wall and the block; the
function of the screws is more to press the block against the wall than
to hold up the weight. Therefore, regardless of where you put the
screws, tighten them as tight as you can get them since the amount of
frictional force you can get depends on the normal force, the force
pressing the wall and block together.

QUESTION:
If two people are carrying a canoe parallel to the ground, how much weight are they each carrying? would it be half of the weight of the canoe, more than half, or less than half?

ANSWER:
It all depends on two things —where the center of gravity is
and where the two people exert their forces. In the figure the weight
W of the canoe acts at the center of gravity
and the two people exert forces F _{1}
and F _{2} at distances at distances
d _{1} and d _{2} from the center of
gravity. The two people must hold up all the weight between them, so
F _{1} +F _{2} =W . The canoe is not
rotating about its center of gravity so the net torque must be zero, so
F _{1} d _{1} =F _{2} d _{2} .
Solving, F _{1} =W [d _{2} /(d _{1} +d _{2} )]
and F _{2} =W [d _{1} /(d _{1} +d _{2} )].
If d _{1} =d _{2} , each holds up half the
weight.

QUESTION:
Why we cannot feel the momentum or the force of the sunlight?

ANSWER:
The most intense light from the sun is about λ =550
nm=5.5x10^{-7} m (yellow); the corresponding frequency is f=c /λ =3x10^{8} /5.5x10^{-7} =5.45x10^{14}
Hz. The momentum of one photon is p=hf /c =6.6x10^{-34} x5.45x10^{14} /3x10^{8} =1.2x10^{-27}
kg·m/s. The
photon flux (photons per second per square meter) for photons of
this wavelength is about Φ =3x10^{21} s^{-1} ·m^{-2} .
So if I estimate your area to be about 1 m^{2} , there are about
3x10^{21} photons hitting you every second. The total momentum
imparted to you over that second (assuming that you absorb the photons)
is about Δp =3x10^{21} x1.2x10^{-27} =3.6x10^{-6}
kg·m/s. The average force you feel is about F =Δp /Δt= 3.6x10^{-6
} N; and this force is spread uniformly over your area. This is
really small; for example, if we take the typical weight of a one dust
particle to be about 10^{-8} N, the weight of about 360 dust
particles spread over your whole body is about what you "feel" from the
sunlight.

QUESTION:
If we jump from the ground where we are standing we fall back to same place, even though earth is rotating or if i want to go to america, i takeoff straight from india and wait for 12 hours and jump from there i should reach america but this does not happen.
Why so?

ANSWER:
Because you are rotating with the earth. Therefore, as seen from outside
the earth, you have a component of your velocity which is horizontal and
the same as the speed of the earth's surface.

QUESTION:
The standard atmospheric pressure is 100 kpa.
Why don't we experience this pressure which is very high?

ANSWER:
Because we evolved in this atmosphere and therefore the pressure inside
every cell of our bodies is also about 100 kPa; also the pressure in any
cavity of our bodies (stomach, intestines, etc.) is about 100 kPa. So
the net force on us is zero.

QUESTION:
How can E=mc2? Since E is in ergs,mass in grams and speed of light in centimeters per second squared which amounts to nearly 900 billion billion.How is it that the person who determined the exact size of the centimeter got it just right? If the size of the centimeter was set slightly bigger or smaller, the equation would fail. If the duration of the second was slightly more or less the equation would fail. if the amount of the gram were slightly more or less, the equation would fail.All three of these were determined long before E=mc2. Even the size of the erg has to be exact for the equation to be true. True there is an exact amount of energy released when an exact amount of mass is gone,but how is it that the erg, gram, centimeter, and second where determined so long ago, so exact?

ANSWER:
How we define the units of mass, length, or time is totally arbitrary.
If the centimeter is defined to be twice as long as it is currently
defined, the speed of light would be a number half as large as it is but
it would still be the speed of light. The energy of one gram would be a
number ¼ as large, but it would still be the same amount
of energy. You would not say that E=mc ^{2} was wrong if
we got two different numbers if we calculated it in Joules (using kg, m,
s) than in ergs (gm, cm, s) would you? E =(1 kg)(3x10^{8}
m/s)^{2} = 9x10^{16} J and E =(1000 g,)(3x10^{10}
cm/s)^{2} = 9x10^{23} erg. In fact, 9x10^{23}
erg=9x10^{16} J is a perfectly good equation.

QUESTION:
A plastic cup is filled with an ideal fluid and held at a certain height. A small hole is then punched on the wall of the cup near the bottom and the fluid leaves the cup at a certain speed. The cup is then dropped. What happened to the fluid that leaves the cup through the small hole? What is its speed?

ANSWER:
No fluid will come out the bottom of the falling cup. The easiest way to
understand this is to apply the equivalence principle —there
is no experiment you can perform to tell if you are at rest in empty
space or free-falling in a gravitational field. Another way to look at
it is that the acceleration due to gravity is independent of the mass
and so the cup and the fluid it contains both fall exactly the same.

QUESTION:
I have a question about light please. I was thinking about one of Einstein's thought experiments. Let me set some the situation first - there is a stationary observer at a train station. Einstein is on a train travelling at 99.99999%c and is about to pass the train station. As the front of the train reaches the train station, the stationary person emits a light beam in the same direction as the train while at the same time, the train turns on its headlight. Based on what I understand, both beams of light should be travelling at C but I am interested in understanding what each person sees. I can understand that the stationary person will see both light beams leave him at C, along with the train at 99.99999%c. But what about Einstein on the train? I understand that you cannot add the velocity of the train to its headlight beam. But I can imagine that he would see the beams moving ahead and away from him at .00001%c since he is already travelling at 99.99999%c. But that seems to violate the idea C being the same for everyone in every reference frame. I can also imagine that due to that rule, he would see both light beams moving ahead and away from him at C. But if that's true, how can he possibly be going 99.99999%c if the beams are moving away from him at C and how could the stationary person see both the light moving at C and Einstein moving at 99.99999%c? I need your insight to understand this please.

ANSWER:
This is a very long question, but the answer is very short. All
observers, regardless of their motions, see the speed of any
electromagnetic radiation to be c . Your statement that the
train rider would see the beams having is speed 0.00001c is
simply wrong. That is not to say that the beams are unchanged —they
would be Doppler-shifted to different wavelengths, but still move with
speed c .
See the faq page to help yourself to accept that this is true.

FOLLOWUP QUESTION:
Thanks! I do accept what you are saying is true. Acceptance was never the issue, understanding was.
Since the doppler is shifted then what Einstein would see with his eyes would be the light around him becoming redder and redder the closer to C he gets?

ANSWER:
I hope you did go to the faq page and now have an "understanding"; if
not, you should look at these links
there . You might also read an
earlier answer about relativistic velocity addition. You are right,
if light with frequency f is moving in the same direction as
you are, you will see a lower frequency light f' ,
red-shifted. The Doppler equation is f'=f √[(1-β )/(1+β )] where β=v /c.
Suppose we take your case where β= 0.9999999=10^{-7}
so √[(1-β )/(1+β )]=2.24x10^{-4} and choose λ =550
nm (yellow light). Then f=c /λ =3x10^{8} /5.5x10^{-7} =5.45x10^{14}
Hz and f' =2.24x10^{-4} x5.45x10^{14} =1.23x10^{11}
Hz=123 GHz. This frequency is in the short-wave radio range, nothing he could "…see
with his eyes…"! The Doppler effect has red-shifted the wave
way beyond visible red!

QUESTION:
Why is that no particles may ever have exactly zero kinetic energy as it would violate the Heisenberg Uncertainty principle.

ANSWER:
The kinetic energy of a particle of mass m may be written as
K=p ^{2} /(2m ) where p is the linear momentum.
The uncertainty principle states that Δp Δx>h
where h is Planck's constant and Δp and Δx
are the uncertainties in momentum and position x respectively.
This tells you that you cannot know either p or x
exactly, they will always be uncertain. For the uncertainty of K
to be zero the uncertainty of p would have to be zero. But, if
the uncertainty of p is zero, the uncertainty of x is Δx>h /0=∞.
You would lose all knowledge of the position of the particle.

QUESTION:
If you could drop a coin from the orbit of the moon to the surface of the Earth, how fast would it be going when it hit the Earth? How much energy would it release when it hit?

ANSWER:
The gravitational potential energy of the coin of mass m is
U=-GMm /r
where G =6.67x10^{-11} J·m/kg^{2 } is the
gravitational constant, M =6x10^{24} kg is the mass of
the earth, r =3.85x10^{8} m is the radius of the moon's
orbit, and m is the mass of the coin. So, U= -1.04x10^{6} m
J. But, when the coin strikes the earth its energy will be E=-GMm /R+ ½mv ^{2}
where v is its mass and R =6.4x10^{6} m is the
radius of the earth. Ignoring air drag when the coin gets near earth
(which will not really be negligible), energy may be conserved: -1.04x10^{6} m=-GMm/R+ ½mv ^{2} =(-6.25x10^{7} +½v ^{2} )m.
Solving for v I find v =1.11x10^{4} m/s. The
kinetic energy it would release would 6.15x10^{7} m J.

QUESTION:
I'm sitting in my bedroom at home and light is coming in the window and
illuminating the whole room. I am able to see objects in the room
because they reflect the light. The problem is that these light waves
must be bouncing off everything in a zillion different directions. How
is it that I can see an object rather that a ball of light fuzz. (I'm
just a 70 year old guy with some dumb questions.)

ANSWER:
Suppose that the sun is shining in the window. Then you will agree that
objects illuminated directly by the rays of the sun are th most
illuminated. As you note, you can see everything else in the room. You
are correct that when the sun shines on something, light is reflected or
reemitted from that something in all directions, and this is one source
of lighting for the rest of the room. But, keep in mind that light
outside is not all coming directly from the sun —the sky is
also bright, just not as bright as the sun, and that light comes in
through the window also and illuminates other parts of the room from
which you could see the sky if you looked out. You will also note that
the wall of the room where the window is located is not as well
illuminated as the rest of the room.

QUESTION:
I'm shopping for a pole chainsaw and trying to figure how heavy a 7 lbs saw at the end of a 8' pole would "feel", at about a 45 degree angle.

ANSWER:
It depends on how you hold it and where the center of gravity is. Refer
to the figure. The weight W acts at the center
of gravity which I have chosen to be at a diatance c from your
right hand. The position of your left hand I have denoted as being a
distance d from your right hand. For simplicity I have assumed
that you will exert a force L with your left
hand perpendicular to the pole; your right hand exerts a force with
horizontal and vertical components H and
V respectively. You specify 45 °, so
the horizontal and vertical components of L
are equal and of magnitude L /√2. The two equations for
translational equilibrium are ΣF _{vertical} =0=-W +L /√2+V
and ΣF _{horizontal} =0=-L /√2+H .
These may be simplified to H =L /√2 and V=W-H .
There are three unknowns and only two equations, so we need a third
equation; the sum of torques must also be zero. Summing torques about
the right hand, Στ=dL-cW /√2=0 or L =cW /(d √2).
Therefore, H =cW /(2d ) and V=W [1-cW /(2d )].
For example, suppose W =7 lb, d =3 ft, c =4 ft:
I find L =6.6 lb, H =4.7 lb, V =2.3 lb. The
force R exerted by your right hand is R =√(V ^{2} +H ^{2} )=5.2
lb. The key to easy handling is to choose the saw with the smallest
c which will minimize the torque you must exert.

QUESTION:
I have learnt that a magnetic field only produces a deflection in the path of a charge, and does not contribute to increase in speed or kinetic energy of the particle. So, if a charge is moving along some curved path with its speed increasing with time, and let us say the space is free space, then will there be any magnetic field (or electric field) present in that region?

ANSWER:
An electric field can change both the magnitude and direction of the
velocity of a charged particle. A magnetic field can only change the
direiction. So, in your case, there could be either an electric field
alone or both magnetic and electric fields.

QUESTION:
I don't know if you know about the device scientist called the torus experiment. I heard about it at UGA thirty years ago. It was to make a round tube with wire around all of it. Then special particles went through the tube, and the device put out a lot of energy. They tried to make one 5 miles in diameter, but they kept having problems with centrifugal force. The machine would make as much electricity as a nuclear power plant, or more. What I thought of is real simple, but maybe it was overlooked or something. My idea is to make the shape like a track around a football field, and make
the straight parts a lot longer. So the particles would have time to realign in the center and all start going the same speed. My teacher said it is like a contained nuclear explosion. It would put out a lot more electricity than it takes to run it.

ANSWER:
What you are describing sounds like a
tokamak , an experimental fusion reactor. Nuclear fusion is the
energy which powers stars. Roughly, the idea is that if you get hydrogen
atoms hot enough (hundreds of millions of degrees) they will have enough
energy to fuse together to make helium, and that releases a great deal
of energy. The problem turned out to be much more challenging than
anticipated and there still is not a practicable fusion reactor which
will give you more energy out than you use to run it. The main problem
was the difficulty of keeping the hot plasma confined long enough to
have a sufficient number of fusion reactions. The current research is
centered on constructing a tokamak by an international collaboration
known as ITER ; the lab
is in France. There is an old cynical joke among physicists: "Fusion is
the energy fo the future and always will be!"

QUESTION:
I have measured the thickness of steel pipes with a thickness meter (pocketmike) with a sound velocity of 5813 m/s.
No I need to change my thickness values with a sound velocity of 5920 m/s.
This because this value is more accurate for this steel pipes I measured...
Which formula I need to use?

ANSWER:
I would assume that the device measures a time. In that case, the
thickness T would be proportional to the product of the
velocity v and the time t , T=cvt where c is
some constant. So, if you change v to v' but keep
t (measured) constant, the corrected thickness T' will be
T'=cv't. Take the ratio of the two equations and solve for the
thickness: T'=T (v /v' )=(5813/5920)T= 0.982T.

QUESTION:
The following
question came up during lunch with the ROMEOS (retired old men eating
out) yesterday: Why is a star's spectrum mainly a black-body spectrum?

ANSWER:
Innumerable photons are produced by innumerable processes in a star.
These diffuse around in the plasma which comprises the star. Those in
the photosphere, the outermost layer of the star, come to thermal
equilibrium. A relatively small number "leak out"; this then is
analagous to the classic cavity with a small hole containing a photon
gas. Thermal equilibrium is maintained by photons diffusing in from the
more interior regions of the star.

QUESTION:
How much energy would you produce, using a controlled mechanical process to drive a generator, if you lowered a 10,000 tonne weight through 100m.

ANSWER:
Use U=mgh for potential energy, m =10,000 t=10^{7}
kg, h =100 m, g =9.8; U =9.8x10^{9} J.
If it falls over a time t =10 min=600 s, the power generated
would be P=U /t =1.63x10^{7} W=16.3 MW.

QUESTION:
What ideas are the basis of the special theory of relativity and what conservation laws does it preserve?
This isn't from homework, but a class discussion that I didnt understand

ANSWER:
This could be called an unfocused question, but the answer is pretty
brief so I will carry on. Most textbooks will tell you that you need to
postulate that the laws of physics are the same in all inertial frames
of reference and accept that the speed of light (electromagnetic
radiation) is independent of the motion of the observer or of the
source. My take on this is that only the first postulate is needed
because Maxwell's equations are laws of physics and they predict that
the speed of light depends only on two physical constants, not on the
motion of the source or observer. All important conservation laws —energy,
linear momentum, angular momentum—are preserved but only if
changes to the definition of these quantities are changed. E.g .,
kinetic energy is no longer ½mv ^{2} and linear
momentum is no longer mv ; the new quantities, however, are
approximately their classical values if the velocity is very small
compared to the speed of light.

QUESTION:
How does a torque applied to a wheel cause translational motion? Please don't say the ground pushes on the wheel.
The wheel is stationary at the road surface. From my understanding friction creates a torque equal to the driving torque so where does the net accelerating force come from and where does it act? If you could explain with a free body diagram showing all the forces present that would greatly enhance my understanding.

ANSWER:
If you are asking someone to explain something you do not understand,
don't tell them what not to say! You have, in essence, asked me to not
answer your question because it is the friction which accelerates your
car forward. Let me see if I can explain it to you. Suppose you have a
car with its brakes on at rest on a road. You get behind the car and
push as hard as you can but, alas, it will not budge. What force is
preventing it from moving? The road exerts a static frictional force on
the car, even though it is "stationary"; then, clearly, you cannot say
that the road cannot push a surface in contact with it just because the
surfaces are not sliding on each other. Suppose that there were no
friction between the tire and the road and the car was at rest. No
matter how hard you push your engine which, through the transmission and
various linkages, exerts a torque on the wheels, you would go nowhere,
the torque just spinning the wheels. But, if there is friction, the
wheel, at the point of contact with the road, would exert a force on the
road (opposite the direction you want to go). But Newton's third law
tells you that if the wheel exerts a force on the road, the road exerts
an equal and opposite (forward) force on the wheel. That is the force
which accelerates that car forward.

QUESTION:
A steel pole weighing 1000kgs and is 9 metres long what would the weight of the pole be on the end if I tried to lift it from the end?

ANSWER:
First, let's get one thing clear—the weight of something is the
force which the earth exerts on it and never changes. Also, technically
a kilogram is a unit of mass but I will treat it as a weight since many
countries do. What you want, I believe, is the force you need to exert
to lift it.
It is not really clear what it means to lift "from the end". Two
scenarios suggest themselves to me:

Lift it
from one end while the other end rests on the ground

In order for the rod to be
just barely lifted off the floor at one end, the torque due to
F must be equal in magnitude to the torque due to W =1000,
9xF =4.5x1000=4500 or F =500 kg. The floor holds up the
other half of the weight.

Grab onto
one end and lift the whole rod keeping it horizontal.

It is
impossible to lift it horizontally using only a force at the end because
you could not exert enough torque to keep the rod from rotating due to
the torque caused by the weight. You might try to do it by using two
hands, one at the end pushing down and the other some distance d
from the end pushing up. In that case (see the figure above), summing
the torques about the end you find that F _{up} =4500/d ;
for example, choosing d =20 cm=0.2 m, F _{up} =22,500
kg. You can also find F _{down} by summing all forces to
zero, F _{up} -W-F _{down} =0=22,500-1000-F _{down}
or F _{down} =21,500 kg.

QUESTION:
Dear Physicist, my question relates to special relativity and time dilation. This question may sound dumb but it continues to bother me. I have a space ship which I am sending to Alpha Centauri, the distance is 4.3 light years. I am able to push the ship to .7 the speed of light.
According to my understanding of time dilation, events on the ship will slow down due to the increase in the mass put into the system because, I am approaching the speed of light (a fixed constant) is it correct events back on Earth, which is moving at a much lower percent of the speed of light, will pass at a faster rate? If so, then doesn't that mean increasing the velocity of the ship to get there faster is really self defeating because instead of reaching Alpha Centauri in 5.59 years, time dilation will cause the actual time measured in Earth time to be much longer?

ANSWER:
I am afraid you have some serious misconceptions. Time dilation says
that if you are moving with respect to me, I observe your clock to
run slowly . Do not mistake that
for saying that your clock appears
to run slowly —it might
appear to run faster or slower
than mine, but is running slowly. Furthermore, it has nothing to do with
mass increasing with speed. But how do you observe things? Your clock
appears to run just the same as if you were at rest; but wait, you are
at rest relative to yourself and you see me moving and therefore find
that my clock runs slowly! Now, the time that I see you getting to the
star is 4.3/0.7=6.14 years. But, you see it as a considerably shorter
time. How can that be if your clock is running normally and you clearly
see your speed as 0.7c ? The answer is length contraction—you
observe the distance to the star to be less than 4.3 light years by the
gamma factor, √(1-0.7^{2} )=0.71, or 3.1 light years so the
time will be 3.1/0.7=4.43 years. If you could get going a speed 0.9c ,
you could get there in less than half a year by your clock but still
6.14 by mine. I have given you a couple of links to the
faq page , but you might want to peruse that page
for more relativity answers which might interest you, particularly the
twin paradox.

QUESTION:
This came up in a discussion about the use electro-magnetic rails guns whose power is contained in the speed of a given mass when it slams into a stationary object as there is no explosive material involved, the energy produces is converting the kinetic energy of the projectile into heat on impact.

How much energy is contained in every stationary object on earth due to the rotation speed of the earth? I understand this this is only Potential energy and would require the object to be shielded from the earth's gravitational pull in order to release it but if a steel block weighing 100 kg were to no longer be held by gravity how much potential energy would it contain as it was flung from the earth rotational spin.

Also, once released from the earth's gravitational attraction, would you also need to factor in the speed of the earth's rotation around the sun to the calculations? The mass was travelling in two different directions (planetary rotational and orbital rotational) at extremely high speeds before the gravitational ties were cut?

And as a final factor, while trying to calculate this someone pointed out that the starting 100 kg piece of mass would instantly become a 0.KG piece of mass once it was no longer had any weight with respect to the Earth.

At that point, we pretty much tossed in the towel.
But, still feel that because we started with a 100 kg item that became weightless, that would not remove the potential energy it contained before gravity released it.
If afterward it were to crash into the moon or some other object in its path, wouldn't the amount of destruction still be equal to the energy it contained when it was "fired from the planet" much as a bullet fired from a gun?

ANSWER:
You have many misconceptions, so I have edited your question(s) by
itemizing the parts of your question. I will answer by parts.

The most
important misconception should be dealt with first. Energy is
something which depends on your frame of reference. If a mass is on
the surface of the earth it has zero kinetic energy if you are also
on the surface of the earth. If you view that same mass from a frame
moving along with the earth in its orbit but not rotating like the
earth is, it has the kinetic energy you envision, that due to the
earth's rotation. Kinetic energy is given by K =½mv ^{2}
where m is the mass and v is the velocity you see
it to have. The speed v of an object on the earth's surface
depends on the latitude λ , v=Rω cos(λ )
where R =6.4x10^{6} m/s and ω =7.3x10^{-5}
s^{-1} is the angular velocity of the earth. Let's just look
at the equator where v is largest, v=Rω= 6.4x10^{6} x7.3x10^{-5} =470
m/s=1050 mph; you should note that this is not really that large.
The kinetic energy of 100 kg flung into space with this speed is
K =½x100x470^{2} =1.1x10^{7 } J.

By now you
should appreciate that it depends on who is looking at your
projectile. Someone who is at rest relative to the sun would need to
calculate the speed and therefore add in the velocity of the earth
around the sun, call it V . But, this would get complicated
because it would depend on time of day the projectile was launched.
If you are standing right on the orbit and the earth is coming
toward you, the speed would be V+v an noon and V-v
at midnight; at other times it would be somewhere between these two
extremes.

You have
this all wrong. Weight is the gravitational force which the earth
exerts on the mass; the weight does not "instantly" become zero when
it is launched, it gradually decreases as it gets farther away from
earth. But, that is beside the point because the kinetic energy
depends on the mass, not the weight, and the mass stays the same
regardless of whether there is any gravity or not.

Whenever
the projectile stikes something, like the moon in your question, all
that matters is what its velocity is with respect to the moon .

QUESTION:
According to third law of motion a bird sitting on a tree branch cannot fly due to reaction then how does it fly?

ANSWER:
Why would you think that? The so-called reaction force is exerted down
on the branch, not on the bird. Let's dissect the physics of this bird.
The eagle on the left is sitting on a branch. There are two forces on
him —his weight W and the force
the branch exerts up on him F . These forces
are equal and opposite because of Newton's first law. But, what
about the reaction partner of F , the equal and
opposite Newton's third law partner? That force is the force which the
eagle exerts down on the branch and is not a force on the eagle
(which is why I did not even draw it in the force diagram of the eagle).
The picture on the right shows the eagle just as he is lifitng off. Now
there is an additional force L due to the lift
generated by the flapping wings. Since L+F>W , the eagle is no
longer in equilibrium and will accelerate upward. As soon as the talons
leave the branch, F disappears.

QUESTION:
How much force is required to pull 8 metal tubes (bound together) that weigh
28 lbs. (total) and are each 20' long off of a truck?

QUERY: What is sliding on what? For example steel rods sliding on steel truck bed.
How are they bundled? In such a way that the bundling material is sliding on the truck bed rather than the rods themselves?

FOLLOWUP:
Steel rods sliding on wooden truck bed.

ANSWER:
The only thing which matters is what the surfaces are and the force
N pressing the surfaces together. The force F to keep your
bundle moving with constant speed is given by F=μN where μ
is called the coefficient of friction and depends on what the surfaces
are. In your case, if the truck is level they are pressed together with
a force equal to the weight, 28 lb. For clean metal on wood, μ
can be anything between 0.2 and 0.6, so F will be somewhere
between 5.6 lb and 16.8 lb.

QUESTION:
If the average bullet travels around 1700 MPH, how far would you have to be for it to stop/slow down to the point of stopping (not including
gravity pushing it into the ground, assuming it was shot completely straight, and there was no wind)? I'm trying to get my friend to admit that it would be insane for someone to be able to survive a bullet shot by the bullet getting caught in a bible. I would imagine it would have to be about 1700 Miles away, True?

ANSWER:
It seems to me you are asking the wrong question. You do not want to
know how far a bullet will travel before it stops, you want to know
whether a thick book would stop a bullet. It is altogether possible for
a 1700 mph bullet to be stopped by a bible but it would depend on the
shape and composition of the bullet and the design of the bible (type of
paper, cover material, number of pages). Also, not all bullets go 1700
mph. Examples of book penetrations by various bullets can be seen
here . Incidentally, if there were no gravity the only thing
slowing the bullet down is the air; it would slow down to the terminal
velocity and then continue moving in a straight line with that speed. In
the real world (with gravity), it would slow down and fall and
eventually be falling straight down with the terminal velocity. To think
that it would stop after 1700 miles with no gravity is absolutely wrong.

QUESTION:
If you are holding up a picture frame, is there work being done?
I think no work is being done because there is no displacement. Is that correct?

ANSWER:
No work is being done on the picture frame because it does not move, but
your muscles are doing work. See an
earlier answer .

QUESTION:
Can energy be extracted from mass using quantum tunneling other that in the case of hydrogen fusion?

ANSWER:
Normally tunneling will release some energy. Alpha decay, for example,
starts with a heavy nucleus and ends with an alpha particle and a
lighter nucleus. The alpha particle has kinetic energy (which I am
calling released energy) and the mass of the lighter nucleus plus the
alpha particle is smaller than the mass of the heavier nucleus. I never
thought of fusion as tunneling.

QUESTION:
I'm in my car and I see a car coming behind me at a high rate of speed, I know the driver isn't going to stop and that I will be rear ended very soon.
Would it lessen the impact on my body if I were to touch bumpers with the car ahead of me? Wouldn't this somewhat add to the weight of my surroundings and create more resistance to forward motion caused by the car hitting me from the back?

ANSWER:
You are essentially increasing your mass and so I would think it would
be advantageous. Let's do a simple example to illustrate.
Suppose all collisions are perfectly inelastic—after the collision
all vehicles are stuck together. I will also assume all three vehicles
have the same mass, M , and that the incoming speed is v ;
I will also assume that the time t which the collision lasts is the same
for either the two- or three-car case. If it is a two-car collision,
using momentum conservation Mv= 2Mu or u =½v ;
therefore you experience an acceleration of a =½v /t
and a force of ½mv /t
during the time t , where m is your mass. If it is a
three-car collision, using momentum conservation Mv= 3Mu or u =v /3
w here u is the speed of the three cars;
therefore you experience an acceleration of a =v /(3t )
and a force of mv /(3t ) during the time
t.

QUESTION:
Can we conserve angular momentum about hinge point just at the instance of collision when a particle strikes hanging rod ?

ANSWER:
Only if the collision time is very small. If the collision is not
instantaneous, the rod will be rotating and both friction in the hinge
and gravity will be delivering impulse to the system during the
collision. So, you could say that it is never exactly conserved, but in
most cases it would be nearly conserved.

QUESTION:
This past weekend I was in Vermont and was walking down a relatively steep gravelly road when I started gaining speed unintentionally and it seemed as thought my feet couldn't keep pace with the top half of my body. I kept accelerating for about 20-25 paces and saw a tree looming in front of me when I fell down flat.
Is there a name for what happened and is there
anyway I could have stopped it once it began? I ask because this happened to me once before in almost the same conditions years ago and, at the time, I ended up with 2 missing front teeth, a black eye, and bruises/cuts on both forearms.

ANSWER:
I have had this happen to me and know the helpless feeling of not being
able to stop running. The red arrow in the picture represents your
weight, acting at your center of gravity somewhere in your trunk. This
force exerts a torque about your leading foot which tends to rotate you
in a clockwise direction. Imagine that this runner were not running but
just standing still. The only way to keep from falling forward (rotating
about that leading food) would be to exert an opposite torque and this
could only be done by your foot/ankle; but the requisite muscles are not
strong enough to exert the necessary torque. Think about it —if
you were just standing on level ground, how far could you lean forward
without falling forward? Not very far! And it is even worse if you are
running because you have a forward momentum which also is tending to
topple you forward if you try to stop. All that I can think of that you
can do to avoid a fall is to try to slow enough that you can get
yourself more vertical so that your center of gravity is vertically
above your feet. If the hill is not too long, you can keep running until
you get to the bottom, but you will be speeding up the whole way.

QUESTION:
How many g forces will a driver experience accelerating from 0 to 200 mph
in 1/4 mile, straight line acceleration.

ANSWER:
No homework. But there are lots of advertisers on this page which will help
with homework.

FOLLOWUP:
This is not homework.
It happens to come from one of the
country's top 100 trial lawyers who would have gone to medical school
instead if he had the math ability to figure this out myself.
Of course, I have lost two trials in 28 years and my time goes for $650 an h)p8
I will just have an associate do it for me tomorrow--if it is still of any concern.
This is the first time that I have ever used one of these "Ask Jeeves" sites and will be the last.
And my new Vette has a g force display so if I really wanted to know I would have gotten in and fkoored it instead of relying on some bottom feeder like you.

ANSWER:
Wow! I am so excited to get a question from one of the country's top 100
trial lawyers. I am so honored, even though I am just a tiny "bottom
feeder" in the presence of such a paragon. Sorry for the sarcasm, dear
readers, but this is a good opportunity for me to emphasize that one of
the important things about Ask The Physicist is that it is not a
homework help or tutoring site. I feel very strongly about this because,
having been a teacher for 40 years, I feel strongly that students should
do their own work and the internet gives too many opportunities for
"cheating" in that regard. I reject what I judge to be homework
questions and, inevitably, I occasionally make a mistake. If you think
my ethical stand on this issue makes me the scum of the earth, so be it.
When I do make a mistake, I usually answer the question and I will do
that here. The equations of motion for uniform acceleration for the
object starting at rest are v=at and d = ½at ^{2}
where, in this case, v =200 mph=89.4 m/s and d =¼
mile=402 m. The time can be eliminated using the first equation, t =89.4/a
and substituting into the second equation, 402=½a (89.4/a )^{2} =3996/a
or a =9.94 m/s^{2} . The acceleration due to gravity
is 9.8 m/s^{2} and so this is just about 1 g of acceleration,
a =1.01g . I figure that anyone who charges $650/hour
for his services can afford to send a little of that my way!

QUESTION:
This relates to Einstein's moving train thought experiment.
Suppose the train is moving at speed q*c relative to a trackside observer. This means the axels are moving at speed
q*c, but the wheel rims are moving at various speeds from 0 at the point in contact with the track to 2*q*c at the opposite end of that diameter. But hold on - if q is more than 0.5 that would result in parts of the wheel rim moving faster than c which is forbidden.
Does this imply that a train powered by wheels rather than electromagnets can only travel at speed less than .5 c ?

ANSWER:
Thought experiment means you do not worry about such details! What about
any q which is not extremely small relative to 1? The whole train would
burn up from the air drag. Regarding the wheels, just ask yourself if a
wheel could be on a fixed axle and rotate such that the speed at the rim
was say 0.8c . I doubt that you could fabricate any wheel which
could have a tangential speed 0.001c and not fly apart from the
stress of the centrifugal force. Think about a space ship in a vacuum!

QUESTION:
If a man pulls a 13 ton truck a distance of 120 ft in 28 seconds , what is his muscle strength?

