Ask the Physicist!

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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_putt²-2gh). However, there is friction which will slow the ball further and the frictional force on a slope (f=μmgcosθ) 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 L₂, the distance it travels is s=L₂/cosθ and so the work done is W_f=fs=(μmgcosθ)(L₂/cosθ)=μmgL₂, 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_putt²-2g(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=½at² , so for this case a=2y/t²=2x3/(1.5)=2.7 m/s². 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/s² 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=½v²/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 N₁+N₂+N₃=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 N₁=N₂and N₃ 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=W₁=W₂; 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:
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: ½MV²-μMgS=0, or V=√(2μgS). Next I applied momentum conservation to the collision.
M_SUVv-M_sedanu=M_SUVV_SUV+M_sedanV_sedan
and solved for v
v=[M_SUVV_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=½mv²+mgh; g=9.8 m/s² 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 E₁=mgh and the final energy would be E₂=½mv²; 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=μmgcosθ. 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 E₂=E₁, we write E₂=E₁+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. ½mv²=mgh-μmgs₁cosθ₁-μmgs₂cosθ₂=9.8x[(1.721+2.536)-0.14(3cos35⁰+6cos25⁰)]=31.6 m²/s². 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(s₁cosθ₁+s₂cosθ₂)]}.

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? This 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/s²)=14.7 N. The angle of the incline is stipulated to be 45⁰. 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=mgR(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=48 mm; 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 20⁰ 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.

QUESTION 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-mgcosφ=0 and ΣF_x=mgsinφ-μN=ma; the solutions are N=mgcosφ and a=g(sinφ-μcosφ). 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=45⁰. 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=mgcosθ-μN=ma and ΣF_y=N-mgsinθ=mv²/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)≈√(2gRsinθ), N≈3mgsinθ, f≈3μmgsinθ. 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/s², v_{flat_ramp}=26.3 ft/s, v_¼pipe≈23.2 ft/s. (v=27.2 ft/s for no 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 [½mv²/(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=(4ωv³cosθ)/(3g²) 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+Ncosθ-fsinθ=0 and ΣF_x=Nsinθ-fcosθ=0. The solutions are f=Wsinθ and N=Wcosθ. 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=v²/R where R is the radius of the track. So now, ΣF_y=-W+Ncosθ-fsinθ=0 and ΣF_x=Nsinθ-fcosθ=mv²/R. Solving these equations, f=m(gsinθ-v²cosθ/R). Now, notice that if v₀=√(gRtanθ), f=0; at this speed the car can go around the track even if it is perfectly slippery. For speeds greater than v₀, f will be negative which means it must point down the slope. As the speed increases beyond v₀, 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+μcosθ)/(cosθ-μsinθ)].

Not every angle of incline or every μ will result in the car slipping, however. Whenever (cosθ-μsinθ)]=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)=63.4⁰. 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 mv²/R and, if it is greater than Nsinθ-fcosθ=N(sinθ-μcosθ) 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. 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/s². 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≈¼Av² where A is the cross-sectional area you present to the wind, let's say about A≈1 m²; 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 v²/R in the positive y direction. Newton's laws are ΣF_x=0=mgsinθ-f and ΣF_y=mv²/R=mgcosθ-N. The solutions are simply obtained: f=mgsinθ and N=m[gcosθ-v²/R]. An interesting result is that when θ=cos^-1(v²/(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=v²+2gR(1-cosθ) and so the centripetal acceleration is a_y=v²/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=mgsinθ and ΣF_y=m(v²/R+2g(1-cosθ))=mgcosθ-N. The solutions are a_x=gsinθ and N=m(v²/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/s². 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=½kx²=½x200x0.01²=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/c²=0.01/(3x10⁸)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 W₁=mMG/R² 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 W₁=mg. But, if the point mass is a distance H above the surface, its weight is smaller, W₂=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 W₃=mg/(1+(L/R)). In your example L=1000 mi is not small compared to R≈4000 mi, so W₃=mg/(1.25)=0.8W₁. 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 N₁ and N₂ 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 N₁+N₂=mg. The frictional force is f=μ(N₁+N₂)=μmg. The torques will sum to zero if the bottle is not tipping, 0=fH+RN₁-RN₂. Now, if the bottle is just about to tip over, N₁=0 and so the torque equation becomes 0=fH-RN₂=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, the 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=mc². The uncertainty principle is ΔEΔt≈h/2π where h/2π=6.6x10^-16 eV≈10^-34 J·s. Since c=3x10⁸ has no uncertainy, Δm≈[h/(2πc²)]/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=gsinθ 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=mgsinθ=-Nsinθ+fcosθ+mgsinθ and 0=Ncosθ-fsinθ-mgsinθ. Solving, I find N=mgcos²θ and f=mgsinθcosθ.
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=-mgcosθ+N and ma=mgsinθ, just the same as the classic object on the frictionless incline. So the scale reads N=mgcosθ.

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/R² 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=R²/(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 F₁ and F₂ at distances at distances d₁ and d₂ from the center of gravity. The two people must hold up all the weight between them, so F₁+F₂=W. The canoe is not rotating about its center of gravity so the net torque must be zero, so F₁d_₁=F₂d₂. Solving, F₁=W[d₂/(d₁+d₂)] and F₂=W[d₁/(d₁+d₂)]. If d₁=d₂, 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=3x10⁸/5.5x10^-7=5.45x10¹⁴ Hz. The momentum of one photon is p=hf/c=6.6x10^-34x5.45x10¹⁴/3x10⁸=1.2x10^-27 kg·m/s. The photon flux (photons per second per square meter) for photons of this wavelength is about Φ=3x10²¹ s^-1·m^-2. So if I estimate your area to be about 1 m², there are about 3x10²¹ photons hitting you every second. The total momentum imparted to you over that second (assuming that you absorb the photons) is about Δp=3x10²¹x1.2x10^-27=3.6x10^-6 kg·m/s. The average force you feel is about F=Δp/Δt=3.6x10^-6N; 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=mc² 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)(3x10⁸ m/s)²= 9x10¹⁶ J and E=(1000 g,)(3x10¹⁰ cm/s)²= 9x10²³ erg. In fact, 9x10²³ erg=9x10¹⁶ 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=3x10⁸/5.5x10^-7=5.45x10¹⁴ Hz and f'=2.24x10^-4x5.45x10¹⁴=1.23x10¹¹ 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=p²/(2m) where p is the linear momentum. The uncertainty principle states that ΔpΔx>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/kg²is the gravitational constant, M=6x10²⁴ kg is the mass of the earth, r=3.85x10⁸ m is the radius of the moon's orbit, and m is the mass of the coin. So, U=-1.04x10⁶m J. But, when the coin strikes the earth its energy will be E=-GMm/R+½mv² where v is its mass and R=6.4x10⁶ 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: -1.04x10⁶m=-GMm/R+½mv²=(-6.25x10⁷+½v²)m. Solving for v I find v=1.11x10⁴ m/s. The kinetic energy it would release would 6.15x10⁷m 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=√(V²+H²)=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 of 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:
se U=mgh for potential energy, m=10,000 t=10⁷ kg, h=100 m, g=9.8; U=9.8x10⁹ J. If it falls over a time t=10 min=600 s, the power generated would be P=U/t=1.63x10⁷ 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 ½mv² 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, √(1-0.7²)=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=½mv² 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=Rωcos(λ) where R=6.4x10⁶ 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=6.4x10⁶x7.3x10^-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=½x100x470²=1.1x10⁷J.
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 at the top above 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 second picture above 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. SinceL+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 where 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 & 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 & 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 if tending to topple you forward is 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=½at² 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/s². The acceleration due to gravity is 9.8 m/s² 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=½at²=½a(28)²=392a, so a=0.094 m/s². 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/m³, so the volume of the steel is about 100/8000=0.0125 m³. If it is a sphere its radius is about 0.144 m and its cross-sectional area is about A=0.0661 m². At sea level a pretty good approximation for the air drag is F=¼Av² (only for SI units), so terminal velocity will be achieved when mg=¼Av², 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_t² (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=πR²=πx0.38²=0.45 m². 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) is some pretty tricky integration. The solutions are v(t)=v_ttanh(gt/v_t) and y(t)=h-(v_t²/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=((47²/9.8)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=½Cρv² 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 m₁ and m₂ 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=√(πμa³/K) where μ=m₁m₂/(m₁+m₂) is the reduced mass, a=d/2, and K=Gm₁m₂. Therefore, t=√[πd³/(8G(m₁+m₂)].

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=mγc²where γ=√(1-(v/c)²). In your case, 2000 lb=907 kg, c=3x10⁸ m/s, γ=1.005, 1.011, and 1.048. The energies in Joules are E=8.204x10¹⁹, 8.253x10¹⁹, and 8.555x10¹⁹ J. There are about 4x10⁹ 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 8x10¹⁹/10^-4=8x10²³ W=8x10¹⁴ 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 mΔv=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. L₁=Iω₁+0=Iv/r where I is the moment of inertia about the center of mass; L₂=Iω₂+mrv'=(v'/r)(I+mr²). Therefore, (v'/r)(I+mr²)-Iv/r=-mv'r or, v'=Iv/(I+2mr²). 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=½mr²; 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/s², 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 t₁=2L/c and along the other length is the same, t₂=2L/c. In the moving system all lengths along the the x' direction are reduced by a factor √(1-β²) where β=v/c; therefore L'=L√(1-β²) so t₁'=2L'/c=2L√(1-β²)/c=t₁√(1-β²). For L" only its x' component is contracted, L"cosθ'=L√(1-β²)cosθ while the y" component remains the same, L"sinθ'=Lsinθ; from these you can easily show that L"=L√(1-β²cos²θ) and tanθ'=(tanθ)/√(1-β²). Finally, t₂'=2L"/c=[2L√(1-β²cos²θ)]/c=2t₂√(1-β²cos²θ)≠t₁'.

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 or 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 0.47²/17²=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/s² 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 v₂=√(2gh); jumping from the third floor would result in a speed of v₃=2√(gh) which is about 1.4 times larger than v₂. 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/s²=5 lb·s²/ft. (The unit of mass in Imperial units is called the slug, 1 slug=1 lb·s²/ft.) The speeds from second and third floors will then be v₂=27.7 ft/s=18.9 mph and v₃=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 t₂=2x3/27.7=0.22 s and t₃=2x3/39.2=0.15 s. Finally, the average forces during impact are F₂=160+5x27.7/0.22=790 lb and F₃=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/m². 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:
We 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, ²H. 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 H₂O⁺, H⁺, O⁺, OH⁺, O₂⁺, and H₂⁺; 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:
If you tow a boat from a tow path which is the best point to tie the rope onto the boat?

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/2π)√(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=mc²? 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=½at². For your question, a=7x9.8=68.6 m/s² and t=10 s, so d=½x68.6x10²=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, ΔEΔt≈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/(2πΔt)=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 4x10⁸ 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 2.5x10⁹ 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 ft²·°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 & 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=mc²=9x10⁹ 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=6.8x10⁵ J. So, the ball would have increased its mass by 6.8x10⁵/9x10⁹=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 H₂O. 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 (6,670²)would be D₂O. The term "heavy water" refers to D₂O 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 D₂0 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% H₂0, and 25% D₂O. Where did all that D₂O 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% D₂O 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 gs. 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/ft³=0.0361 lb/in³. Therefore the gauge pressure 2 ft down is P=ρh=62.4x24=0.867 lb/in²(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:
The 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_Dwhich 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≈N²/2 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. Your 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)=Isin(ωt) and v(t)=Vsin(ωt+φ), so p(t)=IVsin(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:
As 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=10¹⁰. The easiest way to do this is with logarithms: log(2N)=N·log2=log(10¹⁰)=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=MhRω² where R=6.4x10⁶ 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=Rω² and therefore the net force on M is Mg-MRω². Therefore, the net work done is W≈(Mg-MRω²)h+MhRω²=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 9x10⁷/3x10⁵=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/m³and 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+½ρv²=constant; P is the pressure, g=9.8 m/s² 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≈10⁷ N/m²≈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)=(vxE(r,t))/c where the electric field is E(r,t)=[q/(4πε₀)][ρ/(ρ·u)³][(c²-v²)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.

ADDED 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/(4πε₀)][(r-x)/|r-x|³] and B(r,t)=[qμ₀/(4π)][vx(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=QVxB. 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/s²: M=W/g=(200 lb)/(32 ft/s²)=6.25 ft·lb/s². The speed at impact can be determined from V=√(2gh)=√[2x(32 ft/s²)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/s²)(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 & man C. I understand that to man B & 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 & 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 & 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_CAv_BA/c²)] 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+v²/c²); 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-v²/c²). 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_Cmeans 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/s² where g=32 ft/s²is 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/s². 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/√(1-0.8²)=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/√(1-0.6²)=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=v²/(Lsinθ) where L is the length of the string; applying Newton's second law, Tsinθ=mv²/(Lsinθ). There is no acceleration vertically and so you can apply Newton's first law, Tcosθ-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=10⁶). Now, if you let this mass drop over 3600 s, it is losing its energy at the rate of (196.2x10⁶ J)/(3600 s)=5.45x10⁴ J/s=54.5 kW (not kW/hr). For your case, your mass has a potential energy of 10⁴x9.81x5=4.9x10⁵ J. If you deliver this energy over an hour, the power is 4.9x10⁵/3600=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 ke²/d²=9x10⁹x(1.6x10^-19)²/d²=2.3x10^-28/d²=mg=9x10^-31x9.8=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 45º 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 ½kx²=½x41.7x53.3²=59,200 J. This must equal the kinetic energy of the pig at launch, ½mv²=½45.4v²=22.7v²=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=v²/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/s². 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=-½mv² 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=v²/(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/m²) 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=10^6.510^-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/(4π1²)≈2.5x10^-7 W/m². 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/m², twice as loud. As you moved farther away, you would find that the intensity fell off like 1/r² where r is your distance from the source; at 500 m away, the intensity of the "raw" will be at 2.5x10^-7/500²=10^-12 W/m² and the amplified will be at 2x10^-12 W/m². The "threshhold of hearing" is about 10^-12 W/m², 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=y₀+v_0yt-½gt²=200-0-½9.8t²=200-4.9t², so t=6.39 s. The equation for x motion is x=x0+v_xt=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=mrω². 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, mgrcosθ-mr²ω²sinθ=0so tanθ=g/(rω²).

Now, the questioner found that for most balls, θ≈45º so tanθ≈1. Taking g≈10 m/s² and ω²=(10,000 rpm)²≈10⁶ 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=mgrcosθ=4.6x10^-6cosθ N·m. The moment of inertia of the sphere is I=2mR²/5=4.1x10^-5 kg·m². 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 rigid

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=Rω² 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 less 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

a perfectly inelastic collision where the man sticks to the train afterward and
a perfectly elastic collision where the man flies off with no loss of total energy.

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=mv²/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[v²/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=½x390xu²=½x390x(0.92)²v²=170v² 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/s² is the acceleration due to gravity. So, K+W=0=170v²-34,000 or v=14 m/s=31 mph.

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/s² 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/r². Then, using Newton's second law, each mass will have an acceleration independent of its own mass of a=[mMG/r²]/m=MG/r² and A=[mMG/r²]/M=mG/r². 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=½mv² 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=½·60·224²=1.5x10⁶ 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 1.5x10⁶/600 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 & 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 (Tsinθ) and therefore exerts a torque τ=rTsinθ 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, ½mv₁²=½mv₂² or v₁=v₂, 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=mv²/L

=0.05x4π²/0.05=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 m₀v/√[1-(v²/c²)] where m₀ is the mass of the object when at rest. The relation among energy E, momentum p, and rest mass m₀ is E²=p²c²+m₀²c⁴. 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=mc² 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 45º, 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 km², density of water is 1000 kg/m³. I find that the total volume to fall is about 2x10¹¹ m³, the mass is 2x10¹4 kg, and the potential energy of this mass is about 3x10¹6 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, ½Mv²=½(W/g)v², where M is the mass and g=32 ft/s². So, s=(1/(2μg)v². 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 ft²)(1 yd/3 ft)²=100/9 yd² and t=(1 in)(1 yd/36 in)=1/36 yd. So V=(100/9)(1/36)=0.309 yd³. 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=πR²L=LWt where R is the radius of the cylinder; therefore, R=√(Wt/π). So, if your rug is 10x10 ft², 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 & 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=½CρAv²; 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/m³ is the density of water, and A≈1 cm²≈10^-4 m² 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=v₀/(1+kt) and x=(v₀/k)ln(1+kt) where k=½CρAv₀/m≈7000 s^-1 and v₀=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/r² where r is the distance from the center of the earth. If R=6.4x10⁶ m is the radius of the earth, W is your weight at the surface, and W⁺ is your weight 1 mile=1.6x10³ m above the surface, then W⁺/W=R²/r²=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 Foreland historic 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/m³, so the volume you must displace is V=M/ρ=0.15 m³=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 m³, 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/R² 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+μcosθ) 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_inclineis 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=mc² 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 (²H) to one alpha (⁴He) 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 ¹⁴N atom left after the decay has only 6 electrons, not the 7 in a ¹⁴N 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/s²)]. 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 r₀ to be g=9.8 m/s² is ω=√(g/r₀).

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)=P₀-½ρω²(r₀²-r²)=P₀-½(ρg/r₀)(r₀²-r²); here r₀ is the distance from the center to the outer surface, P₀=P(r₀), 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 r₀=1000 m, r=900 m, and P₀=10⁵ N/m² (approximately atmospheric), P(r)=0.99x10⁵ N/m². In a uniform gravitational field there is an empirical equation to calculate pressure as a function of altitude h, P(h)=P₀[1-(Lh/T₀)]^gM/(RL). Without going into detail, I find that P(h=100 m)=0.98x10⁵ N/m²; 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=r₀ω=√(gr₀). 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 r₀ 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=2muxω). 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: mu²/r₀=N-mg+2mu√(g/r₀). So, when N=0, u²+gr₀-2uv=0=u²-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 mu²/r₀ 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:
1. 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!
2. 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=2πr₀/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.
1. 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/s²=g/45 because the time is much longer.
2. 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?

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/d² 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 R₁ at time t₁ and a position R₂ at time t₂, then the displacement vector D is defined as D≡R₂-R₁ and average velocity vector is defined as v_avg≡D/(t₂-t₁). 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/c² will be captured.

QUESTION:
I recently saw my son spinning an object on a wire overhead & 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 & 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=2πRf=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=1.94x10⁴ 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 ½mv²=½ky²+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(y₁+y₂)=y_rest. The solutions for the problem at hand are y₂=+0.096 m, y₁=-0.142 m, y_rest=-0.023 m; note that ½(y₁+y₂)=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=v²/(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/m². 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?

four layers are shown

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=½ΦL². 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=ML²/3. If it has an angular velocity ω₁, its angular momentum is L₁=Iω₁=Mω₁L²/3. If you increase the mass to M+m, the moment of inertia will increase to I'=(M+m)L²/3 and its angular velocity will change to ω₂. But, the angular momentum will not be changed, Iω₁=I'ω₂; you can then solve this for the new angular velocity, ω₂=(I/I')ω₁=[M/(M+m)]ω₁ which is smaller. However, the rotational kinetic energy E of the blade is now lower, E₁=½Iω₁² and E₂=½I'ω₂²=½I{(M+m)/M}{[M/(M+m)]ω₁}²or E₂=[M/(M+m)]E₁. On the other hand, if you wanted to add mass but keep the energy the same, E₂/E₁=1=I'ω₂²/Iω₁² or ω₂=ω₁√[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;
the 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 & 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 & 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, 45º 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≈¼Av² 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 m², you can then solve ¼Av²=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 gt²." 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 S₅=½x32x5²=400 ft; at the end of 6 seconds the position is S₆=½x32x6²=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 mgsinθ doing positive work; if f<mgsinθ 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=¼Av² 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/s², and A≈2x4=8 m² (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_t²/(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_t²/g (derived from the expression v/v_t=0.99=√[1-exp(-2gh/v_t²)]. 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:
What is after death (AD), birth of christ (bc)? Then how we are caluclate the age difference in between AD & BC.

ANSWER:
This is not physics, but easily answered. Actually, AD is anno domini, (in the year of the Lord in 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_Dρv²A 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_Ddepends 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=mc², 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/c².

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=½mv²=½x3.83x10⁷x2.34x10⁶=4.48x10¹³ 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-Wsinθ
ΣF_y=0=F_y+N-Wcosθ
Στ=0=½WLcos(θ+φ)-NLcosφ.

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-Wsinθ+B
ΣF_y=0=F_y+N-Wcosθ
Στ=0=½WLcos(θ+φ)-NLcosφ+BLsinφ.

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=2πRwhere 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% ²³²Th, absorbs a neutron to become ²³³Th which β-decays to ²³³Pa (half life 22 minutes)which β-decays to ²³³U (half life 27 days). The ²³³U 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 ²³³U. As the ²³³U 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=¼Av²; 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 4.5x10⁴ kg. Putting it all together, I find that A≈10⁴ m², 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 2x10³⁰kg, 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/cm³, 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 2x10²⁰ radioactive nuclei it will be twice as big than when you measure R for 10²⁰ 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=N₀e^-λt where N₀ 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 N₀=1, they would all decay at once! You just could not predict when. If N₀=2, they might both go at once or else one might go before the other but not necessarily at τ_½.

QUESTION:
When solving questions involving two identical springs being stretched to points A and B to create a total length and the natural length of the springs are given. What is a consistent way to calculate the amplitude? Is it half of the length of AB subtract the natural length of the spring?

ANSWER:
I assume that the springs are in series—one attached to the end of the other. Suppose that you exert some force F such that the springs are stretched by a distance s. Then each spring will be stretched by ½s. If k is the common spring constant, F=½ks. Therefore the two together behave like a single spring with spring constant ½k.

QUESTION:
How fast are we moving? Besides Earth rotation and revolving, there's our path in the galaxy, and even greater in relationship to all other galaxies, and please include the expansion of space-time.

ANSWER:
First you should understand that it only makes sense to ask what our velocity is relative to some inertial frame of reference. And, also, it depends on things like what time of day it is, what time of year it is, what year it is, etc. because all these things affect your velocity; e.g., relative to the earth's axis, you are moving in opposite directions at noon and midnight. There is a nice qualitative discussion of this question at Curious about Astronomy; adding up all the various velocities, she finds a maximum speed relative to the center of our supercluster of about 900 km/s. Given the roughness of this calculation, trying to fine-tune it by adding in the expansion of space-time is pointless.

QUESTION:
In my AP Physics class, we have had a class-wide debate over a physics problem for the last few days. The problem asked how much work it would take to move a satellite that was orbiting Earth at a certain height to a greater height. Some of us say that the work equals the change in potential energy, while others say that the work is the change in the total mechanical energy. The total energy method gives an answer of exactly half of the amount the potential energy method gives. Who is right? Both orbits are circular.

ANSWER:
Since the speeds in the two orbits are different, the kinetic energies will be different, so the correct way to do this is
½mv₂²-mMG/r₂²=W+½mv₁²-mMG/r₁². Of course you will have to figure out what the velocities are in terms of the radii, but that should be a piece of cake for AP students!

QUESTION:
A car is travelling 45 mph, with the road being at a 5-8% incline. How long would it take a car to slow down without using breaks? This is not homework. I am a claims examiner that is trying to get some information as to the rate of speed in which a car decelerates.

ANSWER:
You are a claims examiner and you spell brakes "breaks"?! I could do this if I assumed no friction whatever, but it would not be predictive of the real world because there is plenty of friction acting on a moving car even without brakes applied. I could make a better estimate if you could tell me how far this particular car, starting at 45 mph, traveled on level ground with brakes not applied. Also, is the car in gear? In neutral? Engine running? If there were absolutely no friction, the car would keep going until it had gone vertically up a distance of about 20 m≈66 ft regardless of the grade of the incline. For a 6% incline, the distance traveled by the car would be about 1100 ft. In the real world it would be way less than this.

QUESTION:
As we know the earth is revolving around the sun. Why don't we have any feeling of revolving?

ANSWER:
In order to have a "feeling" of motion, there must be an acceleration. The acceleration of the earth in its orbit is about 6x10^-6 m/s². For comparison, if you are in a car with an acceleration from 0 to 60 mph in 10 s, your acceleration is about 3 m/s²; this means that you "feel" pushed back into your seat with a force about one third of your weight. The force you "feel" due to the orbital motion of the earth is about 1/600,000 of your weight, far less than you could "feel". The acceleration due to the earth's rotation on its axis is larger than this but still way smaller than you could "feel".

QUESTION:
Trying to understand a relationship between nuclear physics and particle physics. In nuclear fusion 4x H1 fuses to give (ultimately) 1 x He4 + 1 positron a neutrino and energy. However, 4 protons have become 2 protons and 2 neutrons and neutrons are more massive than protons - so we have apparently gained mass - which should require an input of energy. My research finds that the He4 nucleus is less massive than the 4 protons that made it because mass has been converted to 'binding energy'.

Questions:

Since a lot of the mass of a proton comes from energy, doesn't this 'binding energy' add mass to the He4 nucleus?
The protons that formed this nucleus are made up of Quarks held together by energy in the form of the exchange of Gluon particles. To convert to a neutron, an Up Quark has become a Down Quark with more mass. What mass has the He4 Nucleus lost?
Do the Quarks in the He4 nucleus have less mass than those in the H1 since the Quarks appear to be the only mass in the process that can convert to energy? Surely if the energy came from the Gluons it would still just be energy and hence not have the mass/energy conversion ratio?
If the energy did come from the Gluons, are Quarks in He4 less tightly bound than those in H1?
Does this continue to occur as you move up the fusion sequence to higher mass elements?
Has the creation of matter from energy (e.g. the electron produced in this fusion) ever been actually achieved? Or are we working purely on theory based on remote observation of the Sun?

ANSWER:
I should really delete this question altogether because it has ignored the site ground rules stipulating "…single, concise, well-focused questions…" However, I will address at least parts of the question to clear up some major misconceptions. Although the questioner asserts that he is "trying to understand a relationship between nuclear physics and particle physics", I will ignore all questions relating to quarks and gluons because if the nuclear physics itself is not understood, then there is no point in trying to relate it to particle physics. To understand the fundamentals of nuclear physics, you do not need to even know that quarks and gluons exist, only that a strong nuclear interaction exists. In the opening paragraph the mass of the ⁴He nucleus is compared to the mass of four protons. This is a meaningless comparison because the constituents are two neutrons and two protons. (See below* for how protons are converted into ⁴He nuclei.) Indeed, the mass of the ⁴He nucleus is less than the mass of its constituents. If one knows E=mc², the reason for this mass difference is intuitive. Does it take work to pull a proton or neutron out of the ⁴He nucleus? Of course it does, otherwise the ⁴He nucleus would not stay together. So energy must be added to disassemble the nucleus and this energy resides in the greater mass of the constituents. I find this much easier to understand than to say that the excess mass of the constituents is "converted" to binding energy.

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 ³²S nuclei to a single ⁶⁴Ge 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 ⁴He 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.

QUESTION:
The substance which undergoes deformation with a small force will be elastic or inelastic?

ANSWER:
You cannot judge the elasticity of something by how easily it deforms. If it stays deformed after the force is removed, it is inelastic. If it returns to its original shape after the force is removed, it is elastic.

QUESTION:
The particles that make up a rock are constantly moving. However a rock does not visibly vibrate. Why do you think this is?

ANSWER:
The amplitude of the vibration is less than the distance between atoms in the rock, ~10^-11m. The atoms are all randomly vibrating so that there always as many atoms moving in one direction as in the opposite direction. The frequencies of the vibrations are ~10¹³Hz, far above a frequency which you could feel, hear, or see.

QUESTION:
I was just wondering if you have ever thought that dark energy was actually just gravity? Could the gravitational force act differently over large(r) scales? This seems like something I would like to study at a PhD level as this could be an area to reveal the connection between gravity and quantum mechanics.

ANSWER:
The best theory of gravity is general relativity (GR). GR already contains a term, called the cosmological constant, which can add a repulsive force to the attractive gravitational force. I do not quite see how this could be a route to quantizing gravity, but who knows? Good luck!

QUESTION:
A quantity of electrical energy is defined by volts x amps x time. A quantity of mechanical energy could be defined (or is) by force x distance which equates to kinetic energy. When electrical energy is converted into mechanical, a force can be created by applying voltage and current (amps). Is this a paradox? Electrical energy is functionally force x time while mechanical energy is force x distance.

ANSWER:
1 V=1 J/C and 1 A=1 C/s, so 1 volt·amp·second=1 Joules=force·distance. No discrepancy, no paradox. Another way to look at this is that current times voltage is power and power is W=J/s.

FOLLOWUP QUESTION:
I totally understand that 1 amp is 1 coulomb per second. I don't know where 1 volt is equal to 1 joule per coulomb comes from or why that is true.

ANSWER:
The electric field E is defined to be the force F felt by a charge Q divided by Q. The electric potential V is defined as E times distance d over which it acts. V=Ed=Fd/Q=[J/C]

QUESTION:
If gravitons mediate the gravitational force between particles, how does this work for a black holes where gravitons cannot escape the event horizon? Are gravitons predicted by General Relativity?

