Ask the Physicist!

QUESTION:
If a carousel is out of control, spinning at high speed, and suddenly it is stopped, there will be chaos and horses and riders flying out of the carousel structure. How do you explain this in terms of physics? Are they ejected due to the centrifugal force? Loss of centripetal force? And what happens to the kinetic energy? Is it transformed to what kind of energy?

ANSWER:
If the carousel is moving in a circle with constant angular velocity, the only forces horizontally are forces necessary to provide the centripetal acceleration. In the photo, the girl holds the pole and presses on the side of the horse with her leg to provide those forces; the horse is held by the force of the pole to which it is attached. All horizontal forces are toward the center of the carousel because the speed of everything on the carousel is constant. What happens if the carousel suddenly stop. All those radial forces quickly drop to zero. The tendency is for everything on the carousel to continue moving with the velocity they had before, but now in a straight line. But now, although the horizontal radial forces are gone, each object experiences horizontal forces opposite the direction of their velocities. The girl would initially move forward until she smashed into the pole; the horse would probably be held in place by the pole although the force required to stop the horse could very possibly bend the pole; someone standing on the spinning floor would only have the friction of the floor to stop her and would likely keep moving in a tangential direction. Nothing is "ejected", it just keeps going in the direction it was going when the carousel stops, unless something stops it.

QUESTION:
I push on a wall and am accelerated backward. Per Newton's third law, how does the acceleration of the wall manifest itself.

ANSWER:
You push on the wall with some force F and, as you note, Newton's third law says that the wall pushes with a force -F. So the magnitude of your acceleration a is a=F/m where m is your mass. The magnitude of the wall's acceleration A is A=F/M where M is the mass of the wall. For all intents and purposes, the mass of the wall is infinite so its acceleration is zero.

QUESTION:
Are electrons made of quark/anti-quark pairs? If yes, which pair?

ANSWER:
No. To the best of our knowledge electrons are elementary, i.e. they have no components.

QUESTION:
What force pulls the train? I am doing a science project about this simple electromagnetic train. This project looks so simple yet so complicated, here is the video of the project that I am working on.

ANSWER:
The construction details and a brief description of the physics are shown in another youtube video. Essentially, when the magnets touch the copper wire the battery causes a current to flow in the coil between the two batteries which causes a magnetic field in that section of the coil; each magnet, now being in that field, experiences a force moving it forward. Important things:

The same pole, north or south, must point away from the battery.

Be sure to use bare copper wire. Often copper wire has an insulating layer on its surface and this would not allow current to flow

Good luck on your project. It is really not as complicated as you thought.

QUESTION:
I can't grasp how two waves can pass each other on the same piece of string. For a wave to travel on a string each piece of string is, let's say, pulled up wards by the preceding piece of the string, and the wave propagates forwards. If a wave moving in the +x direction meets a reflected inverted wave in the - x direction, a node will be formed as one wave pulls that piece upwards and the other wave pulls it downwards. Therefore the piece of string doesn't move, so how can either wave travel past this point?

ANSWER:
What you need to understand to see why the two waves, moving in opposite directions are able to do so is a little bit of the physics of waves on a string. Any wave which moves through a medium is a solution of a very famous equation, the wave equation:

d²f(x,t)/dt²=v²d²f(x,t)/dx².

Here, x the position on the string, t is time, v is the velocity of the wave, and f(x,t) is the solution to the equation which will describe the shape of the wave at a time t. You may not know calculus and this equation is goobledy-gook to you, but all you need to know is that, mathematically, any function f is a solution as long as it of the form f(x-vt). The most commonly used example of waves is sinesoidal, for example, f=Asin(kx-ωt) where k is called the wave number and ω is the angular frequency of the wave. Note that ω/k=v. Also, to touch base with quantities you might be more familiar with, k=2π/λ and ω=2πf where f is the frequency of the wave and λ is the wavelength. Now comes the the important part: because any f will be a solution to the wave equation, if you have a wave traveling to the right, f_right=Asin(kx-ωt), and an identical shaped wave traveling to the left, f_left=Asin(kx+ωt), their sum will also be a solution to the wave equation. As you can see from the figure, waves traveling simultaneously right (red) and left (green) add up to a "standing wave" (brown) which does not appear to move but is still oscillating*. You are correct, there are nodes which do not move but both component waves go right on by them nevertheless, it just isn't apparent when you look at there sum. The fact that the net motion of the medium (string) is just the sum of all individual waves is called the superposition principle.

*If you're handy with trigonometry, you can calculate f_right+f_left which is apparently not moving even though you know that before you did the calculation you could certainly see that it was two waves.

QUESTION:
Hi, I have been toying with the concept of using gravity as a perpetual energy source, and have had a lot of pushback in the process of trying to ask questions. The conversation usually ends before I can get an answer. The skepticism surrounding perpetual motion is understandable. If I understand correctly the issue is that closed systems lose energy to various forces such as friction. Firstly, does it count as perpetual motion if the energy is supplied constantly from outside forces?Secondly, if not, is there a proper term for what I've described?

ANSWER:
I am afraid that you have it exactly backwards. In a closed system, defined as one which has no external forces acting on it, energy is conserved. In a closed system where friction is present, when kinetic energy is lost (e.g. a spinning wheel slowing down or a box sliding to a stop on a surface), the lost kinetic energy shows up mainly as heat. And, if you do work on a closed system to keep it moving forever, that is not what we mean by perpetual motion. Proper term for what? If you mean motion that you keep going by pushing on it, it has no particular name.

QUESTION:
First, I am way out of my field of understanding here so please keep it simple. I watched some videos on E=MC2 which led to how light reacts differently than matter at high speeds causing time to slow down when moving fast. My question is, if I was to shine a flashlight perpendicular (90 degrees) to the direction traveled am I correct to say if I was moving at half the speed of light the beam would actually be at a 45 degree angle and when travelling at the speed of light the beam would be horizontal (0 degrees). This would also be true whether the beam was inside or outside of the spacecraft, correct?

ANSWER:
No, you have it wrong. The most important thing to remember here is that the speed of light, c, is the same for all observers. If you are in the rocket and shoot a beam of light straight across the ship, it will go straight across the ship in some time t_s and the distance it goes will be ct_s; if the width of the ship is w, t_s=w/c. Now, someone on the ground will see the rocket moving by with some speed v, so the point on the inside of the ship where the light beam will have moved a distance vt_g when the light strikes it; but since this is light, the distance it will travel is longer because the light has to go farther. Now, to find the angle the light relative to the perpendicular to the velocity v, we note that sinθ=v/c (t_g cancels out). In your question, since v=c/2, so θ=sin^-1(0.5)=30°. This example also demonstrates time dilation because the two observers see different times of transit of the light. If you work it out, t_g=t_s/√[1-(v²/c²)].

QUESTION:
The radius of earth is 6440 km . Suppose, the angle between Canada and USA inside the earth is 10 degrees , then what is the distance between them? I have been stuck with this problem . Can't i answer this problem using The Cosines Laws? If i do so....does it would go wrong? please answer my question...? It is not a homework question . i am thinking to solve this one with a different way....

ANSWER:
This is a very strange question because the USA and Canada are adjacent and therefore the angle would be 0°. However, you can easily find the distance between two points which subtend 10° just by knowing the definition of radian measure of angles. The angle θ subtends a distance s for a circle of radius r. This angle, in radians, is defined as θ=s/r. Since the circumference of a circle is 2πr and there are 360° in a circle, there are 2π radians in 360°. So 10°=10x(2π/360)=0.175=s/6440, so s=1124 km.

QUESTION:
Hi! I would like to settle a bet with my father. My questions is that if I was sitting in a chair that was tied to a rope, and the rope was divided by a pulley on the ceiling, would it be possible to lift yourself up just by pulling down on the other side of the rope.

ANSWER:
The figure on the left shows all the forces on you: T is the force which the rope exerts up on you*, N is the force which the seat of the chair exerts up on you, and Mg is the force of gravity (your weight) down on you. Choosing +y up as indicated, Newton's second law in the y direction is, for you, N+T-Mg=Ma where a is your acceleration and M is your mass. The figure on the right shows the forces on the chair: -N is the force you exert on the chair, the same magnitude but opposite direction as the force the chair exerts up on you because of Newton's third law; T is the force which the rope exerts up on the chair, the same magnitude as the tension on you because the rope and pulley are assumed to have negligible mass; mg is the force of gravity (weight) down on the chair. Newton's second law in the y direction is, for the chair, -N+T-mg=ma where a is the acceleration of the chair (the same as yours) and m is the mass of the chair.

Let's first assume you are exerting just the right force T on the rope so that you are at rest, so a=0. In that case, if you solve the two equations you will find that T=(M+m)g/2. In other words, you need to be able to exert a force down on the rope which is equal to half the weight of you and the chair combined to hold the chair from falling. All you need to do is be able to pull just a little harder to move upwards. That answers your question but it interests me to look at a case where a is not zero.

If you solve the two equations for a and N you get

a=[(2T/(M+m))-g] and

N=T(M-m)/(M+m).

A few observations about what these results tell us:

If you let go of the rope, T=0, a=-g, N=0; you and the chair are in free fall.
If m=M, N=0, a=(T/m)-g; the motion of you and the chair are identical even though there is no interaction between you and the chair.
You will accelerate upwards if T>½(M+m)g, half the total weight. Your strength is the only thing limiting how fast you can accelerate upwards.
If 0<T<½(M+m)g you will accelerate downward with an acceleration smaller in magnitude than g. Of course T<0 is not possible for a rope.

*It is important to note that, because of Newton's third law, if the rope exerts a force up on you, you exert an equal and opposite down on the rope. T is a measure of how hard you are pulling. How large T can be depends only on how strong you are and how strong the rope is.

QUESTION:
We define image as the intersection point of reflected rays and the focal point is the point at which light will meet after reflection. So, why is it that we have different positions of image when object is placed at different distances from concave mirror and not always at focal point except when object is placed at infinity?

ANSWER:
Because only rays which come in parallel to the optic axis are reflected through the focal point. This occurs only (approximately) for objects very far from the mirror.

QUESTION:
Does radiation from your phone stay on your hands after you use it?

ANSWER:
Absolutely not. Any radiation from the phone is electromagnetic, essentially radio waves. It is transient and does not "stay" anywhere. Even when you hold the phone the radiation is not "on your hands" but harmlessly passing through them just like the waves from some radio station is passing through your whole body.

QUESTION:
When a uranium or plutonium atom is fissioned, energy is released visa E=MC2. I believe this energy is in the form of photons. What I do not understand is how these photons (light) create such high temperatures in an uncontrolled nuclear explosion. Can you please explain?

ANSWER:
Actually, emitted photons account for only about 2.5% of the total fission energy. Emitted neutrons account for 3.5%. Most of the energy is contained in the kinetic enerngy of the fission fragments, 85%. The other approximately 9% of the total energy shows up later when the fission fragments, which are not stable, decay radioactively, mainly from β-decay. The high temperatures are due to the kinetic energy of the fission products which have speeds typically 3% the speed of light.

QUESTION:
I'm embarking on a horological project. I need to buy a very particular type of spiral spring. I guess it's a type of torsion spring. The only supplier I can find who sells these springs classifies them with two numbers. The overall diameter in mm and the torque in gm/cm/100°. (As noted below, I assume that this means the torque on the spring when the angle is 100° and it should be gm·cm, not gm/cm.) I know the spring I need. But I only know in terms of the CGS System. I'm following an old book and I've ascertained that the spring I need has a "CGS number" of 0.76. So, my challenge is to find a way to convert between these two unfamiliar units of torque measurement. A "CGS number" and gm/cm/angle. There are plenty of calculators for converting between different units of torque. But I don't think it's as simple as that. I have two questions. I know a bit about the CGS system and there is loads of info online. But I'm not clear what it is in the context of classifying a torsion spring. The book only says "the CGS number indicates the restoring couple of the spring when its diameter is 1cm." So what is the CGS number exactly? Is it dyne-centimetres? If so, then that's easy to work with since that's a measure of torque I'm familiar with. So I'm half way there. If it's something else, then perhaps you can help explain. Then there is the question of gm/cm/angle. The problem is that none of the calculators I've seen have a notion of angle. This is specific to torsion springs. The torque, in this case, is the working load when the spring bends 100°. Will I ever find the correct spring? Sadly, I have spoken to the manufacturer, and they were unable to help me.

ANSWER:
I think the "number" of interest is not a torque, rather the spring constant associated with the spring for angular displacement. First, a review of linear springs which when stretched stretch a distance x when a force F is applied to the spring. It is experimentally determined that, to an excellent approximation for a good steel spring, is that the distance stretched is proportional to the force applied, x∝F or F=kx. The proportionality constant k is called the spring constant and is measured, in SI units, as N/m (Newtons per meter) [dynes/cm in CGS units, lb/ft in Imperial units]. k indicates the stiffness of the spring and this equation is called Hooke's law. The form of Hooke's law for angular motion (as in a torsion spring) is that the angular displacement θ is proportional to the applied torque τ or τ=κθ; I believe the quantity you want is κ (kappa) and is measured, in SI units, N·m/radian [dynes·cm/radian in CGS units, lb·ft/degree in Imperial units]*. I have noted that κ is often called, by manufacturers and vendors, the rate of the spring since it is the rate of change of torque per unit angle. I am puzzled by your reference to a constant defined as gm/cm/degree; 1 dyne=1 gm·cm/s², so κ should have units gm·cm²/s²/degree; sometimes the gram is treated as a unit of force (called a gram force where 1 gram force=980 dynes), in which case κ would be gm·cm/degree, not gm/cm/degree. So, there must be somewhere in the reference you are using a definition of what the CGS number is. Suppose that it is 0.76 dynes·cm/radian (which would be the likeliest units if purely CGS system of units is applied) but you want it in lb·in/degree; then 0.76 (dyne·cm/radian)x(2.25x10^-6 lb/dyne)x(0.394 in/cm)/(57.3 deg/radian)=1.18x10^-8 lb·in/degree. Or, suppose you wanted it in (gm force·cm/degree); then 0.76 (dyne·cm/radian)x(gram force/980 dyne)/(57.3 deg/radian)=1.39x10^-5 (gm force·cm)/degree).There are lots of unit converters around, my favorite is here.

The choice of units for κ which it seems most vendors use is lb·in/deg. The best calculator I found to get an idea of the magnitude of these springs can be found here. Until you are able to deduce the units of the so-called CGS value of 0.76 you cannot know what it is.

*Sometimes κ is implicitly given by specifying the torque at a particular angle.

QUESTION:
If you lined up frictionless gears over an exceptionally long distance, could you effectively communicate faster than the speed of light through one person moving the first gear and a second person reading/interpreting the resulting movement of the last gear?

ANSWER:
I have answered this question in many guises many times. In your variation, each gear will experience a force from the previous gear in the line and exert a force on the next gear. The time it takes for the force to travel from the input force location to the output force location is determined by the speed of sound in the gear. The message you send would travel at the speed of sound, much smaller than the speed of light.

QUESTION:
If you are driving side by side at 60mph with another car and throw a can of coke at them would the can be travelling at the same speed so hit the target as if you were parked next to each other or merely disappear behind your vehicle?

ANSWER:
When you throw it out the window it originally has a forward component of its velocity of 60 mph. At this speed, the can will experience a very significant force slowing its forward velocity down because of the air drag. So the can will not keep pace with the two cars and will appear, from either car, to be accelerating backwards. So, no, the can will not hit the other car where it would have if the cars had not been moving.

QUESTION:
Under relative motion, it's experimented that A person sitting on a back seat in a moving bus has a speed with respect to the earth which is the same as that of the bus. But if he now walks towards the driver of the bus,he has a speed relative to the earth which is more than that of the bus. Suppose the bus is moving at 100km/h and the person walks at 5km/h towards the driver, his forward speed relative to the earth is 100+5= 105km/hr. But when he walks back from the driver to his back seat at the speed of 5km/hr his speed relative to the earth is now 100-5= 95km/hr. WHY the increase speed of such person when moved forward and the decrease when moved backward

ANSWER:
The problem you are having is that you do not understand the phrase "relative to the earth". Let's alter your example by having the train moving at 5 km/hr. If you are in the train walking 5 km/hr in the opposite direction, someone standing by the side of the tracks will see you standing still; the same as running on a treadmill in a gym with the treadmill and you moving in opposite directions with the same speed. If you walk in the same direction as the train is going, you will be seen by the trackside observer as moving with a speed of 10 km/hr.

QUESTION:
IF WAVES need a medium to transmit energy but light can travel through a vacuum. Does dark matter have Anything To Do With Lights Ability To Do This?

ANSWER:
There is absolutely no mystery about light waves; electromagnetism is arguably the best-understood theory in all of physics. You may be sure it has nothing to do with dark matter. Dark matter is arguably the least-understood theory in all of physics. There are many features of the universe which do not fit into our current understanding of physics and introducing some kind of particle which interacts with the rest of creation only via gravity would solve a lot of them; all attempts to directly detect dark matter have failed and until there is direct observation we understand nothing. To my mind, failure to observe dark matter may be indicative that we do not understand gravity as well as we think we do.

QUESTION:
Bowling ball vs. billiard ball What would ultimately travel further, a 12 lb bowling ball or a standard billiard ball: if thrown/rolled with the same amount of physical strength/athletic ability, at ideal trajectory for each, on a frozen lake (not without small imperfections but as close to perfect as could be found naturally), under normal atmospheric conditions (Wisconsin in the winter), with little to no wind? Would temperature play a roll in determining a winner? Would the small difference in friction coefficients be significant? Any other variables Etc... My brother and I have been debating this for years while ice fishing.

