Ask the Physicist!

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QUESTION:
As I was driving this week I thought of this question: what would happened if I grabbed a person, who is standing still, while I am moving at 50 mph in a car. Would the person instantly move at 50 mph? The person would not accelerate due to the car moving already at 50 mph.

ANSWER:
Most likely you would break your arm and the person would experience a force as well but since your arm would bend away, the force on the person would modest and maybe break a few ribs. Suppose we think about a really strong steel arm sticking out the side of the car. If it hits in a way that he "sticks" to it he would be accelerated up to 50 mph very quickly. During the collision time his chest would be crushed maybe by a few inches, probably doing enough interior damage to kill him.

QUESTION:
Hypothetically, could gamma rays be used as a method of sending and receiving information from one person to another telepathically. I know gamma rays are harmful to people. But can a gamma ray be altered on a subatomic level to prevent the bad stuff from the gamma ray harming the person but still allow the info to be received or sent?

ANSWER:
One gamma ray cannot carry very much information, not even a single word or letter. You would need a whole bunch of them, the idea presumably being to modulate them somehow like radio waves. Then you are faced with having enough to injure. And, no, you cannot "alter" them to remove the "bad stuff" because the bad stuff is the ability of gamma rays to damage cells. You would also be faced with the problem of how they would be generated in my brain and detected in yours.

QUESTION:
A log is floating downstream. How would you calculate the drag force acting on it?

ANSWER:
The log is at rest relative to the water so the drag is zero.

QUESTION:
If humans were to make an insanely long straight stick, would it go around Earth, or breifly touch earth at the centre before gradually inclining into outer space?

ANSWER:
It can't be "insanely long [and] straight" but still have the possibility to "go around Earth" because straight and around are contradictory. If it remained straight its ends it would "incline" into space. If it were flexible it would curve around the earth but exactly how tightly it was bound to earth would depend on its elastic properties.

QUESTION:
This question comes from vertical rope sports such as rock climbing, rappelling and canyoneering, where there is a recurring debate on the pros and cons of rappelling on one rope, or two parallel ropes. There's times when lives depend on it. Assume a rope of tensile strength 20KN, typical in mountaineering. The rope is doubled back on itself, and snagged around a bollard. A tensile force is applied equally to the two strands (e.g.

a descending rappeller). What is the resultant maximum strength of the system? Is it 40KN (because the tensile force is distributed across two 20KN ropes), or is it 20KN (because there is only one strand at the back of the bollard)? I say 40KN, but I can't explain away the 20KN argument. Assume the bollard is stronger than the maximum system tension, the bollard is frictionless, and ignore the resulting internal tension/compression within the rope as it curves around the bollard. The latter is usually the real-world point of rope failure in a doubled rope system, as more tensile force is taken by the outside of the rope as it curves around the bollard.

ANSWER:
Assuming that the situation is as shown in the figure, i.e. both halves are held as one so that each holds half the total load, up to 40 kN is possible. If indeed the bollard is frictionless, each half of the doubled rope will hold half the load regardless of how it is held. If there were friction, one half could go slack requiring only one rope, up to 20 kN, to take the load. In the real world, ensure that the ropes are held together.

FOLLOWUP QUESTION:
We regularly create loops of rope with lightweight cord, known as prusik loops, for climbing a rope. Tying a knot in rope results in a strength loss. For cord of 8KN strength tied in a loop with 25% strength loss at the knot, what is the tensile strength of the loop when pulled end to end? Assume frictionless endpoints, e.g. perfect pulleys.

ANSWER:
I am no expert in the technology of rock climbing, but when googling around I found many configurations for prusik loops. I will answer the question for the simplest, shown in the figure. So 25% loss I will take as a loss for both sides of the loop, so pulling only on one half will result in 6 kN strength. But the idea apparently is that you distribute the load equally to both halves (hence the frictionless pulley spec); in that case, the total strength should be 12 kN.

QUESTION:
Hi there I just watched a clip on gravity on YouTube. The one where Brian Cox of BBC humanuniverse does the gravity experiment at that NASA lab. So according to Brian Cox Einstein's theory suggested that the bowling ball and the feather in space are not moving at all and in this clip if there were no background that's what it would look like. Does this then mean that the earth is moving towards the objects?

ANSWER:
Actually, I find this a rather poor explanation for what is going on. I suspect what he was suggesting was one of the postulates of the theory of special relativity is that the laws of physics are the same in all inertial frames of reference. This is called the principle of relativity and says that once you have found one inertial frame, one in which Newton's first law is true, any other frame moving with constant velocity relative to that one is also an inertial frame. Another way of stating this is that there is no special frame which is the universal "at-rest frame" since no experiment you can do can tell you which frame is "at rest", all are indistinguishable. The problem is that the frame of the ball/feathers is not an inertial frame; it therefore seems wrong to me to state, as Cox does, that in fact the ball/feather are at rest because they are accelerating. The appropriate way of understanding this is the equivalence principle which may be stated in two ways. First you can say that there is no experiment you can do which can distinguish whether you are at rest in empty space (or moving with constant velocity, which is the same thing because of the principle of relativity) or you are accelerating due to a gravitational field (free fall). Alternatively you can say that there is no experiment you can perform to distinguish whether you are accelerating in empty space (by burning your rockets, e.g.) or whether you are at rest in a gravitational field. The first of these is most applicable to the case at hand: if you were inside the bolwing ball and falling, it would be the same as if you were in the middle of space and not falling. Finally, the equivalence principle, along with the principle of relativity, lead to the principle of general relativity: the laws of physics are true in all frames of reference. You have hit upon the main flaw of Cox's statement: just taking them at rest would imply that the earth is accelerating toward them which is incorrect.

Here is what Cox was probably referring to as Einstein's happiest thought (although it is probably apocryphal): Einstein, bored at his desk in the Swiss Patent Office, was watching a house painter across the square on a ladder painting a building. The painter fell. As he fell Einstein thought "there is no experiment that painter can perform which will distinguish whether he is at rest in empty space or falling in a gravitational field!" You can read a bit more detail here.

QUESTION:
When a positron and electron collide, dont they first accelerate towards each other through electrical attraction? And if so, shouldnt the photons produced by this collision have more energy than the pre collision particles? I.e. electrical potential energy converted into photon energy. Any analysis of energy momentum conservation seems to ignore this.

ANSWER:
You are absolutely correct, the two photons must have the same total energy as the positron and electron had originally. Similarly, the momentum must also be conserved. The usual textbook example of the annihilation assumes that the two are at rest when they collide. This is most often the case because as a positron enters a solid or liquid it loses most of its energy before it finds an electron to interact with. In a high-energy laboratory setting, this is not the case; the general case is generally not treated in elementary texts because the relativistic kinematics can be quite complicated algebraically.

QUESTION:
I was looking into some things on the net and it made me wonder about particles and anti-particles. My question is, if everything is made from particles and every particle also has a anit-particle, what is stopping them from annihilation. From what I have read, it would seem any contact with each other would cause that to happen but it is not always the case. I also seen labs trying to make anti-matter and was thinking, if we can find what naturally does it, couldn't we technically contain it? Also, when particles are contained to create a bigger body, at what point are they kind of cemented in, or am I just feeling the properties of the material itself?

ANSWER:
First, every particle does not have an antiparticle, but most of the most abundant particles—electrons, protons, neutrons—do. (And, "particle has antiparticle" does not mean that each particular particle has its own corresponding antiparticle; it means a that for a particular type of particle in general there exists a corresponding antiparticle.) One of the open questions to which cosmologists do not yet have an answer, is why the universe is composed nearly completely of particles. Whenever you do have an antiparticle, it will promptly find a particle with which to annihilate. When you refer to "make anti-matter" I presume you are talking about, e.g., anti-atoms like antihydrogen with an antiproton with an orbiting positron; now, we can fairly easily contain charged antiparticles with electric and magnetic fields, but an anti-atom has no charge and could not be easily contained. The difficulty in making an anti-atom is that the antiparticles we create usually have very high speeds and the components of the anti-atom need to be nearly at rest relative to each other to form. I do not understand your last question.

QUESTION:
IF the photoelctric effect cause atom to lose electron which produces current, then why does not heat falling on a object not produce the same effect,??
-- Theoretical part of question

Why does do we get heat as a by product in electric machines or during generation of PV panels why cant that heat be used up in production of electricity altogether??.
--- Practical part of question

Can this two different phenemenon influence a object to create electricity at the same time.

ANSWER:
Adding enough heat to something will cause it to emit electrons. An example of this is a vacuum tube where the filamant, called the thermionic cathode, is heated by a current and emits electrons. Heat in machines is generally from ohmic heating by an electric current; a PV panel is heated by some of the incident light increasing the temperature rather than freeing electrons. This heat is not sufficiently high in most cases to be worth the effort and expense to capture it; and if you were to capture it, it would be better suited for supplying heat to a home, for example, not for generating electricity.

QUESTION:
I recently read that a new upper limit has been suggested for the mass of neutrinos. This got me thinking about the ratio of dark matter in the universe, which I understand estimates put it around 70% of all mass is cause by dark matter. Was a mass used for neutrinos in the original estimation of current mass ratios and how does the changing of the upper limit on neutrino mass potentially effect this? For example if the neutrino were twice as massive as previously expected would the ratio of matter to dark matter be more like 68% is dark matter? How significant are neutrinos in this ratio?

ANSWER:
There is a nice summary of this experiment in a recent post. The upper limit for the mass of an electron neutrino is 1.1 eV where eV is electron-volt; 1 eV=1.6x10^-19 J. Note that, although physicists usually call an eV a mass by convention, it is really the rest mass energy of the particle, i.e. mc². So the upper limit of the mass energy of an electron neutrino is 1.6x10^-19/(3x10⁸)²=1.8x10^-36 kg. Now, most of the mass (not counting dark matter) of the universe is in protons and neutrons each of which has a mass of about 1 GeV=10⁹ eV, a billion times bigger than the the neutrino. Electrons have a mass of 500 keV=500,000 eV; we would expect there to be about as many electrons in the universe as protons so electrons would only contribute about .025% of the total mass. There are about 337 neutrinos/cm³=337x10⁶ m^-3 in the universe, so the mass energy density is about 3.7x10⁴ eV/m³; the average ordinary mass density of the universe is about 4x10^-28 kg/m³=3.6x10^-11 J/m³=2.25x10⁸ eV/m³. So the neutrinos contribute less than about 3.7x10⁴/2.25x10⁸=1.6x10^-4=0.016% of the ordinary mass. Keep in mind that all this represents order of magnitude calculations; the numbers should not be taken too seriously but the relative numbers should be right to better than one order of magnitude.

QUESTION:
In the Science Fantasy universe Warhammer 40k, one of the Primarchs (who're super soldiers created by the most powerful being in existence God Emperor of Mankind) Leman Russ throws a spear from a planet to its moon. Assuming it's similar planet and moon with similar distance as Earth and Luna, how much strength would be required to throw a spear to the Moon? Would it be more than what one of the other Primarchs can do which is to lift hundreds of tons?

ANSWER:
I am going to estimate this by assuming the speed to be the escape velocity of a projectile, the minimum speed at the earth's surface for which a projectile will never come back. This will slightly overestimate the speed necessary to reach the moon for two reasons: the moon is not infinitely far away and the moon has its own small gravity. Still, to do it more exactly will not really be necessary to get to the heart of this question. The escape velocity is 1.12x10⁴ m/s, so the necessary energy of a spear of mass m is E=½m(1.12x10⁴)²=6.3x10⁷m J. I will take m=0.8 kg which is the mass of an olympic javelin, so Fd=5x10⁷ J where F is the force necessary and d is the distance over which the force is applied which I will take to be 2 m. Hence, the force is about 2.5x10⁷ N=5,600,000 lb=2,800 tons=3,100 metric tons.

QUESTION:
After the constant velocity of earth revolving around Sun becomes 0, it is pulled towards the sun with massive attarction, i was researching and testing out in this game named Universe Sandbox, sometimes it is crushed and disappear and sometimes it is pushed massivily that it goes out of the solar system with masive acceleraration, i couldn't manage to notice any differnece in environment in both scenarios. Could someone explain...

ANSWER:
If the earth stops dead in its orbit, it will experience a force toward the sun as you state. But its only possible motion (ingoring interactions with other planets while falling) is to fall straight into the sun. I have worked this problem in an earlier answer and find the time the earth will take to get to the sun is about 2 months. I see no reason it should explode; it will simply fall into the sun and be totally incinerated. And there is no way that it will "go[es] out of the solar system with masive acceleraration".

QUESTION:
If you can leave the earth and travel very fast in space and time slows for you, then could the reverse be true? Since the earth is moving us through space very quickly. could someone leave the and travel extremely slowly through space and time would speed up for them?

ANSWER:
Your first question, which is the basis of your question, has it wrong. No matter what frame you are in, your clock runs perfectly normally; someone who is moving relative to you (or relative to whom you are moving) will observe your clock to run slowly relative to hers. Similarly, you will observe her clock running slowly relative to yours. The speed of the earth in its orbit is about 67,000 mph which is pretty fast but extremely small relative to the speed of light, 186,000 mps (a ratio of 0.0001 or 0.01%) so relativistic effects are negligible. Anyhow, as I said, everybody's clock runs normal rate in their own frame of reference.

QUESTION:
How to convert V/A [Volts/Amper] data to T/s [tesla/seconds] I have a list of time domain electromagnetic data that I need to invert, but the equipment provides the data in V/A and the software to invert need it in db/dt [T/s]. Would someone help me, please?

ANSWER:
Something is wrong here. V/A has the same units as an ohm and so 1 Ω=1 kg·m²/(s³·A²). One tesla has units 1 T=1 kg/(s²·A) so one tesla per second is 1 kg/(s³·A). In other volts per ampere is not equilivant to teslas per second. Since I do not know the details of your experiment, I cannot help you further.

QUESTION:
Why does a ring when thrown horizontally with a jerk comes back after travelling a certain distance?

ANSWER:
When you throw it you throw it forward but get it spinning in the opposite direction from the direction its rotation would be if it were rolling; so the ring is slipping on the floor. The forces on the ring are the weight W down, the force up from the floor N, and the frictional force f. The only force in the horizontal direction is f so that force will slow the forward motion, eventually reversing the motion. When it comes back it will likely roll without slipping.

QUESTION:
Just wanting to know the Maths science behind this senecio A weight of 10 kg falling from a height of about 1 metre hitting a person standing in the head What force speed etc Would the actual weight change etc Anything would be great

ANSWER:
From a height of 1 m the speed would be v=√(2gh)√(2x9.8x1)=4.4 m/s. You can't know the force though. If you look at the faq page (search for how much force on the faq page) you will see that what matters is either how far the 10 kg went before stopping or how long it took it to stop. E.g., suppose that it stopped in 0.1 s. Then the force would be approximately F=mg+mv/t=10x9.8+10x4.4/0.1=538 N (about 55 kg-force). So for a tenth of a second it would be like having a 55 kg mass on the person's head. The weight is the force the earth exerts on the mass which is mg=10x9.8=98 N; this never changes.

QUESTION:
I am an arborist and interested in creating a bow/crossbow to shoot a 12-16oz weighted bag in to trees roughly 100 feet up for rope access to climb the tree. How much force would the bow need to create for say the 14oz to shoot, at a high nearly vertical angle, in to the tree say 120 feet?

ANSWER:
If you look at the faq page (search for how much force on the faq page) you will see that it is not the force which matters but rather the force and the distance and/or time over which that force acts. The velocity you would need to get to h=100 ft would be more than v=√(2gh)=√(2x32x100)=80 ft/s=54.5 mph. I say "more than" because a 1 lb object would occupy enough volume that air drag would not be negligible. Now, how much energy would you have to impart to the object? Again for no air drag considered, and therefore more than E=Wh=1x100=100 ft·lb. So, suppose you could have a bow with a draw of 2 ft; the force you would need would be more than 100 lb. (I have modeled the bow as a simple spring which would have a spring constant of 50 lb/ft. See an earlier answer if you are interested in the details.) With the likely air drag plus the drag of the string being, I presume, trailing behind, I would not be surprised if you needed as much as twice the energy I have calculated here or more. Since you have to send a pound, I would think one of those air cannons they use to send tee shirts into the croud at atheletic events would be a better bet.

QUESTION:
how did ships get designed if the builders didnt understand bouyancy until Archimedes discovered it in 250 B.C.?

ANSWER:
The way ancient societies always have done, trial and error and good ideas. Do you think that the Vikings, around 1000 AD, ever even heard of Archimedes or his principle? But they built ships able to cross the North Atlantic.

QUESTION:
A car travels 6kgm/s. What does it mean?

ANSWER:
All that I can think of is that kgm is kilogram-meter and the linear momentum p of a mass m with velocity v is p=mv and so the car has a linear momentum of 6 kg·m/s. Otherwise it is a typographical error and should have been 6 km/s which would be v.

QUESTION:
Does changing the height the weight is dropped from onto the lever, change the distance the marble flies? I am conducting a science experiment in class, and we built a see-saw type contraption. And we want to know if the height we drop a weight from affects how far the marble on the other side will fly.

ANSWER:
Physics is an experimental science. We observe nature and then we try to make sense out of what we observe. You should first do the experiments, make your measurements, then you should come back to me and we can see if we can understand why you see what you see. I will make a few suggestions regarding how you should perform the experiment.

It would be simplest if you could make your marble go straight up; that is not imperative, but it will be easier to analyze. At the very least, the marble should leave your contraption at the same angle relative relative to the horizontal for every measurement.
You will also get better results if you can arrange for the falling weight to stick to the contraption and always hit at the same place.
Use a good number of falling weights so you can get a lot of data.
Plot however you measure what the marble does (height, distance horizontally, or whatever) versus the dropped weight.

I will help you after you have data if you need help.

QUESTION:
I read that if an object is thrown(from earth's surface) at a speed less than the escape velocity, then it may orbit the earth & what I concluded was that this is only possible when the object is thrown obliquely? Can i please know ( my mind is being obsessed thinking this) that with what speed & at what angle must be an object thrown so that it attains the minimum possible orbital radius of earth? ( Or is there even a thing like it, should a question like this even exist?)

ANSWER:
I will neglect air drag to answer this question; anything with a speed large enough to go into orbit at the surface of the earth would burn up due to the atmosphere. I will also assume that "orbit" means a path which returns the object to where it started. If you throw the object, with a speed less than the escape velocity, at any angle relative to the horizontal other than 0⁰, it will end up crashing into the earth before it completes one orbit. The minimum possible orbit would have a radius equal to the earth's radius and have a speed given by v=√(gR) where R=6.4x10⁶ m is the radius of the earth and g=9.8 m/s² is the acceleration due to gravity. Doing the arithmetic, v=7.92x10³ m/s which is about 17,700 mph.

QUESTION:
if you jump out of an airplane (10,000 ft), in the real world, will you continue to accelerate until you hit the ground?

ANSWER:
No. Because of air drag you stop accelerating after you reach what is called the terminal velocity. For a falling person this is about 125 mph. Someone jumping from 10,000 ft would hit at about 545 mph.

QUESTION:
I am in a law class and we are working on a case of an accident. I�m not in a physics class and have never taken one but we are wondering if it is possible to figure out how fast a car was going based on the information given. �Dana�s car was next to the pole when Vanessa�s car made impact with the pole. That impact began Dana�s reaction time. In 0.75 seconds, Dana travelled at constant speed 20 feet, then hit the brakes and came to a full stop 115 feet after braking.�

ANSWER:
The braking part of this question is actually irrelevant. If Dana's car traveled 20 ft in 0.75 s the speed is 20/0.75=26.7 ft/s=18.2 mph.

It is possible to independently get an approximation for the speed from the data for the braking period. To do this I assume that the acceleration is constant and the coefficient of friction between rubber and dry asphalt is about 0.8. Without going into detail, I find the speed of the car to be 76.7 ft/s=52.3 mph. So these data would certainly not hold up in a court of law! Possible sources of error are:

The reaction time was perhaps overestimated by nearly a factor of three; 20/0.26=76.9 ft/s.
I have assumed that the car did not skid (antilock brakes); possible values of the coefficient of friction have a rather large range depending on conditions, but the smallest likely value is about 0.6 which would result in a speed about 66.5 ft/s, still quite a bit larger than 26.7 ft/s.
If the car skidded the coefficient would be somewhat smaller but not enough to yield a speed in agreement with 26.7 ft/s.
If the road were wet or even icy and she skidded, the speed would be quite a bit smaller.

I looked up a typical human reaction time and found that the average reaction time is about 0.25 s. If Dana's reaction time was average, everything would point to a speed of around 50 mph.

QUESTION:
So, I read how the gravity of a blackhole is so powerful that even light cannot escape. I thought I also read that gravity has no effect on light,(no mass to grab). So, maybe it's time that holds light to a blackhole?

ANSWER:
One of the triumphs of Einstein's general theory of relativity was the prediction that a beam of light will be deflected by a large mass. In fact, the first experimental confirmation of the theory was made in 1919 by Eddington who observed a shift in the apparent position of a star during a solar eclipse due to bending of light passing close to the sun. So, gravity does interact with light and bends it. The reason light cannot escape a black hole is pretty straightforward. You can calculate the escape velocity (the speed necessary to move away and never return) from any object as a function of how far away from it you are. At a distance called the event horizon from a black hole the escape velocity is equal to the speed of light and larger for closer distances. Hence, since nothing can exceed the speed of light, nothing can escape once inside the event horizon.

QUESTION:
Is there an average speed or speeds to maintain in a semi truck in high winds to keep your truck up right and smooth. In other words so the wind would have little or no effect on the truck at all. Theoretically or real world.

ANSWER:
No, as far as I can see. By only adjusting speed you could do nothing to compensate for cross winds. If there were a tail wind you would be able to improve your MPG by driving at the same speed as the wind since then there would be no air drag because it would be like being at rest in still air. But, nothing you do with speed would keep "your truck up right and smooth".

QUESTION:
Is it possible for a bomb that its kinetic energy increases when it explode in the middle of the sky?

ANSWER:
Not only possible, necessary. Imagine a bomb which has been launched straight up. It explodes at the top of its trajectory when it has zero kinetic energy; if it explodes all the fragments go flying off with a large amount of kinetic energy. Since kinetic energy is a scalar quantity, they cannot cancel each other if moving in opposite directions. The fragments fly off in such a way that linear momentum (zero before the explosion) is conserved.

QUESTION:
Why is a cavity with a tiny hole a good approximation of a black body and why are the waves inside of such cavity treated as standing waves in the Rayleigh-Jeans law?

ANSWER:
It is not the cavity which is the black body, it is the hole itself. The argument is that any radiation going into the cavity through the hole has a nearly zero likelihood to find its way back out of the cavity; therefore the hole may be considered a perfect absorber of radiation and a perfect absorber must be a perfect radiator. Only standing waves are allowed in the cavity because otherwise the electric field at the walls must be zero; otherwise energy would be lost to the radiation violating the requirement be that the system is in thermal equilibrium.

QUESTION:
I'm tring to understand whether the various 3D representations of the wave function of the hydrogenic atom, varying according to the l,m,n.. properties, are valid for the atom of hydrogen alone or not. If not, what do they represent? Do they represent the progression of the elements beyond the first hydrogen? If not, is it possible to know the wave probability function of other elements, or molecules, as well?

ANSWER:
The wave function for the hydrogenic atom applies only to atomic hydrogen or any other atom which has had all but one of its electrons (ionized atoms). The wave function of any other atom is considerably more complicated because the electrons all interact with each other as well as the nucleus. Closed-form values of the exact wave functions of multi-electron atoms are not possible and one must use approximation techniques to get solutions.

QUESTION:
with my limited understanding of space travel and all i've been taught with respect to space travel (movies and tv), it seems that a vessel returning to earth must have the ability to withstand the intense heat created during the process, due to its speed and the growing density of the atmosphere nearing the earths surface. is it not possible to enter the atmosphere as slow as we please and drift gently to the ground? i know sufficiently thin atmosphere renders a parachute useless at great altitudes. but what about something like a helium balloon? the vehicle reaches a point in orbit where gravity will inevitably pull it to the ground; so a helium tank inflates a balloon to capacity, and releases helium from the balloon gradually as the vehicle falls.

ANSWER:
It certainly is possible to enter the atmosphere with a slow speed, but since orbital speeds are so large, it requires a lot of fuel to do that. I do not understand how your balloon is going to be significantly different from a parachute in creating necessary drag.

QUESTION:
My question relates to the physics of balloons out of curiosity. Why is it initially hard to blow up a balloon? If the volume and pressure increases does this contradict boyle's law? What would a graph of volume against pressure look like when blowing up a balloon compared to when you let it go and it deflates?

ANSWER:
First, let's address the question of Boyle's law. It is a special case of the ideal gas law, PV/(NRT)=Constant, where P is pressure, V is volume, N is some measure of how much gas you have, T is the temperature, and R is a constant the value of which depends on the units you use to measure the variables. Boyle's law is constant only if the temperature and amount of air in the baloon are constants, which could be approximately true for T but certainly not N since you are adding air as you blow the balloon up.

A balloon is made of rubber. For simplicity, let's think first about a rubber band. As you start to stretch it, it acts like a spring, F≈kx where F is the force that you pull with (and therefore the force the rubber band exerts on you) and x is the distance it is stretched. Then, when you pull it far enough, it starts having less and less stretch for each increase of force. Finally it breaks, or if you pull it past where the stretch is linear and then let is go back you find that the rubber band is longer than it was when you started. This property of not coming back to where you started is called hysteresis. The balloon is analogous to the rubber band as it is blown up. The graph depicts this behavior, but needs a little explanation. Instead of plotting length as a function of force, we plot stress as a function of strain. Strain is essentially the volume (analogous to length for the rubber band) but as a ratio to the original volume; stress is the pressure times the surface area of the balloon which is a force. So the stress (pressure) starts out as a straight line, then maintains almost constant stress (pressure) as the balloon gets bigger, and finally it gets harder to blow it up farther so the stress (pressure) increases rapidly but the strain (volume) increases more slowly until it pops.

Finally, we look at the graph where the balloon deflates. Now the stress (pressure) and strain (volume) both rapidly decrease not following the inflation curve, come to a point where the stress (pressure) is nearly costant, then both finally decrease approximately linearly to zero stress (pressure) but with a strain (volume) bigger than before the balloon was inflated.

There is a nice discussion of balloon physics at this link. (I borrowed their graphs.)

QUESTION:
If I am on a treadmill running 100 miles an hour, would my clock be slower then a man sitting next to me?

ANSWER:
No, because you are both in the same inertial frame.

QUESTION:
I just weighed my dog and he is 61 pounds , and I wondered to myself : How much would it hurt if he came running at me and slammed into me from a running start from about 40 feet away?

ANSWER:
See a similar question recently answered.

QUESTION:
According the Merriam Webster dictionary the definition of falling is, "to descend freely by the force of gravity". How can the explanation given by Newton that a canon ball shot at a specific speed will fall, (descend) around the the earth never hitting it due to the earths' curvature. Would not a more true observation be that the centrifugal force acting on the canon ball is equal to the gravitational force being applied ball thus keeping the ball at a specific distance from the center of rotation?