ANSWER:
See an earlier answer. Gravitons are not predicted by general relativity. Physics inside the event horizon is not well understood.

QUESTION:
Why is there a slightly rough surface on a basketball? Does this affect the static friction acting between your hand and the ball during a shot?

ANSWER:
The dimples on a golf ball and the fuzz on a tennis ball reduce air drag and thereby allow the ball to go faster and farther. However, the speeds of these balls are much larger than speeds ever encountered by basketballs and this cannot be the reason for the bumps on a basketball. A little research reveals the purpose is just to make the ball easier to grip and handle, as you speculated.

QUESTION:
Are there any known stable nuclei for witch Z>N? Why are they so rare?

ANSWER:
The only stable Z>N nucleus is ³He. The reason that all stable isotopes heavier than helium have the same or more neutrons than protons is complicated because the nuclear force is complicated. One important reason is that the repulsive Coulomb force as well as the attractive nuclear force is present trying to blow the protons apart and the more protons you add, the more important this becomes; adding neutrons tends to push the protons farther apart to reduce the repulsive force. Neutrons are sort of "insulation" for the protons' electric repulsion from each other.

QUESTION:
Electrons orbit around the nucleus in varying degrees of proximity to the nucleus. Do electrons farther from the nucleus orbit at a different rate of speed than those closer to the nucleus?

ANSWER:
Yes, if you use a Bohr model for the electrons. However, the notion of electrons running around in well-defined orbits is naïve and incorrect.

FOLLOWUP QUESTION:
Would it be correct to apply Kepler's Laws of Motion to the revolution of electrons about the nucleus in the Electron Cloud Model?

ANSWER:
The electrostatic force is a 1/r² force just like the gravitational force, so if the atom were a classical system Kepler's laws could be used for an atom. In fact, the Bohr-Sommerfeld model, the first extension of the Bohr's circular orbit model, essentially does this by including elliptical orbits and appropriate quantization. Of course, the atom is not a classical system and although such models can be instructive, they are not strictly accurate. What you refer to as the "electron cloud model" would be the proper solutions to the Schrodinger equation, not having well-defined orbits.

QUESTION:
I would like to know if one horsepower is equivalent to 33,000 pounds per minute, and for a four cycle engine fires on every 2nd revolution of the crankshaft, and lets just say this 1 hp engine runs at 2800 rpm , so it fires 700 times , is there a formula to calculate how much force was created in the combustion chamber in pounds?

ANSWER:
Power is energy per unit time; some possible units of power are Watts (W), horsepower (hp), and foot-pounds/second (ft-lb/s). So your lb/min is not an acceptable unit for power because lb is a unit for force, not energy. But, you apparently looked this up somewhere because 1 hp=33,000 ft-lb/s. The power of the engine does not tell you the force in the combustion chamber, in fact force is not really what would be of interest. What you want is the energy delivered by each piston for each stroke. If the pistons fire, for example, at the rate of 1000 times per minutes, each piston delivers 3300 ft-lb per stroke.

QUESTION:
I knew that when a car which travels very fast and is brake very sudden, the car will like "fly away". But is there any theory or principle or rule can explain this?

ANSWER:
I do not know what "fly away" means.

FOLLOWUP QUESTION:
To explain, an illustration is made. Imagine a bicycle that travels in very high velocity and it is braked suddenly and hardly, what will happen. The bicycle will like "flying up" caused by the momentum. It is also related to the principle of acceleration and deceleration.

ANSWER:
OK, I get it now. Refer to the figure above. The easiest way to do this is to introduce a fictitious force. If the car is accelerating with acceleration a (which points opposite the velocity v when braking), Newton's first law will be valid in the car frame if a force F=-ma acting at the center of mass (COM) is introduced. The "real" forces on the car are a normal forces up on each wheel by the road, the frictional forces backwards on each wheel by the road, and the weight mg which acts at the COM. In the drawing I have only labeled the weight, the normal force on the rear wheel, and the fictitious force because those are all you need to answer your question. If you wish, you could find both normal forces and the sum of the two frictional forces in terms of a; if all wheels were locked you could find the individual frictional forces if you knew the coefficients of kinetic friction. Now, I will sum torques τ about the point where the front wheel touches the ground, Στ=mah+NL-mgs=0 where h is the height of the center of mass above the ground, L is the horizontal distace from the front axle to the COM, and s is the horizontal distance between the wheels. You can now solve for N, N=(mgs-mah)/L. Now think about N; if N<0, the road would have to pull down on the back wheels to hold the car from rotating forward about the front wheels. N will be zero when the car is just about to "fly up"; therefore, if a>g(s/h), the car will "fly up".

QUESTION:
If you aimed a high powered laser beam in a tangential direction to the surface of the Earth, would that beam travel in a straight line and shoot out into space, or would Earth's gravity be enough to at least bend the light in a measurable amount?

ANSWER:
See an earlier answer. If you calculate the angle for earth instead of the sun, φ=4GM/(c²r)=0.002 arcseconds, just about impossible to observe.

QUESTION:
When speaking of particle accelerators,the accelerators keep adding energy to the particles, even though they cannot speed up any further. But where does the energy go?

ANSWER:
Well, they never really get to a speed where they cannot go any faster because they never reach the speed of light. I have always thought "accelerator" was a misnomer for very high-energy machines because the acceleration (rate of change in speed) is almost zero; I would prefer "energizer". The easiest way to think about it is that the particles acquire mass and that is "…where…the energy go[es]". The energy of the particles is E=mc²=m₀c²/√[1-(v/c)²] where m₀ is the mass at rest. For example, if v/c increases from 0.99 to 0.999, that is only a 1% change in speed; but the energy increases from 7.1m₀c² to 22.4m₀c², more than triple.

QUESTION:
A cylinder is kept in friction less inclined plane (curve face touching incline). Why it does not roll (transnational motion only) although a torque is working on it with respect to contact point due to component of force parallel to incline

ANSWER:
Because the cylinder is accelerating. The point of contact is also accelerating and so you cannot use it as a point about which to sum torques because Newton's laws are not valid in an accelerating coordinate system. An accelerating coordinate system is called a noninertial frame. There is, however, one point where you can always sum torques even if that point is accelerating—the center of mass. The net torque about the center of mass is zero because the weight vector acts at that point and the normal force passes through it.

QUESTION:
I am looking for relationship between viscosity of air and pressure. I want a table listing the viscosity at different pressures. Pressure should start from 0.1 atm to 1 atm.

ANSWER:
The table above comes from an article by Kadoya, Matsunama, and Nagashima. Pressure is given in MPa and 0.10 MPa=0.987 atm≈1 atm. So you want the variation between 0.01 MPa and 0.1 MPa. As you can see the variation is very small; the viscosity is much more dependent on the temperature.

QUESTION:
Two questions for you concerning the recent proof that gravity flows in waves. 1). How fast do these gravity waves move? Do they move at a constant speed or is their speed predicated by the mass of the object that generated the waves? 2). Since the earth is trapped in orbit around the sun by the sun's gravity, how can a wave moving out from an object in turn pull a second object back towards the object that generated the wave? In other words, how can gravity waves moving out from the sun at the same time pull the earth towards the sun?

ANSWER:
The answer to your first question is in a recent answer. In your second answer, you are confusing gravity with gravitational waves. Our picture of gravity is that mass (like the sun) deforms the space around as in the animation above. Something like the earth orbiting the sun is not really feeling a force, it is following the contour of the space. (Do not take this simplistic model too seriously, it is really the four-dimensional space-time which is "deformed".) But, if something accelerates, like the earth going around in its orbit, it will send out ripples and those are the gravitational waves. On the scale of this animation, the waves the orbiting earth is sending out are far too tiny to be seen. But, if the objects orbiting are much more massive, like the two orbiting black holes which were observed in the recently reported observation of the waves, the waves are much bigger.

QUESTION:
Ok so you know how they say, "if you look up at the stars in the night sky, you are seeing light that's taken millions of years to travel to earth, so you are seeing Into the past potentially"? Ok so here's the question. If say I'm on a planet in the Andromeda galaxy, and you're on earth looking at me with a super telescope( to where u can actually see me wave at you) and let's say we agreed on a specific time and date to do this. If I start waving at you at the exact time you are observing me with the telescope, do you ever see me wave or does it take millions of years to see me wave at you ?

ANSWER:
How do you plan to agree on a specific time and date to wave if it takes millions of years to communicate. Just imagine that there is a star exactly halfway between you and me and we have, by some magic, both been instructed to do our things (wave and look) when we observed the star to become a supernova. If you and I are separated by a million light years, I will not see you wave until a million light years have passed. Furthermore, the supernova would have occurred a half million years before either of us saw it.

QUESTION:
Why is it that when I listen to music on my PC through an external speaker set and my cell phone right next to me gets messages, etc... and they cause distortions in how the music sounds?

ANSWER:
Your cell phone is both a receiver and a transmitter of radio waves. The transmitted radio waves may be detected by your sound system.

QUESTION:
I mean to say the angle derived in Einstein equation φ=4GM/(c²r)=1.75 arcseconds is total deflection when light source is opposite side of the Sun?

ANSWER:

QUESTION:
When a television set is turned on it often generates an electrostatic field. You can still feel this charge if you move your arm near the screen shortly after the set has been turned off. Why does this field not disappear as soon as the television set has been turned off?

ANSWER:
You probably have an old cathode ray tv. The screen is coated with a phosphor which glows if you strike it with an electron. Electrons are shot from the rear of the tube to cause a picture to be formed. Some of these electrons are still on the screen when you turn off the power.

QUESTION:
Ships are often built on ways that slope down to a nearby body of water. often a ship is launched before most of its interior and superstructure have been installed and is completed when a float. Is this done because the added weight would cause the ship to slide down the ways prematurely?

ANSWER:
Friction can be a tricky business, but the simplest behavior is that the frictional force increases proportional to the weight. But the force of the gravity trying to slide the weight down the slope is also proportional to the weight. Therefore, doubling the weight of the ship should not increase its tendency to slide down. Besides, if this were a concern you could always temporarily block the path down the slope like placing blocks in front of a vehicle on a slope to keep it from rolling. I suspect the real reason is that the structure of the ramp is probably not strong enough to support the full weight of the ship.

QUESTION:
If one wanted to turn a cylinder 5 feet in outer diameter and 3 feet inner diameter and the reel is supported via a 2 in bar through the center of it Look like this i suppose (more like a reel) A rope is wound around the outer diameter and pulled (think of a yo yo) how much torque would you have to put on the rope to get it to turn? at 35 rpm? Can you show how one can figure that?

ANSWER:
It would depend on how long you pulled, what the mass of the hollow cylinder is, and how much friction there is.

FOLLOWUP QUESTION:
I have asked several people to give an answer on this and no one has been able to answer it for me. So I attached a pic I just drew out for you... On the pic you see that the master reel is 5000 lbs... I want to pull the rope off the reel and put on another reel... I would like to do so at 35 rpm on the small reel... The small reel is just like the large reel but the OD is 24 inches and the center of the reel is 12 inches... with a bar going through it just like the large one... how much pulling torque would be needed to get the large reel to start turning... neglect friction...not enough there to consider...

ANSWER:
First you have a misconception. If friction is truly negligible, any torque, no matter how small, will start the big reel turning. It is a matter how long you want to wait until you get up to 35 rpm on the smaller reel. For several reasons this is a quite complicated engineering problem. The most important problem to deal with, I think, is that as rope comes off the larger reel its weight gets smaller and its radius gets smaller; at the same time the smaller reel gets larger and heavier. So if you want to keep the smaller reel going at a constant 35 rpm the larger reel will have to change its speed if the rope is to remain taught. So, I will only address the question of how you start up just to demonstrate how torque determines what happens. From your drawing, it looks like the rope fills the reel so its outer radius would be about 3 ft≈1 m and its inner radius would be about 1 ft≈0.3 m. Since the mass of the rope is 5000 lb≈2300 kg is large compared to the 200 lb reel, I will neglect the contribution of the reel; this simply says that the torque you apply mainly has the effect of getting the rope spinning. Since it is the smaller reel you want to go 35 rpm, the upper reel would need to have an angular velocity of (0.5/3)x35=5.8 rpm=0.61 radians/s. (Assuming that your idea is to use the empty reel to pull the heavier reel via the rope.) Again, I will neglect the torque necessary to get the smaller reel going since it will presumably be lighter than the already neglected weight of the heavier reel. The basic physics principle to use is the rotational form of Newton's second law, τ=Iα where α is the angular acceleration, τ is the torque, and I is the moment of inertia. In your case, the angular acceleration would be the final angular speed divided by the time to get there; so if you want to get up to speed in 10 seconds, the angular acceleration would be 5.8/10=0.58 rpm/s=0.061 radians/s². You see, now, why there is no answer to your question: the torque will depend on how quickly you want to get up to speed. The moment of inertial of the rope is I=½M(a²+b²)=½(2300)(1²+0.3²)=1250 kg·m². Finally you can estimate the torque τ=1250x0.061=76 N·m=56 ft·lb. A longer spin-up time would need a smaller torque.

QUESTION:
I've got a basic question about signal processing within the discipline of management information systems. Today, some basic signals within management information systems, which people come across are electrical, light (fiber optics), and radio waves. Is it true that radio waves are a type of electromagnetic radiation or many types of electromagnetic waves, which can travel at the speed of light? Is that true? I didn't think it was possible for anything to travel at the speed of light. As I understand it, the speed of light in a vacuum, such as space or other, commonly denoted c, is a universal physical constant that's very important in many areas of physics. The speed of light, it travels (approximately 3.00×10⁸ m/s) or 186,000 Miles per second. Can anything you know of travel at the speed of light especially radio waves?

ANSWER:
When a physicist refers to "the speed of light" she is talking about the speed of electromagnetic radiation in a vacuum. Of all the possible wavelengths of light, visible light is but a tiny fraction.

QUESTION:
Whether light is a particle or wave? Which is right and why?

ANSWER:
Light is not a particle or a wave, it is a particle and a wave. This is called wave-particle duality. If you design an experiment to observe a particle, you will observe a particle; and, if you design an experiment to observe a wave, you will observe a wave.

QUESTION:
If you have a wind up toy car, meaning that when it starts you can wind it up to maximum and have a constant force pushing it forwards, how would you expect the distance in travels to change as you add mass to it?

ANSWER:
The spring will deliver a certain amount of energy. Giving the same amount of energy to a large mass and a small mass will result in the small mass going faster. The small mass will therefore go farther. This assumes that the wheels never slip.

QUESTION:
If all things with mass have a proportional amount of gravity and gravitational waves were recently observed to have been produced by 2 black holes converging, is it correct to deduce that all things with mass produce gravitational waves proportional to their mass?

ANSWER:
Yes. Any object with mass which accelerates should radiate gravitational waves. Wave to someone and you cause gravitational waves. However, gravity is the weakest force in nature so, for the the waves to have a big enough amplitude to be detectable, the masses must be extremely large (as in black holes).

QUESTION:
How fast do gravitational waves move? Is that rate constant?

ANSWER:
The speed of gravitational waves has never been measured. The speed at which a gravitational field propogates should be the same but it has never been measured either; this would determine, for example, if the sun suddenly disappeared how long it would be until the earth stopped orbiting. The theory of general relativity, which predicts gravitational waves, say that the speed of gravity should be the same as the speed of light. The recent observation of gravitational waves determined the distance to the source to be about 1.33 billion light years, but the uncertainty was very large, about 40%, so it really provides no measurement of the speed. Whatever the speed, there is no reason to think that it would not be constant everywhere in empty space.

QUESTION:
A car is moving forward through a road. Which part of the wheel of the car moves fast- the upper part or the lower part of the wheel?

ANSWER:
Inasmuch as the part of the tire in contact with the road is at rest, the answer should be obvious. You should be able to show that the top of the wheel is moving forward with twice the speed of the car.

QUESTION:
Has anybody ever done the double slit experiment on a very large scale. By large I mean physically where the electrons are shot at tow slits several feet wide separated by an even greater distance from a large distance away to see if the electrons themselves will still manage to make it through the slits at all? As long as the electrons are unobserved acting as a wave they should still continue to go throught the slits and produce a wave pattern behind the slits regardless of scale. However does this deteriorate at some point and break down to cause the electron to again act as a particle and simply hit the mass between the two slits and if so could this be somehow used as a physical means to measure the energy of the electron? Could an electron with a greater energy then once again act as a wave?

ANSWER:
As far as I know, nobody has ever done a true double-slit experiment with electrons at all. Given the extraordinarily short wavelength of an electron, the spacing between slits would have to be on the order of 10^-10 m and you just cannot make physical slits that close because that is like the distance between atoms. What is actually done is to shoot electrons at a single crystal and you get a double-slit like experiment. For diffraction to be observable the slit spacing and slit width must be small compared to the wavelength of the wave.

QUESTION:
I am doing an experiment on factors affecting the travel distance of a toy car from down a ramp and thought it would be a good idea to understand how a physicist thinks of things. My question is, how do you think weight of an object affects the distance it will travel after going down a ramp?

ANSWER:
The best discussion I have seen of the physics of pinewood derby races is this youtube video. You will see that what matters more than the added weight is where you put it. (Thanks to my son Andy for pointing me to this video; his son and my grandson Finn placed second in the Cub Scout pinewood derby last year on the strength of the tips here!)

QUESTION:
Hello, question in physics. Why would the speed of a rollercoaster with potential energy of 5MJ have less than that predicted for a perfect frictioless track at the bottom of the slope?

ANSWER:
A roller coaster has wheels and some of the kinetic energy would be in their rotation, not in the speed of the cars. Of course, if the track were perfectly frictionless the wheels would not go around at all. Maybe your problem means no energy lost to friction? Or maybe the bottom of the slope is not at the zero of potential energy?

QUESTION:
What will happen if we fill water in the tyres of our cars instead of air?Does it have any effect?.

ANSWER:
Three important issues I can think of. First, it would add a lot of weight to the vehicle which would hurt your gas economy. Second, the moment of inertial of a wheel would be larger requiring a larger torque having to be exerted for either acceleration or braking. Third, air is compressible and water is not and so the wheel would not have a cushioning effect on the ride.

QUESTION:
We say that time slows as we accelerate. Is time some existent entity that can slow? Or is it the accelerated object [or particles it is composed of] that somehow "ages" more slowly because of the energy applied to it? i.e., is it the change in mass, and not time that slows aging?

ANSWER:
You do not need to talk about acceleration. Time dilation says that clocks which have a high velocity relative to you run more slowly than your clocks. And, it is not that they look like they are running slower, in fact they may look to be running faster, but they actually do run slower; see faq page for links discussing this. The best way to see this, I think, is to consider the light clock discussed in an earlier answer. To understand the light clock, you must accept that the speed of light is the same for all observers; see the faq page. Also, to help you with understanding time dilation, read about the twin paradox.

QUESTION:
What would happen if you threw a baseball at the speed of sound?

ANSWER:
At such a large speed, air drag has an enormous effect on ball. To see the mathematical details see earlier answers for a lacross ball and a baseball. I will make the same assumption that I did in those answers that the amount by which the ball will fall will be very small compared to its horizontal distance and the speed acquired in the vertical direction will be very small compared to its horizontal speed. So I will ignore the small effect which the vertical motion will have on the horizontal motion. As discussed in the earlier answer, the horizontal distance x and speed v are given by v=v₀/(1+kt) and x=(v₀/k)ln(1+kt) where v₀ is the initial velocity and k is a constant determined by the mass, geometry, and initial velocity of the ball. For v₀=340 m/s (speed of sound) and a baseball (mass=0.15 kg, diameter=0.075 m) these become v=340/(1+2.8t) and x=121ln(1+2.8t) and are plotted below.

In one second the ball goes about 160 m and slows down to less than 100 m/s. During the same time, the ball will fall approximately 5 m and so, if launched horizontally from a height of 5 m will hit the ground in one second as shown below. Be sure to note the difference in vertical scales; an insert shows the trajectory drawn to scale. The small distance fallen is the justification for my approximations above.

QUESTION:
Hi! I am new to Quantum Mechanics and I am a little confuse about the representation of the spin. So if I have a 1/2 spin particle I usually put spin up in the positive direction of Z axis and spin down in the negative one and the probability of getting spin up or down is proportional to the cos^2 of half of the angle between spin and the certain direction. But if I have multiple options for the spin (e.g. in spin one i have spin 1,0,-1) how do I visualize them? I mean If i put upward and downward in the z axis, and at z=0 for spin of 0 I don't get the probability one if I apply the above formula.

ANSWER:
I do not know what you mean about the "…cos² of half of the angle…", but you seem to not know what, e.g., spin ½ means. When you say spin s, this means that the spin angular momentum quantum number is s; this means that the magnitude of that spin angular momentum is S=ħ√[s(s+1)] where ħ is the rationalized Planck constant. But, there is another quantum number, m_s=-s, -s+1, -s+2, …, s-1, s which is the quantum number for the z-component of S, S_z=m_sħ. Hence, the angle θ which the vector makes with the positive z-axis is θ=cos^-1(S_z/S). For s=½, θ=cos^-1(½/√(3/4))=54.7⁰ and θ=cos^-1(-½/√(3/4))=125.3⁰; similarly, for s=1, θ=45⁰, 90⁰, 135⁰.

QUESTION:
Does a uniformly charged ring rotating at constant angular velocity about its axis perpendicular to its plane radiate electromagnetic waves? Because the magnetic field produced is constant hence there is no changing magnetic field and hence no em waves.is it ?

ANSWER:
Any accelerated charge radiates. The radiation from a charge moving in a circle is called synchrotron radiation. The theory of synchrotron radiation is very difficult and not appropriate to do here. I can tell you, though, that the energy which the charges have is so low that the power radiated would be immeasurably small.

QUESTION:
I just saw a music video in which a group of performers appear to be in an aeroplane cabin in free-fall for 2 minutes and 40 seconds. The choreography is spectacular, and it appears to have been done in ONE take! How far would the plane have had to descend to maintain zero gravity conditions for 160 seconds?

ANSWER:
This video was shot in Russia in a reduced-gravity jet provided by S7 Airlines. Weightlessness is not achieved by falling but by following the parabolic path which a projectile would follow. Imagine that someone shot you from a cannon with a speed of 500 mph (the typical speed of a commercial jet). If there were no air drag, you would follow a parabolic path. The plane which contains you now follows that exact path and that is how you appear to be weightless. An alternative way would be for the plane to simply go straight down with an acceleration of 9.8 m/s², but I think you can see that this would not be a very safe situation. Anyhow, back to your question, I could not find reference to any such parabolic flight lasting longer than 30 s, so the video must have been shot in more than one take. Because this is such an unfamiliar environment, I cannot believe that, even with a lot of practice, it could be done in a single take without errors. And, if the plane were simply falling for 160 s, it would have to have started at an altitude of about 80 mi (far higher than a plane can actually fly) and would end with a speed of about 3600 mph (far faster than the plane could fly without disintegrating).

QUESTION (submitted by my daughter!):
Is this exciting for you?? I dont really understand it but I'm trying to...it seems cool!

ANSWER:
Observing gravitational waves has been a holy grail of physics since before I was an undergraduate (like 55 years ago, gasp!). So yes, it is pretty exciting news. This is actually just the first direct evidence for gravitational waves. Indirect evidence was found when observing a pair of stars orbiting each other and spiraling in toward a collision. The loss of energy turned out to be exactly equal to the amount of energy they would lose if radiating gravitational waves. A nobel prize in physics was awarded in 1993 for this observation. The animation below shows the predicted gravitational waves from two neutron stars orbiting each other.

QUESTION:
A bit off beat from what you are usually asked. Jews are not allowed to drive a car on the Sabbath. The problem is the internal combustion engine which creates a spark in the piston igniting the gasoline. In a Tesla automobile does the magnetic field around the motor ultimately create a spark driving the wheels?

ANSWER:
It is natural, if you have a little knowledge about electric motors, to think of the magnetic field driving the motor. In fact, a magnetic field never does work because of its very nature. Although the presence of magnetic fields is imperative, work is always done by electric fields they produce. Most electric motors have "brushes", contacts which slide on the rotating armature, and small sparks are inevitable when electrical contact is made or broken. The Tesla electric car, however, employs a type of motor called an induction motor which is brushless and is therefore probably sparkless.

QUESTION:
Hi this may be a hard question but If I wanted to run 2200 gpm through a 2500 foot run with 50 feet of fall what size pvc pipe would this take? This is all gravity.

ANSWER:
At first I just did a calculation with no corrections for viscosity or drag. I found the velocity had to be about 17 m/s and the diameter of a pvc pipe would have to be about 0.1 m≈4 in. But then I worried about the fact that a pipe that long is likely to have significant drag over its length. It is a pretty complicated engineering calculation and I was unfamiliar with many of the parameters. But I did find a web site which seems to have made it easy for me by including a calculator. Frankly, I have no idea what the roughness coefficient is, but it suggests a value of 150 for plastic. The result is below. As you can see, to get a flow rate of about 2200 gpm would require a pipe with diameter of about 10 in.

QUESTION:
What if Einstein's General relativity wrong? Would all physics need to be rewritten?

ANSWER:
Actually, the theory of general relativity, essentially the theory of gravity, has very little effect on most of physics. Furthermore, like all theories, it is an idealization and approximation and is already "wrong" at some level. It is also incomplete because there is no theory of quantum gravity and it does not address the issues of dark matter and dark energy.

QUESTION:
If I place a liquid filled container on a scale and suspend a mass with greater density than the liquid within the liquid and then release the mass, will the scale register the full weight of the mass while the mass is in motion (falling) as compared to when the mass has settled on the bottom? Will the scale read the same while the mass is accelerating as when it has achieved terminal velocity?

ANSWER:
Your second question indicates that you understand the answer will be different depending on whether the falling mass is accelerating or not. The figure shows that the weights of the fluid and the container will act down on the scale. Now look at the falling object. In addition to its weight there are two upward forces, the buoyant force B and the drag force D; these are both forces which the fluid exerts on the object. But Newton's third law says if the fluid exerts a force on the object, the object exerts an equal and opposite force on the fluid. Therefore the scale will read W_f+W_c+B+D. If the object is falling with constant speed, it is in equilibrium and so B+D=W_o and the scale reads the total weight of container, object, and fluid. If the object is accelerating down, B+D<W_o and the scale reads less than the total weight of container, object, and fluid. There is an earlier question similar to yours except the object is rising instead of sinking.

QUESTION:
Have scientists done experiment on what is the value of gravity below the earth surface as depth increases? if done pl. provide chart g vs depth.

ANSWER:
The deepest hole ever drilled is only about 12 km deep. I could not find any reference to attempts to measure g at various depths down this hole. Since the radius is about 6.4x10³ km, you would only expect about a 0.2% variation over that distance. There are models of the density of the earth, though, which have been determined by observing waves transmitted through the earth during earthquakes or nuclear bomb tests; these are believed to be a pretty good representation of the radial density and can be used to calculate g. The two figures above show the deduced density distribution and the calculated g.

Usually in introductory physics classes we talk about the earth as having constant density, but as you can see, that is far from true—the core is much more dense than the mantles and crust. If it were true, g would decrease linearly to zero inside the earth. Instead, it increases slightly first to around 10 m/s² and remains nearly constant until you are at a depth of around 2000 km. There is little likelihood that g will ever actually be measured deep inside the earth because the temperature increases greatly as you go deeper, already to near 200ºC at 12 km. However, if you have detailed information on density distribution, there is really no need to measure g.

QUESTION:
I'm so confused, my question is that if i hit the chair and chair change it's position from it's original position, then where is the reaction of chair and how according to third law of motion

ANSWER:
The force which you exert is on the chair. The reaction force is the force which the chair exerts on you. Only forces on the chair determine how the chair will move.

QUESTION:
I have a question thats driving me nuts. Say on a circle that spins, there are two points, one on the outer rim and one close to the rim. Both are in line with each other and are traveling at the same time but the distance of the point on the outside making one revolution is longer than the distance of the one on the inside. So the speed are the same but the distance of the inner point is shorter. The time they take to complete one rotation is also the same as the two points stay in line with each other. So speed is equal to distance divided by time as far as i know. Something have to give, don't understand it.

ANSWER:
The two points have the same angular speed but different translational speeds. For example, if the distance from the center is 1 m to the rim and 0.5 m to the other point, and the angular speed is 1 revolution/second, the distance traveled in one second is 2π m for the rim and π m for the second point; the second point has half the speed (π m/s) of the first (2π m/s). Don't go nuts!

QUESTION:
Scientists say that centrifugal force is the only thing stronger than the pull of a black hole. What i want to know is what would happen if somehow the material around a black hole stopped spinning around it what would happen and how might the spinning be stopped?

ANSWER:
Funny, I never heard scientists say this. All that I can imagine you are thinking about is that in the right circumstances an object might orbit around a black hole just like the earth orbits around the sun without falling in. But, the black hole exerts such a large force that even light can orbit at a certain radius which is outside the Schwartzchild radius; inside the orbiting light, nothing with mass could have a stable orbit. See an earlier answer. Obviously, if you stopped this orbiting, anything would fall into the black hole.