ANSWER:
I assume that by "…his muscle strength…" you mean what
force does he have to exert (F in the figure). You have not
given me enough information. You need to also know how much frictional
force (f in the figure) is on the bus as it moves. Assuming that the bus accelerates from rest with a uniform
acceleration a , F-f=ma where m is the mass of
the bus. Switching to SI units, m =13 ton=12,000 kg and 120
ft=37 m. Then, from kinematics, 37=½at ^{2} =½a (28)^{2} =392a ,
so a =0.094 m/s^{2} . So, F-f =0.094x12,000=1130
N=254 lb. One could make a reasonable guess about the friction by
assuming that it is mostly "rolling friction" of the tires, with a
coefficient of rolling friction of about 0.03, so f ≈0.03x12,000x9.8=3530
N=794 lb; the man would therefore need to exert a force of approximately
794+254=1048 lb.

QUESTION:
I have a dispute with a friend. Playing baseball as an outfielder, I learned to catch the ball by staying back for as long as I could and still catch the ball.
This was because sometimes (particularly on hot days in the afternoon) the ball would suddenly rise. Sometimes this would be a foot or more and you would look the fool as it went over your head.
I maintain that this was due to thermal currents from the sun heating the large outfield surface. The currents at times were sufficient to raise the already buoyant body(the ball) as it traveled along.
Am I out of my mind?

ANSWER:
How do you conclude that the "currents
at times were sufficient to raise the already buoyant body(the ball)"? I
can guarantee that no thermal is going to cause a baseball to rise. It
could
keep the ball aloft longer, but can you imagine a
ball released at rest in one of these currents actually accelerating
upwards? What is most likely to cause unexpected behavior is spin
imparted to the ball when it leaves the bat. In particular, back topspin
can cause a ball to appear to rise because it is falling less than it
would with no spin. See the earlier answer on the
rising fastball .

QUESTION:
From the physics theories and calculation, how does a crane fall of a bridge when trying to pull out a bus from the water? (crane = 28 ton, bus = 14 ton)

ANSWER:
It's all about the torques. In the figure, if the crane tips over it
will rotate about the rear wheel. When this is just about to happen the
front wheel has zero normal force from the road on it. At this time, the
road exerts a 42 ton force up on the rear wheels (they are holding up
all the weight). The weights each act at the center of gravity of the
crane and the bus. If I sum torques about the rear wheel, they must
equal zero, that is 28D =14d or d /D =2.
So, if d is greater than 2D , the whole system will
rotate about the rear wheels and topple off the bridge.

QUESTION:
I recently saw that the Chinese space station will have an uncontrolled reentry in 2017 and a 100 kg chunk of molten steel(engine part) will survive from the 370 km orbit to the earth surface. I tried to do the terminal velocity calculation via a calculating program. I was guessing at some of the parameters, but the mph speed it came up with seemed very slow for what I expected. I am sure I misunderstood some of the equation. I am assuming the object will be somewhat spherical molten steel by the time it arrives. what would be a good estimate of its speed/velocity?

ANSWER:
The density of steel is about 8000 kg/m^{3} , so the volume of
the steel is about 100/8000=0.0125 m^{3} . If it is a sphere its
radius is about 0.144 m and its cross-sectional area is about A =0.0661
m^{2} . At sea level a pretty good approximation for the air drag
is F = ¼Av ^{2} (only for SI
units), so terminal velocity will be achieved when mg =¼Av ^{2} ,
or v =2√(mg /A )=62.6 m/s. Just an
estimate. Also, since it experiences less drag at higher altitudes
because of thinner air, I cannot guarantee that it would lose enough
speed from its orbital speed of about 8000 m/s to get to the
ground-level terminal velocity.

QUESTION:
I am a writer. A plot idea came to me, and I am perplexed as to how to explain it using physics. I refuse to write something if there is no way it could plausibly happen, so I am hoping you can help me with some explanation as to how this could happen.
The premise is, some celestial event occurs that causes Earth to be repeatedly, and fairly regularly bombarded with devastating asteroid strikes. These would be sufficient enough to destroy our infrastructure, plunging the world into darkness and challenging the heros to survive what would seem to them, a shooting gallery.
I was wondering, if there is some way to explain how this could happen, perhaps an exoplanet disrupts one of the asteroid belts on a regular basis, perhaps by its orbit passing through it.
I know this may seem unusual, but as a writer of hard sci-fi, I refuse to do a project, no matter how intriguing it is, if I can not offer at least some plausible explanation, even if it is a bit stretching the limits, to justify my work. I trust you might take sympathy on me and help me with this, because I would hate to think this idea is a dead end.

ANSWER:
The only possibility I can envision is the one you suggest —a
planet-sized object somehow gets inserted into an orbit which
periodically passes through the asteroid belt. The period of a typical
asteroid in a roughly circular orbit is about 3 years, so I imagine an
elliptical orbit for our rogue planet which has a semimajor axis about
equal to the radius of the asteroid belt as shown in the picture.
Kepler's first law then says that this orbit would also have a period of
about 3 years and therefore pass through the asteroid belt about once
every 1½ years; it would pass through the asteroid belt twice
each orbit spending only maybe a couple of months in the belt. So, we
are off to a good start for your scenario! To get some numbers I found
the a
Physics Stackexchange discussion useful. An estimate of the
distance between asteroids larger than 1 km is shown to be about 2
million miles (about 3 million km). This alone is sort of eye-opening—we
tend to think of the asteroid belt, ala sci-fi movies like Star Wars, a
dangerous densely-packed hazard whereas your rogue planet would be
extremely unlikely to encounter even a single one as it passed through
the belt. With a passage taking only a couple of months, the likelyhood
of even seeing a single large asteroid is small, let alone being close
enough to significantly alter its orbit. But, suppose that a large
asteroid is shaken loose. Now you have to think of the probability that
it will collide with the
earth; this will be even tinier than the probability of its having been
kicked out of the belt in the first place. I have not done any
quantitative probability calculations because I think the main point can
be made with very rough qualitative estimates as I have done here. The
bottom line is that space is unimaginably vast and even in places where
the density of objects is relatively large, it is still
extremely small in absolute terms. If this rogue planet were to appear,
I doubt that, on average, even a single asteroid collision with earth
would occur in a lifetime.

QUESTION:
Given a 15 inch standard automobile tire with an average steel wheel attached. What is its terminal velocity and at what height must it be dropped to reach it before impact?
I am a math teacher trying to inspire higher thinking in a group of Auto Tech high school guys who had a bit of a disagreement about the answer.

ANSWER:
Since you are a math guy I will give you more detail than I normally
would. Since your students are not math guys, you can use your own
judgement regarding how to deliver this to them. Keep in mind that any
calculation of air drag is approximate. I took the tire plus wheel mass
to be 55 lb=25 kg, the radius to be 15"=0.38 m, and the tread width to
be 10"=0.19 m. It is pretty easy to to understand how to get the
terminal velocity. The forces are the weight mg down and air
drag D up and their sum must be the mass m times the
acceleration a of the tire: ma=-mg+D ; this is just
Newton's second law. When the tire is falling with constant velocity (v _{t} ),
a =0 so mg=D . Now, what is D ? For this kind of
situation the drag is proportional to the square of the velocity —double
the speed and you quadruple the drag; a pretty good value of the
proportionality constant is A /4 where A is the area
presented to the onrushing air, so D ≈¼Av _{t} ^{2}
(this works only for SI units). What A is depends on how it is
falling; I will assume it falls with one face down, not edgewise, so
A=πR ^{2} =π x0.38^{2} =0.45 m^{2} .
Now we can estimate the terminal velocity, v _{t} ≈2√(mg /A )=2√(25x9.8/0.45)=47
m/s=105 mph. Now you want to know how high to drop it from. Well,
technically it is infinitely high because the speed approaches v _{t}
and never really reaches it. So, to get a good idea we can ask how long
it takes to get to 0.99v _{t} and how high h it
was dropped from. Solving the problem for y (t ) and v (t )
involves some pretty
tricky integration. The solutions are v (t )=v _{t} tanh(gt /v _{t} )
and y (t )=h -(v _{t} ^{2} /g )ln[cosh(gt /v _{t} )].
So, setting v (t )=0.99v _{t} and
solving for t , I find t= (47/9.8)tanh^{-1} (0.99)=12.7
s. Finally, setting y (12.7)=0, we can get the height, h =((47^{2} /9.8)ln[cosh(9.8x12.7/47)]=442
m=1450 ft.

Obviously, this math will be too much for your guys. It occurs to me
that they should be able to read a graph, though, so I have calculated
graphs of the functions below.

QUESTION:
Due to the ideal gas law/approximation the speed of sound does not change with altitude or barometric pressure. Yet ask any long range shooter and they will tell you, from practical experience, that the lower the barometric pressure or the higher the altitude, the less drag there is on the bullet and therefore the bullet takes longer to slow resulting in less time for gravity to affect the bullet leading to less "drop" for the bullet over long ranges. Why then doesn't the ideal gas law apply to bullet trajectory?

ANSWER:
I do not know about the speed of sound, but I would certainly not assume
that anything about sound could be applied to bullets. The drag force
F _{d} of a bullet of speed v in air is well
represented by F _{d} = ½Cρv ^{2}
where ρ is the density of the air and C is a
constant determined only by the geometry of the bullet. As the density
of the air gets smaller, the drag on a bullet gets smaller, in accord
with your experience. There is really no need to invoke the ideal gas
law at all, but if you want to connect density to the ideal gas law,
PV=NRT , note that the density will be proportional to N /V ,
so ρ ∝ P /T . For
constant temperature, density is proportional to pressure—the
lower the pressure, the lower the drag.

QUESTION:
If you had two point masses m_{1} and m_{2} in space separated by a distance d, how long would it take for them to collide under gravity alone? I'd imagine it's a changing force/acceleration, so some calculus must be involved.

ANSWER:
I have solved this problem many times before but always for specific
masses and distances. To avoid the calculus you refer to, the trick is
to use Kepler's laws for planetary motion. You can find references on the
FAQ page. The one you
might find most useful is this
link . You will see that
the relevant time to collision is t=T /2=√(πμa ^{3} /K)
where μ =m _{1} m _{2} /(m _{1} +m _{2} )
is the reduced mass, a=d /2, and K=G m _{1} m _{2} .
Therefore, t = √[πd ^{3} /(8G (m _{1} +m _{2} )].

QUESTION:
What is the cause of drastic variability among very small nucleons on the plot of binding energy/mass number?

ANSWER:
I think you must mean "very small nuclei", not nucleons. The force felt
by nucleons in nuclei is very short ranged but the nucleus is very
small. Therefore in heavy nuclei each nucleon sees the force from all
the others as an average force and the removal or addition of another
nucleon makes little difference in the force felt. In light nuclei,
however, the addition or removal of a single nucleon can and does make a
significant difference because there are not enough constituents to have
reached what is referred to the saturation of the force.

QUESTION:
Would it be possible for an 'average' human to throw a 500 g weight 40 ft but
with only a 2 ft high corridor of travel without the projectile touching any
of the corridor's sides? This isn't homework, it's a really ridiculous question that we can't physically try at work! We work in a theatre that has two people positioned 40 ft apart and roughly 30 ft up in the air near the ceiling. Ages ago someone said that they could throw a spanner from one person to the other person. Some people agreed, some said that it wasn't possible due to the arc needed. This question comes up every few months and I just wondered if you might have an answer.

ANSWER:
So the problem is, with what speed would you need to throw the spanner
in order that it reach a maximum height of 2 ft above its starting
height when it has gone 20 ft? This is a straightforward kinematics
problem. You are probably not interested in the details, so I will just
give you the final result. The necessary speed would be 57.7 ft/s=39.4
mph. A major league baseball pitcher can throw a fastball with a speed
of around 100 mph, so there is probably somebody in your theater company
who could throw the spanner with the necessary speed, but I would not
want a 1.1 lb wrench coming at me with a speed of nearly 40 mph!

QUESTION:
In some of the more realistic space combat science fiction there is a concept called a 'BFR' (Big Fast Rock) which, mined from a dead world or asteroid, melted to molten and then reformed to a near-perfect density distribution with collars of ferrous metal impressed in it, is shot at some fraction of light speed from a large EML cannon running down the long axis of mile long ships. I would like to know how to calculate the impact force release for a 2,000 lb BFR moving at .10, .15 and .30 C. I'm assuming it's going to be in the high Megaton range and I don't know what the translative math per ton equivalence in TNT.

ANSWER:
What you want is the energy your projectile has when it hits. The energy
in Joules is E =mγc ^{2 } where γ= √(1-(v /c )^{2} ).
In your case, 2000 lb=907 kg, c =3x10^{8} m/s,
γ= 1.005, 1.011, and 1.048. The energies in Joules are
E =8.204x10^{19} , 8.253x10^{19} , and 8.555x10^{19}
J. There are about 4x10^{9} J per ton of TNT, so the energies
are 20.51, 20.63, and 21.39 Megatons of TNT. I might add that this is
not actually very "realistic". Where are you going to get that much
energy (you have to supply it somehow) in the middle of empty space? Or,
look at it this way: I figure that for a mile-long gun the time to
accelerate the BFR to 0.1c is about t =10^{-4}
s. During that time the required power is about 8x10^{19} /10^{-4} =8x10^{23}
W=8x10^{14} GW; the largest power plant on the earth is about 6
GW! Also, don't forget about the recoil of the ship which would likely
destroy it.

QUESTION:
If atoms are in a constant state of flux between particle nature and wave-function is this supposition supported by the
2014 discovery that the uncertainty principle and wave-particle duality are in fact the same thing?

ANSWER:
I find this question puzzling. I have read the article about the
"2014 discovery", but I do not understand how it can be a "discovery".
It has always been obvious to me that wave-particle duality
is an inescapable consequence of the uncertainty principle. (This same
view is stated many times in the comments on this article.) When the
position wave function envelope narrows, the linear momentum wave
function broadens because the two are conjugate variables and thus
Fourier transforms of each other. I also
consider "a constant state of flux
between particle nature and wave-function" to be a misstatement; I
consider duality to mean that the particle/wave is simultaneously
particle and wave, not some dance between the two.
By making an appropriate measurement you can observe either one or the
other aspect.

QUESTION:
I have a question related to ballistics. How would a gun that doesn't recoil and eject gases when the shell leaves function?

ANSWER:
It is not possible to make a gun which does not recoil. The bullet
carries away linear momentum and the gun must recoil, at least a little.

QUESTION:
Does a 1 kg block drop to the ground slower or faster then it did 50 yrs ago?

ANSWER:
Every year the earth's mass increases by several hundred tons because of
the constant rain of meteorites. So, technically, the earth's mass is
increasing and therefore the acceleration due to gravity also increases.
However, over a period of 50 years the increase is so tiny compared to
the mass of the earth that you could never hope to measure the change.

QUESTION:
If any particle drop falling from height h, then velocity increases but in case of photon with light veloctiy c falling drop then what changes occurs, it will increase velocity or not?
If velocity of photon increase then how its possible because no one have
velocity is greater then speed of light?

ANSWER:
A photon gains energy when it "falls" in a gravitational field, just
like a rock does. But, it does not speed up because the speed of light
is a universal constant. The energy of a photon is proportional to its
frequency, so as its energy increases its frequency increases. As light
travels (with constant speed) in the direction of a gravitational field
(down) it shifts toward the blue end of the spectrum and this is called
a blue shift. As light travels (with constant speed) opposite the
direction of a gravitational field (up) it shifts toward the red end of
the spectrum and this is called a red shift.

QUESTION:
A tire rolling on a level surface at a linear speed of 10 MPH rolls on to a conveyor belt which is also moving at 10 MPH in the same direction. How will the tire's speed change? Will it be 20 Mph? 10 MPH? Will it's rotation stop?
Reverse?

ANSWER:
This problem is a little trickier than I had anticipated. On the other
hand, the final answer is much simpler than I had anticipated. Shown in
the figure to the left is (top) the situation when the rolling tire
first touches the conveyer belt. It is rolling without sliding so the
point of contact with the floor is at rest, the center is moving forward
with a speed v (your 10 mph), and the top is moving forward
with speed 2v ; the belt is moving forward also with speed v .
I find this problem much easier to do if I transform into a coordinate
system which is moving with speed v to the right; in that
coordinate system the belt (upper surface) and tire are both at rest and
the top and bottom edges of the tire have speeds v as shown.
Before getting into the hard part, there is a special case which we can
get out of the way first: if there is no friction the tire will continue
on its merry way unchanged, both its speed and its angular velocity
unchanged because there are no net forces or torques on it.

Now,
as soon as it gets on the belt there will be a frictional force
f trying to accelerate it to the right so it will
start sliding along the belt (at rest in the frame we are using). The
frictional force will be f=μmg where μ is the
coefficient of kinetic friction, m is the mass of the tire, and
g is the acceleration due to gravity. Call v' the
speed which the center acquires in some time t . Then Newton's
second law for translational motion is m Δv=μmgt=mv' ,
so v'=μgt.

There
will also be a torque τ=fr=μmgr which acts opposite the
direction the tire is rotating. There will come a time when the bottom
edge of the tire will be at rest relative to the belt because of this
torque. At that instant, we can transform back into the original
coordinate system and the tire will be moving forward with speed
v+v' . Newton's second law for rotational motion is ΔL =τt=- (μmgr )(v' /μg )=-mv'r .
where L is the angular momentum of the system. The angular
momentum is the angular momentum about the center of mass plus the
angular momentum of the center of mass. L _{1} =Iω _{1} +0=Iv /r
where I is the moment of inertia about the center of mass;
L _{2} =Iω _{2} +mrv' =(v' /r )(I+mr ^{2} ).
Therefore, (v' /r )(I+mr ^{2} )-Iv /r=-mv'r
or, v'=Iv /(I+ 2mr ^{2} ). This is
surprisingly simple, particularly surprising that it does not depend on
μ. However, the time to stop sliding does depend on μ ,
t=v' /μg ; the slipperier the surface, the longer it
takes to stop slipping, as expected.

Finally
we need to transform back into the original coordinate system by simply
adding v as shown in the final figure. As an example, suppose
we model the tire as a uniform cylinder of mass m and moment of
inertial I =½mr ^{2} ; then v'=v /5,
or in your case, 2 mph, so the tire is rolling without slipping with a
speed of 12 mph.

ADDED
NOTE:
Note that the new angular velocity of the tire is ω'=v' /r
only a small fraction of the original value of ω=v /r.
This drop in angular velocity can be clearly seen in a
Mythbusters
episode showing a car driving onto a ramp from a moving truck, a
very similar situation to the conveyor belt problem in this question.

QUESTION:
Is most of the kinematics and mechanics I learn in high school and freshman college physics applicable to other planets?
Would any ideas change if I was a citizen on the planet mars or the moon and went to a university there?
Would the kinematics/mechanics content learned on Earth be the same for a student who was living on the moon or mars?

ANSWER:
To the best of our knowledge, the laws of physics are applicable
anywhere in the universe. You do have to be careful to be cognizant of
local conditions though; for example, if you calculated the path of a
projectile on Mars using g =9.8 m/s^{2} , it would be
wrong.

QUESTION:
What would be the result if an object have mass value more than earth and we will throw that object from earth surface?

ANSWER:
Momentum would be conserved. For example, if the mass were twice as
large as the earth's mass and it had an initial speed of v (as
seen by an outside observer at rest) the earth would recoil with a speed
of 2v . But, due to the gravitational attraction between the
two, they would go out to some distance and fall back to where they
started, again at rest.

QUESTION:
According to newton's third law of motion if someone applies force on say, a table, the table exerts an equal and opposite force. How's it possible,doesn't applying force require energy, e.g. muscular energy is used when a person pushes (applies force) on the table.

ANSWER:
A force does not require energy. Just imagine a table sitting on the
floor; the table feels a force down due to its weight; but the table is
sitting at rest on the floor and the floor exerts an upward force on the
table equal to the weight. None of these forces (gravity, floor on
table, table on floor) requires any energy. A force can be used to
deliver energy but only if it is applied over some distance. Now, you
know that if you exert a force for a long time you get tired; so energy
must be being used. You can find links to earlier questions addressing
work done by muscles on the faq page.

QUESTION:
My wife and myself were arguing over whether or not air goes into an air mattress faster if the air mattress is unfolded I said it does because it's meeting no resistance by having to unfold the mattress she said it doesn't matter that the air enters at the same speed no matter what.
Can u help solve this debate before we get divorced?

ANSWER:
Suppose that you do an experiment and time it both ways. Physics is an
experimental science after all. The precise time will depend on how it
is folded, but I believe that you will find that it is fastest if
unfolded. And, no, the air does not "enter at the same speed no matter
what". Imagine that you are sitting on the folded matress —air
would not enter at all.

QUESTION:
since clay doesn't bounce how was clay, on top of a basketball, able to bounce up and maybe hit the ceiling ?

ANSWER:
See an earlier
answer .

QUESTION:
Two frames xOy and x'O'y' are in uniform motion along their x axes. We will
consider for simplicity the first frame to be "fixed" while the second one
moves to the right with a velocity v. From the origin of the moving frame
x'O'y' two rays of light are emitted simultaneously, one along the axis O'x'
and the other one at an angle of 60 degrees with the O'x' axis. Two mirrors
are placed at the same distance l on the two tracks and the light gets
reflected. Obviously, the two reflected rays, as observed in the frame x'Oy'
return to 0' at the same time. I made some calculations for v = c/2 and l =
1,000,000 m. Surprisingly, the two rays, as observed from frame xOy, do not
appear to meet return to the origin at the same time. I can share my
calculations with you. It is possible that I made a mistake. However if my
calculations are correct, this would be a very strange thing indeed.

ANSWER:
First of all, it would not be " …a
very strange thing indeed… " to find what is
simultaneous in one frame is not in another. One of the keystones of
special relativity is that simultaneaty of two events depends on the
frame of reference. I have not included your solution to the question
because I prefer to work it out myself rather than troubleshoot your
work. The figure shows the situation as seen by each observer. The
primed system moves in the +x direction while the red-drawn
distances to the mirrors are at rest in the unprimed system. The rest
lengths of the two distances are L and the angle one makes with
the x -axis is θ . The round-trip time along the
x -axis is t _{1} =2L /c and
along the other length is the same, t _{2} =2L /c .
In the moving system all lengths along the the x' direction are
reduced by a factor √(1-β ^{2} ) where β=v /c ;
therefore L'=L √(1-β ^{2} ) so t _{1} '=2L' /c =2L √(1-β ^{2} )/c=t _{1} √(1-β ^{2} ).
For L" only its x' component is contracted, L" cosθ' =L √(1-β ^{2} )cosθ
while the y" component remains the same, L" sinθ' =L sinθ ;
from these you can easily show that L"=L √(1-β ^{2} cos^{2} θ )
and tanθ'= (tanθ )/√(1-β ^{2} ).
Finally, t _{2} ' =2L" /c =[2L √(1-β ^{2} cos^{2} θ )]/c =2t _{2} √(1-β ^{2} cos^{2} θ )≠t _{1} ' .

ADDED
COMMENT: I see that I misread your question. I see that
you have put the lengths at rest in the moving frame whereas I have them
in the stationary frame. But in special relativity it makes no
difference which is moving and which is at rest. So, you have found that
the times in the frame not at rest relative to the lengths are not the
same, just as I have; I reiterate that this is expected, not surprising.

FOLLOWUP
QUESTION: Thanks a lot, your competent and prompt
answer is much appreciated!

I would like to reiterate that for two events that occur in the same place to be simultaneous in one frame but not in the other frame is really very strange, the non-existence of absolute simultaneity being a central tenet of Einstein's relativity not withstanding. For suppose that two light pulses when they arrive simultaneously to O' (according to a frame x'O'y' observer) they trigger an event, perhaps an explosion for which the simultaneous arrival of the two pulses is a necessary condition. So the explosion is the result of the simultaneous arrival of the two pulses for the x'O'y' observer but not for an observer in the frame xOy (who observes the explosion, too), since the latter sees the two pulses arriving at different times and thus cannot correlate the explosion to the two light pulses. The x'O'y' - observed concurrent emission of the two light pulses that go out from the very same point O' may be in an analogous way correlated to some event in the x'O'y' frame, but not in the xOy frame.

One can think with some justification that this scenario means that the x'O'y' system is somehow more "special" than the xOy system, pretty much in contradiction, if not with Einstein relativity postulate proper, at least with its spirit. In other words, in my opinion, the relativity theory is not devoid of inner contradictions, or at least it leads to conclusions which seem to invalidate the most fundamental notions of cause and effect or of things being what they are irrespective of where we observe them. And to me it is debatable whether a physical theory, however legitimately counter-intuitive it may be understood to be, can lead to this type of conclusions and be nevertheless considered a valid description of the physical world.

ANSWER:
Again, I would like to reiterate that it is neither strange nor
unexpected! First of all, the two events in the moving frame are at
neither the same time nor at the same place; only in the stationary
frame are the two events simultaneous and at the same place. By
"stationary frame" I mean here the frame where the apparatus is. Now,
your followup expands the apparatus to include also a "trigger" and a
bomb. The trigger will be some sort of electronic device to detect the
simultaneaty of the two light pulses' arrival at the origin and will
detonate the bomb in that case. An observer in the moving frame will not
see any contradiction, he will not say
that the bomb will not detonate because the device which detonates it is
in the stationary frame and he will agree that the pulses in that frame
are indeed simultaneous. Is the stationary system in some way "special"?
Of course it is —that is where the apparatus resides and all
frames will agree that the bomb will be triggered in that frame.

QUESTION:
I was wondering if, when an object enters earth's atmosphere from space, does it experience a lateral force as it enters the atmosphere? In other words, does it suddenly get "pushed" sideways due the earth's rotation?

ANSWER:
This had never occurred to me, but certainly the meteorite will be
deflected because of the "wind" of the rotating atmosphere. But, how big
an effect is this? The average speed of a meteorite is about 17 km/s;
and I calculate the speed of the "wind" to be about 0.47 km/s at the
equator. So, even if the meteorite acquired all of the speed of the
atmosphere, it will still be trivially small compared to the vertical
speed. Another way to look at it is to look at the drag forces the
meteorite experiences when it encounters the atmosphere; these are proportional
to the squares of the speeds, so the ratio of the horizontal force to the
vertical force will be about 0.47^{2} /17^{2} =0.0008. Compared to the force slowing down the vertical motion, the force
speeding up the horizontal motion is tiny.

QUESTION:
Suppose two charged particles having same charge (say two electrons) are at finite distance. So initially there is a electric repulsive force between the two particles. Now one particle is moving in a particular direction. At that time what is the force between them? A varying electric repulsive force or a magnetic force?

ANSWER:
You can calculate the electric and magnetic fields of each particle (see
an earlier answer ). If both particles are at
rest, each feels the electric field of the other as you state. If one
moves, the other sees only the electric field of the moving particle,
not its magnetic field; but the electric field of the moving field will
not be the same as when it was at rest so the force felt by the other
will be different; if the speed is small compared to the speed of light,
the change in electric field will be negligibly small.

QUESTION:
I have a question related to determining the force at impact.
Here's the question in two parts:

If a firefighter adds 45 pounds of gear to his overall weight, does it
increase the impact force if he has no choice but to jump out a window and
if so to what degree?

What is the effect on impact force of jumping out a second floor window
vs. 3rd floor and above?

This actually isn't homework. I'm 54 years old and advising a charitable organization that provides safety systems to firefighters. One of the challenges they face is fire departments who don't have high rise buildings and feel they don't need the bailout systems. So I'm trying to figure out what the impact is for a firefighter forced to jump out a second story or third story window. This will help inform how they talk to prospective donors.
If you could help me with that (understanding/identifying the increase in impact force from 1st to 2nd to 3rd floors and then up) it would be greatly appreciated. I've done a lot of googling around calculating impact and g forces, but its not making sense to me (I'm not being lazy, just not understanding).

ANSWER:
The "force at impact" depends, essentially, on the time to stop. If she
has a speed v , a mass m , and stops in a time t ,
the average force F during the stopping time is given by
F=m (g+v /t ) where g= 32 ft/s^{2} is the acceleration due to
gravity. So, yes, if you increase the mass you increase the force
proportionally; I guess I would toss that extra 45 lb over before I
jumped! Regarding your second question, call the height of one floor
h . Jumping from the second floor would result in a speed of v _{2} = √(2gh );
jumping from the third floor would result in a
speed of v _{3} =2 √(gh ) which is
about 1.4 times larger than v _{2} . In general, jumping
from the n ^{th} floor would result in a speed v _{n} =√[2(n- 1)gh ].

Maybe some numerical examples would be useful to you. You prefer
Imperial units, which makes things a little complicated. The quantity
mg is the weight. I will take mg =160 lb and h =12
ft for my numerical calculations, so when I need the mass I will use 160
lb/32 ft/s^{2} =5 lb·s^{2} /ft. (The unit of mass
in Imperial units is called the slug, 1 slug=1 lb·s^{2} /ft.) The speeds from
second and third floors will then be v _{2} =27.7
ft/s=18.9 mph and v _{3} =39.2 ft/s=26.7 mph; these
speeds are independent of the mass. Finally we must approximate the
times to stop. If she lands feet first, she could extend, as
parachutists do, her stopping time by bending her knees into a squat; I
will estimate the distance s she will travel while stopping as
s =3 ft. The time may be shown to be t =2s /v
so t _{2} =2x3/27.7=0.22 s and t _{3} =2x3/39.2=0.15
s. Finally, the average forces during impact are F _{2} =160+5x27.7/0.22=790
lb and F _{3} =160+5x39.2/0.15=1467 lb. These are
approximately 5 and 9 times her weight (g-forces of about 5g
and 9g ).

QUESTION:
I want to ask about the Mandela effect theory. I do not want detailed answers concerning blackholes and cosmology,instead i want to know if it's real. I personally do not believe in time travel and all the "clues" people have been providing don't seem credible enough. So please if it is true what is the exact physics background that proves it? And how can i have further information and dig deeper into this theory?

ANSWER:
I never heard of the
Mandela effect . For readers who, like me, are similarly ignorant it
is a special case of a psychiatric phenomenon called
confabulation ,
defined as "the production of fabricated, distorted or misinterpreted memories about oneself or the world, without the conscious intention to deceive";
in other words, a "false memory". The Mandela effect is a confabulation
where a large number of people have the same same false memory; it is
apparently so called because there are many people who "remember" that
Nelson Mandela died in 1980, not 2013. So, where is the physics in all
this? A "paranormal consultant" named
Fiona Broome
suggested that such cases could be explained as "bleed through" from
parallel universes. She is not a physicist and her speculation, maybe
amusing, has no real basis in physics and there is no way to test it one
way or another; therefore it should not be labeled as a "theory" since
it is neither testable nor falsifiable.
Snopes.com has a long article on the Mandela effect which you can
read if you want to learn more.

QUESTION:
As surface tension means force acting on surface then why not its unit if force per unit area i.e N/m^{2} .
If it is N/m then which length will we have to take here.

ANSWER:
Think of the surface as an elastic sheet. What you want to call the
surface tension is how hard you need to pull on this sheet in order to
make it rip. The way this is usually measured is to create a film which
is actually a very thin volume with surfaces on both sides; a simple
device to do this is shown in the figure. Now, pull on the sliding side
to stretch the film. When the film breaks, measure F . It is
found, by doing many experiments, that F depends on L ,
but that the ratio F /L is always the same.
Furthermore, it makes no difference what the shape of the film to the
left of the slider is. We therefore define the surface tension for this
fluid to be γ=F /(2L ) as determined by
this experiment, the factor of ½ because there are two surfaces
we are stretching.

QUESTION:
If you can rotate a magnetic field in the opposite direction of the earths spin At the same MPH as the earth spins can it or will it counteract gravity and allow Us to overcome it?

ANSWER:
I do not see how you could ever do that. However, if the earth's
magnetic field needs, for some reason, to be cancelled out, you can do
it easily in a limited region of space by simply adding a field in the
opposite direction. This is usually done using Helmholtz coils.

QUESTION:
Does standing with a bag of 10kg need a lot of energy or just a little?

ANSWER:
According to physics, it takes no work (therefore no energy) to hold
something stationary. However, work is done by muscles in holding
something, even if you do not move it. See faq
page .

QUESTION:
Well
I hope this doesn't seem off the wall but I want to know how much force
it would take to tip a 7.62 m tall,6.40 m wide and 22 ton Transformer over? It is for a story I'm doing about a comic book character.