ANSWER:
(You might not want to read this rather long introductory paragraph if you are just interested in the final answer.) This problem was very interesting to me because I initially thought it was pretty easy to understand. The billiard ball starts out with much greater speed than the bowling ball, jumps into the lead. Neglecting air drag is usually what is done initially in considering such problems and there has to be some kinetic friction for the balls sliding on the ice; but because the kinetic friction is proportional to the weight, each ball experiences the same rate of change of its speed as it slows. Therefore the billiard ball is the obvious winner. When I initially solved the problem this way using reasonable numbers, I found that the bowling and billiard balls had initial speeds of 14 mph and 81 mph, respectively; the bowling ball went 686 ft and the billiard ball went 21,700 ft, a little more than 4 miles! When I was writing the final paragraph I discussed the possible effects of two things I had neglected, rolling of the balls rather than sliding and the possibility of air drag. Just for fun I estimated the effect of air drag and discovered that at the beginning, particularly for the billiard ball, the drag force was considerably larger than the sliding friction. I realized that I had to redo the whole problem, now much more difficult and mathematically involved than originally. Because the air drag estimate I used is valid only if you work in SI units, I will do that throughout and convert to imperial units when needed. The mathematics gets very complicated and I will only sketch my calculations and whoever is interested can fill in the gaps.

First I will tabulate quantities needed for the problem:

We need to quantify what is meant by "…the same amount of physical strength/athletic ability…" I will assume that the thrower exerts a constant force F over a distance d on both balls. Physics says that the result is that a kinetic energy K=½mv² is acquired where m is the mass of the ball and v is the speed it has when it leaves the thrower's hand; this kinetic energy is equal the the work W done which is W=Fd; that is, Fd=½mv². If you solve this equation for v you find the speed, v=√(2Fd/m). I will use Fd=110 J (about 25 lb over 1 m) because this value would give the billiard ball a speed of about 80 mph which seemed reasonable to me.
The masses of the balls are 0.17 kg=6 oz for the billiard ball and 5.4 kg=12 lb for the bowling ball.
I will use a coefficient of sliding friction for each ball which is very small, μ=0.01. Each ball experiences a frictional force of f=μmg=ma (Newton's second law) where a is the acceleration and g=9.8 m/s² is the acceleration due to gravity.
For the air drag I will use the approximation D=¼Av² where A is the area presented to the oncoming air (πR²) and v is the speed. The radii of the balls are 0.0286 m and 0.108 m. As noted above, this equation is only valid for SI units because the factor ¼ includes things like air density, drag coefficients, etc.

We are now ready to start doing the physics. Start with Newton's second law

ma=m(dv/dt)=-¼Av²-μmg

dv/dt=-Cv²-μg (C=¼A/m).

This equation is integrable. The result is

t=tan^-1[v√(C/(μg))]/√(Cμg)|_v(0)⁰.

Evaluating this expression between the limits v=v(0) to v=0 and inverting to solve for v(t) is straightforward but messy. The essential results are illustrated in the two graphs below. The effect of air drag on the bowling ball is modest but noticeable, reducing the time to stop by about 10 s. The effect on the billiard ball, though, is enormous; because of its high speed at the start, it is slowed much more than due to sliding friction alone. Still, it stayed in motion about 20 s longer than the bowling ball and was going faster most of the time, so certainly went farther before stopping. Still, it would be useful to find the position as a function of time to determine how far each ball went.

To find the position, x(t), is straightforward but again messy: integrate the differential equation dx/dt=v(t) where v is the function being plotted in the graphs above. The results are shown in the graphs below. The bowling ball stops at t=54 s at a position x=158 m=518 ft; the billiard ball stops at t=77 s at a position x=535 m=1755 ft. The winner is the billiard ball which goes more than three times farther than the bowling ball! I believe that regardless of any details of the friction and drag forces or how we choose the quantity Fd, the billiard ball will always go farther if the initial kinetic energies are equal.

Finally, I want to discuss rolling vs. sliding. It is really hard to get an object rolling without slipping on a very slippery surface. A sliding ball will not roll until it is nearly ready to stop, and that would have very little effect on my calculations. Also, rolling friction is velocity-independent and proportional to the normal force so it would affect both balls the same.

QUESTION:
How can a photon have the properties of a wave if it is not in an element? It is said to be in a vacuum so ostensibly it is not in an element, but if it is a wave it must have an element. Has science explained this or is it still an unsolved question?

ANSWER:
You have struck on one of the most important realizations in physics: light does not need a medium to move through, it moves through a perfect vacuum. Other waves we know need a vacuum: sound through air (or just about any other medium), water waves could not propogate without water, waves on a guitar string could not exist without the string, etc. But light is unique and does not require a medium. The photon is just the smallest possible piece of light and has wave properties. No unsolved question here!

QUESTION:
I have a question about gravity. In order to general relativity mass bends the space. I saw a simulation of gravity that a man put a mass on a surface like silk. that the mass bends the silk. and we put another mass they attract to each other. that 2D silk is an examinition of our #d world. I wanna know why masses attract to each other. is it becuase of bending space?

ANSWER:
That simulation is a good way to understand how space bending can alter how objects move but keep in mind that it is only a cartoon to illustrate the warping of spacetime. See my faq page.

QUESTION:
If repulsion forces between protons is high why hasn't the nucleus exploded

ANSWER:
It is true that the Coulomb force experienced by protons in a nucleus is very strong. However, the nuclear force at very short distances is stronger and wins the "tug-of-war" and holds them inside. However, this strong force is also a very short-range force and if you pull a proton just outside the nucleus, it will be repelled and fly off. It is also interesting to ask why the nucleus doesn't collapse if the nuclear force is so strong. The nuclear force becomes repulsive if the protons get too close together; this is called the saturation of nuclear forces.

QUESTION:
What would be a ballpark estimate of all the mass converted to energy (lost) by all the fusion reactions in all the stars that are or have ever been in the last 13 billion years (lots of assumptions)? Would this lost mass effect the overall gravitational attraction of the universe? Would it be significant enough to be a factor in the current acceleration of the universe?

ANSWER:
I would not presume to make even the roughest estimate you ask for. However, it is not relevant to the questions you seem to be interested in. According to general relativity, gravitation is caused by energy density, not specifically mass. (Don't forget, mass is just a form of energy, E=mc².) So changes of mass will affect the details of the gravitational field in some volume of space but if you get very far from that volume, the gravity you see originating from that volume will not change.

QUESTION:
Our 5-year-old niece wants to know why it's so hard for her to ride her Big Wheel uphill (3% slope) on her family's driveway. My BSN-RN wife would like a friendly, informative way to break that down for her (Daphne), including whether her weight (50 pounds), her position on the Big Wheel, the coefficient of friction on said drive wheel, and other factors play a part. Said aunt would also like a kid-friendly way to calculate the horsepower necessary for her to ascend that grade on her Big Wheel for perhaps 100' (the driveway's length). No, this is in no way homework. Except it's at their home. And, well, it is work.

ANSWER:
A little tough for a five-year old! Here, maybe, are some suggestions:

Get her to understand Newton's first law: an object at rest or moving with constant speed has all the forces on it cancelling each other out.
- For example, have her imagine imagine sitting still on level ground. Her weight (a force pulling her down) is trying to make her fall down but the ground (a force pushing her up) is holding her up. If the ground disappeared she would fall downward.
Get her to understand Newton's second law. If the forces are not cancelling each other out, the object will start moving and continue to speed up in the direction of the uncancelled forces.
- For example, have her imagine sitting on the trike on the incline with her feet off the pedals or the ground. She will start moving down the incline. What is happening is that her force, which is still straight down, must have a little bit of itself pushing her down the incline. (This is basically a component of the weight but the concept of vectors and their components is likely too hard for a five-year old to comprehend.)
Finally, if, on the incline, there is a piece of the weight which she will have to compensate for in order to keep that force from making her go down the hill: she does this by pedaling to create a force up the incline: the force she generates must be greater or equal to the piece of the weight pointing down the incline. On level ground there is no unbalanced weight so the pedaling is much easier.
Friction, I would opine, is an unnecessary complication.
If she were heavier, the piece down the incline would be bigger so it would be harder to pedal.
(This part is for you, the little girl will not get it.) Your question about the horsepower is incomplete because to calculate the power you need to know the time it takes her to go the 100 ft. Suppose she maintains a speed of about 2 ft/s. Now, the component of a 50 lb weight on a 3% grade is about 50x0.03=1.5 lb so the work done is 1.5x100=150 ft·lb. The time it takes her is (100 ft)/(2 ft/s)=50 s, so the power is 150/50=3 ft·lb/s=0.0055 hp.
Friction has little to do with it because the friction is not significantly different on level ground vs. the incline. Position on the trike plays no role since any change would be the ease of pedaling which would be the same in either case.

QUESTION:
what would a light/sound wave look like in 3 dimensions, all diagrams are in 2d and it just came to mind that it wouldn't be like that in real life

ANSWER:
Here is a cylindrical wave. The blue surfaces represent wave fronts.

QUESTION:
I am currently doing an experiment on the metronome pendulum. As you know a metronome pendulum has two types of pendulum which is the inverted pendulum and simple pendulum. It is called as double-weighted pendulum. This type of pendulum has two weight which is one above the pivot and one below the pivot. In my experiment I want to find the relationship between the varying mass of the top mass of metronome pendulum and its angular frequency but I cant seem to find any formula that describes my metronome pendulum. What formula for angular frequency should I use because from my data, I will be looking for its period of oscillation. There are too many angular frequency formula for different types of pendulum so I am confused on which formula I should be using.

ANSWER:
I will show you how to get a general answer. Then you can apply it to the conditions of your experiment. Newton's second law for rotational motion is τ=Iα=I(d²θ/dt²) where τ is the net torque about the pivot (the blue cross in the figure), I is the moment of inertia about the pivot, α is the angular acceleration, and θ is the angle of the system at some time t. If the system is a pendulum with no forces exerting torques except weights (vertically down), all torques are proportional to sinθ where here θ is the angle relative to the vertical. Now our equation may be written as

(d²θ/dt²)=-(|τ|/I)sinθ

where |τ| is the torque about the pivot when the pendulum is horizontal (θ=π/2). The negative sign results from the fact that the angular acceleration and the angle are always opposite, a restoring torque. This equation is extremely difficult to solve, but if the angle is small (much smaller than 1 radian), you can use the small angle approximation, sinθ≈θ, which gives us

(d²θ/dt²)≈-(|τ|/I)θ

This is the simple harmonic oscillator equation for an oscillating system for which we know the angular frequency, ω=√(|τ|/I). So, let's apply it to your situation as illustrated in my figure. There are two point masses and the mass of the rod itself which I will take as m. The rod has a center of gravity (orange cross) a distance of L₃ from m₁as shown; if your rod is uniform, L₃=L/2. (If the mass of the rod is much smaller than both the point masses, you can approximate m as zero.) Now, since all weights are of the form W=mg, you can write |τ|:

|τ|=g(m₁L₁-m(L₃-L₁)-m₂L₂).

The moments of inertia for the two point masses are

I₁=m₁L₁² and I₂=m₂L₂².

The moment of inertia for the rod is trickier, so to avoid a lot of messy algebra I am going to assume a uniform rod where L₃=L/2. I find

I_rod=(m/(3L))(L³-L₁³)

The total I is just the sum of the three moments of inertia. So, finally,

ω=√[(g(m₁L₁-m(L/2-L₁)-m₂L₂))/(m₁L₁²+m₂L₂²+(m/(3L))(L³-L₁³)]

And, if m can be neglected,

ω=√[g(m₁L₁-m₂L₂)/(m₁L₁²+m₂L₂²)]

QUESTION:
So I have just thought about this, let's just say that hypothetically you were able to fall through the earth completely. I was wondering what the gravity would do to you if you went in one side and came out the other

ANSWER:
See the faq page.

QUESTION:
If everything is expanding, and every galaxy is moving away from the Milky Way, why is the Milky Way and Andromeda predicted to collide in 3.5 billion years?

ANSWER:
See an earlier answer. See a simulation video.

QUESTION:
this is for a horror book I am writing. how fast can a bat fly, if it weights 4500 pounds and is 25 feet 6 inches tall? and in 1 flap of its wings, can rise a rise a quarter, of a mile? also how wide would its wingspan be?

ANSWER:
The best I can do is to estimate how the wings would have to be if the ratio of weight (W) to wing area (A) is the same as a real bat. I find that a typical value for that ratio for bats is about R=W/A=20 N/m²=0.42 lb/ft². So, for your case, A=W/R=4500/0.42≈10,000 ft² or one wing has an area of about 5000 ft². If I model a wing to be a triangle with base B=25 ft and a width H, 5000=½BH=12.5H or H=400 ft, a wingspan of about 800 ft. Since it has the same R as for a real bat, it should be able to fly as fast as a real bat which can be as high as 100 mph.

QUESTION:
I am a part time instructor and one of the things I teach is rigging. An important part of the class is calculating sling stress. I was asked a question that I couldn't quite answer. I don't believe it was my high school or college physics teachers that failed me rather the more than two decades since I sat in a physics class. The question relates to vector forces. If I am hoisting a load of a given weight (say 2000 lbs) and I have two slings that are at a given angle (say 45 degrees) from the load. Each sling would be carrying half the load or 1000 lbs straight up. However, the sling in this example would have 1414 lbs of force (inverse sine if I remember correctly). The question I was asked was regarding a crushing or compression force. Since each sling is carrying 1000 lbs vertical load that means it has 414 lbs of force perpendicular to the vertical. Does that mean that there is 414 lbs of "crushing" force or is it 828 lbs since each sling is pulling towards the center of gravity.

ANSWER:
After an email exchange with the questioner, I was able to determine that "sling" is just a rope or a chain or a string etc. In the diagram I have shown all the forces on the load, the tensions in the slings, T₁ and T₂ and the weight of the weight, W. The magnitudes of the tensions are equal, T₁=T₂ =T. Also shown are the horizontal (x) and vertical (y) components of the tensions; they are T_1x=Tcosθ, T_2x=-Tcosθ, T_1y=T_1y=Tsinθ. So the equilibrium equation is 2Tsinθ-W=0; therefore T=W/(2sinθ). So, for your example, sinθ=cosθ=1/√(2), W=2000 lb, and T=1414 lb. Now, your interest is in the horizontal components; each is Tcosθ=W/(2tanθ)=1000 lb "crushing force" on each side. Your main error is that you treated the vector forces as scalars, you assumed T=T_y+T_x where in fact T=√(T_y²+T_x²).

QUESTION:
Since everything in the universe is always moving, is the idea of being still/idle/stationary just an illusion?

ANSWER:
I would not put it like that. If you are in a system where Newton's first law (if the net force on an object is zero, it will remain at rest or moving with a constant velocity) is correct, you may think of yourself at rest. This is called an inertial frame of reference. But the catch is that this is not the only frame of reference in the universe where Newton's first law is true, any other frame which moves with a constant velocity relative to yours is also an inertial frame. In other words, there is no such thing as absolute rest. A frame accelerating relative to yours, however, is not an inertial frame.

QUESTION:
I am a cancer patient about to receive Y-90 treatments. If Y-90 has a half life of 64 hours: 1. When does that time clock begin running and 2. How can they ever have a full strength dose on hand at any given time?

ANSWER:
⁹⁰Y is the decay product of ⁹⁰Sr which has a half life of about 29 years. 90Sr is an abundant byproduct of the fission of uranium and therefore available in the waste of reactors. When the ⁹⁰Y is needed it can be chemically separated from the ⁹⁰Sr. I am sure that the hospital cannot keep a supply on hand and will have it delivered when it is needed for your treatment. It does not matter how long it has been since it was separated, only that the radiation level be correct for the prodedure. It probably arrives with a level too high and they wait for it to be at the needed level.

QUESTION:
I am struggling with the concept of relativity, as it relates to time. Not in terms of the underlying principle or the mathematics. These have been confirmed experimentally and are now in everyday use through technologies such as satellite navigation systems. My concern relates to the way that relativity treats time as a variable rather than an absolute quantity. It seems to me that we can only measure or experience time by reference to some physical process or change, whether that is the oscillation of a quartz crystal or the lifetime of a muon! Any experimental proof of time dilation as a result of the effects of either velocity or gravity really just relates to the way that we experience the passage of time and does not exclude the possibility that for the universe, as a whole, time is absolute and unchanging, passing in a constant fashion regardless of relativity and whether we measure it or not. Therefore, although relativity theory works in practice, this is only because it deals with our limited understanding of, and ability to measure and/or experience, the passage of time. Is this nonsense or is it a possibility? It follows from this reasoning that we can never prove experimentally that time either speeds up or slows down, only that the rate of how we measure or perceive time can vary. It also implies that time (and indeed space) are concepts that we will never fully explain as they ultimately cannot be qualified or quantified by measurement or mathematics.

ANSWER:
You are correct that "…time is absolute and unchanging, passing in a constant fashion…" but only in your frame of reference; any clock at rest relative to you runs at a constant rate. But, what special relativity tells us is that clocks not at rest relative to us do not run at the same rate as ours. I have always felt that the best way to convince someone that this is the case is the light clock. In order that this example is convincing to you, you must accept that the speed of light is a universal constant in all frames of reference; but this is an experimentally well-verified fact and also is a consequence of the principle of relativity—that the laws of physics are the same in all inertial frames of reference.