QUESTION:
As part of our business we bag wrap passengers bags / suitcases prior to flying at the major UK airports. We use and have used for many years a power pre stretch cast film – 17 micron nano with a 300% capability. Recent feedback from Heathrow airport suggests some of the passengers bags are sticking to the conveyer belts and are being miss-directed. I am being asked for the “coefficient of friction” for the film we are using. I have advised our supplier of this, they have sent through the data spec sheet but there is no mention of COF, on speaking with them they have never had this question raised before. Personally, I do not think this Is an issue with our film but more where customers themselves are wrapping their own bags with home use film. However, I need to provide proof that the film we are using does not have any adhesive properties. My question is – would the COF affect this and how do I get the actual information on the film?

ANSWER:
The force of friction f depends on only two things: what the surfaces which are sliding on each other are (conveyer material and your plastic film) and the force N which presses the two surfaces together; normally, on a level surface, the force N is simply the weight of the object (suitcase in your case). There are two kinds of COFs, kinetic and static. The kinetic coefficient, μ_k, allows you to determine the frictional force on objects which are sliding. In that case, f=μ_kN. The static coefficient, μ_s, allows you to determine how hard you have to push on the suitcase in order for it to start sliding; in this case f_max=μ_sN where f_maxis the greatest frictional force you can get. Since you are being asked to prove that it is not too "sticky", it is the static, not the kinetic, coefficient which you need; measuring μ_sis quite easy. The only problem is that μ_sdepends on the surfaces so you must have a piece of the material from which the conveyer belts are made to make a measurement. Once you have that, use it as an incline on which to place a wrapped suitcase. Slowly increase the slope of the incline until the suitcase just begins to slide. Your COF (μ_s) is equal to the tangent of the angle of the incline which (see diagram above) is simply μ_s=H/L.

QUESTION:
I have a hill approximately 25 degree slope and 300 feet in length. Want to install a tow rope, what size of motor would I need at 240V, 3/4" rope and a 3 to 1 gear box. The anticipated speed is approximately 4 feet per second and at most I will have only two people on the tow rope at one time.

ANSWER:
I can only give you a rough answer. I will calculate the power needed to lift two large people (total mass 200 kg) up the hill. I will work in SI units because that is the system in which the watt is the unit of power. Your speed is v=4 ft/s=1.22 m/s so the rate at which the load is rising is vsin25⁰=0.516 m/s. The rate at which the energy of this mass is increasing is mgv=1011 J/s≈1 kW. I guess I would throw in a safety factor to account for friction and other energy losses, so maybe a 2 kW motor would do you. The voltage (240) and gear box are not really relevant in determining the power. Disclaimer: I am not an engineer, so you should get a second opinion!

QUESTION:
A 1500lb 8'x8' box that is 3'6" tall is lifted at one of the four side so that the opposite side acts as a pivot on the ground (like a strong man flipping a giant tire in a a world strongest man competition). How much actual weight is being lifted? assuming that the weight distribution of the box is perfectly even.

ANSWER:
Well, that depends on how you lift it. Let's assume that you lift it so that you cause it to rotate with uniform speed. One way that you could accomplish this is to push in a direction perpendicular a line drawn from where you are pushing and the edge on the box remaining on the floor. Referring to the figure above, the equilibrium conditions are N+Fcosθ-W=0, f-Fsinθ=0, and ½LW-LFcosθ=½W-Fcosθ=0; here F is the force you exert, W is the weight, N is the force the floor exerts vertically, and f is the frictional force exerted on the floor. Solving these I find that N=½W, F=½W/cosθ, and f=½Wtanθ. I suspect that the case you are interested in is when you first lift it off the ground, θ=0. F=750 lb, half the total weight. Note that this analysis is valid as long as the floor is not too smooth, that is the box does not start sliding at some angle; the angle is less than θ=tan^-1(L/H) because at that angle the center of gravity is directly over the pivot side.

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.

QUESTION:
If a tennis ball and a football is thrown from a certain hieght then which ball will land first?

ANSWER:
I will assume that you do not mean an American football, rather a soccer ball. Also, it is not unambiguous what "…thrown from a certain height…" means. Imagine that we just drop each from some height. If air friction is ignored, they hit the ground simultaneously; this would approximately be the case if they were dropped from a few meters. However, air drag becomes more important as the speed increases. The drag force f may be approximated as f≈¼Av² where A is the cross sectional area (πR²) and v is the speed (you must use SI units). So, for a ball, f-mg=ma where m is the mass, a is the acceleration, and g is the acceleration due to gravity. Since f gets bigger as v gets bigger, eventually f=mg and the ball will stop accelerating and fall with a constant speed. The velocity at which this happens is called the terminal velocity v_t=√[4mg/(πR²)]. The mass and radius of a tennis ball are m=0.059 kg and R=0.069 m; the mass and radius of a soccer ball are m=0.43 kg and R=0.22 m. So, I find v_t^tennis≈50 m/s for the tennis ball and v_t^soccer≈11 m/s for the soccer ball. The tennis ball will easily win the race because it continues accelerating long after the soccer ball stopped accelerating.

QUESTION:
I love hearing about new discoveries from particle accelerators, but one aspect of them confuses me. If we accelerate particles to 99% of the speed of light, shouldn't that make the matter appear to go slower to us, a la the twins paradox? Why don't these particles approach infinite mass and compress time from our perspective?

ANSWER:
Stable particles, those usually accelerated, do not carry clocks with them; but, if they did, those clocks would run slowly compared to yours. Some particles do carry clocks, those which are unstable. Suppose that the average lifetime of some particle were 1 s and its speed was 0.99c. You would observe it to live for 1/√(1-.99²)=7.1 s. The accelerated particles do indeed approach infinite mass but they have a really long way to go to get there even in the most powerful accelerators; see an earlier answer.

QUESTION:
My question relates to my own voluntary time involved with the rope/rigging community. It is not of any commercial interest to me, but the answer to my question has eluded me for a long time despite research and attempts to calculate it ! If you could at least explain to me the correct method of attempting to work our the answer I would greatly appreciate it. Question background - horizontal restraint line. A 7.6 m long horizontal rope is anchored at both ends. It is 2m horizontally above ground level (AGL). It is pre-tensioned to 50kg. A 20kg mass is suspended vertical above the horizontal rope and attached to the rope by a leash of 0.7 m length. The object/leash attachment point is mid-way along the 7.6 m horizontal rope i.e. 3.8 m from either anchor. The 20 kg object is then dropped. At the time of peak impact force, the sag in the horizontal tension line is 1.0 m. Despite my best efforts I do not seem to be able to calculate theoretical (i) peak impact forces on the anchors (ii) peak impact force on the object. I do however have peak impact force load cells that have recorded average (i) peak impact forces on the anchor 220 kg and (ii) peak impact force on the object 115 kg. Is it possible for you to detail the correct formula/method for at least working the theoretical approach to calculating these forces ?

QUERY:
It would be helpful if you could tell me the sag and forces when the mass is hanging at rest on its leash. I take it that the 115 kg measurement is the tension in the leash.

REPLY:

ANSWER:
Let me first try to understand the data for the "loads" (essentially the tension in the rope, T, and the tension in the leash, F) in terms of simple physics; then I will try to generalize it. Normally, physicists do not like to use kg to measure force, but I will go ahead and do that here. Focusing on the point where the leash is tied to the rope, the force which the rope pulls up is 2Tsinθ=F. In the case of the hanging mass, sinθ=s/√(L²+s²)=0.41/√(3.8²+0.41²)=0.11, so F=20=0.22T; thus T=91 kg which is in reasonable agreement with the measured value of 100 kg. In the case of the maximum s, sinθ=0.25 and F=0.51T. Taking the measured value of F=115 kg, T=225 kg, again in good agreement with the measured value of T=220 kg.

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=√(L²+s²)-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.

FOLLOWUP QUESTION:
If possible, without the aid of the load cell values which I get by practically performing the drop test. How would you derive the theoretical values for the tension/force at the both the anchor and on the body in the drop test scenario. I understand how to calculate forces etc for the system at rest, but it is the dynamic falling system, deceleration etc that I am unable to clarify. To summarise, if you have the load/object, placed at any height above or below the horizontal line (depending on the leash length), without loading the horizontal line, and then you drop the object, what would be the correct formula method to determine the approximate theoretical peak impact forces on the anchors and body, without the luxury of any load cell info. when it comes to rest after being decelerated by the horizontal line.

ANSWER:
You apparently did not understand my answer, particularly the second and third paragraphs. The first paragraph shows that the data are consistent with the simple model for the tensions in the ropes. I found these very encouraging. From the data for the dynamic point you could, of course, calculate the acceleration of the weight at that point. 20a=(115-20)g so a=4.75g=46.6 m/s².

Then I tried to model the rope as a simple spring, the tension is proportional to the stretch. I failed to do so in both attempts. You do not know the force which this rope will exert given a certain amount of stretch. Physics may be powerful, but you cannot do any predictive calculation if you are ignorant of the force. I have two data points but that is not enough if the force is not linear. The last paragraph suggests the only hope for having a predictive analysis: you need to measure the tension in the rope as a function of what you call sag. To do this you need to use many weights, say at 2 kg steps, and measure load#1 and sag. This would give me the information necessary to do predictive calculations. Life in the real world is not always simple and analytical. It occurs to me you do not really need to hang varying weights because you have the load#2 device; just pull down on it until it reads 2 kg, 4 kg, 6 kg, etc.

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.

QUESTION:
A conductor, even though it is carrying a current, has zero net charge. why then does a magnetic field exert a force on it?

ANSWER:
Magnetic fields exert forces on moving charges. In the conductor, negative charges are moving but positive charges are not. A magnetic field exerts no force on charges at rest.

QUESTION:
Do all five balls on a Newton's Cradle have to weigh the exact same amount in order for it work correctly?

ANSWER:
No. The balls on the ends need to be the same mass. However, you could no longer do the demonstration where if you send 2, 3, or 4 balls in that there would be 2, 3, or 4 balls out.

QUESTION:
If F=Q1.Q1/R^2, does it mean that two unlike charges touching each other have infinite forces of attraction between each other? If so, how are we able to separate 2 oppositely charges objects which are stuck together with our bare hands?

ANSWER:
As often happens, you are applying an equation (Coulomb's law) without asking whether it applies. This law is true for two point charges or for two spherically-symmetric charge distributions; in the second case, R is the distance between the centers of the spheres. So, if you have two charged insulating spheres of radii 1 cm and 2 cm which are touching, R=3 cm.

QUESTION:
What is inertia? From what I understand, the inertia for a sphere is: I = (2/5)MR^2. Why is the radius included? Why is the volume of the sphere related to the resistance to change motion? As long as any 2 objects have the same mass, why should they have different inertias?

ANSWER:
Inertia means the inherant ability of something to resist being accelerated. There are two kinds of acceleration, translational acceleration a (like a car speeding up or slowing down), and rotational acceleration α like the rotational speed of an engine's flywheel speeding up or slowing down. Inertia for resisting a is simply mass which you can see from Newton's second law, a=F/m. If you double the mass on which a given force acts, you halve the acceleration; mass is sometimes called inertial mass for this reason. For rotational acceleration, the rotational inertia depends not just on how much mass there is, it also depends on how it is distributed; for example, it requires much more effort to get a wheel with radius 1 m spinning than it does to spin a wheel with the same mass but a much smaller radius. The rotational form of Newton's second law is α=τ/I where τ is the torque applied to the object whose moment of inertia is I. In rotational physics torque plays the role of force and moment of inertia plays the role of mass.

QUESTION:
If friction acts perpendicular to the direction of motion, if i place an object against a wall and let go, since the object is falling downwards, will friction from the wall cause the object to move away from the wall? Also if I am standing still, gravity results in a downwards force so friction should act leftwards or rightwards. Which one is the correct one (assuming that i am not moving)?

ANSWER:
Friction does not act perpendicular to the direction of the motion but opposite it. When the surface is vertical there is no frictional force. This ignores irregularities in the wall and air drag. If you are standing on a horizontal surface and not moving, the frictional force on you from the floor is zero.

QUESTION:
I am an engineer working on the reconstruction of a traffic accident where it is alleged that a car traveling over a railroad crossing became airborne at a speed lower than the posted speed of the road. The information that I have available includes the type and make of the car and the geometry of the road. Nothing in the engineering literature that I can find addresses this issue. Despite the five quarters of physics that I took long ago, I am having trouble finding the information that I need to model this. Any suggestions?

ANSWER:
This is sort of a classic introductory physics problem. The idea is: when does an object lose contact with the surface on which it is moving, usually taken to be a segment of a circle. I will do an approximation that the shape is a circle and that the car is a point mass. You can then generalize to your case from there or else give me more information. In the sketch above, the radius of the circle is R, the weight of the car is mg, the angle specifying the current position of the car is θ, the force which is causing the car to move with some constant speed v is F, and the normal force of the road on the car is N. Note that although the car has a constant speed, it has a centripetal acceleration a=v²/R toward the center of the circle. Applying Newton's laws, F-mgcosθ=0 and mgsinθ-N=mv²/R. The first equation tells you what force you need to keep it moving at a constant speed, F=mgcosθ which is really not of interest to you; note that the force is in the direction of v on the way up and opposite on the way down since the cosine changes sign at 90⁰. The second equation tells you what N is for any position of the car, N=mgsinθ-mv²/R; note that if N is negative it points toward the center but the road cannot pull down, only push up, so the car could not stay on the road at that speed and angle. What is really of interest is under what conditions would N=0=gsinθ-v²/R or v²=Rgsinθ; this would tell you the speed (angle) at which a car with a particular angle (speed) would leave the road. Note that it is independent of the mass. Notice also that v²/Rg<1 because the sine function cannot be larger than 1.0. For example, for what speed will the car leave the road at 45⁰ if gR=300 (m/s)² (R≈30 m)? v=√(300x0.707)=14.6 m/s=32.7 mph. Or, at what angle will a car with speed 35 mph=15.6 m/s leave the road? θ=arcsin(15.6²/300)=54.2⁰.

One thing which occurs to me though is, since you know the car and railroad crossing, why don't you just do the experiment and drive it over the crossing?

QUESTION:
When i place an object in between my 2 palms, why does it not fall? My palms only supply horizontal force, where does the vertical force come from to hold the object between my palms without falling?

ANSWER:
You are wrong to say that your "…palms only supply a horizontal force…" When surfaces are in contact, the forces they exert on each other have components both perpendicular (normal) and parallel (friction) to the surfaces. It is the friction which holds the object. If the object were very heavy, say 100 lb, and your hands were greased, you would not be able to hold it up because the frictional forces would not be as large as the weight.

QUESTION:
Say there is a cylinder on a ramp and the friction force from the ramp cancels out the parallel component of gravity. Therefore, the cylinder should be in linear equilibrium. However, from the reference point of the center axis of the cylinder, there is a net torque exerted by the friction force. Additionally, there is also a net torque exerted by the gravitational force from the reference point of the point of contact of the cylinder and the ramp. Therefore, it is not in rotational equilibrium and should start to rotate, correct? How is this possible, because if the cylinder starts to roll how can it also be in linear equilibrium?

ANSWER:
There is a simple answer to your question: the frictional force is not equal to the component of the weight along the incline. Rather, -f+mgsinθ=ma where θ is the angle of the incline and a is the acceleration of the center of mass down the incline.

FOLLOWUP QUESTION:
Thank you very much for your response. However, I think you may have misunderstood my question. I was asking what would happen in a case where the frictional force is set to cancel out the parallel component of weight. It seems as if the center of mass cannot move, but the cylinder needs to rotate. Therefore, it would appear as if the only outcome of this situation would be a cylinder rotating in place on a ramp, which does not seem possible. I think that the cylinder would have to roll down the ramp, but I can't see how this would be consistent with linear equilibrium.

ANSWER:
I did not misunderstand your question. You cannot simply adjust the friction to be what you want it to be. You can, however, simulate what you want to happen by wrapping a string around the cylinder and pulling up on the string with a force mgsinθ where θ is the angle of the incline and m is the mass of the cylinder; imagine that the incline is smooth (frictionless). Now there will be a net torque about the center of mass of mgRsinθ=Iα where I is the moment of inertia and α is the angular acceleration of the cylinder. The cylinder will spin in place. Your hand will have an acceleration of a=mgR²sinθ/I; for a uniform solid cylinder with I=½mR², a=2gsinθ.

ADDED THOUGHT:
If the coefficient of kinetic friction is exactly equal to μ_k=tanθ, the cylinder will slide down the incline with constant speed because the frictional force will be f=μ_kN=(tanθ)(mgcosθ)=mgsinθ. So, if you start it sliding and not rolling, it will begin spinning about its center of mass because of the torque due to the friction and have an angular acceleration α=fR/I; it will continue sliding down the plane with constant speed, though.

FOLLOWUP QUESTION:
Thanks again but I am still a bit confused. It makes sense that the center of mass will move at a constant velocity while the cylinder is rolling, but how did it acquire that velocity in the first place if I start the cylinder at rest and not sliding as you wrote in the additional thought. In other words, what happens if the coefficient of static friction is equal to mgsinθ and the cylinder starts with no velocity of any kind?

ANSWER:
If you start the cylinder at rest on the incline and μ_k=tanθ (coefficient of static friction will be larger), the cylinder will roll without slipping. If you solve the dynamics for the cylinder rolling without slipping you will find that (see the figure above) f=(mgsinθ)/3 and a=(2gsinθ)/3 where f is the frictional force and a is the acceleration of the center of mass. Since f<mgsinθ, if you simply let it go, it will roll without slipping. So, if you want it to slide down with constant speed, you must give it a shove to start it slipping.

QUESTION:
Does mass of an object increase its mass exponentially as it approaches infinitely close to the speed of light?

ANSWER:
If df/dq=Aq for some function f(q), where A is a positive constant, f is said to be exponentially increasing with q. For your question, f is m and q is v. The expression for m(v) is m=m₀/√[1-v²/c²] where m₀ is the rest mass. Calculating the derivative, dm/dv=mv/[c²-v²]. Thus, although m increases without bound as v increases, it does not increase exponentially.

QUESTION:
Can sustaining enough angular momentum help us stay in a black hole for longer(near the event horizon)? If yes, how much energy would be required to sustain it for ten minutes.

ANSWER:
What do you mean by "in a black hole"? A black hole is a singularity so if you were "in" it you could not have any angular momentum. In any case, I am assuming that your idea is to orbit the black hole so as not to fall in. If you were able to do this, no energy would be required; the earth orbits the sun in a stable orbit with no energy input. However, it is not possible for any object to orbit a black hole anywhere near the Scwartzschild radius because the speed would be too large. At a radius of 1.5 the Schwartzschild radius, light would orbit a black hole in a circular orbit (see earlier answer). This is a result of general relativity. No stable orbits inside this exist and anything inside would fall into the black hole.

QUESTION:
Hi! I have been told that hypothetically speaking when a metal bar or something other object travels in space at the speed of light it shrinks. So my question is if it is correct and why yes or no?

ANSWER:
First of all, nothing can travel at the speed of light. However, the fact that moving lengths are shorter is not hypothetical, it is simple fact which has been verified by measurements. It is the result of the special theory of relativity. And, it is not because they appear to be shorter, they actually are shorter. You might look at an earlier answer about length contraction.

QUESTION:
We have material that generates electricity when exposed to light, or force. Why haven't not found one that does this when exposed to atomic radiation. Imagine almost permanent batteries.

ANSWER:
There are batteries which get their primary energy from radioactive decay. Atomic batteries are routinely used in heart pacemakers and low-power requirements in space probes.

QUESTION:
My question is about the theory of relativity and time dilation. It appears to only work in one direction and I'm not sure why. For example, take a person on earth and compare to a person in a rocket going near the speed of light. The person on earth observes the rocket man going near light speed and aging slower. The man in the rocket observes earth moving past him near the speed of light except the earth man ages quicker.

ANSWER:
Congratulations, you have discovered the twin paradox! It is true that the earth man's clock runs slow in the rocket man's frame. It also is true that the rocket man's clock runs slow in the earth man's frame. But to make a definitive comparison, they must somehow bring their clocks together to compare. In other words, the rocket man has to come back to compare his clock with the earth-bound clock. To get the full explanation of the twin paradox, look on the faq page.

QUESTION:
Can a ball bounce higher than the height it was dropped? I know that air resistance would slow it down but is it possible?

ANSWER:
Even if you bounced a ball in a vacuum where there would be no air resistance, it could not bounce higher than the height from which you dropped it. The only way to achieve this would be to add more energy to the ball, either throw it down rather than drop it or maybe have a little "kick" from its interaction with the floor.

QUESTION:
An ice-hockey player throws his stick on the ice. The stick translates and rotates. Before it stops, it always rotates and translates - never only rotates or translates. Why do both motions occur simultaneously like this?

ANSWER:
It actually depends on how it is thrown

FOLLOWUP QUESTION:
Suppose it isn't thrown intentionally in such a way so as to just make it spin or just make it translate.

ANSWER:
The motion of a rigid body moving in two dimensions (the ice, call it the x-y plane) may be broken into two components, motion of the center of mass (translation) and motion about the center of mass (rotation). There are, in your example, no forces in the x- or y-direction once the stick has been thrown if you neglect friction. If the force you throw it with is directed through the center of mass, it will not rotate because there is no torque about the center of mass to get it rotating. If the net force you throw it with is zero (you would need to use two hands to do this) it will not translate. If you just randomly grabbed it and threw it, there would be a net force which would result in the center of mass accelerating during the time you were throwing it; and there would be a net torque which would result in the stick having a rotational acceleration during the time you were throwing it. Once you let go, there is no force and no torque on the stick and so both linear and angular momentum would be conserved meaning it would continue moving the way it was moving when you released it.

QUESTION:
Suppose a container full of water is kept in a certain area. Let no external force be applied. Now the question is: Will there be an overall circulation of molecules of water? Or will the rest position of water perfectly 100% at rest?

ANSWER:
There is a difference between no "circulation of molecules" and "100% at rest." The molecules in a glass of water are never at rest, they are always moving around. However, in a glass which has been sitting undisturbed for a long time the average velocity is zero, there are always as many moving in any direction as there are moving in the opposite direction. However, you could have a fluid which has a zero average velocity of molecules but they could be circulating (like a whirlpool). If you start such a circulation, it will often last for a relatively long time but, due to viscosity and fluid drag forces, it will eventually die out.

QUESTION:
This is a brain teaser I have been having trouble with. Lets say a clock is moving towards me at a 50% of the speed of light. When the clock is moving towards me, do I observe its hands to be ticking slower, faster, or at the same rate compared to the rate of ticks that I observe when it's moving away from me?

ANSWER:
A clock will appear to run faster than yours when moving toward you. A clock will appear to run slower than yours when moving away from you. But that clock is actually running more slowly than yours in both directions (and not by the same factor as it appears to run when going away from you). I think you will understand why is you look at earlier answers to a particular faq on the faq page. Keep in mind that moving clocks run slow, they do not just appear to run slow; how things are and how things seem are often not the same.

QUESTION:
Many Science Fiction ships have for protection a shield that stops projectiles and saves the ship from a lethal impact. Such an example would be the energy shields possessed by the faction known as Covenant from Halo. So given the laws of Newton and the third law about there always being an equal force countering the force that was exerted first, wouldn't this just mean that whenever the energy shield takes a hit by a projectile, the force would ''travel'' from the shield to whatever generator generates the shield. Like the UNSC Frigates shoot a 600 ton slug at 30km/s. They're 30ft long and around 7ft wide. So they have a lot of force and momentum behind them. So wouldn't an impact like this just cause the shield generator to be violently thrown off its attachments and ''fly off'' through the compartments of the ship destroying a lot of the ship?

ANSWER:
"So they have a lot of force and momentum behind them." Yes, these projectiles have a lot of momentum, but saying that "…there is a lot of force…behind them…" has no meaning. The momentum a slug has is p=mv=(6x10⁵ kg)(3x10⁴ m/s)=1.8x10¹⁰ kg m/s. When this hits something, it will certainly exert a force, but the magnitude of that force will depend on how long it took to stop. I have no idea how big it is, but suppose it has a radius of 10 km from the ship; I will think of it as very flexible and suppose that it stretches inward just stopping before it hits the ship. I also will assume that the ship is much more massive than the slug; (elsewise how could a comparably-sized ship carry a bunch of them and fire them without huge recoil?) If it stops in 10 km=10⁴ m with uniform acceleration, I can apply simple kinematics, -3x10⁴=at and 10⁴=3x10⁴+½at², and find that t=1.5 s, the time for the slug to stop. The average force felt by the slug is (Newton's second law) the rate of change of the momentum, F=1.8x10¹⁰/1.5=1.2x10¹⁰ N. You are certainly right, this very large force will be felt by the ship because of Newton's third law. But suppose that the ship is 100,000 times more massive than the slug, 6x10¹⁰kg; in that case, the final velocity of the ship after being hit will be found from F=mv/t=6x10¹⁰v/1.5=1.2x10¹⁰ N, so v=0.3 m/s, not so bad! If the shield were very rigid, it would be catastrophe for the ship. I have never played these games but I expect the shield is shown as stopping the slug almost instantly.

A little should be said about the other end, launching the slug in the first place. In this case, unless the cannon has a 10 km barrel, the recoil force on the ship will be huge. You would be interested in similar Q&As along the same line as yours.

QUESTION:
If I were to drop two round balls of different mass under water would they both fall to the bottom at the same velocity or would one reach the bottom first?

ANSWER:
The forces on a ball are its own weight mg down; the buoyant force B up which would be equal in magnitude to the weight of an equal volume of water; and the drag force f up which would depend on the size the ball and its speed. The net force F would be F=-mg+B+f; as the ball went faster and faster, f would get bigger and bigger until eventually F=0 and the ball would go down with a constant speed called the terminal velocity. The larger the mass, the larger the terminal velocity. Without specifying the the sizes of the balls, your question cannot be answered. If they had identical sizes, the heavier ball would reach the bottom first because it would have a larger terminal velocity. Of course, if B>mg the ball would float!

QUESTION:
If you performed the double slit experiment in outer space and were observing the electron particles would you get a diffraction pattern or two rows of particles. Does the earth's atmosphere have any effect on the experiment?

ANSWER:
Any electron diffraction experiment is always performed in a vacuum. The range of the electrons in air is short enough that it would entirely ruin any experiment you tried to do. Electrons interact strongly with any atoms in their vicinity. In outer space you would have a vacuum so you would see a diffraction pattern.

QUESTION:
I'm trying to find a fast/easy way to test whether a sealed, consistently-dimensioned rectangular box is sufficiently "stable" for transport on a (small) 2-wheeled bicycle trailer. The box is pretty tall. If it's over-weighted and top-heavy, it'll flip the trailer around turns (which are sufficiently tight/quick). I figure there might be a quick, static "tip test" with a combination pull gauge, inclinometer and scale, but my math skills are primitive. Is there a simple way to ascertain whether, for a given object, a target stability threshold is met?

ANSWER:
There is an earlier answer about a bicycle making a turn. It would be helpful for you to read that first. The easiest way to do this problem of your trailer turning a curve is to introduce a fictitious centrifugal force which I will call C, pointed away from the center of the circle; the magnitude of this force will be mv²/R where m is the mass of the box plus trailer, v is its speed, and R is the radius of the curve. The picture above shows all the forces on the box plus trailer: W is the weight and the green x is the center of gravity (COG) of the box plus trailer; f₁ and f₂ are the frictional forces exerted by the road on the inside and outside wheels respectively; N₁ and N₂ are the normal forces exerted by the road on the inside and outside wheels respectively; the center of gravity of the box plus trailer is a distance H above the road and the wheel base is 2L (with the center of gravity halfway between the wheels). Newton's equations yield:

f₁+f₁=C for equilibrium of horizontal forces;
N₁+N₂=W for equilibrium of vertical forces;
CH+L(N₁-N₂)=0 for equilibrium of torques about the red x.

If you work this out, you find the normal forces which are indicative of the weight the wheels support: N₁=½(W-C(H/L)) and N₂=½(W+C(H/L)). A few things to note are:

the outer wheel supports more weight,
if C=0 (you are not turning), the inner and outer wheels each support half the weight,
at a high enough speed C will become so large that N₁=0 and if you go any faster you will tip over, and
if the road cannot provide enough friction you will skid before you will tip over.

Now we come to your question. You first want the maximum speed without tipping. Solving for v in the N₁=0 equation gives

v_max=√[RWL/(mH)]=√[RgL/H]

where g=9.8 m/s²=32 ft/s² is the acceleration due to gravity. For example, suppose that R=7 ft, L=17 in=1.42 ft, H=30 in=2.5 ft. Then v_max=√[7x32x1.42/(2.5)]=11.3 ft/s=7.7 mph.

Be sure to note that the assumptions of a level road (not banked) and wheels not slipping are used in my calculations. Also be sure to note that W is the weight of both box and trailer and 2L is the wheel base, not the box width.