ANSWER:
It depends on where and how the force F is
applied; if applied as shown in the diagram, it will be the smallest
possible.
I will assume that the center of gravity of the transformer is in the
geometrical center, so the weight W acts
there. When the force is just about to tip it over, all the force from
the floor, the normal force N and the
frictional force f , act on the edge opposite
where the force is applied. It will be in equilibrium, so you can easily
see that F=f and N=W . But, what you need is F
and to get it you need to sum the torques around the edge where N
and f act and sum to zero: Fh-Ww /2=0 or F=Ww /(2h ).
Using your numbers, F =22x6.40/(2x7.62)=9.24 tons. If you want
the force in Newtons, assuming that the 22 ton mass is metric tons
(22,000 kg), F =9.8x9240=90,500 N.

QUESTION:
What is the decrease in weight of a body of mass 500 kg when it is taken into mine of depth 1000 m? (R=6400 km )

ANSWER:
I am guessing this is a homework question where you are supposed to
assume that the earth is a uniform sphere. But, guess what: the earth is
not a uniform sphere. As shown in an earlier
answer , the gravitational force probably increases slightly or stays
about constant to a distance from the center of about 4000 km.

QUESTION:
W e all know that there is a nuclear force that affects protons and neutrons.
I was wondering which is more stable two neutrons connected together by this force or one neutron and a proton considering the force of
mass attractions. If you can, provide mathematical equations.

ANSWER:
The nuclear force is much more complicated than the electrostatic force
or the gravitational force. Those forces are simple central forces whose
magnitude and direction depend only on the distance between the
particles. Therefore I cannot " …provide
mathematical equations…" I can tell you, though, that a proton
bound to a neutron, called a deuteron , exists in nature. It is
the nucleus of heavy hydrogen, ^{2} H. There is, however, no
bound state for two neutrons (referred to as a dineutron) —it
does not exist; the same is true for the diproton.

QUESTION:
I just came across an ad saying that
the earth's magnetic field keeps us healthy and strong, and saying that if we live in the world with no magnetic field, all life on earth will end (assume that sunlight / water / oxygen exist / and for some weird reason gravity still works)
So my question is, is this valid? If there is no magnetic field, even though we have sunlight, water and air, we would perish?

ANSWER:
The
magnetic field has no direct effect on you. There are many magnetic
products touted as having miraculous beneficial effects on your health;
do not buy them, they are a scam. There are important indirect effects
of the magnetic field which help provide a healthy environment at the
surface of the earth, though. Probably the most important is that the
field extends far into space and protects the earth from the solar wind
which is a stream of charged particles (mainly electrons and protons)
which are deflected around the earth by the field. If there were no
field, the effect of this "wind" would be to strip the upper atmosphere,
including the ozone layer which protects you from intense ultraviolet
radiation from the sun. But I do not think that buying a magnetic
matress pad, for example, would help that issue! Lack of a magnetic
field cannot result in our perishing, though, since it is well known
that the field reverses direction periodically; the reversal would not
be instaneous which would mean there would have been a period at least
hundreds of years long when the field was negligibly small. The most
recent reversal was about 780 million years ago when humans existed and
we survived somehow. One thing about your question is strange: " …for
some weird reason gravity still works …" The earth's
gravity has nothing to do with its magnetic field.

QUESTION:
How does iss withstand 3,600 degrees Fahrenheit In the thermosphere?

ANSWER:
Temperature is a measure of the average kinetic energy (hence, speed) of
the molecules. But, at high altitudes the density is extremely low so
the number of molecules hitting the ISS per second is too low to have
any significant effect.

QUESTION:
I am a retired chemist. I was talking with my 8 year old grand daughter about physical changes and the three states of matter. She was actually very interested... we were working with hot water and chocolate. But later I began to think about the forth state of matter, i.e. plasma. When water changes its physical forms it is still molecular water. My question: if water where heated to plasma levels what does it turn into and what would it be when the plasma cooled to room temperature.

ANSWER:
A plasma is defined as "a highly ionized gas containing an approximately equal number of positive ions and electrons"
(Dictionary.com ).
If you start with a molecular gas like water, the definition is
sufficiently vague that there are many answers to your question. If
temperatures are sufficiently high to start ionizing molecules, you will
also have dissociation taking place so possibilites for the positive
ions present would be H_{2} O^{+} , H^{+} , O^{+} ,
OH^{+} , O_{2} ^{+} , and H_{2} ^{+} ;
the gas also would have neutral molecules present of all these, the
fractions depending on the pressure and temperature. As the plasma
cooled down, there is no reason why all the ions and neutral molecules
would return to being water molecules.

QUESTION:
I f you tow a boat from a tow path which is the best point to tie the rope onto the boat?

VIDEO

ANSWER:
The rope will exert a force on the boat, obviously. This force will tend
to do three things: exert a torque which will tend to rotate the boat
about a vertical axis through the center of gravity (you don't want
this), have a component along the bank which will pull it along the
canal (that is what you want), and have a component which will pull the
boat toward the shore (you don't want this). To minimize the tendency of
the boat to turn, attach the rope close to the center of gravity of the
boat. To minimize the tendency to drift (not turn) to the bank, make the
rope as long as possible so that most of its force will be exerted along
the bank. Some tiller will be needed to make small corrections, but
those should be minimal.

QUESTION:
Does water slow down gradually as it rises vertically out of a fountain? It would seem that it should, but that would mean the water behind would catch up and create a constantly wider jet. However it seems that the jet stays relatively consistent in width (subject to some splashing) until it reaches the top. At which point it splays and falls. This would suggest it is at a consistent speed all the way up until it suddenly decelerates and falls.

ANSWER:
Any "piece" of the water has a downward acceleration (slows down on the
way up, speeds up on the way down) at all times; ignoring friction
effects, each drop of the water follows the same trajectory as a thrown
ball would follow. So, if one piece is slowing down, why doesn't the
piece behind it catch up? Because it is slowing down too at exactly the
same rate. If you threw two balls up, one right after the other, the
second would never catch up. Of course, if the stream of water were
going vertically up, the water falling from the top would collide with
the water rising and, as you say, splay out.

QUESTION:
I am really confused over the spin of neutrino, my teacher says it is always spin 1/2, however with my calculation from beta decay it can be 3/2 also? Please clarify why it can't be spin 3/2?

ANSWER:
I have no idea how you can calculate the spin of a particle from the
decay. Spin is the quantum number of the intrinsic angular momentum of a
particle, one of the properties of the particle itself, and cannot be
anything other than what it is, ½ in the case of a
neutrino (or electron, or proton, or neutron). A spin-½ particle
might also have an orbital angular momentum quantum number L
and its total angular momentum quantum number J=L ±½
could be different from its spin. Maybe that is how you are coming up
with 3/2. For example, if L =1 for the neutrino, J =½
or 3/2. (By the way, if you are just
learning about angular momentum, the actual angular momentum of an
object with angular momentum quantum number N is (h /2 π )√(N (N +1))
where h is Planck's constant.)

QUESTION:
I do not know your opinion on quantum mechanics (I am only just getting into it myself) but
I was thinking if you could put an object into a superposition where it is moving and standing still at the same time. Would e=mc^2 still apply?
In that by standing still it is not using energy therefore not gaining mass while moving at the same time.

ANSWER:
I am a physicist, how could I not have a high "opinion on quantum
mechanics"?! Every object is already in such a superposition because of
the uncertainty principle. Because you cannot simultaneously know both
the position and the momentum of a particle, the wave function is a
superposition of all momenta (hence, velocities) although the "amount"
of at-rest wave function may be very small if you observe the particle
moving approximately with some speed. But why would that invalidate
E=mc ^{2} ? Just as the particle has uncertain
momentum, it also has uncertain energy. Your last sentence makes no
sense since it takes no energy to be moving (Newton's first law).

QUESTION:
What distance would I travel in kms would I travel going at 7 Gs straight up in 10 seconds it is for a book I am writing your help would be greatly appreciated as I cannot block myself working this out

ANSWER:
Starting from rest, the distance d traveled by an object
accelerating at a constant rate a for a time t is
d = ½at ^{2} . For your question, a =7x9.8=68.6
m/s^{2} and t =10 s, so d =½x68.6x10^{2} =3,430
m=3.43 km.

QUESTION:
Since the rest mass of the constituent quarks only makes up about 2% of
the total mass of the proton (or neutron), would you please explain in
layman's terms what it is in QCD that forces all protons (or neutrons) to
have the exact same mass. Does the answer to this question imply that time is quantized?

ANSWER:
One of the things which the uncertainty principle says is that you
cannot know precisely both the energy and the time of a quantum system, ΔE Δt ≈ h /2π
where
h /2π =6.6x10^{-16} eV·s is the rationalized Planck's
constant. In order to measure the energy of a system, you must spend
some time; the more time you have to examine the system, the more
accurately you can determine its mass. But, if the system is unstable,
that is it has some lifetime before it decays to a different state, you
will get a different answer every time you try to measure its mass
(which is, equivalently, its energy). A proton is a stable particle and
therefore, since it lives forever its mass has zero uncertainty and is
why all protons have identically the same mass. However, you are wrong
that all neutrons have the same mass because the neutron beta-decays to
a proton+electron+neutrino with a half life of about 880 s. So, the
uncertainty in the neutron's rest mass energy is about ΔE ≈h /(2π Δt )=7.5x10^{-19}
eV=7.5x10^{-28} GeV. The rest mass energy of a neutron is about
1 GeV, so the uncertainty is about 7.5x10^{-26} %! This is small
but not zero. These results have nothing to do with either the masses of
quarks or time quantization.

QUESTION:
A ray of light, such as from a lighthouse, rotates at 1/sec.. What happens to the ray, i.e. its speed, at points distant 300,000km and beyond from the source?

ANSWER:
I think what you are getting at is that the "end of the ray" moves
faster than the speed of light, right? You do not need to go out 300,000
km for that to happen. For example, the moon is about 4x10^{8} m
away and a beam rotating with a frequency of 1 cycle per second would
have a spot going across the moon with a speed of about 2.5x10^{9}
m/s, nearly 10 times the speed of light. Does this violate the law that
nothing can go faster than the speed of light? No, because this spot is
not anything and you cannot use the spot of light on the moon to
transmit information between two points on the moon.

QUESTION:
I recently encountered a problem while doing my IB physics higher level IA. My topic was 'how does the concentration of salt in water affect the refractive index measured'. Although the refractive index increased as the amount of salt in water increased as per hypothesised, I could not figure out why this is the case at my level of studies. According to the marking guidelines if I could explain why this is the case then I could probably score a lot higher. I really hope you could help me on this.

ANSWER:
My first guess would be that since the index is larger for salt (1.523) than it is for water
(1.333), adding salt to water will increase the index.
However, there is a problem with this. Some years ago on the radio show
Car Talk there was a puzzler
which asked "how much salt is there in salt water?" The answer was
"none". In solution, salt, NaCl, dissociates into Na^{+} and Cl^{-} .
So, it is not such a good answer to simply say that salt has a higher
index. Understanding index of refraction on the atomic or molecular
lever is very difficult.

QUESTION:
I have a question about black holes,:
If they are infinitely dense, how can one black hole be "bigger" than another i.e "stellar" black holes and "super massive" black holes...
Obviously if two black holes merge or obtain mass through eating stars 1 + 1 = 2 but if the mass already equals infinity how can it get any larger. This would assume that it is equal to infinity + 1.

ANSWER:
You might first read an earlier answer . Now,
infinite density does not mean infinite mass; rather the mass is some
finite number and the volume is zero. So, clearly, the rest of your
question does not apply to black holes since their mass is never
infinite. Super massive black holes have masses much larger than the
mass of a star.

QUESTION:
what does a charge q distributed uniformly over a surface mean?

ANSWER:
It means that the surface charge density is a constant everywhere on the
surface. Surface charge density is defined as σ =dq/ da ;
(dq is the charge on an area da ). Another way to put
it is if you look at any area a anywhere on
the surface, you will find an amount of charge q =σa.

QUESTION:
what do you think about particle wave theory? and does the particle wave theory negate heisenberg's uncertainty principle?

ANSWER:
What do I think about it? I think that wave/particle duality is a
fact of nature. Why would you think that it would "negate" the
uncertainty principle. You could actually say that duality results from
the uncertainty principle or vice versa .

QUESTION:
If planets are the part of stars around which they revolve then the stars and the planets must have the same stuff then why the stars do not cool down as the planets do?
Why they continue to burn and expand while planets cool down?

ANSWER:
The solar system formed from a huge interstellar cloud of material. The
material was overwhelmingly composed of hydrogen. A very large fraction
of the material collapsed gravitationally to form the sun and, as this
mass collapsed, it got hotter and hotter until eventually nuclear fusion
of the hydrogen ignited and it became a star. A smaller fraction if the
material from the cloud, because of angular momentum, did not become
part of the sun and, through a long period of condensation and
collisions among clumps, became the planets and their satellites. But
the hydrogen did not ignite because the gravity of the planets was too
small. Eventually some of the hydrogen combined chemically with heavier
elements (for example, water) but most of it escaped back into space
because the gravity was too weak to hold lighter elements in the
atmosphere.

QUESTION:
It's a question about R value, and directed to you partly because it appears you are in New Zealand. I'm looking at a video from an inventor in New Zealand who is recycling plastics into building blocks. I cannot find him on the net to ask what is the R value of these blocks. I know it is a long shot to ask you, but I wondered if there is a way to calculate R value, based on assumptions about physical properties of plastics such as milk jugs, trapped air and density in these blocks.

ANSWER:
There are
tables where the R-value per inch of thickness is tabulated for
various materials. But, these blocks are fabricated from a random mix of
all classes of recyclable plastics. Furthermore, it is not clear how
much air is trapped in them. Plastics seem to have R-values in the range
of 3-5 ft^{2} ·°F·h/BTU/in, so you could
certainly take this range to estimate the value for a block of known
thickness. If you are interested in the details of the physics of the
R-value, you might be interested in an
earlier answer .

QUESTION:
An iron ball made of only iron ,no hollows no any other material,just iron. So first we measure its weight at room temperature and then we heat it up to higher temperatures near to melt but not not melting. Then we measure its weight. Does weights are different or same. If different please indicate which is higher.

ANSWER:
For purposes of calculation, let the ball have 1 kg of mass at room
temperature, 20°C. Because mass is a form of energy, the iron ball
has energy E=mc ^{2} =9x10^{9} J. The melting
point of iron is 1540°C and its specific heat is 449 J/kg·°C,
so the enegy to heat it from 20 to 1540°C is 1x1520x449=6.8x10^{5}
J. So, the ball would have increased its mass by 6.8x10^{5} /9x10^{9} =7.6x10^{-5}
kg. You would be hard-pressed to actually measure so small a change in
mass.

QUESTION:
Heavy water is Deuterium oxide. Water is H_{2} O. So is Deuterium oxide water with one Protium and one deuterium atom, or water with two deuterium atoms. If both hydrogens in water are actually deuterium, how did that come to be?

ANSWER:
Deuterium (D) is a naturally occuring isotope of hydrogen; approximately
0.015% of hydrogen atoms are deuterium, which is about 1:6,670.
Therefore about 1:3,330 molecules of water would be HDO and
approximately 1:44,400,000 (6,670^{2} )would be D_{2} O.
The term "heavy water" refers to D_{2} O and DHO is sometimes
referred to as semiheavy water. You would think it would be a hopeless
task to extract such a tiny fraction of D_{2} 0 from natural
water. I was very interested to learn that there is a much more clever
way to get nearly pure heavy water. There are relatively many DHO
molecules so it is much more practicable to use standard isotope
separation techniques to get enriched DHO. So, for example, suppose that
you enrich the water such that the hydrogen atoms present are 50%
deuterium. Then it turns out that the water is 50% HDO, 25% H_{2} 0,
and 25% D_{2} O. Where did all that D_{2} O come from? It
turns out that in water the hydrogen atoms (both isotopes) do not stay
attached to the same oxygen atom, rather they move more or less freely
around. By repeating whatever you did to get this far, you can get a
higher and higher percentage of deuterium. Enrichments of 99.75% D_{2} O
are routinely achieved.

QUESTION:
Is it possible for a skydiver to not be strapped in to a parachute, just holding onto it,
or even wrapping his arm into the straps. When he pops the schute, is it possible to hold on to it? Would his arm withstand the sudden deceleration or could it be torn from the shoulder socket?

ANSWER:
There is, of course, no simple answer to this question. It depends on
how fast he is falling and the parachute design. It also depends on
"luck" as the graph shows since the two graphs are for identical
parachutes, weights, and speeds. The drag force (the force the parachute
exerts up on him) as a function of time is shown in g s. One
g would correspond to the weight of the parachutist. Both graphs
end up after about 7 s at g _{force} =1 which means that
the force up on him would be equal to his weight. For the "soft"
deployment, about 3g is the maximum, so you would have to be
able to hang from your hands from a stationary bar with three times your
weight attached to you; you could probably do this briefly and it would
certainly not tear the arm from its socket. For the "hard" deployment,
almost 6g is the maximum, so you would have to be able to hang
from your hands from a stationary bar with six times your weight
attached to you. You could probably not do this but it still would
probably not tear the arm from its socket. In any case, I would never
depend on being able to hang on with my hands!

QUESTION:
If I have a .25 inch airtube in
a 2 ft deep fish tank, how can I calculate the force necessary to expel the first bubble of air at the bottom of the tank if the other end of the tube is at the surface?

ANSWER:
The weight density of water is ρ =62.4 lb/ft^{3} =0.0361
lb/in^{3} . Therefore the
gauge pressure 2 ft down is P=ρh= 62.4x24=0.867 lb/in^{2
} (psi). That pressure will just keep the tube filled with air all
the way to the bottom. Any pressure greater than this will expel air
into the water. This is probably the number you want; the force over the
area of the tube is F = 0.867·π· (0.25/2)^{2} =0.0426
lb=0.68 oz.

QUESTION:
T he weight of the person shows more or less than the original weight when
he is in the lift which is accelerating upwards and the person is stand
inside the lift in a weighing machine and why?

ANSWER:
There are two forces on the person, his own weight W which
points down and and the force N the scale exerts up on him. (N
is what the scale will read.) If the person is accelerating upward,
Newton's second law states that N-W=ma where m is the
person's mass. Therefore, N=W+ma and so the scale reads more
than the person's weight.

QUESTION:
I am an avid sports fan and I have often wondered if the experts may be wrong about the myth of the rising fastball in the game of baseball. I played baseball for over 20 years and I can tell you that the ball does appear to rise when certain pitchers throw hard put a heavy backspin on the ball. I have been told that experts say it is nothing more than a visual trick your eyes play on you because a rising fastball is considered to be physically impossible. I can tell you first hand that a softball pitcher I know can throw a ball that rises after being thrown on a straight trajectory. I suspect the Magnus effect may have something to do with the anit-gravitational behavior of the ball. Do you think this could be what causes the ball to appear to rise as it travels, or is it just our perception?

ANSWER:
There is such a thing as a rising fastball, but it does not actually
rise; it simply falls more slowly than a nonspinning ball does. An
experienced hitter knows intuitively what a normal fastball does and
when presented with a rising fastball he will swear that it rose because
it actually fell less. Incidentally, by rise or fall, I am talking about
the direction of the acceleration. So a ball which is thrown at an angle
above the horizontal is obviously rising but it rises at a
decreasing
rate of rise until it reaches the peak of its trajectory and then begins
back down; the rising fastball will actually rise farther. Then why do
we say it is a myth? It is easiest to understand by looking at a ball
thrown purely horizontally. Can spin cause the ball to actually go
upwards? To answer this, you need to think about all the forces on a
pitched ball. There is the weight, F _{G} , which points
vertically down and causes the ball to accelerate downward (a
horizontally thrown 90 mph fastball falls about 4 feet on the way to the
plate); there is the drag F _{D } which points opposite
the direction of flight and tends to slow the ball down (a
90 mph fastball
loses about 10 mph on the way to the plate); and there is the magnus
force F _{M} which, for a ball with backspin ω
about a horizontal axis points perpendicular to the velocity and upward.
If the ball is moving horizontally the only way it could rise is for the
Magnus force to be larger than the weight. Measurements have been done
in wind tunnels and it is found that if the rotation is 1800 rpm, about
the most a pitcher could possibly put on it, the ball would have to be
going over 130 mph for the Magnus force to be equal to the weight. When
you say "thrown on a straight trajectory", you cannot mean it left his
hand horizontally because it would hit the ground before it got to the
plate; a fast pitch like that is impossible to accurately judge the
initial angle of the trajectory.

QUESTION:
I'm learning about ideal gases in class and learning about the ideal gas law, but we also (very briefly) mentioned that at low temperatures and high pressures the ideal gas law no longer holds very well. We didn't go much pass that but I did a bit of research and an explanation included another set of constants calculated experimentally, but again only really explained very briefly that the size of the molecules become much greater and the inter molecular force between the molecules is no longer insignificant (as is a factor of the kinetic molecular theory of gases). I guess my question is, what is so substantial to the aspects of low temperatures and high pressures of ideal gases that causes them to follow a different set of parameters?

ANSWER:
V=NkT/P is the ideal gas law (IGL). I write it like this to
emphasize what happens if T decreases and/or P
increases while N stays the same—V decreases.
(You may write it as V=nRT/P where n is the number of
moles and R is the universal gas constant. For reasons of
clarity in my answer, N is the number of molecules and k
is Boltzmann's constant.) The IGL assumes that the volume occupied by
the molecules is insignificant and that the molecules do not interact
with each other. But, when the volume decreases the molecules get closer
together so that their interactions start to matter and the total
fraction of the volume which the molecules occupy is no longer a
negligible fraction of the volume of the container. (You are incorrect
when you say that "…the size of the molecules become much greater…";
the size stays the same but now occupies a larger fraction of the whole
volume.) The first thing to do to correct the IGL is to replace V
by V-Nv where v is the volume of a single molecule:
P (V-Nv )=NkT. Although this improves the
accuracy of the IGL, it still overestimates the pressure observed
experimentally. The reason is that the effects of the intermolecular
forces have not been included. It turns out by doing measurements that
the pressure is proportional to the square of the number density, P ∝(N /V )^{2} ;
this makes some sense because the molecules interact by pairs and there
are N (N -1)/2 ≈N ^{2} /2
pairs because N is very large. So, finally,
(P-c (N/V )^{2} (V-Nv )=NkT
where c is a proportionality constant which depends on what the
particular gas is. This is sometimes called the van der Waal equation.

QUESTION:
Read yet another explanation how a space traveler went into some other persons future. To me, what they showed was that the traveler did, was get to the mutual meeting point in less time, was less old. But as soon as the travel ended and they met, time was the same for them. At no point did the traveler have access to even one second into the future of the stay at home person with whom he now shakes hands with. Am I totally missing something critical that makes writers say they traveled into the other persons future?

ANSWER:
I totally fail to get your point. Let's take an extreme example where
the traveler is a baby and the stay-at-home is his twin. Speeds and
distances are chosen such that the traveler arrives back on his tenth
birthday and for his brother it is his 70^{th} birthday. It is
true that the traveler did not have any "access to" 70 years of his
brother's life, but that does not negate the fact that he had now,
spending 10 years of his life, been able to arrive at what would have
been his 70^{th} birthday had he stayed at home. From this point
on, he is 60 years into the future because of having taken his trip. He
now has "access to" every second of that future. You may want
to read my discussion of the twin
paradox .

FOLLOWUP
QUESTION:
Thank you for your quick and detailed response. Your language was what I see as happening. That is, while I may have worded my question poorly, my problem was the use of the phrase "into the other person's future".
What I understand and what I think you confirmed is that the traveler traveled to the other persons "present, his now". Yes, they now share the same passage of time and yes she will always be decades younger but she did not have any access to his future - that is anything beyond the time of the handshake. Yes she got to 2170 in 10 years and it took him 70 years, but she never visited, saw, heard anything about what happened or would happen even 1 hour after the handshake. She never even saw the handshake in advance of it happening.
Please, I am not trying to be argumentative. I just am frustrated that so many writers use language that implies the traveler can see another persons future, a future that other person is causally connected to, but which has not yet happened.

ANSWER:
I now understand your point of view better. I agree, that if you look
literally at the semantics of the usual twin paradox explanation, it
could be misconstrued as you interpret. In more than 50 years of doing
physics, I have never encountered this objection to the verbiage before.
All I can suggest is, since the laws of physics forbid time travel
backwards in time, one could never go back in time and therefore could
never know the future from the perspective of the present. Maybe I
should rephrase a sentence from my first answer as "He now has 'access
to' every second of that future, but he has to wait for it."

QUESTION:
If time moves slower at ground level vs high altitude due to gravity. Does time slow under artificial g"s eg: centrifugal forces?

ANSWER:
The equivalence principle, a linchpin of the general theory of
relativity which predicts the time dilation you cite, states that there
is no experiment you can perform which can distinguish between being in
a frame with acceleration g in
empty space and being in a gravitational field in which the acceleration
due to gravity is g .

QUESTION:
When I had solar panels installed on my house the old fashioned meter ran backwards when the Sun was bright. They have now fitted a digital meter which can sense when energy is being sent into the grid. I can see how this could be done with DC, but how does it work with AC? In AC the current is switching direction at 50 Hz. How can the meter sense the 'direction' of the energy flow?

ANSWER:
You are right, the average current is zero. However, the current is not
the power —the power is the product of the current and the
voltage. Both current and voltage are sinusoidal functions of time,
i (t )=I sin(ωt ) and v (t )=V sin(ωt+φ ),
so p (t )=IV sin(ωt )sin(ωt+φ ).
The graph above shows three choices for the phase φ between
i and v . For φ=π /2 the time average of
the power is zero, no energy flow; for φ=π and φ= 0
the time average of the power is negative and positive, respectively.
The motor in a mechanical meter turns in opposite directions for
different signs of the average power; in a digital meter the average
power is determined by an electronic circuit.

FOLLOWUP
QUESTION:
Thank you for your reply to my question about the power direction with energy generated by solar panels. I now understand that phase angles are crucial.
The inverter in the loft takes DC from the panels and converts it into AC. To feed energy back into the grid does the inverter have to deliberately adjust the phase or does this occur naturally when there is an excess of energy?

ANSWER:
All that matters is what the phase is at the meter. If energy is flowing
into your house the phase will be one way, if flowing out it will be the
other.

QUESTION:
Can you explain the physics of the following to me (in layman's terms)?
"On earth, where speeds are small compared to the speed of light and the gravitational field is weak, it turns out that nearly all of our weight arises due to the warping of time, rather than space. What this means in practice is that gravity on earth is "equivalent" to acceleration mostly in the sense that clocks on the surface run more slowly than clocks in outer space."
I'm actually studying architecture and building decay, so it would be ideal if your answer could be related in some way to time and building ruins - which have an obvious connection to gravity!

ANSWER:
I am not sure I understand exactly what is meant by that statement. I
can tell you that clocks at different altitudes run at measurably
different rates because of the gravitational field; this turns out to be
extremely important in GPS technology which relies on extremely accurate
timing to get accurate positions. Still, the difference is very tiny;
for every second that ticks of on a clock in zero gravitational field, a
clock on earth ticks off 1-7x10^{-10} =0.9999999993 seconds. You
may be sure that relativity has nothing whatever to do with observable
building decay here on earth!

QUESTION:
A s a bicycle moves on a wet road, why drops sticking to the tyre leave it tangentially?

ANSWER:
Because that is the direction they are moving when they lose contact.

QUESTION:
If a 6 foot tall man where to shrink half in size every second, how long would it take him to reach the quantum level?

ANSWER:
So, his height is about 2 m and we want to shrink him to, let's say,
2x10^{-10} m. So, if he halves N times, 2/2^{N} =2x10^{-10}
or 2^{N} =10^{10} . The easiest way to do this is with
logarithms: log(2N)=N·log2=log(10^{10} )=10 or
N=10/log2=33.22. N should be an iteger, so the time it takes is about 33
seconds.

QUESTION:
In a rotating object the velocities of all the points are different i.e the relative velocity between two point is not zero. Still if one observer sitting at one point will observe another point it will seem to be at rest. Why?

ANSWER:
Because the rotating object is not an inertial frame of reference
because every point (except the center) is accelerating. In order to go
from the nonrotating to the rotating frame, you need to add a different
velocity to each point, not the same velocity to each point if you were
transforming to some other inertial frame.

QUESTION:
Okay, I have a question, me and my boyfriend are in this dispute:
If we are in a car traveling 35mph and the car in front of us is also traveling at 35mph will we remain at an equal distance apart or will we eventually be side by side?

ANSWER:
If the road is straight, you will always remain the same distance
behind. If you are rounding a curve and you are on the inside lane and
the other car on the outside lane, you will catch up; if you are
rounding a curve and you are on the outside lane and the other car on
the intside lane, you will fall farther behind; if you are both in the
same lane, you will remain the same distance behind.

QUESTION:
In lifting an object to a higher level directly over its original location, the energy I expend increases potential energy. But, does some of the energy used in lifting it also go to accelerating it to higher rotational velocities as the circumference of its "orbit" increases as it is raised over its original position? Does this add kinetic energy and mass to the object whereas increasing potential energy does not?

ANSWER:
(Preface: all my calculations below assume that the height lifted is
much smaller than the radius of the earth. I also neglect the change in
the gravitational force over the distance the mass is lifted. Also, to
simplify things, all my calculations are at the equator.) Yes, work is
done to increase the kinetic energy. As viewed from an inertial frame,
watching the mass M get lifted to h , I estimate that
the kinetic energy changes by ΔK≈ 2hK _{initial} /R=MhRω ^{2}
where R =6.4x10^{6} m is the radius of the earth and ω =7.3x10^{-5}
s^{-1} is the angular velocity of the earth. For example,
lifting 1 kg a height of 1 m requires 0.03 J of work to increase the
kinetic energy. But wait a minute! Once we acknowledge that the earth is
rotating, we have to recognize that the mass, being in a circular orbit,
has a centripetal acceleration a _{c} =Rω ^{2}
and therefore the net force on M is Mg-MRω ^{2} .
Therefore, the net work done is W ≈(Mg-MRω ^{2} )h +MhRω ^{2} =Mgh .

QUESTION:
Can gravity be focused like light thought a lens? Say following the plane of a qalaxy. If so, could stars be traveling around galaxies at the same speed because gravity effects time?
The reason for the question above, IF gravity focuses out along the plane of a galaxy, could it be slowing down the stars closer to the center of the galaxy because gravity effects time?

ANSWER:
Gravitational lensing is light being focused by gravity, not gravity
being focused. So I am afraid that your idea to explain the anomalous
orbiting properties of stars in a galaxy will not work. An example of
gravitational lensing is shown to the right; shown is "Einstein's cross"
which shows four images of the same quasar.

QUESTION:
Why don't liquid pipelines that run downhill rupture from the weight of the liquid in them? For instance, a 14 mile long section of 10" ID pipe carrying crude oil that runs at a 22.5 degree angle has about 301,592.89 gallons of crude in it. That Crude weighs 91,328,360.39 pounds. I calculate that the static weight at the bottom of that run should be about 22,832,090.10 pounds of oil. That exceeds the tensile strength of the pipe wall by a factor of 50 to 100. At first I thought it was because a closed pipe would have sort of a vacuum at the top, but then I realized that would make the pipe crush from atmospheric pressure. I'm missing something simple I am sure, but darned if I can figure out what it is...

ANSWER:
There is something seriously wrong with your numbers. They imply that
the weight of one gallon of crude is about 9x10^{7} /3x10^{5} =300
lb, and 14 miles at 22.5° would take you to 14sin22.5°=5.4
miles, higher than Mount Everest! Also, since we are working with a
fluid, pressure would be a more appropriate quantity to look at than
force. I will start from scratch with new
numbers: the density of crude oil is about ρ =900 kg/m^{3
} and I will take a distance of about y =1000 m between the
top and the bottom, about the height of a small mountain; you need only
the drop, as we shall see below. Like you, I will assume that the oil in
the pipe is static which simplifies things. A good approximate way to
solve this kind of problem is to use Bernoulli's equation, P +ρgy+ ½ρv ^{2} = constant;
P is the pressure, g =9.8 m/s^{2} is the
acceleration due to gravity, and v (=0 for us) is the speed. At
the top, P _{top} =P _{A} and y _{top} =1000
m and at the bottom P _{bottom} =P+P _{A}
and y _{bottom} =0; here, P _{A} is the
atmospheric pressure. Therefore, P _{A} +900x9.8x1000=P+P _{A} +0.
So the gauge pressure is P ≈10^{7} N/m^{2} ≈100
atm=1450 PSI. The brief research I have done indicates that this pressure
is at the extreme upper limit of specifications for pipelines. To make
the oil move, you need to add a lot more pressure. My use of Bernoulli's
equation is a very crude approximation because it assumes a fluid with
no viscosity and no frictional forces with the walls, obviously not a
very good approximation.