QUESTION:
SINCE FREE FALL DONT NATURALLY OCCUR ON EARTH BECAUSE FLUID FRICTION IS PRESENT IF YOU DROP A BOWLING BALL AND A BASEBALL OFF A TOWER OF PISA WHICH WOULD LAND FIRST

ANSWER:
The (upward) drag force on a falling sphere of radius R and speed v in air at sea level may be approximated as F=¼πR²v²=0.79R²v² (this is correct only in SI units); the (downard) force of the weight W on an object of mass m is W=mg. The net force is the mass times the acceleration a (Newton's second law), ma=-mg+0.79R²v². So as the object falls it goes faster and faster until v is large enough that it falls with a constant value speed v_t (a=0), v_t=[√(mg/0.79)]/R. For a baseball, m=0.144 kg and R=0.037 m, so v_t^baseball=9.28 m/s. For a bowling ball, m=8 kg and R=0.12 m, so v_t^{bowling ball}=83.0 m/s, nearly ten times greater than the baseball. So the baseball stops accelerating sooner than the bowling ball and loses the race to the ground.

QUESTION:
My 8 year old asked me if the cycle of the universe expanding and contracting will ever end?

ANSWER:
What makes you and your son think that the universe is in such a cycle? The ultimate fate of the universe is one of the most important questions in cosmology and nobody knows the answer. I believe most cosmologists believe that the universe will continue expanding forever. Eventually the supply of hydrogen and helium in the universe will be exhausted (fused into heaviers elements) and stars will no longer exist. The universe will be dominated by black holes which will eventually decay away by Hawking radiation, so the universe will be a very dilute, cold, dark place filled with radiation. To get more information, check out the Wikepedia article on the fate of the universe.

QUESTION:
My son has a question. A shadow is long or short, wide or narrow. He thinks this seems like area (taking up space). But the real question is... can you have a shadow without matter? Because if you can't have a shadow without matter, then doesn't that mean the shadow is actually matter? (I have to ask because I'm not sure how to answer this; I suppose the shadow is actually the part of the matter that doesn't have light reflecting off of it...? Or how do I answer this for him?)

ANSWER:
A shadow is not matter; it is not anything, it is a region in which light illuminating something is blocked by some obstruction. You might say that a shadow is the lack of anything, light which might have otherwise illuminated where it is. And there are degrees of "shadowness". If all the light illuminating an area is blocked by the obstruction, it is called an umbra; if only part of the light is blocked it is called a penumbra. (Umbra is Latin for shadow.)

QUESTION:
U recently answered a question regards Earth's precession = 26,000 yrs. Can u show what specific equation(s) used to yield this time? I doubt that a single equation was used.

ANSWER:
Just like the precession of a top is due to the torque exerted by its own weight, the precession of the earth is due to torques on the earth, mainly by the sun and the moon. Your equations may be found in this Wikepedia article.

QUESTION:
On an ordinary day as one goes upward from the surface of the ground the electric potential increases by 100 volts per meter, this means that outdoors the potential of your nose is 200 Volts higher than that at your feet. Then why is it that none of us get a shock when we go out in the street?

ANSWER:
As is often the case, the most lucid answer to your question can be found in the Feynman Lectures.

QUESTION:
Why doesn't the earth just stop spinning and orbiting don't you need energy for motion where does the earth gets it's energy and is the earth in perpetual motion?

ANSWER:
You are hitting one of Isaac Newton's most important discoveries—inertia. It says that you do not need to exert a force (torque) to keep an object moving in a straight line (rotating about an axis) to keep it going; this means, also that the energy it has by virtue of its motion will not change. So the rotating earth keeps rotating since there are no torques on it*. The earth is in a circular orbit† around the sun and so its orbital motion keeps going since, even though there is a force on the earth, there is no torque.

*Actually, the moon does exert a torque on the earth which is causing the spinning to get smaller, but the change is so tiny that it takes millions of years to be significant.

†The earth's orbit is not exactly circular. When it moves closer to the sun it speeds up, when it moves farther from the sun it slows down by exactly the same amount, so the orbit just continues forever.

QUESTION:
With the devastating tragedy that happened in Lebanon, there has been some videos surface of the explosion. One such video is of some ladies in a shop, the pressure wave hits the shop and the lady outside the shop doesn't get thrown back but inside the other lady does. What is the explanation here??

ANSWER:
Note that the glass doors are closed. Outside the front of the pressure wave hits, but it takes some brief time to reach its maximum. However, the pressure difference between inside and outside is not large enough to break the glass until it is near its maximum, so the pressure difference changes much more quickly so the air rushes in quickly. I will admit that this is just an educated guess!

QUESTION:
If electricity always take the path of least resistance, why does a lightening discharge form a zigzag path to earth and not straight down which would be the shortest distance?

ANSWER:
Because air is not just a uniform homogeneous medium. There are local fluctuations in temperature, humidity, density, wind speed, and composition which makes the path of least resistance at any instant, not a straight line.

QUESTION:
how momentum is conserved of both objects with same masses moving opposite direction to each other collide to a point and at same time both objects chemically combined to each other?

ANSWER:
In an isolated system, linear momentum is always conserved; this is because the definition of linear momentum is designed to be conserved in the absense of any external force. This is also true in special relativity where the linear momentum is not p=mv but rather p=mv/√[1-(v²/c²)] where c is the speed of light. In the event of chemistry going on, there is either energy gained (endotermic) or lost (exothermic). If lost, the momentum of the radiation has to be taken into account. If gained, the energy will be taken from the kinetic energies of the incident objects but momentum will be conserved.

QUESTION:
Does precession and nutation ( or anything else ) affect the axis of the earth?

ANSWER:
Precession of the earth's axis is caused mainly by torques exerted on the earth by the moon and the sun. The period of precession is 26,000 years. But precession does not cause the actual rotation axis to change. What causes the axis to change (such that the geographical positions of the poles would change) is thought to be three things:

glacial ice melting and sending its water to the oceans,
glacial rebound which is the raising up of the land previously supporting the melting glacier, and
tectonic motion, the drift of large land masses.

QUESTION:
The momentum transmitted to the golf ball comes from the club head and,to a lesser extent, from the club shaft. Golfers are instructed to hold the club very loosely. Thus, it seems, no momentum is transmitted from the arm. What if the golfer instead holds very tightly to the club? Will the momentum of the arm be transmitted to the golf ball?

ANSWER:
The physics of a golf swing is much more complicated than you might think. From the little research I saw, your "hold the club very loosely" is an oversimplification. The club needs to accelerate very quickly on the way to the ball and that acceleration is only going to come from the torque you provide. What should happen is that, just before impact, your wrists must relax. There is an excellent discussion on the golf swing which, for the first half is fairly conceptual and the second half goes into the detailed physics analysis of the swing.

FOLLOWUP QUESTION:
I understand that the standard golf swing depends on uncocking the wrists to generate velocity at the moment of impact. Moe Norman, a Canadian golfer now deceased, was the best ball striker in the history of golf. Moe used a completely different swing that he developed. Among other things, it involved holding the golf club tightly. Thought experiment: The shaft of the golf club is fused to the golfer's arm and runs all the way up to the shoulder. That is, there is no wrist play at all. I understand that such a set-up will generate less velocity than a hinged wrist. Therefore, the 'v' in 'mv' will be lower. My question is whether the 'm' will be higher. Given that the golfer is now swinging a much more massive club, albeit from the shoulder, I posit that the mass striking the ball is greater. Take the center of gravity of the club-arm, calculate its velocity, multiply by the mass of the club-arm, and one has the momentum that is transmitted to the ball (I think). This may or may not be as great as the momentum of the faster wrist-hinged swing. I just want to know whether the 'm' will be greater with this fused club.

ANSWER:
I believe that there is no way to understand the club as a point mass, so to talk about its mass in a collision is meaningless. Think of it as a machine which imparts momentum to a small mass by exerting a force on it over some short time. I think it probably boils down to having the machine achieve the greatest possible velocity when it hits the ball; if you can achieve that by holding the club more tightly, go for it! It was interesting to learn a little about Moe Norman of whom I had never heard.

QUESTION:
When a golf club hits a golf ball, part of the momentum of the golf club transfers to the golf ball. The golf ball, being light, takes off at a velocity greater than the club head speed. What if the golf club is accelerating at the point of impact? Will the ball take off faster (and go further) than when it is hit at the same impact velocity by a club that is not accelerating? Note: amateur golfers tend to reach maximum velocity at the point of impact. Pros reach max velocity after impact. Pros do swing faster, which imparts more velocity to the ball. I would like to know whether their acceleration adds even more velocity.

ANSWER:
What will determine the exit speed of the ball is the speed of the club at impact. The impact time is approximately Δt=0.0005 s. How big would the acceleration need to be in order to significantly change the speed over this time? A typical speed of the club head is about 150 mph, about v=70 m/s. Suppose that the speed increased over the impact time by Δv=1 m/s. Then the acceleration would have to be a=Δv/Δt=1/0.0005=2000 m/s²; this is about 200 times greater than the acceleration due to gravity. We could estimate the acceleration by guessing that the club took about 0.1 s to go from zero to 70 m/s, a=700 m/s². So the speed of the club might increase by 700x0.0005=0.35 m/s and the average speed during impact would be about 70.2 m/s. My feeling is that this would have a negligible effect on the speed of the ball.

FOLLOWUP QUESTION:
I did some more research and I have a follow up. This article states that Force equals mass times acceleration. It indicates that applying a force to an object over time creates momentum. Therefore, it seems to me that a club head that is accelerating at the point of impact transmits Force and, therefore, momentum to the golf ball. And that this momentum would be in addition to the momentum of the club head. A club head that is not accelerating will transmit no force to the golf ball. It seems to me that a club head accelerating is going to hit the ball further, velocity at impact being equal.

ANSWER:
Sorry, but you have this all wrong. (This is what comes from using formulas when you do not really understand what they mean or when they are applicable!) F=ma is Newton's second law and simply says that if you push or pull on a mass it will accelerate. Indeed, if you apply a force over a time you will accelerate it and therefore change its momentum. But these do not imply that the source of the force itself needs to be accelerating. Here is what happens: when the club strikes the ball, it exerts a force on the ball; Newton's third law says that if the club exerts a force on the ball, the ball exerts an equal and opposite force on the club. For the short time that they are in contact (after which the forces disappear), the ball will speed up and the club will slow down if there are no other forces on it (which may be the case*). If the average force they exert on each other is F and the time of contact is t, the ball will acquire a linear momentum of p=Ft; the ball will experience a change in its momentum but by how much is not so simple since other forces act on it. Note that F=p/t=mv/t, so the smaller t, the larger F, all else being the same. In my answer to your original question, using v=70 m/s and t=0.0005 and noting that the mass of a ball is about m=0.046 kg, F=0.046x70/0.0005=6,440 N=1448 lb.

*If the club is accelerating when it strikes the ball there must be some force on it—ultimately that comes from you.

QUESTION:
I am writing a video response to Dr. William Lane Craig, a well known apologist, and so I am answering every position he states from a debate between him and Christopher Hitchens at Biola university some years ago. I write to you now because Dr. Craig says something that I don't understand. He states, "First, when the laws of nature are expressed as mathematical equations, you find appearing in them certain constants, like the gravitational constant. These constants are not determined by the laws of nature. The laws of nature are consistent with a wide range of values for these constants." I don't understand this at all.

ANSWER:
(This is not the whole question submitted, but gets to the heart of what the questioner wants to know. Reminder that site ground rules specify "…concise, well focused questions…") I consider that the most important "laws" of nature are not equations, they are proportionalities. In order to keep this discussion focused on my basic description of nature, I will focus solely on classical physics to make my points. In physics we begin with three fundamental concepts to lay the foundation of describing nature: mass, length, and time. In the SI unit scheme we choose kilograms (kg), meters (m), and seconds (s). With the chosen units we can define velocity as the rate of change of position with units of m/s; then we can define acceleration as the rate of change of velocity with units (m/s)/s=m/s². So if an object changes its speed from 2 m/s to 6 m/s in a time of 2 s, its acceleration is 2 m/s². Mass is trickier to understand but it is a quantity which measures how resistant an object is to being accelerated if you push on it; it is harder to accelerate an object of mass 1000 kg than it is one of 2 kg. Note that we have not yet quantified that push (or pull) which we normally call force. However we can certainly imagine figuring out a way to push or pull on something twice as hard, or three times as hard, etc. Whatever force is, suppose we push on something with a constant force F and vary the mass; we will find that if we double the mass we halve the acceleration, if we halve the mass we double the acceleration, if we make the mass 10 times larger, the acceleration is 1/10 of its original value, etc. We have found that, F being constant, acceleration a is inversely proportional to mass, a∝1/m. Now suppose we vary F but keep m constant; we find that if we double the force we double the acceleration, etc., i.e. a∝F. We can now put the two together and get a∝F/m or F∝ma. This is what I consider to be Newton's second law; it tells us an important connection of how quantities depend on each other, it is a law of physics; and, you really did not have to have defined the meter and second and kilogram for it to be true, you just had to understand conceptually what mass, length, and time are. If you know even a little physics, you do not recognize this as Newton's second law—any textbook will tell you that it is F=ma. Now, how can one turn a proportionality into an equation? It is simple, just introduce a proportionality and voila! F=Cma where C is an arbitrary constant since we have not defined how we will measure F. Don't think of it as a "fundamental constant" because if we had already defined how to measure force, we would have to measure C. For example, if we had defined a unit of force to be a pound but measured m and a in SI units (kg and m/s²), F=ma would not be a valid statement of Newton's second law.

Now let's discuss an example of when a constant is a fundamental constant. Newton's law of gravitation expresses the magnitude F of the equal and opposite forces exerted on each other for two point (or spherical) objects of masses m₁ and m₂ separated by a distance of r: F∝m₁m₂/r². Now, to make this an equation, we must introduce a proportionality constant G: F=Gm₁m₂/r². But can we choose it to be whatever we like? No, because there is nothing in this proportionality which is undefined; we can take two 1 kg masses, separate them by 1 m, and measure the force between them (a really hard experiment!). In other words, we must measure the constant G. This is a true fundamental constant of nature, it measures the strength of gravitation. Of course, its numerical value depends on how we measure mass, length, and time, but it is still a universal constant. In SI units it is 6.67x10^-11 N·m²/kg².

You might be interested in two earlier answers, 1 and 2, along similar lines as yours.

QUESTION:
I wondered what happens if one would make a magnet spin really fast. Could it be theoretically possible that it could create visible light.

ANSWER:
The frequency range of visible light is about 4-8x10¹⁴ Hz, so you can be sure that you cannot do this with any macroscopic magnet. One way to get magnetic radiation is to fabricate something called a split-ring resonator (SRR) but technical problems limit SRRs to about 2x10¹⁴ Hz, near infrared. Another technique uses spherical silicon nanoparticles and these can generate visible magnetic light. The method is too technical to discuss here, but you can read about it at this link.

QUESTION:
i had a doubt, to detect an inertial frame we need to define 0 force, but to define 0 force we need an inertial frame,0 force is defined as the sum of real forces should be 0, and we can identify real forces if we believe that the real forces drop with distance,and this could be a possible solution to the above problem, but it based on an assumption, and even if we consider that we cannot detect an inertial frame, and all the physics upto special relativity is just based on if there is an inertial frame, then how come we could do experiments to confirm the theory and how did physics work uptill general relativity, if we cannot solve this basic conceptual problem.

ANSWER:
Technically, there is no such thing as an inertial frame. No matter where you go in the universe there are always electromagnetic and gravitational fields. That does not mean we cannot define an inertial frame as one in which Newton's laws are true at low velocities. In fact we can do better by defining an inertial frame as one in which the total linear momentum (relativistic definition) remains constant; then you are not confined to low velocities. Then we will find that very good approximations to inertial frames can be found in nature. There is nothing wrong with basing a theory on an idealized (fictional) concept as long as the ultimate theory agrees with experimental results in the real world.

QUESTION:
The idea of dark matter and dark energy is puzzling me. It is understood that dark matter acts like glue to hold galaxies together. On the other hand, dark energy is responsible for expansion of universe. Scientists say inside galaxies dark matter overcomes dark energy so gravity wins and keeps galaxies together but when it comes to intergalactic space, it is said that dark energy overcomes dark matter and wins the fight, so galaxies are moving away. How do you explain this contradiction? Why dark energy wins in intergalactic space but not inside the galaxies. In other words, what makes dark matter overcomes dark energy in galaxies but loses in the space between galaxies.

ANSWER:
Even though I state on the site that I do not do astronomy/astrophysics/cosmology, I will take a stab at this one. Of course, keep in mind that nobody knows what dark matter or dark energy are in any detail. Dark matter, as the name implies, is thought to be some kind of mass and therefore subject to gravitational forces. It therefore would be not surprising to find it more concentrated in the vicinity of large masses like galaxies; so its density is likely to be smaller in intergalactic space. Even less is known about dark energy but, if it is anything like uniformly distributed, it is more likely to lose a "tug-of-war" with dark matter inside a galaxy than outside.

QUESTION:
What happens to the gravitational effect of mass when it transforms into energy, as in e=mc^2?

ANSWER:
In general relativity, our best theory of gravity, the bending of spacetime, which is what gravity is, is not caused only by mass. Any local energy density will cause gravitation.

QUESTION:
I would like to ask about gravitational mass. I know inertial mass is changing by motion (speed) according to m=mo/(1-v2/c2)^0.5 And also that is inertial mass which sits in E=mc2. If the statements above is correct, now how about gravitational mass? Does it change with motion (speed)? And what mass should be used for general gravitational formula F=GmM/r2? should we use mo (rest mass) regardless of speed of the object? Or should we use m=mo/(1-v2/c2)^0.5 to substitute in F=GmM/r2? In other words does mass equivalence principle (inertial mass=gravitational mass) hold in hight speeds?