ANSWER:
First, we need to get clear what "centrifugal force". It is a fictitious force, it does not exist. If you are in a rotating system (which orbital motion certainly is) Newton's laws do not work. Adding fictitious forces can make Newton's law work. Therefore, if you are in the space station, for example, there is only one force on you, your own weight, and yet you are in equilibrium (often called weightless which is a misnomer) which violates Newton's first law. Also, your "more true" explanation works only for circular orbits. If you want to understand the physics you need to view the object in orbit from an inertial frame of reference. In that case what we see is an object subject to only one force, its weight. That fits the dictionary definition of falling pretty well, doesn't it? An orbiting object is in constant freefall. Even elliptical orbits are examples of freefall. The figure shows a planet with velocity v in a very elliptical orbit. The force F which the planet feels; this force can be resolved into two components, centripetal F_C and tangential F_T. The tangential force causes an acceleration which decreases the speed and the centripetal force causes the velocity to turn. In the case of a circular orbit of radius R, F_T=0 and the centripetal force is F_C=Mv²/R; the centrifugal force would be a vector with magnitude Mv²/R pointing away from the sun but it doesn't really exist.

QUESTION:
If a partially submerged ice cube melts completely on water, will it increase the volume? What I thought was, the weight of the liquid displaced by any object is equal to the weight of the submerged part of the body. So the upper part, or the part which is not submerged, when melts, should increase the volume, right?

ANSWER:
Ice has a smaller density than water, otherwise it would not float. Therefore when it melts and becomes denser, its volume has to get smaller because its mass is fixed. Your error is an incorrect application of Achimedes' principle: the weight of displaced water is equal to the weight of the entire ice cube, not just the submerged part of it.

QUESTION:
If in my proper frame I observe an object at near light speed, why does that object not appear to be a black hole? Assuming length dilation, the object should achieve a density, from my frame, such that the object achieves sufficient density. From the object frame I should become a black hole. This is evidently not what happens so I am wrong, but why?

ANSWER:
What happens to the object depends on the conditions in its frame, not yours. (By the way, it is length contraction, not dilation. Dilation would mean the length increased.)

QUESTION:
What happens when you go past your own terminal velocity? I've been trying to ask my physics teacher and google but couldn't find it anywhere.

ANSWER:
I guess it depends on what other forces (besides air drag) are on you. For example, if you are a parachutist and are falling there are two forces on you, your own weight down, W, and the air drag upwards which is proportional to your speed, D=Cv², where C is a constant determined essentially by the area you present by to the onrushing air; so, your terminal velocity is when the weight down equals the drag up or v=√(W/C). Before you open your parachute your speed increases until it is about 120 mph. But that terminal velocity is much bigger than what it will be when you open your parachute. So, when you open the chute you are well past your terminal velocity (as per your question), so the increased drag now slows you down until you reach the terminal velocity, about 20 mph. The answer to your question is that the air drag will slow you down until you are traveling at the terminal velocity.

QUESTION:
I want to solve for angle 'a' in this equation: a - sin(a) = 2pi/3

cannot find a trig identity for "a - sin(a)"

ANSWER:
This is mathematics, not physics. Most equations you are likely to encounter in the real world do not have analytical solutions; they are called transcendental equations. I do not know for sure that this is a transcendental equation, but I believe that it is. In that case you need to do a numerical solution. The simplest way is to do a graphical solution. I plotted sin(a) and a-2π/3. The first graph shows that a≈2.6. If you need it more accurately, zoom in as I have in the second graph; now a≈2.6053. After I did that I found a handy on-line calculator which verified my calculation and went one place farther to a≈2.60533. WolframAlpha is very handy.

QUESTION:
Electric stoves: Nearly all electric stoves I've seen, have circular heating elements (whether the metal restive type or induction) Is there some physics reason that stove tops couldn't be made square? Would that effect the resistivity at the corners, creating hotspots, or something? I ask, because I've thought before, it would be more convenient to have square skillets and fry pans - more surface area to fit, e.g. more burgers, pancakes, or whatever. Yet stoves are almost universally designed with circular cooking surfaces - is there a physics reason for this?

ANSWER:
There is not much information I could find but what little there is does not speak well of square cook pans. They do not generally heat as uniformly across the bottom as round pans do. You might argue that this is because the heat source is generally round; however, it has been pointed out that in an oven, which has the most uniform possible heating at any particular level, brownies in a square or rectangular pan tend to be overcooked at the corners. Of course it would probably be a financial disaster since you have to convince folks to buy all new cookware if they by a square-element stovetop.

QUESTION:
Einstein's theory of relativity that states that E=mc2. The speed of light is supposed to be the universe's limiting speed. So how can it be squared? If it can be squared, can it also be quadrupled? If it can be quadrupled, what happens then? Also, that equation, if correct, also would mean that matter can be created by dividing energy by the speed of light squared? How could that happen? And would this account for the Big Bang?

ANSWER:
The speed of light is the biggest possible speed, it is not the biggest possible number. And, none of c² or c³ or c⁴ or etc. has the units of speed, so it could not possibly have any reference to a superluminal speed. The units of speed are, for example, m/s and c has the units of m³/s³. And, yes, mass may be created from energy; for example, if a very energetic photon interacts with a strong electric field (no mass anywhere there) it is possible that its energy will be converted into an electron-positron pair (called pair production), which has the mass of two electrons. Well, the mass of the universe had to come from something and E=mc² demonstrates that mass is just another form of energy.

QUESTION:
Why doesn't The General Theory of Relativity and Quantum Mechanics not work together?

ANSWER:
The theory of general relativity is a field theory. Therefore, it is natural to want to quantize that field. This has been done with electromagnetism and one of the results is the photon, the "messenger" of the field. Because of the nature of the electromagnetic field, the spin of a photon is 1. The nature of the gravitational field is such that the expected quantum of the field, called a graviton, is spin 2. This alone makes quantization much more difficult to quantize. Another serious problem is called renormalization. Here certain calculations cause infinities which can be worked around in electromagnetism but nobody has been able to deal with them in gravity. So the current situation is that there is no successful theory of quantum gravity and the graviton is a hypothetical particle.

QUESTION:
I have a 2013 Sienna Minivan. I want to load 4 Expedition KAyaks on the Racks to drive to the Grand Canyon THe KAyaks measure 12 feet long 2 feet wide and 1.25 deep. They will be stacked on their sides. Each weighs 50 pounds. THe roof rack siderails are rated for 165 lbs. and the cross bars for 220. We hope to be moving at maximum velocity of 80 mph. Would adding bow and stern lines to the front end and rear frame of the car reduce the load? If so, by how much?

ANSWER:
After googling around a bit, I find that tying on a kayak, let alone four, is not necessarily simple and intuitive. For safety's sake, I encourage you to contact a kayak dealer who can give you better advice than I could.

QUESTION:
Regarding the principle of drag, specifically in group cycling, studies have shown that the rear-most cyclist can experience a 60% reduction in energy expenditure. Does this correspond to a 60% increase in energy requirement parsed put through the other cyclists, in increasing fashion towards the leader? Does the first law of thermodynamics work like this? The bigger overall question would be: does the leader of a group experience additional drag when more riders are lined up behind? Thank you for your assistance in solving this discussion I�ve been having amongst friends.

ANSWER:
This, it turns out, is not a simple question with a simple answer. Part of the reason is that aerodynamics is not simple and to get truly definitive answers to questions like yours takes extremely large calculations on supercomputers. I have answered a question just like yours earlier except for cars. You should look at that answer because it provides the best answer I can give you here.

I did find a lot of sites which address the specific case of bicycles by googling. In one case, somebody in Holland did wind tunnel experiments and found some interesting results, one of which is that the lead biker also gets a modest boost (about 3%).

QUESTION:
I was trying to pose a question to some students based around how high you would need to drop different weighted sports balls in order for them to have enough energy/force to break your finger bones or arm bones as a way of spicing up an introduction to basic motion and energy questions. I've read that 4,000N is about the force required to break a femur, is there a formula that exists or a pathway that allows you to convert from height, velocity and energy into force on impact?

ANSWER:
You can certainly relate height h and velocity v: v=√(2gh) where g=9.8 m/s² is the acceleration due to gravity, h is in meters and v is in m/s; note that it is independent of mass m which will be in kg. The easiest way to relate the speed of the ball to the force it applies is to use energy methods: the work W done on the ball by the bone equals the change in energy of the ball W=E_final-E_initial. The initial energy is ½mv²=mgh and the final energy is -mgd where d is the distance the ball moved before it stopped; the work done is -Fd where F is the magnitude of the force the bone exerts on the ball. (Because of Newton's third law, the force the ball exerts on the bone must be equal and opposite the force the bone exerts on the ball.) Put it all together and you get h=d[(F/(mg))-1]. For example, the weight of a bowling ball is about 50 N (m is about 5 kg) and the force needed to break a femur is about 4000 N. If you observe d to be about 1 cm, h=79 cm. For a baseball, m=0.145 kg, F/mg=2760, so h=.01x2759=27.59 m for d=1cm. But you have to be careful as the mass gets small because air drag, which I have neglected, gets important and the object will never get enough speed to break the bone. E.g., for a ping pong ball with m=2.47x10^-3 kg you find that h=1619 m such that v=180 m/s=403 mph when it hits the bone; but the terminal velocity for a ping pong ball is about 8 m/s so that is how fast it would actually be going when it hit the bone. Unless somebody has a gun which can launch a 400 mph ping pong ball at point-blank range, you do not need to worry about getting a broken bone by being hit by a ping pong ball. Also you have to be judicious about your guess for d because a soft ball, like a volley ball, for example, will compress when it hits so d will be larger which makes F smaller.

QUESTION:
Ok, so, I�m a middle aged housewife, and I'm trying to understand about E=MC². I understand that because matter is energy, by going faster, you are adding energy and therefore effectively increasing mass. Right? Anyone who has experienced acceleration can basically feel this in action, I think. Or if you�ve ever had the opportunity to compare falling down to a bicycle crash at speed, same thing. But why is the increase exponential? Why is it not simply additive?

ANSWER:
Sorry, but you do not have this right. Although when you "experienced acceleration" the mass did increase, the amount was immeasurably small; if you accelerated to 60 mph from rest your mass increased by 0.000000000000004%! Mass (M) increases like M_at.v=M_at.rest/√[1-(v/c)²] where v is the speed and c is the speed of light; only when v is quite close to c is there an appreciable change. What happens at lower speeds is well-described by classical mechanics: your kinetic energy ½Mv² increases while M stays approximately constant. So, what does this have to do with E=Mc². When Einstein developed the theory of special relativity, one of the most important restrictions on it was that any results had to reduce to classical values when v was small compared to c because at these speeds classical mechanics is spectacularly successful. It turned out that when he calculated the kinetic energy assuming that M was a constant he found out that it did not reduce to ½Mv²; but, if he let M depend on velocity, M_at.v=M_at.rest/√[1-(v/c)²], it did. The bonus to this was that the energy of a mass M has energy of Mc² even when it is at rest which means that mass is a form of energy itself and if you figure out how to convert it you can use it to power your car, or your lights, or your refrigerator, and on and on. In fact we began to learn out how to do this millenia ago: fire is essentially chemistry. And, chemistry is gasoline, TNT, coal, natural gas, etc. However chemistry is extremely inefficient and the change in mass when you burn gasoline, for example, is so small that you would be hard put to measure it. That is why chemists have the "law of conservation of mass" which is wrong but works if the change in mass is too small to measure. Nuclear energy is enormously more efficient with a noticeable amount of mass being converted to energy. Since the speed of light is so large, the energy of even a tiny amount of mass is huge. If you have a mass of 1 gram and calculate the energy it has you get E=9x10¹³ Joules; the biggest power plant in the US is about 4 GigaWatts=4x10⁹ Joules/second, which means our 1 gram of fuel could deliver that power for 22,500 seconds=62.5 hours!

QUESTION:
When a ball is dropped on a doormat it doesn't bounce as much as it bounces on floor and the ball doesn't make much sound when it hits the mat unlike when it hits the ground. Where does this energy used to produce sound go or why doesn't a ball falling on a mat nevemakes less noise?

ANSWER:
You are certainly right that the sound produced by an inelastic collision carries away some energy. However, almost always the sound carries a really small fraction of the energy. Most of the lost energy is in heat.

QUESTION:
An object in flight, in this case let's just call it a bullet, suddenly gains mass and weight. If you need specifics a 127 grain bullet turns into a 28,000 grain cannonball. The object has been fired and received propulsion, as it exits the bore of a rifle it suddenly gains mass to become a cannonball. How would this sudden increase in mass affect its range?

ANSWER:
Assuming it gains mass quickly we can approximate the linear momentum (mass times velocity) to be conserved, so m₁v₁=m₂v₂ or v₂=(127/28000)v₁=0.0045v₂. The range of a projectile is proportional to the square of its speed, so the range would be reduced by a factor of about 0.0045²=2x10^-5. (I have neglected air drag.) Of course, I know of no way you could make this happen.

CORRECTION:
I have corrected an error in the original answer: range is proportional to the square of the velocity, not the velocity—R=v²sin(2θ)/g.

QUESTION:
So suppose i have a container full of water and let's assume it's being constantly heated to keep it at the same temperature or it is in a container, that just doesn't transfer heat so quickly. If i poured some of it out, would the temperature of the remaining water be lower, because i took away energized molecules? Or would the remaining water molecules wouldn't feel the effect and they'd still move about with the same velocity as they were? Reason i'm asking this is because i know that it works for single molecules, and it's essentially the reason why blowing on a hot everything, helps it cool down, because the water molecules "jump" out and are essentially carried by air, and so, if you blow on something - you're replacing the dense, moist air with a fresh one. And since those molecules had kinetic energy, the energy of the rest of the system should drop, correct? So the reason i'm asking this is if that's the case for micro objects, does that translate to macro scale as well, or are there some other things at play?

ANSWER:
Trying to answer this I realized that there are a lot of complications like how does air get in to replace the missing water and what effect does that have, so I thought of a couple of possible ways to answer it. First, imagine that we drop a sheet in to cut the volume of water in half. You have essentially taken half the water away from the other half; certainly each half does not "miss" the other. Second, imagine a piston at the top of the volume of water which moves down as water is removed. Again, the temperature would remain constant.

QUESTION:
If you drop a rock in the ocean, does sea level rise everywhere at once, or does the atom displacement propagate in some kind of wave form over time?

ANSWER:
Nothing in nature happens instantaneously. The total area of all the oceans is about 3.6x10¹⁴ m². Suppose your "rock" has a volume of 1 m³, pretty big. Then this volume will be spread out evenly over the whole area (we assume), so we have 1 m³=(3.6x10¹⁴ m²)t where t is the height which the ocean level rose; therefore, t=2.8x10^-15 m; but a water molecule has a size of about 2.8x10^-10 m, about 10,000 times bigger than t! I guess your rock would have to be about 10⁴ m³ in order to add a single molecule film to the whole area. So your question really has no meaning. However, if you added a much bigger rock so that the rise in sea level was measurable, the fastest that the information could be transmitted to raise the surface is the speed of sound in water. (All this assumes we do not need to think about things like tides, storms, etc.)

QUESTION:
Plane at rest with 10kg of mass with birds iside sitting still with 5kg total amount of kg is 15kg. If the birds staring to fly inside the plane is mass still the same?

ANSWER:
I have answered questions like this many times. See faq page.

QUESTION:
Light, for example laser light is able to penetrate a metal and make it to two halves , so why cant we use that power with help of some instruments to move a object in light speed?

ANSWER:
The fact that laser light is able to cut steel tells you that it carries a considerable amount of energy. However, to accelerate an object to light speed requires an infinite amount of energy and the laser does not even come close to that. The laws of physics forbid any object having mass from moving at or above light speed.

QUESTION:
Could a large object orbit a smaller object? For example, could a Jupiter-sized mass orbit an earth-sized mass, if the earth-sized mass was super dense and the Jupiter-sized mass was supper fluffy?

ANSWER:
First of all, one object never really orbits another, rather they orbit each other. Put another way, each orbits their mutual center of mass (COM). And all that really matters is the mass of each object. For situations like the earth and the sun, the sun has a mass so much bigger than the earth that, for all intents and purposes, their COM is at the center of the sun and it is reasonable to assume that only the earth orbits; the ratio of the distance of the COM from to the radius of the sun is 0.000646. The earth-moon system has that same ratio as 0.732, nearly 3/4 of the earth radius from its center; so the earth significantly "wobbles" as it and the moon orbit each other. So, in answer to your question, any two masses may orbit each other as long as they are far enough apart that they do not touch each other. For more detail, see Wikepedia.

QUESTION:
How much weight am I actually pulling when pulling a 175lb. tire across a rubberized floor by hand. I assure you that this is not a home work question, I am an avid power lifter/gym rat. This question is for my own personal use.

ANSWER:

What you really want is the force you have to apply to pull the tire; the weight of the tire is the weight of the tire. First, the simple case which is if you are pulling horizontally. The force you need to apply to move it with constant speed is given by F=μN where μ is the coefficient of friction for rubber on rubber and N is the force which the tire and floor are pushed together, the weight in this case. The coefficient is about 1.2, so the force you must exert is about 210 lb. If you are pulling at some angle θ, it is a bit trickier because you are both pulling forward which moves the tire but also pulling upward which causes N to get smaller. Without going into details, the result is F=μW/[cosθ+μsinθ]. I have made a graph which shows this result for your case. As you can see, the closer to horizontal you can pull, the better workout you will get. Also note that if you pull at around 50� you will exert the least force

QUESTION:
please check this answer on Quora, it says a particle cant be in two (or more) places at once. how can this go with what we read on the internet? for example, there is this study which based on the conception of the particles being in places (parallel universes)�When the team fired a photon at the cavity, the atom�s dual personality caused two things to happen at once. In one of the 'parallel universes' of its superposition, the one in which the atom was resonant with the cavity, the photon did not enter. and many many science articles on the internet for the ordinary readers (not specialized) says it can, and it is. some related informations about Superposition and Wave-Function i found.

ANSWER:
My first response is that just because you see it on the internet does not mean that it is accurate; in fact, you should always be wary. The next thing to appreciate is that the idea of parallel universes is a hypothesis, it is not a fact, and no evidence of any other universes has ever been observed; when someone puts forward this notion (for "ordinary readers") of a particle being in two places at once, it has nothing to do with parallel universes. That said, I have to say that the answer on Quora is excellent. One could actually say that saying that a particle is anywhere at all is meaningless unless you have made a measurement of the position and then you only know where it was at the time you observed it. It is meaningless because all you can specify is the probability of its being inside any volume at any given time. I have answered this question several times in the past and you can see those via the faq page.

FOLLOWUP QUESTION:
does that mean that the electrn is in separated places before a measurement is done? or a wave can't be described by place ?

ANSWER:
It means that a "particle" is described by a probability distribution, not by a point at a particular place in space. The modern way of viewing a hydrogen atom is to not think of an electron going around in a well-defined orbit but rather of being like a cloud around the proton, the details of this cloud being determined by the energy state the electron is in, its angular momentum, etc. The electron is therefore spread out all over the place but you wouldn't say that it is "located" in many places; rather, there are many places where you might find it to be if you went looking for it.

QUESTION:
How are air resistance and friction related? Does the size or shape of a parachute or the materials that a parachute is made from affect these factors?

ANSWER:
Air resistance is the frictional force experienced by an object moving through a fluid. For most cases of everyday interest this force (often called drag) is proportional to the square of the velocity, F∝v² or, F=Kv² where K is a proportionality constant and depends on the geometry of the object and the density of the fluid. The actual drag equation is F=½ρC_DAv², where ρ is the density of the fluid, C_D is the drag coefficient which depends on the shape of the object, and A is the area which the object presents to the onrushing fluid. So, for a parachute, since A is much bigger when open than when packed, the drag is much larger and therefore you survive if you jump out of an airplane with one. There is an approximation to the drag equation for motion through air at atmospheric pressure which is convient for estimating drag for those special cases, F≈¼Av²; this approximation does not work unless you do your calculation in SI units, m² for A, m/s for v, and N for F. Here is an example: a skydiver and his equipment have a weight of 1000. She accelerates to some speed v where she experiences an upward drag of F=�Av²=½v²(I have estimated A=2 m²); but her weight is another force on her, so the net force on her is F_net=1000-½v² and when v gets big enough she falls with constant speed because the net force is zero. This is called the terminal velocity and, for this case, v=√2000=45 m/s=101 mph and she will likely die. If she had opened her parachute and it had an area of 25 m², her terminal velocity would have been about v=√160=12.7 m/s=28 mph and she would have survived.

QUESTION:
I do not understand what a "charge" is made of, with respect to electricity. I have been told that it is simply a characteristic of electrons and protons. But that does not explain it. You could say that a charge is pure energy, which we also cannot define. Feynmann couldn't explain it either. Thanks. (I am an electrician and supposed to know these things, and went to college for 4 years, but don't know the answer to this simple question.)

ANSWER:
There is no answer to "this simple question". In science there are some things which are at the root of something which you have to accept as just the way the universe is. When a force is discovered it is natural to ask, what is the source of this force, what property allows some objects exert or feel this force? For the electric and magnetic fields this property was called electric charge. Historically it was found in ancient times that sometimes we find two objects which seemingly magically exert forces on each other. With static electricity it was discovered that the forces objects exert on each other can be either attractive or repulsive; this led to the conclusion that there is not only one property called charge but charge must come in two different "flavors". Asking what charge "is made of" is like asking what "blue is made of"; blue is just a property some objects have.

QUESTION:
I understand that from Newton's 3rd law, when 2 positive charges interact, there is a repulsive force 'away' on both charges. Now, if i consider this system of 2 positive charges to be a closed system, i would expect the net internal work done to be 0. Lets say i want the work done to bring the charge towards a middle point between them. However if i consider net work done to be the sum of both work dones( work done by one of the charges + work done by the other charge) the math would be: -ve work done added with -ve work done. (Opposite direction to where i want the charge to go==>-ve work done) Why is it not 0 then?

ANSWER:
It is not clear what you are asking. Newton's third law says that the force which one exerts on the other is equal and opposite the force the other exerts on the one. Now you want to consider them an isolated system. Suppose that they are some distance apart and at rest; they have a certain amount of potential energy and no kinetic energy. Now we let them go. After they have each moved a certain distance they now have acquired equal amounts of kinetic energy and they have less potential energy. Since the work done on the system by external forces is zero (because all forces are internal to the system), the total energy must be conserved so the potential energy before is equal to the sum of the kinetic and potential energies after.

QUESTION:
How earth travels faster when It is nearer to sun even though its angular momentum remain constant

ANSWER:
At the point in the orbit closest to the sun the angular momentum is L_close=Mv_closeR_close. At the point in the orbit farthest from the sun the angular momentum is L_far=Mv_farR_far. Since the angular momentum is conserved, Mv_farR_far=Mv_closeR_close or v_close/v_{_far}=R_far/R_close. Therefore v_close>v_{_far}. It is a little more complicated elsewhere in the orbit, but L is always the same.

QUESTION:
If a single neutron can exist outside of a nucleus for around 18 minutes before decaying, can this time be extended indefinitely if that neutron collides with another and remains close enough so that the Strong Nuclear force comes into action? Can groups of neutrons exist without the presence of their charged u neighbors? If we extended this idea to more than two neutrons or the equivalent of larger atoms without electron shells, could stability could be achieved. If so, would we be able to detect this mass at a distance. I don't see how it would interact with photons without electrons present.

ANSWER:
No one has ever been able to observe a bound system of only neutrons. There have been suggestions that perhaps the 4-neutron system or a 2-neutron system might be bound, but neither has been experimentally observed unambiguously in spite of many attempts. Also, a neutron star is predominantly a bound neutron matter, but the binding is due to gravity, not the strong interaction.

QUESTION:
I am doing some of my own research (nothing to do with school) about atoms. I asked my teacher what colour is a nucleus, she said to ask a physicist and my old science teacher is a physics teacher but i don't talk to him. Also, can a proton and neutron have the colours green and purple?

ANSWER:
How do you determine the color of something? It either reflects or emits that color. The problem here is that a nucleus has a size (on the order of 5x10^-15 m) which is much more than the wavelength of visible light (on the order of 5x10^-7 m). It is therefore impossible to "see" a nucleus by imaging with visible light like you might do with a bacterium. In that sense, it makes no sense at all to try to talk about its color. If a nucleus radiates, the wavelength is never in the visible region, x-rays being the longest wavelength you will detect. An individual neutron or proton is smaller than a nucleus so the same arguments apply to them. The bottom line is that none of these things have color.

QUESTION:
I cannot get over Newton�s conservation of energy. It seems clear that the energy spent climbing the stairs of a 50 floor building would not compare to the kinetic energy achieved jumping off the building. Since ever my stair be-it the first, the 20th, or the 50th requires the same force to overcome yet gravity increases your speed exponentially. Even with wind resistance the strongest person could never traverse a 50 story building in 4 seconds at 144mph??

ANSWER:
You got your numbers wrong. A 50 floor building would have about 15 ft per floor, so the height would be about 15x50=750 ft. Dropping from that height would result in a time of about t=√(750/16)=6.8 s and the speed would be v=32x6.85=219 ft/s=149 mph. So, if your mass is m, your kinetic energy is ½mv²=2.4x10⁴m ft lb. Suppose there are 20 steps per floor, so each step is a height 15/20=3/4 ft. So the amount of work you do going up one step is (3/4)32m=24m ft lb. So the work you do going up is 24x1000m=2.4x10⁴m ft lb. So, you see that the energy you put in by climbing was exactly equal to the energy you got out by falling back. There is no such thing as conservation of kinetic energy. You do work to increase your energy (called potential energy in this case), so you end up with this potential energy (and no kinetic energy) at the top. If you jump off, that potential energy is turned into kinetic energy.

QUESTION:
My brother died recently and I took great comfort in the Law of Conservation Mass/Energy. I'm guessing this isn't a typical question you'd get, but want to tattoo this equation on my arm. I've seen a few different variations of this equation, so I'd love to hear from a physicist on the best equation to represent this law.

ANSWER:
My condolences on your loss. Well, the total energy of a closed system is indeed conserved which would be expressed (kinda boring) as E_i=E_fwhere i denotes initial and f denotes final. But, there is no such law as conservation of mass because Einstein showed that mass is just a kind of energy. So the mass of a system need not be the same before and after. Chemists still use a "law of mass conservation" because the mass changes in chemical reactions are too tiny to measure; still, as we know in physics, the mass does change in chemical reactions. Now, you know the famous equation E=mc², so how about m_ic²+E_i=m_fc²+E_f where m_i and m_f are the amount of mass initially and finally and now E means all energy except mass energy.

QUESTION:
Hi, I'm confused about how linear-moment (LM) and angular-momentum (AM) can both be conserved simultaneously. For example, just before a bullet sticks in a stationary beam on a pivot, the system has only LM (that of the bullet) and the system has zero AM (because the beam is stationary). After the bullet sticks in the beam, the whole system will rotate, so now we have only AM and no LM, right? It seems to me that the LM was converted to AM and therefore not conserved. What's wrong with my thinking?