QUESTION:
Energy question. On earth kinetic energy transfer is more or less sound and heat given the little knowledge I have on the subject. In a vacuum sound cannot be heard, but does that mean it does not exist? if it does not exist than where does the energy go? if it does exist, and our ears are simply broken in a vacuum does sound expand faster in the vacuum like light? Every thing I read suggests sound cannot exist in a vacuum, but energy must be conserved, so what form of energy does snapping my fingers in a vacuum take if it cannot take the form of sound?

ANSWER:
First of all, sound, being waves in air, does not exist in a vacuum. By
removing one possible way for energy to be removed from the system
simply means that the energy needs to be taken away by other mechanisms.
Let's do an example. Imagine a plucked guitar string which has a certain
amount of energy. In air, I can think of three ways to take energy from
the string —the radiated sound, the drag of the string
moving through the air, and the friction associated with the string
bending and unbending. A guitar in a vacuum no longer has the first two
ways to lose energy, so only the bending friction would be converting
kinetic energy of the string to heat in the string. It would therefore
take longer for the string to stop than in the air, but the total energy
converted would be the same; if the string did not have a way to get rid
of its energy (it does, it could radiate or conduct to the pegs, but
let's forget that), it would end up hotter than it would in air.

QUESTION:
Is there a law in physics that allows me to calculate the magnetic field at a certain point created by a charged particle moving in a straight line
with constant speed in empty space?

ANSWER:
Yes, there is an equation. I will warn you, though, that this is quite
technical and pretty high-level. Refer to the figure to the left. The
charge q , with velocity v is at
position x (t ) at time t ; we
wish to know the field at position r at time
t . However, since the information about the field propogates at
the speed of light, c , the field at time t is
determined by where the charge was at some earlier time t _{ρ} ;
|r -x (t _{ρ} )|≡|ρ |=ρ =c (t-t _{ρ} ).
After much calculation (see the detail in Chapter 10 of David Griffith's
book
Introduction to Electrodynamics ), the magnetic field is
B (r ,t )=(v xE (r ,t ))/c ^{2
} where the electric field is
E (r ,t )=[q /(4πε _{0} )][ρ /(ρ ·u )^{3} ][(c ^{2} -v ^{2} )u ];
the vector u is defined as u ≡c1 _{ρ} -v
where 1 _{ρ} is a unit vector in the
direction of the vector ρ .
For high
speeds, the fields look like the diagrams below.

A DDED
NOTE:
The above expressions for electric and magnetic field are exactly
correct. You can write approximately correct equations for particles
with speed much less than the speed of light, v<<c :
E (r ,t )=[q /(4πε _{0} )][(r -x )/|r -x |^{3} ]
and
B (r ,t )= [ qμ _{0} /(4π )][ v x(r -x )/|r -x |^{3} ].
These are simply
Coulomb's law and the
Biot-Savart law .

QUESTION:
Why do moving charged particles respond to magnetic fields? I know that every charged particle has it's own magnetic field and permanent magnet will attract/repel these particles, but the force will be so little that it won't be able to be measured at all, but when it comes to moving, a "magic" happens, and I don't understand what is special with moving charges versus stationary charges.
Does the magnetic field between magnet and charged particle increase proprtional to velocity and the force gets noticable? In turbines, when magnet is rotating, how that makes electrons move?
Why are those electrons affected by magnetic field at all? There is Lorentz law but how was that
equation figured out? Does that equation only depend on experiments? Did they just see that electrons start moving when we rotate magnet next to them and that's all?

ANSWER:
Site groundrules specify single, concise, well-focused questions so I
should have just thrown this out. Instead I have edited the question a
bit to make it more focused. There is no way I can fully answer the
questions because they really ask that I give you a full course in
magnetostatics. First of all, disabuse yourself of your second sentence.
All charged particles do not have a magnetic field if they are not
moving; true, most elementary particles (electrons, protons, etc .)
have magnetic moments, but these are incredibly weak and do not normally
react to a magnetic field; if you just have an electric charge Q ,
it experiences zero force if at rest in a magnetic field. Yes, it is an
empirical fact (experimentally observed) that a charged particle Q
with velocity V in a magnetic field
B experiences a force F =QV xB .
It is now understood that there is only one field, the electromagnetic
field, and electromagnetism is a relativistic theory; it is, though, no
longer a vector field like you are familiar with, but a tensor field
with nine components. What this means is that the answer to your
question about there being electrons at rest with a magnet moving is
that an electron moving in a static
field is no different from an electron at rest and the magnet moving;
that, essentially is relativity—all that matters is relative
velocity. Also, once you understand that there is a single field, the
Lorentz force arises naturally.

QUESTION:
If I weigh 200lbs and am riding a kick scooter that weighs 14lbs, and I am riding at a speed of 10 mph and I jump the scooter off a curb, say 6 inches, what is the force in terms of pounds, that I am applying to the scooter as it lands?

ANSWER:
Usually it is not possible to answer this kind of question because what
is needed is to know how long the collision between you and the ground
lasted. In this case, though, we can estimate the time of this
collision. As you may know, paratroupers are trained to not land with
stiff legs, rather to bend at the knees during the landing; the purpose
is to prolong the landing time and this reduces the average force on the
legs during the landing. If a mass M hits the ground with some
speed V and stops in time t , the average force over
the time is F=W+MV /t where W is the weight,
200 lb; to convert the
weight to mass, divide by the acceleration due to gravity, g =32
ft/s^{2} : M=W /g =(200 lb)/(32 ft/s^{2} )=6.25
ft ·lb/s^{2} . The speed at impact can be determined
from V =√(2gh )=√[2x(32
ft/s^{2} ) x(½ ft)]=5.7 ft/s. If we approximate that
the distance S over which your legs bend on landing as S =1
ft, the time to stop is t= 2S /V=(2x1 ft)/(5.7
ft/s)=0.35 s. So, finally, F =200 lb+(6.25
ft ·lb/s^{2} )(5.7 ft/s)/(0.35 s)=302 lb. This is
the force the scooter exerts on you which, by Newton's third law, is
equal the the force you exert on the scooter (but in opposite
direction). If you stop in ½ ft, the force would be 404 lb. Keep
in mind that this is a very approximate estimate of the average force.

QUESTION:
If someone spins a stick in a circle,what happens if the stick is scaled up to such a size that the outer edge of the stick would have to break the Speed of Light Limit?

ANSWER:
See an earlier
answer .

QUESTION:
I have been trying to understand this for years. In Einstein's theory of time verses speed I believe he used the situation of a man (lets call him man A) standing in a train yard. A second man is on a train (lets call him man B) and the train is moving. They both drop an object at the same time. If I standing in the yard with man A to me the object man A drops would appear to me in normal time. Man B's object would appear to take longer causing the difference in time. I understand this. My main question is if a man C was on a different train going the same direction as man B's train the time difference between me and man A to man B and man C. I understand that to man B and Man c they would be the same. What if man C's train was going in the other direction? To me and man A the would seem the same, but man B to man C I believe the difference in the dropped object would be appear to be doubled to them between them. Since I believe it would appear doubled between man C and D. I think there should be a double time difference. From man A's view the age difference would be the same, but between man B and man C there should be double the time difference. How can there be doubled the time difference when they are traveling at the same speed?

ANSWER:
Your question boils down to what is called velocity addition .
In classical physics, v _{CB} =v _{CA} +v _{AB} ;
the notation is "v _{IJ} is the velocity of I relative
to to J". I have written this so that it corresponds to your question —A
is you, B and C are the trains; it is more convenient for you if we
rewrite the equation as
v _{CB} =v _{CA} -v _{BA}
which we can do because v _{AB} =-v _{BA} .
If the trains have speeds v in the same direction, v _{CB} =v-v =0;
in opposite directions, v _{CB} =v- (-v )= 2v— each
sees the other moving with speed 2v . But, this form of velocity
addition is wrong for very high speeds (see an
earlier answer ).
The relativistically correct velocity addition equation is
v _{CB} =(v _{CA} -v _{BA} )/[(1-(v _{CA} v _{BA} /c ^{2} )]
which reduces to the classical equation for the speeds much less than
c . So, for the trains moving in opposite directions, v _{CB} =2v /(1+v ^{2} /c ^{2} );
for example, if v =0.5c , v _{CB} =c /(1+0.25)=0.8c .

Now, you seem
to think that if you double the speed, you double the time difference.
This is not correct— time dilation goes like the gamma
factor, 1/√(1-v ^{2} /c ^{2} ).
So, for your situation with v =0.5c , t _{A} =1.33t _{B} ,
t _{A} =1.33t _{C} , t _{C} =1.67t _{B} ,
and t _{B} =1.67t _{C} . These are
confusing, I admit. They are meant to denote, for example, that t _{B} =1.67t _{C
} means that B sees his clock tick out 1.67 s when C's clock ticks
out 1 s; B observes C's clock to be slow.

QUESTION:
I am having a debate with my brother about climbing on an incline.
I understand the basics of climbing a hill on a diagonal. If you climb diagonally, you can avoid taking larger vertical steps at the cost of more horizontal movement. This makes each step take less energy while increasing the overall work and time needed.
However, it seems as though this rule does not work for stairs. Stairs do not allow for shorter vertical steps (you either make 100% progress on a step or 0%). Do I have this correct? Am I missing something?

ANSWER:
Slaloming up the incline will increase time spent but not increase work
done. This assumes no frictional forces are important, the only work you
do is the work lifting you. Since work done does not change but elapsed
time does, the average power you are generating going straight up is
greater than zigzaging. Going up steps, though, if you go across a step
you do no work, the only work done is lifting you to the next step. The
only way to get the equivalent lowered average power output as you do by
slaloming up the slope is to rest between steps.

QUESTION:
How strong would a man have to be to push a 16,000 lb bus on a flat surface?

ANSWER:
That depends on how much friction there is. And not just the friction on
the bus, but more importantly, the friction between the man's feet and
the ground. Newton's third law says that the force the man exerts on the
bus is equal and opposite the force which the bus exerts on the man (B
in the picture). Other forces on the man are his weight (W ),
the friction the the road exerts on his feet (N ),
and the force that the road exerts up on him (N ).
If the bus is not moving, N=W and f=B , equilibrium.
The biggest that the frictional force can be without the man's feet slipping is
f= μN where μ is the coefficient of
static friction between shoe soles and road surface. A typical value of
μ for rubber on asphald, for example,
μ≈ 1, so the biggest f could be is
approximately his weight W ;
this means that the largest force he could exert on the bus without
slipping would be
about equal to his weight. Taking W ≈200 lb, if the
frictional force on the bus is taken to be zero, the bus would
accelerate forward with an acceleration of a=Bg /16000=200x32/16000=0.4
ft/s^{2} where g =32 ft/s^{2 } is the
acceleration due to gravity; this means that after 10 s the bus would be
moving forward with a speed 4 ft/s. If there were a 100 lb frictional
force acting on the bus, the acceleration would only be a=0.2 ft/s^{2} .
If there were a frictional force greater than 200 lb acting on the bus,
the man could not move it.

QUESTION:
While reading about the twin paradox, I've been told at the end of the traveling twin's journey, he begins decelerating in order to land back on Earth, and he, the traveling twin, observes his brother's clock on Earth to SPEED UP. This makes sense to me except for one problem: This suggests that the light pulse in the Earth clock would be percived by the traveling twin to be moving faster than C. Of course, the traveling twin is no longer in an inertial frame. I thought perhaps that since he feels himself moving now, he would also measure himself and the light having a CLOSING SPEED greater than C, even though he would see the light moving across the ground on Earth to equal C? If so, at what rate would a clock in frame S behind the traveling twin run at, faster or slower? Is it also possible that acceleration, from the point of view of the traveling twin, causes length contraction perpendicular to the the ship's vector, shortening the distance the pulse has to travel?

ANSWER:
There is no need to discuss acceleration to understand the
twin paradox . Acceleration
just makes everything harder to understand. Basically, just assume
necessary accelerations (departing, turning around, landing) occur in a
vanishingly short time. But, you are not really interested in the twin
paradox, you are interested in how things appear in an accelerated
frame. I am sorry, but nothing in your question after " …clock
on Earth to SPEED UP…" makes any sense. First of all, the fact
that any observer will measure the speed of light to be c is a
law of physics and the general principle of relativity states the laws
of physics are the same in all (not just inertial) frames.
Second, to measure a speed in your frame of reference you must use your
clock, not somebody else's. And third, how another clock looks is really
irrelevant because how it appears to run and how it
is actually running are not the same.

I would like
to address how the clocks look from the perspective of the
Doppler effect . The relativistically correct Doppler effect is
usually expressed in terms of the frequency of the light; for our
purposes, it is more convenient to express it in terms of the periods,
T _{observer} =T _{source} √[(1+β )/(1-β )]
where β=v /c and β is positive for
the source and observer moving apart; it makes no difference which is
the observer—each twin will see the other's clock running at the
same relative rate. Let's illustrate with a specific example, β =-0.8,
the traveling twin coming in at 80% the speed of light; T _{observer} =T _{source} √[(1-0.8)/(1+0.8)]=T _{source} /3
so the observer will see the source clock running fast by a factor of
three. But special relativity tells us that moving clocks run slow, not
fast; 1 second on the moving clock will be 1/√(1-0.8^{2} )=1.67
seconds on the observer's clock. How the moving clock looks is thus
demonstrably not a measure of how fast it is actually running. Now, if
the incoming twin puts on the brakes such that he slows to β =-0.6,
T _{observer} =T _{source} √[(1-0.6)/(1+0.6)]=T _{source} /2
so the observer will see the source clock running fast by a factor of
two, apparently slowing down. Now, 1 second on the moving clock
will be 1/√(1-0.6^{2} )=1.25 seconds on the observer's
clock, speeding up compared to when β =-0.8.

QUESTION:
Sir can you please tell me where I am going wrong in this one? (attachment)

ANSWER:
The pendulum bob is not in equilibrium.

FOLLOWUP QUESTION:
I found the second equation in the book "An Introduction to Mechanics" by Daniel Kleppner and Robert Kolenkow.. Which you must be knowing about... So the second equation is definitely correct.. And first equation is just like the second one.. Whereas in second equation I have taken the components of tension force.. In the first one I have taken the components of weight.

ANSWER:
Did you not read my first answer? You are trying to apply Newton's first
law where it does not apply. Is the pendulum in equilibrium? Or will it
accelerate in some way if no other forces are applied? If it is a simple
pendulum (at rest right now), the mass will have an acceleration
perpendicular to the string; as soon as it begins moving it will have a
component of its acceleration along the string. Therefore, both of your
equations will be incorrect. Your second question reveals that this is
not a simple pendulum, but rather a conical pendulum (the mass moves in
a horizontal plane with the string tracing out a cone); you should have
given me that information in your first question. Now I can give you a
more complete answer. The conical pendulum, with m moving in a
circle, has only a horizontal centripetal acceleration toward the center
of that circle, a _{c} =v ^{2} /(L sin θ )
where L is the length of the string; applying Newton's second
law, T sinθ =m v ^{2} /(L sinθ ).
There is no acceleration vertically and so you can apply Newton's first
law,
T cosθ-mg= 0.

QUESTION:
Suppose I have a 10 tons weight hanging 5 meters up in the air. I want to get electricity by lowering the weight against a dynamo (for example).
How much energy do I get?
A 100 W light bulb needs 100 W of power when it's ON. So, if it stays on for 10 hours it will consume 1 KW, am I right?...

Ok, so my question is...
How many 100 W light bulbs can I have ON at the same time with the energy coming from that falling weight? -- while the weight is falling, obviously.

if the weight falls for 1 hour

if the weight falls for 2 hours.

What's the formula?
Somebody asked the same question on some forum on the web. His weight was 200 tons falling 100 meters for 1 hour, and someone said that the solution is:
dU=Fdy =
200,000kg * 9.81m/s2 * 100m =
196.2MJ =
196.2MW/3600 = 54,500KW/h Is the formula right? If yes, how do I apply it? Because I get extremely small numbers if I change his 200,000 with my 10,000, and his 100 meters with my 5 meters.
Plus, I don't know what KW/h means. All I'm interested is knowing how many Watts are available at any given moment while the weight is falling.

ANSWER:
The first thing we need to get straight is what a watt is. The unit of
energy in SI units is the Joule (J); 1 J is 1 N·m where a Newton
(N) is the unit of force and the meter (m) is the unit length. A Joule
is the kinetic energy which a 2 kg mass moving with a speed of 1 m/s
has; or, it is the work you need to do to lift a 1 kg mass to a height
of 1/9.81 m. A Watt (W) is the rate at which energy is delivered or
consumed, 1 W=1 J/s. Therefore, your 100 W bulb consumes 100 J of energy
every second. Incidentally, if you look at your electricity bill, you
will be billed for how many kW ·hr you have consumed; a
kW ·hr is a unit of energy, 1
kW ·hr=1000x3600 J=3,600,000 J.

The example
stated is correct but the units are not. It is fine up to the point
where the potential energy of the mass is 196.2 MJ (mega=M=10^{6} ).
Now, if you let this mass drop over 3600 s, it is losing its energy at
the rate of (196.2x10^{6} J)/(3600 s)=5.45x10^{4}
J/s=54.5 kW (not kW/hr). For your case, your mass has a potential energy
of 10^{4} x9.81x5=4.9x10^{5} J. If you deliver this
energy over an hour, the power is 4.9x10^{5} /3600=136 W; You
could power one light bulb over this hour and have some energy left over
at the end (about ¼ of what you started with). Clearly, the power
delivered over two hours would only be half as much, not enough to power
even one 100 W bulb.

QUESTION:
Imagine free electron is falling towards the Earth due to the gravitational interaction. In order to prevent the electron from falling, you come up with the idea to fix a second electron below on the ground to prevent the first electron from falling. What is the distance between the two charges, such that the top electron is in balance (the net force is zero)?

ANSWER:
I am going to assume that I can write the gravitational force as mg ;
if I find the distance d between the electrons to not be very
small compared to the radius of the earth, I will have to start over
again and use the force as MmG /(R+d )^{2} . The
electrostatic force between the two electrons is ke ^{2} /d ^{2} =9x10^{9} x(1.6x10^{-19} )^{2} /d ^{2} =2.3x10^{-28} /d ^{2} =mg =9x10^{-31} x9.8=8.8x10^{-30} ;
solving, d =5.1 m. Good, I don't have to start over!

QUESTION:
How far would a 100 pound pig go if you put it in a sling shot with a
500 pound pull back with a 5 mile an hour wind? In feet?

ANSWER: There is not enough information. How far do you have to pull the sling
shot to reach 500 lb? With that info I can get a good estimate without
including air drag which means without the wind. Air drag adds quite a
bit of difficulty to the problem.

FOLLOWUP QUESTION:
You would have to pull it back 175 feet.

ANSWER:
This is a truly peculiar question! I will first neglect air drag and the
wind; then I will include an approximate calculation including air drag.
I will assume the pig launches at 45 º and returns to the
same level from which it was launched. You should know that scientists
prefer to work in SI units, so I will convert all your numbers and
convert back to English units at the end. 100 lb=45.4 kg, 500 lb=2224 N,
175 ft=53.3 m. So the spring constant k =2224/53.3=41.7 N/m. So
the energy stored in the spring is ½kx ^{2} =½x41.7x53.3^{2} =59,200
J. This must equal the kinetic energy of the pig at launch, ½mv ^{2} =½45.4v ^{2} =22.7v ^{2} =59,200.
Solving for the launch speed of the pig, v =51.1 m/s. The range
of a projectile launched at 45º is R=v ^{2} /g =266
m=873 ft.

If
air drag is included, the angle for maximum distance is changed. I used
a calculation from a demonstration by
Wolfram . Without going into details, the graph shows the maximum
distance gone is about 120 m=394 ft and the launch angle is about 36º.
The red shows the path with air drag, the blue without. I emphasize that
this is a very rough calculation but it is useful to demonstrate that
air drag is important. To include the wind would be pointless since the
uncertainty in air drag is much bigger than any effects such a wind
would have; however, if you were to do it, you would need to specify the
direction of the wind.

QUESTION:
if the jeep weighs 2000 kg and tricycle weighs 1000 kg. among them? who will difficult to stop when both travelling same initial speed?

ANSWER:
(The numbers you give are masses, not weights. Weight is the mass times
the acceleration due to gravity, W=mg , where g=9.8 m/s^{2} .
But it doesn't really matter for this question because, as you will see
below, the stopping distance does not depend on what the weight is.) The
maximum force you can get when braking is the maximum frictional force
which, on level ground, is μW where μ is
the coefficient of static friction between the wheels and the road and
W is the weight. The work done by friction is equal to the
change in kinetic energy, -μWs =-½mv ^{2}
where s is the stopping distance, v is the starting
speed, and m is the mass. But, the mass is the weight divided
by the acceleration due to gravity, m=W /g ; therefore
s=v ^{2} /(2μg ). As you can see, the minimum
stopping distance does not depend on weight, only on both having the
same wheels on the same surface, e.g ., rubber on asphalt.

QUESTION:
Does amplified sound travel further than unamplified sound,i.e. Two objects emit sound at 65 decibles apiece, one objects sound is a "raw" 65 decibles , while the second object is 65 decibles coming from an amplifier, do the amplified decibles travel further?

ANSWER:
There is no way to unambiguously answer this. A decibel is a measure of
the intensity of sound relative to a standard intensity. It is a
logarithmic measure which means that when dB increases by 10 dB, the
intensity increases by a factor of 10; e.g., 10 db to 20 db increases
the power ratio from 10 to 100. Intensity of sound is energy/area/time
and usually measured in watts/square meter (W/m^{2} ) and the dB
is proportional to the logarithm. So, if you want to specify the "dB of
the source" you need to specify some geometry. Suppose we say that we
will measure all the energy passing through a sphere of radius 1 m.
Then, you would measure for both sources that the total energy per
second passing through that sphere would be the same. Using the
definition of the
dB, 65 dB=10^{6.5} 10^{-12} =3.2x10^{-6} W. Now we
come to the tricky part; you should be able to see that how far we will
be able to hear this depends on how this power is distributed over the
sphere. If the source radiates equally in all directions, this power
will be evenly distributed over the sphere; as an example, let's assume
your "raw" is distributed that way. Then the intensity of the sound is
the power divided by the area of the sphere, 3.2x10^{-6} /(4 π 1^{2} )≈2.5x10^{-7}
W/m^{2} . Now, suppose that your amplified sound comes from a
speaker which only radiates in the forward direction; if you think about
it, the intensity at the 1 m distance will be about 5x10^{-7}
W/m^{2} , twice as loud. As you moved farther away, you would
find that the intensity fell off like 1/r ^{2} where
r is your distance from the source; at 500 m away, the intensity of
the "raw" will be at 2.5x10^{-7} /500^{2} =10^{-12}
W/m^{2} and the amplified will be at 2x10^{-12} W/m^{2} .
The "threshhold of hearing" is about 10^{-12} W/m^{2} ,
so, in principle, you could barely hear both, the amplified being
louder. Beyond this distance, and out to about 700 m, only the amplified
sound would be heard. These specific distances depend on the assumption
that there are no other damping mechanisims in the air. The bottom line
is that it all depends on the pattern of the radiation of both. You
might be interested in an
earlier answer .

QUESTION:
How can two people play catch on a moving train. Without the ball zooming past like if you are in a car?

ANSWER:
Because the ball is moving along with the train so if the two are on the train and the train moves with constant velocity, all laws of physics are exactly the same as for somebody standing beside the tracks. So a game of catch will be exactly the same either in the train or beside the tracks because it is the laws of physics which govern how the ball will move.I

QUESTION:
To help fight the fires, the state uses planes to drop water and fire
retardants on the flames. One such plane flies horizontally over a fire at a
speed of 60 m/s and drops a giant water balloon to help extinguish the fire.
It flies at a height of 200 m.
If the plane released its load when right over the flames, it would
overshoot its target. It must release it a little earlier, marked by d on
the drawing. How far before the fire must it release the water?

ANSWER:
No homework. But there are lots of advertisers on this page which will help
with homework.

FOLLOWUP
QUESTION :
Please acknowledge the following question as this turned into a big argument with my son yesterday. This is not a homework question by any means, but rather the values were just manipulated by me and the question is just bare with no values for any measurements provided in the question.

ANSWER:
OK, I'll take your word for it this time. The equation for y
motion is y =0=y _{0} +v _{0y} t - ½gt ^{2} =200-0-½9.8t ^{2} =200-4.9t ^{2} ,
so t =6.39 s. The equation for x motion is x=x 0+v_{x} t =0+60t =383
m.

QUESTION:
Can you possibly explain to me what the precise nature of the mechanism in matter is which allows it to retain its internal inertial frame of reference even when it is tumbling in its flight? This is something which we take for granted every day, yet I have never been able to see into matter to understand why this is so. I am more than happy to make a meaningful donation if you can possibly explain this to me.

ANSWER:
I do not understand your question. Something "tumbling" "in its flight"
is not an inertial frame of reference. If you are inside an airplane
which is doing violent maneuvers and not buckled in, you will be thrown
around because the frame of reference is not inertial.

QUESTION:
I received a novelty gift that purports to find the "balance point" of a golf ball by spinning it up to 10,000 rpm. After 10-20 seconds the ball reaches an "equilibrium" spin and a horizontal line is marked that indicates the "balance axis". The assumption that on tees and greens you orient the ball to put the line vertically along the intended path so the
center of gravity (CoG) is rolling/spinning over the target line and thus minimizing potential effects of the CoG being on the side and potentially causing a "wobble". Putting the ball at different starting orientations in the device doesn't matter. It does tend to find the same "equilibrium" spin after a time so it is consistent.
I decided to mark a dozen balls using the device and then put them in a container of salt water to compare it to finding the CoG using a buoyancy test. Put in enough salt and eventually the golf balls float and
will reorient to put the CoG at the lowest possible position in the solution. Out of 12 balls only 1 had the previously marked axis running through the top of the ball. The others were all off by somewhere in the 40-45 degree range. The physics of the buoyancy test seem pretty straightforward and understandable to me. What is happening in the spin device is less clear. Can you explain the difference in the two tests? Is there a different "balance point"/CoG that is being located by the spin device? And to preempt your first obvious statement, Yes, I understand that none of this has a significant impact on my golf game compared to all of the other variables at play.

NOTE FROM
THE PHYSICIST:
This question has a history of numerous exchanges between me and the
questioner. It took, as you can see below, several hypotheses before I
was finally able to understand the results of his experiments. I have
elected to put the final answer first, labeled as "NEW ANSWER". But, I
have left the original answers in, beginning with "OLD ANSWER", because
there is quite a bit of interesting physics there (just not really the
solution to this question) and because it is of interest to see how a
series of ideas eventually can lead to the right idea in science.

NEW
ANSWER:
First, let's consider the physics of the spinning method. Clearly
the idea is that if the CoG is not at the geometrical center of the
ball, if you spin the ball about a vertical axis, the CoG will be
"thrown out" horizontally. But, will it end up in a horizontal plane
passing through the center of the ball? What I show here is that the
answer is no. If the CoG is a distance r from the
geometrical center of the ball, has a mass m ,
and is spinning with an angular velocity of ω , the CoG experiences
(in the rotating frame) three forces: the weight mg , the force T
holding it in place, and the centrifugal force C=mrω ^{2} .
In spite of all
earlier
published explanations, I now realize (see the earlier answers below) that the material is sufficiently rigid to
maintain its shape, i.e. r remains constant as the ball spins.
There will be an angle θ where the sphere is in equilibrium. Summing the
torques about the center of the sphere, mgr cosθ-mr ^{2} ω ^{2} sinθ =0^{
} so tanθ=g /(rω ^{2} ).

Now, the questioner found that for most balls, θ ≈45º
so tanθ ≈1.
Taking g ≈10 m/s^{2} and ω ^{2} =(10,000
rpm)^{2} ≈10^{6} s^{-2} ,
I find that r ≈10^{-5}
m. This is very small, 1/100 mm, but demonstrates that the spinner will not find
the correct plane for very small r . Technically, it never finds the
correct plane, but for r >1 mm, θ <0.6º. Since it is my understanding that
modern golf balls are very homogenous, this device is not useful for most balls.
It would work better if the axis of rotation were horizontal.

Next I should address the question of whether such a small r will be detectable by the
floating method. We can use the same figure above except with C =0.
The mass of the ball is about 0.046 kg and the radius of the ball is about R =0.021
m. Taking r =10^{-5} m, the torque about the center of the ball is τ=mgr cosθ= 4.6x10^{-6} cosθ N·m. The moment of inertia of the sphere is I =2mR ^{2} /5=4.1x10^{-5}
kg·m^{2} . So, the angular acceleration is α =τ /I= 0.11cosθ
s^{-2} . This means that, if you start at θ= 0º, after 1 s
the angular velocity would be about 0.11x180º/π ≈6º/s,
easily detectable I should think.

The bottom line here is that the questioner discovered that only one
of the 12 balls was not essentially perfectly "balanced".

All this is truly academic, though, since I am sure nobody
really thinks that a ball with its CoG less than 1/10 mm from
the center of the sphere will behave in any measureable way
differently than a perfect ball. So the Check-Go Pro does no
harm to your game, it just does no good unless you happen to
have a really off-center CoG. If you do both measurements,
you can locate surprisingly precisely where the CoG is with θ
giving you r and the vertical giving you the direction
of the line between the CoG and the center of the sphere.

OLD ANSWER:
You have devised a simpler way to find the "balance axis" than the fancy
gizmo you received. And I see no way that your method would not work—the
CoG should be vertically below the geometrical center of the ball. The
spinning gizmo seems like it ought to work also with the CoG being
forced to seek the plane perpendicular to the axis of rotation. So, why
are the two experiments different? I believe that the the two
experiments would yield identical results if the golf ball were a
perfectly rigi^{} d
body. But look at the picture above. A struck ball experiences a force
of about 2000 lb during the collision with a club; the weight of the
ball is about 0.1 lb, so this is about 20,000 times the weight, or about
20,000g . (One "g " of force is equal to the weight of
the object.) Now, if your gizmo gets the ball spinning with an angular
velocity of ω =10,000 rpm ≈1000 s^{-1} ,
the centripetal acceleration is a=R ω ^{2}
where R is the distance from the axis of rotation. A point only
1 mm from the center would have a =1000 m/s=100g and
for a point on the surface where R ≈2 cm, a =2000g ;
the resulting forces on the ball will surely cause the ball to deform
into an oblate spheroid, flattened along the axis of rotation. (The
figure shows a much exaggerated flattening compared to what the actual
would be.) However, unless the CoG is at the center of the ball,
changing the shape of the ball will change the location of the CoG and
after the ball stops spinning the CoG will move back to its original
location. The CoG will never be very far from the center, so I would
guess that the errors in locating the plane in which it lies could be
quite large. I can see no reason why your floating method would not
work. Maybe you ought to market it!

FOLLOWUP
QUESTION:
One follow up question (again purely academic) if I might. Since a golf ball spin rate could vary between 2500-9000 rpm depending on the club, would the "spinner" axis be more accurate/appropriate for a ball spinning at a high rate and the buoyancy axis more accurate for a ball rolling on a putting green?
Also, the buoyancy test wasn't my idea. I wish I were that clever. It's been around for a long time. I'm told that Ben Hogan checked his golf balls that way back in the 1950s when golf balls were quite a bit les s uniform than those manufactured today.