ANSWER:
Read the answer to the previous question. The moving mass has more energy than the stationary mass and therefore has a stronger gravitational field as measured by a stationary observer. Newton's gravitational equation is amazingly accurate for speeds small compared to the speed of light. However, if you are at very high speeds you should not use it at all. The whole notion of force is, itself, not useful in relativistic circumstances. We might ask why force is a useful concept in classical physics. Newton's second law, the central concept in classical mechanics, is F=ma. Why? If someone is moving past you and measures the acceleration a' of something in your frame, she will measure exactly the same value as you. Therefore you will both conclude the same force is being applied to m, if a=a' then F=F'. But if relative speeds are not small, a≠a' so Newton's second law is no longer true. Instead we rely more on conservation principles more in relativity theory. In particular, we want to redefine momentum so that it is conserved in a closed system.

QUESTION:
I have recently learnt about the LIGO detector and watched a few videos on how it works, I understand how they detect the stretching and contracting of space time due to the gravitational wave only stretching and contracting a few hundred times per second and the new photons entering this space must travel farther or shorter deviating from the normal time travelled. I would like to ask how these new photons enter this stretched space time since the gravitational wave itself is travelling at the speed of light does that not mean the stretched space will either move with a certain packet of photons, and since the wave travels at light speed, there wont be enough contractions to allow detection to occur.

ANSWER:
You just make this too hard to visualize thinking of photons. Think instead as electromagnetic waves. (Although the speed of the gravity wave does not matter for the interferometer to work, it is just as if somebody were jiggling the mirror ends, I want to note that it is only assumed that the speed of gravity is c; the speed of gravity has never been measured, we just know that it is very fast. Since the gravitational field has never been quantized, we can only assume that the graviton has no mass. Note that for decades it was assumed that neutrinos had no mass, something we now understand is not true.)

QUESTION:
Who first derived the relativistic energy-momentum relation E²+p²c²=m²c⁴ and in what publication? Wikipedia says that it was Dirac in 1928, but I cannot find evidence of this in his famous 1928 paper, "The Quantum Theory of the Electron."

ANSWER:
As far as I can tell, Dirac's 1928 paper used the energy-momentum relation as a jumping off point for his derivation of the Dirac equation. Certainly this equation was known long before 1928 since special relativity itself was introduced in two papers by Einstein in 1905. The first paper did not mention momentum at all. In the second paper he shows that if a particle of mass m radiates energy L that the mass changes its kinetic energy by L/c², hence E=mc², but does not mention momentum. Finally in 1907 he writes an article in which he shows that if relativistic momentum is p=mv/√[1-(v/c)²], then it is a conserved quantity as in classical Newtonian physics. From here, and knowing E=mc²/√[1-(v/c)²], it is straightforward to get the energy-momentum relation.

QUESTION:
Why would lifting my feet off the ground make me heavier. Let's say I were to do a push up while my friend is on my back for extra weight. If the friend was to plant her foot on the ground, why would the push-up become easier.

ANSWER:
In the first figure I show all the forces on the man in yellow. They are: N is the force which the ground must exert up on the man, W is his weight, and F is the force which the woman exerts down on the man. (I have drawn N as one vector, but it is really distributed among his two feet and two hands. Nevertheless, the size of N is proportional to the force he must exert to lift himself.) The man is in equilibrium, so the forces on him must add to zero, N-W-F=0 or N=W+F. In white are the forces on the woman: w is her weight and f is the force which the man exerts up on the woman which by Newton's first law equal and opposite to the force which she exerts down on the man so F=f. Since she is also in equilibrium, the sum of the forces on her must add to zero, so f-w=0 or f=w=F. So, finally, we see that the total weight the man must cope with is the sum of his and the woman's weight, N=W+w.

So now I put a blue brick under the woman's foot so that it rests on the brick. This is a new force n up on her, shown in red. Each person's weight is the same as before, your weight is just the force down due to gravity. But now there are three forces on the woman adding up to zero: n+f '-w=0 or f '=w-n. The force f ' exerted up on her by the man, which has the same magnitude of the force F ' which she exerts down on the man, is smaller than before. So now, N '-W-F '=0 or N '=W+F '=W+w-n. He has a smaller total force to contend with than before.

The mistake you made is that you assumed that the force down on the man was the woman's weight. But the woman's weight is not a force on the man, it is a force on the woman. This is accidentally correct in my first example because if only the man is supporting her, it just happens that the force she exerts down is equal to her weight in magnitude. But in the second example the brick is supporting part of her weight so the man supports less. She does not become lighter!

QUESTION:
I am writing a fantasy novel currently, and at the ending a giant Dragon, with scales made out of stone is falling into the sea infront of a city and I'm imagining a tsunami destroying the city. The dragon weighs around 2,000 tons, how high would the wave be and how far would it reach inland? (the city is basically a flat plain)

ANSWER:
Well, I have to make some wild approximations to address this at all. I will perhaps get you within an order of magnitude of how big the wave might be. Maybe I should sort of do a rough general case first and then I can throw in some reasonable numbers. If the dragon drops from a height h and has a mass m, then the energy she starts with is E₁=mgh, where g is the acceleration due to gravity. When she hits the water she will have a smaller energy because of air drag falling down and when she comes to rest she will lose some energy to heat and sound and damage to her body; I will say that a fraction f of the original energy goes into the energy E of the wave created, E=fE₁. Now I will assume that all this energy goes into a circular pulse which has a width of w and a length of 2πR (circumference of a circle) where R is the radius of the circle at the time of interest, namely when the wave hits the shore; so R is the distance from shore where the dragon hit the water. Now, I found an expression on Wikepedia which relates the energy of a wave to its height: E/A=ρgH²/16 where A=2πRw is the total horizontal area of the wave, H is the height of the wave, and ρ=1000 kg/m³ is the density of water. If you put it all together you will find H=√[8fmh/(ρπRw)]. Suppose h=5000 m, m=2x10⁶ kg (2000 metric tons), f=½, and w=20 m; then I find H≈25 m. This is in the same ballpark as the largest tsunami wave ever observed which was about 100 ft. There is no way I could estimate how far inland the wave would travel; it would depend a lot on the terrain, obstacles, etc.

QUESTION:
This question is in reference to quantum entanglement. When a Super positioned photon is measured to determine spin does it's spin stay in that orientation as long as they a measuring it or does it immediately go back to a Super positioned state? In other words if you determined the spin of a quantum entangled particle at say 12:00 pm and constantly measured it, would it's spin continue to point in the same direction at say 12:02 pm. or is the measurement only good for the exact moment it was initially measured?

ANSWER:
When you observe it, you put it in a single state (and simultaneously, put the other entangled photon in a single state). Unless it interacts with something else later, they remain in states you put them, they do not reëntangle.

QUESTION:
What is the maximum distance between two electrons where they will notice each other? They would both be at rest to each other in a vacuum with no external stimuli or forms of energy. At what distance do the electrons "See" each other.

ANSWER:
This question has no meaning because "notice each other" is not quantified. In principle, the Coulomb force extends all the way to infinity. You could, for example, ask at what distance would the force each electron feels would result in an acceleration of 1 mm/yr²=[1x10^-3m]x[(1 yr)x(365 day/yr)x(24 hr/day)x(3600 s/hr)]^-2≈9x10^-22 m/s². In general, the force is F=ke²/r²=ma, so r=√[ke²/(ma)] where k=9x10⁹ N·m²/C², e=1.6x10^-9 C, and m=9x10^-31 kg. I calculate, approximately, that r≈5x10¹¹ m≈300,000,000 miles which is about 3 times the distance to the sun. Of course, you could never do such an experiment since you would need to be in a universe which contained nothing but these two electrons.

QUESTION:
Aren't "constants" simply a "fudge number" to make equations work? It seems to me, constants are values of a concept we have yet to identify or understand, but need to make equations with known concepts and values work out. Since these equations work correctly with a variety of known variables, they are generally accepted. It seems to me, non-electrical magnetism (i.e. that which is retained in magnetic material like iron) and gravitation have a variety of descriptions and equations which predict their effects--yet we still are devoid of actual cause.

ANSWER:
Constants are not just "fudge numbers" as you suggest. They can be, but in general constants play a very important role in physics. I will give you a few examples.

The laws of physics are stated most elegantly as proportionalities rather than equations. For example, Newton's second law states that the acceleration (a) of an object is directly proportional to the applied force (F) and inversely proportional to the mass (m), a∝F/m; so you double the acceleration if you push twice as hard and halve it if you double the mass. Now, to calculate we prefer to have equations rather than proportionalities, so we introduce a proportionality constant, call it C, a=CF/m. What does C mean? Since we know how to measure a (m/s²) and m (kg), your choice of C determines how we will measure force. I would like the force to have the magnitude of 1 if a=1 m/s²and m=1 kg, so I choose C=1. The constant C is not a fudge factor, rather its choice defines how we measure forces. (We call the unit kg·m/s²a Newton, N.)

Newton also discovered the universal law of gravity: if two masses (point masses or spheres) m₁ and m₂ are separated by a distance d, they exert equal and opposite attractive forces on each other the magnitude of which is F; then F∝m₁m₂/d². So the equation for universal gravitation would be of the form F=Gm₁m₂/d². where G is some constant. But we know how to measure mass and distance and force, so we cannot choose G to be any old thing we want as we did for Newton's second law, we must measure it. It turns out to be G=6.67x10^-11 N·m²/kg². This is one of the most fundamental constants in all of nature and most certainly not a "fudge factor".

Constants are often used to quantify a particular situation. For example, the frictional force f of one object sliding on another is found, approximately, to be proportional to the force which presses one object to the other (often called the normal force, N). This is not a physical law, just an approximation of experimental measurements, and it obviously involves only the magnitudes of the forces since their directions are perpendicular to each other. So, including a proportionality constant μ, called the coeficient of kinetic friction, f=μN. This constant tells you how slippery the surfaces are; its value for rubber sliding on ice will be much smaller than for rubber sliding on asphalt.

Sometimes, as you suggest, constants are added to fit data and their physical meaning is unknown, you might label such constants as "fudge factors".

You may be interested in learning about Planck units discussed in an earlier answer. where new units are expressed by combinations of the five fundamental constants of nature.

QUESTION:
In a example of a electrical generator. Central in the generator are rotating magnets. I understand that as the magnet moves the cores of the surrounding coils will react producing a magnetic field. Also the copper wire winding around the coil will react. My question is does the copper wire winding react to magnetic field according to the field produced by the core or the moving magnet ? In a lenz law sense. Plus would I be correct in thinking if there is no circuit on the copper winding there no organised field would form on the copper wire winding itself.

ANSWER:
The best way to understand the priciples is to look at the simplest possible generator. The figure shows a generator consisting of a single loop rotating in a magnetic field which is approximately uniform. There would be no difference in the results if it were the magnets rotating around the coil as you describe. Since the induced EMF in the loop is proportional to the time rate of flux through the loop and the flux is proportional to the cosine of the angle θ between the field and a normal to the area of the loop, the induced EMF will be sinesoidal. The angle may be written as θ=ωt where ω is the angular velocity of the loop (or magnets) and t is some arbitrary time, so the voltage is proportional to cos(ωt). If there is a load across the terminals, and therefore a current through the loop, there will be a field due to the current which does not affect the output voltage if the source of power maintains a constant angular velocity. If you are drawing power from the generator, it is harder to "turn the crank" than if there is no current flowing. If you need detailed information on "real-life" generators, it is more an electrical engineering question than physics.

QUESTION:
How did Einstein derived general relativity math when he understood that the gravity is just the space bending around an object how did he applied this to the world of math?

ANSWER:
You can learn about how Einstein struggled with the mathematical methods necessary to fully describe general relativity and with which mathemeticians he consulted in his biography by Walter Isaacson, Einstein: His Life and Universe.

QUESTION:
I have been thinking about clocks lately and found myself in a sort of a pickle over the velocity of a dial. (I am not native so sorry if my description makes it very complicated). Suppose we have a disc with a dial starting at its center. The dial draws two circles on that disc as it turns. The first circle has its radius equal to r, whereas the second circle has a radius of 2r. Now suppose that 1 full turn takes 1s (which would be the same for both circles since they are being drawn by the same dial). Now because of the difference of radii, there is also a difference in their circumference c. And so we've got t=1s, c=2πr and c'=4πr (2π2r). Now here's the question - if the time is the same, but the circumference or distances are different does that mean the velocity of these two points on the same dial is different? How is that possible since the dial turns with a set velocity which in my understanding would be the same for each point on that dial?

ANSWER:
It is simply because all points on a rotating rigid body have the same angular velocity, (revolutions per second, e.g.) but not the same translational velocity (miles per hour, e.g.). In general, v=rω where v is the translational speed, r is the distance from the axis of rotation, and ω is the angular velocity in radians per second.

QUESTION:
I was just wondering why does the centripetal acceleration formula work a=v^2/r?

ANSWER:
Why does it work? It "works" because it is correct. Maybe you wanted to ask "where does it come from?" or "how is it derived?" I will give you a brief standard derivation.

On the left of the figure I have drawn the particle moving with constant speed v around a circle of radius R. In some time interval Δt the particle moves through an angle theta and has velocities v₁ and v₂ but both speeds are equal to v, v₂=v₂=v. On the right I have placed the velocities tail-to-tail and drawn the difference vector v₂-v₁=Δv. So now I see two isoscoles triangles, each with the same angle between their equal sides; they are therefore similar triangles and so we can write v/Δv=R/Δs. If I now rearrange and divide each side by Δt, I find Δv/Δt=(v/R)Δs/Δt. Now, Δv/Δt is the magnitude of the average acceleration; to get the instantaneous acceleration, we must take the limit as Δt approaches zero. But, when Δt approaches zero, Δs/Δt approaches v. So, finally, a=v²/R.

QUESTION:
Is Planck time a concept derived from the theory of relativity or its premises?

ANSWER:
No, that is not where the concept came from. The origin of Planck units came from the desire to have a system of units which are based only on constants of nature, not on the specific units like the meter, the second, etc. You take the five universal constants:

the speed of light in vacuum, c,
the universal gravitational constant, G,
the rationalized Planck's constant, ℏ ,
the Boltzman constant, k_B, and
the Coulomb constant, k_e=1/(4πε₀).

You now form combinations of these constants which have the correct dimensions of the units you want:

length,
mass,
time,
temperature, and
electric charge

I have copied the table from Wikipedia showing the values of these Planck units in SI units:

It is hypothesized that, at distances smaller than l_P=1.6x10^-35 m, the usual physics as we know it will not work. I think you might say that the origin of the idea of Planck units is quantum field theory.

QUESTION:
A toy top is a disk-shaped object with a sharp point and a thin stem projecting from its bottom and top, respectively. When you twist the stem hard, the top begins to spin rapidly. When you then set the top's point on the ground and let go of it, it continues to spin about a vertical axis for a very long time. What keeps the top spinning?

ANSWER:
There is a property of spinning objects called angular momentum. For example, the angular momentum of your top is a vector approximately equal to ½MR²ω where M is the mass of the top, R its radius, and ω is the rate at which it is spinning; the direction of the angular momentum vector is straight up if it is spinning counterclockwise if viewed from above, straight down if clockwise. There is a physical law which states that if there are no torques on a spinning object, its angular momentum never changes. There are no torques on the top if its spin axis is vertical and so it does never stops.

QUESTION:
The third dimension is 3D (spheres, etc.). The first dimension is a dot. Everything that has a front and back, has an edge. So isn't a dot 3D?

ANSWER:
A dot has zero dimensions. The formal definition of the dimensionality of a space is an answer to the question "how many numbers does it take to specify the location of a point in that space?" You may heard of Cartesian coordinates, (x,y,z); they were named in honor of the 17th century philosopher and mathematician René Descartes. Legend has it that he was once watching a fly buzzing around his room and asked "what is the minimum number of numbers I must specify to describe the position of the fly at any time?" He figured it was three: the distance up from the floor, the distance from the back of the room, and the distance from one of the side walls; his room is said to have three dimensions. One dimension is a line; here you simply measure the distance from one point on the line to wherever the particle is. The line need not be a straight line. A circle is one-dimensional and the location of a point on the circle is usually specified by an angle relative to some point we call zero. Two dimensions is a surface. The surface of the earth is two-dimensional, the position being angles, latitude and longitude. One surface of a flat sheet of paper is two-dimensional and the two numbers are usually Cartesian coordinates (x,y). The spatial universe in which we live is three-dimensional, just like Decartes' room. A solid sphere has three dimensions usually measured by the distance from the center and two angles (longitude and latitude).

QUESTION:
Why there is only time dilatation and no time shrinkage? For me I think there should be time dilatation regarding objects traveling away from each other and shrinkage regarding those closer to each other. As what makes one gets older and the other still young for both of them are moving with same speed relative to each other.

ANSWER:
Thinking how you think something should be does not make it true! You should read my answer about the twin paradox. How fast a clock runs and how fast it appears to actually runs are two different things; see earlier answer.