One more thing is that you might not know how to find the COG of the trailer plus box. If the COG of the trailer is H_trailerabove the ground (probably close to the axle) and the COG of the box is H_box above the ground, then H=(H_boxxW_box+H_trailerxW_{_trailer})/W.

ADDED THOUGHT:
When just about to tip, all the weight is on the outer wheel and so N₂=W and f₂=μN₂=μW, where μ is the coefficient of static friction. If you work it out, the minimum value the μ must have to keep the trailer from skidding is μ_min=L/H. For the example worked out above, μ_min=0.57. For comparison, μ for rubber on dry asphalt is about 0.9, so the trailer would not skid.

QUESTION:
If a 20 pound rock and 100 pound rock or drop from 1000 feet which will hit the ground first.

ANSWER:
If air drag is neglected, they would hit simultaneously. If air drag is considered, it would depend on the geometry of the two objects. If they have identical shapes and sizes, the heavier rock would hit first; otherwise, you would need to know the shape and size of each to calculate the times to fall. See the faq page.

QUESTION:
A starship pilot wants to set her spaceship to light speed but the crew and passengers can only endure a force up to 1.2 times their weight. Assuming the pilot can maintain a constant rate of acceleration, what is the minimum time she can safely achieve light speed?

ANSWER:
This question completely ignores special relativity. It is impossible to go as fast as the speed of light. Furthermore, acceleration is not really a useful quantity in special relativity and you must use special relativity when speeds become comparable to the speed of light. I have earlier worked out the velocity of something which would correspond to occupants of your spaceship experiencing a force equal to their own weight due to the acceleration which I will adapt to your case later. (See the graph above.) First, though, I will work out the (incorrect) Newtonian calculation which is presumably what you want. The appropriate equation would be v=at where v=3x10⁸ m/s, a=1.2g=11.8 m/s², and t is the time to reach v; the solution is 2.5x10⁷ s=0.79 years. For the correct calculation, you cannot reach the speed of light; from the graph above (black curve), though, you can see that you would reach more than 99% of c when gt/c=3. To make this your problem, we simply replace g by 1.2g and solve for t. I find that t=7.7x10⁷ s=2.4 years.

QUESTION:
Why do wavefunctions need to be normalized?

ANSWER:
So that the absolute square of the wave function can be interpreted as a probability density. The integral of ψ*ψ over all space is then the probability of finding the particle somewhere which, of course, must be 1.

QUESTION:
Will a heavier ball roll down a small slope faster than a lighter ball? Or will a lighter ball roll down a small slope faster than a heavier ball?

ANSWER:
If both are solid balls of the same radius but of different masses, they will take equal times if air drag is neglected as might be appropriate for a "small slope". If they get going fast enough that air drag becomes important, the heavier ball will win the race.

QUESTION:
During refraction why does a light ray bend...and how we can account for conservation of momentum?

ANSWER:
This is not simple. See this article.

QUESTION:
The following situation occured with my son, would you mind sending me the formula or the resulting impact force? Car A 1430kg rear-ends car B 1143 kg. Car B (Automatic gear is on park) is shoved 2 meters (6.56168 feet) before stop. What is the impact force on Car B Tons?

ANSWER:
No way to calculate this without much more information. See the FAQ page for other impossible to calculate the force situations.

FOLLOWUP QUESTION:
Thank you very much for your reply. I went to your site faq and understand what you mean (speed + stop time), but on this site it gives me force on the car 7. Tones with just the weight of car one and 2 meters of course it refers to 1 car hitting a tree, but the force of impact must be close. Does it sound reasonable to you?

ANSWER:
Maybe I could make a very rough estimate if I knew whether the incoming car was skidding, if the pavement was dry asphalt, was there any notable compression of the cars (like a bumper moved in by 2 cm)? Would a speed of about 15 mph be reasonable?

FOLLOWUP QUESTION:
Thank you very much for tryng to help find the force of impact. It was a mass pile up 6 cars (cars 1 to 4 Totalled) 01 Car 1950kg (estimated speed 55/60mph) ->02 Car 1172kg no brakes->(((Car 03 A 1430kg no brakes ->Car 04 B 1143 kg in Park gear)))->C 05 1270Kg (+ 4 occupants +- 280Kg no brakes)-> Shoved 1 meter to Car 06 Irrelevant - To simplify the calculations I reduced it to Car A and Car B - It was on dry asphat. The incoming (A) car was shoved 1meter (neutral gear no brakes) before hitting B and Car B in Park was shoved 2 meters into Car C which was shoved 1 meter into Car D - B car rear bumper was compressed by 25cm (lower body frame steel bent 10cm) - The estimated speed 25 mph of impact (Car 03 A to Car 04 B)

ANSWER:
Good grief! If I understand things correctly, it really is impossible now to do any meaningful calculations. I had thought from your earlier question that car B was at rest after going 2 m but now there are far too many unknowns. With two cars there was a possibility that I could have done a very rough estimate, with this situation it is truly hopeless. There is no reasonable way to estimate car A’s speed or car B’s speed after 2 m.

FOLLOWUP QUESTION:
Yep. That's why I simplified it to 2 cars. According to the lady in car A 1430kg she thinks her car was travelling at about 40Km / (I would say 25 miles) and shoved car B 1143 kg for 2.5 meter until it stop. (Stop means B hit C but this is irrelavant to a rough estimate) Dry asphalt Car B 25 cm rear bumper compression Without making it complicated with the compounded forces from other cars etc, all I would like to have a rough idea is, what aproximatly was the force of impact exerted from Car A onto car B. Do you that is possible to roughly calculate the force of impact Tons?

ANSWER:
OK, I give in! It is not irrelevant that car B was moving when it hit car C, but we can make a rough estimate now that I have the speed of A and the compression of the damage on B without any information about after the collision. I will assume that the time of the collision (compression of 25 cm=0.25 m) is very short so that I can ignore friction of the road, the collision is perfectly inelastic (A and B stick together), the speed of car A was 40 km/hr=11 m/s, and the compression of car B is negligible. Because this is a very rough estimate, I will retain only two significant figures throughout. Then the speed v after the collision has finished of the cars may be estimated from momentum conservation: 1400x11=(1400+1100)v, so v=6.2 m/s. The average force on car B is the rate of change in its momentum F=(6.2x1100)/t=6800/t where t is the time of the collision. Since we know the acceleration a is 6.2/t, we can use the kinematic equation for distance to get the time: 0.25=½at²=3.1t, so the time of collision is t=0.081 s; therefore F=84,000 N=19,000 lb=8.5 ton.

QUESTION:
Why must the trajectory of the reduced-gravity airplane to be parabolic. Why not arc of the circle or anything else?

ANSWER:
The reason is that the path a projectile follows is a parabola. If you were in an airplane and it suddenly disappeared, you would follow a parabolic path determined by what your velocity was when the plane disappeared. So, if the plane flies to follow that path, you would be in free fall. For more information about this, see an earlier answer.

QUESTION:
I am writing a blog and had my friends to ask of what they would want to talk about.. I get this one. I have no clue with physics, so, I am asking. Explain why a photon particle- which is a very small bundle of energy and travels at the speed of light- seems to defy the laws of physics by never losing speed or velocity?

ANSWER:
There is no law of physics which says an object naturally loses velocity. To change the velocity of something you have to apply a force to it, it must interact with something. If you had a bowling ball which had no forces on it it would continue going with a constant speed in a straight line forever. This is just Newton's first law. However, Newtonian mechanics is not valid for a photon, but it behaves like any other particle when it experiences no interactions which would change it. There is one important difference—regardless of how it interacts with something else, it never speeds up nor slows down; the speed of light in a vacuum is a constant of nature. If you look on my faq page you will find links to discussions regarding why the speed is constant and why it has the value it does. When you throw a ball up in the air, it slows down; you throw it down from a tall building, it speeds up. Photons don't do that but they do change their energies by changing to a longer, redder (shorter, bluer) wavelength when going up (down) in a gravitational field.

QUESTION:
Gasoline contains 40 megajoules of energy per kilogram and gasoline trucks have around a hundred tons of it. So how does a gasoline truck exploding not produce an explosion similar to a small nuke?

ANSWER:
Since it is always a little ambigous what is meant by a ton, I did my own calculation using the volume of a tanker truck of about 10,000 gallons and the density of gasoline of about 2.7 kg/gallon. I got the total energy content of about 10¹²J. The energy of the Nagasaki bomb, a small bomb by today's standards, was about 10¹⁴ J, 100 times bigger. Two things to consider are:

The bomb number represents total energy delivered whereas I would guess that likely less than half the energy content of the gasoline would actually be delivered.
The time over which the energy is delivered is likely much longer for the gasoline explosion than the nuclear explosion. As an example, just to illustrate the importance, suppose the bomb exploded in 1 ms=10^-3 s and the tank in 1 s. Then the power delivered by each is 10⁸ GW for the bomb and 10³ GW for the tank. As a result the destructive power of the bomb would be much bigger.

QUESTION: Centrifugal force seems so much confusing to me. I read on the books that centripetal and centrifugal force never act on the same object (Halliday-Resnicks' "physics" and several of our intermediate text books) and centrifugal force never acts on the moving object. But in many cases (effects of earth's rotation on g, roller coaster's loop etc) the centrifugal force are said to act on the moving object (as our text books say!). So what's right and what centrifugal force actually is?

ANSWER:
Newton's laws are true only in what we call an inertial frame of reference. Any frame which is accelerating is not an inertial frame of reference. Here is an example: suppose you are in a car going 80 mph and the driver slams on the brakes. If you are not wearing your seatbelt, you suddenly experience an acceleration toward the windshield. But, there are no forces on you which suddenly impel you forward, so Newton's first law is false for you because you are accelerating in your frame even though there are no forces on you. Of course someone watching this from the roadside has no problem understanding what is happening: because there are no forces on you, you continue moving forward with a constant speed while the car is stopping. There is a very neat trick you can use to make Newton's laws work in an accelerating frame: you just add a force which is the negative of the acceleration of the frame times the mass of the object you are calculating; such a force does not really exist and we call it a fictitious force. So, for you in the accelerating car we add a force, let's call it the "impulsive force" in the forward direction, which is your mass times the magnitude of the car's acceleration. You can then say that you accelerate forward with the same acceleration as the car is accelerating backward because you are being pushed by the impulsive force.

Centrifugal force is a fictitious force. If you are on a merry-go-round you feel a force trying to push you off; there is no such force. But if you pretend there is, you can do use Newton's first law: there is a centrifugal force outward which is mv²/R and you are at rest; the force which is keeping you from accelerating outward is friction with the floor, so it must be equal to mv²/R. If someone on the outside analyzes the problem she will say that you have an acceleration v²/R inward so there must be a force (centripetal) inward to provide that acceleration, F=mv²/R which is the same frictional force we found on the merry-go-round. Clearly the centrifugal and centripetal forces in these two cases are acting on the same body—you—so your claim that texts say otherwise is strange. In 40 years of teaching I have never heard such a claim. It is true that both centripetal and centrifugal forces never act in the same solution to a problem; if you choose a person on the merry-go-round to analyze, you never have both a centripetal and centrifugal force acting on that person. Perhaps you misunderstood and that is what is meant by never acting on the same object.

You can learn more about fictitious forces in an earlier answer and the links it leads to.

QUESTION: Two balloons, one with He gas is attached to the lower part of car and one with ordinary air is attached to roof hanging downward. When brakes are applied in what direction both balloons will move?

ANSWER:
The helium balloon will move backward; see an earlier answer for an explanation. (Note that the earlier answer has acceleration rather than braking.) The air balloon will move forward because it feels no buoyant force and therefore its inertia will cause it to want to crash into the windshield.

QUESTION: If a man could travel with a near speed of light, what would happen to him if he'd run though an elephant. Would that man be demolished or could he survive that impact?

ANSWER:
Are you kidding? A man going 200 mph, far below the speed of light, would be instantly killed in this scenario. A 100 kg man going at 80% the speed of light would bring in about 5x10¹⁸ J of energy to the collision which is equivalent to about 10,000 Nagasaki atomic bombs. Not only would the man and elephant be obliterated, also anything within miles would be.

QUESTION:

What exactly is momentum and how is it different from force?
I understand that p = mv and F = ma but they seem so similar in application that I haven't fully wrapped my head around it.
Additionally, what does it mean for something to have momentum in the first place, and why must it be conserved?
How does light have momentum if it, by nature, has 0 mass (I presume E =mc^2 comes into play somewhere here)?
Finally, what is the significance in the fact that light does have momentum (if it does)?

ANSWER:
You have lots of questions, really. I have rearranged your question to delineate it into parts:

For your first question, just say that p=mv. My answer to #2 should clarify how force and linear momentum are related. Force and momentum have to be different because they are not even measured in the same units—momentum is mass*length/time and force is mass*length/time².
What is acceleration? It is rate of change of velocity, a=(v₂-v₁)/t where t is the time to change speed from v₁ to v₂. So one could write that F=(mv₂-mv₁)/t=(p₂-p₁)/t; so force may be thought of as the rate of change of momentum. Newton actually stated his second law this way, not as F=ma. It is the second law which is a fundamental law of physics, momentum is just defined because of its simple and natural relationship to the second law.
It doesn't "mean" anything for something to have momentum, it is just a definition. However, consider the second law if there is no force acting; then F=0=(p₂-p₁)/t. In other words, p₂=p₁which simply means that momentum does not change (is conserved). The condition for conservation of momentum of a system is that there be no external forces on it. For example; suppose you look at an isolated galaxy which has billions of stars in it all interacting with each other and there are negligible forces from the outside; if you sum up all the momenta of all the stars today and in 10 years from now, you would get the same answer even though the shape and orientation of the galaxy would change.
It turns out that if v is very large, comparable to the speed of light c, Newtonian mechanics is incorrect. (You could say that Newtonian mechanics is wrong but a superb approximation for low speeds.) If you say that p=mv it turns out that momentum is not conserved for an isolated system at very large speeds. However, since momentum conservation is such a powerful way to solve problems, we redefine momentum (in the theory of special relativity), to be p=mv/√(1-(v²/c²)), momentum is conserved again and we still have p≈mv for small v. It also turns out that, as a result of this new definition of p, we can write that E=√(p²c²+m²c⁴) where E is the energy of a particle of mass m. So if the particle is at rest, p=0 and E=mc²; if the particle has no mass, p=E/c.
No more significant than if a billiard ball has momentum—it just does.

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

QUESTION: I am a high school student interested in relativity and I recently read an article about relativity . The article stated that the "The combined speed of any object's motion through space and its motion through time is always precisely equal to the speed of light." For an object moving at 90% the speed of light, it should only be moving at 10% the speed of light through time. I assume that means that time should pass only at 10% the actual speed of time for the object (correct me if I am wrong). However based on the Lorenz factor, time passes at 0.43 the actual speed of time for the object. Why is this so? Furthermore when time dilation occurs it is only seen by someone else in a stationary frame of reference. In the moving object's frame of reference time does not slow down at all. Does this mean that the combined speed of the moving object's motion through space and time can be more than the speed of light in the moving object's frame of reference? Can the moving object go beyond the speed limit in its frame of reference?

ANSWER:
Look closely at the example given: "To get a fuller sense of what Einstein found, imagine that Bart has a skateboard with a maximum speed of 65 miles per hour. If he heads due north at top speed—reading, whistling, yawning, and occasionally glancing at the road—and then merges onto a highway pointing in a northeasterly direction, his speed in the northward direction will be less than 65 miles per hour. The reason is clear. Initially, all his speed was devoted to northward motion, but when he shifted direction some of that speed was diverted into eastward motion, leaving a little less for heading north." The speed in the northward direction will now be v_N=65cos45⁰=46 mph; his velocity in the eastward direction will be v_E=65sin45⁰=46 mph. His total velocity will be √(v_N²+v_E²)=65 mph, not (v_N+v_E)=92 mph. In relativity N and E now are x and t for the stationary observer and x' and t' for the position and time of the moving observer as seen by the stationary observer. The catch, though, (which I presume is why the author of your article avoided these details) is that time and space are slightly different from space (N) and space (E) in that to get the length of the vector you calculate the square root of the difference of the squares instead of the sum. So the speed through spacetime would be √(c²-v²). So, the stationary observer will see a "spacetime speed" of v_spacetime=√(c^2-v²) and the "space time distance traveled" by the moving observer as seen by the stationary observer would be d_spacetime=v_spacetimet' where t'=t/√(1-(v²/c²)). If you do your algebra you will see that d_spacetime=ct indicating that the stationary observer sees the moving observer having a speed of c through spacetime.

QUESTION:
Bodies of water bend with the curvature of the planet. How large would a body of water have to be in order to measure a difference of 1 inch from one end to the other.

ANSWER:
I often get questions like this. I have used this figure many times before. Here R=6.4x10⁶ m is the radius of the earth, d=1 in=0.024 m is your 1 inch, θ is the angle which subtends the arc length, call it s, you seek. From the triangle you can write cosθ=R/(R+d)=[1+(d/R)]^-1≈1-(d/R)+… where I have done a binomial expansion of [1+(d/R)]^-1because (d/R) is extremely small. Now, because θ is also very small, I can represent cosθ by the expansion cosθ≈1-½θ²+… so (d/R)≈½θ². Finally we can write that θ=s/R. If you now solve for s you will find s≈554 m.

QUESTION: Hey! So, somebody is standing on Earth, and they measure the distance from Earth to some star to be D. Now, they hop in a spaceship, and travel 90% the speed of light towards the star. How will they measure the distance between Earth and the star now? <D, D, or >D? I initially thought they would measure to be <D, because I've heard of how length contraction works. After I thought about it more, I changed my guess to be D, because I thought ONLY the spaceship would experience length contraction, but I could be wrong there (I'm not 100% familiar with the phenomenon of relativistic length). Now I'm stuck between D and <D, and I'd greatly appreciate if you could clear this up!

ANSWER:
Length contraction says that a length which is moving is shortened. From the perspective of the spaceship the distance between earth and the star is a length which is moving backwards relative to the the spaceship. It is therefore <D. A meter stick on the spaceship, however, will be seen by the passengers to be 1 m in length.

QUESTION: magnetic force on charge q moving with velocity v =qV x B if i observe this charge from a car moving with same speed and direction as that of q than it velocity as observed by me will be 0 so the force will be 0.i am not able to understand this dilemma at one time force non zero and at other time it is 0

ANSWER:
The problem is that the electric and magnetic fields in one frame of reference are not the same as in another moving frame. (This is special relativity.) In your case you first start with a magnetic field and zero electric field. Suppose that the magnetic field is in the y-direction, B=jB, E=0, and the velocity of q is in the x-direction, v=iv. Then the force would be F=kqvB in the z-direction. In the moving frame the new fields would be B'=jγB and E'=kγvB, where γ=1/√(1-(v/c)²). Note that E'=vxB'; therefore the force, as seen in the moving frame is F'=qE'=qvxB'=kγqvB, as you would expect. Note, however, that F'≠F , they differ by a factor of γ; this is because force is said to be not Lorentz invariant and it is not really a useful quantity in relativity.

QUESTION:
We can see things billions of light-years from Earth. This means that light photons have traveled for billions of years at the speed of light to reach us. Why don't those photons slow down or stop? Is there no energy consumed over that vast distance in order to keep that photon moving forward?

ANSWER:
It is a law of physics that photons in a vacuum can only exist moving with the speed of 3x10⁸ m/s. So, nothing has to "…keep that photon moving forward…" But, space is not really a vacuum and not 100% of photons make it over such vast distances. Some interact with the occasional molecule they might encounter and are changed or scattered. Some find themselves striking a star or planet and being removed from the stream coming from the source. Some are deflected by the strong gravity near galaxies. Some encounter a black hole and disappear altogether. Because space is so empty, though, the majority come through unchanged.

QUESTION:
The light emitted from a light bulb is thermal radiation. Probably the rushing electrons cause the atoms to vibrate and this vibration will cause photons to be emitted with a Gaussian distribution in the EM spectrum. Still i wonder. Why is the vibrating of the atoms causing EM radiation in the visible spectrum and not lets say microwave or x-ray part of the spectrum. Also, I still not sure where the photons pop into existence. Is it from all the electrons , in all shells and perhaps also the vibrating nucleus?

ANSWER:
All objects are always radiating and absorbing thermal radiation. It is electromagnetic radiation. Many objects are well-represented by black-body radiation which you might want to research a little bit. As you can see from the plot of black-body radiation at 5000 K above, all wavelengths are present, not just visible light. It may appear that there is no very short wavelength radiation, but at this temperature it is just too small to register on the graph. (Incidentally, this is not a gaussian curve.) Atoms or molecules in a solid or liquid have electric charge distributions (nuclei and the surrounding electron clouds), most importantly dipole moments, which oscillate in ways which depend on the temperature. An oscillating charge distribution looks like a tiny antenna and radiates the energy. These "antennas" are the source of the radiation. Historically, trying to understand the black-body radiation classically was what was the first indication of quantum physics. The oscillators must be quantized which means that only discrete packets of radiation are emitted—your photons.

Q&A OF THE WEEK, 2/5-2/11 2017

QUESTION:
I work in a high school where this question was posed by one of the pupils in a class I support. The question is this.... If you could attach a rope to a rocket, which would also be attached to earth, and sent it into space (out of our atmosphere) until the rope went taught and then cut the string. Would it stay in space or would it fall back to earth?

ANSWER:
This is one of those problems which I had fun with and I hope it will not be too exhaustive an answer. I will make the following assumptions:

the rocket always goes straight up;
the rocket stops moving vertically when the rope is taught;
the rope is cut the instant that the rocket stops;
fuel and the weight of the rope are not issues; and
the launch is from the equator. This makes things much simpler and I will briefly talk about a similar launch from some latitude at the end.

My view of this problem, therefore, is the same as if the rocket were on top of a very long stick vertically straight up and the stick is suddenly removed.

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.

So, if the rope happens to be 5.6 times larger than the radius of the earth, the rocket will remain (apparently) stationary above its launch point; it is actually going in a circular orbit with a period of 24 hours. (The animation above illustrates this, although not to the correct scale.) For any other L the orbit will be an elipse with the center of the earth being at the focus. Visualize.
If L>5.6R, the starting point of the rocket will be the perigee (closest point to the earth) of its orbit. Visualize.
If L<5.6R, the starting point of the rocket will be the apogee (farthest point from the earth) of its orbit. Visualize.
For L<5.6R, though, there will be some critical distance L=L_c when the perigee of that orbit is exactly equal to R; in that case the orbit will just skim the surface of the earth. After some laborious algebra I found that L_c≈3.7R, about 2.2 earth radii inside the geosynchronous orbit. Visualize.
For L<L_c, the rocket will crash back into the earth but not where it was launched from because it is a projectile which has a horizontal speed greater than that of the earth's surface. Visualize.
Finally, if the horizontal speed of the rocket ω(L+R) is greater than or equal the escape v_e=√[2MG/(R+L)], where M is the mass of the earth, the rocket will escape the earth and never come back. I calculated this to be when L=7.6R, two earth radii beyond the geosynchronous orbit. Visualize.

For any other latitude θ the speed of the satellite will be ω(L+R)sinθ. The resulting orbits, while all elliptical, are much more difficult to visualize and maybe we should save that for another day! However, the simplest launch of all would be from the north or south pole (θ=0⁰ or 180⁰) because it acquires no horizontal velocity (ω(L+R)sinθ=0) where the rocket would fall straight back down regardless of how high it went.

QUESTION:
Assume there are 2 black bodies, 1 with a mass double of the other. If both black bodies reach a temperature of 5000K, do both emit blue light as black body radiation, with the heavier body emitting a brighter light. Or does the heavier body emit white light while the lighter body emits blue light? Why?

ANSWER:
A black body does not radiate one color but rather a spectrum of colors. The spectrum for 5000 K is shown above. The most intense color of this spectrum is at 580 nm which is the green color shown to the right. The intensity of the light emitted does not depend on the mass, it depends on the surface area of the black body; if you double surface area, you double the amount of energy being radiated.

QUESTION:
What force causes your hands to warm up when you rub then together?

ANSWER:
Friction. But the energy to warm your hands comes from you. When you rub your hands together you are exerting a force so you are doing work. But the energy derived from your work does not speed up your hands, rather it is lost to friction and shows up as heat.

QUESTION:
I'm a science geek and an IT person who just found out that there are sixteen types of water ice. I've been googling the phenomena, and can only find discussions of the physical make-up at the atomic level. Can you help me with some discussion of the various forms of ice at a macro level? Like, what does it look like, how does it act, etc. Everything I've found is in science speak, which I really can't envision myself.

ANSWER:
Refer to the figure above which came from the Wikepedia article on ice. You can get a clearer picture there. The thing to notice is that all the ices except for ice I and ice XI occur at extremely high pressures—1000 atmospheres or higher. So, you cannot really "look" at them and say what they look like since they are formed in a containment vessel of some sort. Ice XI occurs at atmospheric pressure but only at very low temperatures. If you look at the Phases section of the article, you will find links to separate articles for all 16 forms of ice where you can get information about their properties. Although the terminology for the crystal structure is pretty "science speak" as you say, you will find usually pretty comprehensible pictures of these crystal structures. I think you would find these articles about as understandable as you will find.

QUESTION:
If a good absorber is a good emmiter also then why don't we prefer to wear black clothes in summer or in hot conditions?

ANSWER:
Because we are better off if we do not absorb any heat in the first place. We wear a poor absorber.

QUESTION:
Can electric field and magnetic field collide to each other? If yes what will happen when they collide to each other.

ANSWER:
Two fields "colliding" really does not have any meaning. Certainly an electric field and a magnetic field can exist at the same time and location. If either is changing at that time, it will cause the other to change.

QUESTION:
How much physical energy is exerted in a "typical" gym and how does that compare to the amount of energy necessary to operate that same facility. I am wondering if there have been any investigations into redirecting this type of activity towards more productive ends.

ANSWER:
A person doing vigorous exercize may expend about 10 kCal/hr≈2.8 W. The gym would probably consume a few kilowatts so you would probably need a couple of thousand folks to generate that much power. I don't think you are on to something here!

QUESTION:
If vision works when we receive the reflected wavelength of white light from a particular object. Also we knew angle of reflection depends upon incidence angle, how so many people could possibly see the same object unless they stood in the path of reflected wavelength.

ANSWER:
Because the surface is not smooth like a mirror, the angle of reflection is not equal to the angle of incidence.

QUESTION:
Why is it easy to accelerate tiny sub-atomic particles like electrons close to the speed of light compared to large macroscopic objects like spaceships?

ANSWER:
There are lots of ways you could explain this. For example, Newton's second law states that a=F/m where a is the acceleration of an object of mass m when acted on by a force F. Suppose you exert a 1 N force on an electron (m≈10^-30 kg) and a space ship (m≈10⁶ kg). Then the acceleration of the electron (initially) is 10³⁰ m/s² and that of the space ship is 10^-6 m/s².

QUESTION:
If atoms at room temperature move around at roughly the speed of a jet airplane, how fast do atoms that make up a jet airplane move when said airplane is moving (at room temperature)?

ANSWER:
I assume you are talking about molecules in a gas. It is much more complicated to talk about molecular speeds in solids. So let's talk about the air in the airplane cabin. The speeds are distributed from very low to very high, but the most probable speed would be around 1000 mph, about twice the speed of a commercial jet. But, that does not mean that the most probable speed of molecules in a jet flying by would be around 1500 mph because the molecules measured inside the airplane would be moving in all directions so the result of adding 500 mph to all the molecules going 1000 mph would yield anything from 500 mph to 1500 mph. Measured inside the airplane, the average velocity of all the molecules would be zero because for every molecule going in one direction with some speed v there is always another going in the opposite direction with speed v. The average velocity of the air in the airplane as measured by an observer on the ground would be 500 mph. It could be thought of as like a wind in which there is a net flow of air in the direction the wind is blowing.

QUESTION:
Hi, my question relates to time dilation and the twin paradox/moving clock theory. A worked example of time dilation with a moving clock (one clock taking a relative tonitself 1 hour round trip at near the speed of light and one clock staying as a reference) shows it seemingly loosing time due to time dilation. Wouldn't this only seem to be correct until the observation of the moving clock ended? For example, the observation point which watched the moving clock is getting its informatiom from the trail of light from the journey, which shows slower time due to the speed at which it is travelling (as light is effectively taking longer to reach the observation point). However, when the travelling clock lands back on earth after an hour (relative to itself) and you were to end the observation of its journey and simply look at its landing spot, it should be there. Not only that but it should be showing the same time as the reference clock. So an hour long trip at the speed of light would only seem to take longer if you only seen it through your observation of its journey. Which would make any worked examples true up until observation ended. If this would in fact be true, wouldnt it be a more suitable explanation that our time is relative to light and that time dilation is no more than a light-trick as we only see what we see when light reaches us? Or that our time is only relative to the centre of our universe?

ANSWER:
I think you can clear up your confusion by reading the faqs for the twin paradox and for how fast clocks appear to run.