ANSWER:
(I will assume that the beam is like a uniform beam and hinged frictionlessly so that the beam can swing in a horizontal plane.) Your main mistake is that the bullet does have angular momentum before the collision. Consider the bullet as a point mass; a point mass has a moment of inertia I_bullet=mr² where m is the mass and r is the distance to the axis about which you are computing the angular momentum, in this case the hinge of the beam. The angular momentum of the bullet at the instant of impact is L=I_bulletω=(mr²)(v/r)=mrv. Because the hinge has no friction, there are no external torques on it and therefore angular momentum will be conserved. The beam has a moment of inertia I_beam=ML²/3 where M is its mass and L its length. After the collision the beam plus bullet have an angular velocity ω'. Conserving angular momentum, mrv=(ML²/3+mr²)ω' which can be solved for ω'.

What about linear momentum? It is not conserved because the hinge exerts a force on the system, the absense of which is the condition for linear momentum conservation. If this were all happening in empty space, both angular and linear momentum would be conserved; afterwards the beam+bullet would rotate about their center of mass and the center of mass would move in the same direction the bullet came in with a speed v'=mv/(m+M).

The bottom line is that linear momentum is conserved if there are no external forces on the system and angular momentum is conserved if there are no external torques on the system

QUESTION:
Sir, the average density of nucleus of an atom is very high in the range of 10^17kg/m^3.However the density of elements which are made of these atoms is low why

ANSWER:
Because the volume occupied by the nuclei is extremely small.

FOLLOWUP QUESTION:
my question has been misunderstood. Let me put in this way. There is a wooden empty box having some mass. Now if in this box we put a small volume of dense material say mercury. Then the mass of the box with mercury will be more even if mercury occupies small volume. The overall density will be greater than the density of empty wooden box. It will simply add the density of wooden box. Similarly in case of nucleus of atom being denser up to the level of 10 raise to power 17 kg/m3 and the matter consisting of millions of atoms, how it does not reflect in the over all density of that matter. I mean there is a huge difference. In 1 cm cube of water the mass comes to 1 gram In that 1 gm of water there will be more than 6.023x10^23 no of atoms having these massive denser nucleus, each with density of 10^14 gm/cm^3, then how is it that water in that 1 cubic cm volume comes to be 1 gm only.

ANSWER:
I have not misunderstood your question. You misunderstand how to calculate density. suppose the box has a volume of 1 m³ and a mass of 1 g=10^-3kg. Its density is 10^-3 kg/m³. Now, put into that box an object whose mass is 1000 kg and whose volume is 10^-3 m³; the density of this object is 10⁶ kg/m³, but the density of the box plus object is 1000.001/1 kg/m³, 1000 times smaller. So, now let's do the same thing for a single atom (which can be approximately scaled to a macroscopic object). I will take your number for the nuclear density, 10¹⁷ kg/m³; I am sure that you know that the mass of an electron is about 2000 times smaller than that of a proton or neutron, so the nucleus mass is roughly 4000 times greater than the mass of all the electrons; therefore I will approximate the mass of the electrons to be zero. The ratio of the radius of the atom to the radius of the nucleus is about 10⁵ and therefore the ratio of their volumes is about 10¹⁵. Therefore the density of the atom will be on the order of 100 kg/m³; this number is comparable to typical macroscopic densities. This is just a rough approximation but the average density of an atom is simply its mass divided by its volume, and that is what matters when you determine the density of some object made up of a great many atoms, not what the density of some little part of it is.

QUESTION:
State the condition at what height and what depth the acceleration due to gravity remains same?

ANSWER:
I have previously answered this question.

QUESTION:
I have a bit of doubt relating to a scenario and I would really appreciate your assistance. If a person of known mass jumps off a wall of known height, but his body comes to rest a while after his feet strike the ground (perhaps 0.15 seconds), when determining the force, impulse,momentum and stopping distance, how do you take that 0.15 seconds into account?

ANSWER:
You can calculate how fast he is going when he hits the ground. Then you need to assume that the force he feels from the ground is a constant. Knowing the time to stop you can deduce the force needed. Once you have that force, you can calculate the acceleration, impulse, and distance. Don't forget about his weight.

QUESTION:
I am a layman, with a strong interest in natural Sciences. Lately light has caught my attention, and I am starting to understand the complexity of the topic and how far we are from a definitive answer on its real nature. Is EM waves energy dependent only on amplitude or amplitude and frequency?

ANSWER:
When you say "…how far we are from a definitive answer on its real nature…" does "we" refer to you or to everyone? Just asking because it is important for you to understand that electromagnetism is perhaps the best and most completely understood field theory in all of physics. Regarding light (electromagnetic radiation, in general), I would recommend that you read something about the photoelectric effect to understand where the notion that the energy of an electromagnetic wave was proportional to the square of its amplitude was found to fail. The solution was simply that the EM field is quantized which means that the radiation has a minimum amount of energy, called a photon, in which it can exist. In essence this means that if you have a particular wave and wish to reduce its energy, you cannot do it smoothly but only in steps of the energy of one photon at a time. This had never been observed in a laboratory because the energy of a single photon is so tiny compared to the energy of a macroscopic wave. So, if the wave interacts in a macroscopic way, for example a radio wave interacting with a radio antenna, you find that the energy is indeed dependent only on its amplitude. On the other hand, when you try to make that wave interact with an individual electron, the electron sees and interacts with individual photons of energy hf as you note in your original question. How can we reconcile these two facts? Simply that if the radiation interacts as a wave, amplitude matters, but if the radiation interacts as a photon, frequency matters. If you have a blue wave and a red wave with equal amplitudes, they carry the same energy; but the blue photons have a higher energy than the red photons, so there are more red photons in the red wave than there are blue photons in the blue wave.

So, you asked whether the energy of light is "…dependent only on amplitude or [on] amplitude and frequency…"; the answer is neither, it depends on amplitude or frequency!

QUESTION:
Why do the planets rotate is a disk pattern and not a spherical one?

ANSWER:
Imagine a huge cloud of gas and dust, trillions of miles across. It has formed randomly, gradually growing because of the gravitational attraction for material in its neighborhood. It is almost impossible to think that this cloud has no angular momentum, that there is not some net spin. Even if the net flow of particles around the center of mass of this cloud were, say, 1 cm/s in its far reaches, because it is so far away it has a large angular momentum. Now the cloud begins to get smaller, being pulled together by its own gravity. However, if that cloud has angular momentum, the result is that it will spin faster and faster as it gets smaller and smaller. But, imagine a squishy ball which is spun faster and faster—it flattens out and eventually it is more like a pancake than a ball. The center of this squished ball is the most dense and eventually is dense enough to ignite to become a star; the outer reaches of the squished ball contain the matter from which planets will eventually coalesce and form.

QUESTION:
Regarding Kinematics, I understand that the outer edge of a car tyre is travelling between 0mph where it is contact with the road to 2x the speed of the car when that point is furthest from the road. My first questions is: If the car was driving at the same speed on two wheels, so right side was off the ground, would the drive wheel tyre still continue to have the velocity change if it was not in contact with the road? The reason I ask is that someone mentioned that we, on the Earth should feel the velocity change due to the rotation and orbit of the Earth. I'm sure it is due to the fact that the velocity has to be relative to a point, ie the road in the case of the car tyre, but I have been looking online for an answer to this problem and so far no luck!

ANSWER:
Actually, your first question has nothing to do with your second question. I am assuming that the wheel off the ground is spinning at the same rate as the wheel on the ground. Then, relative to the road, the bottom point of the off-ground wheel will be observed to be at rest just as if it were touching the ground. For your second question, you should, in principle, be able to feel the effects of the earth's motion; however, the effects are too small for you to be able to "feel" them. But those effects are due to the acceleration you encounter, not the velocity. One simple example would be the centrifugal force you feel due to the earth's rotation. If you were standing on the equator you would find that you weigh about 0.35% less than if the earth were not rotating; if the earth spun fast enough, you might just fly off into space.

FOLLOWUP QUESTION:
I realise now that I may not have fully explained my question. I understand your answer as to why we don�t feel the rotation. As the Earth�s translational motion of the orbit around the sun is an average of 30 kilometres per second, would there be a change from 0Km/s to 60Km/s? And if so, why do we not feel that acceleration/deceleration? I have attached a small diagram incase I have not explained it properly!

ANSWER:
Ah, now I understand! But, no, this is not the same situation as the car wheel. If the earth were rotating on its axis such that the equator had a speed of 30 km/s, it would be so. However, the actual speed is 0.46 km/s. Therefore the far side is going about 30.46 km/s and the near side about 29.54 km/s.

FOLLOWUP QUESTION:
One final part, could you explain why it is not the same situation as the car wheel? I need to understand the difference between the physics of rolling without slipping and the orbit of a rotating planet.

ANSWER:
In my original answer I said "I am assuming that the wheel off the ground is spinning at the same rate as the wheel on the ground." That is the necessary condition. In other words, if the earth were spinning such that the equator moved with speed 30 m/s relative to its axis, the sun-side of the planet would be moving 30 m/s backward relative to the 30 m/s orbital speed forward.

QUESTION:
I have a hose that is 4 inches in diameter and a thousand feet long. I want to roll it up on a reel. It wieghs 4375 lbs. It is stretched out on a concrete road. I am going to reel it around a drum that it 6 feet in diameter after i attach it to the drum. How much torque will i need to reel this thing up? I have looked at the coeffiecient of rubber/concrete and it is .65 That would mean it would take 2843 ft lbs of torque to reel it in. I don't think this is right because the whole 4" of rubber is not sitting on the ground. I would think that you have to factor in the area of hose that is in contact with the ground.

ANSWER:
First of all, the way we often calculate friction, F=μN, is not a physical law, it is just an approximate method for getting the frictional force if an object is sliding and based on experimental measurements. N is just how hard the two (sliding and stationary) surfaces are pressed together and μ depends only on what the two surfaces are. Note that the area of contact is nowhere in this equation, so that is not relevent for your question. Secondly, the number you calculate is not a torque, it is the force you would have to exert to just pull the whole hose. The torque is determined by how long the handle you crank with is. I envision a crank of length L attached to the axle of the drum on which you exert a force F at its end and perpendicular to its length; I imagine a ratchet which will lock the crank to the drum when turning counterclockwise (taking in hose) and slipping when the crank is repositioned by moving it clockwise. The resulting torque about the axle of the drum is FL. Meanwhile the hose is exerting a force T, as shown, on the drum and its torque is -TR. (This is negative because it is trying to spin the drum clockwise whereas the torque due to F is trying to turn it counterclockwise and we have called that torque positive.) If the hose is at rest or moving with constant speed, the sum of the torques is zero, so FL-TR=0 or F=T(R/L). In your case, T=2843 lb, so if, for example, L=5R, F=569 lb. The resulting torque would be 17,058 ft lb.

Practical notes:

The more hose you bring in, the smaller the friction will be;
as hose piles up on the drum, R gets bigger; and
static friction is bigger than kinetic friction, so it will be harder to get it moving than to keep it moving.

QUESTION:
The question in which I asked keeps me up for hours. There's a lot more to my question than I put which also doesn't add up. Could Einsteins theory of general relativity be flawed, at least when it comes to gravity. I'm ot saying I know how gravity works but I just have a niggly feeling that the whole warping of spacetime by a massive object just doesn't work in so many ways. I am not an argumentative person so don't worry. I'm just tired of this thought pestering me constantly for years. Just your thoughts would be much appreciated and maybe I could finally be rid of this thought wish bothers me so much.

ANSWER:
It is not at all clear what it is that is bothering you, your "niggly feeling". I also have been bothered by the notion that gravity is apparently just geometry, not really a force, yet one of the most important efforts in physics today is the attempt to quantize it as if it were a force. You must admit, however, that considering it as "flawed" is overstatement inasmuch as it has passed every experimental test over the last century. I recently answered a question along these lines in which I found that "warping of spacetime" is just one possible interpretation of the nontrivial mathematics which is the framework of general relativity; this visualization is attractive because it allows persons like yourself to comprehend the implications of the theory. However, the mathematics of the theory may also be interpreted as a field theory, and fields can be quantized. The essay linked to is short and not too technical; reading it may put you in a state let you might be able to get a good night's sleep!

QUESTION:
On Reditt I made the staement "emissions from Earth are limited by temperature" in regards to it's continuous radiation. Given that the average temperature is 15C the shortest continuous wavelength of radiation would be IR at about 10 μm. I've been inundated by people who insist that Earth radiates continuously at all wavelengths right up x-rays. Every source says that's not possible. Would you please settle this and post the answer so that I can link to it?

ANSWER:
The figure shows satellite measurements (solid line) of the radiation spectra over northern Africa at midday. The temperature is near 320 K (about 115⁰F). If there were no atmosphere, this spectrum would look like black body radiation, the dashed line labeled 320 K. Note that its peak is near 15 μm. But there is atmosphere and it absorbs a lot of the radiation resulting in large discrepancies between observed and predicted spectra. Most notable is CO₂, the prime culprit in global warning, which reradiates this infrared (heat) back to the earth. Also ozone (O₃), methane (CH₄), and water vapor absorb a lot of enery. I do not understand what the argument is; I do not know what is meant by

limited by temperature,
continuous radiation,
radiates continuously, and
Every source says that's not possible.

I do know that data clearly show that the earth radiates at all wavelengths, some of which do not fully escape the earth because of interactions with the atmosphere; wavelengths far from the maximum black body intensity for a given temperature will be small enough intensity to be negligible.

QUESTION:
My question may be too simpple for you and it is this: My dog, 22 Kg weeight, whle at play with another dog crashes against my ribs at almost 20 kilometers per hour breaking two of my ribs in the process. What force in pounds/feet was exerteed against my ribs? I am 75 years old so this is not a 'homework' question but mereely onee of intereest to me. My dog, Dylan, had done nothing wrong other than not look wheree he was going

ANSWER:
There is no way to calculate this in any exact way. You need to know how long the collision lasted and how elastic it was. You could make some simplifying assumptions like the collision was totally inelastic (essentially that you stick together immediately after the collision) and maybe the collision lasted 0.2 seconds. I will take your weight as 78 kg so you plus Dylan weigh 100 kg. The dog hits you going a speed of 20 km/hr=5.6 m/s. Conserving momentum, 5.6x22=100V where V is the speed of the two of you right after the collision, so V=1.23 m/s. Now, the rate of change of your momentum was equal to the force on you in Newtons, so (1.23x78-0)/0.2=480 N=108 lb. You see that the time of collision is crucial, for example if you used 0.1 seconds the average force during the collision would have been 216 lb. For 0.01 seconds, the force would be over a ton.

QUESTION:
How much energy (heat/cool) is lost through a 6� X3� piece of privacy glass 2� thick? I�m considering this for home construction.

ANSWER:
It seems to me that what you really need is not energy but the rate of energy loss. The equation for thermal conduction is R=kAΔT/d where R is the rate of energy loss (typically Watts in SI units), k is the thermal conductivity of the material, ΔT is the temperature difference, A is the area, and d is the thickness. For glass, k is about 1 W/(K⋅m). For your case, A=18 ft²=1.67 m²and d=2 in=0.0508 m. Therefore R=32.9ΔT. For example, if ΔT=(20-0) ⁰C=20 ⁰C=20 K (about (68-32) ⁰F), R=656 W=0.622 BTU/hr.

QUESTION:
I understand that the universe has no absolute frame of reference and that the velocity of any object must relative to some other object. I also understand the following thought experiment to be valid: If superman were to fly to a distant star (say, 5 light years away) at very close to the speed of light, and then fly back to Earth also at close to the speed of light, then when he arrives back on Earth, all of us would all be much older (around 10 years in this case) while superman would be still be almost the same age as when he left, all due to time and space dilation. So here is my question: If there is no absolute frame of reference, then why does superman get to be the one who didn't age?

ANSWER:
Go to the faq page and look for twin paradox.

FOLLOWUP QUESTION:
Well I don't see the answer to my question under "twin paradox".

ANSWER:
Your question is precisely a simple variation of the twin paradox. Below I have worked out the answer to your specific question (using 99% of light speed for "close to the speed of light"). This follows exactly the method of my earlier answer on the twin paradox but using your numbers, although I have not redrawn the graph for your numbers because that would be impossibly tedious.

My favorite way to understand the twin paradox is to suppose that Superman and someone on earth, each using his own clock, sends 100 light pulses per year to the other. Each receives all the pulses from the other, but Superman sends fewer since, because of length contraction, he has less far to travel in his frame than the earthbound person observes; Superman travels to a star and back which is 5 light years away with a speed of 99% the speed of light. Since Superman, because of length contraction, sees a distance to the star of only 5√(1-0.99²)=0.71 light years, he sends out 71 light pulses on the way out and 71 on the way back, and all are received by the earthbound observer, so both agree that he has aged 1.42 years. The earthbound someone sends out 990 pulses and Superman receives them all, so both agree that he has aged 9.9 years.

I have done this in the frame in which both the earth and the star are at rest. If you want, you can do it in the frame where Superman is at rest, but the result would be the same because the distance between the earth and the star would still be contracted to 0.71 light years for Superman.

QUESTION:
I was playing a game called war thunder and it said that boats took longer to stop and start because they have more inertia than things on land. Is this correct? My take on a boat taking longer to start and stop is because of the lack of friction to slow the boat down and the lack of traction to propel the boat. Am I correct or is the game correct?

ANSWER:
When talking about acceleration, it is pointless to just talk about inertia (mass) or about the force applied. After all, Newton's second law is a=F/m. If you compare two objects with the same mass, the one with the largest force applied wins. Similarly, if you compare two objects encountering the same force, the one with the smallest inertia applied wins. Maybe your game is designed such that boats are always much heavier than land vehicles.

QUESTION:
I have a question about Quantum reconstruction. But first, I'm a junior in high school and don't possess enough knowledge of the Quantum Theory or Quantum mechanics to base my theory on, if you can even call it that. Theoretically speaking, what if we had a computer that could store data on a microscopically low scale, like down to an atom. What if you found a way of coping that same data to another position using the stored atoms as reference. It might sound like an unintelligent theory, but it's all I got. Am I saying that this could be a beginning into teleportation... NO. I believe that it's purely fiction. An example; Say we have a meatal cube. We take the cube and use some sort of device to break it down to its atoms, store that information and then transfer the atoms through an electromagnetic cable. We can use the cable to manipulate the electrons, protons, and neutrons in the said atom. Then use a device to reconstruct the atoms into the form that was deconstructed. Like I said this is not based on any test that I know about, just got curious.

ANSWER:
"Beam me up, Scotty!" So, is teleportation possible? There is a very entertaining book, The Physics of Star Trek by Lawrence Krauss, which has a lengthy discussion of teleportation.

QUESTION:
If I am traveling at 99% the speed of light, and I have a special flashlight with a very high powered light on it, and I flashed a message in Morse code to an anticipating target, will the Morse code message arrive there faster than if I sent it out while stationary? I understand that the speed of light is constant, so the light will leave your flashlight at its usual speed relative to you. But since you're moving at 99% the speed of light, will this reduce the time it takes the light to reach its target?

ANSWER:
Any given pulse you send will not get there any faster than it would if you were standing still at the position from which you sent that pulse, so you are definitely not going to get your message there faster. However, the pulse you just sent was from a position closer to your target than the previous pulse, so if they were sent 1 s apart by you, the target would receive them much closer together. So, if your whole message were sent in 1 minute, the target would receive it in a shorter time, but not earlier. You might like to read the answer to a question similar to yours but with considerably more detail

QUESTION:
Does CERN effect earths gravity. i am talking about when the protons are at full speed but before they crash. I am not talking conspiracy . but the mass of the protons and the magnetic field used to control them. would it be possible to get the American gravity sensing satellites to record the earths gravitational field when its at full speed. I am only a cleaner who likes to learn and this question has been bugging me.

ANSWER:
Any effect from the increased mass of the high-speed protons would be immeasurably small. See an earlier answer.

QUESTION:
It is commonly taught that the B (magnetic) field and the E (electric) field, propagate together in a way that shows them as equivalent in magnitude. If a B field is greater when I (current) is greater, and E is greater when the charge differential is greater, then I come to the logical conclusion that the B field should be 0 when the E field is max and vise versa. Is my logic flawed here? I ask because I am trying to study and understand what is actually happening when an EM wave reaches an electron. The entire interaction between the two is usually summed up as �photon hits electron� but this situation implies that the photon strikes the electron in some perfect manner. I know that EM waves have amplitudes of significant size compared to the size of an electron and the entire interaction baffles me.

ANSWER:
There is no way that any picture can accurately represent the relative values of the fields since the two have different units which measure them. If you take the ratio of the two fields in any electromagnetic wave at any time and any point in space, you find that E/B=c where c is the speed of light. Since this is a constant, your conclusion that the E and B fields are not in phase must be wrong because two waves out of phase have different ratios at different places. Where you went wrong is thinking of B as just being created by some current; in fact, another origin for a B field is a time-varying electric field. Similarly, a time-varying magnetic field causes an electric field. In some sense, the E and B fields are keeping themselves going, sort of a perpetual feed-back loop.

The second part of your question involves the photoelectric effect. What happens when you send electromagnetic wave into matter cannot be described by thinking about waves. Understanding that it could only by acknowledging that sometimes these waves fo not behave like waves at all, but must be thought of as particles called photons. Do not waste your time trying to understand the photoelectric effect in terms of electromagnetic waves.

QUESTION:
Me and my friends were arguing about a planes carry weight and someone asked what would happen if you had 15 pounds of feathers and circulated them through the air of the cargo hold would it still weigh down on the plane?

ANSWER:
Look at one feather. The feather will be moving with a constant velocity because it has a very low terminal velocity (that is, if you drop a feather it will at first accelerate down but rapidly stop accelerating and fall with a constant speed called the terminal velocity). An object with a constant velocity will have, by Newton's first law, a net force of zero. What are the forces on the feather? There is its own weight W and an upward force D due to the air drag and they must be equal and opposite because the net force, as noted, must be zero. Now, Newton's third law says that if the air exerts a force up on the feather, the feather must exert an equal and opposite force down on the air. Therefore the feather exerts a force of magnitude D down on the air, but since D has the same magnitude as W, the feather exerts a force equal to its own weight on the air. Finally, since the air is also in equilibrium, the force it exerts down on the floor of the airplane will be the weight of the air plus the weight of the feather. In other words, the 15 lb of feathers floating around is still felt by the plane just as if they were all sitting on the floor. For a really exhaustive examination of this kind of problem see an earlier answer. For another question similar to yours, see another earlier answer.

QUESTION:
What would happen if a solid sphere of electrons ( a sphere of electrons so densely packed, that it's not only large enough to be visible, but also to weigh a considerable amount) was suddenly brought into existence?

ANSWER:
This question is so vague that I should have just discarded it. What does "large enough to be visible" mean? What is "a considerable amount" of weight? What I can tell you is that whatever happens will be, since electrons repel each other, that this sphere will probably blow apart more quickly than "suddenly". Also, "suddenly brought into existence" could be thought to violate site ground rule forbidding "questions based on unphysical assumptions". Anyhow, I can talk a bit about uniformly charged spheres and demonstrate how unphysical this question really is. The total energy it takes to assemble a uniformly charged sphere of radius R to a total charge Q is E=3kQ²/(5R) where k is Coulomb's conatant and is about 9x10⁹ N m²/C². Now, the mass of an electron is about 9x10^-31 kg and its charge is about -1.6x10^-19 C. So I will call 1 gram=10^-3 kg "considerable" and will therefore need about 10²⁷ electrons; their total charge will be Q=-1.6x10⁸ C and Q²=2.6x10¹⁶ C². Suppose we jam them into a sphere of radius 1mm=10^-3 m, plenty big to "be visible"; then the energy to assemble these electrons would be E=1.4x10²⁸ J. If "suddenly" means 1 s, this is a power requirement of 1.4x10²⁸ W. The total power consumption of the earth is 44x10¹², miniscule compared to that the create this electron ball. Furthermore, there is no force to bind this ball and it will blow itself apart with the energy you put into it; you would not want to be around it!

ADDED THOUGHT:
I am aware that the energy to assemble a sphere full of discrete charges is different from the energy to assemble a uniformly charged sphere. Doing the problem this way would essentially lead the same conclusions.

QUESTION:
I have a thought experiment that I've been puzzling about lately. Say, that we make a narrow hole from the surface and vertically down to the centre of the earth, and fill it with air. If we put a human in this centre spot, he/she will experience weightlessness as the net graviation will be zero, i.e. equal in all directions. But what will the air pressure be in the centre point given it is 1 atmosphere at the surface. We may assume a homogeneously dense earth with equal mass per m^2 all the way through.

ANSWER:
First of all, you should acknowledge that the density of the earth is not uniform, far from it; for more details about this, see an earlier answer. So what we are doing is an academic exercise not really very close to reality. Still, it is an interesting exercise. Above the earth the prediction of the pressure as a function of altitude is fairly standard. I urge you to look over that derivation since I will follow the same procedure underground. The result there is P=P_Aexp(-mgh/(kT)) where h is the altitude, m is the mass of a molecule of air, g is the acceleration due to gravity, k is Boltzmann's constant, P_A is atmospheric pressure, and T is the absolute temperature. This derivation assumes that g is constant over the range of h that contains most of our atmosphere; that is not a bad approximation since h is much smaller than the radius of the earth. It is important to note that there are really two unknowns here, P and T. This model is usually used assuming that T is constant; since T actually decreases with h, if you measure P and then solve for T, you will find that the measured pressure is larger than the predicted pressure.

So, to predict the pressure in your hole I will follow the derivation given in the hyperphysics site and start with ΔP=-ρg(r)Δr where ρ is the density of the air, r is the distance from the center of the earth, and g(r) is the acceleration due to gravity at r. For a uniform sphere it is easy to show that g(r)=gr/R where R is the radius of the earth and g=9.8 m/s². Now use the ideal gas formula to write ρ=mP/(kT) so that the differential equation for P is dP/dr=[mPg/(ktR)]r; integrating, I find that P=P_Aexp[-(C(1-(r²/R²))] where C=mgR/(2kT). Choosing T=300 K, I find C≈350 so P/P_A≈exp[-350(1-(r²/R²))]. I have plotted, below, the predicted pressure both inside (red) and outside (blue for constant g, black for g(r)=g(R/r)²) the earth; note that the graph covering 0<r/R<2 is logarithmic because the pressure is almost zero very rapidly. Both inside and outside the pressure falls very quickly away from the surface, faster for inside because of the quadratic rather than linear term in the exponential function. The other graph shows the curves 1% above and below the earth's surface; inside the pressure is nearly zero, outside 80% below atmospheric pressure.

The center of the earth has a temperature of about 6000 K which would result in a much larger pressure of about 10^-8 P_A, but still very small. Just as the case of the prediction of P above the earth's surface has difficulties related to the temperature profile, so does the prediction of P below the surface.

FOLLOWUP QUESTION:
I thought a uniform-earth assumption would make the calculations easier, but still give a reasonable idea. I guess the earth density and temperature will follow a similar pattern as the air pressure; the physics are similar?

ANSWER:
You are right, the uniform-earth assumption does make it easy to calculate and will give you a very good estimate of what happens in the first few kilometers. The second graph is the one which gives you the most information; what you can say about the pressure deeper than a few tens of kilometers is that the pressure is trivially small. The same is true for the pressure profile above surface: after a few tens of kilometers the atmospheric pressure is, for most purposes, zero. For both cases, knowing the temperature profile would improve the accuracy of the model but there is no likelihood of knowing that more than about 12 km below the surface where the temperature is alread at about 500 K. It is harder to know what the temperature profile actually is because we have not way to get instruments down there. The model we have been playing with would be correct if we could perfectly insulate the walls of the tube. The 6000 K comes from the fact that we know that the core is molten iron.