QUESTION:
While we use the combination of rear and front brakes to stop a motorcycle, I mean we can chose between the the front and rear without any effort, why do cars primarily use the front brakes to stop? Why does the brake pedal engage only the front brake? The rear brake is only seem to be used when the car is parked

ANSWER:
All modern cars have four-wheel braking. So where did you get the idea that the brake pedal engages only the front-wheel brakes? Parking brakes engage only rear wheel brakes. Your confusion may be that the braking causes the front-wheel tires to wear more than the rear-wheel tires; this is due to the car slightly "rocking forward" when stopping and also because in a front engine car more of the weight is likely to be supported by the front tires.

QUESTION:
This is a kinematics question about the motion of wheels; specifically, about how friction between a wheel and a surface causes forward motion of the entire wheel. There's something unsatisfactory to me about how this forward motion is typically explained. Consider the rear wheel of a bicycle; it is driven by torque applied to the wheel (via chain drive connected to pedals, etc.). In the classical treatment of this scenario, it is claimed that the friction between the wheel and the road surface, which is a forward-pointing force, causes forward motion of the wheel. But this frictional force is tangential! So wouldn't this tend to be a rotational force instead of one of displacement? It would seem to me that somehow the tangential rightward force of friction at the point of contact needs to be transformed into a rightward force applied at the *center of mass* (i.e., at the axle) of the wheel in order to explain an induced forward/rightward motion, but no source that I have found has attempted to explain this gap.

ANSWER:
The figure shows all forces on the rear wheel of a bicycle:

C, the force exerted by the chain on the sprocket (radius r),
W, the weight of the wheel (radius R),
N, the vertical force exerted by the ground on the wheel,
f, the frictional force exerted by the ground on the wheel, and
F, the force exerted by the rest of the bike on the wheel.

Note that I have resolved F into its horizontal, H, and vertical, V, components. For any body which cannot be approximated as a point, Newton's second law (N2) has two parts, translational and rotational:

The sum of all forces on a body is equal to the product of the mass (m) and the acceleration (a) of the center of mass of the body;
The sum of all the torques on the body about any axis is equal to the product of the moment of inertia (I) of and the angular acceleration (α) about that axis.

You seem to think that a force which exerts a torque about the axis you have chosen (I will choose the axle for the bike wheel) only can cause angular acceleration; in fact, every force on the object contributes to the translational acceleration. Suppose now that you are standing on the pedal but also applying the front brake such that the bike remains at rest; both translational and angular acceleration are zero. So the translational N2 yields two equations, N-W-V=0 and C+f-H=0; the rotational N2, choosing the axle as the axis, yields Cr-fR=0. (F, W, and N exert no torque about the axle.) If you now release the brake, H will get much smaller and the wheel will start moving forward and rotating.

If you are interested in the entire bike, not just the rear wheel, the force C has no contribution to the motion because it is an internal force; the chain exerts a force C on the rear sprocket but a force -C on the front sprocket netting zero net force.

FOLLOWUP QUESTION:
You say, "in fact, every force on the object contributes to the translational acceleration" but the force, f, in your answer, is not directed at the center of mass of the wheel. I guess my question is simply, "why is f treated as a force directed at the center of mass and not as a torque on the wheel"? I don't understand where the forward acceleration of the *center of mass* of the wheel is coming from.

ANSWER:
You did not read my answer carefully. The translational N2 is "The sum of all forces on a body is equal to the product of the mass (m) and the acceleration (a) of the center of mass of the body". Note that it does not say …the sum of all forces which are directed through the center of mass… Let me give you a simpler example. Suppose that you are in the middle of empty space and you have a wheel which has a string wrapped around its circumference. You pull on the string so that the tension in that string is the only force on that wheel and it is tangential. Surely you do not think that your pulling will cause the wheel to only spin and not accelerate toward you too. Here is another example, a yoyo. There are two forces on it, its own weight which passes through the center of mass and points down and the tension of the string which you are holding which points up and does not pass through the center of mass. If the string does not affect the translational acceleration, the only force on the yoyo would be its weight so it would fall with acceleration due to gravity, g, which it clearly does not do.

SECOND FOLLOWUP QUESTION:
In classical billiards problems, for example, when a force is applied to a ball, the force has to be decomposed into its normal component directed at the center of mass (call it f_n) and the tangential component (call it f_t). Only f_n will cause translation, right? It's my understanding that f_t will only contribute a torque, but not to translation. So why is the bike case any different? Contact friction between the wheel and the surface is a tangential force, so, according to my understanding, it can not cause translation, because it has no component directed at the center of mass of the wheel...

ANSWER:
Look at the figure*. It is much better to resolve F into its horizontal and vertical components rather than normal and tangential components. It is wrong to insist that a force must be directed toward the center of mass for that force to contribute to tranalational motion; my examples above should have convinced you of that. Now, a billiard ball is confined to move on the table, so there will be no translational motion in the vertical direction. So the ball is in equilibrium in the vertical direction, so N-V-W=0. But in the horizontal direction there will be an acceleration, ma=H-f. Only f and F will exert torques and it will be a little tricky to calculate the torque due to F but just geometry. I have drawn the line of force to be below the center of mass; if it is through the center of mass, it will exert no torque and only f will cause angular acceleration.

Here is the take-away for you: just use N2 as I have given it to you, always look at all forces on the object, and do not continue insisting that a force which exerts a torque does not affect translation.

*I have drawn the forces on the ball, which are

the ball's weight W,
the normal force N the table exerts vertically on the ball,
the frictional force f the table exerts horizontally on the ball, and
the force F exerted on the ball.

QUESTION:
A hammer hitting an anvil creates energy (sound) which moves in all directions at about 700 mph in atmosphere. Sound energy is also created by a large fire. In space, vaccuum, that same energy must also be present if a hammer hits an anvil in space, same as a star (a large fire ball) should also create sound energy, along with all the other energies it creates, so if sound energy in space, a hammer hitting an anvil, does not have atmosphere to slow it down, or to make noise, how would one measure that energy, its speed and pressure. Could this be what is called "Dark Energy".

ANSWER:
Sound is a wave that exists in a solid, liquid, or gas. If you are in a vacuum, there is no sound. The energy which the sound carried away in air is instead retained as sound in the anvil (and hammer). But this sound eventually damps down and where that energy goes is an increase in the temperature of the anvil (and hammer). This energy eventually radiates away, not as sound but as electromagnetic waves (mainly infrared) until thermal equilibrium is achieved. It certainly has nothing to do with dark energy.

QUESTION:
So my question is a bit out of reality question as its not possible to do this in real life. But if you were able to turn a cannon ball into a musket ball and fire it from a flintlock, having the ball turn back into a cannonball apon exiting the gun, would the cannonball retain its velocity after changing or would it lose all energy and drop to the ground?

ANSWER:
Essentially what you are asking is if an object with a certain mass m and speed v suddenly loses or gains mass, what happens? The mass either disappears without trace or appears without a source. Suppose your musket ball has a mass m and speed v and your cannon ball has a mass M. Since there are no external forces on the ball, linear momentum must be conserved, that is mv=MV where V is the speed of the cannon ball. So V=mv/M, much smaller than v. However, the energy is not conserved: E₁=½mv² and E₂=½MV²=½mv²(m/M)=(m/M)E₁; a large amount of energy is lost.

QUESTION:
How many pounds per inch does a lacrosse ball exert when it hits a surface? For info: Ball is traveling 90mph. A lacrosse ball weighs 8 ounces. The ball is 5.25 inches in circumference. As a sidebar, does it matter if the surface the ball hits is stiff like a wall or springy like a bounce-back? If so, could you expand on the difference in pressures against the two different surfaces?

ANSWER:
If it is pressure you want, you should ask for pounds per square inch. But I do not think that is what you really want; it is more straightforward to estimate the average net force the ball exerts on whatever it hits. Then you could think about pressure, how that force is spread over the area which the two surfaces have touching; pretty hard to do that, though. If something with mass m with a velocity v₁collides with a much more massive object at rest and rebounds with velocity v₂, the average force F can be estimated as F≈m(v₂-v₁)/t where t is the time the collision lasts. ("Much more massive" means the struck object does not recoil significantly.) You seem to want to get answers in imperial units (lb, ft, s); without going into details, m=0.5 lb/(32 ft/s²)=0.016 lb·s²/ft=0.016 slug and v₁=90 mph=132 ft/s. Suppose that v₂=0, called a perfectly inelastic collision; then F=-132x0.016/t=(-2.1 lb·s)/t. So, the smaller the time, the larger the force. (The fact that the force is negative means that this is the force the struck object exerts on the ball because I have chosen the incoming velocity to be positive. The force the ball exerts on the object is positive and equal in magnitude.) That addresses your second question: the ball will certainly stop in a much shorter time when hitting a hard surface than a soft one. Of course that should not be too surprising to you because you know that if you fall onto a mattress it hurts much less than if you fall on a concrete floor. So, if t=1 s, F=2.2 lb whereas if t=0.01 s, F=220 lb. Now, if the ball hits a hard surface it does not penetrate very far and therefore the force is spread over a small area and pressure is large. If the ball hits a soft surface, it penetrates deeper and therefore the force is spread over a larger area and pressure is smaller.

QUESTION:
Since this isn't technically a homework question and was a challenge put out to the class to see if anyone can do it, I hope you will consider it. I'm attempting to do some fun physics maths as a challenge from our professor. It may be a trick question though since she's done that in the past. The problem assumes that c is only 30000 kmph. The problem was to determine the time to reach 29999 kmph from 25000 kmph with a total of 6.48MN of force in a vehicle that weighs 168000kg (168t) in the vacuum of space, too far away from any celestial body to have gravity noticeably affect it. So just to clarify, there is no air resistance or gravity or time dilation. Relative mass increases when reaching said 30000kmph, and the goal is to find the time it takes to accelerate to 29999kmph from 25000kmph.

ANSWER:
I don't see the point of this odd choice of c; it makes the problem neither easier nor harder, just unphysical! You will still have to do some work! I am going to point you in the direction of solving it yourself. First, go to an earlier answer; you will see that I have solved your problem but starting at rest. If you start at some initial velocity v₀, the result is that Ft=p-p₀. Here, p is the relativistic linear momentum, p=mv/√[1-(v/c)²] where m is the rest mass, 1.68x10⁵ kg in your case; similarly, p₀=mv₀/√[1-(v₀/c)²]. So now you know everything except t, so just solve for it; this is just algebra/arithmetic. I would advise you to convert everything to SI units (kg, m, s) before doing your calculation. Then t will work out to be seconds.

I do not understand what you mean by "…there is no…time dilation…" since time dilation is simply not of any interest here but would come into play if the whole thing were seen by an observer in a different frame. Also, I almost never apply the idea of relativistic mass because I view the linear momentum as being the quantity redefined in relativity, not mass. Read my discussion of this in one of the links in the earlier answer.

QUESTION:
Atoms are mostly empty space. How can they form solid objects?

ANSWER:
First of all, atoms are not mostly empty space. See an earlier answer. Although almost all the atom's mass is in a volume much smaller than the volume of the whole atom, the electrons fill the rest of the space in a "cloud". Bohr's model of the atom, tiny electrons in tiny orbits, is wrong. If you need a simpler explanation without reference to the "electron cloud" think of all the atoms connected to their nearest neighbors by tiny springs. Each atom connects to its neighbors by some interaction which can be approximated as a spring. The atoms close enough to interact with each other will stick together and thus form a solid object.

QUESTION:
2 balloons, one filled with 1 cubic meter of air, and the other with 1 cubic meter of Helium, are lowered to a depth of 50 feet under water and simultaneously released. Will they reach the surface at the same time?

ANSWER:
With the information you gave me, I will have to put stipulations on the problem. The baloons are rigid so they do not get compressed under the water, they have identical shapes, their masses (without being filled) are the same, and both are filled to atmospheric pressure. There are three forces on each balloon: its own weight -mg, the buoyant force of the water B, and the drag force opposite the velocity v which has the form -Cv² where C is some constant which depends only on the shape of the balloon. So the net force on each balloon is -mg+B-Cv²=ma where I have put this equal to the mass times the acceleration, Newton's second law. When you release each balloon, each will accelerate up and, as v increases, eventually a=0 and it will have a constant speed v_t, called the terminal velocity, thereafter. It is easy to show that v_t=√[(B-mg)/C] The only difference between the two is m which is smaller for the helium which therefore has the larger terminal velocity. The helium balloon reaches the surface first.

QUESTION:
does a bouncing ball possess simple harmonic motion?

ANSWER:
A bouncing ball is not an example of simple harmonic motion (SHM). By definition, the ball's motion must be describable by simple sinesoidal functions of time t, i.e. like y(t)=Asin(ωt)+Bcos(ωt) where A, B, and ω are constants and y is a variable describing the motion, for example the height above the ground in your case. If the ball were perfectly elastic, the motion would be harmonic, but not SHM, because y(t) would be a train of identical downward-opening parabolas. If it were a real ball, it would not even be harmonic because each subsequent bounce would be a little smaller that the previous bounce.

QUESTION:
What exactly is the 4th dimension? How do we know it's real? How are space and time related?

ANSWER:
I have recently answered this question in some detail.

QUESTION:
As I understand it, electricity moves at the speed of light. Conductivity is where I'm getting stuck on. Are we able to measure speed from it? I tried searching "siemens/meter to kph" with no results. My questions here are these: If silver has a conductivity of 63x10^6 siemens/meter, and copper has a conductivity has a conductivity of 59x10^6 siemens/meter, does that mean energy in silver is travelling faster than copper, or something else? But, silver can't go, say 101% the speed of light. So, there's obviously something else I'm not understanding. I guess the point I'm getting at is this: What does conductivity mean, and can you convert siemens/meter into a speed calculating how many particles move through 1 meter of wire in x amount of time, thus making copper's speed x time/meter, while silver is y time/meter? I'm sorry if this is a bit confusing, or not even your field. But, I figured electricity is, in some part, in the field of physics.

ANSWER:
I am going to give you a qualitative overview since you seem to not have any understanding of what goes on in a conducting wire. If you want a more detailed discussion of the simple model of electric currents, you can find it in any introductory physics text book. A conductor has electrons which are moving around at random in the material, not really bound to any of the atoms; these are called conduction electrons and they can be thought of as a gas of electrons. But there is no flow of current because the electrons move randomly so there are just as many going in one direction as there are going in the opposite direction. When this wire is connected to a battery an electric field is established inside the wire which points from the positive terminal to the negative terminal. When you say "electricity moves at the speed of light", what that means is that the establishment of the field moves with speed c; the electrons do not move at that speed. So all electrons see the field turn on almost immediately. All electrons, having negative charge, begin accelerating in the direction opposite the field. If the only force seen by the electrons was due to the field, they would accelerate until they reached the positive terminal and left the wire. But, even though they are not bound to atoms, they certainly see the atoms and when they hit one they are momentarily stopped or scattered. So each electron is constantly being accelerated, then stopped, then accelerated and the net result is that, on average, all the electrons participating in the current move with a very small velocity, on the order of one millimeter per minute; this is called the drift velocity. All the electrons are continually gaining and then losing kinetic energy and that lost energy shows up in the heating up of the wire. What determines the conductivity is the number of conduction electrons per unit volume of the material as well as its crystaline structure and how individual atoms scatter the electrons.

QUESTION:
The classic escape velocity formula takes in consideration an uniform acceleration of gravity, g. What would be a mathematical model for escape velocity where the force of gravity varies as the body distances from the planet?

ANSWER:
"The classic escape velocity formula" that I know does not assume a uniform gravitational field, it assumes the correct field. Let's see where it comes from. The potential energy for a particle of mass m a distance r from the center of a large spherical mass M is, choosing zero potential at r=∞ is V(r)=-mMG/r where G is the universal gravitational constant. If m has a speed v at r=R, where R is the radius of the earth, the total energy is ½mv²-mMG/R=E. Now, suppose we want v to be big enough that m will go all the way to r=∞ before it finally comes to rest; then E must be zero. Therefore v_e(R)=√(2MG/R); but, MG/R=gR, so v_e(R)=√(2gR). That is the escape velocity at the surface of the planet (ignoring any other forces, notably air drag) where g is the acceleration due to gravity. Now, you want it everywhere else. That's easy, just replace R by r, v_e(r)=√(2MG/r). You could use g=MGr if g now means the acceleration due to gravity at an altitude of r-R. (Incidentally, this does not work anywhere inside the planet.)

QUESTION:
I have a bit of a debate with colleagues and looking for a professionals take. I'm a plasterer and when mixing up the plaster introducing too much air to the mix makes it less workable. Some people I work with think mixing counter clockwise introduces less air to the process than clockwise, I think it would make no difference.

ANSWER:
Well, I just so happen to have a plaster/paint stirring tool which is powered by a drill. Examine it carefully and you will see that clockwise and counterclockwise are not the same in their effect; if the direction is clockwise (as viewed from above), the blades push down on the mixture and if the direction is counterclockwise, the blades push up. I do not know for sure which, if either, of these would introduce more air. My best guess would be that for the clockwise rotation there would be a whirlpool into the surface, whereas for the counterclockwise rotation there would be more of a hump on the surface; I would think the whirlpool would pull in more air. I must also note that not all stirring tools are necessarily of the same design as mine.

QUESTION:
Why does this toy move in a straight line? Shouldn't the toy rotate due to gyroscopic precession? https://youtu.be/hs4iv7IHvGg

ANSWER:
Well, it does not move in a straight line. One thing I noticed is that the toy starts moving immediately when released; this indicates to me that the surface it is on is not level. Just because it is a gyroscope does not mean it will precess. There must be an external torque acting on it; imagine an axis running between the two legs about which you would calculate torque. In the figure that I have clipped from the video you can see the motor and battery and the cds which are rotating; the toy is leaning such that I would certainly expect the center of gravity to be beyond the being above the axis and, were the discs not spinning the toy would certainly fall. If you could put the center of gravity directly above the axis, it would not precess. You can see several instances in the video where the toy is moving in a curved path, and in one case the path is a pretty tight circle; these are due to precession because of the torque due to the weight.