QUESTION:
I am trying to understand how to predict stopping distances at other speeds if I know a stopping distance at one speed. Let's say a car going 25 mph with studded tires on packed powder stops in 20 meters. (I got that from a study on various types of tires on various surfaces- it's realistic). What would be a good approach to finding the the stopping distance of the same car with the same tires on the same surface, if it is going 20 mph? We do not know the mass of the car, but perhaps it is accurate to create an acceleration constant that will apply for other speeds?
a=v/t and v=d/t
so therefore by combining these we get
a=v^2/d
25mph is 11.176 m/s deceleration constant:
a = (11.176m/s)^2 / 20m = 6.245 m/s^2
then can we apply the deceleration constant to the other question? What will be the stopping distance at 20 mph? 20mph = 8.94 m/s; stopping distance is 12.8m Is this valid? My applications include predicting future stopping distances at other speeds if one has experienced a stopping distance at a particular speed, and also, finding a reasonable speed for a traffic situation near where I live, where there is limited visibility.

ANSWER:
Your final answer of 12.8 m is right, but your method is wrong. Your answer for the acceleration (and stopping time if you had calculated it) is wrong. Your principle error was that d/t is the average velocity, not the starting velocity; the velocity is changing the whole time. In this case, uniform acceleration, the average velocity is ½v₀so your acceleration is wrong by a factor of 2; a=-3.12 m/s². The correct expression for the stopping distance is d=½v₀²/a where v₀ is the initial velocity and a is the magnitude of the acceleration. The force which stops the car depends only on the nature of the rubbing surfaces and the weight of the car. But the weight of the car cancels out when you calculate the acceleration, so the acceleration is independent of both the weight and the initial speed. Therefore your guess that you can use one datum to determine the acceleration and then use that for all other speeds is correct; the added bonus is that it is also independent of how heavy the car is as long as it is the same road surface with the same tires.

QUESTION:
An interesting question came up amongst my friends and me resulting in differences of opinion that I hope you can settle. Question: If an object is dropped from a fixed position while travelling in a moving jetliner(say 300+ MPH) would the same object dropped from the same position while not moving land on precisely the same spot? ie not a micron different. Of course all variables being equal.

ANSWER:
Absolutely they would not land in the same place. The way you describe it is a little confusing: "dropped from a fixed position while travelling" is kind of ambiguous isn't it? I assume that it is like dropping a bomb from a plane; the plane and bomb are both moving forward. Neglecting air drag, the bomb will land directly under the plane if the plane continues moving with constant speed in a straight line. The object dropped from rest exactly where the plane was when it dropped the bomb will land directly below the drop point. If air drag is taken into account, the bomb will land somewhat behind where the plane is. See an earlier answer.

FOLLOWUP QUESTION:
The scenario involves an object dropped from a fixed point inside a moving airliner and hitting a point under it but still inside the aircraft. No bomb dropping. Then a repeat of this operation while the airliner is sitting still. Does it hit the same spot in each case?

ANSWER:
Sorry, I misunderstood. In an elementary physics class, it is assumed that the only force which acts on the object is its own weight, a force toward the center of the earth. And actually that is true. However, it is also assumed (although not always stated) that the plane is moving in a straight line with constant speed; that is not true because if the plane maintains a constant altitude it is actually moving in a circular path above a spherical earth and the earth itself is spinning on its axis. If the observer is in the airplane (which she is), Newton's laws are not true because in either case the experiment is being done in an accelerating frame. However, there is a neat trick to make Newton's laws work in an accelerating frame by introducing fictitious forces; the best known fictitious force is the centrifugal force. The force (real+fictitious) which acts on the falling object depend on the direction the plane is flying, its speed, and its latitude. These are different for your two experiments and therefore the two objects would not land precisely at the same spot. To read more detail on these fictitious forces, see an earlier answer.

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.

QUESTION:
An object travels at a very high speed relative to some planet. Its velocity is normal to the gravitational field lines of the planet at a particular instant. At that instant, does the gravitational attraction between the two bodies increase due to the object's relative motion or does it stay the same as if the object was not moving at all?

ANSWER:
The object moving at high speed has an increased mass and therefore feels an increased gravitational force.

QUESTION:
What causes the curvature of spacetime/gravity to be so strong around black holes? Is it due to their density (mass density/energy density)? If not what is it due to? And why dont other extremely dense stars like the sun exhibit properties like black holes

ANSWER:
For any spherically symmetric mass distribution the gravitational field is proportional to the mass m and inversely proportional to the square of the distance from the center 1/r². Because a black hole is a singularity (zero size) it is possible to get very close to the center where the field will be huge. Also, many black holes are very much more massive than a typical star. A regular star might have a very big mass, but it is spread out over a very large volume; inside the star itself the field decreases linearly to zero at the center.

QUESTION:
If you were able to drive your car with forward acceleration in outer space and you rolled the window down and stuck out your arm, would it be pushed back like it would here on earth? This isn't a homework or test question, it is a question that came up in a fun discussion at work the other day...

ANSWER:
You had better have your space suit on because outside the car there is no air in outer space. And the fact is that the main thing which causes your arm to be pushed back here on earth, particularly for high speeds, is the air drag, the apparent wind you perceive to be blowing from in front of you. That is totally absent in space. However, you have specified that the car is accelerating. When you accelerate forward, it feels as if there is a force pushing you (and your extended arm) backward. There is no such force (which is why we call it a fictitious force), but it feels like it because your accelerating arm needs a forward force (your shoulder provides it) to accelerate it and your brain interprets this as there is something pushing your arm backward which your shoulder counters. Here on earth, unless your acceleration is very large and/or your speed is very small, this force is pretty small compared to the air drag. So the answer to your question is that your arm will be pushed back but not "…like it would here on earth…"

QUESTION:
I was wondering what would happen if an object, say a cube, were travelling toward you at a relativistically interesting speed. Its motion is strictly in the x component in an XYZ coordinate plane. Would you see a length contraction in the dimensions of the cube in all three dimensions or just the x component? Also I was curious about the effects of the length contraction on the observed density of the cube. Observed mass goes up as velocity does, and you observe a length contraction in one or more dimensions (depending on previous answer). Therefore the density would go way up. Is this a proper interpretation of relativity?

ANSWER:
As I have said before, saying that mass increases with speed is a matter of taste. It turns out that in special relativity the linear momentum p is not conserved if momentum is defined as p=m₀v where m₀ is the inertial mass at rest. If you redefine p to be p=m₀v/√(1-(v²/c²)), momentum of an isolated system is constant. You can interpret this as simply a redefinition of p or that the mass has increased so that m=m₀/√(1-(v²/c²)). That said, let's answer your question assuming that mass increases.

Length contraction occurs only along the direction of motion. Since mass increases by a factor 1/√(1-(v²/c²))) and volume decreases by a factor of 1/√(1-(v²/c²)), density increases by a factor 1/(1-(v²/c²)).

QUESTION:
if a cat fell from 100 feet how fast was the cat in mph falling when it hit the ground

ANSWER:
The terminal velocity of a cat is about 60 mph and that speed is achieved after falling about 50 ft. Therefore falling from 100 ft the speed would be about 60 mph. See earlier answers (1 and 2) for more details. Keep in mind that all cats are not the same size, weight, or fuzziness, so this can only be an approximate answer. It is interesting that about 10% of cats falling from 5 stories are killed but fewer from higher stories because, once reaching terminal velocity, they can relax, spread out (to slow down) and get ready to hit.

QUESTION:
E=MC^2. It implies that energy and mass are exchangeable and consistently so. My problem comes with c^2. How is this possible? c= the speed of light, if I am not mistaken and one of the fundamental laws of the universe is its speed limit, that of c. So how is c^2 possible?

ANSWER:
Consider the following: you have a few hundred square tiles, and they have varying sizes. I will use d to specify the length of the side on any particular tile. However, due to some fundamental law of the universe, none of these tiles is allowed to have to have d>d_max=4 cm. Now suppose that I write an equation for the area of any of these tiles, A=d². Now one of these tiles has a size of d=3 cm (not violating our fundamental law) and so its area is A=9 cm². But wait, how can that be? Since d² is more than twice as large as d_max, this must violate our fundamental law of the universe, right? Of course not. The area of a square is simply a different thing from the length of the side of a square and does not even have the same units (cm vs. cm²).

QUESTION:
I am a weight lifter and want to calculate how much "work" I do when performing the deadlift. I did the math and want to ensure my concepts and calculations are correct, as it has been a while since I took physics. Here is what I did: I am lifting 225lb (1000.85 newton) a distance of 0.613m. So the work (in joules) = 1000.85n X 0.613m = 613 joules. I did 32 repetitions of this in total, so the total work would be 19,632.64 joules. So, if I convert that to watt hours, it is about 5.45 watt hours. Does this mean that to do that same amount of work, 5 watts for 1 hour would be required? Or 20 watts for 15 minutes? Or 300 watts for 1 minute? Did I use the right calculations? Ultimately I want to compare several of my lifts to see how much total work I performed (for example 10 sets of 5 reps on Bench Press at 150lbs VS 6 sets of 7 reps on Deadlift at 225 lbs)...

ANSWER:
All your calculations are correct (allowing rough rounding in "__ watts for __"). The problem is that simple physics like this is not really a good approximation to how biological systems are actually working. If you simply hold a weight with a horizontal arm, physics says no work is being done, but you and I know that sugar is being burnt to provide energy to hold the weight there. Likewise, physics says you are doing negative work when you lower the weight (cancelling out the positive work you did lifting it) but I suspect this lowering also requires you to actually use energy, that is do positive work. The explanation of how work is being done is given in an earlier answer.

QUESTION:
This is quite for a long time that I'm thinking about a question in Galilean kinematics which brings me to a paradox, which I cannot solve myself. Consider a body with a mass m and velocity V₁ in one reference system and V₁ + V' in another one. Now let's say this body starts slow down and its velocity change is ΔV in both systems. Simple algebra shows that the kinetic energy change in the 1st system will be ΔE₁ = mΔV/2(2V₁+ΔV) and in the 2nd system ΔE₂ = (mΔV/2)(2V₁+ΔV+2V') i.e. its different in different reference systems. Now if we imagine that this slow down and kinetic energy change was due to the friction it follows that the amount of heat released is different for different reference systems! Nonsence! It should be a mistake somewhere.

ANSWER:
The first thing to appreciate is that energy is not invariant under Galilean transformation. This is easy to appreciate since a mass with speed v has kinetic energy K=½mv² but zero kinetic energy in a frame moving with it. Therefore, the change in kinetic energy will be different in different frames. However, the work-energy theorem (ΔK=W where W is work done by external forces) will be true in all frames, just not the same numbers in all frames.

In the first frame in your example, where m has speed v₁, the floor is, I assume, at rest. To make the algebra simpler and the explanation easier to grasp, I am going to let m come to a complete stop, Δv=-v₁. Then, when m moves some distance s and the frictional force is f, the work done is W=-fs and so ΔK=-½mv₁²=-fs. You can deduce that the heat generated is ½mv₁²; all other frames will see this amount of heat. Now, in a frame with speed v' moving in the same direction as v, you can easily show that ΔK'=-½mv₁²+mvv'. You already know that -½mv₁² is the heat and mvv' is simply the amount that ΔK is different in this frame. If you calculate the work done in this frame, it is different because the distance m slides is different; but, if you do that you will find, as expected that
W=-½mv₁²+mvv'.

One more special case. Suppose v'=½v. 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=-½mv₁²+½mv₁². 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.

QUESTION:
I'm trying to come up with an equation at work on how to figure out how much a parent roll of toilet paper weighs just by measuring the distance from the core to the outside of the roll without taking it out of the machine. For instance 1inch of paper on the out of the roll will not weigh the same as 1 inch of paper from the core. The measurements I know are the weight of the parent roll when we start using it, the core diameter, core weight, and distance from the edge of the core to the outside edge of the roll before it's started. As it is running we can measure the current distance from the core edge to the outside of the roll while it is running. My question is how do I calculate a equation to determine the weight of the roll at any given time based on the distance from the edge of the core?

ANSWER:
Not really physics, but I can do it. Notation:

W₁ is the weight of the parent,
W₂ is the weight of the core,
W₃=W₁-W₂ is the weight of the paper in the parent,
R₁ is the radius of the parent,
R₂ is the radius of the core,
L is the length of the roll,
V₁=πL(R₁²-R₂²) is the volume of the paper on the parent,
W is the weight of the paper on the partially used roll,
R is the radius of the partially used roll, and
V=πL(R²-R₂²) is the volume of the paper on the partially used roll.

You may be interested only in the final answer, but I will outline the solution for anyone interested.

The density of the paper is ρ=W₃/V₁=W/V.
Therefore, W=W₃(V/V₁)=W₃(R²-R₂²)/(R₁²-R₂²).
What you want, is the weight of the whole partially used roll, W+W₂=W₂+W₃(R²-R₂²)/(R₁²-R₂²).

So, just plug in your measured values and you will have the weight of the partially used roll. Incidentally, you say "distance from the edge of the core". If this (call it D) is more conveniently measured you can replace R in the equation above by (D+R₂) which results in
W+W₂=W₂+W₃(D²+2DR₂)/(R₁²-R₂²).

QUESTION:
I'm a writer. I'm working on a story. It's sold actually, but they want some changes and I'm trying to figure out the science..

IDEA: A distant planet picks up broadcasts from Earth, so the inhabitants have some knowledge of us, and know that life is out there. A "man" on the alien planet gets part of his consciousness encoded into a light beam that is sent out into space. When it hits earth -It is able to project itself and appear to be human. When the light fades the man vanishes, leaving a message that he is long dead. (It's a love story.) But the question is what broadcasts would he have received time wise. His knowledge of earth would be limited, ending at a certain time...

This is the end/explanation. Thank you in advance for your help!

"My people developed a holographic beam so powerful it could be projected through space. Through light-years. It is more advanced than anything on Earth, so advanced that some actual particles of my being were intertwined with the particles of light being projected. So it was like I was actually here, next to you, talking to you, laughing with you. But I was not. I am not. "When we sent the beam of light traveling into space my particles were woven into it, not all of them, of course. But enough particles of my consciousness were intertwined with the particles of light so that I could interact with anyone I discovered. ...So that I could learn more about the world we'd previously only known through television broadcasts. .. "I traveled in a spaceship made of light. Now that the light has died so have the particles of my consciousness that were interwoven with the beam.... The real 'me, ' the corporeal 'me,' died HOW MANY? light years ago." Long ago, men went to sea. And women waited for them, peering out into black waters, searching for a tiny speck of light on the horizon. Now I too wait, looking out into the vast blackness of space, searching for my love, my heart, HOW MANY? light-years dead.

ANSWER:
For starters, a light year is a measure of distance, not time, so you want to say 'years' not 'light years' in your concluding sentences. (A light year is the distance light will travel in one year.) It would be better, I think, if the traveler could have learned about earth from radio broadcasts as well as television because that could take us back about 100 years ago compared with only about 50 for tv alone. His consciousness travels here at the speed of light, so he could not have come here from a distance of greater than 100 light years away from earth which is a pretty small region of the universe. (The diameter of our galaxy, the Milky Way, is about 100,000 light years.) So, the most extreme case would be that it takes 100 years for the signals to first reach his planet, then it takes his consciousness another 100 years to get here; so when his 'consciousness' arrives on earth, the stay-at-home him would be at most 100 years older and he arrives on earth 200 years after the first radio signals he received were sent.

One other detail which probably isn't that important would be you could not really carry 'particles' of consciousness (whatever that might be!) on a light beam if they had any mass at all. Nevertheless you could envision the information regarding his 'consciousness' could be carried by light.

ADDED THOUGHTS:
I did not answer one of your questions. How much information he had would simply depend on how long he waited after first receiving signals. Since the light beam would not be able to receive and interpret later signals, he would have no further information. However, if you were to put his 'consciousness particles' aboard a spaceship going 99.999% of the speed of light, all the broadcasts could be intercepted and interpreted en route and incorporated into his consciousness somehow; other timing would be pretty much the same. Or, if you simply put him on that spaceship, he would age only about half a year during the trip which the earth would see as 99.999 years. But then you would not have your dramatic sad ending!

QUESTION:
I've been given a query by a friend, and I'm having trouble finding answers! Consider the following: The tallest building in the world is 250m tall (completely fabricated stat). At the top of this building is a light detector and a clock. At the bottom of the building is a laser and identical clock. At t = 0 the laser fires a single beam of light at the light detector atop the building. Exactly 5 seconds later the laser fires another beam of light at the light detector atop the building. Therefore, at the bottom of the building 5 seconds has passed in between laser shots. How much time has passed between the light detector detecting the first and second laser beams at the top of the tower? The same, more or less? I've ben given a query by a friend, and I'm having trouble finding answers! Consider the following: The tallest building in the world is 250m tall (completely fabricated stat). At the top of this building is a light detector and a clock. At the bottom of the building is a laser and identical clock. At t = 0 the laser fires a single beam of light at the light detector atop the building. Exactly 5 seconds later the laser fires another beam of light at the light detector atop the building. Therefore, at the bottom of the building 5 seconds has passed in between laser shots. How much time has passed between the light detector detecting the first and second laser beams at the top of the tower? The same, more or less?

ANSWER:
Since you acknowledge that the height of the building is immaterial, let me make it 1.0 light seconds tall. (A light second is the distance light will travel in one second.) So the first pulse is observed at t=1 s and the second at t=6 s, an elapsed time of 5 s.

FOLLOWUP QUESTION:
the identical clock statement simply means that if they were switched around and the experiment performed again, the result would be the same. The answer lies in general relativity.

ANSWER:
OK, I did not realize you wanted to get so exacting! I am still going to use a building 1 ls=3x10⁸ m high to keep the numbers from becoming impossibly small. Indeed, general relativity tells us that the stronger a gravitational field is, the more slowly a clock will run. This is called gravitational time dilation (and is directly related to gravitational red shift). The rate at which a clock runs slower in a gravitational field (MG/r²) is √[1-(2MG/r)] where M is the mass of the source (the earth in your case), G=6.67x10^-11 N⋅m²/kg² is the universal gravitation constant, and r is the distance from the center of the source. If some time t elapses on a clock in zero gravitational field, the time which will elapse for other clocks is t/√[1-(2MG/r)]. For our purposes I will assume that the zero-field clock will measure 1 s for a pulse traversing your building.

After doing some arithmetic I find that if t elapses on the zero-field clock, t'=t[1-(4.4x10^-3/r)] elapses on another clock. So, for the first pulse where both clocks have begun at t'=0, the elapsed time of the two clocks will be t'_bottom=(1-6.9x10^-10) s and t'_top=(1-1.5x10^-11) s. The upper clock will will measure a shorter time by about 6.8x10^-10 s=68 ns.
Now, we need to specify who determines when 5 s have elapsed. The most sensible choice would be the guy at the bottom of the building who will be triggering the laser. When his time is t'_bottom=5 s, the zero-field time is t=5/(1-6.9x10^-10)≈(5+3.5x10^-9) s and (this is tricky) the top clock will measure t'_top=(5+3.5x10^-9-6.8x10^-10x5)=(5-1.0x10^-10)s.
We know from the first pulse the additional time on both clocks, so we can now calculate what each clock reads when the second pulse hits: t'_bottom=(5+1-6.9x10^-10)=(6-6.9x10^-10) s and t'_top=(5-1.0x10^-10+1-1.5x10^-11)=(6-1.2x10^-10) s.

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 250/3x10⁸=8.3x10^-7 times smaller.

QUESTION:
Now, I know the centre of gravity can change depending on how close to uniform the gravitational field is, and it is where the weight appears to act from. But, what is the centre of mass? Ok, it is a constant; doesn't change; and in an uniform gravitational field the two centres overlap; but how do you define the centre of mass without referencing centre of gravity? An online source said it is the point from which the mass is 'equally distributed in all directions' for an object; but I can have irregular shapes. What does that source means? Or is the centre of mass defined in another way? I am really confused!

ANSWER:
Actually, the center of mass is the fundamental quantity, not center of gravity. The the position r_cm of the center of mass for a collection of point masses is defined as r_cm=[Σm_ir_i]/Σm_i where r_i is the position of the point mass m_i. If you have a continuous mass distribution with mass density ρ(r) you have to integrate over the whole volume to get r_cm: r_cm=[∫rρ(r)dr]/∫ρ(r)dr. Notice that in both cases the denominator is just the total mass.

QUESTION:
When I was very young, I used a hand-held fish scale to try and measure my own weight by attaching the hook (normally placed in the gills of the fish just caught) to my belt and then pulled up on it. I'm interested in the physics behind my folly and, expanding it further, if the scale was attached to a light (not heavy) bar (that would support more than my weight) and the scale could measure a fish weighing more than me, and I was strong enough to lift more than my weight above my head, how close could I expect to get to seeing my weight on the scale (attached to the bar I'm pushing up, and a belt-device around my waist)?

ANSWER:
The way you always solve a statics problem like this is to first choose a body, then apply Newton's first law which states that the sum of all forces on the body must equal zero; I choose the scale as the body. What are the forces on the scale itself? Your hand pulls up with a force F_hand, your belt pulls down with a force F_belt, and the earth pulls down on the scale (its weight) with a force W_scale. These three have to add to zero which means that F_hand=F_belt+W_scale. Now, the scale is calibrated so that it reads zero when F_belt=0, so the scale will read whatever force your hand exerts up. Note that when you do this analysis, the desired quantity, your weight, never appears; everything would be just the same if you weighed 1000 lb. Regarding your second question, it is really no different from the first question: simply replace 'belt' by 'bar' everywhere. In either case you would observe the scale reading your weight when you pulled up with a force equal to your own weight.

FOLLOWUP QUESTION:
Thank you! So it sounds like it wasn't a folly after all (except for the fact that the scale I used wouldn't support my weight at that age).

ANSWER:
Of course it was folly, total folly! Doing this experiment gives you absolutely no information about your weight. Only if you already knew your weight and were able to pull with a force that hard would the scale read your weight.

QUESTION:
I'm having a discussion on a website that came up because of another post. That post asked the question (paraphrased) If you connected a wire to your mouth and another (unstated) orifice, how fast would you have to travel around the earth to create enough current to electrocute yourself? One of the other posters wanted to know if the voltage would be the same "going" through your body as through the wire. And wouldn't your body have current induced also? The answer for the induced current in the wire I can deal with, But haven't enough electro/biology (I just made that up) to explain why ones body wouldn't have current induced in it. I know intuitivly why but can't explain it satisfactorily to him. Can you help me out here (in colloquial English)? I'd appreciate it if you can and willl

ANSWER:
The body is certainly not a perfect insulator or else you would never have to worry about getting electrocuted. The physiology is very complex and variable, and it doesn't seem you want to get into too much detail either, so I will just talk qualitatively. Two important features:

A voltage large enough to drive a large enough current through your heart will kill you.
A conductor moving through a magnetic field will have a voltage induced across its ends. (Details depend on direction of the field and the direction of the velocity of the conductor. Just assume that you have a straight wire moving perpendicular to the field.)

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?

QUESTION:
Would it be possible physically to encapsulate a black hole with solar panel type devices and use its energy to power a civilization? Like when we reach trans-galactic civilization status and run across a black hole and we want to utilize it.

ANSWER:
Well, that is a pretty crazy idea because a black hole is an energy sink, not an energy source!

QUESTION:
im struggling to get a grasp of this. say car A and car B are travelling at a constant speed, maintaining same distance in between. form car A you shoot a gun 90 degrees up. which vehicle is the bullet likely to fall in. is car A or B?

ANSWER:
Before you fire the gun, the bullet in it is already traveling along with you in car a with the same speed as the car. When you fire it up it goes up with some speed but it continues moving parallel to the ground with car A also. Since there are no forces in the horizontal direction acting on it (neglecting air drag), it will go straight up and straight back down as seen by car A. Someone on the ground will see it go in a parabolic path beginning and ending in car A.

QUESTION:
I don't quite understand escape velocity. I know rockets have to reach 7mi/sec to leave earth's gravity but If I set off in my hypothetical spaceship and I moved at a constant 7 miles per minute on a trajectory perfectly perpendicular to the earth and directly above its launch pad (until I hit a geosynchronous orbit). If I can maintain this velocity away from and perpendicular to the earth what could possibly be stopping me from escaping its gravity?

ANSWER:
The escape velocity is defined as the speed an object must have at the surface of the earth (neglecting air drag) to escape. Obviously, as you note, if you could could continue pushing on something to keep it moving a speed of 1 cm/s, it would eventually get as far away as you liked. For more on escape velocity, see the FAQ page.

QUESTION:
I was wondering if the charge of a particle increases as its mass does (at relativistic velocities). For example, if a quark is going very fast, does its charge increase the same proportion as its increase in mass?

ANSWER:
No, electric charge is strictly conserved.

QUESTION:
When a photon loses energy (Redshifts) climbing up thru a gravitational field, does the photons decreased energy go into the mass of the gravitational field itself? According to GR, the gravitational field itself contributes to the mass of the system. For earth, it is very small. When I elevate a massive object in a gravitational field, I can say the "potential energy" is in a very small increase of the mass of the object. But photons have no mass! And are never at rest! The only thing I can think of is that the photon "transfers" energy to the gravitational field itself, which appears as a small increase in mass of the field.

ANSWER:
Let's think of a star emitting a photon of frequency f. Initially an energy of hf is removed from the star. But, by the time that the photon is very far away the star, it has lost some amount of energy Δhf, so the net loss to the star and its field is hf-Δhf. At the instant the photon is created, the mass of the star is reduced by hf/c². But when the photon, which is losing its energy to the field, is far away, the energy of the field will have increased by Δhf/c²; I would not use the terminology "mass of the field" since the field has energy density, not mass. But now, it seems to me (not a cosmology/general relativity expert) that there is a bit of a paradox: we always associate a given mass with a particular gravitational field, so the field should have the energy content associated with a mass M'=M-hf/c². But, in fact, the field would have energy content associated with a mass M'=M-hf/c²+Δhf/c². I am guessing that the field and the mass somehow "equilibrate" so that the final mass of the star is consistent with the energy of the final field. (I could easily be wrong! Perhaps it is only meaningful to look at the total energy of the star and its field. Whatever the case, I would not talk about mass of the field.)

QUESTION:
Considering a lack of strong ambient light sources, such as the sun, is the exterior glass of a spacecraft, the part that comes into contact with the vacuum of space, cold, neutral, or hot? Would it depend on the temperature inside the craft itself, or would the vacuum have enough of an impact on it to change it?

ANSWER:
It is just the same as a window in your house in winter. The inside surface will be in equilibrium with the air in the room and the outside surface will be in equilibrium with the air outside. Then heat will be conducted from inside to outside at a rate proportional to the temperature difference. If the sun is shining directly on the window, absorption of the radiation would raise the temperature of the outside surface somewhat.

QUESTION:
Why isn't Hawking radiation trapped in a black hole's gravitational pull and eventually sucked right back into the black hole?

ANSWER:
What happens is that pair production occurs near the Schwartzschild radius. One particle of the pair is outside that radius and can therefore escape.

QUESTION:
"IF" we violate the laws of conservation, what would happen?

ANSWER:
There are so many such laws that no single answer could cover it all.

FOLLOWUP QUESTION:
But is it possible to violate the laws even tho we don't know how since its just a law and some laws are meant to be violated...sorry for my curiosity, I just some opinions from a Physicist

ANSWER:
I would say that if a "law" is properly stated, it is not "meant to be violated". Again, there are far too many conservation laws to discuss. But let me give you just a couple of examples of how a law can be misstated but, if properly stated, it cannot be broken.

Energy conservation:
1. Incorrect: the total energy of any system is conserved. In fact you can change the energy of a system by doing work on it.
2. Correct statement: the total energy of any system is conserved if no external forces do work on it.
Electric charge conservation:
1. Incorrect: The total electric charge of a system must remain constant. In fact, you can always add or subtract charge from a system.
2. Correct statement: The total electric charge of an isolated system must remain constant.

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, ΔEΔt~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.

QUESTION:
Are there any similarities between light waves and sound waves? For example, When I pluck a guitar string while placing the guitar parallel to the tv, I can see waves that I previously couldn't see under visible light. The color blue seems to work best.

ANSWER:
The picture on your TV flickers at a rate of 60 Hz (times per second). Therefore it is like a strobe light. If the string is vibrating with a frequency of near 60, 120, 180,…it will appear to be frozen; sound has frequencies around these frequencies. So, you are probably seeing this strobe effect. The frequencies of light are enormously larger, hundreds of THz (10¹² Hz).

QUESTION:
What was the first thing Einstein did for the Mathematics of General Relativity? I just mean to say how did he start General Theory of Relativity?

ANSWER:
There is a story about how Einstein got the idea for general relativity, possibly not true. He was sitting at his desk in the Swiss patent office watching a workman across the square on a ladder painting a building. The workman fell off the ladder and Einstein thought to himself, "There is no experiment he can do which could distinguish whether he is in free fall in a gravitational field or is at rest in empty space." This is a statement of the equivalence principle which, along with principle of relativity (the laws of physics must be the same in all reference frames) form the basis of general relativity.

QUESTION:
If i am walking around the earth. Gravity keeps me on the earth but i'm moving around a sphere so the question is. By how much does the curve of the earth drop to keep me on the surface of the earth after 8000 meters? So that i don't just fling out into space.