ADDITION:
There was an error in my original post in the calculation for r>R which has been corrected in the graphs. I also did the analogous calculations for r>R where the variation of g was included.

QUESTION:
What determines an object�s speed when falling with an open parachute?

ANSWER:
What determines any objects motion is the forces acting on it. A freely falling object has two forces on it, its own weight down mg and a drag force up D. D is proportional to the square of the speed v². So, if you drop something it starts with very little drag and the weight is the main force which causes the object to accelerate downward; eventually v gets big enough so that the drag is equal to and opposite the weight so the object then falls with a constant speed, called the terminal velocity, for the rest of the fall. There is a well-known expression for the drag force which works well in most every-day situations, D=½ρv²C_DA. You can learn more about it at the link given above, but for our purposes here, for an object which presents an area A to the onrushing air at sea level, a fairly good approximation is D=¼Av². (Warning: this equation only works if you work in SI units.) So, we can estimate the terminal velocity, v_T=√(2mg/A). Let's estimate the mass of a parachutist plus unopened parachute as 100 kg and her area about 1 m²; then you could estimate v_T for a sky diver plus unopened parachute as about 45 m/s=100 mph. Now, when the parachute is opened maybe the area increases to 20 m²; now v_Tis about 10 m/s=22 mph. These are just rough approximations to emphasize that the area is key. Actual numbers will depend a lot on the design of the parachute.

QUESTION:
Ok this is gonna sound weird but say your falling from an airplane and you have your phone in your hand. If you throw your phone upwards right before impact would it still break or at least suffer less damage? I can�t tell if this is like the jumping in a collapsing building situation or if it would actually help.

ANSWER:
The terminal velocity of a human body is about 120 mph. Do you think you could throw your phone with that speed? Certainly not—a big-league pitcher can throw a baseball only about 100 mph and even he could not do that with his feet not on the ground. If you could give the phone that speed straight up it would have zero velocity relative to the ground. I doubt that you could possibly give the phone more than about 20 mph vertical velocity meaning it would hit the ground with speed around 100 mph.

QUESTION:
We have a large puddle outside our back door. Yesterday, the sun was casting a reflection from the puddle onto the ceiling of our kitchen through our sliding doors. It was pretty so we were watching it for awhile. What we noticed was that although the puddle was very still to our eyes, the reflection on the ceiling showed constant movement. We saw ripples and swirls and lines of movement, but the puddle appeared completely still. Little bits of dust/pollen/leaves which were floating in the puddle weren't moving at all. If it matters, the water was very clear, it was about an inch at it's deepest, and the puddle was sitting on a very white roofing liner which was just installed last week. (Our back door leads to a roof.) Is the reflection showing movement under the surface that we cannot otherwise see?

ANSWER:
The rays from the sun come in parallel. If the puddle were perfectly flat, like a mirror, the reflected rays would also be parallel. But if there are very small ripples in the water, too small to notice, the reflected rays would not be parallel. Even if the angle between two rays was very small, their separation would be greatly enhanced over the large distance they have to travel to the ceiling. This magnifies any "nonflatness" of the puddle.

QUESTION:
I am writing scifi fanfic and I need help in physics. In the fic planets are protected by a sort of a future version of coastal artillery when waships are not available. So the problem I have is the needed quantity of the guns. Assuming a gun is placed at the height of a height of 25 meters on a turret tower, and the gun can aim straight up or straight ahead effectively having a 180 degree firing angle, how many guns would be required on minimum to cover the entore space around the Earth with no blind spots? Would a thousand suffice?

ANSWER:
This is not physics. I find description of the way a gun can point to be not clear; any direction above horizontal seems to be what you can aim at or see. The 25 m height seems to be superfluous unless the point is to extend the horizon; but then horizon to horizon would be greater than 180⁰ and the change in horizon would be trivially small. I assume that you are looking for something coming in from far away, that is it is ok to have blind spots near the earth as long as anything far from the earth can be

seen. Imagine one gun at each pole; they see anything above a plane tangent to the earth at the pole. Imagine one gun at the equator, another at the equator on the opposite side of the planet; they can see anything above their tangent planes. Finally station guns 900 longitude from the other two on the equator. You have made an imaginary cube around the planet using only six gunners into which no one can enter undetected. You are screwed, though, if somebody gets inside. It just occurred to me that you could put the planet in a tetrahedron with only four gunners.

QUESTION:
If you could make a radio with a frequency inside the visible spectrum, would you be able to see it?

ANSWER:
Yes, that is what our eyes evolved to do—see light in that wavelength range. If the radio were designed to operate at a given frequency, like AM radio is, that is the color you would see. You would have trouble creating an antenna, the dimensions of which are usually the same size as the wavelength.

QUESTION:
Hello! I have a really quick question: what are the simplest forms of the formulas for each of Newton's 3 laws of motion?

ANSWER:
It breaks the heart of a physicist to think that students think that the best way to learn physics is to memorize "formulas". Newton's laws are monumental concepts and should be not thought of as formulas. I will try to specify these conceptually, not mathematically.

An object which experiences zero net force will remain at rest or moving with a constant speed in a straight line.
The acceleration (change of its velocity) of an object is proportional to the net force applied to it and inversely proportional to its mass.
If one object exerts a force on a second object, the second object exerts an equal and equal force and opposite on the first.

QUESTION:
So I work on boats and we entered a problem with barge drafts in the Mississippi river. And we have a 9 foot 6 inch draft on our barges but we generally have a 9 foot draft. The river was running at 4 mph. We stalled out and eas going 2mph backwards and the captains were blaming a 9 foot 6 in draft. Had me curious what the differnece in drag does 6 in make in water against a 4 mph current.

ANSWER:
The operative equation is the equation for drag force D moving with speed v in a fluid with density ρ and speed v_R m/s, D=½ρ(v+v_R)²C_DA; C_D is called the drag coefficient and depends only on the shape of the barge and A is the area presented to the oncoming water. I have assumed the barge is moving upstream and this force, of course, is downstream. Now, what you want to know is what is the increase in drag when you change the draft. The only thing changed by the the draft is A and A is proportional to the draft. So, if we take a ratio D_9.5/D₉=9.5/9=1.06; so the drag increases by about 6% if the two draft situations have the same speed. If they have different speeds, then you would have to calculate that, D_9.5/D₉=1.06[(v_9.5+4)/(v₉+4)]². For example if some force (tow boat) pulls the barge and the speed is 5 mph for the 9.5 ft draft and 7 mph for the 9 ft draft, D_9.5/D₉=1.06[(5+4)/(7+4)]²=0.71; there is about 29% less drag for the 9.5 ft draft because it is going slower. I can't get any more quantitative without more information about the barge and the way it is being towed.

QUESTION:
Is there a way to find initial velocity with just the time and the angle of the projectile motion?

ANSWER:
Suppose we choose our coordinate system so that the projectile starts with speed v₀ at the origin. Then we have four equations describing the subsequent motion:

v_x=v₀cosθ
v_y=v₀sinθ-gt
x=v₀tcosθ
y=v₀tsinθ-½gt²

What do you know? t, θ, and g. What do you not know? v_x, v_y, x, y, and v₀. Since you have more unknowns than equations, you cannot solve the problem. The answer to your question is no.

QUESTION:
If you drop a 10 pound cooking pot on your foot is it just 10 pounds?

ANSWER:
See the figure. When the pot hits your foot there are two forces acting on it, its own weight W and the force F which your foot exerts on it. In order for the pot to slow down and stop, there must be a net force upward. Therefore, F is greater than W. Finally, we invoke Newton's third law which says that if you exert a force on the pot, the pot exerts an equal and opposite force on your foot and so you experience, for a brief time, a force larger than 10 lb.

QUESTION:
Have you seen the movie 'Avengers:Endgame' In this film Bruce Brenner says that if you go back in time and do something there it will only affect that particular timeline But in the back to future series its not as such? What will really happen if I go back in past and do something

ANSWER:
The laws of physics, as we currently understand them, forbid backward time travel. Forward time travel is allowed.

QUESTION:
A gecko was clinging on my windshield and my friend who was driving the car decided to accelerate. 60 mph gecko was still on the windshield but at 64 mph poor little guy went airborne. I was upset of course and now wondering if the gecko would have a chance to make it ... being so light. Does mass matter? There was no wind and gecko most likely landed on the road.. yes the car is a moving object but the mass of the reptile is so light.

ANSWER:
Geckos come in quite a range of sizes. I am going to guess yours was some where in the middle, say his mass is about m=2 oz=0.055 kg and his length is L=5 in=0.13 m. The initial speed is v₀=64 mph=29 m/s. Estimating his width as 1 cm=0.01 m, his area is roughly A=0.01x0.13=1.3x10^-3 m². So you can estimate the force on him as F=-¼Av²=ma=m(dv/dt) or (dv/dt)=-v²x6x10^-3. Integrating this differential equation, 6x10^-3t=(1/v)-(1/29) or t=167(1/v-.035) seconds. Oh dear, I started this answer expecting to reassure you that the little fellow would be fine. But, suppose that we wanted him to have a speed of 2 m/s to be ok when he hit. Then the time to slow to that speed would have been about 78 s and he would have hit long before that. So, I calculated his terminal velocity, the speed he would attain if he had simply been dropped from some large height. That would be v_terminal=√(2mg/A)=28.8 m/s, just about the speed he already had when he became airborne! If there had been no air drag at all and you dropped him from about 3 m, he would have hit with a speed of 7 m/s=15 mph and it would only have taken him about 0.4 s. In this time, there is no hope of his slowing down appreciably. I am afraid that the gecko met an unhappy end.

QUESTION:
This question is driving me nuts. Please answer it if you can. Here it comes. Why doesn't the direction of the vibration of a vibrating tuning fork affect the direction of the soundwave it produces? As you probably know, sound obeys the inverse-square law, which means sound waves spread from the vibrating object like a growing sphere, radiating out in every direction with equal velocity and intensity. As I'm sure you also know, sound traveling through air is believed to be a longitudinal wave. That is, as the sound wave travels up from the tuning fork, that air vibrates up and down. As the sound wave travels left from the tuning fork, the air vibrates left and right, etc. Significantly, it makes no difference in what direction the tuning fork pushes and pulls the air around it, the direction in which the air moves depends only on its orientation to the tuning fork. Air to its left will move left and right, the air above will move up and down, etc. In other words, where sound is concerned, the direction of the force does not equal the direction of the acceleration. How can this be?

ANSWER:
You state that "As you probably know, sound obeys the inverse-square law". What I definitely know is that this is true only if the source is a point or a sphere; a tuning fork is neither. You also state that "sound traveling through air is believed to be a longitudinal wave"; that implies that we are not sure—trust me, sound is a longitudinal wave. A longitudinal wave is one where the medium vibrates parallel to the direction of propogation. The animated figure shows a representation of a vibrating tuning fork which, when vibrating, compresses the air in front of it and that becomes a high pressure (higher density) part of the wave and propogates forward. The figure represents the waves as plane waves which is not, strictly speaking, accurate, because of the finite extent of the vibrating tines; however, it is certainly more accurate than representing them as spherical waves which would be the case if the inverse square law applied. The second figure shows roughly what the distribution of intensity of sound is in a horizontal plane (viewed from directly above the tuning fork) sounds like if your ear is very close. Contrary to what you state, there are places where the sound is quiet (off in diagonal directions) and where it is loud (left and right where you would expect it, as well as forward and backward where you might not expect it). In fact, if you move the tuning fork farther from your ear you find that the forward/backward loudness dies off more quickly than the left/right intensity. The reason for this is too technical to go into here, but not unexpected intuitively. The next two figures show the measured intensities in the near and far observation positions; the distance from the center of the circle of the curves at any angle represents the loudness of the sound.

I really have no idea what you are trying to get at when you talk about the motion of the air. It is important for you to understand that there is no net motion of the air itself is any direction. Focus on any air molecule and it simply moves back and forth parallel to the direction the wave is moving there. The wave moves in some direction, the air does not.

QUESTION:
If I have two barbells that weigh exactly the same but one is 7' in length and the other is 9' in length would there be a difference in how they felt to the lifter?

ANSWER:
If the lifter has his hands the same distance apart for both barbells, I can see no reason why you would notice any difference. The longer barbell would have a larger moment of inertia because the major part of the weight is farther from the center than for the shorter barbell. What that means is that the shorter would be slightly easier to get rotating in a vertical plane; so if the lifter applied a net torque to the barbell, for example by lifting with a slightly larger force with one arm than the other, the longer barbell would be more stable against rotating. I doubt that this would be noticeable to the lifter. That is the only difference I can see. (I have assumed that negligible bending occurs.)

QUESTION:
Considerable water mass is stored in Earths polar ice-caps. If the ice-caps melted, and the stored water released, would this raise the sea-level more in equatorial regions? If so, would this in turn slow the earths rotation and thus reduce the total days in say a century?

ANSWER:
Indeed when glaciers and ice sheets on land melt the redistribution of all this mass will result in an increased moment of inertia which would slow down the the earth's rotation rate. The most reliable and most quoted estimate (by a Professor Steven Dutch of UW, Green Bay) is that the length of the day would increase by about 2/3 of a second. I also found, interestingly, that removing the weight from polar regions would result in the crust spring back upwards by some amount, analogous to removing a bowling ball from a trampoline; similarly, the increased weight at lower latitudes would tend to push the crust down there and toward the poles. At least one site I visited stated that the resulting decrease in moment of inertia would approximately nullify the increase in day length due to the melting.

QUESTION:
I've got a question that bothers me, please pardon my level of physics ignorance. So: Let's assume that lightning strike has 100MV. If it strikes me directly, 100MV will go through my body into the ground. Let's say that I hold conductive rod with R₁ = 0.01 Ohm that's let's say twice as tall as me (doesn't really matter in my question, it has to

strike the rod first) and lightning strikes the rod. Let's assume that human body in this situation has resistance of R₂ = 1,000,000 Ohm. If the lightning strikes the rod it will go through parallel resistor (me + rod) hence the total R will be 0.00999 Ohm. If we calculate total current it will be 10,000,000,100 A. 10,000,000,000 A will go through the rod, 100 A will go through my body. I know I simplified it a lot, I'm not even a decent at that, but the question is - there's a saying that electricity takes path of the lowest resistance - but why would I prefer to hold super conductive rod than to hold anything less conductive, if the current through my body will remain unchanged? I total = I₁ through R₁ + I₂ through R₂, right? So that would mean lower R₁ means lower I₂, but it doesn't, because with lower R₁, R_total is lower hence I is higher, so I₂ remains unchanged. I believe it has to do something with power, I'm just not sure.

ANSWER:
As best as I can tell, you are asking what happens if the rod has zero resistance. The resistance of you and the rod is R_net=R₁R₂/(R₁+R₂). If R₁=0 then R_net=0, all the current will flow through R₁. No rod will have zero resistance, but as R₁⇒0, I₂⇒0.

QUESTION:
Imagine that you are trapped in a room in a space station that rotates so that you are subject to a centrifugal force exactly to one G. You don't know if you are on planet earth or in deep space. Is it possible to tell if you are subject to gravity or centrifugal force? If yes, how do you do that, what instruments do you need to have?

ANSWER:
The answer is no; the equivalence principle, an important cornerstone of the theory of general relativity, states that there is no experiment you can do which allows you to determine whether you in empty space accelerating with an acceleration a or you are in a gravitational field where the gravitational field causes an acceleration a at the point where you are performing the measurement. Usually, like your rotating cylindrical room, we think of it as having an angular velocity ω and a radius R and if we adjust these so that g=v²/R=(Rω)²/R=Rω², then it will be just like being on the earth. But that is wrong because the field close to the surface of the earth is very nearly uniform—g is very nearly the same at both your head and feet. That is not the case in the rotating room because the acceleration depends on R and therefore your head "weighs more" when you are standing up than when you are lying down. If you could simulate a gravitational field which increased linearly with R as you moved away from some straight line (analogous to the rotation axis of your room), there would be no way you could determine whether you were in that field or in your rotating room. So for the rotating space station to be a good simulator of gravity on earth, it would need a very large radius. If you want to read much longer discussions of rotating gravity simulators, there are several earlier answers to questions similar to yours.

QUESTION:
Imagine that you are trapped in a room in a space station that rotates so that you are subject to a centrifugal force exactly to one G. You don�t know if you are on planet earth or in deep space. Is it possible to tell if you are subject to gravity or centrifugal force? If yes, how do you do that, what instruments do you need to have?

QUESTION:
I was wondering how do we observer galaxies complete, If they are 100 light years in diameter then the light would be scattered in a trialing effect not a complete shape, yet we see them as entire complete shapes, as through images we see on Hubble. To me that would be impossible as relativity would suggest. Would it not?

INTERPRETATION
I believe the questioner must not be a native English speaker since I had difficulty understanding what he was asking. I believe that the question is essentially the following. If we observe a galaxy, the plane of which is not perpendicular to our line of sight, how can the image be correct if the light we see from the far side came many years before light from the near side?

ANSWER:
As stated on the site, I normally do not answer astronomy questions, but this one intrigued me. The questioner has made an error in the size of a galaxy. Since spiral galaxies are as large as about 400,000 light years in diameter, the Milky Way being 100,000 light years across, how can an image of a galaxy be even approximately correct unless it is normal to our line of sight? The far side will be what the galaxy looked like tens of thousands of years earlier than the image of the near side, greatly distorting what it must "really look like". The answer, I believe, is simply that the galaxy does not really change much over times on the order of tens of thousands of years. For example, although our galaxy is 100,000 light years across, its rotational period is on the order of 250,000,000 years; so the angle which galaxy has rotated in the time it takes light to traverse its diameter is 360⁰(100,000/250,000,000)=0.14⁰.

QUESTION:
How would I put the following statement in an equation to solve? if an event occurs every 40 seconds how long would it take to reach 300,000,000 events? I am a technician at a casino and working out an issue.

ANSWER:
The the time T required is T=(300,000,000 events)x(40 seconds/event)=1.2x10¹⁰ seconds. That is the answer, but maybe we should put it into different units of time, years, let's say: T=(1.2x10¹⁰ seconds)x(1 hour/3600 seconds)x(1 day/24 hours)x(1 year/365.25 days)=380 years.

QUESTION:
Please understand I have very little understanding in physics. Because of best one third law if I jump and land in the ground, I would apply force to the ground and it would go back up at me, so if I go to the ground jumping with more force than my bodyweight gravity would hit back with more than my bodyweight. I would even if for less than a second weigh less right.

ANSWER:
First of all I want to get something well understood: weight is the force which the earth exerts down on you and no matter what else might be happening to you, your weight is always the same. Therefore your statement that you might "weigh less" is wrong. So, I hope that I can help you understand the physics of jumping. Look at the picture. One woman is in the air and there is one and only one force on her, her own weight W. Because of Newton's second law (N2), there is an acceleration in the same direction as the net force which slows her down if she is going up and speeds her up if she is going down. The other woman has her feet in contact with the box and is either just landing of just about to go up. There are now two forces on her, her weight W and the force which the box exerts up on her which I will denote as N. If she is just landing, she is slowing down and so the acceleration needs to be upward because she is moving downward; in this case the net force must point up, so N>W as shown. If she is just about to go up, she is speeding up and so her acceleration also needs to be upward because she is moving upward; in this case the net force must also point up, so N>W as shown. Newton's third law (N3) now states that, if the box exerts an upward force of magnitude N on you, you exert a downward force of magnitude N on the box.

Here is what you said correctly: you said that the force with which you would have to push down on the box would be bigger than your weight. Here is what you said incorrectly: you said that "gravity would hit back with more than my bodyweight". You were apparently thinking that W and N would have to be opposite (correct) and equal (incorrect) because of Newton's third law. But W and N are not a N3 pair because they are both on you. N3 pairs are always on two different objects which exert forces on each other: if woman exerts a force on box, box exerts an equal and opposite force on woman.

If the woman is just standing on the box, N and W are equal and opposite but N3 has nothing to do with it; rather, because she is in equilibrium (not accelerating) all forces on her must add to zero, Newton's first law (N1).

QUESTION:
Would there be any change in the theories of relativity or other related theories if the speed of light was slower, like say, 30000 m/s?

ANSWER:
There would be no change in the theory of relativity, but it would make a huge difference for most of the technology we are used to which often requires very exact measurements of and comparison of times or distances. The number you suggest is 10,000 time smaller than the actual speed of light, so it would take about 2 months for light from the sun to get to us. Also, since c plays an important role in quantum mechanics, atomic physics would be different, perhaps making the formation of atoms impossible given the other physical constants. And, since E=mc² would still be true, stars, if any existed at all, would burn out approximately 10⁸ times sooner; for example, the sun will have a total life time of about 10 billion years but would only have lasted about 100 years with your value of c. I believe that for atoms, nuclei, elementary particle physics, etc. to work at all, you would have to adjust all the other physical constants to also be different.

After writing the above I googled around a bit and discovered that it is a very complex situation. A thread on Physics Forums is enlightening. This should give you some food for thought!

QUESTION:
I was wondering if you could help me out here. what unit is used to measure radioactivity around you(like if you're in a radioactive neighborhood/city, etc), I've heard about Roentgens and Curies but those aren't used anymore, also there are Becquerels, Grays, Sieverts and RADS but i'm kinda confused which unit is actually used to measure radioactive levels in an area, both in SI Unit and Non-SI unit.

ANSWER:
I agree it is confusing, but there are many kinds of radiation of many different energies and effects. The Curies and Becquerels basically measure the rate at which the sources are decaying, therefore counting everything. For most purposes these are not useful in determining how dangerous the radiation is. Maybe the source is only emitting low-energy alpha particles which would have a very short range in air and not penetrate past the surface of the skin. Maybe 1000 Curies would be the level but nothing to worry about. There is a very good summary from the CDC of which units are useful for which situations.

QUESTION:
When the train makes a turn, its outer wheel and inner wheels covering a different distance at the same time. How come the train does not come off the tracks?

ANSWER:
Basically, it is because the wheel shape is not like a cylinder but a section of a cone. As the train goes around the curve, centrifugal force moves the train outward making the effective radius of the outer wheel larger than that of the inner wheel. There are lots of web sites which explain this. Here are three of them:

https://www.quora.com/How-do-trains-turn

https://www.quora.com/When-a-train-makes-a-turn-isnt-its-outer-wheel-covering-more-distance-than-the-inner-one-How-come-the-train-doesnt-come-off-the-tracks

QUESTION:
How come glass store front windows eventuality sag but bricks can be laid into a wall thousands of feet high if matter is mostly empty space ? strong nuclear force ?

ANSWER:
Glass does not flow if that is what you are referring to; this is a myth and you can read about it at this link. It is incorrect to say that matter is mostly empty space. Although true that most of the mass occupies a very small volume, the intervening space is occupied by a "cloud" of negative charge which the electrons occupy. Bricks are stronger than glass because they are "proper" solids, i.e. the atoms arrange in regular crystaline patterns which tend to be strong, whereas in glass they do not. Macroscopic properties of matter have nothing to do with the nuclear force, they are determined by atomic forces among atoms in the material.

QUESTION:
Our weight in more in day or night .I saw Vsauce video weight of shadow in which they said weight is more in day than weight in night but it is opposite in Google . Please explain to me?

ANSWER:
I generally define weight to be the gravitational force which the earth exerts on me, but let's define it for this question to be the net gravitational force which the earth and the sun exert on me. In that case, during the day these forces are opposite and during night they are in the same direction; therefore you would have less weight during the day. However, the moon also exerts a force on you so you really should define weight to be the combined gravitational forces of earth, moon, and sun. Then it would also depend on the phase of the moon. A little googling and I found that the force on a 70 kg person would be 686 N due to the earth, 2.4x10^-3 N due to the moon, and 5.9x10^-3 due to the sun. So I will estimate the effect on the total weight for four situations, new moon (moon between the earth and the sun) midday and midnight, and full moon (earth between the sun and the moon) midday and midnight.

New moon day: 686-2.4x10^-3-5.9x10^-3
New moon night: 686+2.4x10^-3+5.9x10^-3
Full moon day: 686+2.4x10^-3-5.9x10^-3
Full moon night: 686-2.4x10^-3+5.9x10^-3

So you are "lightest" with a new moon during the day and "heaviest" with a new moon at night.

QUESTION:
I know this is more along the lines of nuclear chemistry and engineering, but I figured it was worth a shot. In Marvel comics, Iron Man's suit is powered by an "Arc Reactor," which is about the size of a fist and glows blue. Assuming that the technology could be perfected and miniaturized, is it plausible that such a reactor could be made by a self-sustaining hydrogen fusion reactor that gets its fuel from the moisture in the air by electrolysis? In more direct terms, a fuel cell harvests hydrogen from moisture in the air, feeds it to a fusion reactor, which, once kick-started with initial energy, can feed its own power needs plus extra, which is then sent to the suit.

ANSWER:
There is an earlier answer also addressing the possibility of a very small fusion reactor. Regarding getting the fuel from moisture from the air, any practical fusion reactor cannot operate on single-proton hydrogen, ¹H, but requires deuterium, ²H and/or tritium,³H. There is very little deuterium in plain water and virtually no tritium.

QUESTION:
How much time (s) would it take a 40 kg object to accelerate from 0 m/s to 40 m/s if 100 N of force is applied (assuming 0 friction). I can't figure out how to solve the formula F = ma for time (s). [I just submitted a question to you about acceleration and time. I am an 8th grade science teacher and trying to remember how to solve for time. Please don't ignore the question. I would like to teach it to my students tomorrow but need to remember how to do it myself first haha!!!]

ANSWER:
OK, Newton's second law gives you the acceleration, a=F/m=100/40=2.5 m/s². There are two equations you need to work with for uniform acceleration if you start at x=0 from rest, x=½at² and v=at. In your problem you know that v=40 m/s so you can solve for t, t=40/2.5=16 s. You can also find the distance traveled, x=2.5x16²/2=320 m.

QUESTION:
Assuming:

That an open vessel is filled with water and fitted with a submersed pump;
That the water in the vessel is under normal atmospheric pressure, and at all times, remains so;
That the vessel is immersed in, and is level with, a surrounding pool of water;
That the surrounding pool of water is infinite in size;
That the surrounding pool of water has a pressure of 2 atmospheres;
That friction losses are ignored;
That the pump operates at 100% efficiency.

I need to know how many joules of energy would be required to pump 1 litre of water, from the vessel, horizontally, into the surrounding pool of water.

ANSWER:
I shall assume that the volume of the low-pressure fluid is very large compared to 1 litre so that the level of the fluid does not change appreciably when the fluid is pumped out. I imagine moving water out through a small hole of area a between the two fluid containers which is at a depth d below the surface of each; the volume of moved water is La, imagined as the volume shown in the diagram. The pressure at depth d is P_A+ρgd inside the hole and 2P_A+ρgd outside, so the pressure difference is ΔP=P_A and the force necessary to push water through the hole is aΔP=aP_A. The work to expel the fluid is then W=aLP_A=VP_A. Taking P_A=10⁵ N/m² and V=1 l=10^-3 m³, W=100 J. It becomes a more difficult problem if the volume of the low-pressure water is not very large; in that case, as the level fell the pressure difference would become larger.