QUESTION:
Would you be marginally taller when on top of a tall building/mountain or at the bottom of the ocean. Does the pressure/gravity change and if so which one makes you taller?

ANSWER:
If you are in the the vicinity of the earth there are two forces which are different between your head and feet, tidal force and pressure force due to the fluid you are in. The tidal force is due to the fact that the gravitational field at your feet is slightly larger than at your head, both pointing down; therefore the net force on you tends to stretch you taller. We can estimate the tidal force because the gravitational field falls off like 1/r²: F_feet/F_head=(r+h)²/r²=[1+(h/r)]²≈1+(h/r)² where r is the distance to your feet from the center of the earth and h is your height; I have used the binomial expansion because h/r<<1. So, F_feet-F_head≈F_head(h/r)². Clearly, either force is going to be approximately equal to your weight, so the tidal force is about mg(h/r)²; if I take h=2 m, mg=1000 N, and r=6.4x10⁶ m, this is about 10^-10 N (2.2x10^-11 lb).

The pressure force is down on your head and slightly larger and up on your feet; so the pressure force tends to compress you shorter. The net force on you is simply the buoyant force which equals the weight of the displaced fluid. If I take your volume to be about 0.02 m³, the buoyant force force is about 1000 N in water and 1 N in air. So, as you can see, the stretching due to the tidal force is negligible to forces due to fluid pressure. So, you will be shorter than your "real" height, but more so in water than air. Therefore you are taller up high in the air than down low in the water.

QUESTION:
I know that at sea level, atmospheric pressure is 14.7 pounds per square inch. I know that the higher in elevation you go, the thinner the air gets and a corresponding decrease in the amount of oxygen such that trying to survive above 16-18k feet is not very easy. My question is this: If you where to drain all of the oceans so that the existing atmosphere before draining was now occupying the void left by the water, would you be able to survive at what was sea level? My thought is that the atmosphere would become so diluted with the oceans gone that you would actually have to travel way below what was sea level to able to survive the atmospheric pressure change and the oxygen dilution. Wondering about this has always bugged me lol.

ANSWER:
The volume of all the oceans is about 1.4x10¹⁸ m³ and their average depth is about 3700 m (about 12,000 ft). Now, dealing with the volume of the atmosphere is tricky because the density of the atmosphere gets smaller with altitude so you have to be careful how you describe volume. The mass of the whole atmosphere is pretty well-known, though, to be about 5.15x10¹⁸ kg; so you can calculate the volume if the whole atmosphere were at sea level pressure where the density is 1.225 kg/m³, 5.15x10¹⁸/1.225=4.2x10¹⁸ m³. So, if the atmosphere were uniform, about 1/3 of it would flow into the emptied oceans. But, it is not uniform so less than 1/3 would come; there would be some base pressure 3700 m below the sea level (since there is no sea now, when I say sea level I mean what it was) which I would estimate to be about the same as the pressure at sea level was because 3700 m is small compared to the radius of the earth, 6.4x10⁶ m (0.06% smaller). Since the oceans occupy about 70% of earth's surface, sea level pressure would be about what the current pressure is at about 12,000 ft, the height of a modestly high mountain and certainly habitable. The highest habitable place on earth is about 16,000 ft, so lots of places (like Denver) habitable now would not be. And certainly nobody would be climbing Mt. Everest in this new world!

QUESTION:
What are the signs that show scientists that some particles have a quark structure and others do not?

ANSWER:
Quarks are something which were hypothesized after most of the elementary particles and their properties were known. There are three kinds of particles, hadrons, leptons, and field quanta. Leptons are few, electrons, muons, and neutrinos. Field quanta are also few, photons, gluons, Z bosons, W bosons, and the Higgs boson. Hadrons are many, the proton and neutron being the best known, but what has been referred to as a "zoo" of other hadrons. To try to systemize and explain this array of particles, first group theory and then, building on that, introducing hypothetical particles called quarks was extremely successful. There is no way I could answer "what are the signs" because the "signs" are all the hadrons and their properties.

QUESTION:
What is the difference between a force and weight

ANSWER:
A force is a push or a pull. Weight is the name for a specific kind of force, one which is caused by gravity. For example, your weight is the force which the earth's gravity pulls you down; the gravity on the surface of the moon is smaller and therefore your weight is smaller there. All weights are forces but not all forces are weights.

ADDED COMMENT:
You should be aware that a few textbooks define weight as being what a scale measures. So if you are in an upward accelerating elevator you have a larger weight. I find this to be total nonsense.

QUESTION:
If time slows down near massive objects, how fast is time in the space between Galaxies ?

ANSWER:
Suppose that your clock ticks off a time t if you are in a region of space where there is essentially zero gravitational field. Now imagine bringing your clock a distance R from a spherical, nonrotating object of mass M. The same time interval will now be given by t'=t√[1-(MG/(Rc)] where G=6.67x10^-11 m³kg^-1s^-2 is the universal gravitational constant and c=3x10⁸ m/s is the speed of light. For example, let the location be on the surface of the earth, M=6x10²⁴ kg, R=6.4x10⁶ m; then, t'=t√[1-7x10^-10]≈t[1-3.5x10^-10]. So, for example, the clock in space would click off 1 second while the clock on the earth would tick off 1-3.5x10^-10 seconds. To see significant gravitational time dilation requires a larger M, for example a black hole; for a black hole with a Schwartzschild radius R_S, the time dilation equation becomes t'=t√[1-R_S/R] so time stops when R=R_S. If R=2R_S, t'=0.7t.

FOLLOWUP QUESTION:
Previous answer did not address this question : If time slows down near massive objects, HOW FAST IS TIME IN THE SPACE BETWEEN GALAXIES? Could you do the math at a distance of 10,000 light years from any mass, please.

ANSWER:
There is no answer to that question because time depends on the observer, it is relative. Any observer in any frame of reference and in any gravitational field will say that it takes any clock in his frame 1 s to advance 1 s. Let's take the case of the black hole example above. The observer in empty space sees his clock advance 1 s but observes the clock of the girl near the black hole to advance in 0.7 s in that time. The girl near the black hole sees her clock advance 1 s but observes the clock in empty space to advance 1.4 s in that time.

QUESTION:
In the theory (?) of quantum entanglement, maybe you could clear up some basic things for me: a) in order for two particles to be entangled - no matter how far apart in space they are - must those same identical two particles have been physically 'partnered' up at some point in the past? b) If a particle in a spin up/spin down position is OBSERVED, it chooses (?) to become spin up or spin down. WHY does the act of observation force it to make that choice? Or is that even the right question? I understand that this idea of observation causes change goes to the heart of quantum physics, but I'm still trying to get my head around the WHY.

ANSWER:
The answer to your first question is yes, the particles become entangled by interacting with each other. So, for example, if you had two electrons, each with a wave function which is half up spin, half down, unless they have been put into those states by a mutual interaction, they are not entangled. For the second question, the particle does not "choose" anything. Rather, the measurement puts the particle into the state which you observe. The result of a measurement is to observe a single state, not an admixture. So if you have an electron whose wave function is half up, half down, your measurement will either give one or the other, not both; if you observed a large number of electrons, all with the half and half wave functions, you would observe spin up for half your measurements and spin down for the other half.

QUESTION:
The longer the garden hose the less the pressure at the end - Is this a true statement, if so, how much less? It's not a homework problem, it's a heated discussion with spouse, we're 76

ANSWER:
There is a pressure drop for a fluid flowing through a pipe. The main reason is that there is frictional loss due to viscosity of the water and the rubbing of the water on the walls of the hose. The loss depends on the size and length of the hose, the density of water, the flow rate, the viscosity, a constant characterizing the friction for the material, and the change in height if the hose is not horizontal. The physics is complicated, but there is a handy on-line calculator where you can get at least a qualitative estimate of how big an effect it is. I did a calculation for water in a 1 inch diameter, 100 foot rubber hose, and for a flow rate of 3 gallons per minute; I assumed that the hose is horizontal and straight. The results are shown in the figure. The calculated pressure drop is 0.38 psi which you can compare with a typical in-city water pressure of about 40 psi. So the drop is only on the order of 0.1%. If the water flow were much faster, say 20 gallons per minute, the drop would be about 10 psi, a 25% drop.

QUESTION:
I have a question, if the pulling force of mass or acceleration or electromagnetism can cause time dilation... would a pushing force of the repelling force of electromagnetism or the expansion rate of the universe cause reverse time dilation?

ANSWER:
Only speed (magnitude of velocity) matters for time dilation. It is irrelevant how that speed was acquired, attractive or repulsive force.

QUESTION:
What is the meaning behind multiplication in physics? Is multiplication in physics purely mathematical or there is a physical explanation to it? How do we explain the product for example, s=v.t? Is there any meaning behind this? For example, I can say that "Distance is defined as the product of velocity 'times' time"? But what does this even mean?

ANSWER:
Suppose you are in 6th grade. Multiplication is explained as simplified multiple addition, the product of 5 and 3 is 5 plus 5 plus 5. Then we learn all our multiplication tables to not have to figure out all the simple products we might need. This is arithmetic and, indeed, we are taught that it often means something, like in the example above, if we have three baskets, each containing 5 apples, we have 15 apples. And then it starts touching on physics. The area of a rectangle is the height times the width; or, like your example (but not exactly) the distance traveled is the product of the speed and time. But the second example is sort of ambiguous if you think about it. What does "distance traveled" mean? It might mean the distance between the starting and ending points; but if you get there on a winding path, that is not the really the distance you traveled. The problem is that before you get to physics, there is only one kind of number, called a scalar quantity, which you can specify using only a number. Examples of scalars are time, speed, length, area, etc. But, in the physical world, some quantities require more than one number to adequately describe them. The simplest are vectors which require a specification of both magnitude and direction. Examples are force (e.g. 10 lb straight up), velocity (e.g. 25 mph due north), displacement (e.g. 5 ft straight down). Vector quantities are denoted as bold-face, for example F=10 lb straight up. Now multiplication takes on new and different meanings, and the ways we multiply all have a meaning. The easiest to grasp is how to multiply a scalar times a vector: as in arithmetic, multiplication is just repeated addition, so 3F=F+F+F. But, what is the multiplication of two vectors? There are two ways we could imagine multiplying vectors, namely the product is also a vector or the product is a scalar. The scalar product or dot product is defined as A·B=ABcosθ_AB where θ_AB is the angle between the vectors A and B and A and B are the magnitudes of the vectors. An example of a scalar product is the work done by a force F which acts over a displacement d, W=F·d· The scalar product is commutative, i.e. A·B=B·A. The vector product or cross product is defined as A×B=uABsinθ_AB where u is a dimensionless vector of magnitude 1 which is perpendicular to the plane defined by A and B. You can do additional research on your own if you want to understand whether u is "up" or "down" relative to the plane, but vector products are not commutative, A·B=-B·A. An example of a vector product is torque τ caused by a force F a displacement r from an axis, τ=r×F.

QUESTION:
Something that confuses me is that scientists say it takes approx. 8min for light to travel from the sun to earth. So if the sun were to explode we wouldn't know till 8min after. However Einstein gathered that the faster an object travels the more time slows down so as you approach the speed of light time stops. But isn't this just relativity? If you are moving that fast everything seems like it's not moving cause your moving so fast. If time were to actually stop how could it take 8min for us to realize the sun exploded wouldn't we realize it instantly? So why can't you move faster than the speed of light? Wouldn't that just mean things relative to you would seem to be stopped for longer, meaning time doesn't stop when you reach the speed of light?

ANSWER:
See the faq page for an explanation why you can't move faster than the speed of light. Suppose that someone was moving with speed 99% of the speed of light from the sun to the earth, a distance of about 8 light minutes. Because of length contraction the distance would, for her, be shortened 8x√(1-0.99²)=1.13 light minutes and the time for her to get to earth would be 1.1/0.99=1.14 minutes; so she would observe the sun's demise to hit earth in a time of 1.13 minutes, and she would arrive 0.01 minutes later. But, just because she observed the time to be 1.13 minutes does not mean someone on earth does; the earth-bound do not see the distance to the sun to be shorter but see it as 8 light minutes.

QUESTION:
Suppose you have two persons watching universe expanding, one of them is on earth, the other is moving fast towards the end of universe. Both of them are watching it all the time. Does this mean, that the "end" of universe exists twice and each is in a different "position"?

ANSWER:
You miss the whole point of special relativity, that the time and position of some event do not have absolute values. That the traveler will observe the end of the universe to be in a different location and at an earlier time than you do does not imply that the universe ends twice at two different locations; rather, it implies that time and position are different for him than for you.

QUESTION:
I was telling my kids that time is space and explaining the block universe and they asked me, if time is space, then how many minutes are in a mile? It sounds stupid, like it does not have an answer, but the more I think about it, the more it seems like if space is time, it should have a specific answer, and if not, why not?

ANSWER:
I will first give the simple answer without much detail since you are dealing with kids. But your kids must be fairly sophisticated to be thinking about space-time at all, so I will then give a little more background on how space-time (four dimensions) arises. When special relativity is mathematically represented as suggesting a four-dimensional space, the fourth dimension is not represented by time (t) but by x₀=ct where c is the speed of light. So the fourth dimension has units of length just like the original three, x₁=x, x₂=y, x₃=z. So, if x₀=1 mi=t·186,000 mi/s, then t=1/186,000 s=5.38x10^-6 s=9x10^-8 minutes, approximately one nanominute.

Before Albert Einstein developed special relativity, it was generally assumed that time and length are universal and independent. One second for me is the same for anyone else in the universe; similarly, one meter is the same regardless of the position or motion of the meter stick I am measuring. If someone is moving with a speed v along my x axis and I measure the length of his meter stick, I find it is shorter; similarly, if I measure the rate his clock runs, I find that it is slower than mine. We are talking about meter sticks and clocks which, if located in my reference frame, are identical to mine. Mathematically this is expressed in what we call a Lorenz transformation,

x'₀=γ(x₀-βx₁)

x'₁=γ(x₁-βx₀)

where the primed coordinates are in the moving frame as measured by the stationary frame, β=v/c, and γ=1/√(1-β²). So, two of the coordinates in our 4-space get mixed up together if the other frame is moving. Does anything analogous happen in everyday 3-space? The answer is yes; if you take a coordinate system with three axes, (x,y,z) and rotate it through an angle θ about the z-axis, the new x'- and y'-axes both depend on what θ is and they depend on both x and y are:

x'=xcosθ+ysinθ

y'=-xsinθ+ycosθ.

Hence the Lorenz transformation may be interpreted as a rotation in the x-t plane. This is the impetus of mathematically working in a 4-dimensional space (space-time).

QUESTION:
I teach high school physics, and I recently performed a lab to investigate Lenz's law. Given the current state of things, this was done in my own home, with students on video chat timing my experiment with their phones. My experiment involved dropping a cylinder of neodymium magnets (about 3 cm long and 1 cm in diameter) down a 1.5 m long copper pipe about 2 cm in diameter. I marked out different lengths on the outside of the cylinder, and used a steel nail to hold the magnet in place at these various “starting lines.” So I could directly drop the magnet 153 cm from the top, and 130, 110, 90, 70 and 50 cm using the nail. For each distance, we did three trials. For each of those trials, seven students recorded through the Google Meet. I expected a bit of a hectic disaster, but the precision of the results was amazing. My idea was to use the graph of distance vs. time to discuss what was happening (they haven’t learned about Lenz’s law yet) in terms of forces. I talked about different types of graph shapes for different situations and how they relate to what is happening with forces: uniform acceleration, uniform velocity and free-fall with air resistance (terminal velocity). My fingers were crossed that our data would approach linearity like the latter, and it did! A valuable learning experience for all. The one hitch is that the z-intercept of our linear trendline is positive, not negative. When an object goes from rest to terminal velocity, I had shown them that the z-intercept of the linear portion of the graph is negative, because the speed only increases to this section of the graph. With our results, I am forced to conclude that the magnet slows down before achieving linearity, because the average slope of the graph is steeper prior to the domain of our results.

ANSWER:
The questioner said in his initial message that the graph of the data would be included but it was not. I have not been able to get him to reply to any messages, but I spent some time working this all out, so I thought I would go ahead and post it. In doing my calculations I approximated some of the variables with reasonable guesses. I had three neodymium approximately 1 cm magnets which when stacked were about 1 cm; the mass of three was 10 grams so three cm would be about m=30 grams=0.03 kg. I took the acceleration to be about g=10 m/s². I made a wild guess at the terminal velocity by watching videos on the internet and chose v=20 cm/s=0.2 m/s. I also note that many videos seem to indicate that terminal velocity is achieved very quickly. The resistive force is given by F=-bv (not proportional to v² as most air drag problems are) where b is a constant to be determined. I choose the coordinate z to increase in the positive vertical direction and the initial position to be at z=0 and the initial velocity to be v=0. This is a classic problem and I will just state the results which result from solving the Newton's second law equation, -mg-bv=m(dv/dt):

v=(mg/b)[e^-bt/m-1]=0.2(e^-50t-1)

z=(mg/b)[(m/b)(e^-bt/m-1)-t]=0.2[0.02(e^-50t-1)-t]

Note that, for large bt/m, v≈-mg/b=-0.2 which is the terminal velocity. This can be solved for b, b=1.5 N/(m/s).