ANSWER:
I think you are just asking about geometry. Certainly, the curvature of the earth does not keep you from being flung "out into space"! You seem to be asking about the distance d in the figure here which is easy to calculate. If you go 8000 m along the surface, that is the arc length subtended by the angle θ. Therefore, θ=8000/R=8000/6.4x10⁶=1.25x10^-3 radians=0.0716º. Finally, you can write cosθ=R/(R+d) or d=R(1-cosθ)=5 m.

QUESTION:
What role do gravity or the weak nuclear force have in keeping the human body alive? Clearly, the strong nuclear force keeps us from dissolving into particles, but if you were suited up in between galaxies and gravity "disappeared", would the strong force be enough to keep us intact and alive? Also, electromagnetism is necessary for brain function (and heat transfer?), but do we use the weak nuclear force when making energy from food or anything like that?

ANSWER:
We evolved in a gravitational environment, so the proper functioning of many of our biological systems is dependent on gravity. Astronauts come back after long missions with weakend bones and muscles, for example. The electromagnetic force is probably the most vital because all of chemistry is electromagnetic. I cannot think of any biological functions which are directly dependent on the weak interaction.

QUESTION:
If g=GM/d² then why does acceleration due to gravity decreases as we go deeper into earth?

ANSWER:
That equation is only valid for the acceleration outside a spherically-symmetric mass distribution. If you go inside the mass distribution, the equation is g=GM'/d² where M' is the mass inside d.

QUESTION:
I did read one of your answers. I didn't quite get it. Electricity is electrons flowing through a electric field right? I had the problem, if electrons lose energy as heat etc. then some of the electrons energy must be lost. I thought it must be the electrical potential energy, since the electron is not losing kinetic energy. Because if it did lose kinetic energy, then the speed would reduce and that would reduce the current in different places of the circuit, which does not happen. Current in all parts of the circuit is the same. But you say, that the electron keeps bumping and has to start again and again, and that he moves real slowly. I thought the electrons were moving at light speed?

ANSWER:
I'm not sure which answer you read. This one is probably the best. Another good one to read is this one. So, to understand current you have to appreciate that it is statistical; there are such a huge number of electrons moving that all it makes sense to talk about is the average velocity of the electrons which is referred to as the drift velocity. When a collision with an atom occurs, kinetic energy is turned into heat energy. On average, electrons have constant kinetic energy but lose potential energy as they move along in the electric field. This energy lost to heat is supplied by the power source which is why you have to change batteries now and then or burn coal in a power plant somewhere to supply the energy. The drift velocity is, as explained in the second reference above, extremely slow, not the speed of light. When you turn on the potential difference, the electric field establishes itself along the wire at about the speed of light which is why a light seems to come on instantaneously.

QUESTION:
Why it is important for fission reactions to emit neutron ?

ANSWER:
Because a neutron can cause a fission to happen in another nucleus and therefore you have the possibility for a chain reaction.

QUESTION:
Do all weak interaction produce neutrinos ?

ANSWER:
Here is one which does not produce a neutrino. It does, however, involve a neutrino.

QUESTION:
Will two objects traveling in the same direction ever collide? Assume the objects are on earth, unmanned and their mass, volume, weight, density and speed are the same. All variables that can come in to play should be assumed that they are equal, for example no hills, curves bumps and no change in surfaces to have any effect on the coefficient of friction. Just a question my wife and I were wondering about.

ANSWER:
I'm not really sure what you are getting at here. If you are just wondering whether two parallel lines ever intersect, the answer is no. But you seem to want to know about material objects originally moving on parallel lines. To make the situation simpler, let's just have the two objects, originally with parallel velocities, move in otherwise empty space. If the original velocities are equal and they are traveling side by side, they will eventually collide because there is a gravitational attraction between them which will eventually bring them together. However, if they are originally side by side and traveling with unequal speeds, they would not collide if their relative velocities were greater than the escape velocity. The escape velocity for equal masses m originally separated by a distance r is v_escape=2√(mG/r) where G=6.67x10^-11 N⋅m²/kg² is the gravitational constant.

QUESTION:
A 400lb person jumps up 2-inches on earth. If same person jumps up on the moon, how high would the jump be?

ANSWER:
I will assume that whatever the jumper does will add the same energy on both earth and the moon. The gravitational potential energy U at the highest point must be equal to that added by the jump, and U=mgh where m is the mass (400 lb), h is the height (2 in on earth), and g is the acceleration due to gravity (9.8 m/s² on earth, 1.6 m/s² on the moon). The mass is the same both places, so h_earthg_earth=h_moong_moon. Solving, h_moon=12.25 in.

QUESTION:
Hi, I was wondering what are the chances of survival from falling from the ninth floor of a building, going over the science of that how does surface affect the fall, body weight and trajectory. What is the difference from falling from a third story window as opposed to a higher up one?

ANSWER:
Someone else also asked this question; apparently it refers to a recent actual incident of a student falling out a dorm room window about 85 ft≈26 m high; the student survived without serious injury. The second person also wanted to know if I could estimate the force experienced on impact. First I will calculate the speed he would hit the ground if there were no air drag. The appropriate equations of motion are y(t)=26-½gt². and v(t)=gt where y(t) is the height above the ground at time t, and g is acceleration due to gravity which I will take to be g≈10 m/s². The time when the ground (y=0) is reached is found from the y equation, 0=26-5t² or t=√(26/5)=2.3 s. Therefore v=10x2.3=23 m/s (about 51 mph). The terminal velocity of a falling human is approximately 55 m/s, more than double the speed here, so the effects of air drag are small and can be neglected for our purposes of estimating. (If there is air drag, terminal velocity is the speed which will eventually be reached when the drag becomes equal to the weight.)

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=mΔv/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 m², I find that this results in a pressure of about 82,000 N/m²=12 lb/in². 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.

ADDED NOTE:
A rough estimate including air drag would have his speed at the ground be about 21 m/s rather than 23 m/s as above. Given the rough estimates in all these calculations, this 10% difference is indeed negligible.

QUESTION:
Does the gravity of the earth work through the moon? I.e. If you are on the "dark" side of the moon, do you weigh more than you would standing on the side facing the earth?

ANSWER:
Yes, the net gravitational field you see is due to all masses, not just the one you are closest to. On the far side of the moon the earth's field is weaker than on the near side, but it points in the same direction as the moon's field; so, you are right, you would be slightly heavier on the far side.

QUESTION:
Gravity makes Earth orbit sun, Milky Way and Adromeda collide with each other,etc... My question is, what can provide the necessary energy for a system like this for billions of years?

ANSWER:
The earth orbiting the sun has energy, but it takes zero energy to keep it orbiting. The Milky Way and Andromeda galaxies have energy but no additional energy is added as they move toward each other and eventually collide.

QUESTION:
If I use some magnetic bars, cut them perfectly so that I can put them together to form a globe, with the same pole pointing outwards, and the other pole pointing inwards, do I get a Magnetic monopole object?

ANSWER:
Think of your bars as dipoles of positive and negative magnetic charges (monopoles) separated by a distance d. The magnitude of the magnetic field B of a monopole is inversely proportional to the square of the distance r from the charge, B=kq/r² where k is some constant. In the drawing above the field at point p is B=B_-q+B_+q=kq[(1/(r-d)²-(1/r+d)²]=4kqrd/[(r+d)²(r-d)²]. Now, look at the field when r>>d: B≈4kqd/r³. The field does not look like a monopole because it falls off like 1/r³, not 1/r².

QUESTION:
It might look like a homework question, but it is not. Please help me. I have asked this question everywhere I could, but everybody seems to ignore it. So, the problem is: Let us say we have two bodies A and B in contact with each other, with A lying at the back of B, and the system is on a friction-less horizontal surface. Let A have mass 5 kg and B 10 kg. Now let's say I apply a force of 45 N on A with my hand, then the system begins to accelerate at 3 m/s^2 and the net force on B by A is then 30 N, and B in reaction applies a net force of -30 N on A. Thus, the net force on A is 15 N. What I do not understand is why A is not applying a force of 45 N on B? If it is due to the reaction of B on A, how does A know in the first place that it is to exert a force of 30 N on B so that it receive a reaction of -30 N from B? Is not the reaction force of B on A some kind of a function of the action of A on B, and if it is, then how is the magnitude of the action of A on B is first determined? What is it that I do not understand about Newton's Third Law of Motion?

ANSWER:
OK, I will take your word for it that it is not homework. It is important to be able to solve these kinds of problems. My method is to choose a body and look only at that body. I choose first (as you did) to choose both masses as the body, so M=15 kg and F=45 N and therefore a=F/M=3 m/s². Next I choose B as the body. The only force on it is the force which A exerts on it, F_BA. Since we know that m_B=10 kg and a_B=3 m/s², F_BA=m_Ba_B=30 N. Finally choose A as the body. Two forces act on A, F=45 N and the force which B exerts on it, F_AB=-F_BA=-30 N; its mass is m₅=5 kg and therefore a=(F+F_AB)/m₅=(45-30)/5=3 m/s². The reason that there is not a 45 N force on B is because your finger is not touching B, only A is touching B. The reason A "knows" to exert a force on A is that it has no choice since A's acceleration and mass are already fixed. Once you know the "reaction" force you automatically know the "action" force because of Newton's third law, F_AB=-F_BA=-30 N for this problem. You could also have chosen A as the body before you chose B as the body. Two forces act on A, F=45 N and the force which B exerts on it, F_AB. Its mass is m₅=5 kg and its acceleration is 3 m/s²; therefore (45+F_AB)=m₅a=15 N or F_AB=15-45=-30 N.

QUESTION:
I'm bothered by the seeming interchangeability in physics of the terms 'measurement' and 'observation'. Scenario: Doing a double slit experiment while firing one electron at a time - in a darkened room with a person "observing the experiment" and one without an observer. And also doing another set of experiments with and without a detector - are the results different? if so, how?

ANSWER:
There can be confusion but it is mainly semantic. If you simply "observe" the double slit experiment by counting electrons which hit the screen and determining where they hit, no "measurement" has been made regarding the double-slit experiment itself, you have simply observed how the results play out after the electrons pass through the slits. However, if your observation includes some apparatus to determine through which slit each electron passed, that is a "measurement"; in that case, the measurement will destroy what you previously observed—the interference pattern.

QUESTION:
Are star wars blaster weapons possible? i know lasers wouldnt work because its just light but maybe plasma energy or energy projectiles could work? i just need to know if the blaster rifles/pistols as pictured in star wars and the way they function could EVER work in the distant future and what way you think they could possibly work, assuming we had all the technology necessary to do so. thank you in advance for your knowledge.

ANSWER:
First of all, they are fictional with not much information available regarding their power. However, a quick search of the web reveals that many Star Wars enthusiasts have tried to make estimates of their properties from detailed examination of the movies. First observation is that they are not lasers because you can see their trajectories and Han Solo is seen to jump out of the way in time to not get hit; light is just a lot faster. (You should not write them off just "…because it's just light…" light, though; lasers can melt steel.) Also, if the pulses are any kind of material (super-heated plasma, charged particle beams, etc.), they could not propogate very far in air which they do. But the thing that makes them, in my view, impossible is the energy each pulse carries on the highest setting; one site I found estimated the energy of a single pulse to be on the order of several megajoules. Suppose that a single pulse lasts 1 ms=10^-3 s and carries an energy of 1 MJ=10⁶ J; the power carried by that pulse would be 10⁶/10^-3=10⁹ W=1 GW. This is the power output of a typical power plant. Where are you going to get this kind of power in some compact form you can carry around with you? No way!

QUESTION:
When in the early universe did life become possible? I'm assuming it would have had to have been at some point when the temperature had cooled down enough so that the protons and neutrons of basic elements of life like carbon and oxygen could bond and come together. Also, what do you believe is the purpose for intelligent life in the Universe?

ANSWER:
I usually do not answer questions about astronomy/astrophysics/
cosmology as stated on the site. I can give you a little information here, though. The universe just after the big bang was almost exclusively hydrogen. No elements which could be used to form planets where life might develop or the material for life itself. The first stars started forming about 100 million years after the big bang, basically pure hydrogen stars. These had to be much larger than the sun (perhaps 300-1,000 times heavier) because of the lack of heavier elements and consequently their lifetimes were too short (a few million years) for life to evolve (it has taken about 5 billion years to reach our stage of evolution, about half the sun's lifetime). Inside these stars nuclear fusion caused heavier and heavier elements up to about iron to form; then when those stars used up all the fuel, they exploded (super novae) scattering the heavier elements to eventually be part of clouds of dust and hydrogen which subsequently formed new stars with the possibility of planets. (As you can see, it is not "cooling down" which is responsible for heavier elements which are created in stars which are super hot.) So, we are now at least several billion years into the age of the universe. More detail on early stars can be read in a Scientific American article. The "purpose" for anything is not within the purview of physics.

QUESTION:
I want to design a home-built experiment that will physically display the actual curvature of the Earth, and I would appreciate validation that my methods are (at least theoretically) - scientifically accurate. This test is based on the geometric relationship of a tangent line (globally 'straight' line) when it is perpendicular to a radial line of the sphere. . I am going to set 51 fenceposts fifty feet apart for a total length of 2500 feet. I'm going to use a water linelevel to mark a "locally' level line on the first two posts. Then, using a laser light as a straight-edge, I will extend this straight line for the full 2500 feet. Then I will take the water linelevel to the last two posts, hold the level to the straight line marks, and note that the globally straight line is now an inch and 3 quarters out of "local" level, due to the Earth's curvature. . This inch and 3 quarters divergence is derived from the formula for the Earth's curvature that is eight inches per mile squared: 2500 feet is .4734 of a mile. Squaring that, then multiplying by 8 inches, gives 1.7931 inches. . Any advice or criticism or insight you could offer would be well appreciated. For brevity's sake I kept this desription quite short, but I will be happy to elaborate in greater detail if you want me to.

ANSWER:
Here is the problem: You assume that the earth is a perfect sphere. It most certainly is not, particularly at the half-mile level. You can see some detail in a calculation I did in an earlier answer about the Bonneville Flats where a similar question to yours was looking to observe curvature at the 10-mile level. It takes several hundred miles to be able to do an accurate measurement of the earth's curvature; the ancient Greeks did this around 200 BC.

QUESTION:
If I am running at average sprinting pace for an 17 year old male and I jump off of a 80 meter drop, how far forward will I land from the jumping point? Assuming I am around 11 stone.

ANSWER:
80 m is pretty high, so you will be going very fast when you hit; therefore, neglecting air drag might introduce significant error. But air drag is pretty tricky to calculate and I will neglect it; the answer I get will be somewhat bigger than what would really happen. I will take your speed to correspond to running a 100 m dash in 15 s, about v ≈6.7 m/s. The equations of motion are x=6.7t and y=-4.9t² where x and y are the horizontal and ver tical positions relative to the edge of the cliff and t is the time after jumping. Solving the y equation for t when y=-80 m (the ground), t=√(80/4.9)=4.04 s, so x=6.7x4.04=27.1 m. That is the answer neglecting air drag. Note that it is independent of the mass m.

QUESTION:
We represent waves by crest and troughs like ( ups and downs ) but in the slit interference process we represent it in the form of other figures like ))))))) . can u tell me how crests and troughs are represented in this form ?

ANSWER:
"...figures like )))))..." are called wave fronts. Each front is meant to represent the locus of points at some instant where the wave crests are and halfway between are the wave troughs.

QUESTION:
According to Einstein,gravity is the bending of space-time by matter which is unlike Newtonian gravity.But there are boson particles called gravitons which are thought to be the cause of gravity. Aren't these theories conflicting?How can gravitons exist if there is no gravitational force?

ANSWER:
It is generally believed that gravity, like any other field in physics, should be quantized. So far, there is no successful theory of quantum gravity. If there were such a quantized gravitational field, there would be a boson "messenger" like the photon for the electromagnetic field or the gluon for the strong interaction field; that would be the graviton. Since there is no theory, there is no graviton, only speculation.

QUESTION:
Many explanations in physics which I have seen proceed with a 3D Euclidean space, empty but for a population of objects. Lines representing electromagnetic radiation are drawn from emitter to sensor to explain various phenomena. The apparently discreet nature of the representations leaves me wondering - When I see a star, I see it "everywhere", there seem to be no black spots . Spokes radiating from a wheel grow farther apart as they proceed, why not light from a star? Be it particle or wave or some probabilistic combination, does not some m/e travel from the star to strike my eye? Why does this not put the burden of infinite m/e on any light source?

ANSWER:
Those "rays" you see drawn in geometrical optics are not meant to represent the only place light is. There is no rule about how many rays to draw but the direction of a ray is always the direction that light is traveling at any point on the ray and the intensity of the ray is implicit in the density of the drawn rays. The rule for how to draw a ray between two points is that you find the path which will take the shortest time. For a point source of light (like a star) in a vacuum (approximately what is between you and the star) that will be a straight line. In the picture above you could have drawn twice as many rays and that drawing would still have conveyed the same information about the light. An alternative way to represent the waves is to draw wave fronts which are concentric about the point source and equally spaced; these are meant to represent the location of the crests of the light waves at some instant and are imagined to be expanding outward from the source. The rays are always normal (perpendicular) to the fronts. This drawing is really a cutaway because the fronts are really spheres, so you see that wherever you stand you will see light. Because a wave front is expanding, its area (4 πR²) is getting bigger so the energy of that wave is getting more and more spread out meaning that the light is getting dimmer.

QUESTION:
I have a question about the properties of light. If I am traveling at 99.99999... The speed of light and shine a light in the exact opposite direction, how would that light behave to an observer? Would it stay still? Would the light still travel at c?

ANSWER:
Every observer measures the same speed of light, c. That does not mean that the light is identical. If the source has a huge speed, the light has a huge red shift, its wavelength is greatly larger than that observed if at rest relative to the source. It would likely be invisible to the unaided eye.

QUESTION:
I have 300 grms of baked beans that I want to heat in a microwave. Wil it take longer or more energy to cook these beans in one dish or in separate dishes of 150grms each?

ANSWER:
The rate at which the beans will be heated by the radiation (R_H) is proportional to their volume. But, as soon as they start heating up, they start cooling off by radiating from their surfaces; the rate of cooling (R_C) is proportional to the area. Roughly speaking, the volume is proportional the the L³ and the area is proportional to L²where L is a measure of the size of the pile of beans. The ratio R_H/R_C=KL³/L ²=KL determines the net rate of heating; K is some constant. For example, suppose your beans were contained in a spheres of radii L and 1.26L; I chose those because the larger radius sphere would have twice the volume of the other. Then, the larger sphere will heat up 1.26 times faster than the smaller (two spheres). Assuming that the microwave energy is constant with time, the longer time it takes the more energy is used.

The problem with my analysis, though, is that it depends on the power of the microwave. If it has a relatively low power, the radiation will be mostly absorbed in the beans and the assumption that R_H is proportional to the volume becomes untrue. Only if a small fraction of the radiation is absorbed by the beans will my analysis be approximately correct. You should do an experiment, try it!

QUESTION:
Why only magnetic moment due to spin is considered in proton nmr spectroscopy. what about magnetic moment due to orbital motion??

ANSWER:
You are right, a proton in a nucleus will have both spin and orbital magnetic moments. But the whole idea of NMR is to put magnetic moments in a magnetic field and they tend to align. To flip the spin requires a certain amount of energy. So you send photons of the right energy, typically radio frequency radiation) and they will flip the proton spin. You need not worry about other moments because the photons are tuned only to the right frequency for spin.

QUESTION:
In understanding that two objects (no matter their phisical properties) will accellerate at the same rate towords the same gravitational force in a perfect vacuum, do we take into consideration the gravity of the falling objects themselves? Would the larger gravitational force of the larger falling object be perfectly offset by the larger ammount of inertia or tendency to resist motion?

ANSWER:
The force F between masses M and m whose centers are separated by a distance R is F=GMm/R². But, the acceleration a of m is a=F/m=GM/R². So, as you ask in your last sentence, the inertia (m in a=F/m) cancels out the mass (m in F=GMm/R²) in the acceleration. This is actually a profound finding for the following reason. In Newton's second law, mass is, as you say, inertia (inertial mass, m_I); in Newtons universal law of gravitation, mass is the ability to feel or create gravity (gravitational mass M_G). The fact that gravitational accelerations are the same implies that m_I=m_G, not something you would necessarily expect. Why these are equal was not really understood until the general theory of relativity was proposed by Einstein.

QUESTION:
i am a metal worker and maybe it's dumb but i have thought up an idea for renewable energy. I need to build a huge tank capable of holding 10000 of water, it will have a coned base with a hose coming off it and running back up into the tank, returning the water to it's point of origin....question is...with 10 tons of water down force, 1 meter from the ground....would a decent amount of water travel up that hose pipe and back into the tank???..

ANSWER:
It makes no difference if you have a whole ocean of water, the water at the bottom can never push water higher than the surface of the tank, can never cause a continual current like you envision. There is a very detailed answer for a very similar question to yours that you should look at.

QUESTION:
In the event horizon of the black hole, a pair production of a particle and an anti particle and the falling of one into the black hole and the releasing of the other from it gives us the Hawkins radiation, but my question is , isn’t it possible for both the particle and the anti particle to fall into the black hole. If that happens defiantly the vacuum energy of the space decreases how can the universe compensate this decrease in energy to validate the law of conservation of energy.

ANSWER:
There are two ways you can imagine the creation of the particle-antiparticle pair (PAP):

Vacuum fluxuations where a particle-antiparticle pair is spontaneously created just outside the event horizon, violating energy conservation. However, the uncertainty principle (UP) allows this violation but only for a very short time. If one escapes, it adds energy to the universe so the captured particle must have negative total energy thereby decreasing the mass of the black hole and maintaining energy conservation. If both particles are captured, they are required by the UP to recombine thereby keeping the mass of the black hole constant.
When a virtual pair is created just outside the event horizon, the intense gravitational field of the black hole can provide the energy for the PAP to become real rather than virtual. In this case, the energy acquired by the pair will be lost by the black hole reducing its mass. If only one of these is captured, the energy of the escaped particle will be added to the universe but the energy of the captured particle will be added to the (already reduced) energy of the black hole resulting in a net loss of energy of the black hole equal to the energy of the escaped particle.

Energy of the whole system is conserved.

QUESTION::
If the speed of sound is inversely proportional to the density of a material, why does sound travel faster in solids (it is the most dense). I have read that it takes more energy for sound to travel in dense materials so it takes longer but then neighbouring molecules are closer so sound does not have to travel that far, making it faster. This doesn't make any sense because it says the more dense a medium is, sound is both faster and slower. Also, how does bulk modulus affect the speed of sound.

ANSWER:
You are oversimplifying because the density ρ is not the only thing which determines the speed of sound. In general, the speed of sound v may be written as v= √(K/ρ) where K is a parameter which specifies the "elasticity" of the material. (Note that v is inversely proportional to √ρ, not ρ as you state.) For example, just qualitatively, you might say that steel is much denser than air but it is also much more elastic. The way K is specified for gasses is quite different than for solids. For an ideal gas, K=γP where P is the pressure of the gas and γ=5/3 for monatomic gasses and 7/5 for diatomic gasses. On the other hand, for solids K=B+4G/3 where B is the bulk modulus and G is the shear modulus. For solids K is enormously bigger (~10¹¹ N/m²) than γP for gasses (~10⁵ N/m²). That answers your question. So, let's just do a rough calculation to see if I get reasonable values for air and steel:

Taking air as mainly diatomic γ≈ 7/5 , atmospheric pressure P≈10⁵ N/m², ρ≈1 kg/m³, v ≈374 m/s; a little high, but close.
For steel, B ≈1.6x10¹¹N/m², G≈7.93x10¹⁰ N/m², ρ≈7900 kg/m³, v≈5800 m/s; this is exactly the speed of sound in stainless steel!

QUESTION:
The earth's radius varies depending on latitude. How many times greater is the acceleration of gravity at the poles of the earth than the equator if the radius of the earth is 99.5% smaller at the poles than the equator?

ANSWER:
To calculate exactly would be impossible because the earth is not a perfect spheroid and its mass is not uniformly distributed. But you can get a good idea by calculating the difference in gravitational force for spheres of the same mass but having the two different radii; these radii are 6378 km (equator) and 6357 km (poles), about 0.33% larger at the equator. Since the gravitational force is inversely proportional to the square of the radius, this would give about a 0.11% smaller g at the equator. This effect is smaller than the effect due to the earth's rotation which results in about a 0.34% reduction at the equator (due to centrifugal acceleration).

QUESTION:
What is the purpose of utilizing a percentage of body weight to determine how much weight to bench press/push/whatever? I know this seems like a fitness question and not a physics question, but what I am interested in is WHY weight would be used to determine how much one could (or should be able to) lift/push? For example: A gym teacher wants to grade his students on their strength. He decides to use abililty to push a weighted sled across the floor as the measure. He wants to make the task equally difficult for every student in order to make the grading fair. So, he decides that each student will push 2x his/her body weight for 5 minutes and the grade will be based on how FAR the student is able to push. So, Student A weighs 170 pounds and pushes 340 pounds (including the weight of the sled) for a total of 160 yards. Student B weighs 240 pounds and pushes 480 pounds (again, including the weight of the sled) for 80 yards. Student A pushed farther and gets a better grade, but Student B complains that he had to push much more weight so he should not get a worse grade. Does Student B have a legitimate complaint or does his heavier weight contribute somehow to his ability to push that doesn't have anything to do with his strength? As in, does his weight help push the sled in some way? Sorry, I don't know enough about physics to ask this question using proper physics terms like force, mass, etc. I hope you will still answer my question!

ANSWER:
I cannot comment on the rationale for correlating weight to strength. I can certainly comment on the physics of your particular example of sled pushing. I would first of all comment that this example is certainly not one solely of strength because, since it is a timed activity, endurance as much as strength is being tested; if one student, for example, were a heavy smoker, he would likely become exhausted more easily. As a physicist, I would equate "strength" with force. The specific example you give, though, seems to me to be more related to energy (work done by the student) or power (rate of energy delivered) than strength; purely in terms of strength, the heavier student exerts more force. The force F which each student must exert depends on the weight w he is pushing and the coefficient of friction #956; between the sled and the ground, F= μw. The work W done in pushing the sled a distance d is W=Fd= μwd. The power generated if W is delivered in a time t is P=W/t=μwd/t . Both students have the same μ and t , so W_A/W_B=P_A/P_B=d_Aw_A/d_Bw _B=(160x340)/(80x480)=1.42. So student A did 42% more work, generated 42% more power, than student B. From a physics point of view, B demonstrated more strength, A demonstrated more power. I would judge that this is not a fair way to assign a grade. It would be interesting to see if A (B) could move B's (A's) sled 80 (160) yards.

QUESTION:
Say I have 2 factories producing an object, we'll be arbitrary and say bottle caps. Each produces them at a rate of 1 per second. Now ignoring things like raw materials etc... One stays on Earth, producing its bottle caps constantly. The second one (somehow) gets accelerated to (again being arbitrary) 90% the speed of light and remains at that velocity until completing a round-trip 100 years long. When the moving factory returns, 100 years later from the perspective of the factory on Earth...will there be a difference in the total number of bottle caps each has produced?

ANSWER:
This is just a variation of the twin paradox. The number of bottle caps produced by the moving factory will be N √(1-0.9²)=0.44N where N is the number produced on earth.

QUESTION:
Okay, so for a while I've been wondering why bass travels through certain materials that higher frequency sound doesn't. I have a theory which could be wrong but what I thought is that bass, having a larger wavelength, passes through particles that are bigger, and that shorter wavelengths can't get through. Somewhat like red light and blue light traveling through the atmosphere. Okay, so I looked it up everywhere (almost) and couldn't find a straight answer, even my physics professor didn't know surely why this happens. So, maybe you could. Why can I hear just the bass when someone blasts music in their car?

ANSWER:
The reason is that the attenuation of sound intrinsically depends on the frequency of the sound, the higher the frequency the greater the attenuation. The theory of attenuation is rather complicated, so I will just sketch the results. For a Newtonian fluid, Stokes showed that A(d)=A₀e^- ^αd where A(d) is the sound amplitude a distance d into the material, A₀=A(d=0), and α= 8 π ² η f² /(3ρv³); here η is viscosity of the fluid, f is frequency of the sound, ρ is the density of the fluid, and v is the speed of sound in the fluid. The important thing to notice is that the amplitude decreases like exp(-βdf²) for any Newtonian fluid. For example suppose you compare f₁=20 Hz with f₂=1000 Hz for a thickness of d=1 cm=0.01 m and A₁(0.01)=јA₀=A₀e^{-β�0.01�400}=A₀e^-4β which can be solved for β, β=0.35 s²/m; then, for f₂, A₁(0.01)=A₀e^{-0.35�0.01�1,000,000}=A₀e^-3500 = A₀x 9x10^-1521≈0. Finally, for any material, Stokes' law of attenuation still may be written as A(d)=A₀e^- ^αd except α=βf ^εwhere 0<ε<2. Water and most metals have ε≈2. But even if ε is smaller, it is always positive which means that the attenuation is always greater for a higher frequency.