QUESTION:
Suppose you are driving at speed v and find yourself heading straight for a brick wall that intersects the line of your path at 90�. Assuming that the coefficients of friction for stopping and for turning are the same, are your chances of avoiding a crash better if you continue straight ahead while braking or if you simply turn along a circular path at a constant speed?

ANSWER:
Suppose that the coefficient of static friction is μ and the car has a mass m, weight mg, and the frictional force is at its maximum, f=μN=μmg for both crash-avoidance options.

The straight-ahead option starts with kinetic energy K=½mv² and has work done on it W=-fd=-μmgd before it stops with no energy, where d is the distance to stop; therefore, ½mv²-μmgd=0 or d=v²/(2μg).
The turning option moves in a circle of radius R and the centripetal force is μmg=mv²/R. Therefore, R=v²/(μg).

Since d=R/2, straight ahead is a much better option!

QUESTION:
I am a caregiver. I need to figure out how many pounds a force it takes to push someone up a ramp. I can't find anything online that will give me a simple answer. May I please give you the parameters and could you please figure it out for me. The ramp length is 6 ft long ,the height of the ramp is one foot tall, the objects I am pushing up the ramp is a 40 lb wheelchair and a 200 lb woman. What I need to know is how many pounds a force am I pushing. Also when I bring her down the ramp I am walking backwards and pulling her down the ramp using the same parameters can you figure out how much weight is on me taking her down the ramp. I can't find anyone else to ask. Can you help me. My livelihood as a caregiver depends on it at this point in my life.

ANSWER:
You are probably not interested in the mathematical details, just the final answer, but I always include the physics in my answers. If you want, you can just skip to the end and see what the final answer is.

Assuming you go up with approximately constant speed, this is a Newton's first law problem�all the forces on the wheelchair plus patient must add up to zero. There are four forces on the wheelchair plus patient�the force F which you exert up the ramp, the weight W of the patient straight down, the normal force N the ramp exerts perpendicular to the ramp, and the frictional force f down the ramp. The angle of the ramp is θ=tan-1(1/6)=9.46⁰, so the component of the weight down the plane is W_x=-240sin(9.46⁰)=-39.4 lb; the component of the weight normal to the plane is W_y=-240cos(9.46⁰)=-237 lb. The most problematic force is the friction; the most important thing is that f is proportional to N, that is f=μN where μ is called the coefficient of friction. What this constant is depends on things like the quality of the bearings in the wheels, the nature of the surface you are rolling on (a rug would have a bigger μ than a hard surface, for example), how inflated the tires are, etc. Wheelchairs are presumably designed to minimize friction so I will make a guess that μ≈0.1. Now, the physics:

y forces add to zero, so N=237 lb;
therefore, f=-0.1x237=-23.7 lb;
x forces add to zero so
F=23.7+39.4=63.1 lb;
if there were no friction, F=39.4 lb.

For going down the ramp, the force diagram is just the same except now f points up the ramp rather than down�friction now works with you, not against you. N and W are the same but now F=39.4-23.7=15.7 lb. I noted that you said you are "pulling her down the ramp" but with my choice of coefficient of friction I find you are actually "holding her back" from accelerating down the ramp. If you need to pull her down the ramp, then my estimate of μ is too small which would mean you have to push more than F=2x39.4=78.8 lb when going up.

FOLLOWUP QUESTION:
I truly don't mean to look completely ignorant but.....from your answer,are you saying that I am applying 63.1 lbs or 78.8 lbs of force onto my spine when I am pushing my client in her wheelchair up her ramp. Also are you saying it's only 15 lbs of force when I am guiding her and her wheelchair down the ramp. Also how do I know this part of your answer. "find the force (call it G) necessary to move her with constant speed on level ground (with the same kind of surface as the ramp) " I'm guess I'm going very slow and steady as to not allow her to fall off the side or myself to lose my balance as her weight pushes on my body. You can probably imagine that my legs are slanted far back to keep my balance,if that helps at all.

ANSWER:
Let me clarify what I mean by all those forces F (e.g. 63.1, 15.7, 78.8 in various situations). F is the force which you must apply on the wheel chair to move it up or down the incline with a constant speed. Newton's third law says that if you exert a force on the wheel chair, the wheel chair exerts an equal and opposite force on you. This is a force on your hands. There is no way I can deduce what this implies about your spine, your shoulders, your legs, etc.; the body is a very complicated machine and how you push might be different from how somebody else might push, so there is no simple answer. [For example, consider the simpler example of lifting of a box. If you ask me for the force you need to lift it, I could tell you that you must exert an upward force equal to the weight of the box. You have probably heard "lift with your legs, not with your back". So for two diferent people doing the exact same thing but in different ways, one might end up with an injured back and the other not.] Regarding the 78.8 lb force I derived, you can probably simply ignore it because it is only for the special case of very high friction in the wheelchair which I think unlikely. You can easily test whether the friction is very high by carefully letting go of the wheelchair at rest: if it begins begins moving down the ramp, the friction is pretty low, but if it just sits there, the friction is large. High friction would mean that you have to exert a force up the ramp on the chair to move it up the ramp of at least 78.8 lb (pointing up the ramp). The condition for high friction would be μ≥39.4/237=0.17. Keep in mind that I am only guessing about the coefficient of friction μ; if you do the test above and find the chair will roll down if you do not hold it, at least you will know that μ<0.17. I have removed the part about a force G in my original answer because it is too confusing.

ADDED THOUGHT:
Look at the second picture. Is that the way you bring her down the ramp, you being below the chair? If not, I will need to redraw that picture with you above the chair on the ramp. The numerical answers will not change, though.

QUESTION:
Hello, my question is regarding radioactive decay. How exactly does an individual atom decay into byproducts? What is the actual cause of this decay? Is it the strong nuclear force becoming weaker over time and then at a certain point the nucleus can no longer be kept together? Perhaps like with the earth and the moon, with the moons orbit slowly decaying. There must be a cause and there must be an order to why atom x decays at this time even if we can�t know how.

ANSWER:
Radioactive decay is a sudden event, not one that slowly occurs like the moon's orbit. Suppose that there is an atomic nucleus which is bound together. It is stable, technically only if there is not some allowed process which would result in energy release if it occurred. There are lots of nuclei which are bound but not stable. Let me give you just one type of decay which can happen and discuss it. A proton in the nucleus can spontaneously turn into a neutron, an electron, and a neutrino. This is called beta decay; it cannot happen to a free neutron since you would have to add energy to a proton to make it happen, but it can happen inside a nucleus. The example I will use is the decay of ¹⁴C which is commonly used for carbon dating of archeological items. Now, the decay is a statistical process and since this nucleus is bound you might have to wait for a while for the decay to happen. But, somehow the nucleus knows that if it can turn one of its protons into a neutron, making ¹⁴N, 0.156 million electron volts of energy will be released; depending on the detailed structured of this nucleus, you may have to wait a long while. In this case you will have to wait 5730 years for half the ¹⁴C nuclei to beta decay to ¹⁴N. ¹⁴N is stable against beta decay which would be to ¹⁴O, which means that if you want it to happen you would have to add energy. Incidentally, there is another type of beta decay where a neutron turns into a proton, a positron (a positively charged electron), and a neutrino. So, if a nucleus has too many protons it will beta minus decay, too many neutrons it will beta positive decay. A neutron, unlike a proton, will also beta decay when outside a nucleus.

QUESTION:
If quantum theory states that shining red/blue lights on metals knocks out electrons, can these be used as a shrinking ray?

ANSWER:
Actually, visible light does not knock electrons out of a metal; you really need ultraviolet and beyond to see the photoelectric effect. But why would you think it would shrink something? Removing electrons just result in the metal becoming positively charged and it would eventually have a high enough voltage to cause a breakdown cascade in the air around it, called a spark, to replentish the lost electrons.

QUESTION:
I understand that a Rocket moves upwards, because it's an opposite reaction of the gases escaping, but why is that? Can we explain the Newton's 3rd law of motion using any other theories? Is it because electrons in the gas repel from each other, thus creating this repulsive force upward? Best i can visualize it is a rocket going up, by building a platform underneath it to repel from, but i would like to know how accurate that is!

ANSWER:
It is conservation of linear momentum. To simplify things, just consider a 100 kg astronaut floating in empty space holding a 20 kg ball. He tosses the ball and it flies away at the speed of 5 m/s. The astronaut recoils with some velocity -v in the opposite direction (the negative sign denotes velocity opposite the ball). Linear momentum is given as mass times velocity, so the ball has momentum 5x20=100 kg m/s and the astronaut has a momentum 100v. But the ball plus astronaut had zero momentum before she threw the ball and they must have zero momentum afterwards, so 100-100v=0 or v=1 m/s. That's all a rocket is: you throw stuff out and recoil in the opposite direction. It has nothing to do with platforms repelling or electrons in the ejected gasses. (Incidentally, conservation of linear momentum is easy to derive just using Newton's second and third laws. It is only true if there are no other forces acting. But the recoil idea is true of a rocket going up in gravity, it is just that a given amount of fuel ejected in space will have a larger acceleration than when leaving earth.)

QUESTION:
If I am at the lake on a boat, drop a line 50 ft straight down with 4oz weight and the boat is traveling at 4mph at what depth will the 4oz weight be. And with an 8oz. If I want the weight to be at 25 ft in depth while traveling at 4mph what weight should I be using?

ANSWER:
You cannot do this calculation, even approximately, without knowing the size and shape of the weight.

FOLLOWUP QUESTION:
Please find pictures of the weight Used. This is a 2oz.

ANSWER:
This requires a calculation of the drag force on an object moving through water. It should be appreciated that such calculations are always approximations and generally not reliable to better than perhaps 20% or more without doing sophisticated aerodynamic calculations on a super computer. In that spirit, I will push forward to get estimates; I enjoy puzzles like this. In the end, however, I will advise you to determine the answer by trial and error! I will, as most scientists do, work in SI units but will return to English units in the end. The drag force D on an object with a cross-sectional area A presented to the oncoming fluid and moving with speed v through a fluid with density ρ, is given by

D=½ρv²C_DA

where C_D is the drag coefficient, a constant which depends only on the object's geometry and orientation relative to the fluid flow. In our case, ρ=10³ kg/m³ and v=4 mph=1.8 m/s. Now we come to your weight; it is clear that it is made of lead which has a density of 1.13x10⁴ kg/m³. The problem is that the drag coefficient for this shape is not known and you really do not know what angle it would assume relative to the direction of the line, so you would either have to do extensive measurements or make simplifying approximations; since we just want a rough approximation, so I am going to assume that the weight is a sphere of radius R, C_D=0.47 and A=πR². (You may be amused by "consider a spherical cow".) R for a solid sphere of mass m may be calculated using the density of lead. Putting this all together, I find that D=9.2m^2/3.

So that takes care of the fluid dynamics part of the problem, the part where approximations are made. The rest is pretty simple first-semester Newtonian physics. The weight W is in equilibrium so W=Tsinθ and D=Tcosθ. For your desired situation, sinθ=25/50=0.5, so θ=30⁰. Doing all the arithmetic and algebra here I find that m=0.067 kg=2.4 oz.

Finally, I emphasize again that this is a rough calculation. My advice would be to go out in your boat with a protractor, drop the weight, and go 4 mph trying varous weights. Since you want the sinθ=½, get the angle to near 30⁰. Choose the weight which is closest. I am guessing it will be greater than 2.4 oz because I think the drag coefficients for your geometry will be smaller than for a sphere. If you do this, I would be interested in what you find.

QUESTION:
I am a high school forensic science teacher, and could use your help. My students do a lab in which they use a laser to measure the diameter of hair, using diffraction. The problem is that I�ve seen conflicting instructions for doing this and I am hoping that you can tell me which is correct. We do this by shining a laser on the strand of hair, them taking measurements from the diffraction pattern. The set-up is laser�card holding hair—(1 m)—screen. The diameter of the hair is calculated using the formula d=λL/x, where λ is the wavelength of the laser (around 632 nm) and L is the distance between the hair and the �screen.� (This formula is derived from d=λ/sinθ.) The confusion comes from measuring x. According to some sources, we should measure the distance between the center of the central bright band and the 1^st order maximum (the center of the first bright area on either side). Other sources say to measure from the center to a minimum, either the center or an edge of one of the dark bands! Here�s my question: Which is it (to a maximum or a minimum), and why? What measurement do we use?

ANSWER:
Here is the confusion, I think. Some references say that mλ=dsinθ locates maxima; those authors believe the pattern is the same as a double slit with spacing d. Other references say that mλ=dsinθ locates minima; those authors believe the pattern is the same as a single slit with width d. The first of these is flat-out wrong. There is a well-known rule, Babinet's principle, which states that the diffraction pattern from some obstruction is the same as the diffraction pattern from an aperture of the same shape. The only difference is that the central area of the pattern is much brighter since the part of the incident beam which would have been stopped by missing the slit carry right on through. Approximating sinθ∼x/L, d=mλL/x_m where x_m is the location of the m^th minimum relative to the central maximum. I have never done this experiment, but it occurs to me that it would be difficult to get a good estimate of the center spot because it is so bright and broad. Rather, I would try to get the distance x_m,m+1=x_m+1-x_m, the distance between two adjacent minima, and write d=λL/x_m+1,m.

QUESTION:
I'm trying to solve a little problem related to two vessels held under pressure. Specifically, where one vessel (a fluid dispenser ... like a beer keg) has a volume of 64 ounces and must be held at gauge pressure of 40 psi and the other vessel is the compressed air source with a volume of 26 in^3. What is the equation or procedure to determine the psi that the compressed air source must start at to maintain the fluid dispenser at 40 psi through the dispensing of a total volume of 448 ounces.

ANSWER:
How can the dispenser have a volume of 64 oz but dispenses 448 oz?

FOLLOWUP:
It�d get refilled after it emptied. My crack at it uses Boyle�s Law and because you�d have to open the dispensing vessel after it empties, you�d effectively have to stuff 448 oz x 2 of the gas at STP into the compressed air tank and because the vessels must remain equalized after pushing the full volume of liquid I added 40 psi to my answer. So my semi-educated guess is 954 psi. Am I right?

ANSWER:
(This turned out to be much more involved than I originally thought. Get ready for one of my very long-winded answers!) Boyle's law will not work here because the amount of gas in either of the containers is changing. Where does Boyle's law come from? It is

a special case of the ideal gas equation, PV/(NT)= constant, where N is some measure of the amount of gas in a container of volume V, pressure P, and temperature T; N can be anything you like—number of molecules, mass, number of moles, etc.�because, as you will see, only ratios matter in my analysis. Boyle's law assumes that N and T are constant which is not the case here. In these calculations it is important that T be absolute temperature (Kelvins) and P be the absolute pressure which is the gauge pressure plus atmospheric pressure which is 14.7 psi. I am assuming that there is a valve or other mechanism between the two tanks which keeps the pressure at 40 psi in the beer keg as the beer is dispensed.

I think it is instructive to first do the case where only 64 oz of beer is dispensed and both vessels are at 40 psi at the end. First, I need to convert all volumes to in³. 64 oz=116 in³ and 448 oz=809 in³. In the compressed air tank the initial absolute pressure is P, the final absolute pressure is 40+14.7=54.7 psi, the initial amount of gas is N, and the final amount is N-n where n is the amount of gas filling the dispenser after all the beer has been dispensed. Therefore, 54.7/(N-n)=P/N. Rewriting, 54.7/P=1-(n/N). Now the ratio of the volumes is 116/26=n/(N-n) because when all the beer is out, this ratio is the same as the ratio of the amounts of gas in each volume; don't forget that the pressures are now equal. Solving I find that 1-(n/N)=0.183 and so P≡P₁=299 psi (gauge pressure is 284 psi).

Suppose now that we want to have enough pressure so that two 64 oz tanks of beer can be dispensed. The previous paragraph now represents the second of the two dispensings, so the first dispensing would finish with P₁=299 psi in the tank and start with an unknown pressure P₂. The calculation proceeds the same as for the calculation of P₁ except that the ratio n/(N-n) and the ratio for the volumes are no longer equal because the pressures are not the same. Following the same procedure as for P₁, we now have 299/P₂=1-(n/N). I will now denote the pressure and volume in the beer tank as p=54.7 psi and v=116 in³; similarly, P₂and V=26 in³ for the compressed air tank. Therefore from the ideal gas equation, P₂V/(pv)=(N-n)/n. So, 1-(n/N)=1-pv/(P₂V+pv)=P₂V/(P₂V+pv). So, finally, 299/P₂=26P₂/(26P₂+6350)=P₂/(P₂+244), a quadratic equation for P₂: P₂²-299P₂-73,000=0. Solving, P₂=458 psi (gauge pressure 444 psi).

The above procedure can be generalized as

P_j-1/P_j=1-(n/N)=P_j/(P_j+244)
P_j²-P_j-1P_j-244P_j-1=0
P_j=[P_j-1+√(P_j-1²+976P_j-1)]/2

So,

P₃=[458+√(458²+976x458)]/2=634 psi (620 psi gauge)
P₄=[634+√(634²+976x634)]/2=822 psi (807 psi gauge)
P₅=[822+√(822²+976x822)]/2=1019 psi (1004 psi gauge)
P₆=[1019+√(1019²+976x1019)]/2=1220 psi (1205 psi gauge)
P₇=[1220+√(1220²+976x1220)]/2=1428 psi (1414 psi gauge)

ADDED THOUGHT:
When I first began to think about this problem I thought that just doing the problem by finding the pressure required for a 448 oz=809 in³keg ala the calculation for P₁ above. I was never able to convince myself that this would be equivalent, so I trudged through the step-by-step solution (which wasn't so bad after I figured out P₂). When I had finished I did the one-step calculation: 809/26=n/(N-n). Solving I find that 1-(n/N)=0.0321 and so P₇=1702 psi which is not correct.

QUESTION:
I need to lower a 500lb barrel.straight down from a 2 story building and I have 1 minute to do it. Are there any mechanical tools that could do this?

ANSWER:
Block and tackle.

QUESTION:
How fast would Lady Valarian be traveling to go 17 light yrs in nine minutes?

ANSWER:
There are about L=9x10⁶ light minutes in 17 light years. The speed v must be adequate to shrink, via length contraction, L to L'=9 light minutes. So, L'=L√[1-(v/c)²]. I find v=√(1-10^-6)c≈(1-0.0000005)c=0.9999995c.

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

QUESTION:
Why does value of small g maximum on the surface of the earth and decreases with height and depth ?

ANSWER:
Newton's Universal Law of Gravitation states that the force F of attraction between two spheres (masses M and m) is inversely proportional to the distance r between their centers, F=MmG/r² where G is the universal gravitational constant. In the case you are asking, M is the mass of the earth and m is the mass you are measuring F with. If g denotes the acceleration of m, then g(r)=F/m=MG/r²; g(r) indicates that the value of g depends on what r is. Note that if r=R where R is the radius of the earth, g(R)=9.8 m/s². So this explains why g decreases for r>R—as r gets bigger, g gets smaller. But shouldn't that now imply that as r gets smaller, g gets bigger? The answer is no because of something called Gauss's law which says that only the mass inside where you are contributes to the force, not the whole mass of the earth. If you assume that the density of the earth has a constant value of ρ=M/(4πR³/3), it can be shown that g(r)=4πρGr which decreases linearly to zero at the center of the earth.

It is interesting to note, though, that it is a poor approximation to assume the earth has uniform density. In fact the core of the earth is much denser than the mantle and crust. See an earlier answer for more detail on this.

QUESTION:
When I was in 9th grade, I had completed an assignment early in science class and went up to the front desk of the teacher for grading as he would grade assignments immediately. Out of boredom I was messing around with the faucet on his desk and had a stream of water flowing out at a rate just enough for it to not be separate water drops, but a constant stream. Being a silly 9th grader, out of curiosity I pushed the tip of my pencil (the end with the led) into the faucet where the water stream was coming out of and to my surprise, instead of the water spraying out everywhere or evenly coming out around the pencil, it instead spiraled around it almost as if it were a spring. I asked my teacher about this and he said it was happening because the earth is spinning which I don't know is correct or not, and if it is, how? I'm 23 now and this still bothers me today because I've never been able to find an answer that can fully explain why the water reacted this way

ANSWER:
For starters, it certainly had nothing to due with the earth's spinning. He was perhaps thinking of the Coriolis force but it is far too weak in this situation to be the reason for any phenomenon easily observed. Most likely the explanation was that your pencil had a static electric charge. See a video demonstrating this on YouTube.

QUESTION:
For hydrogen with one proton and one electron, what keeps them from collapsing onto each other due to the strong attractive electric force at that short distance? I would guess that the answer would NOT be the centrifugal force on the electron (like twirling a ball at the end of a string around you) spinning around the nucleus as this would assume the validity of the outdated Bohr atomic model. If that is indeed not the case that what is the answer?

ANSWER:
See an earlier answer. This also discusses electromagnetic theory and quantum theory as applied to the H atom. By the way, simple models like the Bohr model can be useful in getting an understanding of the crux of something; even if they are "outdated" it does not necessarily mean that everything about them is wrong.

QUESTION:
If you have a large wheel and a small wheel can the small one be the force to spin the bigger wheel. Lets just say the small one is offset and a spring is pushing it for the force. Will it spin as long as the force is constant?

ANSWER:
This is the way gears work. In the little animation, a constant torque (forget the "force" you allude to) is being applied to the smaller gear which then drives the larger gear. Note that the smaller gear is spinning with a larger rpm than the larger gear. This would be like low gear in a vehicle where the engine spins fast but the wheels spin slower. It is just the same for wheels provided there is enough friction to keep them from slipping on each other.

FOLLOWUP QUESTION:
This pertains to my previous question: If you have a large wheel and a small wheel can the small one be the force to spin the bigger wheel. Lets just say the small one is offset and a spring is pushing it for the force. Will it spin as long as the force is constant? So the force I speak of is the torque=radius x force. I am trying to make the torque.

ANSWER:
The only spring I know of which would work for you would be something like a watch spring, a spiral spring like the one shown in the figure. However, as it unwinds it does not exert constant torque; there are ways to get around this but they are pretty tricky. If you want a constant torque, why not just have a weight falling weight as in grandfather clocks? Torque and force are not the same thing. It would maybe help me answer your questions if you gave me some idea of what you are trying to do.

QUESTION:
[Note: This is a condensation of several communications with the questioner. It is an accident where a car continued driving on a closed road under construction.] The picture with the orange piece of machinery is setting, just to the right of that is where the highway ended, she drove off the highway at that point, she flew 82 ft horizontally until she hit the bridge support where you see the circled black mark on the support.

ANSWER:
OK, the pictures helped. However, I still can only make a guess as to how far she fell before hitting the bridge support. Assuming that the point she left from was horizontal and level with the top of the bridge support she hit, I am guessing that she dropped 6 ft. In that case, the equations to be solved are 6=16t² and 82=vt where v is the speed she left the road and t is the time to hit. I find that t=0.61 s and v=134 ft/s=91 mph. Other guesses would be drops of 7 ft, 8 ft, and 9 ft giving speeds of 85 mph, 79 mph, and 74 mph, respectively. Hope this helps. There is no additional information which I can add.

QUESTION:
I understand that when something's in orbit around the earth it basically travels far enough in the x axis each second that the surface of the earth 'curves away' under it by the same amount it's fallen towards it. What I don't quite intuitively understand is why if objects dropped from say a cliff continue to accelerate and gain vertical velocity, those in orbit around the earth seem to stay at their initial 1/2 gt squared speed in the y axis and don't eventually spiral into the earth?

ANSWER:
When an object moves in a circle with constant speed it is still accelerating. Acceleration is a vector and velocity is a vector. Acceleration is defined as the time rate of change of velocity. There are two ways velocity can change: by changing speed (what you probably think of as acceleration) and by changing its direction. In the case of a satellite in a circular orbit of radius R and with a speed v, the acceleration points toward the center of the earth and has a magnitude a=v²/R; this is called the centripetal acceleration. Of course there has to be a force which provides the acceleration and that is simply the weight of the satellite. If you are close to the earth's surface (compared to the radius R_Eof the earth) the weight is mg and so F=ma or mg=mv²/R_E; so the speed you must have to orbit close to the surface of the earth is v=√(gR_E). The value of the acceleration g gets smaller as you get farther away from the surface.

QUESTION:
If i am driving 55 mph and throw trash weighing 4 grams out of my passenger window, how far will the trash travel?

ANSWER:
Depends on the speed and direction with which you throw it. 4 grams is quite small so air drag would also be an issue; to calculate that you would also need to know its geometry. But why are you littering anyway? Wait to find a trash can.

QUESTION:
How large would a sphere made of lead need to be to atract a 1cm sphere of lead in deep space so as to capture it in it's orbit?

ANSWER:
There is no limit. Two 1 cm spheres of lead could be in orbit around each other but since the gravitational force is so small, their speeds would be very tiny. A space probe, Rosetta, was sent to a comet and orbited it; the escape velocity from the comet is only about 1 m/s so the probe orbitted with a smaller speed than this.

QUESTION:
If you roll a ball down a gradient it will accelerate due to gravity, but is there a theoretical shape of slope that will allow the ball to roll at constant speed? (disregarding friction or drag.) The only thing I can think of is a banked spiral track.

ANSWER:
There must be zero net force in the direction down the slope. With no friction and drag, the only shape which seems to me to work is a horizontal flat plane.

FOLLOWUP QUESTION:
Thank you for your reply. I've been thinking about this problem some more and if the ramp was made high enough and long enough and parallel to the curvature of the earth, the ball would continue to roll/fall at a constant speed for ever. It would simply be in orbit around the earth.

ANSWER:
Very clever! Actually, our answers are the same but you were thinking globally while I was thinking locally. A horizontal plane is the same as your ramp because it would have to be bent as it gets longer in order to remain "horizontal". However, it does not need to be high and most folks would not call this being "in orbit around the earth". Orbiting usually means an object has only the gravitational force (weight) acting on it. With your ring around the earth there is also a normal force N in addition to the weight W as shown in the figure. Newton's second law says mv²/r=W-N or N=W-mv²/r. So, when v gets big enough N=0; so the ring could be taken away and this would be the right speed for a "real" orbit at this r. And the ring would not work for speeds larger than the orbital speed, the object would leave the ring; any smaller v would result in the speed remaining constant as the mass rolled around.

QUESTION:
We manufacture metal roofing panels and many orders have lots of different lengths of panels. For this reason the longest panels in a stack are rolled out first so the pieces get shorter as the stack grows. I need a formula to find the center-point of balance for any given stack. The panels weigh 2.08 pounds per linear foot and the length and quantity of each panel will feed into the formula. They are always aligned at one end so the stack is much heavier at one end than the other.

ANSWER:
Maybe you just want the final formula, but I will outline how it is derived. Each panel (numbered i) I will denote to have a length L_i, a weight W_i, and a center of gravity r_i. Since the panels are uniform and have weight density of 2.08 lb/ft, r_i=½L_i and W_i=2.08L_i. I will also specify the number of panels with the length L_i to be N_i. The center of gravity of a stack of panels of length L_iis r_i=½N_iW_iL_i/(N_iW_i). Now, suppose there are two different lengths stacked. Then the position of the center of gravity is

r_CG=(½N₁W₁L₁+�N₂W₂L₂)/(N₁W₁+N₂W₂)

=½x2.08x(N₁L₁²+N₂L₂²)/(2.08x(N₁L₁+N₂L₂))

=½(N₁L₁²+N₂L₂²)/(N₁L₁+N₂L₂).