ANSWER:
I believe that the driven ball will experience almost no sensitivity to
the location of the CoG. The reason is that the axis about which the
ball is rotating will always pass through the CoG and there will be no
"wobble", unlike the rolling ball. An extreme example of the CoG far
from the geometrical center is shown in the hammer projectile above.
Although parts of the hammer are often far from the trajectory of the
CoG, it moves smoothly overall. It might have a very minor effect on the
ball by virtue of the lift generated by the spin, but I cannot imagine a
ball where the CoG is more than a millemeter from the center and any
such effect would likely be unmeasurably small.

FOLLOWUP
QUESTION:
Your response to the second question makes this even more academic (if possible) because if the rolling ball is the only one affected then clearly the spinner is superfluous.
One last followup question. I promise. After your first response, I wondered if the change in ball shape is measurable at all. I mounted a laser on a level and shot it over the top of the ball at rest so that it barely touched the top of the ball. After spinning the ball up to full speed I couldn't perceive any change at all in the amount of light touching the top of the ball. Is the expected change in shape that small? If so, does that still explain a 45 degree axis differential in the measuring methods?

ANSWER:
I cannot really predict how large the actual deformation will be other
than I expect it to be quite small. Still, the spinning could cause the
mass distribution to change. Suppose that the solid rubber core is
enclosed in a very rigid spherical shell; keep in mind that the core,
made of rubber, is compressible. At high rpm, the ball will act like a
centrifuge and the rubber core will be squeezed out toward the surface
making the density of the core far from the rotation axis larger than it
is closer to the axis. So, even if it does not change shape, the mass
distribution will change and the CoG would not (necessarily) be in the
same plane as it is when not spinning. Again, if the distance of the CoG
from the center is very small, even a very small change of its location
while spinning could introduce a large error in determining the plane in
which it lies. I cannot understand why 11 out of 12 balls would be off
by 40º-45º, though.

I have to admit, though, as I get deeper into this problem, that I have
some reservations about my answers. Most modern balls use an inner core
made of
polybutadiene rubber which is what superballs are made of. It turns out that
this rubber with a bulk modulus K =1.5-2 GPa is nearly as
incompressible as water with K =2.2
GPa. Assuming your buoyancy results are reproducable, I am puzzled. My final conclusion would have to be that if the ball is very close to homogenous, it is impossible to make a really accurate measurement of the plane in which the CoM resides but not so hard to find the line on which it resides.

There are lots
of videos on youtube about this spinning device compared to the
saltwater method. Most are promos for a golfball called RealLine .
There is one
with a result similar to your results.
One shows a kid using it to mark a ball and the line is clearly off
the equatorial position (i.e. would not cut the ball in equal
halves). If you read the
comments on the spinner on Amazon, there are some enlightening
comments on what might go wrong. Unfortunately, I could not find any
comprehensive comparisons of the two methods like you have done.

QUESTION:
If me and my friends went on a camping trip up in the mountains and place a pot on a fire with water in it with one egg in and and cooked it the same length of time we would of if we were home but the egg was not done what would be the cause of that

ANSWER:
At high altitudes the pressure is smaller (less air per unit volume). At
low pressures water boils at a lower temperature.

QUESTION:
Object A (lets call it a train which is moving) has a mass greater than object B (a man named conner who is stationary (and with abnormal strength)) but object B outputs more Force than object A. What would happen to the objects when they collide?

ANSWER:
Your phrase " …outputs more Force…" does not really
mean anything. During the collision, Newton's third law requires that
the force the train exerts on the man must be equal and opposite to the
force the man exerts on the train. However, you could imagine the man
pushing with his very strong arm during the collision time such that he
adds a certain amount of energy to the system; this would result in a
different situation from if he just stood there. Nevertheless, Newton's
third law would still apply. The extremes of his just standing there are

The speeds of
each may be found easily and are classic introductory physics problems.
Because the mass of the train is much larger than the mass of the man,
the results are pretty simple:

For the
inelastic collision, the man and train continue with approximately
the speed that the train came in.

For the
elastic collision, the train continues at approximately the speed it
came in with and the man proceeds in the same direction with
approximately twice that speed.

If the collsions last some short
time t , the average forces on the man during that time will be
approximately mv /t for the inelastic collision and 2mv /t
for the elastic collision; of course, the forces on the train will
be the same.

Now suppose
the man (mass m ) adds just the right amount of energy to
stop the train (mass M ) coming in with speed v ; I
assume that there is no other energy lost or gained. Then the man's
speed after the collision would be equal to (M /m )v ,
a huge speed! In this case the average force on the man would be Mv /t ;
he will have to be amazingly strong to endure this! The energy he would
have to add to stop the train would be approximately (M /m ) times the
energy the train came in with.

QUESTION:
On a recent BBC program I watched it said that the universe could 'borrow' energy from a vaccum so long as it gave it back quickly enough - that electrons and positrons would spontaniously form in a vaccum, then anialate each other to return the energy.
My question is this. If you had a vaccum inside an incredably strong magnetic field, could you pull these particals appart before they anialate? and if so what would happen, where would the energy for their creation come from
etc.

ANSWER:
A magnetic field would not be a good choice to try to do what you want
because the particles would be nearly at rest and experience little
force. However, a very strong gravitational field around a black hole
could add the energy of one particle's mass and it would escape while
the other was absorbed into the black hole; the black hole would get
lighter by the mass of one particle. This is called
Hawking
radiation .

QUESTION:
Hello, I was wondering about the good old water in the bucket example in a verticle circle and why does the water stay in the bucket at the top of the circle, when it is up-side down (at the top).
I'm actually having a bit tough time visualizing the forces acting, i mean there should be a tension acting downwards on the bucket (due to my hand)? Which could be minimum since at the top, the centripetal force could all be provided by mg (weight). (Is there a good reason why mg can provide all of the centripetal force?)
So mg is a constant, as per N3 law, the water should exert a force on the bucket upwards, and the bucket would exert a force on the water downwards, so there is also a normal reaction force pointing downwards towards the centre of the circle, should this be a minimum too? But wait, aren't we supposed to consider the bucket and water as a "one" object?
Moreover, what's the relation of the speed to this? Why should there be a minimum speed for the water not to fall? Also I'm guessing the radius in this case, would be the length of my arm?

ANSWER:
You are making this way too hard. The most important thing in solving
mechanics problems is to focus on one body at a time. You want to
understand when and if the water will fall, so choose to look at the
water as the body. The tension in the rope and the force of your hand
are irrelevant because they are not forces on the water. The only forces
on the water are its own weight, mg down, and the normal force
which the bucket exerts on the water, N . So, Newton's second
law is mg+N=mv ^{2} /R where v is the
speed and R is the length of the rope; I have chosen down as
the positive direction. Solving, N=m [v ^{2} /R )-g ].
Now note that if v <√(gR ) then N is
negative which means that it would be a vector which points up since I
chose down as positive. But the bucket is unable to exert a force up on
the water and so the water would not stay in the bucket if going too
slowly.

QUESTION:
[It took several exchanges with this questioner to get all
the information I needed, so I have paraphrased the question.]
An 850 pound 4-wheeler (including the weight of the occupants) is sitting still stopped, and gets pushed 32 ft from getting rear ended by an
800 lb motorcycle (including the weight of the rider) how fast is the motorcycle going?
After the collision, the motorcycle is at rest and the 4-wheeler has its
brakes locked and skids on dry asphalt.

ANSWER:
I can only do a very approximate calculation and will only retain 2
significant figures throughout. I prefer to work in SI units, so the
input data are 800 lb=360 kg, m =850 lb=390 kg, s =32
ft=9.8 m. I will use momentum conservation for the collision, 360v =390u
where v is the incoming speed of the motorcycle and u
is the outgoing speed of the 4-wheeler; so u =0.92v . As
the 4-wheeler slides, the friction of the brakes does work which takes
the energy of the 4-wheeler away. The energy is K = ½x390xu ^{2} =½x390x(0.92)^{2} v ^{2} =170v ^{2}
J; the work done by the friction is W=-μmgs =-0.9x390x9.8x9.8=-34,000
J where μ= 0.9 is the approximate coefficient of static
friction of tires on dry asphalt and g =9.8 m/s^{2} is
the acceleration due to gravity. So, K+W =0=170v ^{2} -34,000
or v =14 m/s=31 mph.

it!

FOLLOWUP
QUESTION:
31 mph? That's it, really?

ANSWER:
With the data you gave me, that is the best possible estimate. The place
where there might be a problem is the fact that the motorcycle is at
rest after the collision. Since the two masses are very close, this
implies that the collision was very nearly perfectly elastic, almost no
energy lost in the collision which would be surprising to me; given your
data, only 6% of the energy was lost. If the cycle were going much
faster and tried to brake to avoid the collision and continued on a
little way after the collision before stopping, the answer would have
been faster than 31. In any case, I can tell you that the speed of the
4-wheeler immediately after the collision had to be in the neighborhood
of 30 mph.

QUESTION:
I'm confused? One moment I'm reading about "inertial reference frames" and that "acceleration due to gravity" is unaffected by mass. This is followed up with examples such as the Bowling ball and feather. All good. Maths seems clear enough.
But then we start talking about a particular body/objects "acceleration due to gravity" at or near the surface of the Earth as being 9.8m/s^{2} which is calculated using the masses of earth and the "falling" body and Newton's Law. Similarly I read that if I go and stand on the moon the "acceleration due to gravity" will be different because the masses are different? Again seems to be clear.
What am I missing? How is it that the mass of two object does not affect the acceleration one moment but the accelerations of the moon and the earth on a body have differing values due in part (large part) to their mass?

ANSWER:
Look at the figure. Two masses, M and m , are separated
by a distance r . M exerts a force F _{mM}
on m and m exerts a force F _{Mm}
on M ; because of Newton's third law, these forces are equal and
opposite, F _{mM} =-F _{Mm} .
Because of Newton's law of universal gravitation the forces have
magnitude F _{Mn} =F _{mM} ≡F=mMG /r ^{2} .
Then, using Newton's second law, each mass will have an acceleration
independent of its own mass of a =[mMG /r ^{2} ]/m=MG/r^{2}
and A =[mMG /r ^{2} ]/M=mG/r^{2} .
Note that I have been viewing this from outside the system; this frame
of reference is is called an inertial frame of reference. It is
important that we view the system from an inertial frame, because
otherwise Newton's laws are not correct.

So, what you
have been taught is correct only if you view things from outside the
two-body system. But what you have also probably been taught is that you
measure this constant acceleration relative to the surface of the earth
and that is technically incorrect because the earth is accelerating up
to meet the falling mass and is therefore not an inertial frame.
However, the earth's acceleration is extremely tiny, too small to
measure. If M is much much larger than m , which is
certainly the case for the earth and the moon, what you have been taught
is, for all intents and purposes, correct.

QUESTION:
My question has to do with the conservation of angular momentum and black holes. If a rotating object enters a black hole, would that objects angular momentum not be converted to mass-energy increasing the mass and momentum of a black hole? As I see it, this would indeed violate the law of conservation of angular momentum. However an alternative theory had occurred to me. If indeed the angular momentum of the object is destroyed, it might impart a rotation on the universe as a hole, in the opposite direction of the spinning object. I do not know the mathematics or physics well enough to know whether or not that is a feasible hypothesis, but if it is, could it be possible that this universal rotation which was initially essentially non existent has continued to steadily increase such that today the centripetal force resulting from this rotation might explain the effects of dark Energy which opposes gravity over great distances. Please let me know if there is something basic I am misunderstanding, and as for the dark energy hypothesis, I am sure there is some reason I am wrong but if not and that is a valid hypothesis please let me know.

ANSWER:
As I state on the site, I do not normally answer questions on
astronomy/astrophysics/cosmology and I give some links to sites to which
it would be more appropriate to pose such questions. First of all, there
is no doubt that some black holes, perhaps all, have angular momentum.
Second, any angular momentum which an object has once it has entered the
event horizon, must become added to the black hole it already had
because no information may be transmitted from inside that radius.
Regarding whether angular momentum could be transferred to the "universe
as a [w]hole" (interesting typo on your part!), the universe is not a
rigid object, so any angular monentum lost (via interactions with other
nearby objects) before the object passes the event horizon would be
transferred to objects nearby. But the angular momentum of the whole
universe would remain unchanged since the object is also part of the
whole universe and what the rest of the universe gained, it lost. I urge
you to find another more authoritative source!

QUESTION:
Please explain how kinetic energy
affects the human body in flight.

ANSWER:
I do not understand. Kinetic energy of what? In flight?

QUESTION:
My apologies. I was asking if/how kinetic energy in the airplane
affects my human body while in flight and after.
Perhaps I know too little to even pose a question given that I am already assuming that this 'kinetic' energy is freely existing in the 'air' inside the plane.
That I just love the beauty of physics while remaining totally 'illiterate' to the laws
saddens me so I truly thank you for your kindness.

ANSWER:
Kinetic energy is the energy something has by virtue of its motion and is not something which
"affects" you. It is something you have or don't have, depending on your motion. And it is something which is relative, it depends on the frame
in which you calculate your energy; you have kinetic energy relative to
the ground but not relative to the airplane. The kinetic energy may be
written as K = ½mv ^{2} where m
is the mass and v is the speed. For example, if you have a mass
of 60 kg (about 130 lb) and are in an airplane going 500 mph=224 m/s,
your kinetic energy relative to the ground is K =½·60·224^{2} =1.5x10^{6}
Joules, a million and a half Joules! How did you get that energy?
During the takeoff the airplane pushed on you with some force to speed
you up and give you that energy. Suppose that the time to speed you up
to that speed was 10 minutes=600 seconds; then the power that the plane
had to deliver over that time would have been 1.5x10^{6} /600
J/s=2,500 Watts. The power delivered by the airplane to you would be
enough to power 25 100 Watt light bulbs. During the 10 minutes, the
airplane would have been pushing on you with a force of about 22.4
Newtons=5 lb; that is not much of an effect on you. But suppose that you
had accelerated to 500 mph in 1 second instead of 600; then the force on
you would have been 13,400 N=3000 lb which would have crushed you. So
you see, it is the force when you are acquiring your kinetic energy
which "affects" you, not the kinetic energy itself.

QUESTION:

Where happens all the matter that enters a black hole?
Is all the matter in a black hole crushed into a "singularity"
What are the current theories you suggest I study?

If a black hole can consume matter without limits, could a super massive
black hole also consume the mass our universe?

So, could the "Big Bang" be from this "singularity", a singularity that
holds all the mass of our universe?
(This makes sense to my way of thinking and boggles my mind.)

What are White Holes? Have they been proven to exist yet?
Or like "worm holes" are they only theoretical today?
And does current thinking say White Hole evolve from Black Holes?

ANSWER:
Well, I certainly appreciate your donation —almost nobody
does, even for my best, most creative answers. I do, however, recommend
that people wait for an answer before donating unless you like my site
so much that you just want to support it and I greatly appreciate that!
(Also, I have no way to "refund" contributions.) In your case, you did not read enough
on the site before submitting your
question(s). Two quotes from the site: "If your question is clearly astronomy or astrophysics, particularly detailed questions about black holes, stellar evolution, dark matter or dark energy, the big bang,
etc ., areas in which I am not expert, I may not answer" and "Please
submit single, concise, well-focused questions". Because of your contribution, I
will cut you a little slack this time.

The simplest idea of a
black hole is an object with mass, charge, and angular momentum
which has infinite density and where time has stopped. I believe the
latest thinking is that it is not really of infinite density.
Predictions of singularities are from classical general relativity;
if you view the situation quantum mechanically, then uncertertainty
ideas do not want zero size or stopped time because spacetime itself
should have a granularity. A discussion which you might find
accessible is at
Physics Forums .

This depends on how
things happen to be moving around. You could imagine a much older
universe composed of nothing but black holes and the radiation they
emit when they undergo
Hawking
radiation but moving in such a way that they will never
encounter each other.

There is a very
interesting book by Lee Smolin,
Time Reborn ,
where he reprises an idea originally by Wheeler and deWitt that
black holes are the seeds of new universes. There is a nice
interview with Smolin on
space.com and a
lecture by
him that you can watch.

White holes are
hypothetical and have never been observed. I know nothing about
them.

I hope I have earned your
donation!

QUESTION:
is there any gravity on space? infomations are random on internet, like gravity is every where, there is no zero gravity concept, but as we go far from earth gravity start decreases as black holes are in space, and they have such a strong gravity that nothing escape of it, so where black hole got such a strong gravity, i know with increasing mass gravity increases, but black holes are not near earth surface so from where they gain gravity???

ANSWER:
Every object in the universe with mass causes gravity. The bigger the
mass, the bigger the gravity. You cause a gravitational field, but it is
tiny compared the the earth's field. The earth's field is tiny compared
to the sun's. The sun's field is tiny compared to a black hole's. The
entire universe is permeated by gravitational fields. And gravity is a
very long-range force; although the force gets smaller as you get
farther away, it extends all the way across the universe.

QUESTION:
I have calculated that a 50 g marble attached to a 1 m string wrapping itself around your finger held above your head at a rate of one revolution every second until it reaches 5 cm from your finger will end up applying a centrifugal force of the equivalent of the force of gravity on 1600 kg. Does this prove that conservation of angular momentum is a fallacy?

ANSWER:
This is a variation of the tetherball problem, a classic in introductory
physics courses. If you want to talk about angular momentum
conservation, you have to ask under what conditions angular momentum is
conserved. The angular momentum of your marble relative to the center of
your finger is conserved if there are no forces which exert a torque on
it. The only force is the tension in the string and, as you can plainly
see, the tension T has a component perpendicular to r
(T sinθ ) and therefore exerts a torque
τ=rT sinθ and therefore angular momentum is
not conserved; this does not make it a "fallacy", it just does not hold
for this particular problem. What is conserved, though, is energy.
Because the displacement (along the velocity vector) is always
perpendicular to the tension, the tension does no work so energy is
conserved, ½mv _{1} ^{2} =½mv _{2} ^{2}
or v _{1} =v _{2} , the velocity stays
constant. So, if the m =50 g=0.05 kg marble starts at about 1 m
away from your finger with a frequency of ω =1 rev/s=2π
radians/s, its speed is v ≈ωL= 2π
m/s; then when L =5 cm=0.05 m, T=mv ^{2} /L =0.05x4π ^{2} /0.05=39.5
N=4.0 kg-force.

THE
PHYSICIST :
The questioner submitted followup questions. To see these,
link here to the
Off-the-Wall Hall of Fame.

QUESTION:
Since electrons occupy discrete energy levels in an atom , shouldn't the electrons that are more energetic (are in a higher energy level) need less energy to escape the atom ? When light is shone onto the material and is above the threshold frequency electrons are emitted , but are those electrons from the highest energy level ?? It confuses me , because if that's true , then that would mean that the work function would differ for each energy level which doesn't make any sense , because work function is defined as the minimum energy for an electron to be released.

ANSWER:
You should not look at electrical properties of a conductor by looking
at atomic structure of its constituents. A solid composed of huge
numbers of atoms does not behave the same way as a single atom does. In
a conductor, atoms all interact with their neighbors in such a way that
at least one electron per atom (called conduction electrons) moves
around inside the solid pretty much freely like an electron gas. The
photon strikes the surface and gives all its energy to a single
electron. For that electron to be ejected from the metal, it must have
more kinetic energy than the work you would have to do to just pull it
out of the metal, and this is not how much work you would have to do to
pull it out of a single atom because it is not bound to any atom; the
former is called the work function , the latter is
called the ionization potential . The idea is that when an
electron is removed from the metal, one positively charged atom is left
behind resulting in an attractive force trying to hold the electron in.
A point charge in front of a plane ideal conductor is a classic
electrostatics problem, usually solved using the
method
of images . If you then integrate from the size of an atom (~10^{-10}
m) to infinity you can get a ball-park estimate of the work function,
W ≈3.6 eV.

QUESTION:
My question is that throughout The Principia Newton uses the fact that for a body continuously moving in a uniform circular motion with constant velocity, the body falls a small distance towards the center of the circle. Had there been no centripetal force, the body would move along a straight line(tangent), but due to the force the body falls towards the center, Due to its velocity the body does not completely fall but moves in a circle. Now as Newton, Chandrashekhar and Feynman have shown in their respective books, and also quite obviously, in a very short interval of time the deviation produced by the force or more precisely the distance fallen by the body will pe parallel to the radius at the initial point and parallel to the central force at the point. Since the displacement will be parallel to the force ,some finite work would be done, because work is defined as the product of the force and the displacement caused by the force in the direction of the force. Important thing is that the displacement to be taken is that which is caused by the force and not due to initial velocity. Therefore some finite , very small amount of work will be done by the centripetal force since the fallen distance and and the force are parallel. When this work be integrated over half a circle or 1/4 or 3/4 of circle it will give an increase in kinetic energy showing that the velocity has increased. This is contradictory. So what is wrong over here or is something correct ?

ANSWER:
A short answer to a long question: It is true that an object in a
circular orbit is constantly "falling". It is not true that it is moving
toward the center, rather that it has an acceleration in that direction.
Although the change in velocity is centripetal, the displacement is
always tangential and the force does zero work. By the way, the body
does not move "with constant velocity" as you state, but with constant
speed. The direction of the velocity vector is constantly changing.

QUESTION:
we have defined momentum as the product of mass and the velocity of a body, we say that photon is a mass less particle,i find it really confusing how can a massless particle still have the momentum?

ANSWER:
Linear momentum p was, indeed, defined as mv before
the advent of the theory of special relativity. If v is much
less than c , the speed of light, this is an excellent
approximation, but not exactly true. The correct expression for p
is m _{0} v / √[1-(v ^{2} /c ^{2} )]
where m _{0} is the mass of the object when at rest. The
relation among energy E , momentum p , and rest mass
m _{0} is E ^{2} =p ^{2} c ^{2} +m_{0} ^{2} c ^{4} .
So, you see, even if a particle has zero mass (like the photon) it still
has momentum if it has energy, p=E /c . I am also often
asked how a photon can have energy because E=mc ^{2} and
m =0 for a photon. You can find links on the
faq page to answers which
discuss this question.

QUESTION:
A wheel rolls without slipping with angilar velocity ω and radius r what is the angular velocity of a point in the rim at the same level as the centre ?

ANSWER:
I believe that you are asking the wrong question; you must be asking
what the velocity of the point is, not its angular velocity. It is
important to understand how the wheel is rotating. Since the point of
contact with the ground is at rest at any instant, the whole wheel is
rotating about that point at that instant. The angular velocity ω
about this point is v /R where R is the radius
of the wheel and v is the speed which the center of the wheel
is moving forward with speed v . Now, the point on the rim also
has the same angular velocity ω , but its distance from
the axis of rotation is R' . Because the angle which R'
makes with the ground is 45 º, it is easy to show that
R'=R √2. Therefore,
ω=v /R=v' /R'=v' /(R √2),
and so v'=v √2; the direction of the velocity
v' is 45º below the horizontal, as shown above.

QUESTION:
Is it possible to create a gravitational lens without a black hole or dense object? If so, can gravitational lensing be practical for use? I've been thinking about it for quite some time now and tried thinking of possible uses.

ANSWER:
By definition, gravitational lensing is the result of strong
gravitational fields which are caused by large masses. So, without a
large compact mass, appreciable lensing will not occur.

QUESTION:
Gravity Well question that's been puzzling me since I watched Interstellar...If an object is orbiting in the gravity well of a massive planet/star, it is falling (conforming perfectly to the curved space locally). Other than its orbital velocity, why would its depth in the gravity well cause its clock to tick any slower than another object floating more distantly from the same gravity well.Seems to me that gravity well time dilation only applies if you're fighting the pull of that well (e.g., by standing on the planet's surface)Thoughts?

ANSWER:
Gravity does not just warp space, it warps spacetime. The larger the
gravitational field, the more strongly spacetime is warped. Therefore
one finds that clocks run slow in a gravitational field, the stronger
the field, the slower they run.

QUESTION:
What is the physics behind a baseball curving?

ANSWER:
See an earlier answer .

QUESTION:
If a sun's gravity attracts a planet, is it possible it would repel an anti planet?

ANSWER:
No, that is not possible. There is only one kind of gravitational mass
and therefore only attractive gravitational forces are possible. Every
experiment ever done with antiparticles indicates that the mass is
identical to that of its particle counterpart.

QUESTION:
I am trying to explain to my brother why on a spinning wheel a point farther out is going faster than a point closer to axis, though the wheel is spinning at the same rpms. But he just cant figure it out. Could you give an explanation a 2 year old could understand?
PS My brother is 21.

ANSWER:
I could probably not convince a two-year old, but if your brother is
just a little smarter than one, I can probably convince him. The speed
of something is defined as the distance traveled divided by the time it takes to
travel that distance. For example, a car going around a circular race
track which has a total circumference of 2 miles takes 2 minutes to go
around once, its speed is (2 miles)/(2 minutes)=1 mile/minute=60 mph.
Suppose a wheel has an angular speed of 10 rpm and has a circumference
of 2 m. Then the distance a point on the rim will go in 1 minute is 20 m
because the wheel goes around 10 times; the speed of that point is
therefore 20 m/min. Now look at a point halfway from the axle to the
rim; it will move in a circle of circumference only 1 m so the distance
it travels in 1 min is only 10 m so its speed is therefore 10 m/min. In
a nutshell, a point near the center travels a shorter distance than a
point far from the center in the same time.

QUESTION:
my question is about speed, if our earth is traveling at 1600KM per hour and the Milky way is more then several Million Km Per Hour etc, how our equmileting speed is not reaching the speed of light, or at least breaking the sound barrier..
and if i jump i move little from my original spot?

ANSWER:
See an earlier answer . Also, the sound barrier
is irrelevant because there is no sound in space. Although not really
related to your main question, if you jump vertically upward in a
rotating coordinate system (like the earth spinning on its axis) you
will indeed not land exactly where you launched but the difference is
very small.

QUESTION:
If gravity is not a force but just a curvature of spacetime then how does a massive object (like earth) affect a much less massive object (like a tennis ball) when they are not in motion relative to each other?
For example, if I were able to travel out to space far enough away from earth and then stop so that I am not in motion with respect to earth and let go of a tennis ball it will stay stationary.
But if I traveled to say 100,000 ft above sea level and were able to hover there so I am not in motion to earth and then let go of a tennis ball it will immediately begin to move towards earth.
In other words, how is it possible for the curvature of spacetime to affect bodies that are not in motion to other?

ANSWER:
You have some misconceptions here. First of all, no matter how far away
you get, there will always be a small force toward the earth. It may be
so small that you would have to wait a millenium to see it move a
millimeter, but it is still there. The curvature of spacetime justs gets
smaller as you get farther away. It is certainly true that if you drop
something from an altitude of 100,000 feet it will accelerate toward the
earth. But, what does that have to do with motion? Even if you gave it
some motion, say throwing it horizonatlly, it would still accelerate
toward the earth just the same as dropping it, but now it would also
have a velocity component parallel to the earth as well. Maybe you have
never seen the "trampoline
model " which illustrates (in a simplistic, not literal way) how
warping the space (the surface of a trampoline) by a massive object
(bowling ball) will cause a light object (marble) to be attracted to the
massive object.

QUESTION:
Engineers in the Bay of Fundy are trying to harness tidal power to generate electricity. It's claimed that they can generate enough electricity to power the entire Atlantic Provinces of Canada.
This got me to thinking, if that much power can be harnessed, where has that energy been going? I'm assuming it goes to heat and warms the water. If that's the case, would harnessing the power therefore cool the water and potentially harm aquatic ecosystems?

ANSWER:
My research shows that all the power projects on the Bay of Fundy will
generate about 20 MW. These will be powered by underwater turbines. I
did a very rough estimate of the total power available by the falling
water: The average rise in sea level is 15 m, area of the bay is 13,000
km^{2} ,
density of water is 1000 kg/m^{3} . I find that the total volume
to fall is about 2x10^{11} m^{3} , the mass is 2x10^{14}
kg, and the potential energy of this mass is about 3x10^{16} J. If
you deliver that much energy over the course of a day, the average power
is about 350 GW. So, the power plants will only reduce the energy of the
tidal flow by less than 0.01%, a truly negligible amount.

QUESTION:
I recently read Michio Kaku's book that discusses impossible technologies and how they can be achieved and even possible. He discusses that a perpetual motion is next to impossible, because energy is always wasted, attributed to the laws of thermodynamics. Now, when talking about using anti-matter as energy, it is described as having 100% efficiency. Wouldn't that violate the laws of thermodynamics, or is it one of the few cases where we see the laws of physics bend?

ANSWER:
The relevant "laws of thermodynamics" are essentially just the
conservation of energy. Energy is always conserved in an isolated
system, it is just that in the macroscopic world some of the energy at
the end is not useful. We normally specify efficiency of a machine as
the relation of energy added to a system to the work we get out; any
difference between the two shows up elsewhere, usually as heat. In
particle-antiparticle annihilation energy is conserved just like the for
macroscopic machine, but there is no work, just electromagnetic energy
(two photons); if you think of it as a machine, you need to convert
those two photons into useful work and that conversion would likely not
be 100% efficient. On a microscopic level, any interaction among
elementary particles is "100% efficient" if you compare energy before
with energy after. Shoot an electron at an atom and have it excite the
atom; the electron leaves with less energy than it came with, the atom
recoils because of having been hit by an electron, and then the atom
deexcites by emitting a photon. Add up the energy of the electron, the
photon, and the atom after the collision and you get the same total as
the energy of the electron and the atom before the collision.

QUESTION:
If I were to fall straight through the earth, from one side to the other, how long would that take if it were possible?

ANSWER:
If the earth were a uniform sphere, constant density throughout its
volume, this is a classic elementary physics problem; a solution may be
seen on the
hyperphysics web site. The answer is about 42 minutes. However, as
discussed in an earlier answer , the earth is
by no means a uniform sphere so the answer would be different but not
easy to calculate.

QUESTION:
If there's less gravity on the moon, then why do astronauts move slower? Why don't they have more spring if gravity's pull is weaker? I'm thinking that this should be a matter of resistance, but obviously there's a factor I'm not accounting for.

ANSWER:
Sure, with less gravity (about 1/6 of earth) an astronaut can jump much
higher, but it will take longer to get to the top and back down than on
earth; that would look like slow motion, right? Or, just think about
taking a step. Your center of gravity is forward of your trailing foot
and so you rotate about that foot. The rate of rotational acceleration
is only 1/6 that on earth, again like slow motion.

QUESTION:
Some car drivers reason that since tractor trailers have more tires, they
should be able to stop quicker. How much work does each car tire need to do
to stop it, compared with how much work each truck tire needs to do to stop
it? For ease of reference let's say a car is 4000 pounds and the truck is
80000 pounds while the speed involved is 65 miles per hour. Finally is the
work done proportional to the size difference of truck brakes versus car
brakes?

ANSWER:
I replied that this is not a homework
solving site.

FOLLOWUP
QUESTION :
I didn't realize I had posed this in the form of a homework-ish question. I'm a 47yo truck driver, tired of hearing all the lame-@$$ rhetoric of the motoring public. Long, quiet, empty night highway helped me come at this from a different angle. Without too much detail, one truck wheel wrangles slightly more than a single car's worth of mass; is why trucks do NOT stop better despite having more wheels.

ANSWER:
I have never heard this but it is pretty nonsensical. How quickly any
vehicle can stop is determined by the maximum force which can be applied
by braking; if the coefficient of static friction between the tires and
the road is μ , the maximum force on a level surface
is μW . where W is the weight of the vehicle. The
distance s traveled will be determined by the work done by this
force, μWs , which will equal the initial kinetic
energy, ½Mv ^{2} =½(W /g )v ^{2} ,
where M is the mass and g =32 ft/s^{2} . So,
s =(1/(2μg )v ^{2} . So here is the
interesting thing: the distance traveled does not depend on the weight
of the vehicle, only by the initial speed and the condition of the road.
The force applied does depend on the weight and the force per wheel
would be 0.7x4000/4=700 lb for the car and 0.7x80000/18=3111 lb; but the
larger force is simply because of the larger weight and the only thing
which is important is the distance to stop. Of course, this also depends
on whether you have antilock brakes which let you get the maximum amount
of friction when you are just on the verge of skidding; if you lock the
brakes and you skid, you go farther before stopping. For your example,
v =65 mph=95 ft/s and μ is approximately 0.7 for
tires on a dry road, so s ≈200 ft. I have neglected all
other forms of friction like rolling friction and air drag, but these
should be relatively unimportant. Bottom line—stopping distance
does not depend on number of tires nor on weight.