QUESTION:
We know the relative size of the solar system measuring thru the distances of the planets circling our star. Do we know the thickness of this rotational disc? Is the gravity of our star concentrated only on this single orbital plane as our star rotates? If not, why do planets not circle verticaly as well as on horizontal planes? As far as I know we have only sent space craft thru our orbital disc towards other planets. Do you believe if directed a flight up from our orbital plane would the suns gravity lessen where our space craft could fly faster?

ANSWER:
I should first address your speculation that gravity is stronger on the disc near which the planets orbit. In fact, gravity depends only on how far you are from the source, so if you went 100,000,000 miles from the sun in any direction you would find the same gravitational force. There is not really any well-defined disc whose thickness you can measure; all the planets are nearly all orbiting in the same plane and the inclinations of the orbits relative to say, the plane of the earth's orbit, are all less than 3 Ѕ⁰. But that does not mean that everything is near that plane. Many of the dwarf planets are much out of the plane of our orbit; Pluto, for example, is about 18⁰ inclined relative to our orbital plane. Outside the most distant planets is the Oort cloud, a collection of rocky and icy bodies which is distributed roughly equally in all directions; comets are thought to originate in the Oort cloud and when they enter the inner solar system they come from all directions relative to the orbital plane. The reason that the planets are all nearly on the same plane is that when the sun first formed from a huge cloud of dust and rocks, it had angular momentum and as the system collapsed under its own gravity, it spun faster and faster and the resulting centifugal force caused it to flatten like a giant pizza dough.

QUESTION:
An image is formed when the reflected ray meet each other at a point. But why does an image is not formed by the intersection of incident rays? After all they are the same light rays and it's not like there's some kind of reaction which takes place or something.

ANSWER:
The light rays from each point on the object all diverge away from that point. Unless there is something to change their directions, e.g. a mirror or a lens, they never meet.

QUESTION:
If I were to throw a pencil up into the air, would it come down at the same speed as I threw it up?

ANSWER:
It would if there were no air. However, air drag takes energy away from the pencil and so it would arrive back at your hand going more slowly. For speeds you are likely to be throwing the pencil, though, this effect is small and often neglected in elementary physics classes.

QUESTION:
The demonstration wherein one pours water from a clear vessel and one shines a laser pointer (a red laser pointer in my case) through the water and it appears the laser beam bends with the water pouring out of the clear vessel; What causes the beam to bend with the water? Would this same type of effect work with other fluids than water, such as air or glass?

ANSWER:
It is caused by total internal reflection. If the angle which the light hits the surface between the water and the air is glancing enough, the light will be reflected back into the water rather than be refracted into the air. Total internal reflection can happen whenever light strikes the interface with a material of smaller index of refraction; the minimum angle for which it can occur depends on the relative indices of refraction.

QUESTION:
I was reading a introductory book on particles. I read that neutrinos only respond to weak and gravitational forces, isn't that fact that it responds to gravitational forces prove that it has mass? So why was this years Nobel Prize awarded for figuring out that neutrinos had mass?

ANSWER:
No, responding to gravity does not imply mass. Photons, which are known to be massless, are deflected when passing near a large mass. See FAQ page.

QUESTION:
If a train runs at 100km/h and if in this train somebody drops a rubber ball and catch it, let's say it drops down 1m and up in one second, it means that the ball has travelled 2 meters in one second for the person in the train. If I am at the station looking at the train when it passes me by and let's say it is made of glass so I can see what is happening, I can see the ball dropping down and up but for me as the train is moving the ball will have travelled 2 meters plus the distance travelled by the train in one second so more than 2 meters. I don't think that space time dilatation plays at this sort of speed so how is it explained?

ANSWER:
Actually, you have not calculated the distance traveled correctly. But because the ball is speeding up on its way down and slowing down on its way up, you have set up a problem with unnecessarily difficult mathematics to calculate the distance seen by the person at the station. I will alter your problem somewhat to make it easier to calculate: the ball is made to first move down at constant speed and then to move up with the same constant speed. If the time is still 1 s, the necessary speed is v_y=2 m/s. The train is moving with speed v_x=100 km/h=27.78 m/s so the ball and train move a distance 27.78 m horizontally during the one second the ball is moving. But the distance the ball moves is not 27.78+2=29.78 m but rather 2√( 13.89 ²+1² )=27.85 m because the ball will take a zig-zag path. But the ball also has a different speed as seen from the station, v=√ (v_x²+v_y²)=27.85 m/s. We therefore conclude that the time it takes the ball to travel 27.85 m is 1 s, so everything hangs together. Your error was to ignore the fact that the ball has a horizontal component of its velocity in the frame of the station. You are right that relativity plays no role here.

QUESTION:
Suppose I have a charge +q and there is a point P , Suppose I place a conductor between the charge +q and P . Since there are free electrons in it , Negative electrons move towards +q and equal positive charge inside the conductor near P , The conductor has charge distribution like a dipole. So If I want to calculate E field at P . I could use superposition principle to find E at P due to +q and E due to dipole. But Gauss's law says that dipole doesn't contribute anything to E field at P. Can you explain me 'intuitively' (Not in equations) why the dipole wont contribute anything to the field at P ?

ANSWER:
Gauss's law does not say that the dipole contributes nothing at P. If you put a spherical Gaussian surface enclosing both q and point P the net charge enclosed is q but that does not mean that the field due to the dipole is zero everywhere on that surface. All you can say is that the net electric flux passing through that surface is q/ ε₀. Gauss's law is usually useful in determining a field only if the field can be argued to have constant magnitude and normal to the surface everywhere. Superposition, not Gauss's law, should be used for this problem.

QUESTION:
If I was in a spaceship moving at half the speed of light (in normal space) and I measured the time it took the beam of light from a laser pointer to travel from its source to two points inside my ship, one in the direction of travel and one in the opposite direction, wouldn't C demand that I observe different time measurements? If the time measurements are the same, then when I pointed in the direction of travel, the light would have to be traveling at C plus the speed of the ship from observations made outside of the ship. If the measurements are different, then shouldn't we be able to observe the same effect on earth because of the direction of travel of our planet, solar system, Galaxy, and group of galaxies?

ANSWER:
I am not sure what you are asking here. Does this earlier answer help you? One thing is clear is that you do not really appreciate that the speed of light in vacuum is independent of the motion of either the observer or the source. This is one of the main postulates of special relativity. See the FAQ page for questions related to this.

QUESTION:
Sir,I am a mechanical engineer by profession but very interested in reading physics fundamentals. Recently I went through the fundamentals of electro magnetism and I got this doubt. Consider two charges each of charge +q rigidly fixed in a train moving with a constant velocity, V. Let the train speed be negligible compared to speed of light so that we can treat the problem in non-relativistic terms. The fixity condition ensures that it overcomes electrostatic forces and remain motionless. A traveler in train sees both of them at rest and there wont be any magnetic forces developed between the charges. Now consider an observer in platform. For him, both charges are moving with a constant velocity, V equal to the train velocity. Each charge will develop magnetic field according to this observer as per Biot-Savart law and there will be mutual attraction as each charge is moving under the magnetic field of other.Thus, an observer in train sees no magnetic forces whereas an observer in platform sees mutual magnetic attraction. How do you explain this?

ANSWER:
There is no rule which says that the either the electric or magnetic field must be the same in all frames of reference, even slowly moving frames like you specify. The real root of your problem is that electromagnetism is intrinisically relativistic, even at slow speeds; the electric and magnetic fields of classical electromagnetism are really both components of the electromagnetic field which is a tensor and when you change inertial frames, you cause a transformation of that tensor into another where both the electric and magnetic field pieces of it are different. In your second case you would also find that the electric fields were slightly different from their original values but the differences would be very tiny; the magnetic fields, though, are nonzero but small, but small is very big compared to the original magnetic field of zero.

If you are interested, I will give here the electric and magnetic fields for one of the charges moving with velocity v in the +x direction. E'=iE_x+ γ(jE_y+kE_z) and B'=-(vxE')/c²where γ=1/√[1-(v/c)²] and i, j, k are unit vectors; the vector E is in the frame where q is at rest and E' and B' are when q is moving.

QUESTION:
I see the equations for computing the slowing of time, based on speed compared to light speed. Do we always use c=3^10 cm/sec? I am asking since light speed is slower, sometimes much slower, in mediums other than air or space. If we use a smaller value for c, the time dilation factor is enhanced.

ANSWER:
The significant thing in the theory of special relativity is the speed of light in vacuum. What matters in the theory is the universal constant c, not the speed in the medium you happen to be looking at.

QUESTION:
Is it true that a person flying in an airplane is actually living a shorter amount of time, than a person standing on the ground?

ANSWER:
You must specify " … living a shorter amount of time … " with respect to whom . The person in the airplane sees time progressing at a normal rate. However, the person on the ground will see the clock of the person on the airplane run slowly, so he will perceive the traveling person to be " living a shorter amount of time", i.e. she will have aged less when she returns to earth. You should read FAQ Q&As on the twin paradox, how clocks run, and the light clock.

QUESTION:
If I was travelling in a bus and the bus was travelling at around 60 mph and I jumped in the air, how come I don't fly to the back of the bus. Like with flying, if I was in an airplane travelling at around 500 mph, and I was floating without an aid in the cabin why do I travel with the plane not just get smashed to the back?

ANSWER:
It is impossible for you to be " floating without an aid ", so let's assume that you are standing on the floor of both the bus and the plane. In both cases it is because of Newton's first law. Someone watching you from the ground would see both you and the vehicle as moving with the same speed. From that point of view you are in equilibrium because the only forces on you are your weight down and an equal force up from the floor. Newton's first law says that if you are moving in a straight line with constant speed you will continue to do so if all forces are in equilibrium. When you jump up, the floor no longer exerts a force on you, so you are no longer in equilibrium vertically and you will fall back to the floor. But you are still in equilibrium horizontally (there are no forces on you horizontally) so you will continue, right along with the bus or plane, to move horizontally with the same speed. Actually, I did this the hard way because Newton's third law will still be valid in the moving bus or plane, so when you jump up you will not move horizontally because there are no forces which act horizontally on you to accelerate you (make you start moving horizonatlly).

QUESTION:
If I have two soccer balls the same size, yet one is heavier than the other, which one will go farther with the same kicking force involved? How does momentum and Newton's Second Law come into play here?

ANSWER:
I will assume the balls fly through the air, not roll or slide on the ground. This question has no single answer. Before getting too deeply into this, I need to modify your question a little. The force does not determine what happens to the soccer balls (see FAQ page), so let us change " same kicking force" to "same kicking impulse"; impulse is essentially force multiplied by the time it was applied. That is also more convenient because the change in linear momentum (mv) is equal to the impulse (J). So, with equal impulses, the lighter ball has a larger initial speed: v_light=(m_heavy/m_light)v_heavy. So, if both balls are launched at the same angle and if air drag can be neglected, the lighter one will go farther because it started with a larger speed. Maybe that is all you wanted. But, it is altogether possible that air drag will not be negligible. The force of air drag may be roughly approximated as F=јAv²for a sphere where A is the cross-sectional area of the balls. (This is true only for SI units since ј is not dimensionless.) So now you see that the ball launched the fastest experiences the larger air drag slowing it down. But the heavy ball has even more advantage since Newton's second law says a=F/m so even if the two balls experienced the same force, the less massive ball would have a larger magnitude of acceleration, slow down faster. Now it becomes complicated because you have to know the masses, the impulse, and the angle of the launch to calculate which would go farther.

QUESTION:
Let’s say you have a very long strong tube uncapped at both ends, and laying at sea level. It would have one atmosphere pressure inside. Put a cap on one end and stick that end down to near the bottom of the Mariana Trench. Now pop the cap off. Wouldn't this create a perpetual geyser from the end at the surface? It seems to me that it would come out with enough pressure to run a generator. Would it be considered perpetual motion or just hydraulics? Why wouldn’t this set-up solve the world’s clean energy needs? Please pop my bubble. I can't be the first to think of this. What am I not accounting for? Inside the tube I don’t think gravity would be a player. Neither would the pressure fading outside the tube.

ANSWER:
Once the water reached the top of the tube the column of water would be in equilibrium, just the same as if you had put the tube down without capping the bottom. You might get a spurt at first but then you would just be left with a column of water in equilibrium. I guess you could analyze it roughly by ignoring viscosity and the drag of water moving on the pipe. Suppose the cross-sectional area of the tube were A, the depth of the bottom is D, and the density of water is ρ . At some time after uncapping, the distance up to the current level of the water in the tube I will call z. The force up on the column of water would be F_bottom=(P_atm+ρg D) A, the weight of the column of water is W=ρgzA, and the force down on the top is F_top=P_atmA. The net force on the column is F= ρgA (D-z)=ma= ρAza where a is the acceleration and m= ρAz is the mass. So, a= ρgA (D-z)/ ρAz=g [(D/z)-1]. Suppose we take g≈10 m/s² and D≈10⁴ m; then a≈10[(10⁴/z)-1] m/s². The graph of the acceleration is shown to the right; note the logarithmic scale for acceleration to show the huge variation from bottom to top. Note that it is huge, about 10⁵ m/s² for 1 m and falls by four orders of magnitude when the column of water is halfway to the surface, finally falling to zero when the pipe is full. Suppose we start with z=1 m and the column at rest (where a≈10⁵ m/s²). After 10 milliseconds (10^-2 s) the column will have acquired a speed of approximately v=at=1000 m/s and the water level would have risen approximately to z=1+Ѕat²=6 m. Now, the speed is almost immediately huge and so the assumption of no frictional forces becomes impossible. The drag of due to the water moving in the tube for the column of water going 1000 m/s will be huge and keep the water from speeding up so rapidly. Also, the air above the column of water will not just get pushed up will get compressed and therefore push down on the column with a pressure greater than atmospheric. It is therefore clear that my simple calculation above is only a best case scenario and the drag and added pressure will take energy away. The drag might get big enough that the water in the tube would boil and release steam to futher impede its progress. I believe that you will end up with column of water slowing down as it approaches the surface of the ocean and it will just come to rest there. In any case, when the tube was filled, even if still moving, the column of water would have zero net force on it. If you tried to have it do work on your turbine, it would soon loose all its kinetic energy and stop. Alas, no clean energy solution!

ADDED THOUGHT:
Robert M. Wood pointed out to me that the most energy you could possibly get out of the rising water would be the amount of work you performed to get it down there. If you neglect the weight of the tube itself (or, imagine first lowering the tube open at both ends and then pushing the water down with a piston), this work would be W=ЅρgAD². Neglecting any energy loss, this would have to be equal to the kinetic energy of the column of rising water when it got to the top, K=ЅρDAv², so v=√(gD). For the case above, v=316 m/s, pretty modest. And keep in mind that all the water in the tube is now in equilibrium, so as the water goes above the surface, it is slowing down as it gains potential energy. And, if you imagine the tube projecting high above the surface, you can calculate how high h the water will rise before falling back. The center of mass of the column of stopped water is y=Ѕh and its mass is ρhA, so mgy=Ѕh²ρgA=ЅρDAv², so h=v√(D/g)=D. Therefore, with no friction, the water would rise up and oscillate back and forth for ever. But, as argued above, friction will be extremely important so this oscillation will quickly damp out and will never rise up to a height equal to the depth of the ocean. See the video in the answer to the followup question below as an example for a miniature of this.

FOLLOWUP QUESTION:
Thank you for finding interest in my question and for being able to let the nuts and bolts slide. The calculations you created from my questions were like a body punch to my self esteem. I didn't understand a lick of it. And now I'm feeling like the Dickens character asking for more gruel. Your response knocked me for a loop. You pointed out two things I wasn't accounting for. The weight of the water in the tube and the air resistance. The air resistance I've resolved/incorporated. The equilibrium (if it occurs) from the weight of the water in the tube is a stopper and I hate it. Because it seems so common sense. Then I thought of this,(1000 ft.depth= 30 atmospheres = 420 psi.~), in your model if you bring the tube up at a 70 degree angle instead of the assumed 90, because of the longer tubing, there wouldn't be enough pressure to fill the tube. Also, if you put an uncapped tube from the surface to depth at a 70 degree angle as soon as you reached depth the tube would have too much water in it for the pressure and water would have to be expelled from the bottom until it reached the length of the 90 degree tube length. (I can't even do the baby geometry needed to figure out how much longer the tube would be.) Neither of these seems right to me. What do you think? What I picture happening is the moment the cap is removed at depth the whole ocean rushes to fill the void at the top and the water pressure at the outlet would be the same as at the intake. The entire tube would become an extension of the 420 psi zone at 1000 ft. If inside the tube was a vacuum and capped at the top wouldn't this be the case? Wouldn't the pressure be 420 psi at the cap? Here is my main reason for not being able to let this go. (this is my first version of this train of thought) Same long strong tube only this time it is assembled and capped in the vacuum of space 500 miles above earth and the capped end brought down to the surface. When the cap is removed I expect the atmosphere at that point to start rushing up the tube, through it, and out the end in a very long term geyser. And even though the tube is 500 miles long equilibrium would never be reached. Aren't both examples using the same principal? It seems to me that it is the same principal and that the water at depth would view this one atmosphere pressure at the surface as the next best thing to a vacuum and continue trying to fill it. Just as the tube from space would leak away our air. Doesn't pressure equal stored energy? This thought is like a song stuck in my head. I'm tired of thinking about it. It's been months now. Validation would be best for all but, at this point I would settle for a kill shot. Can you help (either way)?

ANSWER:
You have so many misconceptions that I am not even going to try to convince you with the physics arguments. A couple of comments: The reason that the pressure is so high deep in the ocean is that the water deep down must hold up the weight of all the water above it; to to try to ignore weight of the water in the tube misses the whole point. Suppose you had a tube 10 km long full of water (not underwater). The pressure at the bottom would equal to the weight of the tube of water divided by the area of the tube (+P_atm). Also, I have previously answered your question about the air in a tube in the atmosphere—it would not suck out all the air in the atmosphere for the identical reason your water tube would not create a perpetual geyser. Since you do not know any physics, let me convince you another way; remember physics, at its heart, is an experimental science, so let's do an experiment. Your idea should work equally well if the ocean were only 5000 m deep, maybe not just so robust a geyser, right? That is, your idea should scale. So, try the following experiment. Take a straw and put a little cork to plug the bottom and push it down into a glass of water; pop that cork out some way and see what happens. Does the water come up the straw and continually squirt out the top (tiny geyser) or does it come up to the level of the water surface and stop? Maybe it will rise a bit above the surface but it will settle back to the level of the water in the glass.

QUESTION:
Suppose we have a horseshoe magnet. Now we bend it in such a way that it becomes doughnut shaped and poles remain in contact with each other. In this situation what will happen to the magnet? Will it behave as a magnet? Where would be its poles? What will happen to domains inside the magnet?

ANSWER:
A horseshoe magnet is just a bent bar magnet, so let's start there. I have shown a short stubby bar magnet, but you can imagine that as it gets longer relative to its width the field inside will get more and more uniform. Now, bend it around into a circle and it will look just like a toroid. The magnetization inside will remain pretty much the same as before you bent it and there will just be field inside the torus, not outside. The domains will all stay pretty much the same.

QUESTION:
the Coriolis effect.. on a freely falling body... is to east...? motion of earth is towards east right.? so we must feel that falling body deflects to the west isn't it..???

ANSWER:
I cannot give the full derivation of the motion of a particle in a rotating coordinate system, it is much too involved. I can give you the results, though. The coordinate system we will use is the coordinate system (x',y',z') shown to the right. The Coriolis force is given by 2mv'x ω where v' is the velocity of m and ω is the angular velocity of the earth ( ω≈ 7.3x10^-5 s^-1). Now, for a body dropped from some height h, the direction of the velocity is in the negative z' direction, so the direction of v'x ω is east. This is not what you would intuitively suspect (as you note), but it is true. An expression for how far eastward it would drift before hitting the ground is x'=[( ω�cosλ)/3]√(8h³/g) where λ is the latitude. For example, at the equator where λ=90⁰, and you drop it from 100 m, the deflection would be x'=2.2x10^-2 m=2.2 cm.

QUESTION:
What is impulse? My text book explains this as force acting for a short interval of time. Can you please explain this concept?

ANSWER:
What is the effect of applying a constant force F to a mass m? Newton's second law tells you that you cause the particle to accelerate, that is if you exert the force over some time t the velocity will change from v₁ to v₂. Writing Newton's second law, F=ma=m(v₂-v₁)/t. Rearranging this equation, Ft=m(v₂-v₁); so you can see that the amount which the product mv (which is called the linear momentum p) changes is numerically equal to the product Ft which is called the linear impulse J. If the force is constant, the time need not be short; t could be an hour and J would still tell you the change in p. I hope you can appriciate that J=p₂-p₁ is simply an alternative way to write Newton's second law. If F is not constant, you can still calculate J but it is just harder. In essense what you do is calculate Ft for many vanishingly small values of t all along the path the particle takes and sum them to get the total J. (If you know integral calculus, this is just integrating Fdt over the time interval.) Of course, Newton's second law is a vector equation, so J and p are vector quantities, J=p₂-p₁. If you have a system of particles which interact only with each other, the net force on the whole system is zero because of Newton's third law, so J=0 and p does not change as the system moves around; this is called momentum conservation. Here p means the momentum of the whole system, the vector sum of all the momenta of all the particles.

QUESTION:
I've started reading an elementary particle textbook and it doesn't explain why a moving charged particle radiates an EM field. I was hoping you could help with that.

ANSWER:
The semantics of "radiates an EM field" is somewhat ambiguous. A charged particle at rest creates a static electric field. A charged particle moving with constant velocity (constant speed in a straight line) creates both an electric field and a magnetic field and both change with time. But neither of these situations are said to radiate electromagnetic fields. Electromagnetic radiation propogates through space as waves; visible light, for instance, is an electromagnetic wave. Electromagnetic fields can be created when a charged particle accelerates. E.g., an antenna radiates radio waves when electrons in the antenna are made to oscillate back and forth (accelerating). The derivation of how accelerating charges radiate is a topic in an intermediate E&M course and beyond the scope of this site.

QUESTION:
In the famous book by H.G Wells - "The Invisible Man" - the invisible man explains that his invisibility is due to light getting passed through a membrane (his skin/tissue) clearly , i.e not reflecting or refracting. Is this true? Can a object me made invisible by this method? Further, has there been some actual research in this topic and any advancements?

ANSWER:
First, it is not "true" —this is a book of fiction! More pertinently, is it possible? I would say, given what we know today, it is not possible because for the light to not be refracted the speed of light in the body would have to be the speed of light in a vacuum, and there is no material for which this is true. Research regarding invisibility is focused on "invisibility cloaks".

QUESTION:
A charged parcticle when accelerated radiates electromagnetic radiation or light. All atoms on earth are accelerated therefore the electrons and protons which are charged are also accelerated so why isnt light radiated?

ANSWER:
First of all, atoms are neutral, have zero net charge. Second, the accelerations involved because of the earth's motion are very small.

QUESTION:
Why does centripetal force do what it does? It is a force so why does the object start moving in a circular path and not simply in the direction of the force? Also it should provide acceleration...so even if some how the object move in circle why doesnt it spiral in??

ANSWER:
If an object is at rest and you exert a force on it, that force will not cause the object to "start moving in a circular path". You cannot simply label something a centripetal force. If an object is moving in a circle with constant speed, it is accelerating because the velocity is constantly changing. The velocity vector can change either by changing its magnitude (speed) or its direction. An object moving in a circle therefore requires some force to accelerate it, for example the sun's gravitational force on the earth, which points toward the center of the circle. If you stopped the earth in its orbit, the gravitational force would indeed cause the earth to crash into the sun. You need to simply look up a derivation of centripetal acceleration in any elementary physics text you would understand this better.

FOLLOWUP QUESTION:
unhuh...
how can it change direction without changing its magnitude??

therefore in time dt the velocity in downward direction becomes adt

where v' is equal to the vector sum ie v'=((v)^2+(adt)^2)^1/2, so how is the velocity supposed to be constant when the object is not decelerating tangentially?

ANSWER:
Like I said in my original answer, look this up in an elementary physics text — I am not a tutor. The reason your argument is invalid is that dt is infinitesimal so if adt is perpendicular to the velocity vector v, the result is to change direction. (Limit(as dt —> 0){ √[v² +(adt)²]}=v.) Just use your common sense: a car going in a circular path has a constant magnitude of velocity, right? And nonconstant direction or velocity, right?

QUESTION:
Two bodies A and B are at rest and in contact with each other. Now if some arrangements made body A exerts a pressure of 10 NEWTON on body B then according to NEWTON'S 3rd law of motion body B will also exert an equal force of 10 NEWTON in opposite direction so the resultant force should be zero, while practically we see that there will be a resultant force of 10 NEWTON acting on body b and if mass and other conditions of body B are such that it moves by applying 10 NEWTON force on it. Then it will start moving. HOW?

ANSWER:
You have this all confused. But, you are not alone! When doing this kind of problem, you must choose a body to focus on; only forces on that body affect its motion. So, if you are interested in what body B will do, you look only at that body. The force which B exerts on A is not a force on B. So, if B has a mass of, say, 2 kg and there are no other forces on B, it will have an acceleration of a_B=10/2=5 m/s². Since the 10 N force continues, A and B remain in contact, so a_A must also be 5 m/s². One force on A is the 10 N force from B which is in the opposite direction as the acceleration. But the net force on A must surely be in the direction of the acceleration, so there must be some other force (probably due to you pushing on A) on A which is bigger than 10 N and points in the direction of the acceleration. For example, if the mass of A is 4 kg, F-10=4x5=20 N, so F=30 N. Finally, you could look at A and B together as the body. In that case the forces they exert on each other do cancel, the acceleration is 5 m/s², and the total mass is 6 N; therefore, there must be some external force (you again) causing this 6 kg to have an acceleration of 5 m/s²: F=ma=6x5=30 N. All three ways of looking at this problem are consistent with each other.

QUESTION:
Can I make a simple bike reflector using 20 thru 40 degree angles? Everything I have read uses 45 and 90 degree angles.

ANSWER:
I do not know how 45⁰ would work. The point is for the incident light to be reflected in exactly the opposite direction from the direction it came in. The only way I know to achieve this is with corner reflector which is three mirrors connected as in the inside corner of a cube.

QUESTION:
Why simple harmonic motion is called simple?

ANSWER:
Take a look at the figure above. The red curves are all pure sine waves, just having different frequencies. These are all called simple harmonic motion because they may be expressed as a single sine wave. Now, take all four of these and add them together. The result is the black curve. This curve is still harmonic (which means that it is periodic, it repeats itself after some time, called the period, has elapsed) but it is not "simple" because to describe it you must use four different sine waves.

QUESTION:
I am a Martial Arts instructor who also happens to have a BS in mathematics (well, statistics, sorta like math). I like to explain to my students the science of the art in order to help ground them in reality from a lot of the BS involved in Eastern Mysticism, such as board breaking. My question is this, wooden weapons, such as nunchaku are usually either round/cylindrical or octagonal. A common technique is to place the weapon lengthwise down the arm and block an incoming attack with the wood. Traditional tonfa are octagonal, but modern police batons (with the side handle) are round. Which would protect the blocking arm from a baseball bat strike (for example) to a greater degree? Would an octagonal shape disperse the incoming force more than a round shape? I believe the octagon would spread the applied energy to a greater degree than a cylindrical/round shape and allow less energy to travel through the wood and into the arm, but I cannot find the math to back this up.

ANSWER:
The only reason that I would say that the octagonal shape would be better is that it has a larger area of contact with the arm and therefore would distribute the applied force over a larger area. For example, you would not just want a point in contact with the arm since all the force would be applied at a single place. Even better in this regard would be a rectangular shape. This is probably a negligibly small effect, though. I cannot see how the shape would have any effect as you suggest to " … allow less energy to travel through the wood …"

QUESTION:
How much energy is required to move the electron sufficiently far away from the proton such that it does not experience the proton's electric field. I am kind of confused by this question. The effect of electric field will never end as far as we take the electron from the proton. And if we use the relation V=E.L then if E=0, then L will be infinity?

ANSWER:
Technically, yes the field will be exactly zero only at L= ∞. But, that is only if your proton and your electron are the only things in the universe! You can calculate the ionization energy, the energy necessary to move an electron from the ground state of hydrogen to infinity; it is 13.6 eV=2.18x10^-18 J. You can also calculate the energy necessary to move it to any other distance. For example, the energy to move it to about 0.5 μ=5x10^-7 m (which is about 100 times the radius of the hydrogen atom) is 13.599 eV, essentially the same as to infinity. So you can see that most of the energy is supplied in close to the proton because that is where the field (and therefore the force) is strongest.

QUESTION:
So I've been trying to figure this for a while, if a positively charged and a negatively charged quark orbit each other or actually come in contact, also could they form a functional "atom" and be able to have electrons interact with them?

ANSWER:
Two-quark particles are called mesons (one quark plus one antiquark). If you had a positively- and negatively-charged quark you would have an uncharged meson and so it would not bind an electron and you could not have a meson+electron atom. However, if you had a positively charged pion, it could bind an electron. For example, a positive pion π⁺ is composed of an up quark (charge +2e/3) and a down antiquark (charge +e/3) and would have a charge +e and be able to form an "atom" which you might call pionium.