So, if you have j different lengths, the final answer is

r_CG=½(N₁L₁²+N₂L₂²+...N_jL_j²)/(N₁L₁+N₂L₂+...N_jL_j).

The more compact way to write this is

r_CG=½ΣN_iL_i²/ΣN_iL_i.

Here Σ denotes summing from i=1 to i=j. Note that the denominator is the total weight of the whole stack divided by 2.08 and that the weight per foot is not needed to find the location of the center of gravity.

QUESTION:
With increased bad weather, especially windy weather, I wondered. Does windmills, take any energy from storms, if only a tiny tiny tiny bit. If I recall, storms have more energy in them than nuclear bombs, so I know it would be insignificant. However, would you theoretically be able to have an infinite amount of windmills on earth to reduce storms?

ANSWER:
If anything increases in energy, including windmills, that energy (in equal amount) must supplied by another source. The source obviously must be the wind. Now, the site groundrules forbid unphysical questions, and an infinite number of windmills is obviously unphysical. However, let's investigate whether a reasonable number of windmills could have a significant impact on the weather. At the end of 2018 the total global wind power was about 600 GW, that's 0.6x10¹² W; the total world power production is about 44x10¹² W; an average hurricane has about 1.5x10¹² W; the Nagasaki atomic bomb was about 10¹³ W (10¹⁴ J and I guessed the time of the explosion was 10 s). It has been estimated that if we put wind farms covering all land and all off-shore water, 80x10¹² W would be generated. That same article estimated that you would need to have enough wind mills to generate more than 250x10¹² W to have a noticeable effect on weather, about four times the number feasible.

FOLLOWUP QUESTION:
I guess I wasn't clear enough, in my question, though your answer is really cool. Does windmills decrease the energy of a storm? Does windmills have any effect on storms?

ANSWER:
I think my answer does answer that question: "The source obviously must be the wind." If the wind is the source of the energy taken by the windmills, that means that the windmills take energy away from the storm. It is, as you speculate, an extremely small amount of energy. I have earlier estimated the fraction of energy which is taken out by turbines in the Bay of Fundy during tidal changes; it is harder for the windmills because it is not easy to estimate the energy of a "typical" storm, so the following calculation is a very approximate one. A typical windmill generates about 3x10⁶ W and guess that a typical strong storm is 10 times smaller than a hurricane, 1.5x10¹¹ W; so the percentage by which the energy of the storm is reduced is 0.02%, so 5000 windmills would use up most of the wind energy.

QUESTION:
We like to jump rope for exercise. Many times we use a weighted rope (the heaviest being 3 lbs). I don't think I have this right but if I swing the rope around my body (standard jump, nothing tricky) at say 10 mph and it's a 3 lb rope it has a force of 30 lbs if it hits something. However that is not the same as the rope having a weight of 30 lbs while swinging it. What would be the equivalent weight of a rope swinging at 10 mph? The faster you swing it, the heavier it feels (centrifugal force).

ANSWER:
You are right, you don't have this right! There is actually no way you can tell the force the rope will hit without knowing the details of the collision. To say the force is the product of the weight in pounds and the speed in miles per hour makes no sense; the units of force should be pounds but your 30 would have the units of lb·mi/hr. What you ask for is quite hard to calculate accurately. First, what does "swinging at 10 mph" mean? In the figure I show one piece of the rope in blue; this piece of the rope is moving in a circle of radius r about the axis with some frequency f (f times around every second) and, if r is in ft, the speed it travels is v=2πrf ft/s. For example if you are doing 2 revolutions per second and r=2 ft, v=25 ft/s=17 mph. But, since the speed depends on r, every point moves with a different speed, the top being the fastest. I have shown also the weight w and centrifugal force c for the piece we are examining; presumably, you want to know the forces F with which the rope exerts on you; w is the same for any piece of the rope the same size but c depends on r, the largest being at the very top in the figure and also at the very bottom as. When the rope is straight up as shown, the forces F are their smallest because c and w are in opposite directions; when the rope is straight down, the forces F are their largest because c and w are in the same direction. But to actually calculate the force is a formidable problem because we need to add all the pieces which are all different (this adding is integration in calculus if you have ever studied it). Just to give you some idea of the answer you sought I will assume that the rope has a negligible weight and the whole 3 lb is located in the middle of the rope and is a distance of 3 ft from the axis. You want v=10 mph=14.7 ft/s, and c=mv²/r=6.75 lb. (Here m is the mass which is 3/32 lb/(ft/s²), but you don't need to worry about that.) So, at the top, the force to hold it is 6.75-3=3.75 lb and at the bottom it is 6.75+3=9.75 lb. This is what you are calling the "equivalent weight", I guess.

QUESTION:
How much power is needed to throw a 20 pound ball 14.5 ft straight up in the air?

ANSWER:
Power is not what you want. Power is the rate at which energy is delivered. You must deliver 20x14.5=290 ft lb of energy. The power depends on how long you are pushing; for example, if you deliver the energy over a half second, P=290/0.5=580 ft lb/s=1.055 horsepower=786 W.

QUESTION:
If you had two incredibly large pieces of paper on the same plane that were just out of parallel to each other. One of these pieces is translating toward the other so they begin to overlap, and the point of overlap progresses from one end to the other. The question is if this physical point of overlap can travel faster than the speed of light? It does not seem to violate general relativity.

ANSWER:
Look at the figure. The two solid blue lines are the edges of the paper at some time, and the dashed line is the moving paper's edge a time Δt later. The moving paper moves with velocity v perpendicular to the not-moving edge and the intersection point moves with velocity u along the not-moving edge. The distance moved is vΔt and the distance between the earlier and later edges of the moving paper is s=vΔtcosθ. During this time interval, the intersection point has moved a distance uΔt and s=uΔtsinθ. Therefore,vΔtcosθ=uΔtsinθ or u=v/tanθ. Suppose we want u=2c and cause the paper to move with a speed of v=c/2; then θ=tan^-1(v/u)=tan^-1(0.25)=14⁰. So, indeed, the point of intersection can move faster than the speed of light. How can this be? Isn't c the universal "speed limit"? It is the limit for transmitting information or for any object with mass; this intersection neither has mass nor transmits any information. There is a similar example where something can have a speed greater than the speed of light without violating special relativity. Imagine pointing a laser at the moon and sweeping the spot on the moon across the surface by rotating the laser; because of the long arm, it would be pretty easy to get the spot going faster than c but, just as in your example, the spot has no mass and conveys no information.

FOLLOWUP QUESTION:
You make the comment in your answer that this point of intersection does not contain any information/data, this is what I am questioning... What if instead of sheets of paper they were just lines (as you drew in your figure) and instead of just one line on the top there were a number of parallel lines. The distance between these parallel lines would not be equal to one another would represent some value (lets say a distance of 1 or 2). So it seems like you could transmit some data string i.e. 1212221. It seems like this difference in line spacing could be measured when the points of intersection reach the other end thus passing data faster than the speed of light? Obviously could be digital or analog....

ANSWER:
I was a little puzzled by this one myself, so I opened a thread on PhysicsForums.com. You can see the discussion there.

QUESTION:
I don't understand the first thing about light. If there was a single photon emitted from a light source, would it be seen only by the person it was "aimed" at, or by any and everyone no matter where they were? If the latter, how could that possibly be?

ANSWER:
A photon is not the "first thing about light", more like the last thing! I would say the first thing about light is its wave nature. It was not until the 20th century that it was discovered (by Einstein) that light really is both waves and photons; this is called wave-particle duality. So, the single photon in your question still is wave-like; therefore, in spite of how you might aim your source, there is no guarantee that it will end up where you aimed it. So anybody in the room could see it, but once it has been observed nobody else could see it too.

QUESTION:
To me(observer from earth), since your speed is greater than c , the light from your flashlight illuminates the back of the spacecraft. please reply sir this is from the book of sir arthur beiser book=modern physics (page no 8) .Thier is a diagram which confuse me?

ANSWER:
Reply to what? Nothing can have a speed "greater than c", so this is a nonsense statement. The author is likely trying to make a point ad absurdum.

FOLLOWUP QUESTION:
Sir, I know that anything can�t travel faster than light. but Only I need to now why flashligh�s light going in the backward direction(sir see the figure which I told you).if you have read the book then ok but If you not I am sending the book so that sir you can understand I am asking a valid question or not.

ANSWER:
It is exactly as I said�the author is using an impossible example to demonstrate that you get ridiculous results if something has a speed faster than the speed of light. Usually I will not answer questions which have something moving faster than the speed of light, but now I see that the point of the author is to demonstrate that it is impossible. I can see now how you got confused; I was also confused at first. If you apply Galilean relativity to the example, you would say that the light was going forward with a speed v+c, still illuminating the front of the ship. But that is not what he is doing. Instead he is saying that there will be Cerenkov radiation (from the flashlight, I guess) which is like the shock wave of a supersonic aircraft. This is in the form of a cone behind and that is the light supposedly illuminating the back. I find this a very far fetched example and would advise you to simply ignore it!

QUESTION:
If you have an Entangled particle That falls into a black hole would it communicate With it�s entangled partner outside the black hole.

ANSWER:
First of all entanglement does not mean the particles "communicate" with each other. But, once anything is inside the Schwarzschild radius of a black hole, it cannot communicate with anything. Furthermore, since you have no access with the captured particle, there is no measurement you can make on the uncaptured particle which would tell you anything about its entanglement. A much more detailed discussion may be found here.

QUESTION:
Why earth has gravity?

ANSWER:
Every object with mass has gravity. The earth has a very large mass so it has very noticeable gravity.

QUESTION:
How birds fly by overcoming the gravity of earth even with a large mass difference?

ANSWER:
The gravity exerts a force down on a bird which is called the weight of the bird. For the bird to fly, there must be an upward force greater or equal to its weight. When the bird flaps its wings down they exert a downward force on the air; because of Newton's third law, the air exerts an equal force up on the bird. It is this upward force which allows the bird to fly.

QUESTION:
Please help me settle a long-running argument: A motorcycle carrying two men is traveling, under engine power, in a straight line at a constant velocity on a flat road. The passenger falls off, and the driver continues applying the same constant amount of throttle to the engine. Will the motorcycle accelerate? The f=ma equation has been tortured and abused by both sides of the argument to prove their point, with no resolution in sight.

ANSWER:
I have shown the most important forces on the motorcycle plus men in the figure. These are W₁, the weight of the rider; W₂, the weight of the driver; W₃, the weight of the motorcycle; D, the air drag force on the system; f, all other friction forces retarding the motion, rolling friction by the wheels, friction in wheel bearings, etc.; N₁, the force up on the rear wheel by the road; N2, the force up on the front wheel by the road; and F, the static friction force between the rear wheel and the road which is the force driving this system forward. Since the system is in equilibrium, N₁+N₂=W₁+W₂+W₃ and F-f-D=0. The most important thing to appreciate is that f is proportional to the normal force, f=C(N₁+N₂) where C is some constant. So when the rider falls off, W₁ disappears and so N₁ and N₂ get smaller and so f gets smaller. F, however, does not depend on N₁ because it is static friction; it is determined by how much power is being delivered by the engine so it does not change. I will denote the new frictional force as f' and note that it is smaller than f. The sum of the forces is no longer zero, so F-f'-D=Ma where M is the mass of the rider plus bike, M₂+M₃=(W₂+W₃)/g. where g is the acceleration due to gravity. Finally, the bike has an acceleration a=[(F-f'-D)/(W₂+W₃)]g.

QUESTION:
It occurs to me that a 'station-keeping'scenario to maintain a fixed altitude above a fixed point, (say several hundred miles and above the atmospheric envelope), above the Earth should still allow for normal gravity on the vessel. My question: How much thrust in terms of g-factors would be required to maintain that fixed position?

ANSWER:
If you want something to hover a few feet above sea level, the upward force on it must equal its weight. By weight I mean the force which the earth exerts on something. But, this force (weight) gets smaller as you get farther away from the surface. So, the weight of a mass m at sea level is given by F₀=mMG/R²=W where R is the radius of the earth, M is the mass of the earth, and G is the universal gravitational constant; if you go up to an altitude h, the weight gets smaller, F_h=mMG/(R+h)². Therefore, if you want to hover at an altitude h you must exert an upward force of F_h=[R²/(R+h)²]W where W is the weight at sea level. Even several hundred miles is small compared to R which is about 4000 miles, so we can make an estimate by doing a binomial expansion: R²/(R+h)²=(1+(h/R))^-2≈1-(2h/R)+… So suppose that h=R/20; then F_h≈0.9W. (The exact calculation would be 0.907W.) You asked for the force "in terms of g-factors"; that would be the factor of W, [R²/(R+h)²].

QUESTION:
I've just been reading up on LightSail2, a craft that unraveled a big "sail" made of aluminized Mylar to test the viability of light propulsion. The theory goes, though light doesn't have mass, it does have momentum, and can therefor be used to propel the sail in space. They say the experiment produced about the same force as a paperclip over an entire orbit of the earth. My question is. How can they be sure the propulsion force is being generated by the light striking the sail, and not by the solar winds that are striking the sail?

ANSWER:
The solar wind carries a much smaller momentum flux than that of the solar radiation. The force which it exerts on the sail is about 1% of that exerted by the radiation. By the way, light having momentum is not a "theory", it is a well-measured fact.

QUESTION:
I am an elementary physical education teacher. I would like to have my third through fifth graders record how many jump squats they do each semester. Then at the end of the semester I would like to ask the question: How long have you spent suspended in air this semester? (Assuming that there is a way to figure out how much time an object traveling against gravity spends transferring their direction from up to down???) Is there in easy math equation using the amount of jumps in a semester & maybe their weight to figure this out.

ANSWER:
Because the acceleration due to gravity is constant no matter what the weight is, the weight is only peripherally important; a light student is likely to be able to jump higher than a portly student and will therefore spend more time per jump in the air. So each student needs to do several squat-jumps for some adult, maybe a parent; that person would then estimate the typical height the student's feet go above the ground. So let's call the average height reached by a particular child H and the number of times she jumped during the semester N. The total time, in seconds, that child spent in the air would be T=0.072N√H if H is measured in inches and T=0.045N√H if H is measured in centimeters. For example, if the child typically jumped 8 inches and jumped 200 times, the total air time is T=0.072x200x√8=40.7 seconds. I would only worry that an elementary school student might not be familiar with a square root. (You only asked for an equation; I can also post the derivation of the equations I posted here if you are interested.

QUESTION:
If I am on a moving platform travelling at 20 mph in one direction and a platform is moving towards me at 20 mph, I understand that the relative speed is 40 mph but what would happen if I stepped from one platform to another as they crossed? Would the speed of the platform I moved to cancel my speed in the opposite direction and therefore simply move me at 20 mph in the new direction or would there be a process of braking where I would fall off the back of the platform unless it was long enough to sustain the braking time?

ANSWER:
Imagine you are a passenger in a car going 40 mph. You decide to step out onto the road. That is what it would be like to step onto your crossing platform.

QUESTION:
I have a photo. It is a nature shot in which the sun is going down behind a hill and yet the full (sphere of the) sun is visible in the image. My question is how is this possible? Some sort of light refraction?

ANSWER:
I believe that this is a problem with focus. If you examine the photo, it appears that the objects in the foreground are in focus. Looking at a blowup of the area near the sun, there is no evidence of an image of the sun above the horizon because the whole region is saturated, too bright to image. Nevertheless, the sun itself is brighter than the region around it even though you cannot see it because of the saturation. But there is an out of focus image of the sun and it is this larger image which is visible below the horizon. The sun itself is probably totally above the horizon and with a smaller raduis than the radius of the "bite" out of the hill.

QUESTION:
You are the captain of a pirate ship and you're shooting a cannon at land. If all parameters(air density, drag coefficient, etc.)are constant, and you could shoot the cannon ball from the exact same spot on the water, one shot while the ship is stationary and the other while the ship is in motion would the cannon ball land in a different spot or the same?

ANSWER:
See a recent answer. Think of the passer as your pirate ship and the receiver as the target on the ground. If the passer (ship) in the first figure is not running (moving), the ball will go to the location of the receiver (target) who will receive (which will be hit.) it if not running. If the passer (ship) is running (moving), he (it) will give his (its) velocity to the ball (cannonball), in addition to the velocity he (it) imparts by throwing (shooting); if the receiver (ship) is not running (moving), he (it) will not catch it (be hit) because it will go to the position he (the target) would have been if he (it) were running (moving) too.

QUESTION:
There is a great debate in mountain biking that when going into a turn at a high speed whether to have the outside pedal down, so that if I lean my bike down into a right hand turn my left pedal will be down and my right pedal up. OR to go into the turn with level pedals as the bike leans. Some things to consider here is that weight must be centered over the bike left to right and front to back to avoid tire washout. The leaning of the bike allows the outside knobs of the tire to grip the terrain. While centered weight allows for proper downward force maximizing grip. So I guess the question stated another way is where is your weight more centered when the bike is leaning, pedals flat or outside pedal (pedal opposite of the direction of the turn) down. (Also please try to respond in layman�s terms. I�m far from a physicist). :).

ANSWER:
I am assuming that "tire washout" means wheel slipping because of loss of adequate friction with the ground. This would occur if one or the other wheel was holding up significantly less weight than the other. The front-to-back location of the center of gravity of the bike plus rider needs to be halfway between the two axles for the wheels to have equal friction. The left-to-right location would have nothing to do with having equal friction, so that can be ignored for your purposes; however, since the the legs are moving up and down and forwards and backwards, not sideways, the left-to-right centering never changes anyway. The first figure shows one pedal up and one down; the second figure shows level pedals. In the first case, the right knee is quite forward and the left leg is straight down; in the second case the right knee is less forward than in the first case and the left knee is slightly less forward than the right but substantially forward from the first case. To my mind it is a tossup as to which has the center of gravity more forward; and besides, where is the center of gravity of the bike rider system? Furthermore, in my opinion, any difference between the two cases will make a trivial difference regarding whether one wheel or the other is more likely to lose traction. In addition, if you have an extreme lean or you are on terrain which is sloping upward in the direction you are turning, you are less likely to catch your pedal on the ground if it is all the way up. Do whatever you are most comfortable with; this issue does not merit "a great debate"!

ADDED INFORMATION:
Since you asked for layman's terms, perhaps I should be clear what center of gravity is. It is the point where, if you suspend the object from a rope, it will remain balanced; like the fulcrum of a teeter totter with equal-weight kids. In the figure below, the biker on the right is hanging from a rope attached to her center of gravity (the cross). The biker on the left is hanging from a rope attached to her front axle; note that the center of gravity is directly below the point of suspension. Only if the center of gravity is halfway between axles will the total weight be equally supported by each wheel.

QUESTION:
I know the solar system is traveling thru the galaxy at a certain speed. My analogy is the solar system is in a car, the car is moving down the highway. Say you throw an object out the window, that being a satellite going outside into interstellar space. My question is, will that satellite fall behind us? Kind of like the object being thrown out of a moving car on a highway.

ANSWER:
First of all, the only reason that the object you throw straight out of the car only "fall[s] behind us" is the air drag; if there were no air drag the thrown object would move away from the car but keep pace with it as you will understand below. The first diagram shows the situation in the frame of the solar system: we have "thrown" the satellite so that it is moving with some velocity to the left in the figure. Now, if there is some outside observer, she will see the solar system moving in some direction indicated by the red vector attached to the sun in the second figure. (The sun (car) is moving perpendicular to the velocity of the satellite (object).) But the satellite will have its same velocity relative to the sun but also the velocity of the sun in the second figure; these velocities (shown in blue) have to be added to get the satellite velocity which will be seen from outside the solar system. (Since you chose throwing something through the car window as an analogy, I have chosen the satellite velocity to be perpendicular to the motion of the sun as seen by the outside observer. Other angles will give different results.)

QUESTION:
I was sitting behind a thick double glass door that coused light comming from other side of the bulb to be light coming from multiple bulbs and you can even distinguish each and every bulb and even count the number of bulbs but when the double glass was made to be single it seemed crystal clear that it came from single bulb . Why this happened , what will be the ray diagram of the cenerio . This happed at night.

ANSWER:
When light strikes the glass, a small fraction of it is reflected but most goes through; the amount of reflection depends on the angle the light strikes the glass. If there is a single glass, the reflected light just is not seen on the other side. However, if there is a second sheet of glass, the light is reflected back toward the inside as I show in the figure. You should also have noticed that the images you see get dimmer as you go farther from the main first image.

QUESTION:
Acceleration is rate of change of velocity, then why a body is said to have a constant acceleration due to gravity when a body is at rest at a plane horizontal surface?

ANSWER:
A body at rest does not have any acceleration. Nevertheless, the value of acceleration due to gravity g plays an important role for the body which, if its mass is m, has a weight W given by W=mg.

QUESTION:
If I can throw a tennis ball at 92 mph, how fast would a baseball travel with the same force acting on it?

ANSWER:
If you exert the same force over the same distance for each ball, you impart the same energy E to each, E=½mv², where m is the mass and v is the speed. Therefore, ½m_tv_t²=½m_bv_b² or v_b=v_t√(m_t/m_b). Putting in the masses of the tennis ball and the baseball, m_t=0.058 kg and m_b=0.145 kg, I find v_b=58 mph.

QUESTION:
If I have 1.5 ft H x 25 ft L x 8.6 ft D mass that is suspended by 4 cables, 6 ft long and the mass weight 3000 lbs. How hard does the wind has to blow before the vertical cables reach to 15 degrees?

ANSWER:
It is a lot easier for me to do this in SI units, so W=3000 lb=13,345 N, 1.5 ft=0.467 m, 8.6 ft=2.62 m, 25 ft=7.62 m, and 6 ft=1.83 m. The answer depends on which way the wind blows. So, I will make the calculation for the wind blowing directly on the largest surface whose area will be A=7.62x0.467=3.56 m². There is a pretty good approximation for calculating the drag force F on an object of cross section A in a wind with speed v, F=¼Av², so v=2√(F/3.56). (N.B., this works only in SI units.) The forces on your mass are shown in the figure. When I do the analysis I find F=3576 N=804 lb, and v=63.4 m/s=142 mph. Please keep in mind that this is an approximation as are all calculations of air drag.

QUESTION:
Air conducts heat and goose down is an effective insulator because it prohibits airflow. It seems to me that it may also be effective at keeping cold things cold in a warm environment. Would high-fill down be an effective insulator to keep cold things cold? If not, why not? Do water evaporation effects come into play?

ANSWER:
Any material which keeps something hot in a cool environment would be equally effective in keeping something cold in a warm environment. There is a caveat, though. Imagine a hollow container which has surface area A, thickness t, and temperatures T_in inside and T_out outside. The rate of heat flowing through the material depends on four things: what the material is, what A is, what t is, and the temperature difference ΔT=|T_out-T_in|. (This assumes that conduction through the container is the most important means of having heat flow, not convection or thermal radiation.) So take two examples: inside cold water at 1⁰C, outside room temperature at 20⁰C, ΔT=19⁰C; and inside hot soup at 90⁰C, outside room temperature at 20⁰C, ΔT=70⁰C. So the soup would cool down faster than the cold water would heat up. Why would evaporation have anything to do with it?

QUESTION:
My questions is, how do we rationalize the fact that the speed of sound has to do with modulus and density. As we know that solids have greater density than liquid and air, does the space between each particles effect how fast sound propagates?

ANSWER:
To illustrate your primary question, "modulus and density", let's just consider solids. A simple model of a solid is that the molecules are connected by by tiny springs. Sound passes through this solid by one molecule moving and pushing, via one of the springs attached to it, on its neighbor which then propogates the push to the next neighbor, etc. Clearly, the speed of this sound, displacements of molecules, will depend on the spring constant of the springs; but the spring constants are an approximate measure of the appropriate modulus. If we think of the spacing of the molecules as being roughly constant regardless of the composition of the solid, the mass of a molecule is going to be proportional to the density. But, the mass of a particle on a spring determines the speed of the motion of the particle—the greater the mass, the more slowly the particle moves when attached to a spring. By this reasoning, the speed of sound depends on both modulus and density.

QUESTION:
I have a question about space travel. Since there is little matter in space, and it would have very little slowing effect on a spacecraft. So, can a space craft increase in speed incrementally? What I mean is, we launch a spaceship from Earth. It takes a straight line trajectory once leaving Earth's gravity field. It reaches a certain speed, and cuts it's engines to coast. It wouldn't slow down any real amount while coasting. So, if it fires it's engines again for an extended period of time, would it not go faster than it was while coasting? And if this process was repeating dozens of times, wouldn't that spacecraft have basically a limitless top speed until it ran out of fuel for the burns? I would think that it would.

ANSWER:
Of course, burning more fuel would give you more speed. There is a limit, however, to how fast you can go, a limit set by special relativity—the speed of light.

QUESTION:
if I take a photo-voltaic cell/panel and connect both of its terminals to a load (say an LED or bulb), without any battery in the circuit, and irradiate with light, will the photo-voltaic cell be able to drive the load? Or do I need a battery in the circuit for the photo-voltaic cell to deliver power?

ANSWER:
A photo-voltaic cell is a battery. As long as it can supply the current needed at the right voltage (power), no other battery need be added.

QUESTION:
Assume that I stood at the boundary line of a football / baseball / cricket ground(without any turf/pitch/grass in it) - just a plain ground - like play ground. I have a ball that can bounce when thrown on this ground. My question is - When I throw the ball at an angle to ground/horizontal plane, the ball takes a trajectory- which is very clear. However, after hitting the ground (that is at first bounce), it takes another trajectory and so on and so forth. Are all these trajectories are same in their shape, except for the amplitude/ maximum height it can reach?

ANSWER:
If you consider the effect of air drag to be negligible, each segment is a parabola with the equation y=xtanθ-(½g/(v₀cosθ))x² where the acceleration due to gravity is g=9.8 m/s², θ is the launch angle, and v₀ is the launch speed. In fact, since speeds are relatively modest, air drag will probably be negligible. If you include air drag, each bounce segment will be a different shape because the drag depends on the speed of the ball.

QUESTION:
According to coloumbs law every body attracts other body but how come the electrons arranged in a magnet in such a way that they have such a great attractive force that we notice the attraction of other bodies towards it ?

ANSWER:
You misunderstand Coulomb's law. This law applies to electrically charged bodies, not "every body". And the force between two charged bodies can be either attractive (if the charges are of opposite sign) or replulsive (if the charges are of the same sign). Inasamuch as a magnet has no net electric charge, you can conclude that Coulomb's law is irrelevant. The reason the magnet can exert forces on other bodies is because the way the electrons are "arranged" in such a way that a different force (magnetism) is coming into play because of the magnetic field caused by the electrons.

QUESTION:
In rugby, you cannot pass the ball forward, only backward. We often witness passes that appeared to travel forward across the ground, but are deemed not to be forward passes because the ball was not thrown forward by the passer, but rather it drifted forward due to the momentum it carried by virtue of having been thrown by someone running forward. My question is, how much forward speed/momentum would a runner need to have to affect the trajectory of the thrown ball enough to cause this to happen?