The plot above for v shows that it takes only about 0.1 s to approximately reach terminal velocity. The plot for z shows z(t) in black and z(t) if the magnet moved with the terminal velocity the whole time; the linear portion of the curve is shifted up by 4 mm, possibly responsible for the "positive z-intercept" observed by the questioner. Of course these plots are not what is being plotted by the questioner because each new datum starts at rest; but every datum plotted has the same behavior as the one I have shown, so all the individual runs are shifted by the same 4 mm. The shift is so small compared with the total fall distances, the non linear time is so small compared with the total fall times that the experiment should be able to extract a reasonably good measurement of the constant b. Unfortunately, I do not have the times of the various data points, so I cannot be more specific; the time of fall from the top if the terminal velocity is 20 cm/s would be 7.5 s.

QUESTION:
Is following the line of steepest topographic descent (as stream would do) an example of conservation of energy?

ANSWER:
Let's take a stream as the specific example to discuss the question. The reason the water flows downhill is that there is a force acting on it, the downhill component of the force of its own weight. Generally a stream of water will tend to follow the path where it finds the greatest force impelling it down, that is the steepest terrain. Now, if that were the only force on the water, it would accelerate and the water would be going faster at the bottom. But, in order to do this, the stream would have to get narrower as you went down because water is basically incompressible and if you have, say, 100 gallons per minute flowing at the top of the hill you need the same flow rate at the bottom where it is flowing faster. (Think of a hose nozzle where the water flows much faster out the nozzle than in the hose.) But this is not what happens because there are drag forces which impede the flow. What you generally find is that, if the stream is about the same depth and width as it moves down the hill, the speed is pretty constant. Therefore the kinetic energy at the bottom is the pretty much the same as at the top but the potential energy has decreased; mechanical energy has decreased, not been conserved. However, if you consider the whole system of the stream, the stream bed, the banks, etc., the energy is conserved because the lost potential energy of the of the water will show up as heat in the environment and the water. However, this would be true regardless of whether or not the water flowed down the steepest path. So, I would not use the choice of steepest path to be an example of energy conservation.

QUESTION:
I tried posting this question on a different website and didn't get much of a response. For some reason there doesn't seem to be a lot of information available about this subject, or maybe I just can't find it. Anyway, I am hoping you can help shed some light on the subject. Please forgive the formatting - I've posted a couple of different questions, but if you need a single, concise question to answer, the main one I have is this: What's going on here?

I have a bottle of water that I put in a freezer for a period of time, until the water becomes super-cooled and is still liquid. Now we know that when we shake or disturb the bottle, some of it crystallizes. But it doesn't turn into solid ice. Instead it turns into slush - a mixture of tiny ice crystals and liquid water. Here are my questions:
What determines what parts of the water will become ice, and which will stay liquid?
What is the temperature of the slush mixture?
Is the water part slightly warmer than the ice part?
or is it all pretty much the same exact temperature?
If I poured out the liquid water into a different bottle, and left the slush / ice in the original bottle, would there be any chemical differences between the two?
Is the explanation that the liquid water has impurities such as minerals, and the frozen water is pure H20?
If that is the case, then if we performed this experiment with pure water, then it should turn directly into solid ice, and there'd be no slush, right?
I have read about Fractional Freezing which seems relevant. But this doesn't seem to fully describe what is happening. I am hoping for more of a general explanation rather than a list of individual answers.

ANSWER:
I recommend that you google supercooled water videos and watch a few. It seems pretty clear to me that your problem with slush is that your water is not pure enough. Possibly not cold enough either so that generated heat warms the water just above zero before it freezes. This is a very touchy sort of experiment, one of the videos has the teacher admitting it took him a number of tries to get a good enough take to post. By the way, I thought the answer on Reddit was pretty good.

QUESTION:
Hi, I have a bee in my bonnet about a real world home improvement matter concerning wall tiles and the surfaces they are stuck to. There is much ado about the supposed weight of tiles per square metre that the sub surface can support and arguements are made for using concrete backer boards which can take greater weights than plasterboard (or drywall if you prefer). So plasterboard can take 30kg/m2 and backer boards 50kg/m2 or more. The base of wall tiling always finishes at a horizontal surface, such as a floor etc. So given there are no gaps between tiles and that the wall is truly vertical, what is the relevance of the supporting weight figures for the subsurface board? I can't see that there would be any force acting that would cause any board to fail assuming the tiles end on a horizontal surface that can take the weight. I follow that there will be compressive force down through the tiles and thus through the backing board by virtue of the adhesive to some degree. But surely this compressive force can't affect the backing board as the weight is carried down through the tiles to the horizontal surface that they connect with.

ANSWER:
Well, this isn't really physics, but having had a recent experience (a backsplash behind our range), maybe I can make a few comments. If the wall is sheetrock and sealed or painted, it is best; this is because the surface is just paper and if it is necessary to provide vertical force, it could be too weak. But, as you state, if the tiles extend all the way to the floor (or other support like a cabinet as in my backsplash), there should be no reason to apply the concrete boards. On the other hand, I would not put tiles on a bathroom sheetrock wall because if moisture gets behind the tiles you would have a mess. The same thing goes for bare wood. I once had laid large terra cotta tiles on my kitchen floor on top of an old pine floor; it turns out I had a slow leak in the ice maker which I did not discover until the tiles started coming up from the saturated spongy wood floor. I had to rip up half the floor and leave it to dry for several weeks before reinstalling (too late for a subfloor because it would have raised the floor too far). When I tiled our laundry room I did use the backing board and it is still in pristine condition after 40 years.

QUESTION:
This is really not a question about astrophysics but something simple I am missing. I know that as two massive objects orbit each other they lose energy by emitting gravitational waves and that eventually the orbit decays and they will collide. However as the Earth and Moon rotate around each other and the system loses energy by tidal forces the Moon moves AWAY from the Earth. Why the difference?

ANSWER:
It is understandable that you would think of this because it would appear that the moon is gaining energy rather than losing energy. In fact, it doesn't just appear to gain energy, it is gaining energy. The reason is rather subtle. The moon causes a bulge in the oceans which is responsible for tides. The bulge is actually two bulges, one toward and one away from the moon as shown in the figure (greatly exaggerated); this is because of the tidal force. Because there is friction between the oceans and the ocean floors, the rotation of the earth drags the bulge forward, so it is slightly ahead of the moon as shown in the figure (again greatly exaggerated). Now the bulge exerts a force on the moon which causes it to accelerate a tiny bit; that's where it gets the added energy to increase its orbital radius. Similarly, the moon pulls on the bulge which causes the earth to slow down. If this were the whole story, the energy would be conserved, the earth losing energy and the moon gaining the same amount. But, as noted above, there is friction between the earth and the oceans which takes energy away from the system, giving a net loss of energy but the moon ends up with more than it started with.

QUESTION:
If subject A is traveling through space at a high rate of speed and subject B is traveling at a much slower rate of speed would you expect to see a difference in the rate at which either subject can process information. Example: Would it be easier for subject A to process information than subject B or vice versa?

ANSWER:
This is too vague. What does "process information" mean? What do you mean by "easier". Now, whatever the answers to these questions are, if each subject is presented with the same information stream in his frame of reference and has the same equipment to process it, they will be equally easy to process. But if you mean that A sends an information stream which he processes to B to process (or vice versa), they will not process the information at the same rate. For simplicity's sake, I will choose the "slow" frame B to be at rest; in relativity, only the relative velocity is important. Let's choose a simple concrete example of information stream: the information is a stream of ten pulses separated by 1 second as seen in the reference frame B; B will process them in 10 seconds. Now, how does A process that same information which B processed in 10 seconds? Suppose A is moving away from B; then the time between pulses will be longer than 1 second because he moves some distance away while waiting for the next pulse to catch up with him—the information is processed more slowly. You should be able to see that if you interchanged A and B everything else would be the same so B would process the information more slowly. Suppose A is moving toward B; then the time between pulses will be shorter than 1 second because he moves some distance toward B while waiting for the next pulse to arrive—the information is processed more quickly. You should be able to see that if you interchanged A and B everything else would be the same so B would process the information more quickly. An earlier answer would be of interest to you.

QUESTION:
Matter wave of a particle depends on its momentum,doesn't that mean different observers with different speed will measure different wavelength?

ANSWER:
Yes. This is called the Doppler shift.

QUESTION:
If I am driving in a vehicle in a straight line at a constant speed and activate a drone in the vehicle so the drone is hovering inside the vehicle, open the back of the vehicle and increase my speed so the drone drifts out the back of the vehicle, what then happens to the drone once it is not longer inside the vehicle?

ANSWER:
There will be a period when the drone is leaving the car when there will be lots of turbulence and other currents of air; I am answering neglecting this transition period and also assuming the air outside the car is still. You have adjusted the drone so that, even though it is moving forward with the car at speed v, it hovers in still air. When it suddenly finds itself outside it sees a headwind of v, not still air. This headwind will exert a force which will tend to slow the drone down so that eventually the drone will be hovering at rest relative to the ground.

QUESTION:
Once a spacecraft breaks free from earth's gravity, how much propulsion does it actually need to reach a destination, say Mars? Would the craft have to continually accelerate through the vacuum of space, or once it reaches its maximum velocity, would its propulsion be turned off?

ANSWER:
Technically, you never "break free from earth's gravity". If you acquire a speed greater than the escape velocity, then you will never fall back to earth, but you will slow down continually forever (assuming you have no other objects exerting forces on you). Nevertheless, the force gets very small pretty quickly. If you are 100 earth radii away from earth, the weight (force of earth gravity) is 1/100²=10^-4 times smaller than on earth. And 100 earth radii is not all that far in the whole scheme of things; the distance to the moon is 60. Anyhow, a spacecraft always coasts nearly all the time on a mission somewhere in the solar system. Propulsion is used mainly for getting up to speed and then slowing down to orbit or land on your destination; a much smaller use is to steer, changing direction.

QUESTION:
I have had this question in my mind for over 10y, and I would be very grateful if you could help me to get an answer. The theoretical experiment goes as follows:
1) Bring to outer space 1 round glass chamber, 1 pure iron ball, and a source of heat. 2) Place the iron ball in the center chamber, and heat it until it gets bright red.
3)Go out of the chamber, close it, and generate a total vacuum.
As far as I have been able to go:
Q1: Can we see the bright red iron bar?
A1: Yes, and considering that the ball is in an absolute vacuum the light must be behavior as a particle. *Quantum mechanics dictate that given the duality of light a media is not necessary for light to travel, thus radiation.
Q2: If the light behavior as a particle it must have a mass. Where does this mass come from?
A2. Given that the iron bar is in an absolute vacuum, my logic dictates that the only source of mass for the photons must be the mass of the iron bar itself... Is this correct?
Q3. After an X amount of time when the iron bar has finally cooled down. Will the iron bar be lighter than at the beginning of the experiment? ** Consider that the energy in the iron bar can only be released in the form of light.
A3: I have been stuck in here for several years.....
Thank you very much for your time, please notice I’m not a physicist but a Biology’s with a deep interest in physics.

ANSWER:
I do not see why all the stuff about a glass chamber, heat source, etc. is necessary to ask your question which is, simply: if an object is hot and cools soley by radiation, does the object have less mass after cooling?

First, your A2 is wrong. Photons do not have mass even though they do have momentum and energy. Therefore, your conclusion that the object will be lighter because the photons carry away mass is wrong. However, because mass is just a form of energy, we can say that the photons carry away energy (even though they have no mass). But, where did that energy come from? The object had to provide it. But there are just as many atoms after the cooling as before, so the object evidently changed mass into energy. This change in mass is very difficult to observe in the case of a red hot iron object cooling down. Inasmuch as E=mc², the change in mass would be Δm=ΔE/c² where ΔE is the energy carried off by the radiation. Suppose that ΔE is a real big number, say a 10,000 Joules; then Δm=10⁴/(3x10⁸)²=10^-13 kg=0.1 nanograms! This is approximately the mass of a single yeast cell.

QUESTION:
For a space traveler that leaves Earth headed for a distant star @ 0.9c, wouldn't his onboard clock (that's @ rest relative to him) read the time passes normally......i.e. as if he were still on Earth? He would see no time dilation nor length contraction!

ANSWER:
Wrong. the distance to the star is measured from the earth; think of it as a long stick. The stick is moving with speed 0.9c as seen by the traveler, so he sees the distance shorter by a factor √(1-0.9²)=0.44. Since, as you correctly stated, her clock runs normally, it only takes her 4.4 years to get to a star 10 light years away. It would be worth reading my explanation of the twin paradox which is linked to from the faq page.

QUESTION:
Which is the correct term to refer to the speed of light (scalar) or the velocity (vector) of light. They are both used in books & literature. I think its should be the "speed of light", but just asking.

ANSWER:
It depends on the context. One context is "the speed of light is a constant in all frames of reference and independent of the velocity of the source or observer". Another would be "when a ray of light passes close to a massive galaxy, it is bent and so its velocity changes although its speed does not."

QUESTION:
How do I calculate the kinetic energy of an object moving submerged under water? Specifically if water pressure is involved. Like if a cannonball was fired at 110 m/s under 9,000 feet underwater. Or If a submarine is moving 10 m/s at a depth of 800 feet. I know kinetic energy is factored in and pressure at that depth and possibly water drag. I'm just not sure if there is a specific equation for this. Can you please help me with this?

ANSWER:
The kinetic energy of something depends only on its mass m and its speed v, K=½mv². It has nothing to do with its pressure or its drag force. If there are any forces on the object, those forces do work which might change the kinetic energy. Because of the near incompressibility of water, the depth will have almost no effect in how something moves. The effect of the pressure is simply the buoyant force which is a force equal to the weight of the displaced water which is nearly the same 100 ft down as 10,000 ft down. Similarly, the drag depends only on the density, speed, and geometry, not pressure, the drag will be about the same at any depth.

QUESTION:
What's the physics behind "walking ladders" that move down slopes by themselves? I'm thinking maybe it's the same for slinkys that 'walk' down steps. Gravity and the weight of the ladder being moved around the different legs. I tried Googling for an answer but can't find any scientific answers.

ANSWER:
This was a fun problem to figure out. First, let's think about all the forces on the ladder when it is slightly tipped so that two legs are off the ground. There is the weight W, normal forces (not shown) N_front on the front leg and N_rear on the rear leg, and frictional forces (not shown) f_front on the front leg and f_rear on the rear leg. The weight acts at the center of mass of the ladder which is centered relative to the two sides, but is closer to the front than the rear because the front is heavier. That means that the normal force on the front leg is greater than on the rear leg; this is important to remember as we go further. Now, I want to examine the torque which the weight causes. I have resolved W into components down the incline (W_x) and normal to the incline (W_y). W_y causes a torque about the axis I have labelled DROP AXIS causing the currently-off legs to drop to the ground. The inertia causes the ladder to keep going lifting the other legs off the incline; if none of the legs ever slipped (lots of friction), the ladder would rock back and forth without going forward. Now, the other component W_x exerts a torque about the axis I have labelled TWIST AXIS; if no legs slip, this torque has no effect on the motion. However, for a couple of reasons, the rear leg may slip:

Because the normal force is smaller for the rear leg, the maximum static frictional force will be smaller than for the front leg.

The material contacting the ground may be different for the rear legs than for the front, e.g. the rear legs may have a smaller coefficient of static friction and could slip even if the normal forces were the same.

So, for the video, the rear legs must slip and the ladder, with each "step" twists about a normal axis. There are frictional losses which would cause the ladder to lose energy and therefore eventually stop, but since it is going down an incline, this energy is replenished by the work done by W_x as it moves down the incline. It seems to me that it would be pretty tricky to get this set up since everything must be just right for it to work.

ADDED THOUGHT:
A different possibility occurred to me. I have treated the ladder as if it is rigid. However, if it is kind of rickety the lifted legs will tend, if they can, to move forward because of gravity, even if the two down legs do not slip. So when they come back down they will have moved forward a bit.

QUESTION:
My Dad loves physics and always referred to the speed of light as 186,000 miles per second, per second or 186,000 miles per second sq. that is how he always spoke of it. Is that speed of light an excelleration? (even though nothing goes faster than light.)

ANSWER:
The dimensions for speed are length/time, not length/time/time. Therefore, 186,000 mi/s/s is wrong.

QUESTION:
E=m(c)2
What is the reason for twice the speed of light? Einstein tried everything and found it?

ANSWER:
Oh dear, it is not twice c it is c squared. E=m·c·c=mc². And Einstein did not find it by trial and error; how could he, since it was not really appreciated at the time that mass was, in fact, just another kind of energy? It was a natural result of his theory of special relativity.

QUESTION:
According to Enistine space time is like a fabric and all masses round around the sun so why Moon revolve around Earth not sun?

ANSWER:
Not just the sun, but every object with mass, warps spacetime. It is much easier to think about this in terms of Newtonian gravity for which the equivalent statement is that all objects with mass have a gravitational field. The force which a mass M exerts on a mass m a distance r away is F=MmG/r² where G is the universal gravitational constant. You can calculate the ratio of the force the sun exerts on the moon to the force the earth exerts on the moon:

F_sun/F_earth=(M_sun/M_earth)(R_earth/R_sun)=(2x10³⁰/6x10²⁴)(3.8x10⁸/1.5x10¹¹)²=2.1

So the sun exerts a force more than twice the force the earth exerts on the moon. But if you think about it, the moon does revolve around the sun since it moves right along with the earth; but because of the force the earth exerts, it also revolves around the earth.