ADDED NOTE:
Whoops! Pionium is an "atom" composed of one π⁺ and one π^-. I cannot find any reference to this pion+electron system, so I guess you can call it whatever you like.

QUESTION:
i love physics and this question i asked to my teacher and principal but they couldn't answer it so my question is about third law of motion "every reaction has an equal and opposite reaction" so when a truck moving with constant speed hits a stationary car so according to newton's 3rd law of motion truck shoud be stopped after collision because car applies equal force on truck which it have during collision.

ANSWER:
Why should it be stopped? Certainly the truck experiences the force which the car exerts on it, but every force does not have the effect of stopping the object which experiences a force. If you are running toward a fly which is hovering at rest in your path, you feel the force of the fly but it does not stop you.

QUESTION:
Hi ...here is a link to Felix Baumgartner's freefall jump from space - https://www.youtube.com/watch?v=vvbN-cWe0A0 135,890 feet - or, 41.42 km (25.74 mi) In the video you see speed at which he is supposedly travelling, 729 mph (1173km/hr), 46 seconds after he jumped. The footage is cut then seconds later speed strangely falls to 629 mph...anyway after 4 mins and 18 seconds of free falling he releases parachute. My questions are What speed would he be going at a 4 mins and 18 seconds into freefall? Also how do you explain the deacceleration when it was at 729 mph then down to 629 mph?

ANSWER:
(You might be interested in an earlier answer.) There is a better video at https://www.youtube.com/watch?v=raiFrxbHxV0; this video shows data acquired by instruments on Baumgartner. Here some clips from that video:

The first three are the times also showing speeds; these are roughly the speeds you refer to in your question. The second three show speeds and altitudes. The third graph shows the whole history of altitude, speed, and mach speed. The speeds are all larger than indicated in your video. The speed at 4:20, 113 mph, answers your first question. The speed at 0:50 is about the maximum, 847 mph or mach1.25. The speed at 1:01 has dropped to 732 (about the same change as your video); I suspect this deceleration is due to two factors: there is getting to be much more air resulting in higher drag and, probably after he broke the record he oriented his body to increase drag and slow down. The official maximum speed after everything was calibrated was 843.6 mph.

QUESTION:
Will the drag coefficient be the same in air and water?

ANSWER:
There is no simple answer to this question. The drag coefficient C_Dis a constant characterized by the geometry of an object used to calculate the drag force F_D if it is proportional to the square of the velocity v: F_D= Ѕ C_D ρAv ² where A is the area the object presents to the fluid flow and ρ is the density of the fluid. Whether or not this equation is true depends on a quantity called the Reynolds number Re=L ρv / η where L is a length along the direction of flow and η is the viscosity of the fluid. It turns out that only if Re>1000 is the velocity dependence approximately quadratic. If Re<1 the drag force is approximately proportional to v. Anywhere between these extremes the drag force is a combination, F_D ≈Ѕ C_D ρAv ²+kv. C_D ≈constant only for the case Re>1000; the drag coefficient would then be the same for identically shaped objects. It is interesting, though, to get a feeling for what the relative speeds in air and water corresponding to Re=1000 are. The necessary data at room temperature are ρ _air≈1 kg/m³, ρ _water ≈1000 kg/m³, η _air≈1.8x10^-5 Pa�s, and η _water≈1.0x10^-3 Pa�s. Then v_water/v_air≈(1.0x10^-3/1000)/(1.8x10^-5/1)=0.056 (which is true for any Re). Taking a sphere of diameter 0.1 m as a specific example for the critical Re=1000, C_D=0.47, A=7.9x10^-3 m², L=0.1 m, v_air=1000x1.8x10^-5/(0.1x1)=0.18 m/s and v_water=1000x1.0x10^-3/(0.1x1000)=0.01 m/s. Since the speeds are relatively low, you can conclude that for many cases of interest the quadratic drag equation holds for both water and air. Therefore, for many situations you can approximate that the drag coefficient in air and water are about the same. For a specific case you should check the Reynolds numbers to be sure that they are >1000. The bottom line is that if the drag force depends quadratically on velocity, the drag coefficient approximately depends only on geometry, not the properties of the fluid. Keep in mind that all calculations of fluid drag are only approximations.

ADDED COMMENT:
As I said above, for 1<Re<1000, F_D ≈av+bv²where a and b depend on Re; e.g., a≈0 for Re>1000 and b≈0 for Re<1. So, for Re<1, a is just a constant and F_D≈av≡Ѕ C_D ρAv ² so C_D=2a/( ρAv )=2aL/(A ηRe )≡c/Re where c is a constant. If F_D is proportional to v, C_Dis inversely proportional to Re; this is called Stokes' law. As we saw above, Re>1000 leads to C_D=constant. Ferguson and Church have derived an analytical expression which very neatly reproduces data for the transition from Stokes' law (1/Re) to constant C_D. Calculations for a golf ball are shown at the left (constant C_D, the green line, is called turbulent in the legend). Note, as we have stated, that the transition occurs in the region 1<Re<1000. Note the sudden drop in the data around Re=500,000. I believe that this must be due to the dimples on the golf ball which reduce drag.

QUESTION:
In induced magnets, why does the end nearer the magnet have the opposite polarity to that of the magnet?

ANSWER:
Think about a piece of iron. It is like a whole bunch of tiny bar magnets, each of which has a N and S pole; but the iron is usually not a magnet macroscopically because all the tiny magnets are oriented in random directions. But, if you bring the north pole of a permanent magnet up close to the iron, the south poles of all the tiny magnets turn to point toward the north pole of the permanent magnet.

QUESTION:
I think if an object is turning, it has more gravity than an object which is not turning

ANSWER:
You are right. However, it really has nothing to do with the turning, per se. When something is turning it has rotational kinetic energy and therefore a spinning planet has more energy than an otherwise identical planet not spinning. In general relativity you usually hear about gravity being caused by mass warping spacetime. However, mass is just the most obvious source of gravity and what it is which really warps spacetime is energy density and mass has a lot of energy (E=mc²). The energy due to the turning is infinitesmal compared to the mass energy of the object so you would never be able to distinguish the difference due to the turning by looking at the gravity.

QUESTION:
I've heard that a bullet falls to the ground ~~at the same speed~~ [in the same time] no matter the charge pushing it forward. More powder and it moves faster but vertical speed is the same. I imagine two equal weight gliders of 20/1 and 30/1 lift/drag ratios would also express the same vertical speed? My question is, wouldn't this gravity effect also hold true underwater? An underwater glider expressing 30/1 L/D would travel 50% further, faster than a 20/1 L/D in a given vertical?

ANSWER:
I will assume that we are talking about horizontally fired bullets. All bullets do not hit the ground with the same speed, rather they all take (approximately) the same time to get to the ground; I have corrected your question because I believe that is what you meant. Your comparison to gliders with different lift/drag ratios is not appropriate, because the thing which results in the times of vertical fall being the same is the assumption that there is no lift at all. In other words, the vertical speed of a bullet will always be much less than the terminal velocity so there is no appreciable component of the drag in the vertical direction. (This is not true if the gun is fired from a very high altitude since there will be enough time for the verticle component of the velocity to approach the terminal velocity.) Underwater (I am just talking about bullets, not gliders) the bullet will not behave the same because the terminal velocity is much smaller than in air, in other words the drag is much larger.

QUESTION:
Quantum mechanics has popularized the idea of very small particles existing in two or more points in space during the same instant in time. Is this to be taken at face value as a fundamental aspect of quantum particles, or is this simultaneous existence just a concept used to express our inability to observe these particles without inadvertently altering their speed or location?

ANSWER:
See an earlier answer.

QUESTION:
I read from my history book that when nukes were first created we didn't know that the radiation their explosions produce is harmful to one's health, even to one's life maybe. So my question is, how come neither physicists or medical personnel ever figured out that radiation can be dangerous until we had first victims of radiation?

ANSWER:
Ever hear the expression "hindsight is 20-20"? It is unrealistic to expect that when something new is discovered that we should somehow know all the effects that might have on anything else. At the time radioactivity was discovered, nobody even knew what atoms were composed of or what their structure was. And it was found to be very tiny bits of matter (e.g. what we know as electrons today) and who would have thought that getting hit by something trillions of times lighter than a speck of dust could be harmful? It took experience before it was appreciated how dangerous it could be. One of the best known examples of such experience is the case of radium watch dials. Radium, mixed with a phosphor, was painted by women onto watch dials and it would glow in the dark. The workers were encouraged to point their brushes with their lips to make the fine lines required. Subsequently many became ill with cancer and other radiation sickness. It seems stupid now, but the dangers were not known until they were discovered. Also many of the health effects were long-term effects, the effects not appearing until years or decades after exposure. One example I know about from personal experience is the effect of radiation treatment for acne which was popular in the 1950s. I very much wanted to have this done but my father forbade it, making me furious with him. Years later people who had had this treatment started coming down with cancer —thanks, Dad! Yet another example is the shoe-fitting x-ray machine. When I was a kid it was really fun to buy new shoes because you could look down at an x-ray of all the bones in your foot and how well the shoe fitted you; again, when more was learned about radiation, these machines were all sent to the scrap heap. When it began to be appreciated that there were dangers, studies began to be done to try to set allowable dosage levels. But, it is unethical to do such experiments on people, so lab animals had to be used which always presents problems with scaling and other variables. Much of what we know today was learned after-the-fact by doing long-term statistical analyses over decades of medical records.

QUESTION:
Magnetic flux according to my book is total no. of magnetic field lines passing through a given area in magnetic field. ok but why there are not infinite no. of magnetic field lines, because magnetic field line are defined as the path that a magnetic north monopole would take if left in north part of magnet, so if i take a monopole and leave one atom away from the previous position then it should take a slightly different path and that path should be considered as magnetic field line,so in this way i can draw millions of line.

ANSWER:
Magnetic flux is well defined: Φ_M≡∫∫B�dA, the area integral of the magnetic field. If the integral is over a closed surface, the flux is zero; this is the famous situation which tells you that there are no point sources (called magnetic monopoles) of magnetic fields like there are for electric fields. If you like, you may interpret this as a number, but that is not really fundamental. If you talk about uniform magnetic fields which are perpendicular to a plane surface, Φ_M=BA , is the flux through an area A. Then, if you interpret Φ_M as a number, you would simply say, for example, there were 10 lines through an area of 1 m² if the magnetic field is 10 T. You could ask the same question about electric flux, Φ_E≡∫∫E�dA . This is perhaps a little easier to understand because you can have point charges. For example, if you have a 1 C point charge, the electric field 1 m from it is E=Q/(4 πε₀r²)=1/(4x3.14x8.85x10^-12x1²)=9.27x10⁹ N/C. The flux at a distance of 1 m from the charge would therefore be Φ_E =EA= 9.27x10⁹x4x3.14x 1²=1.17x10¹¹ Nm²/C. Incidentally, this is the flux regardless of where you measure it because the area of the sphere (4 πr²) surrounding the point charge appears both in the denominator of the field and in the numerator of the flux and cancels. Hence, you could say that there were 1.17x10¹¹ lines of electric field emanating from a 1 C point charge. You can see this from Gauss's law, Φ_E=∫∫E�dA=Q/ε₀ if the integration is over a closed surface enclosing the charge Q.

QUESTION:
Why degree of monochromaticity is always non zero?

ANSWER:
The practical reason is that you cannot make a laser which is perfectly free from electronic instabilities and noise. But even if you were able to make a perfect laser, you could not know its frequency perfectly. The more fundamental reason is that monochromatic light means light with a single frequency. But, the linear momentum of the photons is proportional to the frequency of the light. Because of the Heisenberg uncertainty principle, the only way you can know momentum of the particle perfectly is to be completely ignorant of its position; you need an infinitely long wave to know its frequency perfectly.

QUESTION:
This question is about relativity. If person leaves earth and heads toward a star that is 10 light years away, and he travels at near the speed of light, time will slow down for him. The round trip to the star may take him only one year, but when he returns, maybe 100 years have passed on earth. When he started out to the star, he knew it was 50 [I think you meant 10, right? The Physicist] light years away. It seems that for him, he will have traveled 20 light years in one year. That is faster than the speed of light. What don't I understand?

ANSWER:
First of all, time for the traveler will not slow down; rather, her clock will run slower as observed by an observer on the earth. To her, time will run perfectly normally. However, she will observe the distance to the destination shortened because of length contraction. You have rounded things to what they would be if she were going the speed of light, but let's do the whole problem as if she were going with a speed v=0.999c. The earth observer would see an elapsed time of 100x0.999=99.9 years (your 100 years). The traveler will see the distance shrunk to D'= 10 √(1-.999²)=0.447 light years, so the time of travel is t=2x0.447/0.999=0.895 years (your 1 year). You should read the earlier answer on the twin paradox.

QUESTION:
Supposing a weightless container is filled with water. I am sure the pressure at the bottom of liquid, P1 = atmospheric pressure + height of liquid x density of liquid x g = Patm + hdg, where Patm is atmospheric pressure and d is density of liquid. We can calculated this pressure as if liquid in region A and Region C does not exist. But how about the the pressure at the base of the container, that is P2. Is P2 same as P1? For P2, do we need to consider the whole weight of the liquid, that is inclusive of the weight of water in region A and C?

ANSWER:
It depends on what the force on the bottom is. If the container is in equilibrium, imagine it sitting on a table. The table would exert an upward force equal to the weight W of all the water, so P₂ would be W/A_bottom where A_bottom is the area of the bottom of the container. This assumes that atmospheric pressure is the same everywhere in the vicinity of the container; in other words, I have ignored the buoyant force due to the air on the whole container because it will surely be much smaller than W.

FOLLOWUP QUESTION:
Indeed the container is resting on the table. For P1, I use h x d x g. But for P2 you use W / Abottom, So can I say P1 not equal to P2 ?

ANSWER:
Yes, but I have to admit that my answer was misleading in that I gave you the gauge pressure, the pressure above atmospheric. So I should have said that P₂=P_atm+W/A_bottom. There is no problem that P₂ ≠ P₁ because the force which the container exerts on the table is not P₁A_bottom. Think about it —the sides of your container exert a downward force on the bottom of the container.

QUESTION:
Think of words (incapital letters) that can be read properly both with a mirror and without a mirror. What are these words?

ANSWER:
Well, this is not physics and I usually do not answer such questions. However, it is a cute puzzle. First, any letter in the word which is symmetric upon reflection about a vertical axis will look the same in the mirror: A, H, I, M, O, T, U, V, W, X, Y. Second, any such word must be a palindrome composed of the allowed letters, e.g. AHA, OTTO, WOW, YAY, MOM, TIT, TUT, … . But, wait, it is somewhat more difficult than that! Look at the pictures of the Camel cigarette package and its reflection, in particular, the word CHOICE on the side. So words composed of letters with symmetry upon reflection about a horizontal axis will also reflect the same: B, C, D, E, H, I, K, O, X will have a reflection which is the same as the original, but only if turned upside down. There is no restriction for these words to be palindromes; besides CHOICE, some others would be BED, HEX, BOX, BIKE, HIKE, CHIDE, …

QUESTION:
What is meant by virtual image?

ANSWER:
It is an image formed by light which appears to be coming from somewhere but is not actually coming from there. When you look into a mirror, it appears that the light from the image of what you are looking at is coming from behind the mirror, but it is actually coming from the mirror itself.

QUESTION:
Supposing that initial velocity is 0 Is a displacement vs. time (squared) equivalent to a Velocity vs. time graph? I tried to find the slope of displacement-time^2: d/t^2 = V/t which equals acceleration But what I read everywhere is that: d/t^2 = a/2 according to the equation d= (initial V)(time)+ (1/2)(at^2) I don't what is wrong with my calculations?

ANSWER:
If you plot d vs. t², the slope of the line will be Ѕa. You know that d=Ѕat², so suppose that I call t²=u. Then clearly the slope of the line d=Ѕau is Ѕa. Your equation d/t²=V/t is clearly wrong: d/t²= Ѕat² /t²= Ѕa and Ѕa≠v/t.

QUESTION:
Two of us disagree on part of a sol'n given by two people with Physics background, and I want to know if I am correct, or if I am missing something in the analysis of the problem...in case I have to explain it to a student. Question concerning Forces/impulse..... 50kg person falling @15m/s is caught by superhero , and final velocity up is 10m/s. Find change in velocity. Find change in momentum . It takes 0.1 sec to catch them.....ave Force is ? answers are: vel = 25 m/s change in mom.. 1250 kg*m/s, and ave Force = 12,500 N. Here's where we disagree: Person B says that 12,500 N is equiv. to 25 g ????? They try to explain that 250 m/s^2 accel. corresponds to 25g.... I said it makes No sense at all, [ I know the accel. is 250, but that doesn't in any way imply a 25 g "equivalence" to me ]. They then went further to "prove" their point........Here is their argument... 500 N/g = 12500N / ( )g ..... I agree the ( ) = 25, but say there is No justification for the 500 N / g in the first place...... any ideas where it comes from , or how to justify that value ? BTW I teach physics on and off at the HS level.... person B is an Engineer , I think

ANSWER:
Person B is wrong but has the right idea. (As you and your friend have apparently done, I will approximate g ≈10 m/s².) We can agree that the acceleration is a=250 m/s² and that is undoubtedly 25g. Now, we need to write Newton's second law for the person, -mg+F=ma=-500+F=12,500, so F=13,000 N. This is the average force by the superhero on the person as she is stopped, so the answer that the average force is 12,500 N is wrong. When one expresses a force as "gs of force", this is a comparison of the force F to the weight of the object mg, F(in gs)=F(in N)/mg=13,000/500=26 gs; this simply means that the force on the object is 26 times the object's weight. So neither of you is completely right, but if there is any money riding on this, your friend should be the winner because the only error he made was to forget about the contribution of the weight to the calculation of the force. I am hoping that superman knows enough physics to make the time be at least 0.3 s so that Lois does not get badly hurt!

QUESTION:
It is said gravitons are the expected particles to exchange the gravitational force. Do these particles exist in real?

ANSWER:
There is no successful theory of quantum gravity, so gravitons are "expected particles" but I would not call them "real" at this stage; hypothetical would be a better word.

QUESTION:
Why does it take more time to go to Dubai from Mumbai than to come from Dubai to Mumbai?

ANSWER:
There are winds at high altitudes called jet streams. The prevailing direction of these winds is easterly and their speeds are as large as 100 mph. Therefore when you travel east (as from Dubai to Mumbai) you have a tail wind, and when you travel west (as from Mumbai to Dubai) you have a head wind. Since the greatest speed an airplane can fly, say 600 mph for commercial jets, is airspeed, that is relative to the air, the ground speed with a 100 mph wind would be 700 mph going east and 500 mph going west.

QUESTION:
Why is the direction of angular displacement along the rotational axes?

ANSWER:
There are two ways in which you can rotate something about some axis, either clockwise or counterclockwise (as viewed from one side or the other). So, an angular displacement has two possible directions only. It is like a displacement in one dimension, like a bead on a long straight wire. You then call one direction on the wire plus and the opposite direction minus. The only way to get a similar directional assignment for a rotation is to note that the rotation axis is just like a one-dimensional axis, one way being assigned to be positive, the other negative. Usually this is done with the right-hand rule such that counterclockwise is a positive vector on the axis.

QUESTION:
Recently I came across such an article that Gravity isn’t actually a Force, It's the bending of space-time that cause objects to move on curve paths. If two objects of equal shape and masses are there they will bend the space-time in same way by same amount. Then according to that there shouldn't be any kind of force or something acting between them which isn't true i guess. Gravity forces will act b/w those two. Could you please explain me this contradiction.

ANSWER:
I do not understand your logic that because they both have the same effect on spacetime, they will feel no force. Although it is just a cartoon to help you qualitatively understand warping of spacetime, in this case it is illustrative. If you place a bowling ball on a tranpoline, what happens? The surface of the trampoline is warped. Now, suppose that you place two bowling balls a few inches apart from each other. What happens when you let go of them? They will behave as if there were a force between them and roll toward each other but what they are really doing is responding to the shape of the trampoline.

QUESTION:
I have a question about gravity, or more specifically how to calculate it for the purpose of a scifi fanfiction I am making. So Earth has the density of 5.5 grams per cubic centimeter, total mass of 5.97219Ч10²⁴ kilograms and radius of 6371 kilometers. So how do I put these together in a formula to get the 9.81m/s gravity?

ANSWER:
The density is irrelevant. You need to use Newton's universal law of gravitation for two point masses. The force F which each feels is F=GMm/r² where G is the universal gravitational constant, G=6.67x10^-11 N � m²/kg², M and m are the masses of the two masses (take M as the mass of the earth), and r is the separation between M and m. This also works if the masses are spherically symmetric. But, of course, you also know in your case that F=mg. So, if you solve for g, g=MG/r² ≈ 6.7x10^-11x6.0x10²⁴/(6.4x10⁶ )²=9.8 m/s².

QUESTION:
Why do electrons fall back to the ground state after jumping into a higher orbital?

ANSWER:
The simple answer is that a system will always seek a way to move to a lower energy, like a ball rolling down a hill and not up it or sitting still. The more complicated answer addresses whether there is, indeed, "a way" to achieve this. For atoms, the radiation of photons in the transition is usually electric dipole radiation and there is a corresponding quantum-mechanical operator, let's call it O_E1. Then, you have to look at what is called the transition matrix element for this operator, < Ψ| O_E1 |Ψ'>=∫Ψ� O_E1 �Ψ'dτ which is an integral over all space and Ψ' and Ψ are the excited and ground states, respectively. If this is nonzero, then the decay will decay with some half life determined by the value of the matrix element. If it is zero, the state will still probably decay to the ground state but via a different kind of transition.

QUESTION:
We all know about second law of motion. The example is about the coin which is placed above the cardboard and cardboard is placed above the beaker... if the cardboard is quickly pulled then the coin falls in the beaker but if the cardboard is slowly pulled then why does the coin come with the cardboard? Why does this happen?

ANSWER:
This is usually used to demonstrate inertia, Newton's first law; since you are interested in why it does not work if you move the cardboard slowly, the second law also must be used. To move the cardboard from rest it must experience some acceleration. If the coin is to experience an acceleration, that is move with the cardboard, some force must be exerted on it. The only force on the coin horizontally is the static friction and the greatest it can be is μmg (where μ is the coefficient of static friction) so the greatest acceleration the coin could have is μg; for example, if μ=0.3, and you caused the acceleration of the cardboard to be any greater than about 3 m/s², the cardboard would slip under the coin. But, if you gave the cardboard an acceleration of 0.2 m/s² the coin would move with it.

QUESTION:
How can it be said that the universe is only 13.7 billion years old? This seems ridiculous to me that anyone can make this assumption. Taking in the fact that there are black holes almost as old as the universe itself how is this possible? Stars live billions of years so the first group of black holes would be billions of years after the "big bang". Also due to the fact that the universe is expanding way faster and not slowing down like predicted it would seem to me the current model is wildly inaccurate. Can you please explain this to me? My understanding is that stars form and die out all the time so to be able to accurately determine the age of the universe another method would need to be devised. It is my understanding that galaxies form with the help of super massive black holes.(Which aren't even holes at all but super compacted spheres of matter so dense light cannot escape created by countless other bodies of mass.) Which attracts matter into tight clusters of stars we call galaxies. These galaxies would have to take billions of years to form after the creation of the first stars ever created and after the first black holes were created from some of those stars. Does the current model explain this?

ANSWER:
As clearly stated on the site, I do not do astronomy/astrophysics/
cosmology. I can at least say a little about your question, though. The age of the universe is not an "assumption", it is based on careful analyses of many measurements. The stars in the early universe were much more massive than the sun; this resulted from the fact that the early universe was essentially all hydrogen and a bit of helium. Stars began forming only 100-200 million years after the big bang. Very large stars burn much hotter than smaller stars and therefore have much shorter lifetimes, a few million years rather than a few billion years for less massive stars (e.g. the sun with an expected lifetime of about 10 billion years) which would explain why many black holes are very old. The rate of expansion of the universe is not inconsistent with the model of a big bang about 14 billion years ago (in fact, the rate is one of the measurements used to estimate the age of the universe) even though we do not understand all the details (e.g. dark energy and accelerating expansion).

QUESTION:
I would like to ask this question because I do not agree to this but it is written in my school handout. It says that work needs only two things to be calculated which is weight and height of displacement. Now , there is this question that says that two persons who have the same mass went to the top floor of eiffel tower. One used the stairs and the other one used an elevator. The thing is that our handouts states that they both exerted the same amount of work done because they have the same weight and they both have the same height distance covered. I do not agree simply because the man using the stairs carried his own weight up the distance while the the other one was being carried by an elevator and I immediately thought that the elevator was the one doing the work and not the person which is why they cannot have the same work done. If I am wrong then I would humbly accept it. However , if our physics teacher is wrong, I would need a legit source stating about the problem and saying that it was wrong or similar problem like the one given with an answer so I can correct my grade. I would really appreciate anyone's help right now.

ANSWER:
The point is that the net work done on each person was the same. Where the energy came from is different. On the stairs, the energy is supplied by what the guy ate for lunch. In the elevator, the energy is supplied by the motor lifting the elevator. Regardless where it came from, mgh of energy must be somehow supplied.

QUESTION:
How surface tension overcome insect weight and help it to stay on liquid surface, as surface tension is along liquid surface?

ANSWER:
The weight of the insect causes the water to be depressed where the legs touch the surface, so the surface tension has a vertical component.

QUESTION:
Can we make something like anti-gravity on Earth, or something similar to that? And i was wondering if we could make something like anti-gravity with magnets, to actually reduce the weight of some object. I was thinking to make a chamber with a magnet at the bottom, and take magnetized objects and put them on the top, so it will levitate.

ANSWER:
Your device will not work. The magnet on the bottom can be made to exert a repulsive force on a magnet at the top which tends to lift your chamber. However, Newton's third law requires that the top magnet exerts an equal and opposite force on the bottom magnet, so the net force on the chamber is zero.

QUESTION:
Is rotation a reason for earth being round?

ANSWER:
No. The reason that all large astronomical objects are nearly spherical is because the force which holds them together is spherically symmetric. If you have a point mass, the gravitational force another mass feels depends only on how far apart they are, not on the direction. If you want a lot more detail, see an earlier answer about gravity of a cylinder which shows how the tendency is toward a spherical shape.

QUESTION:
Using a spring balance, weigh (separately) a piece of wood and a container with water in it. Then weigh them again, but with the wood floating in the water. Does the reading on the balance change? There will be an upthrust on the wood (Archimedes' Principle), causing it to weigh less. I guess there will be an equal down-force exerted by the wood, resulting in the same weight being shown in both cases. This seems a fairly easy question, but I can't find the answer clearly stated anywhere.

ANSWER:
The measured weight will be the sum of the individual weights. The buoyant force is the force which the water exerts on the wood; Newton's third law requires that the wood exert an equal and opposite force on the water. As you correctly "guess", these two forces cancel out.

QUESTION:
Good afternoon. I started learning quantum mechanics and I have a question about Heinsenberg's principle. I was thinking about a photon created by an electron-pozitron annihilation. As I understand (and I hope I am not wrong here) all observers, no matter their state, can specify the point (and all will specify the same point) where the photon appears, and I think in a Wilson chamber we can do this thing. Anyway, we can specify its position with an error less than infinity. But at the same time we know it's speed with absolute precision, as, assuming we measure it in vacuum, it's speed is invariably c. So the product between (delta)x and (delta)p will be 0, as (delta)x is not infinity and (delta)p is 0. In other words we can predict it's position with a certain accuracy knowing at the same time it's speed (I am not sure, but i think that the (delta)p is not a vector but a scalar quantity). What is wrong about this as it seems the Heinsenbrg's principle is violated?

ANSWER:
The mass of a photon is zero but it has momentum. Therefore your notion that linear momentum p is mass m times velocity v must be wrong; the relativistically correct expression for momentum is p=mv/ √[1-(v/c)²] where c is the speed of light. For a photon, this is a little tricky because p=0/0, but you can also write that E²=p²c²+(mc²)² where E is the energy of the a particle of mass m, so p=E/c if m=0; since E=hf for a photon, where h is Planck's constant and f is the frequency, the linear momentum of a photon is p=hf/c . Therefore the frequency of the photon must be uncertain according to the uncertainty principle. Since the energy of a photon is hf, there will be an uncertainty in the energy of the photon.

QUESTION:
The units of the Hall coefficient is m^3/C that's, cubic meter per coulomb. How so we get this?

ANSWER:
You just need to know the definition of the Hall coefficient R_H, R_H ≡V_Ht/(IB) where V_H is the Hall voltage, t is the thickness of the sample, I is the current, and B is the magnetic field. The Hall voltage may be shown to be V_H=-IB/(nte) where n is the charge carrier density (units [m^-3]) and e is the charge (units [C]]) of a charge carrier. Therefore R_H=-1/(ne) so the units of R_H are m³/C.