ANSWER:
In the diagram I show the velocity of the passed ball in blue as seen in the frame of the passer. The red vector shows the velocity of the passer in the frame of the ground. The diagram in the prasser's frame shows a legal pass since the receiver is behind the passer and the ball is thrown directly at him. However, this pass, as observed by a fan watching from the the ground frame, will have the velocity of the passer added to the velocity of the ball in the passer's frame; the velocity of the ball in the ground frame is shown in green. The answer to your question is that the runner needs to be running forward with a velocity greater than the backward component of the ball's velocity in the passer's frame.

QUESTION:
Hello. I was watching a show about the moon landing last night where they said that the moon's gravity is 1/6 that of the Earth's. I mentioned that to a coworker today, and he said "So if a 180 pound man can bounce like that on the moon, then I would bounce on Earth if I was 1/6 that weight at 30 pounds." I mentioned the fact that you don't see all the 30-pound children floating around in their front yards, but he countered with "I mean, if I had my current muscle mass and weighed 30 pounds then it would work" and I really can't see how that's possible - but I also couldn't come up with the right argument to defend my position. We both know there's a difference between weight and mass, but this is something different. Which one of us is correct, and what's the best way to explain this concept? If no one could bounce on Earth like they do on the moon regardless of how much mass they have, then how do you phrase the reason why not?

ANSWER:
It is pretty absurd to suggest that a 30 lb (14 kg) tot could have the same muscle strength as a 180 lb (82 kg) man. But, let's talk about how you jump straight up. You squat down, a distance maybe half the length of your legs (I would estimate about 0.5 m for the adult, 0.25 m for the child) and spring up. What springing up means is that you exert some force F down on the floor in addition to your own weight W so that the floor exerts an upward force on you which is equal to F'=F-W and it is F' which accelerates you upward so that when your legs are straight again, you leave the floor with some speed v. Your coworker is right that the force which provides the acceleration, F', is greater if F is the same for both adult and child because W is 6 times smaller for the child than for the adult. The calculations are straightforward but a bit tedious, so I will just give you the results for the situation where F has a magnitude twice the weight of the adult, approximately 1640 N (360 lb), so F'=820 N. If this force acts over 0.5 m for the adult, he launches with a speed of about 3.2 m/s, while the force acting over 0.25 m launches super child with a speed of about 7.3 m/s. With these initial speeds, the adult goes up about 0.5 m while the child goes up about 2.7 m. Now, we must calculate the height the adult will rise on the moon exerting the same force over the same 0.5 m; but now the launching force is F'=F-mg/6=1640-137=1503 N. I find the launching speed to be about 4.3 m/s and the height to be about 5.5 m. So the adult on the moon goes about 5.5/2.7=2 times as high as child with equal leg strength does on earth, and about 11 time higher than he could jump on earth; the reason for this large factor of 11 is twofold—first, the fact that the weight is much smaller means that the given leg strength will be much more effective in accelerating during the launch; second, once launched the weaker gravity cause him to slow down more slowly. So your coworker's guess about how the compared jumps would be is pretty far off, a factor of two in terms of height achieved. However, he got it qualitatively correct that a smaller person with equally strong legs could jump higher on earth; it just doesn't work out quantitatively when comparing to the jump on the moon.

But, what is the real difference? If you look at the videos they look like they are taken in slow motion, but they are not. Because the gravity is much smaller, it will take much more time to get up to and back down from a given height; even if the height achieved by the super child on earth were exactly the same as for the adult on the moon, the two would appear very different because the child would be back to ground level much faster than the adult would be.

QUESTION:
A cone is standing inverted on a friction less surface at equilibrium. (tip of the cone touching the ground.) Now a perfectly horizontal force acts "at a point on the surface of the cone" such that line of force passes through the center of mass of the cone. Will it topple or just move in horizontal direction?

ANSWER:
Be sure to understand that this is a gedanken, a thought experiment; the cone is at the point of unstable equilibrium and you would not be able to actually balance a cone on its tip in the laboratory. Any force which has no vertical component and has a line of action through the center of mass will cause horizontal translational motion but no rotational motion. In fact, any force passing through the center of mass will cause no rotational motion. The more general case is shown in the diagram. The effect of the horizontal component (F_x) on the cone is to provide a horizontal acceleration. The effect of the vertical component (F_y) is to reduce the normal force (N) which would otherwise be equal to the weight (mg). If F_y>mg, the cone would have an acceleration in the y direction as well, fly off the surface. The components of the acceleration would be a_x=F_x/m and a_y=(F_y/m)-g. In no case would the cone have any rotational motion.

QUESTION:
I have been trying to understand why an inflated tire does not lose energy and deflate. I am not trying to understand how air isn't lost through holes or permeation of the tire material. What I am trying to understand is the air molecules in a tire are energized they are deflecting outward causing pressure on the tire, and the car is providing downward pressure via gravity. So why don't the molecules "run out of energy" if energy is being used to keep the tire inflated?

ANSWER:
The air in the tire starts out at some temperature, volume, and pressure. The temperature is an indicator of how much energy the air has. I will assume the volume stays about constant. Now, if the environment around the tire is at a lower temperature than the air in the tire, the collisions of the air molecules with the inside of the tire will, on average, slow down the molecules' speed thereby reducing the energy of the gas. When the temperature goes down, the pressure in the tire will go down too. But when the temperature of the air in the tire is the same as the outside temperature, the collisions will on average leave the speed of the molecules the same. So, you are right, sometimes the air does lose some energy but that loss ends when the everything has the same temperature. Conversely, if the air inside the tire is cooler than the outside temperature, the air inside will gain energy until the two temperatures are the same.

QUESTION:
On 9th May 2019 Jeff Bezos made a presentation at the Air & Space Museum in which he said," it takes 24 times less energy to lift a pound off the Moon than it does on the Earth". Since the escape velocity of the Earth is 11.186 km/s and that of the Moon is 2.38, the ratio of their squares is 22.09. Apparently the Earth's escape velocity only includes gravitation, and not air or magnetic drag. But air drag is too small to account for the discrepancy between 24 and 22.09. One reference claims that magnetic torque is 10% of the gravitational torque on a satellite - which would take care of the discrepancy. But when I try to calculate the magnetic drag for a spinning rocket I get micro-newtons, which is too small.

ANSWER:
This discrepancy seems too small to worry about. Bezos was probably just giving an approximate number some researcher gave him. I find it difficult to believe that the earth's magnetic field has any significant effect on escape velocity. However, the air drag will not be small. The projectile will be moving with a terrifically large velocity at the beginning of its flight where the atmosphere is most dense and encounter a very large drag force; this could very easily account for the discrepancy.

QUESTION:
I was wondering if someone were to jump off with hoover dam inside a 100m wide ball if they would live or not. I believe that they would live personally because the hoover dam in meters is roughly 225m. You might consider this off the wall but I'd personally would like to know and to prove my coworker wrong.

ANSWER:
What does being inside a ball have to do with it? I envision you standing on the inside surface of the ball positioned on top of the dam. Now, you each start falling with the same acceleration. You hit the bottom with the same speed as you would have without the ball. Maybe you are thinking the ball would act as a parachute? Because it has so much surface area, if it was not very heavy it might set you down gently.

QUESTION:
My understanding of physics and science in general is limited so please bear with me. Is it possible to create a controlled interaction between electrons and direct the force in a specific direction to create a type of propulsion? If so, could the force created between this interaction be significant enough to move large objects through the vacuum of space?

ANSWER:
Your question is confusing but I think you are asking if electric charge can be used somehow for propulsion. The answer is yes, and it is called an ion propulsion. You can read a nice little article about ion propulsion in this link by NASA.

QUESTION:
When a full bus takes a sharp turn, could it ever tip over? Are there conditions under which this could happen, everyone sitting on one side, wind, etc.? I am aware of the Mythbusters episode debunking the scene from that Sandra Bullock movie, but I'm interested in the theoretical point at which it could happen.

ANSWER:
The problem in examining this question is: what constitutes a "sharp turn". The best way to characterize the turn is to specify its radius. That is the way I will proceed, considering a 50 mph turn and calculating the radii for which the bus will skid, or begin to tip, or actually tip over. I will use 50 mph for the speed and make educated guesses for the distance between left and right wheels, the location of the center of mass (which I will assume is in the center across the bus and some distance above the road), and the coefficient of static friction.

The diagram above shows the forces on the bus if it is turning right (as seen by the driver) around a circle of radius R (center to your left), traveling with speed v; the coefficient of static friction between the tires and the road is μ. The bus is just about to skid and so the frictional forces are at their limits, μN. The driver's side tires have a normal force N₁ and a frictional force f₁=μN₁; similarly, the tires on the other side have a normal force N₂ and a frictional force f₂=μN₂. The weight of the bus, W=mg, acts at the center of mass of the bus; I will assume that the center of mass is halfway between the left and right wheels. The bus is in equilibrium in the vertical direction, so N₁+N₂=mg. The bus has a centripetal acceleration to your left a=v²/R so f₁+f₂=mv²/R. The greatest that f=f₁+f₂=can be is f=μ(N₁+N₂)=μmg=mv²/R. Therefore the smallest radius that the bus can have if its speed is v is R_min,slide=v²/(μg). In your case, v=50 mph=22 m/s, g=9.8 m/s², and μ≈0.8 for rubber on dry asphalt, R_min,slide≈62 m=203 ft.

But you are interested in the radius when the bus will start to tip over; this happens when N₂=f₂=0, N₁=mg, f₁=mhv²/R. Now we need to calculate the torque about some axis and set it equal to zero, but that axis must pass through the center of mass because the bus is accelerating. Now, you need to know the height above ground h of the center of mass and the distance 2d, between the left and right wheels. Doing that, the torque equation is (mhv²/R)-mgd=0 or R_max,start=hv²/gd. This is the radius at which the bus will just start to tip. For smaller radii the inner wheels will come off the ground but the bus will not tip over until you reach some critical smaller radius. So let's just guess that approximate values of h and d are

equal at about 2 m each: R_max,start=v²/g=49 m=161 ft. This choice of h and d result in the bus never tipping because it would slide before it started to tip; to reach this point, f=μmg=mv²/R, μ=22²/(9.8x49)=1.01.

Consider the bus when it has tipped through some angle θ relative to the horizontal as shown in the second figure. Now the moment arms to compute torques are D=dcosθ-hsinθ for N=mg and H=hcosθ+dsinθ for f=mv²/R. (I must admit that it took me a while of floundering with my trig to figure these out!) So now the torque equation is mgD=Hmv²/R or, R=Hv²/(gD). In the graph below you can see that, for d=h, the radius gets smaller and smaller as the tip angle gets larger; all these calculations assume that the bus does not skid, but unless there is something else besides static friction to provide the centripetal acceleration it will likely skid before it tips. Note that if the bus tips until the center of mass is directly above the outer wheels, where D=0 and θ=45⁰ for d=h, it is on the verge of fully tipping over. The bus will roll over when the angle it makes with the ground is greater than tan^-1(d/h). If you manage to get the bus on two wheels and at the angle just before it tips over you can go essentially straight (making small steering corrections), not turning as the video linked to above. The bottom line is that a bus is extremely unlikely to roll over in a turn because it will likely skid before it rolls. I think my choices of d and h are very reasonable and probably, because so much of the weight is in the frame, suspension, axles, transmission, and engine of the bus, the center of gravity is likely to be lower than 2 m making a roll even less likely. It would be a different story for a train on tracks where the "friction" is almost limitless because the train is constrained by the track from moving sideways ("skidding") at all.

ADDED NOTE:
You could probably, by suddenly steering violently to one direction, cause the inner wheels to come off the ground before it skidded much, but I doubt that you could flip it over this way at 50 mph unless the skidding wheels "snagged" on something like a curb or a hole. If you watch a movie of a high-speed car chase you will note that the turns are usually executed by skidding around the curve.

QUESTION:
What percentage of the time in motion is a pendulum or piston motionless?

ANSWER:
Zero.

FOLLOWUP QUESTION:
Thank you, but, they change direction without stopping? That sounds impossible. What am I missing? If it is just that the real number is so small that it might as well be zero, ok, but I was hoping to know what the real number might be.

ANSWER:
Well, dealing with an engineer, I guess I have to consider the "real world"! A physicist will consider an idealized zero friction system for which the pendulum does stop but only for zero time. But an engineer will have to take into account that a real pendulum might stick for a brief time, maybe a microsecond, at the extrema. But I cannot give you an answer to your question because the amount of stick time will depend entirely on the details of the construction of the pendulum. (I believe that the main reason for sticking would be the difference between static and kinetic coefficients of friction.)

QUESTION:
I�m not a physicist, but am curious about �Planck's� constant. Everything seems to have a "smallest" inadvisable value. Given that doesn�t it follow that numbers like "pi" would have an absolute value and not continue toward infinity? I�m sure this must be pretty basic, but seems the "small" has a limit so "infinite series" numbers or values would have a limit also?

ANSWER:
Planck's constant, h, plays more roles than just setting the smallest value of something. Most famous of this role is that, if you have electromagnetic waves with a frequency f, the smallest amount of energy, called a photon, which such electromagnetic waves can have is hf. But, this idea of minimum amount possible does not extend to everything, certainly not to something like the number of digits in any irrational number like π. The idea that something has a minimum value usually has to do with what we call "fields"; when the electromagnetic field is quantized, the photon is the "messenger" which conveys the field.

More interesting, I believe, is the role played by Planck's constant in the structure of systems of particles like atoms or atomic nuclei. The great hypothesis by Niels Bohr that the hydrogen atom can only exist in particular energies owes its inception to the idea of quantization. He assumed that the electron in a hydrogen atom moved in circular orbits but not in any old orbit but only in those for which the angular momentum was quantized, L=nh/(2π) where n=1, 2, 3, 4…; the angular momentum is the product of the electron mass, the orbit radius, and the speed of the electron. This led to quantized energy levels; this model excellently explained everything known about the hydrogen atom at that time. See the Wikepedia article on the Bohr model. (The Bohr model, while easy to visualize, is a very simplistic and incorrect compared to the correct modern quantization of the atom.)

If you think about it, energy quantization is an astounding surprise. It is like saying that a pendulum may have an amplitude of 5⁰ or 7⁰ but not 6⁰. But nature does hand us surprises sometimes.

But energy quantization does not apply to everything. If a system of particles is not bound together it can have any amount of energy.

What is not known but speculated about is whether space and time are quantized. Is there a shortest possible distance? Is there a shortest possible time? Nobody knows but speculations called the Planck length and the Planck time are extraordinarily small, far smaller than current measuring capabilities.

Finally, look at the following Q&A for another example of the role of Planck's constant.

QUESTION:
in quantum world, i know that particle doesn't have a definite position before measurement as per copenhagen interpretation, but what about momentum and energy, does a particle have definite momentum(or energy) before measururement?

ANSWER:
In quantum mechanics there are what are called conjugate variables; these are pairs of observable quantities which are coupled together such that you can never know the exact value of either one. This leads to the famed Heisenberg uncertainty principle. On such pair is position x and linear momentum p, and the product of their uncertainties cannot be smaller than the rationalized Planck constant, h/(2π). So, the more accurately you measure either, the less accurately you can know the other. Another pair are energy E and time t. So, for example, if a radioactive system has a half life of Δt, the most accurately you can measure the energy is ΔE>[h/(2π)]/Δt.

QUESTION:
if action and reaction are always equal in magnitude and opposite in direction, why don't they always cancel each other out and leave no net force for accelerating a body?

ANSWER:
Because they act on different objects. If object A exerts a force F_AB on object B, object B exerts an equal and opposite force F_BA=-F_AB on object A. F_AB tends to accelerate object B and F_BA tends to accelerate object A.

QUESTION:
if action and reaction are always equal in magnitude and opposite in direction, why don't they always cancel each other out and leave no net force for accelerating a body?

QUESTION:
I have a 4ft long ramp, 1ft tall at the take off. My 3 year old hits this on his bike and clears 4ft when he lands. How fast is he going at take off?

ANSWER:
You may not be interested in the details, but I will include them anyway for the interested reader. The equations of motion for horizontal and vertical position are:

4=vtcos14.5=0.968vt
0=1+vtsin14.5-½32t²=1+0.25vt-16t²

These are two equations for two unknowns, v and t, and can be solved using simple algebra. The solutions are t=0.563 s and v=7.33 ft/s=5 mph.

QUESTION:
Consider 2 cases:
Case1: a moving train with constant velocity of v hits a stationary person.
Case2: a person running with constant velocity of v and hitting a stationary train.
In which case is the damage to the person more severe? And by how much?

ANSWER:
What determines "damage"? The laws of physics, in particular Newton's laws of motion. Newton's laws are Galilean invariant. What this means is that the laws are exactly the same to two observers who are moving with constant velocity relative to each other. In your question, Case2 is the same as Case1 if viewed by an observer running along the train with speed v. There is no difference in the "damage".

QUESTION:
I read that during the 70 mega ton atomic detonation, the blast wave was felt 300 miles away in L.A. How is this possible since the Earth is curved, wouldn't it pass over them?

ANSWER:
OK, the earth has a radius of about 5,000 miles. The angle θ subtended by 300 miles is θ=300/5000=0.06 radians=3.44⁰. Looking at the diagram, cosθ=R/(R+d)=[1+(d/R)]^-1≈1-(d/R); therefore d≈R(1-cosθ)=9 miles. Now, you are probably thinking that all the sound waves move in straight lines, that the sound waves coming from the detonation are radiating out as shown in my second figure thereby missing the earth's surface except very close to the source. But, since sound is composed of waves, it is subject to diffraction. Huygen's principle says that the wave a short time later is determined by treating each point on the wave front as a new point source. Finally, I have drawn one of the wave fronts which is off the earth in the earlier picture and how Huygen's principle predicts the wave a short time later. Note that the wave where it originally ended is now bending and lengthening; this is diffraction and allows the sound to follow the contour of the earth. This is the same principle where when you speak to someone on the other side of a sound-absorbing wall with an opening in it (but she is not visible to you), you can be heard just fine; sound diffracts around corners.

QUESTION:
Is water pushed down a river or sucked down by water in front of it?

ANSWER:
To understand what is happening, you need to focus on one tiny volume of the water and find all the forces on it. I will focus on the center cube of water in the figure; these cubes are, of course, imaginary but if I can understand the motion of one tiny piece of the river I understand the whole thing. There are three forces on this cube which have components pointing in the direction of the flow: the weight mg of the cube, the force f_down which the upper cube exerts on it in the direction of the flow, and the force f_up which the lower cube exerts on it opposite the direction of the flow. Notice that all cubes exert pushing forces on their neighbors, not pulling, so "sucked" is never what is going on. Now, the cubes are flowing downstream with constant velocity, so the sum of all forces in that direction must add to zero (note that only the downstream component of the weight is mgsinθ pushes the water downstream): f_up-f_down-mgsinθ=0. So, f_up=f_down+mgsinθ. So, you see, gravity and the water upstream push the water downstream while the water downstream tries to hold the water back. There is a greater force upstream from the water itself than there is a force downstream. Without gravity or with no slope to the riverbed the water would not flow at all.

QUESTION:
What force of magnet do I need to suspend a 36 to 40 inch plastic ball 2mm thick and weighing less than 15 pounds? I am creating a home art item to be suspended about 4 feet off the ground.

ANSWER:
You need a magnet which can exert a force of 15 lb; I would suggest building in a safety factor of 5 lb or so. If you go to a supplier, the catalog should tell you the force rating for each magnet. Here is a nice explanation of how magnets are rated.

QUESTION:
I'm in the middle of watching a show that deals with a massive hole in the ground that the people in the show call "The Abyss". They don't know how deep it is. I was wondering if you were to travel into a hole like this would the Atmospheric pressure increase or decrease?

ANSWER:
What is atmospheric pressure? It is the force (per unit area) which the air pushes on you. This force is the weight of all the air which is above you. When you go up a high mountain, the pressure decreases because there is less air above you than when you are at sea level. If you are deep in the earth there is more pressure on you because there is more air above you than when you are at sea level. Here is a real-world example.

QUESTION:
If you have a basketball that is filled with air, and a basketball filled with cement and you are in a weightless environment, and you throw each one as hard as you can; would they each travel the same distance?

ANSWER:
There is no gravity, but is there anything else? Like air? Let's assume you are in empty space when you do the experiment. "…as hard as you can�" implies that you will exert the same force (your best) on each over the same distance (the length of your arm). But the cement ball has much more mass (inertia) so the effect of the force will give it a velocity smaller than the air ball would get, and both would move with constant speed in a straight line. The air ball would go faster, not necessarily farther.

QUESTION:
This is a practical but I think possibly a complex physics question. I like to climb mountains. On the descent when the grades have reduced to walking speeds but still require retarding input, perhaps a 10 percent downhill grade, my experience indicates to me that it requires less energy to descend at for example 4 mph than 2 mph. Am I just imagining this?

ANSWER:
The human body is very complex and I doubt there is a simple physics answer to your question.

FOLLOWUP QUESTION:
Thank you for your answer, which I am sure is right on the money. One follow up question if I may. Would a ball rolling down a frictionless ramp, 10 degree grade, require less retarding force to obtain zero acceleration at 4 mph as opposed to 2 mph? Yes? No? Maybe?

ANSWER:
First of all, a ball will not roll down a frictionless ramp, it will slide. But that does not matter regarding your question. The unbalanced force which accelerates the ball down the ramp is the component of its weight along the ramp; if the weight is mg, that component is mgsinθ where θ is the angle of the ramp, 10⁰in your case. The force pointing up the ramp which must be applied is therefore mgsinθ. Note that the speed is irrelevant.

QUESTION:
Since Dark Matter/Energy makes up most of the known universe, how can we know if the light reaching us from distant parts of space is an "unaltered version"? More specifically, what are the chances that light reaching us has been gravitationally lensed (along the way), giving inaccurate trajectories to the stars and to their actual positions in the sky? (And if this error cannot be corrected, does that mean that much of todays deep-space astronomy could very well be flawed)?

ANSWER:
First, unless the mass is very large and the light passes very close to it, deflections will be very small. There are numerous examples of gravitational lensing. See an earlier answer. I am sure that astronomers are perfectly aware that masses will alter the position of the images they observe relative to the objects they are observing are not perfect. However, the position recorded is simply the position in our sky at this time. Positions of objects can change simply because they are moving.

QUESTION:
Physics in schools teaches two contradictory and mutually exclusive things: (1) That the upward lift force on an airplane in flight equal its weight (Lift = Weight = mass x gravity). This is based on applying Newtons 2nd law of motion (F = ma) to the airplane in flight. (2) However, modern commercial airplane like the Boeing 747�400 can fly with thrust-to-weight ratio as low as 0.3. Here the engine thrust is only 0.3x the weight of the airplane, but this thrust is sufficient to push the airplane forward and generate enough lift to fly. Therefore the upward force required for lift and flight must be a lot less than 0.3x the weight of the Boeing 747-400. (Lift < Thrust < Weight). Both statements cannot be true. Which is correct?

ANSWER:
You are assuming that all the lift comes from the thrust. That is true for a rocket but not for an airplane. The thrust generates very little lift; its main function is to provide horizontal force to overcome drag and accelerate. Lift for an airplane comes from the way that air passes over the wings; see this link.

FOLLOWUP QUESTION:
Can I follow up your response? If so: The principle of conservation of momentum means that your response is impossible. Upward lift force canot appear out of nowhere, for example by air passing over a wing. In this case, lift will arise due to the wing converting horizontal wing airflow (thrust) into vertical lift. I also note that you did not provide a scientific / reliable source as evidence for your response, but an uncredited / verified popular science video on youtube.

Kind regards

ANSWER:
Kind regards? After such a rude and ignorant response? Really? Conservation of momentum is only true for an "isolated system". If a cannonball is fired horizontally from a cannon on a nearly frictionless track, the momentum of the ball is clearly not conserved; but the ball is not an isolated system, there was the cannon also. The ball acquired momentum but the cannon recoiled backwards and acquired an equal amount of momentum in the opposite direction; momentum was zero before and after the cannon was fired. If air passing over a wing starts horizontally and ends up with a downward component, it does not mean that vertical momentum has materialized out of nowhere, it means that something exerted a downward force on it—the wing. The air is not an isolated system because the wing is exerting forces on it. But if we take the wing and air together as an isolated system (approximately), then Newton's third law says that if the wing exerts a force on the air, the air exerts an equal and opposite force on the wing and that force is called lift. Lift is, as you would probably figure out, the force which lifts the airplane. Regarding "scientific/reliable" and "uncredited/verified", I am not writing an encyclopedia here where footnotes and references are required. I am presenting myself as someone who has a degree of expertise in physics who can answer questions about the subject; it is your responsibility, since there can be lots of BS on the internet, to convince yourself that I do indeed have some physics credibility. To that end I post information about myself, on the page "The Physicist" where you can get the information you need to judge my qualifications. Incidentally, the reason I chose the video which I did was because the Bernoulli aspect of wing lift is presented as the whole story in a great many introductory physics textbooks and only when I was taking flying lessons years ago did I learn that "angle of attack" is much more important in actually creating lift; I like the way this video presented both of these important concepts.

QUESTION:
I am writing a sci-fi book based in the future. I would like to know what would happen to the other planets, such as Venus and Mars, if Earth was to be destroyed by a massive implosion.

ANSWER:
The effect would be minimal.

QUESTION:
Would supporting a barbell on your back weighing equivalent to your body weight be the same as feeling 2G forces downwards?

ANSWER:
No, it would not be the same. Look at your head, for example. In a 2G environment it would be twice as heavy and it is not if the weight is on your back. The only part of your body which would be the same are your feet which have to support twice your normal weight in either case.

QUESTION:
What force does a 1 pound rainbow trout swimming at 23 mph exert on a fishing line. I'm thinking about the shock exerted on a fishing line as quoted breaking strains are irrelevant when a trout strikes, or bolts at speed. I've read that a trout only takes one second to reach maximum velocity.

ANSWER:
(I will work in SI units, preferred by scientists, but will convert to Imperial units, lb, when forces are calculated.) What determines the average force is the time Δt over which the fish is accelerating and the change in linear momentum Δp=mΔv; F=Δp/Δt. In your case, the mass is m=1 lb=0.45 kg and the speed is v=23 mph=10 m/s so the change in momentum is 4.5 kg⋅m/s if the fish starts at rest and accelerates to 23 mph or vice versa. I will assume that the line does not break and then compare forces with the pound-test of the line; then I will compare with a four pound test leader. If your rod had zero flex, was perfectly rigid, the fish would be unable to move at all unless the line broke, so assuming that he could exert a force of four pounds or more which he probably could, you would lose that fish. If he hit the fly going 23 mph and stopped in 0.1 seconds, the average force would be 4.1/0.1=41 N=9.2 lb—bye bye fish! But, that is unrealistic because what really happens is that the rod bends and what that does is lengthen the time the strike lasts; that means that the force on the line will be smaller. Suppose that it takes 2 s for the fish to get up to his maximum speed while hooked; during that time the average force exerted will be 4.5/2=2.25 N=0.51 lb, and you have caught that fish. Now you understand why fishing rods are designed to bend.

QUESTION:
Can a bridge made with two I-beams hold more weight than it's designed for if the object moving over the bridge goes at a high rate of speed where the weight load is just seconds?