QUESTION:
I am trying to figure out how to calculate the Cubic yards of a relatively flat surface. Specific example: A carpet or rug (let's say it is 1 inch thick and 100 square feet). How many cubic yards (3ftx3ftx3ft) would it be when rolled up? Or what is the equation on how to get the answer for varying sized carpets?

ANSWER:
If the area is A and the thickness is t , the volume is
V=At . Of course, you have to be careful to have A and
t measured in the same units. V is the volume
regardless of how you change its shape. Since you apparently want your
answer to be in cubic yards, you should probably convert everything to
yards at the outset. For the example you state, A =(100 ft^{2} )(1
yd/3 ft)^{2} =100/9 yd^{2} and t =(1 in)(1 yd/36
in)=1/36 yd. So V =(100/9)(1/36)=0.309 yd^{3} . You can
get more detailed if you know the length L and width W
of the rug so A=LW and V=LWt . If you roll it up to a
cylinder of length L , its volume will be V= πR ^{2} L =LWt
where R is the radius of the cylinder; therefore, R = √(Wt /π ).
So, if your rug is 10x10 ft^{2} , it will roll up with a radius
of R =√[(10/3)(1/36)/3.14]=0.172 yd=0.52 ft. All this
assumes that the rug does not compress when you roll it up.

QUESTION:
Kind of an odd question with a potentially obvious answer, but i was wondering, if you were on the moon and you jumped forward in a superman like pose, would you keep going? like, would you gain speed or would you still eventually fall back to the moon? and would the spin of the moon have an affect on this?

ANSWER:
The escape velocity of the moon is about 2.38 km/hr which is more than
5000 mph. I do not think you can jump that fast. A low altitude orbital
velocity is about 3700 mph, so I don't think you can jump that fast
either. The gravity on the moon is about 1/6 that on the earth, so you
could certainly be able to jump farther than on earth; but you would
never "keep going". And you would certainly not "gain speed". And the
spin of the moon on its axis would be of no use since it only goes
around its axis once a month.

QUESTION:
I work with a product by the name of Stereotaxis. It's a magnetic navagation system designed to move catheters inside a patients body by people in a control room (as to shield the operator, or physician, from the effects). Unfortunately, my colleagues and I (registered nurses) are not as fortunate. We typically sit about 3 feet in front of the Stereotaxis magnet for most of the day (can be 10 hours or more). There are 2 magnets that produce 800 gauss each (we sit in front of one magnet, not both). The other magnet only effects us when it is deployed for use with the patient. Is this dangerous to us? We all work 40 hours per week; at least, sometimes much more.

ANSWER:
I have just done some cursory research on the effects of magnetic fields
on biological systems. It would appear to me that there are virtually no
effects for fields smaller than about 10,000 gauss and your exposure is
a full order of magnitude below that level. And, if the magnets are
rated at 800 gauss the field where you are is probably less than that.
Also, the effects of a magnetic field would not be cumulative like
exposure to radiation would be—when you go home there would have
been no damage to your cells. The reasons for the people being in the
control room would certainly not be to isolate them from these fields. I
have had two atrial ablations myself and navigation was done, I believe,
by x-ray for which long-term exposure would much more a concern; maybe
the control room is a vestige from earlier days. Also, the control room
likely provides a better environment for concentration by the
professionals doing the procedure.

QUESTION:
If a charged particle is at rest on the surface of the earth, it
emits no light visible in the inertial frame of someone standing on
earth (both accelerated at the same rate, equivalence). Consider a
charged particle at rest on the earth, as viewed by an orbiting
observer, far above the earth, or viewed by an observer in a rocket
hovering thousands of miles above the earth (where gravitational
acceleration is less than on the surface). Will the observer see light
emitting from the charge since it is accelerating (in gravitational
equivalence) relative to the observer?

ANSWER:
Someone standing on earth is not in an inertial frame, but that does not
really matter here because the charge is at rest relative to that
person. And, yes, anyone accelerating relative to the charge will
observe a radiation field. An
earlier question is
very similar to yours and you should read that answer to get more
detail.

QUESTION:
i was studying something on the internet where i need to know one question i.e If an object Travelling at about 2500 km per hour or may be higher collides with other object in the ocean around 90 meters below the ocean level is it possible that the debris of that object will go around 2 km approx ahead in the water itself?

ANSWER:
Extremely unlikely. Let me show you a very rough calculation which will
demonstrate this for an extreme case. If one object collides elastically
with another mass whose mass is very small compared to the incident
mass, the collided-with mass will recoil with a speed twice the incident
speed, in your case that would be 2x2500 km/hr=5000 km/hr ≈1400
m/s. The drag force on an object with speed v in water may be
approximated as F _{drag} =½CρAv ^{2} ;
here I will choose C≈ 1 (order of magnitude for most
shapes) is the drag coefficient which depends only on the shape, ρ ≈1000
kg/m^{3} is the density of water, and A ≈1 cm^{2} ≈10^{-4}
m^{2} is the cross sectional area of the object. I want this
object to go as far as possible, so I have chosen A to be
relatively small; an object this small will have a relatively small
mass, certainly no larger than m ≈100 gm=0.1 kg. Knowing
all this, one can calculate the velocity v and position x
as functions of time t , v =v _{0} /(1+kt )
and x =(v _{0} /k )ln(1+kt )
where k =½CρAv _{0} /m≈ 7000
s ^{-1} and v _{0} =1400 m/s is the
starting velocity. I find that at t =½ s, v≈ 0.4
m/s and x≈ 1.6 m; the object will have lost nearly all its
speed after having traveled only about a meter and a half. I can think
of no earthly way that any debris could propogate 2 km!

QUESTION:
Would my weight be the same on the surface of earth and one mile underground? How about one mile in the atmosphere?

ANSWER:
Above the surface of the earth the gravitational force falls off like 1/r ^{2}
where r is the distance from the center of the earth. If R= 6.4x10^{6}
m is the radius of the earth, W is your weight at the surface,
and W^{+} is your weight 1 mile=1.6x10^{3} m above the
surface, then W ^{+} /W =R ^{2} /r ^{2} =0.9995;
your weight would be about 0.05% smaller. If you assume that the mass of
the earth is uniformly distributed throughout its entire volume, the
gravitational force falls off like r as you go toward the
center. Then W ^{-} /W =r /R= 0.99975;
your weight would be about 0.025% smaller. Given other factors like
local variations in the density of the surrounding earth, these would be
unmeasurably small; one mile (about 1600 m) is, after all, extremely
tiny realtive to the size of the earth. I should also note that
approximating the earth's density as uniform is a very poor
approximation; see an
earlier answer . If you had asked your weight 100 miles below the
surface the answer would be that your weight is nearly the same as at
the surface.

QUESTION:
How is the light red shift, observed in the universe, a factor of distance AND speed and the Doppler affect (in which it is often compared) is only a factor of distance? Galaxies farther away have a greater red shift so we are told they are farther away and moving away faster than nearby galaxies. With Doppler the farther the object is away the greater the shift, but the speed of travel could be constant, say using the historic train example. I could buy into the universe is expanding, I'm not convinced we can predict how fast relative to time.

ANSWER:
The Doppler effect does not depend on distance, only speed. Many years
ago astronomers discovered that distant stars and galaxies exhibited
red-shifted spectra which they could use to determine the speed at
which those objects were traveling. Later methods were developed which
allowed an independent measure of the distance of those
objects. When the two separate measurements were compared, it was found
that the velocity of the objects was a linear function of their
distances. The graph shows an example where distance was determined
using the brightness of type Ia supernovae. I do not understand your
question beginning with " …With Doppler the farther…"

QUESTION:
People say that Universe is expanding, right?
We all know that Gravitational force is an attractive force and it is predominant at the planetary level among all other forces. So shouldn't the universe come closer together as time passes? Not at a very fast rate, but at least very slowly.
I don't understand why still people do say it expands, Can you please help me reasoning it out?

ANSWER:
Right after you throw a ball straight up, it is moving upward
("expanding" relative to you); the harder you throw it, the farther it
goes before you and it "come closer together". The universe is still in
the early stages where it is still "going up". But, if you throw it hard
enough (called the escape velocity ), it will never come back.
And, if you throw it with a velocity greater than the escape velocity,
it will never stop going. So, the ultimate fate of the universe is that
it will either expand forever or everything will turn around and fall
back together (the big crunch ). All that is out of date,
though, because it was discovered almost two decades ago that gravity is
not purely attractive or else there is some other force which is
operative as well (called
dark energy );
this results in the fact that the distant objects in the universe are
not just moving away from us, they are actually speeding up. This
repulsive force is not yet fully understood and the ultimate fate of the
universe is not known.

QUESTION:
I am a volunteer guide at South Foreland2historic lighthouse in the UK. We have an optic weighing approximately 2 tons, floating in a close fitting trough containing only approx. 28 litres of mercury. What is the theory which enables this optic to float as it does not appear to fit within the basics of Archimedes principle.

ANSWER:
To float 2 metric tons (2000 kg) you must displace M =2000 kg of
mercury. The density of mercury is ρ =13,600 kg/m^{3} ,
so the volume you must displace is V=M /ρ =0.15 m^{3} =150
l; this, I presume, is what is bothering you since only 28 l are used.
Suppose that the reservoir for the mercury is a cylinder of radius 1 m
and depth d ; to contain 0.15 m^{3} , the depth of the
container would have to be d =0.05 m=5 cm (estimating
π ≈3). So, I will make a container 6 cm high for an
extra 20%, 180 l and I will buy that 180 l to fill it up. Now, let's
make the pedestal on which the lens sits be a solid cylinder of radius
99 cm so that when you put it into the reservoir there will be 1 cm gap
all around. So as you lower it into the reservoir, mercury will spill
out the top and you will be sure to capture it. When you have captured
150 l, the whole thing will be floating on the mercury. You return the
extra 150 l and have floated the lens with only 180-150=30 l. Of course,
in the real world you would only buy 30 l, put the pedestal into the
empty reservoir, and add mercury until it floats. (I realize that the
shape of the pedestal and reservoir are probably not full cylinders,
since you said "trough", but my simple example wouldn't be so simple
with more complicated volumes. The idea is the same, though.)

QUESTION:
In a controlled environment with no crosswinds, with a tailwind of 10mph (or any speed really), would it be possible for a person to move fast enough to equal the effect the wind has on his clothes, hair, etc...? I.E. get to a point where his clothes are essentially motionless as if a perfectly calm day?

ANSWER:
If you move with a velocity equal to the velocity of the air, you are at
rest relative to the air. Therefore the answer to your question is yes.
An example is a hot air balloon; the balloon will be pushed in the
direction of the wind and attain the speed of the wind. If you are
riding in the balloon, the air around you will be still. This becomes a
problem if you wish to steer the balloon because to change direction of
any aircraft you need to be moving relative to the air.

QUESTION:
If the world was a cube would we fall off?

ANSWER:
No, you would not fall off because the cube would still have a
gravitational field which would attract you. You would weigh the most
(least) at the center of each face (the corners) because you are closest
to (farthest from) the center of the cube. At the centers your weight
would be "straight down", i.e. perpendicular to the ground. Elsewhere
your weight would have a component parallel to the ground as shown in
the figure above. Note that if you released a ball at an edge, it would
roll to the center of the face; or if you were to walk from the center
to an edge, it would be like walking uphill. (By the way, you should use
the subjunctive—If the earth were
a cube… )

QUESTION:
A stone and a feather are falling in a vacuum chamber on earth and if left to fall, which would fall first? I still say
the stone because the gravity force is still a function of mass. so basically more mass, more attraction, if it weren't true, jupiter and feather would reach at the same time.
What is right?

ANSWER:
An object of mass m a distance R from the center of
the earth feels a force F=mΦ where Φ=GM/R ^{2}
is the gravitational field and M is the mass of the earth.
But Newton's second law tells us that F=ma where a is
the acceleration of m ; therefore a =Φ
regardless what m is. All this assumes that the earth, which
also feels the same force as m , has a mass much larger than the
object, M>>m so that the acceleration of the earth is
negligible. If m were comparable to or larger than M ,
the earth would accelerate toward the object so the two would meet
earlier; but the the heavy object's acceleration would be just the same
as the feather's even in this scenario. You are right that the stone
feels more force but it also has a greater inertia which results in
identical accelerations.

QUESTION:
The below question I have found in an old text book. I am soon to be a teacher in training so brushing up on my mathematics, but this one has stumped me. I am not sure if I am going along the right lines.
A uniform rod ab of weight W and which is 20 cm long is suspended by two vertical springs
X and Y attached to the ends of the rod. The upper ends of the springs are attached to a horizontal beam. When the springs are unextended they have the same length. The tension in X is given by
T_{X} =Kx and the tension in Y is given by T_{Y} = 3Ky, where
K is a constant and x and y are the extensions of X and Y respectively. At what distance from A must a body of weight 5W be attached to the rod if the rod is to be horizontal?

ANSWER:
Each spring is stretched by the same amount, so if the tension is X is
T then the tension is Y is 3T . W acts at a
distance 0.1 m from X and 5W acts at a distance d from
X. You now need to apply the equilibrium conditions (Newton's first
law). The sum of the forces is zero, 4T -6W =0, so T =1.5W .
The sum of the torques about the left end is zero, (3x0.2)T -0.1W -5dW =0=0.8W -5dW
or d =0.16 m. (The solution which you attached, not shown here,
got the relative tensions in the springs correct, but you never applied
Newton's first law and never thought about torques at all.)

QUESTION:
one, is there anything we know of faster then light?
two, could I say that the light being pulled into a black hole is the speed in which light cant escape therefor making that faster or is that a measurement of some kind of force. or does it not matter the force speed, speed is speed?

ANSWER:
One: no. Two: This sentence makes little or no sense to me. What I
can tell you is that light falling into a black hole is gaining energy
but it is not gaining speed. Energy of light increases by increasing the
frequency (decreasing the wavelength).

QUESTION:
Is there a way to calculate the weight equivalent of pulling a load of X pounds up and incline of Y degrees? If I tow a 1000 pound load up a 5 % grade, it's like pulling how many pounds on a level?

ANSWER:
There a couple of things ambiguous about this question. First, "weight
equivalent" is not defined; it seems to me that what you want is how
much force you must exert to pull a weight up an incline. Comparing this
force, as I will show below, with how much force it takes to pull it
horizontally is not really meaningful. Second, you really cannot answer
this question without specifying the friction between the weight and the
incline/floor. Suppose that the coefficient of kinetic friction is
μ , the weight is W , and the angle of the incline is
θ . Then
the force to pull it with constant speed up the incline is F _{incline} =W (sinθ+μ cosθ )
and to pull it along the floor is F _{floor} =μW .
If there were no friction (like if the weight were on very well
lubricated wheels) and your 1000 lb weight were pulled up a 2.86º
(5%) slope, F _{floor} =0 and F _{incline} =49.9
lb. With μ= 0.5, F _{floor} =500 lb and F _{incline} =549
lb. The reason F _{incline } is bigger is that, in
addition to pulling the mass in a horizontal direction, you are also
lifting it.

QUESTION:
i should know this already, but i can't completely understand. in chemical processes, a reaction is either exothermic or endothermic. if you put heat into it, the heat can be extracted. why is fusion and fission not the same? i mean if you break the strong bonds, i can understand a release of energy, but when you create new strong bonds in fusion? gravity provides energy for the fusion in stars, but in reactors or weapons, i don't understand

ANSWER:
There is only one rule you need to know for exothermic/endothermic
reactions: If the sum of the masses after the reaction is smaller than
the sum of the masses before the reaction, it is exothermic; otherwise,
it is endothermic. Chemistry is a really poor source of energy, so the
mass changes are very tiny (essentially unmeasurable), so this was not
known for chemical reactions until we knew that E=mc ^{2}
and is not taught in elementary chemistry courses. And, you should avoid
the notion that if you must add energy to make the reaction go it is
endothermic because you might get more energy out than you put in.
Fusion is such a case: in order to fuse two deuterons (^{2} H) to
one alpha (^{4} He) you must overcome their mutual Coulomb
repulsion; but if you compare the masses before and after the fusion,
you will find that their is much less after (an alpha is much more
tightly bound than a deuteron). Similarly, if you compare the masses
before and after a fission of a heavy nucleus, the total mass afterward
is measurably smaller. Fusion (fission) is exothermic for nuclei lighter
(heavier) than iron. That is why stars do not make elements heavier than
iron; heavier elements are created in supernova explosions or other
exotic astronomical events. You might have a look at an
earlier answer .

QUESTION:
Could you please explain why the mass and energy of the Carbon 14 decay into Nitrogen 14 do not seem to add up? Carbon 14 has a mass of 14.003241u, Nitrogen 14 has a mass of 14.003074u, and a beta particle (e-) has a mass of 0.000548u. It appears that the mass after the decay is larger than the mass before the decay. In addition, the beta particle is supposed to have energy of up to 156 keV. Doesn't this energy come from the conversion of mass into energy? How is it that there is more mass after this decay as compared to before the decay?

ANSWER:
The masses you use are atomic masses. The ^{14} N atom
left after the decay has only 6 electrons, not the 7 in a ^{14} N
atom, so you do not add the mass energy of the β ^{-
} (or, if you wish, you add it and subtract the mass energy of the
missing electron).

QUESTION:
In the old days, many types of tags were made out of paper. Nowadays, lots of tags are made out of film, b/c the film is waterproof, grease resistant, and stronger than paper. One application
where film tags are being used is in the grocery store for items like hams, turkey, and fish.
Our company, YUPO, makes film that is used in grocery stores to tag hams, but unfortunately, the grocery store is complaining b/c the tags are failing in the store. I am writing to ask for some help determining how much strength is required from my film to actually work for this application. I hope you can help.
The tag dimensions are 5" long x 2" wide, and 250 microns thick. A key point: of the tag design is that in the middle of small dimension there is a 5/8" wide extension (we call it a neck, b/c it extends out). It extends 2" away from the rest of the tag, so the total length of the tag is 7" long.
The 2" extension gets looped over a string and back onto the body of the tag. This is how the tag attaches to the ham.
Now for the set up to the question. Customers pick-up the ham using the tag. They raise it to about shoulder height. They lower it into the cart holding onto the tag. Then, suddenly, at the last second, they jerk the tag upwards so the ham doesn't crash into the other items in the cart.
When the customer jerks it up at the last second, the tag snaps. The hams weigh 10 pounds. I am wondering how much force is required to prevent the tag from snapping. I think this can be explained through physics, but I don't know how to do it. Our lab has equipment that can be used to test tensile strengths, elongation, etc, but I am asking your help figuring out how much strength will be required. Can you help me?

ANSWER:
Every customer is going to lower it differently, so it is, or course,
impossible to give a definitive answer. I have worked out the force
F which the neck would have to exert on an object of weight W
(lb) if the customer simply dropped the object from a height h
(ft) and then stopped it completly in a distance s (in): F =W [1+1.06 √(h /s ^{2} )].
For example, if a 10 lb ham dropped from 2 ft and stopped in 2 in, F =17.5
lb. That would be a pretty extreme case, though, since I would guess
most people would probably lower it at a lower speed than it would
achieve in free falling 2 ft. I think engineers like to insert a factor
of 2 safety factor. Overall, I would guess that the tag should to be
able to handle at least twice the weight of the product.

QUESTION:
I sent you a idea about putting microscopes end to end, or it's called a compound microscope. I thought you could use the scratch proof stuff they put on glasses, to fill in the cracks. But, I think there would still be a seam, so I thought you could make a mold and put the scratch proof stuff on the mold. It still would not be perfect, but it would get rid of the scratches and only make it blurry. What might be new that I thought of is, the blurriness would have a pattern, and a machine could look at the blurry image and draw a picture of what it would look like if it was focused. You could do it if you knew the exact shape of the lens, and look at a colored grid through the lens and that would help see how to focus the picture. Of course you could try a lot of lenses and find one area that is almost perfect. The lenses today are all scratches if you look up close, and I was taught that is the problem with looking at a virus up close. A microscope that enlarges 1000 times could be enlarged 1000 times, and that would be 1000,000 times. I would take like two microscopes.

ANSWER:
The reason that you cannot image a virus with an optical microscope has
nothing to do with the design of the optics (how you arrange lenses like
your idea of using a second microscope to look at the image of the
first). If the size of the object is comparable to the wavelength of the
light diffraction of the light causes the image to be very blurred,
losing all detail. So, if you enlarge a little fuzzy image all you will
get a big fuzzy image.

QUESTION:
Payne Stewart's Learjet was said to have hit the ground at 300-400 MPH. If terminal velocity is 120 MPH, how is this possible?

ANSWER:
Where did you get the idea that the terminal velocity of a Learjet in an
uncontrolled dive is 120 mph? I am sure that it is much larger than
that. 120 mph is often given as a typical terminal velocity for a
skydiver, but a jet has much more mass and less drag than a human body
which would imply a much larger terminal velocity.

QUESTION:
I've tried asking Maths folk this question but have yet to receive an answer so I'm thinking perhaps it's more of a physics question:
If I'm standing on Earth how do I calculate the size and distance of the celestial bodies (Moon Sun Planets stars etc) without knowing either the size or distance of any of them?
I've been driving myself crazy.

ANSWER:
Don't go crazy! There are many ways you can make these measurements.
Objects in the solar system are pretty easy. If you know the law of
gravitation, you can relate the period (time of rotation) to the size of
the orbit and thereby deduce where they are relative to earth. Once you
know the distances, you can deduce the size by measuring the angles they
subtend. Outside the solar system is a bit trickier. For stars or
galaxies not too distance, if you measure the position in the sky at one
time and then another time 6 months later (one full diameter of our
orbit apart) you can deduce the distance by
parallax . For more distant
objects, it gets a bit harder and we are getting into the realm of
astronomy in which I am not an expert. By measuring the
redshift of the
spectra of stars we can deduce the distance if we know the
Hubble constant
because the farther away an object is from us, the faster it is going.
Another way to determine distance is to have a certain type of
astronomical object which has the same total lumninosity regardless of
where it is. Such an object is called a
standard candle
and an example is a type Ia supernova; knowing how bright something is
and how bright it looks can be used to determine how distant it is.

QUESTION:
my question is based on newton's 3rd law . An elephant pushes a rabbit with 250 N force , then the rabbit should also apply 250 N force but the rabbit is only capable to apply 25 N force . WHY THIS HAPPEN . according to newton's 3rd law the rabbit should also apply this much of force .

ANSWER:
The rabbit does not need to use his muscles to exert a force. Suppose we
had a toy rabbit made of steel. You cannot ask it to exert a force! But,
if an elephant comes up and pushes down on it with a 250 N force, it
will surely exert an upward force on the elephant. If it were a real
rabbit, the force might squash the rabbit and kill it, but it would
still exert an upward force on the elephant. A classic example is a car
hitting a mosquito, mass of about 2 mg=2x10^{-6} kg. If the
collision lasts 10 μs=10^{-5} s and the car is going
25 m/s, the mosquito felt an average force of 5 N; the car will feel a
backward force of 5 N and you know a mosquito could not lift a ½
kg mass!

QUESTION:
Please explain in layman's language, what exactly is time dilation? I solve questions on STR and twin paradox but never understood how time can be delayed? According to twin paradox, after completion of journey one will be younger than other.... How can this be possible?? I mean at biological level :-\
Also in interstellar movie, 7 minutes in another universes is equal to 23 years in our universe and when that hero reached home, his girl become old while he was still young.

ANSWER:
There are two kinds of time dilation. In the movie Interstellar ,
the time dilation is called gravitational time dilation which is a
feature of general relativity, not special relativity; here time runs more slowly because of a strong
gravitational field. In the movie a planet orbits close to a spinning
black hole and time is slowed. It is a bit of a stretch, though. See an
interview with
Kip Thorne who was the science advisor of the movie. The other kind
of time dilation is easier to understand. First, you need to understand
why moving clocks run slow. The best way to understand this is to use
the
"light clock" as an example which I have done in many earlier
answers. In order to find the light clock useful, you must accept that
the speed of light is the same in all reference frames; see my
faq
page for answers which address this. Finally, the twin paradox is
easy to understand; again, see an
earlier answer . Actually, the twin paradox relies more on length
contraction than time dilation to understand because the traveling twin
sees the distance to the distant star as smaller than the earth-bound
twin does although both agree on the traveling twin's speed.

QUESTION:
Would it be possible to have a platform in geosynchronous orbit that was tethered to the surface of the Earth, upon which we could build an elevator into orbit?

ANSWER:
A space elevator is not that simple. But it is not impossible either.
See the
Wikepedia article .

QUESTION:
How much force would be needed to tip over a 31 foot tall,22 ton and 12 feet wide.

ANSWER:
You don't want the force, you want the torque. Also, it depends on where
the center of gravity is. I will give you a general solution and you can
calculate the numbers yourself. The green cross is the center of
gravity, h above the ground and L to the right of the
near side of the object. The
smallest force F you can apply is at the top, H above the ground (31 ft
for you, but if you want to solve for F at a different height, just use
a smaller H in the final equation I will give you). The weight
of the object is W , the normal force is N and the
frictional force is f . The object, just about to tip, has N
acting at the right hand lower edge of the object, so N and
f are acting there. Newton's first law tells you that N=W
and f=F . Finally, sum the torques about the right hand lower
edge: FH-W (d-L )=0 or F =W (d-L )/H .
For your case, if the weight is uniformly distributed, i.e .
h=H /2 and L=d /2, then F=Wd /(2H )=22x12/(2x31)=4.26
tons. You must push it until the center of gravity is over the right
hand lower edge and from there it will fall on its own. Also note that
if the floor is too slippery it will slide before it tips.

QUESTION:
If i am traveling along a path with the velocity of 300km/h,and fired a bullet in opposite direction with the speed of 300km/h,what will happen to the bullet?

ANSWER:
There is an old
Mythbusters
episode which perfectly answers your question.

QUESTION:
Imagine a cylinder in space, rotating at the appropriate velocity so that objects objects resting on the rounded surface of the interior experience a centripetal motion as if they were experiencing normal Earth gravity. Except, with Earth level gravity and atmosphere, (ignoring wind) a pressurized balloon with enough lift to rise in the atmosphere will travel "upwards," relative from the surface; while a balloon ascending from the surface inside a rotating "space station" probably wouldn't behave exactly the same, due to the angular momentum of the rotating surface. Would the observer piloting the balloon find himself rising at a vertical angle relative to the surface, rather than straight upwards? Would he eventually collide with the surface at another point? In short, what would be the path of the rising balloon in the centrifuge?

ANSWER:
I love questions about "artificial gravity" in rotating space stations;
there are sometimes surprising results unless you are standing still on
the interior surface and the radius is very large compared to your
height. Rather than going over a lot of preliminary stuff that I have
done before, I recommend that you carefully read through an
earlier answer where I
examine the problem of whether you could play catch in such an
environment. Another
earlier answer addresses what happens if you jump "straight up".
Perhaps the most important thing to take from those answers is that this
is not really like gravity at all because the instant you lose contact
with the surface you experience no forces at all so you move in a
straight line; nevertheless, with a large enough radius, the behavior of
the path as viewed by an observer on the surface can be quite similar to
how it would be on earth, particularly for the straight-up jump. Also we
will need to know that the angular velocity ω for the centripetal
acceleration of a cylinder of radius r _{0} to be g =9.8
m/s^{2} is ω =√(g /r _{0} ).

Now, why does a hot-air baloon
rise? It is because of the buoyant force and the buoyant force arises
because the pressure on the bottom of the balloon is larger than the
pressure on the top. So the first question we need to answer is whether
there is a pressure gradient in the space station (i.e. pressure
decreases with distance from the surface) and how that compares with the
pressure gradient in a gravitational field. In the space station there
is certainly a gradient because (as you note) it is simply a giant
centrifuge so air will tend to move to larger radii. For an
incompressible fluid there is a centrifuge equation for pressure a
distance r from the center, P (r )=P _{0} -½ρω ^{2} (r _{0} ^{2} -r ^{2} )=P _{0} -½(ρg /r _{0} )(r _{0} ^{2} -r ^{2} );
here r _{0} is the distance from the center to the outer
surface, P _{0} =P (r _{0} ), and
ρ is the fluid density. Technically, this is not correct
because the density varies with r , but we are really only
interested in situations where the density is fairly constant; for
example, if r _{0} =1000 m, r =900 m, and P _{0} =10^{5}
N/m^{2} (approximately atmospheric), P (r )=0.99x10^{5}
N/m^{2} . In a uniform gravitational field there is an
empirical equation to calculate pressure as a function of altitude
h , P (h )=P _{0} [1-(Lh /T _{0} )]^{gM/(RL)} .
Without going into detail, I find that P (h =100
m)=0.98x10^{5} N/m^{2} ; we can therefore conclude that
the buoyant force in the space station will be very similar to that on
earth, at least for the first few hundred meters.

I want to look at what the balloon
does from two perspectives—an observer outside the space station
and the guy inside who releases the balloon.

From outside, the instant that
the balloon is released it experiences no force except the buoyant
force B . At that instant it has a
tangential speed of v =r _{0} ω =√(g r _{0} ).
The buoyant force is constant in magnitude but changes in direction
since it always points toward the center of the space station. So as
the balloon is carried in the direction of the initial tangential
velocity and rises due to B , the direction
of both B and
v changes and the magnitude of v
changes.
If the radius is very large (obviously not the case in the figure
above), the direction of B will not change
much and the path followed will approximately be a parabola.

When looking from inside the
rotating cylinder there are two fictitious forces (see
earlier answer for
more detail), the centrifugal force F _{r} =mg
and the Coriolis force F _{c} =2mv √(g /r ),
in addition to B . In the figure below I
have shown the balloon shortly after liftoff. As you can see, the
Coriolis force will deflect the balloon to the right and
F _{c} will get larger and change
direction as v turns to the right and gets larger and r
decreases. So, if there were a ceiling (space station like a torus)
it would definitely not strike directly above the launch point.
Again, if
r _{0} were large enough, the effect would be
minimal because F _{c} would
never get large enough to cause a significant deflection because of
the 1/√r term.

Note that I have not considered another possible
force, air drag. I believe this would have a small effect and would not
alter the conclusions above. You were right to speculate that the result
would be different from on earth but angular momentum has nothing to do
with it; it is simply due to the Coriolis force.

FOLLOWUP
QUESTION:
What would happen if an observer got into a car and accelerated beyond the rotational velocity of the
cylinder in the same direction, once the vehicle lifted from the surface? Would it then glide for a certain period until, from the perspective of another observer inside the centrifuge, descending back onto the surface at a slightly slower speed? Also, the second part to
the question is, would the driver experience weightlessness if he was slowed enough while airborne? Would any airborne object, for that matter, potentially experience weightlessness if slowed or misdirected?

ANSWER:
I will first look at the situation from inside the space station, the
rotating frame. The diagram below shows the car moving with some speed u
in the same direction as the cylinder is rotating as you stipulated.

There is one real force acting on
the car, the normal force N from the "floor".
There are two fictitious forces acting, the centrifugal F _{r}
and the Coriolis F _{c} , both pointing
radially outward (the Coriolis force is F _{c} =2mu xω ).
So, you see, it has gotten "heavier" rather than "lighter" because of
the Coriolis force and will not
"lift off". So, what happens if you travel opposite the rotation
direction as in the figure below?

Now the Coriolis force points
radially inward, so there will be a speed where N =0. Now, it gets a little tricky to
understand what happens as u increases. The first thing you
have to understand is that the car, in the frame of the rotating
cylinder, is going around the circle; so the sum of all the forces must
add to the centripetal force: mu ^{2} /r _{0} =N-mg +2mu √(g /r _{0} ).
So, when N =0, u ^{2} +gr _{0} -2uv =0=u ^{2} -uv.
So, when u=v , N =0. This makes total sense because
if you look at the car from outside, it is at rest and, since this is
taking place in empty space, there are no forces whatever on the car.
In this situation, any small force with a radial component could launch
the car and, as seen from outside, it would simply move in a straight
line with constant speed (no forces on it once it loses contact with the
wall) until it encountered the wall; the car and all its passengers are
weightless. But there will never be a "lift off" situation;
if you view the situation from from the inertial (outside) frame, for
any u≠v the car will be traveling in a circle and therefore
require N ≠0.