QUESTION:
according to law of gravitation less mass object attract toward heavy mass object so why we don't attract toward wall because wall have greater mass then a human mass ?

ANSWER:
Actually, the law of gravitation states that two masses attract each other with equal and opposite forces. So the wall does exert a force on you and you exert an equal and opposite force on the wall. However, since gravity is such a weak force, you do not feel it at all. For example, if you and have a mass of 100 kg and the wall has a mass of 10,000 kg, the force you experience, if you are 10 m away, is about 10x10,000x6.67x10^-11/10²=6.67x10^-8 N=1.5x10^-8 lb.

QUESTION:
I'm trying to calculate what effect hitting a 1,500 lb water barrel at 30 mph in a 15,000 vehicle would have upon the truck's momentum? Also, if I increased the truck's weight to 65,000 and drove it at 50 mph, what difference would this make on the momentum too? I'm working on a road safety prototype in my shed and to be honest this kind of math is way beyond me, so any help would be much appreciated.

ANSWER:
I will assume that the barrel and vehicle move off together after the collision; this is called a perfectly inelastic collision. In a collision the total linear momentum, the product of the velocity times the mass, remains constant. Although lb is not a unit of mass, it is a measure of mass. So, before the first situation you state, the momentum before the collision is 15,000x30=450,000 lb·mph and after it is 16,500v where v is the speed after the collision; equating the two and solving for v, I find v=27.3 mph. For the second case, 65,000x50/66,500=v=48.9 mph.

QUESTION:
I have watched Human Universe with Brian Cox. If you haven't seen it, in part one he goes to Russian to meet with the Soyuz spacecraft. He tells his viewers that only knowing two of Newton's equations (F=ma and the law of gravitation F=GmM/r²) he can calculate that the spacecraft only needs to reduce velocity by 128 m/s to safely reenter orbit. I have looked and tried to figure out how the calculations work but can't. Do you have any clue how he would have done the math?

ANSWER:
I will assume that the Soyuz is returning from the International Space Station. First I will give some information needed later:

R=6.4x10⁶ m (radius of the earth)
h=4x10⁵ m (altitude of the space station)
G is the universal gravitational constant (will not need it)
M is the mass of the earth (will not need it)
m is the mass of Soyuz (will not need it)
g=9.8 m/s²=MG/R² (acceleration due to gravity at r=R)
√(MG/R)=√(gR) (this will be useful later and is why I do not need to know G or M)
MmG/r² is, as you note, the gravitational force on m a distance r from the center of the earth.

If orbits are circular, a=v²/r and you can show that v=√(MG/r). I am thinking that we need to find the difference of speeds between orbits with r=R+h and r=R:

v_R-v_R+h=Δv

=√(MG)[√(1/R)-√(1/(R+h))]

=√(MG/R)[1-(1+(h/R))^-1/2]

≈√(MG/R)[½(h/R)]

=½h√(g/R)

=247 m/s

Note that, because h/R=0.0625 is small compared to 1, I have used the binomial expansion (1+x)ⁿ≈1+nx. My answer is about twice what Green got. Of course, in the real world, air drag when entering the atmosphere would significantly reduce the required speed change due to the additional braking it would cause; but you cannot estimate that using only those two equations. So, I am at a loss to understand the difference.

ADDED THOUGHT:
Maybe the Soyuz dropped to a lower orbit before beginning its reentry, h~200 km, after leaving the ISS, but I could find no reference to its altitude in the video.

QUESTION:
i am currently designing a fully adjustable boom arm for a microphone my prototype is fine and it is functional but i am interested on getting more weight suspension from my joints the joints im using are friction based and use clamping force to lock the joints in the desired position what material should i to use as washers to create more friction in my joints

ANSWER:
it is hard for me to be specific since i do not know what the joints are made of what you need to do is find a material with a larger coefficient of static friction with the material currently in the joints i think the very best way would simply apply trial and error try different materials to see how they work i would start with obvious choices like rubber why dont you use punctuation or capitalization it is very annoying reading your question

QUESTION:
In a fire where does the photons come from? Are photons in a superposition of wave and quanta particles?

ANSWER:
Fire is many chemical reactions going on. The reactions are, for the most part, exothermic and most of the energy comes out as heat and light. Photons are particles which have wave properties.

QUESTION:
I am confused in the formula of orbital angular momentum of electron. According to quantum physics the formula of orbital angular momentum is L=l√(l+1)h/2π B.M. while according to Bhor's postulate the formula of angular momentum is L= nh/2π. Both are giving different result. For example for hydrogen putting n=1 in Bohr formula and l=0 in quantum mechanics formula gives h/2π and 0 respectively.

ANSWER:
I believe there is an error in your question, L=√(l(l+1))h/2π. The answer to your question is that the Bohr model is simply incorrect. As luck would have it, it describes the energy levels of the hydrogen atom well, but it is incorrect regarding angular momentum of the levels.

QUESTION:
Why would a heavy object like a bowling ball not float on top of water, but a wooden telephone pole, which is much heavier, would float? I'm sure the shape and the material the object is made from plays a role but I would just like to know the science behind it.

ANSWER:
The science behind this is over 2000 years old. Archimedes was a Greek mathemetician who discovered what we call today Archimedes' Principle. It says that there is an upward force (called the buoyant force) on an object in a fluid which is equal to the weight of the fluid which it displaces. It turns out that, applying this principle, anything with a density smaller than water floats and anything with a density greater than water sinks. Wood is less dense than whatever bowling balls are made of. You might then ask why a steel battleship can float. It is because it is hollow and therefore the total density is much lower than the density of steel; a hollowed out bowling ball would float.

QUESTION:
I read that metals conduct heat well because the electrons of metallic elements have free electrons in its outer shell that move quickly when heated. These quickly moving electrons transfer their kinetic energy to other electrons which vibrate quickly and collide with other electrons and transfer the heat energy and so and so forth. I also read that metals conduct electricity well because the electrons of the outer shell are free to move (as is the definition of electricity - the movement of charged particles). So it would reason to stand that heating a metal should generate electricity or at the least facilitate the production of electricity as both heat conductance and electricity involves the movement of electrons in terms of metals. To my surprise after a google search, I learned that heating a metal does not generate electricity, and furthermore hampers (not facilitate) electric conductance. Please dear physicist help me to understand why hot metals are poor conductors of electrons.

ANSWER:
In a conductor, there are "free electrons" which essentially behave as an electron gas. There is normally no net current because, just as there is no wind in still air, the electrons move in random directions and the net flow is zero. If you apply a voltage across a conductor, these moving electrons, on average, flow from the negative to the positive terminals. Since they are now experiencing a force due to the electric field, they accelerate; but, obviously, they do not really speed up the whole time that they move from one end to the other of the conductor but, on average, move with some constant speed called the drift velocity. The reason is that they are not really free; rather, they accelerate for some short distance and then collide with one of the atoms in the solid which scatters them, then accelerate again, then collide again, etc. Now, all the atoms in the solid are constantly vibrating as if attached to tiny springs, and the hotter the metal gets, the larger the amplitudes of their vibrations becomes. This means that they present bigger targets for the electrons to hit and so collisions happen more often which reduces the drift velocity and therefore the conductivity.

QUESTION:
I was confused about the strict physics definition of a wave. Considering it is a traveling disturbance that carries energy, would wind on a field, an audience created "wave", and dominoes falling be waves? I thought the audience would not be a wave, and I was unsure about the dominoes because there is no wavelength or period.

ANSWER:
The "strict physics definition of a wave" is that it is something which satisfies the wave equation. The solutions of a one-dimensional wave equation are always of the form f(x-vt) where f is any function; for example where x is the position, t is the time, and v is the speed at which the disturbance travels through the medium which supports it. For your examples: dominos falling and people in a stadium sitting and standing are both waves, but wind is not. Wind is an example of the entire medium moving but that would not be described by a wave equation. The stadium wave would not be a wave if the fans all all just ran around together, like a hurricane of people.

QUESTION:
In my understanding of time dilation if one has a clock in constant motion relative to me I will measure the time ticking off on that clock as slower. However from the other clocks view it is just as valid to say that I AM in motion so an observer traveling along with that clock would measure my clock as moving slower. How is that reconciled? Especially if at some time in the future both clocks are brought together at rest with each other to compare clocks? Am I missing the part that acceleration plays here?

ANSWER:
There is no need to reconcile that each observer measures the other's clock to run slower unless you bring the two of them back together into the same reference frame. Acceleration has nothing directly to do with it. So this is the "twin paradox" which we know is not really paradoxical at all. Go to the faq page and find "twin paradox".

QUESTION:
We have been taught that the big bang created all heavenly bodies as we understand them. I'll disagree with that. Were the exo-planets we have discovered thus far created by the same creation event that brought forth us, Mars and Venus or was it something else that lead to the Gliese's and the Kepler's?

ANSWER:
Good grief, nobody claims that the big bang created everything as it is right now. The big bang created a huge amount of energy (from where, nobody knows) and physical laws. From that point the universe evolved over time eventually atoms, stars, galaxies, planets, etc. appeared. To see a timeline of the universe, click here.

QUESTION:
I'm designing a special suction tube for dental practitioners (I'm an orthodontist) and have a basic question I feel I should know. I want to maintain as MUCH air flow through the tip of the suction tube(end of the tube by the patient's mouth) as possible. The lumen of the suction at the dental unit has a cross sectional area of 75mm2. This tube sucks the air from the patient's mouth into the dental unit. If I use a tube that has a much smaller lumen EXCEPT at the patient's mouth where the cross sectional area would be 75mm2, will I lose air flow? Another way of asking this is: am I required to keep the cross sectional area of the ENTIRE tube at 75mm2 in order to maintain the same flow rate at the tip of the suction (patient's mouth) that I have at the dental unit?

ANSWER:
I learned something new: in physics, lumen is a measure of luminous flux. Biolgists, though, use it as a measurement of inner space of a tubular structure; apparently it is just the cross-sectional area of the tube since the questioner specifies the area of the opening. If you were dealing with an ideal fluid, incompressible, no wall drag, no viscosity, you could argue that the flow would be constant, the speed in the narrower part of the tube would be larger, but the volume per unit time flow would be the same. But air is not incompressible and energy loss effects are not negligible. In order for you to maintain the same volume flow rate you would need a more powerful pump which is pulling the air. Think of trying to drink through a pinched straw–you have to suck harder to get the same amount of drink.

QUESTION:
I have a question for you. So from what I've learnt, Einstein's Theory of Relativity says that an object with mass will bend light towards it by way of its gravity and the source and direction of the light will therefore appear differently to an observer. Besides that, Einstein also says that E=mc^2 and thus m=E/C^2 so something like light, with measurable energy, would have a measurable mass. My question is, just like that object with mass pulls light towards it with its gravity, can light also pull that object towards it with its gravity? Similarly to how an object falling to Earth pulls on the Earth itself? Or does light not have gravity in the same sense as objects with actual mass on account that its mass is relativistic and not actual, classical-physics mass?

ANSWER:
For the first part of your question, see an earlier answer. (You could also have found this on my faq page.) The answer to your second question is that the theory of general relativity finds that any energy density warps spacetime, that is causes a gravitational field; a photon has energy and therefore creates gravity; keep in mind how small this is when compared to any macroscopic mass.

QUESTION:
This is not a homework question, I am actually a teacher and saw this question in a textbook and would like to know if this is a valid question to ask a third-grader. I may be wrong but I do not believe that the pound and ounce are units of measurement for mass. I have tried looking it up online and did not find anything useful. Which two units of measurement are used to measure mass?
Gram
Pounds
Ounce
Kilogram

ANSWER:
Kilograms (kg) and grams (g) (1 kg=1000 g) are units for mass. Pounds (lb) and ounces (oz) (1 lb=16 oz) are units for force. This is very confusing because of the ways weight is measured in different countries. In countries where Imperial units are used, weight (the force which the earth pulls down on something) is measured in pounds (which is correct) and we think that this must also be a measure of mass (the amount of stuff there is). In countries where SI units are used weight is measured in kilograms (which is incorrect) and we think that this must also be a measure of force. I will leave it up to you to decide whether third-graders can understand this subtle difference!

QUESTION:
I know that in a solid the particles are still in motion, but I was wandering if their is any point in the state of matter where the particles are completely motionless? (something like before solid)

ANSWER:
The Heisenberg uncertainty principle stipulates that you cannot know to absolute precision both the position x and momentum p (mass times velocity). Stated mathematically, ΔxΔp≤ℏ where ℏ is the rationalized Planck constant (a very tiny number and Δ denotes the uncertainy of the quantity. So if the velocity (momentum) uncertainty is zero (particle perfectly at rest), the uncertainty of the position of the particle is infinite—the particle could be anywhere in the universe.

QUESTION:
how can the calculations for the curvature of space-time under general relativity be correct without the mass-energy associated with dark matter taken into account? I am just a retired attorney trying to educate my self about our physical world.

ANSWER:
Keep in mind that I specify on the site that I do not do astronomy/astrophysics/cosmology, so you may wish to take my answer with a grain of salt. When you refer to the "curvature of space-time" you are actually talking about the gravitational field. And, you are certainly right, dark matter, if it actually exists, would affect the field. I believe that it is actually the other way around: by studying the gravitational field you can infer something about the distribution of dark matter. The most well-known such inference has to do with rotating galaxies and how the angular velocities of the stars varies with distance from the center. If the only gravity in our Milky Way galaxy were from stars and gas which we know are there, the galaxy would not be able to hold together, stars at larger radii would have speeds far too large. The Wikepedia article on dark matter halos would be of interest to you.

QUESTION:
If a toy car hits a wooden block and pushes it along the track, do the wooden block and the car have the same level of energy or does the energy of the car decrease and the energy of the block increase?

ANSWER:
You have described a perfectly inelastic collision, one in which the two colliding bodies stick together after the collision. I am assuming that the block is at rest before the collision. So linear momentum is conserved and energy is not conserved. I will denote the velocity of the car before the collision to be V, the velocity of the car/block after the collision to be U, the mass of the car to be M, and the mass of the block to be m. Conservation of linear momentum gives MV=(M+m)U or U=V[M/(M+m)]. Note that U<V, so the car has lost energy and the block has gained energy.

If you are interested you can calculate the change in energy (ΔE=E_after-E_before) for each:

ΔE_car=½M(U²-V²)=½MV²[ (M²/(M+m)²)-1]

ΔE_block=½MU²=½MV²[M²/(M+m)²]

For example, if M=m, ΔE_car=-(3/4)[½MV²] and ΔE_block=(1/4)[½MV²]. The car lost three fourths of its initial energy and the block gained one fourth of the initial energy. In this case, half of the total energy was lost. In this case, half the speed was also lost.

QUESTION:
The question is about Maximum gradient a vehicle can climb. The maximum friction force that can be transferred to the road cannot be more than μmgcosθ. As far as I know, μ cannot be greater than 1 (irrespective of contact patch size) for a Rubber tyre on concrete road. At 45 degrees, the drag due to gradient (mgsinθ) is equal to the Normal force (mgcosθ). Any more than 45 degrees, and the drag force increases and Normal force decreases, thus making the vehicle's tyres slip due to loss of traction. Now, I see videos of RC cars on youtube climbing gradients well over 54 degrees. Could you explain this? Is the μ value greater than 1?

ANSWER:
Your analysis is fine if the vehicle is a point mass. But, the video you refer to has the failure to climb a steep incline due to the vehicle tipping over backwards rather than slipping. I will assume the truck shown above is at rest because it is easier to visualize and exactly the same as if it were moving up with constant speed. The whole weight mg may be considered to act at the center of mass of the truck (white cross) which is located a distance h above the ground, a distance d₁ behind the front axle, and a distance d₂ in front of the rear axel, as shown. There are three equations, sum of the forces along the ramp equals zero, sum of the forces perpendicular to the ramp equals zero, and sum of the torques about the point where the rear wheel touches the road equals zero:

f₁+f₂-mgsinθ=0

N₁+N₂-mgcosθ=0

(d₁+d₂)N₂+hmgsinθ-d₂mgcosθ=0

Now, we are interested in when the truck will start rotating backward about the rear axle; when it is just about to do that, the normal force N₂=0. So, the third equation tells us that tanθ=d₂/h; therefore the maximum angle before tipping is θ_tip=arctan(d₂/h). For example, if d₂=1.5 m and h=1 m, θ_tip=56.3⁰. What determines tipping is just where the center of mass is; you want it as far forward and as close to the ground as you can.

But, if you lose traction, slip, before you rotate, you do not need to worry about rotating. As you correctly surmised above, the truck will start slipping when θ_slip=arctan(μ) where μ is the coefficient of static. I agree with you that μ=1 is a reasonable estimate for rubber on dry concrete. However, the video you sent is not done on a concrete ramp, rather the surface appeared to be carpet. Since the tire tread can press into the carpet pile, it is not surprising that μ>1 for those videos.

QUESTION:
How does a single speaker unit play an audio clip consisting of sound frequency of different magnitude? It's the same speaker and has only one 'vibrating membrane' to induce the sound in the air to propagate. How does this single membrane generate sound waves of different frequency and different amplitude simultaneously?

ANSWER:
Sound waves are linear, i.e. if several are acting on a point in space, the net action is simply the sum of all the soundwaves. As an example, I have plotted three waves representing the individual sounds, each with different frequencies and different amplitudes, at some point in space as a function of time. The net sound at this point in space is simply the sum of the three, shown by the black curve. Your speaker need only be driven to vibrate like this; this is pretty easy to do, usually with a magnet attached to the speaker cone and inside a coil, the current in which varies like the black curve.

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