I must confess that the fact that
all the forces on the car must add to the centripetal force mu ^{2} /r _{0}
eluded me at first. If you set N-F _{r} +F _{c} =0
you find u =½v for N =0 and this just
did not seem right to me. I therefore posted a question on
Physics Forums and got a great answer from Simon Bridge which
set me straight.

FOLLOWUP
QUESTION:
This is my final question about being airborne in a large atmospheric centrifuge in space: If you decided to locate yourself in the very center of the sphere, having no rotation, and you were then pushed toward the rotational equator at a walking pace, what would happen to you before you reached the surface, or rather how do you figure it out? Toward the surface, wind current would be rotating in the centrifugal direction, and the resistance would surely begin to alter your trajectory from "vertically, straight down" to "diagonally," from an observer on the equator, on a curve toward the surface. Assuming that the sphere is large enough that the accelerating effect of wind has enough time, would you ultimately be able to meet the surface on a landing curve that merges into the circumferential curve? And, if you accelerate too much, would you skip like a stone or a soccer ball as you collide with surface, slowing each time until you reach the rotational velocity? Ideally, you would be wearing a protective suit.
(Note from The Physicist: in earlier parts of this discussion I have
edited out "sphere" in the questions many times and replaced with
"cylinder" because a sphere is not a good model for pseudo gravity
because only at the equator of a sphere would the apparent gravity mimic
earth's. My answer below replaces "sphere" by "cylinder" and "center" by
"axis")

ANSWER:
If the initial push gave you a speed of 5 mph=2.2 m/s and the radius of
the cylinder were 1 km, here is what would happen. The speed of the rim
would be about v =√(gr _{o} )≈100
m/s=≈220 mph; this would be be the "wind speed" at the rim as
seen by an observer not rotating with the cylinder, including you
starting at the center; this speed would decrease linearly to zero on
the axis. It is easiest to view your gedanken from outside, but
I will discuss both observers. I will also discuss two scenarios, with
and without air.

If there were no air in the cylinder:
If viewed from outside, you would move
in a straight line away from the axis with a constant velocity
2.2 m/s until you hit the rim (in about 1000/2.2=455 s=7.6 min)
moving tangentially relative to you with speed 100 m/s—bad
news for you!
If viewed from inside, the centrifugal force would would be
radially out and the Coriolis force would push you azimuthally
in the direction opposite the direction the cylinder rotates.
You would still hit the rim with a relative tangential speed of
100 m/s and a radial speed of 2.2 m/s (which implies your
landing speed in the rotating cylinder frame is about 100 m/s)
after 7.6 minutes. Since the period of the cylinder's rotation
is about 63 s (T =2πr _{0} /v ),
your path would spiral around 7.2 times before landing.The paths
might look something like the picture below, the blue being the
outside view and the black the inside view.
But, there
is air, so you would have two real forces acting on you, the buoyant
force acting toward the axis and the air drag acting opposite your
velocity. Earlier in this discussion it was concluded that the
buoyant force would be comparable to that on earth, and you
certainly would not expect the buoyant force to have much effect on
you here; it would be a very small force pointing toward the axis
causing you to slightly slow down vertically. But the wind is a
different story.

If
viewed from outside, as you started falling air drag would be
very small because the wind speed would be small, your speed is
small, and because the air density would be low. It is too difficult to try to calculate
this with any precision, but clearly what is going to happen is
that the outside observer would see you spiraling out from the
axis in the same direction the cylinder was rotating. Given the
large speeds near the end (more than a category 4 hurricane), it is indeed likely that you would
land with a speed close to the speed of the rim; this would be a
much better scenario on landing than if there were no air. You
might think that the
force due to 220 mph winds would hurt you, but as you
acquire more and more tangential speed, the wind speed you see
becomes smaller and smaller. The total fall time would be much
longer than 455 s because of the large tangential velocity you
acquire; the average acceleration over the whole trip would be
much less than 100/455=0.22 m/s^{2} =g /45
because the time is much longer.

If viewed from the inside,
the centrifugal force would
give you an acceleration g radially out (toward the
rim). Early in the
descent the Coriolis force would accelerate you in the direction
opposite the direction of rotation; as the velocity acquired an
azimuthal component, the Coriolis force would have a radial component
radially inward, opposite the centrifugal force. The air drag
would always act opposite the velocity vector, so you would
never reverse the direction (you never rotate with the
cylinder). It is interesting that the inside observer sees you
going against the rotation but the outside observer sees you
going with the rotation; this would mean that your angular
velocity is always smaller in magnitude than the angular
velocity of the space station.

QUESTION:
If there are two objects, object a (a ball) and object b (a wall), how is it possible that the ball can get closer to the wall for all infinity and never touch the wall? What i mean...say the ball is ten meters away from the wall, then 5 meters, then 1 meter, then 99 cm, 50 cm, 1 cm, 50 mm, 1 mm, and the cycle can go on and on forever. Because numbers are infinite the ball can get closer to the wall forever without ever touching it. What is interesting about this question to me is that the ball and the wall are moving at different speeds in the same direction. The ball is always moving closer to the wall, so that means that the ball must be traveling faster than the wall. Now if the ball is getting closer to the wall infinitely (1 m, 1 cm, 1mm, and on and on forever, that means that the ball is traveling faster than the wall forever. So how can object A, the ball, travel faster than object b, the wall, for an infinite amount of time in the same direction but never touch or pass the wall?

ANSWER:
First of all, it makes no difference whether the wall is moving or not;
all that matters is the relative velocity, the rate at which the ball is
approaching the wall. Secondly, your question is just a variation of
Zeno's paradox. See an
earlier answer .

QUESTION:
myself and a friend were discussing this one day when we were bored if you were to fire a projectile and suddenly remove all the wind resistance (think of driving your impossibly fast car alongside said projectile at an identical speed with the window down and moving sideways into the bullets path so that it is now inside the car ) what effect would this have upon the projectile??
we came up with two hypotheses
frst being that the reduction of wind resistance that the windshield affords means the projectile suddenly has a lot less holding it back and would speed up and the second and more likely outcome is that the projectile would continue to decelerate but at a much slower rate owing to the fact that it is now travelling in a pocket of air that is traveling at the same speed as itself. and cannot possibly speed up due to it already being fired thus it has no way of gaining any energy from anywhere

ANSWER:
So, I assume your magic flying car is keeping pace with the bullet
before it comes through the window. Due to the air drag on the bullet,
it is slowing down and so you must also be slowing down. When the bullet
gets inside the car is no longer slowing down, but your car is.
Therefore the bullet will appear to you to accelerate toward the
windshield but you and I both know that what is really happening is that
the car is decelerating toward the bullet.

QUESTION:
Four forklifts are parked in a square like configuration, so that each forklift has its forks under the other. If all four were to be commanded to lift simultaneously, what would be the outcome?

VIDEO

ANSWER:
All four would mysteriously levitate above the ground. Just kidding!
Your forklift would feel four forces: its own weight, down, the force
the floor exerts up, the force which one neighboring forklift exerts up
trying to lift yours which is up, and the force which your other
neighboring forklift—whom you are trying to lift—exerts on
yours. Now, what is the direction of that last force? You are exerting
an upward force on him and therefore he is exerting an equal and
opposite force on you, down ; this is due to Newton's third law.
You can conclude that all the forces which the four exert on each other
cancel out if you take the ensemble of 4 as the object to focus on. As
long an nobody tips over, nothing will happen.

QUESTION:
What happends to a sound wave after it is created? How does it end up? Or does it last forever?

ANSWER:
Any sound wave loses energy as it travels because of internal friction,
viscosity in fluids. Also, the intensity of the sound decreases
approximately by 1/d ^{2} where d is the
distance from the source. Eventually, the intensity will drop below the
sensitivity of the ear. At some intensity the medium will no longer be
able to support the sound wave, but when that happened would be
complicated to understand and depend on the medium properties and the
frequency of the sound.

QUESTION:
I work on a maintenance team in a production facility and i am trying to make a conveyor that moves pallets of cardboard. I have an A/C motor with a sprocket and chain connected to a driven shaft turning the conveyor. My problem is that the pallet is too heavy and my only options are to buy a 3 phase motor or mess with the sprockets to change the gearing. Now to my question, will increasing sprocket size on the driven shaft, therefor decreasing speed, actually increase the torque and therefor increase the amount of weight it can move, or just lower the speed?

ANSWER:
The problem is friction, not weight. If there were no friction, you
could move any pallet with any motor; and, once you got it going (it
might take a while to bring it up to speed for a large mass and small
motor), you could turn off the motor and it would keep moving. The
friction gets larger as the load on the conveyor gets larger. Suppose
that the motor is capable of exerting some torque τ _{motor} ;
if the radius of the gear or pulley attached to the motor shaft is r,
this results in a tension in the chain or belt of F =τ _{motor} /r.
Now, if the gear or pulley on the drive shaft of the conveyor is R,
the resulting torque on that shaft will be τ _{conveyor} =FR =τ _{motor} (R /r ).
So, yes, increasing the size of the drive gear increases the torque.
This is the physics of an ideal situation. I am not an engineer!

QUESTION:
My text book says the direction of displacement and velocity are always the same. that's incorrect isn't it? Cos direction of velocity depends on the direction of movement, and the direction of displacement depends on the initial position right?

ANSWER:
No, that is not incorrect. In fact, the definition of velocity
incorporates this very equality. Start with average velocity. If an
object has a position R _{1} at time
t _{1} and a position R _{2}
at time t _{2} , then the displacement vector
D is defined as D ≡R _{2} -R _{1}
and average velocity vector is defined as v _{avg} ≡D /(t _{2} -t _{1} ).
As you can see, the average velocity and the displacement are in the
same direction. To get instantaneous velocity you find the limit of the
average velocity as the time difference approaches zero. If you know
calculus, the instantaneous velocity is v =dR /dt ;
again, velocity v and displacement dR
are in the same direction.

FOLLOWUP QUESTION:
If a boy moves from position A, towards the East direction 100 m, then turns around and moves 60 m in West direction. The displacement would be 40 m in East direction. The velocity would be (lets say he was moving at 30 ms-1 instantaneous velocity) in the west direction right? Or lets say we draw a displacement time and a velocity/time graph each for the movement. The displacement would have a positive value at the end of the movement (+40 m). But the velocity wud have a negative value, (if we considered velocity in East direction as positive). am i correct?

ANSWER:

You are correct but your example is
meaningless. It makes no sense to compare the displacement over two
different times with the instantaneous velocity at the second time—you
could get any answer. Read my first answer carefully. The relationship
between velocity and the corresponding displacement over any time
interval is clearly defined, even in the limit that the time interval
shrinks to zero.

QUESTION:
How do you convert sound energy into electrical energy ?

ANSWER:
There are many ways. I suggest you read the Wikepedia article on
microphones .

QUESTION:
I have come across this interesting fact about black holes just about every time i research on them: their gravitational force is so strong that even light cannot escape. However, what i do not understand is that in school, i learnt that gravitational force F= GMm/r^2. if light photons are packets of energy with NO mass how is it that they experience this gravitational pull? is it to do with mass-energy conversion?

ANSWER:
What you learned in school is only part of the story. The theory of
general relativity is the modern explanation for gravity, Newtonian
gravity is empirical, originally designed to explain the motion of
objects in our solar system. In general relativity, the presence of mass
actually deforms the space around it and this results in light passing a
heavy object being bent, apparently feeling a force. I would suggest you
read some of the earlier answers about
general relativity linked to
on the faq page. Incidentally, only photons inside the
Schwartzchild radius R= 2GM /c ^{2}
will be captured.

QUESTION:
I recently saw my son spinning an object on a wire overhead and wondered how fast is the object at the end moving in mph? Obviously there are many factors to consider, but let's say a very athletic man was spinning an ideal weight (let's say a marble or ball bearing of equal size) on an ideal length of wire (let's say 3 feet) in a circular motion similar to a sling and a stone of days past, what approximate range or max speed could the marble/ball bearing reach?

ANSWER:
I cannot imagine a frequency of more than about f =5
revolutions/s. If the radius of the circle is R =3 ft, the
velocity is v =2 πRf= 2x3.14x3x5=94 ft/s=64
mph.

QUESTION:
1 light year equals to how many earth year?
For example one of the orion's belt star is 900 light years away. Is this mean that 1 day in the star equal to 900 years on earth?
I assume that time would be slower on earth? Judging from Einstein's gravitational time dilation?

ANSWER:
You have this all wrong for the following reason: a light year is not a
measure of time, it is a measure of distance. A light year is the
distance that light travels in one year, so if you aimed a light source
toward this star, it would take 900 years to get there. Likely time
would run at about the same rate in the close vicinity of this star
because gravitational time dilation is extremely small unless the
gravitational field is very strong —like in the vicinity of
a black hole or neutron star.

QUESTION:
Rod tied to a string tilts vertically but when it is rotated it becomes horizontal Why ???
Can't seem to find any answer

ANSWER:
The easiest way to see this is to introduce the (fictitious) centrifugal
forces shown in the figure above. As you can see, both exert a torque
about the suspension point which will tend to make the rod horizontal.
When the rod becomes horizontal, the forces are still there but no
longer exert torques.

QUESTION:
Hi, I am a mathematician (graduated some years ago) and I am currently studying physics because I will take admission exams in Physics Departement of my local university (in about 9 months).
While studying (from the book "Physics for Scientists and Engineers" by Serway and Jewet) I found a very interesting problem:

I'm not asking you to solve this. I did it myself.
But the physical meaning eludes me.
[Here the questioner adds a bunch of mathematics discussion which is not
really relevant to the physics, but the crux of what eludes him is,
since this is a quadratic equation, what are the physical meanings of
the two solutions and does their average have any physical
significance.]

ANSWER:
Normally, I discard this kind of question because the purpose
of the site is not as a homework helper or tutoring service. I emailed a
glib answer without really analyzing the problem. But when I looked more
closely I realized that the problem did not really seem to make sense;
in particular, where did the 2.80 m come from and why is the potential
energy term involving x on the same side of the equation as the
kinetic energy term?

The equation clearly is intended
to represent modeling the problem of a m =46 kg object (child)
moving with a speed v= 2.4 m/s when landing on an ideal spring
with spring constant k =1.94x10^{4} N/m. This is a
standard introductory physics energy conservation problem. If you choose
y =0 at the top of the spring and the +y direction as
vertically up, the energy conservation equation is ½mv ^{2} =½ky ^{2} +mgy
where y is the position(s) where the mass is at rest. This
says that the kinetic energy of m when y= 0 (initial
energy) equals the potential energy of the spring plus the gravitational
potential energy of the mass when the object is at rest. Note that this
equation assumes that the mass is always attached to the spring and so
the two solutions are how far the mass travels in the negative direction
(y <0 solution) and how far it rebounds in the positive
direction (y >0 solution). These two solutions are the extremes
of the resulting oscillation of the mass and the oscillations are
symmetrically about the location where the mass would be at rest if not
moving, y _{rest} =-mg /k . Since this is
the center of the oscillation, ½(y _{1} +y _{2} )=y _{rest} .
The solutions for the problem at hand are y _{2} =+0.096
m, y _{1} =-0.142 m, y _{rest} =-0.023 m;
note that ½(y _{1} +y _{2} )=y _{rest} .

If you solve the equation in the problem as stated, you find that the
positive solution is greater in magnitude than the negative solution and
that the average is not the equilibrium condition. I believe that what
the author intended was that the stack of mattresses was 2.8 m high, the
floor was chosen as the zero of gravitational potential energy, and the
final gravitational potential energy was moved to the left side for
compactness. This would mean that the final potential energy assumed by
the author was mg (-x ) and therefore x must
mean the distance that the mass went down. But that is wrong—the
final potential energy should have been mg (2.80-x ).
This would then reduce identically to my equation except that x =0
at the position of the unstretched spring and the +x direction
is vertically down. The problem as stated is incorrect.

QUESTION:
How high off the ground would a vehicle weighing 6200 lbs have to be to reach a speed of 45 mph before it impacts the ground?

ANSWER:
If you neglect air drag, the height above the ground, h =v ^{2} /(2g ),
would be about 68 ft. If air drag is included, I estimate that the
terminal velocity (see earlier answer ) would be
about 262 mph. Since this is so much larger than 45 mph, I judge that
air drag would not be important and 68 ft would be your answer.

QUESTION:
How is the gravitational force directly proportional to the product of masses and inversely proportional to square of distance between the masses? i just want to know how they proven it ?

ANSWER:
It is simply an experimental fact. This hypothesis describes the
motion of the bodies in the solar system almost perfectly.

QUESTION:
What factors affect light intensity and how?

ANSWER:
I guess you want a definition of intensity. It is simply the
amount of energy passing through an area per second divided by the area.
Physicists prefer to measure this as watts per square meter, W/m^{2} .
But, to discover how convoluted the measurement of intensity can be, see
an earlier answser .

QUESTION:
Why don't oranges being carried in a semi-truck get crushed by all the oranges on top?

(The figure to the right shows parts of four layers.)

ANSWER:
It is because of the way that spheres tend to
pack
together. As you can see by carefully examining the figures above, each
orange is pushed down by three oranges from above, is pushed up by three
forces from oranges below, and is pushed horizontally by six oranges in
the same layer. These twelve forces all are directed toward the center
of the orange and all add up to zero net force, but there is a
net pressure over the surface of each orange approximately trying to
squeeze it into a smaller orange. But an orange is mostly water which is
nearly incompressible so the orange does not get crushed. Think of a
nicely packed snow ball: if you try to crush it into a smaller ball by
squeezing with cupped hands you will fail; to crush it you will have to
flatten it by pushing it with diametrically opposed hands. You might
think the bottom-most layer would get crushed because there is just one
force pushing up; but, each orange has six others in the same layer
pushing toward its center and these keep it from getting flattened.

Here is a little more about sphere packing. There are two possible
packings which achieve the maximum density of π /(2√3)≈74%:
hexagonal close-packed (HCP) face-centered cubic (FCC); these are
compared in the figure above (HCP on the left).

FOLLOWUP
QUESTION:
What would happen to the oranges once the force exerted upon them reached a critical strength which they couldn't bear? Does the entire group of oranges burst simultaneously in a flood of juice?

ANSWER:
Because the forces are not spread uniformly over the surface
area, there would be a tendency to be squeezed to a different shape but
still approximately preserving the volume of each orange. And I would
not expect it to happen all at once because the oranges on the bottom
are certainly experiencing greater forces. I would expect the
out-of-layer forces (from above and below) to tend to flatten the
lower-layer oranges but the neighbors in the same layer to cause the
oranges to have a hexagonal shape, so the oranges would tend toward a
hexagonal-prism shape. Each orange would occupy about the same volume,
but the amount of empty space (previously 26%) would decrease and each
layer would get thinner. This ignores the possibility of orange peels
rupturing, but I would think things would tend toward this shift before
much juice flowed!

QUESTION:
What is the effect of mass on torque? A wind turbine fan's blades are commonly very long to increase torque and to decrease speed.
How can I decrease speed using MASS? Or, can I increase torque, by increasing of mass, without increasing length of the blade? (without losing the energy.)
What is the formula applicable here?

ANSWER:
You are asking many questions here with no simple answers.

The simplest
place to start is your first question: does the mass of the rotor have an
effect on the torque on it? Typically, the turbine has three blades. I
will just analyze a single one and the same arguments could be made for
the other two. Call the length of the blade L and assume that
the force on it due to the wind is approximately uniform along the
length of the blade (the force on a tiny piece of the blade near the
center is the same as the force on an identical tiny piece near the
end). Then the total force F due to the wind will depend on the
length of the blade, but the force per unit length, Φ=F /L
will be more useful because it will depend only on how hard the wind is
blowing. It is now pretty easy to show that the torque due to the wind
is τ _{wind} =½ΦL ^{2} .
So, the answer to your question is no, mass does not affect the
torque; the torque depends only on how hard the wind is blowing and how
long the blade is.

Your second question is how can
you decrease speed by changing the mass M . If I model the blade
as a uniform thin stick of length L , its moment of inertia is
I=ML ^{2} /3. If it has an angular velocity ω _{1} ,
its angular momentum is L _{1} =Iω _{1} =Mω _{1} L ^{2} /3.
If you increase the mass to M+m , the moment of inertia will
increase to I' =(M+m )L ^{2} /3 and its
angular velocity will change to ω _{2} . But, the
angular momentum will not be changed, Iω _{1} =I'ω _{2} ;
you can then solve this for the new angular velocity, ω _{2} =(I /I' )ω _{1} =[M /(M+m )]ω _{1}
which is smaller. However, the rotational kinetic energy E of
the blade is now lower, E _{1} =½Iω _{1} ^{2}
and E _{2} =½I'ω _{2} ^{2} =½I {(M+m )/M}{[M /(M+m )]ω _{1} }^{2
} or E _{2} =[M /(M+m )]E _{1} .
On the other hand, if you wanted to add mass but keep the energy the
same, E _{2} /E _{1} =1=I'ω _{2} ^{2} /Iω _{1} ^{2}
or ω _{2} =ω _{1} √[M /(M+m )];
in this case, the angular momentum will have changed.

Your third question is moot since we have established that torque does
not depend on mass.

QUESTION:
If I stood beside a small operating hovercraft with a sail built into the front of it and blew air into the sail with a leaf blower I know that the craft would move forward. Now the question.If I sat down on the hovercraft with the leaf blower in hand and we became one with the hovercraft and I then blew air into the sail would we move forward or would action and reaction of the leaf blower neutralize the forward motion?

ANSWER:
It is easiest to understand if you think first of using a stick
instead of a leaf blower. Standing on the ground and pushing with the
stick on the sail, there is an unbalanced force acting on your
hovercraft (the stick). Now, if you stand on the hovercraft, the stick
exerts a backward force on you (part of the hovercraft, now) and the
stick exerts a forward force on the hovercraft and these cancel out. Or,
if you like, the only forces which have any effect on a system are
external forces and by becoming part of the system what you do is no
longer an external force. The leaf blower is a little trickier, but I
believe even worse! The leaf blower will exert a backward force on you
(like a little jet engine) and the stream of air will exert a forward
force on the sail; but some of the force from the stream of air will be
diminished by the air slowing down on its way to sail because of
interaction with the still air. So, the net effect would be for the
whole hovercraft to move backwards; probably not noticible because of
friction and the smallness of the loss of power due to the still air.

QUESTION:
Helium. Where do we get it from if it is lighter than air and doesn't react with any other elements in the normal human tolerant environment?

ANSWER:
Good question. Even though it is the second most abundant
element in the known universe, there is virtually none in the atmosphere
(because it is so light that its average speed is greater than escape
velocity and it shoots off into space) and is not tied up in rocks,
water, or other chemicals (because it is inert) like hydrogen is, for
example. This element was not even discovered until 1868 as a spectral
line in the sun (where untold zillions of tons are being produced every
second from nuclear fusion) and not found on earth until 1895 when trace
amounts were found coming from uranium ore; the source was as a nuclear
decay product in α-decay. The first large amounts were
discovered in 1903 as a byproduct mixed with the methane in natural gas
wells; today large scale amounts come only from helium trapped
underground.

QUESTION:
Why nucleon number is the sum of the number of protons and neutrons instead of the sum of the number of electrons and neutrons or between the number of protons and electrons , explain the logic ?

ANSWER:
It is very simple. Nucleon means either proton or neutron; a
proton is a nucleon and a neutron is a nucleon. Nucleon number
means number of nucleons. Electrons are not nucleons.

QUESTION:
If basketball (A) weighs 1 lb and is tossed upward to goal (A) at a height
of 8 ft (5 ft above my daughter's head) and basketball B is 6lbs, At what
height should goal (B) be to generate the same force to toss?

FOLLOWUP QUESTION :
Actually this is not homework, let me explain the situation. My daughter is 5 years old and playing Kindergarten basketball. Only one person on her team can toss the ball high enough to score, the ball weighs roughly 1 lb and the goal is roughly 8 ft high. I purchased a weight trainer ball that is exactly the same diameter as the regulation ball she uses but it's a 6 lb ball.
My theory is that I can build her a goal in the house that is not as tall but would require the same energy to make the basket. therefore making her stronger. And when it comes time to shoot the lighter ball in the taller goal, she shouldn't have any problems.

ANSWER:
For the 1 lb ball, the energy which must be supplied is 1x5=5 ft·lb,
assumning that she releases the ball at the level of the top of her
head. The 6 lb ball, if sent vertically with an energy input of 5 ft·lb,
will rise to a height of h =5/6=0.83 ft=10 inches above her head. All this
assumes that the ball is thrown straight up. Note that I have not really
answered your question because you asked for force and the energy input
depends both on force F and the distance s over which it is
applied. The energy input W could be written as W=Fs ,
So, if you assume that she throws it the same way and pushes as hard as
she can, the force need not be known.

QUESTION:
I am curious about a topic. In golf, if I hit a ball very hard and then I hit one very softly, is the one
hit very softly more likely to move or sway in its straight path?

ANSWER:
You refer to "its straight path". No golf ball goes in a
straight path, so I presume you mean that it does not curve left or
right; such a ball, if not curving, would have a projected path on the
ground (like the path of its shadow) which is straight. For a
right-handed golfer, a ball which curves right is called a slice
and one which curves to the left is called a hook ; these have
opposite spins. Neglecting the possibility of wind, the reason that a
ball curves is because it has spin. But now it gets complicated because:

the hard-hit ball is in the air much longer than the softly-hit ball;

t he
lateral force causing the curve depends on both the rate of spin and
the speed of the ball, so the hard-hit ball will experience more
lateral force than the softly-hit one if they have the same spin;

even if the slow ball
has a bigger lateral force, the fast ball is likely to be deflected
a greater distance because of its longer flight time;

a lateral wind will
exert the same force on both, but the fast ball will be deflected
farther because of the longer time.

So, you see, there is no
simple answer. To avoid curving, learn to hit the ball without imparting
significant spin!

QUESTION:
We had a phenomenon happen recently about 8:30 PM that is inexplicable to us, but there must be some explanation. My wife and I were in separate rooms when we both heard an extremely loud noise from the living room! The noise sounded like a large glass that just hit a hard tile floor, but loudness was magnified. As it turned out we came into the living room to find a glass platter that we had sitting on the coffee table for about a year just shattered. It broke completely by itself as there was no one in the room.
Do you know how this may have shattered/blew apart all by itself?We used the platter to put 3 little oil lamps on.
Inside our house the next morning at 6 AM we heard thunder outside so thought it might have had something to do with the barometric pressure.Very low barometric pressure and the type of glass it was made up combined just right to explode it like that? The temperature was a constant 68 degrees as it was for the months that was on the table.
If you have any idea about this, we would appreciate it.

ANSWER:
Glass, as you know, is manufactured at very high temperatures.
It has a quite large coefficient of theremal expansion (a large change
in size for a small change in temperature) and is a poor conductor of
heat. This means that as it cools it does not all cool at the same time.
This can result in very large stresses being "frozen in" at some
locations. What causes it to spontaneously break is usually difficult to
determine; most likely it had recently been bumped or your oil lamps
might have caused hot spots on the glass. Such things could have caused
a tiny fracture to begin and the final shattering could easily come at
some unpredictable later time. Unusual but not unexpected.

QUESTION:
On a skate board going down a .5 mile hill at 45 degrees slopes if I weigh
187 lbs how many mph would I be going by the bottom of the slope.

FOLLOWUP
QUESTION:
This isn't homework I'm a 35 year old heavy equipment operator and my son had a accident on skate board and we are curious how fast he was going when he wiped out.
I just wasn't sure how accurate I was when I said about 30-35mph. Please if you don't know just tell me so I can find someone who does—we got bets on it now amongst the family.

ANSWER:
Wow, 45 º is pretty darn steep! A half mile would
correspond to his having dropped by about 0.35 miles≈560 m. If
there were no friction at all his speed would have been v= √(2gh )=√(2x9.8x560)=105
m/s=235 mph! Back to the drawing board! There is some friction due to
the wheels and bearings and I estimate that this is probably not more
than about 15 lb; this would only slow him down to about 99 m/s=220 mph.
Back to the drawing board! Finally, since the speed is going to get
pretty big, we need to take air drag into consideration because the drag
force is proportional to the square of the speed. A rough estimate would
be that the force is about F _{drag} ≈¼Av ^{2}
where A is the area his body presents to the onrushing wind. When F _{drag}
is equal the net force down the incline (component of weight minus
friction, which I estimate to be about 117 lb=520 N), he will stop
accelerating. Taking his area to be about 2 m^{2} , you can then solve ¼Av ^{2} =520 to get
v =32 m/s=70 mph. This is all very rough but should give you an
order-of-magnitude estimate. (I still find it hard to believe that he
went down a half mile, 45º slope without braking at all!)

QUESTION:
I am curious about generating power in space. Why do they always use solar instead of the windmill type of generation?
A coil/magnet rotating. It seems to me, once the rotation is started, it would continue forever? Thus if you used a rocket to start the rotating part of the generator, and it kept spinning, could you use the magnetic field to protect say, an astronaut inside the generator? If it was big enough. Would you get perpetual energy if you used the electricity created in say, a microwave rocket engine or electromagnet. Or does the magnetic force alone cause the spin to lose momentum?

ANSWER:
So, you start something rotating in a vacuum and it never stops
because there is no air drag. You could even imagine making extremely
low-friction bearings so you could mount this on the side of your
spacecraft and it would at least spin for a very long time before
slowing down. But, the minute you hook it up to a generator you are
asking it for energy so it immediately begins to slow down, giving its
kinetic energy to you to power a light bulb, maybe. There is no free
lunch in this universe, and if you want energy you need something to
give it to you and the sun is the most convenient source in our
neighborhood.

QUESTION:
Is dark energy real? If all matter in the universe expanded from a single point the size of an atom (the big bang) wouldn't things be moving faster from each other because of geometry? If you start with a sphere the size of an atom and it expanded outwards over billions years even a Planck length size degree difference would be immense. It would cause things to move farther apart at faster rates as time went on. Are things moving faster away from each other because of geometry and angles, and not from hypothesized dark energy?

ANSWER:
I do not usually answer questions in
astronomy/astrophysics/cosmology but think I can answer this one. If a
collection of objects interact only via an attractive force
(gravity in this case), any one of them can only speed up when moving
toward their center of mass. The details of the motion would be
determined by the initial conditions. If all the objects were moving
away from some common point at some time, the only possible motions
would be

for all to move forever away from each other, but forever slowing
down;

for all to slow down and eventually turn around and speed up back
together; or

for some to come
together and some to keep going.

The simple reason is that
the potential energy of such a system increases as the objects get
farther apart so the kinetic energy must decrease to conserve energy.

QUESTION:
But I read somewhere that the faster you travel through space, the slower you travel through time, and if you reach light speed, time stops. If that's true, why does light have a speed, instead of just being instant? Does time not stop for light when it travels?

ANSWER:
First, you cannot reach light speed, no object with mass can; so
let's not talk about how fast your clock would be running if you went
light speed. What happens at high speed is that your clock will run
slow when measured by an observer you are passing. To you, time would
seem perfectly normal; however you would observe distances along your
line of travel to be shorter and therefore you would take a shorter time
to get there. Now, if you are traveling at almost the speed of light,
say 99.99% of it, I would still see you traveling at that speed
regardless of what your clock is doing; so, if the photon had a
(nonchanging) clock on it, I would still see it going at the speed of
light. Regarding whether time stops for a photon, my stock answer is
that a photon does not have a "point of view" and it is pointless to ask
how fast a photon's clock is moving because a photon does not carry a
clock with it.

QUESTION:
Does the existence of gravitational waves imply the existence of gravitons?

ANSWER:
No, gravitational waves have nothing to do with gravitons.
Gravitational waves are predicted by general relativity, the best
current theory of gravity. Gravitons would be the quanta of the
gravitational field is a successful theory of quantum gravity is ever
devised. You can look in the faq page for earlier answers about
gravitons and
general relativity .

QUESTION:
I've been dealing with a false prophet who says that a comet is coming and is going to skim the earth, as if to skip off of it, like a stone skipping on water. Is this even possible? She says it will skip off of the earth and keep going into space. Please let me know if this is even possible?

ANSWER:
Yes, I believe this is indeed possible. You might recall that
during the Apollo 13 failed moon mission there was concern that if the
spacecraft
reentry angle were too small that they would "skip off" the
atmosphere into space.

QUESTION:
If it's true that oceanic tides can be caused when the
moon's gravity pulls the molecules of ocean water up and away from earth by
a certain distance, and if it's also true that earth's atmospheric tides
can, likewise, also be caused by the moon's gravity pulling the molecules in
the atmosphere up and away from earth by a certain distance, then what stops
the atmospheric molecules, once they have accelerated even just a tiny
distance in the direction away from earth and towards the moon, from
continuing on their journey to the moon? (Let's, for example, say that the
atmospheric molecules in question are-to simplify matters-the ones at the
very top of earth's atmosphere, so that no other atmospheric molecules are
between these particular atmospheric molecules and the moon, as the moon
pulls on them.)

ANSWER:
This can get very complicated because the molecules in the air
have a whole range of speeds from very slow to very fast, but I do not
think that that complication is important to answer your question. The
reason that the molecules do not fall to the moon is the same as if you
throw a ball straight up and it does not fall to the moon —the
earth is pulling on it harder than the moon is. How anything moves is
determined by the net force on it and its weight (the force the earth
exerts on it) is bigger by far than its "moon weight".

QUESTION:
I need to figure out the force of impact from an object weighing 3.25 lbs falling from 2'
I have on reference that a five lb object falling two feet creates a force of impact of approximately 319 lbs.
This is not a school question. I got hit in the head with this object.

ANSWER:
I always try to emphasize that you cannot know how much force an
object exerts when it hits unless you know how quickly it stops (or,
equivalently, how far it goes while stopping). I find that the object
was going about 11 ft/s when it hit your head. Suppose that it stopped
in about 1 inch; in that case, the average force during the time of
stopping would have been about 75 lb. Had it stopped in ½
inch, the force would have been about 150 lb.

QUESTION:
If photons are the charge carriers for the electromagnetic force, then why don't magnets and power lines glow in the dark? Is it because those photons are not in the visible part of the spectrum or is it something else?

ANSWER:
Your terminology is a little off. Photons are the quanta of the
electromagnetic field; we then think of them as the "messengers of the
force" communicating the force among charged particles. However, they
are virtual photons which means that they pop into and out of
existence very quickly, too quickly for you to observe them —hence,
no glow! Also, if you think about it for a minute, if you saw a glow and
the fields did not change, that would violate energy conservation.

QUESTION:
If the planet earth was perfectly smooth and spherical will the water cease to flow?

ANSWER:
Not if everything else stayed the same. If the earth were
completely isolated, not rotating, and without atmosphere, water would
flow until it formed a uniform layer over the earth; eventually any
currents would damp out due to the viscosity of the water. The fact that
the earth is rotating and heated by the sun and has an atmosphere would
mean that the water would try to distribute itself mostly uniformly but
with an equatorial bulge; however heating and cooling of the atmosphere
would cause weather patterns and the resulting winds would move the
water around just like what happens today. Also, the moon causes tides
which are, by definition, motions of the water. You probably could think
of many more reasons the water would not become totally static.

QUESTION:
I am trying to figure this out for a dear Uncle on Vancouver Island as a challenge. I have suffered a concussion so trying is difficult. He used to live in S. Wales and dropped stones done old coal shaft. He was a teacher.
Wonder if you could help me please.
"A stationary rock is allowed to drop down an 800 foot shaft. Without compensating for air resistance, how far does it fall during
the sixth second of its descent? This is the formula. Assume gravity value to be 32 feet per second per second.
Please set out your answer clearly showing your thought process, line by line. Use words as well as numbers.
I'm afraid your answer so far is incorrect.
If needed, the formula we used was S = ut + half gt^{2} ."
Thank you very much. I am in Gr 5!! I want to by a physicist.

ANSWER:
I will assume that this is not a homework problem (forbidden on
this site!); at least if it is you went to a lot of trouble to disguise
it! Your equation is right except since we will start the clock (t =0)
when you let go of the stone, u =0 because u in your
equation is the speed at t =0. Also this equation assumes that
S =0 at t =0 and that S increases in a downward
direction. So, at the end of 5 seconds (the beginning of the 6^{th}
second) the position is S _{5} = ½x32x5^{2} =400
ft; at the end of 6 seconds the position is S _{6} = ½x32x6^{2} =576
ft. So the total distance traveled is 176 ft. I trust you will not
present this work to your uncle as your own.

QUESTION:
I have a really general question regarding the concept of work in terms of Physics. I'm aware that if work is negative, it means that the displacement and force act in opposite directions. However, does negative work also always imply that the the speed of the object is decreasing, or is this only true when looking at objects moving on a horizontal plane.

ANSWER:
The acceleration of any object is determined by all the forces
on it. If only one force acts on an object and the work done is
negative, it must be slowing down. If any other forces are present, all
bets are off. For example, a box sliding down an incline has friction
f doing negative work and gravity mg sinθ
doing positive work; if f <mg sinθ
the box will be speeding up.

QUESTION:
I'm a physics teacher in South Australia.
My question is related to the He-Ne laser and has bothered me for some time as to the actual mechanics.
As He is raised to 20.61eV and then transfers to Ne with 20.66eV for population inversion etc...
where has the extra 0.05eV appeared from? the quanta of energy is lower therefore Ne electrons should not be excited to that state.

ANSWER:
It comes from the kinetic energy of the collision between the He
and Ne.

QUESTION:
Why is water used to cool car engines?

ANSWER:
Because it is cheap, readily available, nontoxic, minimally corrosive, and can be kept from boiling with pressure. Perhaps most important,
though, is that it has a high specific heat which means it can absorb a lot of heat without a large temperature increase.

QUESTION:
What would be the estimated terminal velocity be of a 4,300lb car falling from 30,000ft above sea level be?

ANSWER:
The terminal velocity, v _{t} , does not depend
on the altitude from which you drop your car. This can be a very tricky
problem because v _{t} does depend on the density of the
air which changes greatly from sea level to 30,000 ft. So to get a first
estimate, I will just assume sea level density everywhere. There is an
estimate for the drag force in sea level air which is good for a rough
estimate, F _{D} = ¼Av ^{2}
where A is the cross sectional area and v is the
speed. From this you can show that v _{t} =2 √(mg /A ).
In SI units, m =4300 lb=1950 kg, g =9.8 m/s^{2} ,
and A ≈2x4=8 m^{2} (estimating the car as 2 m wide
and 4 m long). Then v _{t} ≈100
m/s=224 mph.

I guess we should now ask
whether we expect it to reach terminal velocity before it hits the
ground. Actually, it will technically never really reach terminal
velocity, only approach it—see an
earlier answer . I will
calculate how far it falls before it reaches 99% of v _{t} .
In the earlier answer , I
show that the height from which you must drop it for it to reach
terminal velocity with no air drag is h ^{no drag} =v _{t} ^{2} /(2g ),
and the height from which you drop it for it to reach 99% of terminal
velocity with air drag is h ^{drag} =1.96v_{t} ^{2} /g
(derived from the expression v /v _{t} =0.99=√[1-exp(-2gh /v _{t} ^{2} )].
So, for your case, h ^{no drag} ≈510 m and h ^{drag} ≈2000
m. At 2000 m (around 6000 ft) the air is about 85% the density of
sea-level air, so I believe that my approximation assuming constant
density is pretty good and the car would probably reach 99% of the
terminal velocity by the time it hit the ground. To actually put in the
change in density with altitude would make this a much more difficult
problem.

QUESTION:
W hat is after death (AD), birth of christ (bc)? Then how we are caluclate the age difference in between AD and BC.

ANSWER:
This is not physics, but easily answered. Actually, AD is
anno domini , (in the year of the Lord i n Latin) and BC is
before Christ . The dividing line is, supposedly, either the
birth or conception of Jesus. Times before are labeled BC and those
after are labeled AD. There is no year zero, so the first year after
this time is labeled 1 AD and the first year before is labeled 1 BC.
Hence, the time from, e.g ., January 1, 10 BC to January 1, 10
AD is 20 years; but, the time from January 1, 1 AD to January 1, 20 AD
is 19 years.

QUESTION:
We're trying to design a vessel for working in the vacuum of space. We have a vacuum chamber that can pull a 0.01 atm partial vacuum, so the question is : How does the force difference compare on the walls of a vessel, with 14 psi inside to outside chamber or space, i.e. force comparison between 0.01 atm in the chamber and the 10^{-14} in space?
My guess is that we are capturing 99% of the effective force differential using the chamber, so not much more to expect from the vacuum of space.

ANSWER:
Ok, the 14 psi inside your chamber is about 0.953 atm and the
pressure outside is 0.01 atm making the net pressure difference 0.943
atm. The pressure outside in space is, for all intents and purposes,
zero, so the net pressure difference would be 0.953 atm. So, the percent
difference is 100x(0.953-0.943)/0.953=1.05%, about what you guessed. In
other words, the force on any part of the walls of your chamber is
about 1% smaller than it would be in space.

QUESTION:
Not sure if this is physics or not but how can you record silent sound ? I found the patent for it on google it's
5159703
and I wanna know how you can record it because a regular microphone doesn't pick it up.

ANSWER:
The idea here is essentially the same as AM radio where the
high-frequency radio wave is modulated by the audible message. For this
invention the radio wave is replaced with sound of a frequency larger
than is audible but modulated by an audible signal. You could certainly
make a detector (call it a microphone if you like) to detect these
high-frequency sound waves; ultrasound imaging in medicine does just
that. Then you would need some electronics to extract the audible signal
from the carrier, just like you need a radio receiver to extract the
audible signal from the AM radio carrier.

QUESTION:
Does a round and square object, the same weight, fall the same?

ANSWER:
If air drag is negligible, like if you drop them from a few
feet, yes. If they fall fast enough for air drag to be important, they
will fall differently and, if they are about the same size, say a sphere
and a cube, the sphere will fall faster. That may be all you want, but I
will go on and explain in a bit more detail. The drag force F _{D}
on an object may be approximated for every day objects, masses, and
speeds as F _{D} = ½C _{D} ρv ^{2} A
where ρ is the density of the air, v is the speed,
A
is the cross-sectional area presented to the onrushing air, and C _{D}
is called the drag coefficient. C _{D } depends only on
the shape of the object. C _{D} ≈0.5 for a sphere
and C _{D} ≈0.8 for a cube (falling with one face
to the wind). So, if their areas are about the same, the drag on the
sphere will be smaller and it will go faster. Of course if the sphere
area were ten times bigger than the cube area, that would be more
important than the somewhat smaller drag coefficient and the cube would
win.

QUESTION:
What would happen if light traveled at normal speed?

ANSWER:
I recommend the
Mr Thompkins books by George Gamov.

QUESTION:
Is it correct to say that nuclear fusion violates the law of conservation of mass since a portion of the mass is converted into photons?

ANSWER:
Well, you could say that if there were such a law as
conservation of mass. Since 1905 when Albert Einstein showed us that
mass is just another form of energy, the only valid such law is
conservation of energy. Even in chemistry where conservation of mass
appears to be correct, the ultimate source of energy is mass being
changed into kinetic energy of the chemical reaction products (heat);
chemistry is such an inefficient source of energy that the mass changes
are miniscule.

QUESTION:
Can you tell me why muons are extremely unstable (lasting only fractions of a second) while electrons and neutrinos are pretty much stable? I just don't get why muons are so unstable since they're just leptons like electrons and neutrinos just more massive than the other two.

ANSWER:
If you look at the decay of the muon, you will see that the mass
of the products, an electron and two neutrinos, is much less than the
mass of the muon. This means that the decay is energetically possible,
energy is released by the decay so the decay products have kinetic
energy afterwards. In nature, almost always when a process is
energetically possible it will occur. Only in cases where a decay would
be prohibited by some selection rule will decay not occur. For example,
a proton cannot decay into three electrons because charge conservation
would be violated, even though it is energetically possible.

QUESTION:
I think that it is possible to travel faster than speed of light. I explain it in this
link .
Is it correct?

ANSWER:
No, it is not correct. In your example, an earth-bound observer
observes the traveler to travel 4.24 ly in 4.28 years.
The traveler, as you correctly surmise, observes the trip to
take 0.604 years. However, you have not taken length contraction into
account. The traveler observes the distance she has to travel to
be 4.24/γ =0.598 ly, so she perceives
her speed to be 0.99c .

QUESTION:
So if black holes suck in everything in including light that must mean everything is getting pulled in as fast or greater then the speed of light. So if light is weightless and it is sucked in. What happens to any mass as it is sucked in. Would the mass of the object then cease to have mass? Because im pretty sure anything traveling at the speed of light has to be mass-less correct? And how does gravity effect something with no mass? I dunno if it is a good question or not but i couldnt find a whole lot on the subject.

ANSWER:
Nothing ever goes faster than the speed of light and only light
can go at the speed of light. When an object merges with the black hole,
its energy, E=mc ^{2} , is not lost and the black hole
becomes more massive by the amount of the mass energy. When light is
swallowed by the black hole, the mass increases even though the light
does not have any mass because light does have energy and the energy
shows up as increased mass of the black hole, m=E /c ^{2} .

QUESTION:
What happens gravitationally when the center of mass can no longer be considered a point but is instead an area? Specifically, suppose the Sun was to "explode" or supernova; ignoring the obvious destruction of the solar system, what would happen to the planetary orbit of Earth? I presume it would be roughly akin to letting go of the string at the end of which I have a ball spinning around me.

ANSWER:
The center of mass is always a point. If the sun were to
"explode", the center of center of mass would continue to be at the
center of where it was before the explosion. A star explodes
approximately isotropically, that is, material goes out at the same rate
in all directions. So, until the material reached the earth's orbit, the
orbit would be unchanged. But, as material gets outside the earth's
orbit, only the material inside would contribute to the force felt by
the earth (this is Gauss's law). So the earth would behave as if there
were a star of constantly decreasing mass at the original center of
mass.

QUESTION:
okay, so i wanted to ask what would be the KE of a tungsten rod of length 70 m with a conical end of height 10 m, 6 m in diameter at one end and 6 cm at the other , weighing 38307.5 tons and falling at terminal velocity which i think is about 2340530 m/s?

ANSWER:
The speed, while large, is still less than 1% the speed of
light, so you can use the classical expression for kinetic energy, KE= ½mv ^{2} =½x3.83x10^{7} x2.34x10^{6} =4.48x10^{13}
J. Note that the composition, size, and shape are irrelevant. Since
everything else was in SI units, I assumed that ton is metric tonne, 1
tonne=1000 kg. This must be some kind of projectile in a computer game.

QUESTION:
Can an electric current flowing in a wire be stopped by a magnetic field? If so, how?
I need to stop it from distant.

ANSWER:
The magnetic force on a moving charge is always perpendicular to
its velocity. To stop a moving object you must apply a force
antiparallel to its velocity.

QUESTION:
I bought a 400lb gun cabinet and need to pull it on a 2 wheel hand cart up
a 12ft ramp at about 35degrees to the horizontal How much load does 1 or 2
people have to carry and how much is borne by the wheel. I am trying to make
sure we can be comfortably safe!

ANSWER:
I could make a rough estimate but would need to know the dimensions of the
cabinet, if the center of gravity is near the geometrical center. I would
assume that the cabinet was parallel to the ramp when being pulled.

FOLLOWUP QUESTION:
It is 20X29X55. It would not be parallel to the ramp but about 20 degrees from the ramp (which is about 35 degrees to the ground (thus avoiding 4 steps).

ANSWER:
Since only an approximation can be reasonably done here, I will
essentially model the case as a uniform thin stick of length L
with weight W , normal force N
of the incline on the wheel, and a force F
which you exert on the upper end. In the diagram above, I have resolved
F into its components parallel (x )
and perpendicular (y ) to the ramp. Next write the three
equations of equilibrium, x and y forces and the
torques; this will give you the force you need to apply
to move it up the ramp with constant speed.

ΣF _{x} =0=F _{x} -W sinθ ΣF _{y} =0=F _{y} +N-W cosθ
Στ= 0=½WL cos(θ+φ )-NL cos(φ ).

I summed torques about the end where you are
pulling. Putting in W =400 lb, θ= 35º, and
φ= 20º, I find F _{x} =229 lb, F _{y} =206
lb, and N =122 lb. Note that you do not need to know the length
L . The net force you have to exert is F =√[(F _{x} )^{2} +(F _{x} )^{2} ]=308
lb. If someone were at the wheel pushing up the ramp with a force
B , that would reduce both F _{x}
and F _{y} . This would change the equations to

ΣF _{x} =0=F _{x} -W sinθ +B ΣF _{y} =0=F _{y} +N-W cosθ
Στ= 0=½WL cos(θ+φ )-NL cos(φ )+BL sin(φ ).

For example, if B =100 lb, the solutions
would be F _{x} =129 lb, F _{y} =169 lb,
and N =159 lb; so your force would be F =213 lb.

QUESTION:
Imagine if you wrapped a rope tightly around the earth. How much longer would you have to make the rope if you wanted
it to be exactly one foot above the surface all the way around?

ANSWER:
I hope you don't think that the rope would spontaneously rise up if
it were longer than the circumference of the earth; you would have a
slightly slack rope laying on the ground. You are specifying the difference
in radii between one circle with a circumference C and another of
circumference C+ δ ; that
is not really physics. But, it is easy enough to do. If C is the
circumference of the earth, then C= 2πR _{ } where R
is the radius of the earth and
C+ δ =2πR' where
R' is the radius of the circle your rope would make if δ =1
ft. Then δ= 2π (R'-R )=2π =6.28 ft.
Note that δ depends only on how high the rope is above ground,
not how big the earth is: if the earth were 1 ft in radius and you increased
the length of the rope by 6.28 ft, the rope would be 1 ft above the surface!

QUESTION:
I'm writing a research paper for my college english class and the topic is Thorium Reactors. My question is "Are thorium based reactors such as LFTR
fusion or fission and based?" I was just wondering because the it seems from what I've learnt that the reactors use the thorium to produce a reaction that makes another element such as uranium 233 which I assume is fusion because I'm sure they're using the energy put off from that initial reaction to power something. But after the uranium 233 is used and to produce energy as efficiently as possible I would think that you would implement a system that would immediately and directly use said produced uranium into some form a fission reactor.

ANSWER:
Fusion always involves light nuclei and there is no such thing
as a fusion reactor, only ideas for them. So a thorium reactor must be a
fission reactor. It would be inaccurate, though, to call thorium the
fuel because thorium is not
fissile . If a thermal neutron is absorbed by a fissile
nucleus, it will fission and result in more neutrons leading to more
fissions and the reaction can be self-sustaining. Thorium, which is 100%
^{232} Th, absorbs a neutron to become ^{233} Th which
β -decays to ^{233} Pa (half life 22 minutes)which
β -decays to ^{233} U (half life 27 days). The ^{
233} U is fissile and is nearly stable (α -decay half
life 160,000 years). Thorium is said to be fertile , absorption
of a neutron results in production of a fissile fuel. So a thorium
reactor is a breeder reactor, a reactor whose purpose is create fuel. At
the startup one needs a starter fissile fuel as well as the
thorium to provide neutrons to create the ^{233} U. As the ^{
233} U builds up, it becomes the fuel.

QUESTION:
Would it be physically possible to create a parachute capable of delivering a main battle tank safely to a theatre of war?
Like how huge would it have to be?

ANSWER:
A reasonable estimate of the force F of air drag on an
object of mass m , speed v , and cross sectional area A is F = ¼Av ^{2} ;
this works only in SI units. The speed v when F=mg is called
the terminal velocity. I estimated a reasonable terminal velocity would
be the speed the tank would have if you dropped it from about 10 m,
v ≈14 m/s. The mass of the
MBT-70 (KPz-70)
is about 4.5x10^{4} kg. Putting it all together, I find that
A ≈10^{4} m^{2} , a square about 100 m on a side or a
circle of radius about 30 m. You could do a more accurate calculation
but this gives you a reasonable estimate.

QUESTION:
If you stand holding a box but the box is not moving, are you doing any work?

ANSWER:
See an earlier answer .

QUESTION:
Why is it necessary to have a minimal temperature of 150 million degrees Kelvin for nuclear fusion on earth if the sun does nuclear fusion at a temperature of 15 milloin degrees?

ANSWER:
There is more than one reason that I can think of. The mass of
the sun is 2x10^{30 } kg, quite a bit bigger than the mass of
fuel in a fusion reactor. Therefore the rate of fusion in the sun can be
low but the energy output would still be huge. Increasing the
temperature would increase the rate. The density of the sun's core is
about 150 g/cm^{3} , 150 times more dense than water. In a fusion
reactor, the practical densities are many orders of magnitude below
this. Again, the rate of fusion would be dependent on the density of the
fuel.

QUESTION:
My name is Justin and my friend Richard blew my mind when he told me that nothing doesn't have mass. Again. Nothing doesn't have mass. Is this true? I was taught that everything has an inherent mass. Is he wrong? Am I wrong? Does sheer nothingness have a mass or not.
Please help me. My friend just blew my mind and I'd like it to be back together.

ANSWER:
It is true that the vacuum does not consist of nothing. See a
recent answer for more detail.
However, the vacuum does not contain mass in the sense you normally
think of it —you cannot "weigh" a vacuum. Also, photons, the
quanta of light, do not have mass. Until recently it was thought that
neutrinos have no mass; they actually do have very tiny masses. All the
double negatives in your question and my answer make this a little
confusing; I hope I have answered what you meant to ask!

QUESTION:
in classical physics angular momentum is a variable factor, but why in quantum mechanics angular momenta are quantized. and how its related to spin.
I have read that Paul Dirac showed, how changing some relativistic factors in Shroedinger's equation can spontaneously lead to the "spin"
concept. but what the spin actually tells us and how we can visualize it.

ANSWER:
The reason why orbital angular momentum is quantized is that,
when you solve the Schrödinger equation for the atomic wave
function, the wave function is not normalizable unless the angular
momentum is
ħ √[L (L +1)] where L is an
integer. (For our purposes here, "not normalizable" means that the wave
function becomes infinite somewhere.) It also turns out that L
is called the orbital angular momentum quantum number and its being an
integer has nothing to do with spin except that spin is also an angular
momentum. To visualize spin, read two earlier answer (#1
and #2 ). In relativistic quantum mechanics, the
Dirac equation replaces the Schrödinger
equation. When you write the Dirac equation for an electron, it turns
out that spin is indeed predicted to have an angular momentum quantum
number of ½ as is observed experimentally.

QUESTION:
I've just found out that a proton isn't made up of two up and one down quark but also contains zillions of other up and down quarks along with their anti matter equivalents. Here's a
link
to an article from the L.H.C people at Cern that shows this.
I'm amazed and confused because I thought that matter and anti matter particles would annihilate each other. Why don't they? And, if they do, how is the 'zillions of other quarks' balance maintained?

ANSWER:
The crux of what is going on here is that a vacuum is not really
a vacuum as we generally think of it —nothing.
Particle-antiparticle pairs are continuously popping into existence and
then annihilating back to nothing after a short time. This is called
virtual pair production. Also, if a particular particle experiences a
particular force, the messenger of that force (gluons for the strong
interaction, photons for the electromagnetic interaction) are
continuously being emitted and reabsorbed. All this is called vacuum
polarization. So, I could give you a similar description of the hydrogen
atom: the hydrogen atom is not really a proton and an electron, it is a
proton and zillions of electrons and positrons and photons. If you
really want to understand the hydrogen atom in detail, you need to take the
effects of vacuum polarization into account (see the
Lamb shift , for
example). The CERN explanation should have included zillions of
positrons and electrons and photons inside the nucleus also. Just as a
hydrogen atom is pretty well described as a proton and an electron as a first
approximation, a proton is pretty well described as three quarks as a
first approximation.

QUESTION:
Why the whole matter of radioactive sample does not disintigrate at once or Why it always take half life to disintegrate half of initial value?

ANSWER:
Because decay is a statistical process. Whenever you have a
large number N of anything they will, at some time t ,
have a time rate of change R =dN /dt . If R <0,
N is getting smaller (as in radioactivity) and if R >0,
N is getting larger (like bacteria growth). For a great many
cases in nature, it turns out that the rate is proportional to the
number, R ∝N . Radioactivity turns out to
behave that way: if you measure R for 2x10^{20}
radioactive nuclei it will be twice as big than when you measure R
for 10^{20} nuclei. You can then solve for what R is for a given
situation: dN /dt=-λN where
λ is the proportionality constant, called the decay
constant; the minus sign is put there so that λ will be
a positive number since the rate is negative. If you know differential
equations, you will find that N=N _{0} e^{-λt}
where N _{0} is the number when t =0. The decay
constant is related to the half life τ _{½} =ln(2)/λ.
However, this all depends on there being a large number to start.
In the extreme case, if N _{0} =1, they would all decay
at once! You just could not predict when. If N _{0} =2,
they might both go at once or else one might go before the other but not
necessarily at τ _{½} .

As you move
up to create heavier and heavier nuclei via fusion, the fusions continue
to release energy but in decreasing quantities until the final product
is iron. Thereafter, if you want to fuse nuclei, you must add energy.
So, elements heavier than iron are not produced in stars. Heavier
elements are produced in supernova explosions.

You ask if "…creation of matter from energy…" has ever
been observed. It happens all the time. But that is really the wrong
question because matter and energy are not different things, matter is
simply a form of energy. So, for example, if you fused two ^{32} S
nuclei to a single ^{64} Ge nucleus, the Ge would have more mass
than the two S. Any time you have an endothermic chemical reaction you
end up with more mass than you started with (although almost immesurably
small because chemistry is so inefficient).

*As can be seen in the figure above, the proton-proton cycle requires an
input of 6 protons and ends with one ^{4} He and 2 protons. So the net input
is 4 protons. However, there are numerous outputs—2 positrons, 2
neutrinos, and 2 photons, all of which carry energy away. That is why
comparing the net input with the final product is not meaningful.

Of course there are lots of other ways you could
lift it which would be more efficient if your aim was to tip it over;
for example, you could start pushing horizontally once you got it off
the ground so that the floor would hold up all the weight rather than
half the weight. There is an
old answer very
similar to yours that you might be interested in.

At this point what a physicist normally does is to try to understand the
situation in terms of a simple spring model. If the rope is like a
simple spring, i.e. its tension is proportional to its stretch,
you can usually make approximations which would result in the simple
model that F≈ks where k is a constant. For the
equilibrium situation, then, 20≈ 0.41k
or k≈ 49 kg/m. This would then predict that the force
when s =1 m would be F =49 kg which is too small by more
than a factor of 2.

The previous try indicates that the rope is probably not approximated as
an ideal spring. My last attempt is to try to treat the rope more
explicitly as an ideal spring, not using small approximations used
above. So, looking at one half the rope, I will write T =50+kδ
where
δ= √(L ^{2} +s ^{2} )-L
is the amount by which this half of the rope is stretched relative
to L . Note that k is not the same as in the
approximation above where s was the stretch parameter rather
than δ. Calculating δ for the at rest
situation where T =100 kg, k =2270 kg/m. Then,
calculating T for the maximum s using this value of
k , T =50+2270x0.129=344 kg which is too large by nearly 60%
compared to the measured value of 220.

I conclude that the rope is poorly approximated as an ideal spring and
that to do any more detailed analysis of this problem would require
measuring s as a function of the load by varying the load for
the equilibrium situations.

It is, of course, simple to calculate the speed the weight has just
before the leash goes taught: v =√(2gh ) where is
the distance fallen.

I have deleted
your question about angular momentum—it is off topic.

The
thing to appreciate is that even though the rocket goes straight up, it
will have the same angular velocity ω as the earth so its
speed will be ω (L+R ) where L is the
length of the rope and R is the radius of the earth. The angular
velocity is ω =[(2π radians)/(24 hours)]x[(1 hour)/(3600
seconds)]=7.27x10^{-5} s^{-1} . If L is just
right, the rocket will assume an orbit like the
geosynchronous communication satellites ; this turns out to be if L =5.6R .

Finally, there are two places on earth where the fictitious forces are
zero and therefore the two objects will move identically—at the
north and south poles.

One more special case. Suppose
v' =½v _{1} . You should be able to convince yourself that
m originally is catching up with the moving frame but then stops
and turns around and ends up where it started from; thereafter it moves
with constant speed v' . The total distance it has traveled in the
moving frame is zero and so W =0 and therefore ΔK'= 0.
So the heat is not anywhere evident here even though you would feel it. In
this case you would find W =- ½mv _{1} ^{2} +½mv _{1} ^{2} .
Now you can see where that heat is hiding, right?

So, if you want
to most conveniently determine the amount of heat generated, you apply the
work-energy theorem in the frame where the surface is at rest. This is
because to the moving observer the surface is moving as well as m . The
moving observer has to correct for that motion by calculating m 's motion relative to the surface.
m 's motion relative to the second observer is not
really relevant in calculating the heat generated.

Thanks to R. M.
Wood and A. K. Edwards for helpful comments.

The top clock
will see 57 ns more time elapse than the bottom clock. I believe that if
you use the 250 m tall building, things would simply scale and time
differences would be 250/3x10^{8} =8.3x10^{-7} times
smaller.

So the wire
will have a voltage across its ends and will therefore look like a
battery. Similarly, the body will have a voltage of the same polarity (say
positive at the mouth and positive at the end of the wire in your mouth)
but with a much smaller voltage than the wire; this would create a circuit
looking like two batteries, one weak and one strong, which would drive a
current through the weaker battery (your body). So, the idea works if you
move fast enough. But, the earth's magnetic field is really weak and the
speeds would be impossible to achieve without burning everything up in the
atmosphere I would bet. Let's just do a rough calculation. The voltage
V is about V=BLv where B is the field, L
is the length, and v is the velocity. I will take L =1 m,
v =18,000 mph≈8000 m/s (low earth orbital speed), and B ≈5x10^{-5}
T. So the voltage would be less than half a volt! Have I been colloquial
enough for you?

These laws are laws because they
are always found to be true and it is rather pointless to ask "what if".
If energy conservation were not true, then energy could suddenly appear
out of nothing.

Only if there
is some overriding physical law could another be broken, but only within
certain constraints. For example, the Heisenberg uncertainty principle
says that it is impossible to precisely know the energy of a system for a
short enough time, ΔE Δt ~10^{-34} J⋅s
where ΔE is the amount by which you change the energy of a
system and Δt is the time during which the energy is
changed by that amount. For example, 10^{-20} J could appear out
of nothing for as long as 10^{-14} s; after the time had elapsed,
though, it would have to disappear again.

Estimating the force this guy experienced when he hit the ground is a bit
trickier, because what really matters is how quickly he stopped. Keep in
mind that this is only a rough estimate because I do not know the exact
nature of how the ground behaved when he hit it. The main principle is
Newton's second law which may be stated as F=m Δv /Δt
where m is the mass, Δt is the time to stop, Δv= 23
m/s is the change in speed over that time, and F is the average force
experienced over Δt. You can see that the shorter the time,
the greater the force; he will be hurt a lot more falling on concrete than on a
pile of
mattresses. I was told that his weight was 156 lb which is m =71 kg
and he fell onto about 2" of pine straw; that was
probably over relatively soft earth which would have compressed a couple of
more inches. So let's say he stopped over a distance of about 4"≈0.1 m. We
can estimate the stopping time from the stopping distance by assuming that
the decceleration is constant; without going into details, this results in
the approximate time Δt ≈0.01 s. Putting all that into
the equation above for F , F≈71x23/0.01=163,000 N≈37,000
lb. This is a very large force, but keep in mind that if he hits flat it
is spread out over his whole body, so we should really think about
pressure; estimating his total area to be about 2 m^{2} , I find
that this results in a pressure of about 82,000 N/m^{2} =12 lb/in^{2} .
That is still a pretty big force but you could certainly endure a force of
12 lb exerted over one square inch of your body pretty easily.

Another possibility is that the victim employed some variation of the
technique parachuters use when hitting the ground, going feet first and
using bending of the knees to lengthen the time of collision. Supposing
that he has about 0.8 m of leg and body bending to apply, his stopping
distance is about eight times as large which would result in in an eight
times smaller average force, about 5,000 lb.

Falling from a third story window (about 32 feet, say) would result in a
speed of about 14 m/s (31 mph) so the force would be reduced by a factor
of a little less than a half.