ANSWER:
I cannot make any recommendations; if I was wrong, I might be held responsible. Engineers always make the structures they design to be stronger than the specified maximum load; for example, if the maximum specified load is 2000 lb, it might be 2500 lb before the structure started to fail. And the design, I suspect, always is made assuming static equilibrium, so a very brief load above the maximum might very well not cause the bridge to fail; nevertheless, I would be wary.

QUESTION:
While metal detecting I found a strip 1/4� thick 2� wide by 4� long I thought to be lead as it is heavy! Upon doing the marking test it does not write or mark anything! It is pliable and weighs 156grams It is also non magnetic Could it be silver or other valuable metal?

ANSWER:
I calculate the density to be about 4.76 g/cm³. The closest metal to this is germanium with a density of 5.32 g/cm³; but germanium is brittle and would not be pliable. Lead has a density of 11.34 g/cm³ and silver 10.49 g/cm³, both about twice as dense as your sample. So it must be something else. To sort it out would require a chemical analysis, I guess.

QUESTION:
Why does an astraunaut experience weightlessness in outer space?

ANSWER:
Because he is in free fall all the time when he is in orbit. If he is just in space, far from any sources of gravity, it is because weight is the force which earth exerts on you and there is no such force in space.

QUESTION:
Why are there so many types of baryons but the only nuclides that form atoms ( and I suppose molecules too for that matter ) are protons and neutrons?

ANSWER:
The simple answer is that the proton is stable and the neutron is almost stable, so they are the only baryons which hang around long enough to contribute to the formation of nuclei. The more complicated answer is that, once bound in a nucleus, they tend to somewhat lose their identity and form a "quark-gluon plasma" from which many different types of baryons can pop into and out of existence.

QUESTION:
How fast does a 3000 pound vehicle have to go to launch the vehicle in the air and travel 130 ft. I know the angle of the ramp you would have to know and I don't know that so use the optimum. My grandson was driving a 2014 Mazada 6 and lost control and went in the ditch and totaled the car The police officer said he went in the grass on a rural road hit a driveway and launched him a 43 yards

ANSWER:
Assuming he lands on the same level as he launched and that air drag is negligible, the range is R=v²sin(2θ)/g where R=130 ft, v is the speed, θ is the launch angle, and g=32 ft/s2 is the gravitational acceleration; the "optimum" launch angle is 45⁰ for maximum range. So v=√(130x32)=64.5 ft/s=44 mph. For a smaller launch angle the speed would be larger; e.g. at 30⁰, v=50 mph.

QUESTION:
Could a spaceship traveling very,very fast suddenly stop and make a reverse without harming the occupent inside if somehow the occupent was suspended in a liquid,or a jelly substance?

ANSWER:
If the space ship experienced a very large acceleration, greater than on the order of 50 Gs, you would have to cause the acceleration experienced by an occupant to be much smaller. This could be achieved by making the change in velocity occur in a much longer time than for the ship. I suppose that being suspended in a viscous fluid could achieve this provided that the tank is large enough; if the tank were not long enough, you would smash into the end before you had lost enough speed.

QUESTION:
Hello, my question involves interstellar navigation. Assuming that such travel becomes a reality, how would one navigate accurately to a target star? We see a past instance of a star (or other distant object) based upon the light that can be observed from our distance to it. That object may no longer exist. If you were targeting the object by line of sight, it may not exist once you get to its location. I am assuming that light alone would not be enough for accurate and reliable navigation. Also, there is distortion to consider, so that, the object might not necessarily be in the observed location. What mechanisms would compensate for this? Also, since we see an older representation of an object due to the distance the light has to travel, how close must one be to an object before the image becomes "current", or, real-time?

ANSWER:
For a lengthy discussion of the problems of space navigation, see an earlier answer. The issues you raise could mostly be handled by making a judicious choice of target. Stellar evolution is reasonably well understood and the life time and age of a star could be decuced from its properties. For example, if the star you have in mind is within a million years of running out of fuel and is a million light years away, it would foolish to plan a journey there. Regarding distortion, the earlier answer emphasizes that problem but the most important distortion is that caused by the high speed of the vehicle itself; "distortion" due to motion of the target star itself as well as gravitational lensing could be adjusted for in flight if you could actually see and lock onto the star at high speeds. Your last question has no answer because you need to specify how small a lag you would call "current"; we are quite close to the sun but light from it takes about 8 minutes 20 seconds to get to us.

QUESTION:
How long does it take one ounce of water to hit the ground from twelve feet high, in Denver vs Los Angeles, in June?

ANSWER:
A very peculiar question! The only thing which will make the answer different is a significant effect from air drag which will be greater in LA because the density of the air is higher there. I did a rough calculation assuming the water fell as a sphere. The terminal velocities were 29.54 m/s in LA and 31.90 m/s in Denver. If there were no air drag the speed the water would have when it hit the ground would be about 8.5 m/s; since the water is never anywhere close to the terminal velocity, air drag can be neglected. Extraordinarily accurate measurements would find that the time is shorter in Denver but it is a pretty involved computation to get those times and the theortical differences are only approximations anyway.

QUESTION:
When a ball is thrown into the air, how long does it spend at its maximum height, not moving up or down? Some friends have told me the change from upward motion to downward motion is instantaneous. If that is true, how would it be possible to take a picture of the ball (or catch it) at its maximum height considering that zero time would be spent there?

(To give a background on why I asked the question. I have a habit of clicking my pens on my desk so they shoot a few inches into the air. I always try to catch them right at the height of their motion. I started wondering how long I actually had to do that. When I asked some engineering friends they said the pen instantaneously goes from traveling up to falling down. I�m not sure if that�s true but if so it introduces a bit of a crisis. If the pen spends zero time at its maximum height, is it ever actually there? And would I ever be able to actually catch it at is maximum height? Can something be somewhere but spend zero time there?)

ANSWER:
According to classical physics, it spends zero time at any particular velocity, including zero. So, is it ever somewhere if it is only there for a zero amount of time? Of course it is, and I see no "crisis"; the fact that it spends zero time at one location simply means that it is "passing through". Only the center of gravity of your pens will necessarily be at rest. You can take a picture of your pen over the time interval of your shutter and, because the speed is very small during the times just before and after the maximum height, you can get a pretty clear picture with a pretty slow shutter speed. When the pen is moving faster, you need to use a faster shutter speed.

To get more technical, what is the probability of finding the object at a particular point? It is zero. That is probably what is bothering you. So you could ask for the probability of catching the pen during the interval 1/1000th of a second before and after the zero speed time. I will not get into quantum physics here, but you cannot know precisely both the speed and location of your pen at any time; but this truth implies an incredibly small uncertainty, certainly not important for a macroscopic object like your pen. One final thought, we do not know whether or not there exists a shortest possible time or length, sometimes referred to as the Planck length and the Planck time; if it exists, the Planck time is expected to be about 5x10^-44 s, so perhaps your pen stands still for that time!

QUESTION:
If a small nail from little distance is thrown towards a man, it hurts. But why don't we feel hurt if a rain drop falls on our head with more speed and from long distance

ANSWER:
The terminal velocity of a raindrop is about 20 mph. You could probably throw a nail with a speed greater than 50 mph. Perhaps even more importantly, the force which you feel when something strikes you is inversely proportional to the time the collision lasts—the longer the collision lasts, the smaller the force. The nail, being solid, stops very quickly and all the force is on a very small area, so there is a large pressure over that area. The raindrop spreads over a larger area and takes longer to stop, so the smaller force is spread over a larger area. (Read the next question and answer to see what terminal velocity is. You also can find terminal velocity on the faq page.)

QUESTION:
In uga physics class we learned that you can go up in the sky and drop an object. It will accelerate up to about 120 mph, and then it doesn't get any faster. If you drill a hole from the surface of the earth down to the center of the earth, and then drop the object down the hole, how fast of a speed will it get up to?

ANSWER:
Actually, your instructor was getting you to see that, because of air drag, objects dropped do not continue accelerating at 9.8 m/s² until they get all the way to the ground. The air drag is a force on the falling object upward (opposite the velocity) which gets bigger as the object falls faster; eventually the drag force becomes equal in magnitude to the weight which is a force down, so the object falls with constant velocity after that. That speed is called the terminal velocity and that depends on both the mass and the geometry of the object. The speed of 120 mph is the approximate terminal velocity of a falling person. A bowling ball would have a much larger terminal velocity and a feather a much smaller one. If the earth had no atmosphere, anything would continue accelerating until it hit the ground. If you want to learn more about air drag, you can find links to many of my older answers on the faq page. If there were no air in the hole you want to drill and the earth were a uniform sphere (which it isn't), the speed would be about 18,000 mph at the center.

QUESTION:
Are there different types of plasma? What I mean is, do chemical differences (i.e., water and mercury) result in different plasmas just as they make different liquids? The largest variation I have yet to see in plasma is between the blue electrical stuff in desktop toys and the pink haze in the core of that fusion reactor.

ANSWER:
Plasma is a state of matter, a vapor with a significant fraction of the molecules ionized. Of course a plasma depends on what it is composed of as well as other distinguishing properties such as the average fraction of singly ionized molecules, doubly ionized molecules, etc. Is a block of iron a "different type" of solid from a block of tin? Yes. Similarly, a plasma composed of iron ions is a "different type" of plasma than one composed of tin irons.

QUESTION:
So, I am a bit of a martial artist and I fancy my self a bit of a scientist. However I am studied in the fields of anatomy and biology, not physics. So my question is this. If I were to punch or kick a person, would more force be transferred through my blow into their face if my whole body was in mid air or would there be just as much force or more if I leave my feet grounded. Please feel free to leave a lengthy answer.

ANSWER:
Applying simple physics to situations involving the very complex human body can lead to incorrect conclusions, so my response should be taken with a grain of salt. The simplest thing to try (and I mean really simple) is to compare collisions between two objects which has one at rest and the other moving with speed v. I will assume that the time which the collision lasts, Δt, is the same for any situation I consider. There will be two extremes for a collision—perfectly elastic where energy and linear momentum are conserved, and perfectly inelastic where the objects are stuck together after the collision and energy is not conserved but linear momentum is. The impulse received by the object originally at rest (your opponent) is equal to the linear momentum he has after the collision and is equal to FΔt; since I specified that Δt is the same for all collisions, the force experienced is F=mv'/Δt, where v' is the speed of the struck object after the collision. First consider a collision between two equal masses m. For the elastic collision the incoming mass is at rest after the collision and the outgoing mass has a speed v; for the inelastic collision both masses (stuck together) have a speed ½v. Therefore the force experienced by your opponent is given by ½mv/Δt≤F≤mv/Δt. Now suppose that your mass M is much larger than your opponent's mass m, M>>m. In this case, the inelastic collision will have both objects moving with approximately a speed of v; the elastic collision will have M moving approximately with the speed v and m moving approximately with the speed 2v. Therefore the range of possible values for F experienced by m is mv/Δt≤F≤2mv/Δt, which is twice the force for equal masses. So my simplistic view is that being anchored to the floor is like increasing your mass allowing you to deliver more impulse to your opponent as compared with if you are flying toward him which is more like the collision of equal masses. As I said, it is much more complicated than this but this might give a little insight. I would welcome any comment on this treatment. (Lengthy enough answer for you?)

QUESTION:
If a gyroscope is undisturbed, possible mounted in a 3-axle gimbal the direction of the angular velocity vector remains constant? That makes them useful for navigation. But, the angular velocity vector is constant relative to what? Not a point on the earth, or in the solar system, or the galaxy? What point does it remain fixed upon? Does it define some absolute location within the ponderable mass of the universe?

ANSWER:
You can point it in any direction you like. Now, as long as the gyroscope encounters no torques, it will remain pointing in that direction.

QUESTION:
DEAR SIR WE HAVE A PRECAST CONCRETE MANUFACTURING UNIT IN COIMBATORE. NOW WE ARE IN A NECESSITY TO CAST A 7 METER BY 4 METER CONCRETE SLAB. THICKNESS 150 MM . SO THE TOTAL WEIGHT COMES AROUND 10.5 TON. MY QUESTION HERE WHAT WILL BE THE FORCE ACT ON THE BOTTOM SURFACE IF WE LIFT THAT SLAB. WHY I ASK THIS IS , WE USE RUBBER MATERIAL BETWEEN THE MOULD AND CASTING FOR THE SMOOTH FINISH . SO WHEN WE LIFT THE RUBBER MATERIAL COMES WITH THE SLAB INSTEAD OF STICKING TO THE MOULD. SO IF I KNOW THE PRESSURE THEN I CAN DECIDE HOW TO FIX THE PROBLEM.

ANSWER:
The pressure is the total weight divided by the area. The total weight is (10,5000 kg)x(9.8 m/s²)=102,900 N and the total area is (4 m)x(7 m)=28 m². So the pressure is 3675 N/m². Other units used for pressure which you might prefer are 375 kg/m² or 76.75 lb/ft².

QUESTION:
Why is gravity considered a Fundamental Interaction? I would think that gravity is a feature of the geometries of space-time rather than a fundamental force of nature like the Weak, Strong and Electromagnetic Interactions.

ANSWER:
I believe that the main reason is that most theorists feel that any field should be quantized and that general relativity is a field theory. Although attempts to find a theory of quantum gravity have been unsuccessful, it remains one of the holy grails of physics. The best explanation to reconcile these two viewpoints may be seen in an earlier answer (skip down to the last paragraph).

QUESTION:
What is the reason of collapsing of an empty plastic bottle when air is sucked out of it?

ANSWER:
Before evacuating the bottle, atmospheric pressure, about 15 psi, acts both inside and outside the bottle, so the net force on the bottle is zero. When you take the air out, there is now a pressure of 15 psi on the outer surface of the bottle which crushes it.

QUESTION:
When a charge radiates energy,does it radiate in all directions at the same time? (like when a pebble is dropped in water and it creates wave in all directions)

ANSWER:
A charge will radiate when it accelerates. Electrons oscillating back and forth in a wire radiate; this is what an antenna is. What the radiation pattern is depends on how the charges are accelerating. One of the simpler systems is when you have an electric dipole p=qd, equal positive and negative charges ±q separated by some distance d; imagine these as vibrating sinesoidally along the direction of the vector d with some frequency. They will radiate most strongly in the direction perpendicular to p and radiate not at all along the direction of p. A nice animation of this can be seen here.

QUESTION:
On page 114, in Hawking's book, (Brief Answers to Big Questions) it is stated that a chemical rocket with a thrust velocity of 3 km/sec, that jettisons 30% of its fuel, attains a velocity of 'about half a km/sec'. Using Tsiolkovsky's rocket equation I get a value of 1.07 km/s. Have I made a mistake?

ANSWER:
I have used this equation in an earlier answer, v=uln[(M+m)/M] where M+m is the initial mass, m is the fuel burned, u is the exhaust speed, and v is the speed acquired by the rocket. So I find v=3ln(0.7^-1)=1.07 km/s. I do not know how Hawking came up with his answer, but it might hinge on what he means by "30% of its fuel"; you and I have interpreted that to mean the exhausted fuel has a mass equal to 30% of the mass of the rocket before the fuel was burned, hence M/(M+m)=0.7. Also, maybe he was thinking about a launch from earth which would require some of the energy from the rocket to overcome the gravitational force and the energy loss from air drag. Or maybe he was guesstimating the effects of efficiency of the engines. In any case, you did not make a mistake.

QUESTION:
What would happen if a plastic bottle was taken up in the air 10 000 meters?

ANSWER:
At sea level the pressure both inside and outside the bottle is about 15 psi, so the net pressure is zero. At 10 km the pressure outside drops to about 4 psi, so the net pressure is about 11 psi. One site told me that, for the plastic in a 2 liter bottle to fail, the net pressure would be about 150 psi. So, nothing much would "happen". Actually, since the temperature at 10 km is about -50⁰C, if the air in the bottle chilled to that temperature, the pressure in the bottle would be smaller than 15 psi. Using the Gay-Lussac Law, P/T=const., I find the pressure would be about 11 psi, a net of about 7 psi. (If you want to check this, you must use absolute temperature, not ⁰C.)

QUESTION:
what happens to a satellite when it reduces its mass, or to any rotating object in an orbit? Is angular momentum conserved? velocity of the satellite does not depend on its mass (gravitational force = centripedal force) so if the velocity doesn't change when mass is reduced, how is angular momentum conserved, does the radiues change? if angular momentum is not conserved, what causes a net torque?

ANSWER:
Mass does not just disappear. If the satellite jettisons some mass, that mass will carry away both linear momentum and anglular momentum. Depending on how the mass was ejected, the orbits of both parts will now likely be different from the original orbit.

QUESTION:
If I'm holding a flashlight and I'm running with it turned on, does the light from it move at my speed + the speed of light?

ANSWER:
No, the speed of light is a universal constant. Its value is completely independent of the motion of the source or the observer. On my FAQ page you will find many links to earlier answers which relate to this.

QUESTION:
Why does the smoke of a burning incense stick first follow a straight path and then disperse?

ANSWER:
The answer is sort of conceptually simple but quantitatively very complicated. Just about anything involving fluid dynamics in the real world is pretty difficult to calculate quantitatively and accurately. The simple concept is that a fluid, under the right circumstances, is laminar which means that if you focus your attention on a small volume of the fluid it will move along a smooth line and be followed and preceded by many other small volumes following the same smooth line. Under other circumstances, the fluid will become turbulent, moving chaotically. So something happens which causes the smoke from your incense stick to change from laminar to turbulent at some point. This is usually quantified by a dimensionless variable called the Reynolds' number, Re=ρuL/μ where ρ is the density of the fluid, u is its speed, L is a length which describes the geometry, and μ is the dynamic viscosity of the fluid. So, Re changes in such a way that the original laminar flow becomes turbulent after rising a certain distance from the source of the smoke.

QUESTION:
How is buoyancy affected by zero gravity? Specifically, if I put a steel ball-bearing in a jar of water in zero g, where will the ball bearing rest in the jar?

ANSWER:
Buoyancy is a purely gravitational effect, does not exist in zero gravity. Your ball bearing will stay exactly where you put it in the water. (One proviso is the equivalence principle which says that there is no experiment you can perform to determine whether you are in a uniform gravitational field with a gravitional acceleration of a or in zero field but having a uniform acceleration a; buoyancy would be observed in an accelerating rocket ship in empty space.)

QUESTION:
This is not a homework question, I'm just trying to sharpen my knowledge for my ap test on Tuesday. The question puts two identical blocks, block A and B. These blocks have the same mass and velocity, except Block A's velocity is directed straight down, where Block B's velocity is directed right or parallel with the ground in the positive x direction. The answer, according to Khan Academy, is that both blocks will hit the ground with the same speed, but I just can't wrap my head around this since Block A will have the advantage of an initial velocity when using the kinematic vf^2=vi^2 + 2ay, whereas block B's velocity wouldn't count for Vi since it is perperndicular to the accerleration. Please help me to understand.

ANSWER:
The easiest way to do this is with energy conservation. E_i^A=½mv_i²+mgh, E_f^A=½mv_f²and so v_f=√(v_i²+2gh). Exact same calculation for E^B.

QUESTION:
Let's assume a rocket (or flying saucer) could hover above an atmosphere-free planet - (that avoids all the added concerns that air would add for a helicopter). I'm asking whether one might consider the work done by thrust and gravity to be offsetting work that results in no Net Work. Also, I believe thrust adds KE at the same rate gravity converts KE to PE when traveling up at Constant Velocity (or when at hover). Am I correct in assuming that once you reach constant velocity, you do no NET work as long as you maintain CV (again, noting that the work done by thrust offsets the work done by gravity)? Lastly, I often see the comment, "Work adds energy to the object", yet if we drop an apple, gravity does work on it (W = f x d and there is a displacement in the direction of gravity). But gravity adds no energy of course - it simply converts KE to PE (when traveling UP) and PE to KE (when the object is traveling down). Should the definition of work be expanded to include energy conversions?

ANSWER:
For physics at this level there is no need to introduce potential energy at all; I always tell my students that PE is a clever bookkeeping mechanism to keep track of the work done by some force which is always present (like gravity in your examples). (It should be noted that you cannot introduce a PE function for some forces, but gravity is not one of them.) In your flying saucer example, no work is being done by either gravity or thrust because neither are moving and so d=0 in your (simplistic) W=f x d definition of work. If the saucer is rising with a constant speed v, then when it has gone some distance d, the thrust T has done W_saucer=Td and gravity has done work W_gravity=-mgd because T=mg for it to be rising with constant speed. The net work W on the saucer is still zero, W=W_gravity+W_saucer=0. Your last question "…simply converts KE to PE…" is again using PE unnecessarily. Gravity does work on an object as it falls because there is a force mg acting down.

QUESTION:
Which trip takes longer when you throw a ball upwards,the trip down or the trip up?

ANSWER:
On the trip upwards, the air drag and the weight are both down, whereas on the trip down the weight is down but the air drag is up. So the speed decreases more rapidly going up than the speed increases on the way going down. It follows that the trip up will be shorter than the trip down.

This is the most popular way to explain the answer without resorting to very opaque mathematics, but I find it just a bit unsatisfying even though it is certainly correct. I found a more convincing argument put forward by Eltjo Paselhoff on the website quora.com. Suppose we plot the velocity v as a function of time t for the upward trip. At t=0 the magnitude of the slope must be larger than g because the air drag is slowing the ball down as well. As the ball gets higher, its speed decreases so the air drag gets smaller so that finally at the top of its trajectory it is at rest and the magnitude of the slope is equal to g. Now, the ball falls back down and speeds up because air drag is smaller than weight; eventually, though, the two forces become close to equal and the speed becomes constant at the terminal velocity v_T. One more thing you must understand now is that the area under the velocity curve is the distance traveled by the ball, so since up and down go the same distance the green and red areas need to be the same. The only way this can be is if the time to fall is larger than the time to rise.

QUESTION:
In a tug of war, if a team exerts a force on the second team, then the second team exerts an equal and opposite force one the on the first. Then how is it possible that one team wins?

ANSWER:
To simplify things, each team is one person. How does the red-shirt guy exert a force on the blue-shirt guy? Via the rope. So, the forces which the two exert on each other are equal and opposite because of Newton's third law (F=-F in the figure). Are there any other forces on the guys? Yes, each is pushing against the ground using the friction between their feet and the ground. I have drawn it so that the guy in the red shirt is able to exert a bigger frictional force on the ground than the guy in the blue shirt, so the net force on everything is to the right—the guy in the red shirt wins. You could think of this happening because the red-shirt guy has stronger legs or because the ground under the blue-shirt guy is slipperier. (Incidentally, the lower case fs are the forces which the ground exerts on each guy; not shown are the equal and opposite forces which the guys exert on the ground, Newton's third law again. Similarly, the uppercase Fs are the forces which the rope exerts on each guy, not the forces which the guys are exerting on the ropes. In other words, the only forces shown are the forces on the guys, both of whom will accelerate to the right.)

QUESTION:
I'm wondering why there is no drag in frequency response in an electromagnetic loudspeaker. Given a fixed input frequency, say 1KHz audio sine wave, given the friction of the air, the inherent mechanical resistance in the paper of a speaker cone and the constantly changing momentum of the vibrating cone, why isn't the frequency of the sound coming out lowered some amount by all those factors? This is something that has puzzled me for a long time. BTW, I have a B.S. in math and have taken college and univ. physics, so you can go into a little depth if you must.

ANSWER:
This is a classic intermediate-level classical mechanics problem, the driven damped oscillator. Your having been a math major should make it pretty straightforward to follow the details. I found a very good web site which gives the details of the solution of the physics problem but is still quite readable. For the casual reader, I will include here an overview. Newton's second law is an inhomogeneous second-order differential equation. What makes the equation inhomogeneous is the presence of the driving force, the signal your amplifier sends to the speaker; with no driving force, the solution of the problem would be x_t(t)=A_te^-γtsin(ωt+φ_t) where ω is the natural frequency the speaker would vibrate with on its own, γ is a parameter which describes the damping of the speaker, and A_t and φ_t are constants determined by the initial conditions. If the driving force is present, F=F₀cos(ω₀t+ψ), the solution is a linear combination of the what is called the steady state solution, x_s(t)=A_ssin(ω₀t+φ_s), and the solution for the undriven oscillator x_t(t) which is referred to as the transient solution: x_t(t)=A_te^-γtsin(ωt+φ_t)+A_ssin(ω₀t+φ_s). The constants A_t and φ_t are determined by the initial conditions (speed and position of the speaker) and the constants A_s and φ_sare determined by the driving force (F₀ and φ_t). Because of the damping term e^-γt, the transient term dies out in a very short time (assuming the speaker has been sensibly designed) and you are left only with the term which has the same frequency as the output from the amplifier.

Below are shown a couple of examples: The figure below shows the transient (red), steady-state (blue), and total (black) solutions separately.

These show the total displacement for different initial conditions. In both cases the speaker started at rest. The first shows the starting position at zero, the second shows the starting position at 0.5. The important thing is that they both end up, after the transient has died, as the same.

QUESTION:
I have been searching for help/answers with no success, so here goes: My dog competes in AKC dog diving distance jumping. Yesterday�s event my dog was jumping a distance of 21�, (basis being where the base of the dogs tail hits the water), with the height of the dock being 16� over the water. However AKC rules state the dock height is supposed to be 24�. Seems logical to me that the increase in dock height would increase the distance to impact. If not, ok, and sorry for wasting your time. If so, can you tell me how you would calculate that, and what the increase would be?

ANSWER:
There are three important factors which determine how far the dog will go: the height of the dock, the speed the dog leaves the dock, and the angle relative to the horizontal of his initial velocity. Of course, I have no idea what these are but if the dog goes horizonatlly 21 ft I have to assume that he has a running start, not from a standstill. Also, you probably know that if the dock were level with the water, the maximum distance for a given initial speed would be maximized if the angle were 45⁰. I found a little calculator you can play with; a screenshot is shown below. All you would need to vary are the velocity, the angle, and the height of the dock. I found that if I chose the initial speed to be 26 ft/s (about 18 mph) the distance the dog would go for a 45⁰ jump would be about 21 ft from a zero-height dock. Now, when I do the calculations for the 24 in=2 ft and 16 in=1.33 ft, I find the best angles and distances (for 26 ft/s) are (43.5⁰, 22.9 ft) and (42.5⁰, 22.3 ft), respectively. All in all, there is just about no difference.

Note that I have not included any air drag. I believe the speed of the dog is low enough to make drag negligible.

QUESTION:
We know that looking at the stars is like looking at the past because we are seeing the light that left the star many years ago. However, the faster something travels, the slower time passes for that object. So, while for that light 1.000 years have passed since it left the star, for us, here on earth, many more years have gone by. My question is: when we see the light from a star 1.000 light years away, are we seeing 1.000 years in the past or actually, that number is a lot bigger because of relativity?

ANSWER:
No, that is wrong. Light emitted from a star 1 light year away will arrive, using earth's clocks, in 1 year.

QUESTION:
Why does distance increase torque? I am aware of the equations, and how a lever behaves, but why does it behave that way? Why does me putting my force on a point farther away from the fulcrum multiply my force?

ANSWER:
I think it is easiest to think of a static situation where the system is not experiencing any motion whatever. Consider the little setup shown in the figure. A weightless stick of length D is hinged to a wall. A force F is applied down on the end of the stick; clearly this exerts a force which, if there were no other forces acting (except the hinge) would cause the stick to experience a clockwise angular acceleration, in other words a torque around hinge. Now suppose that we exerted a force T a distance d from the hinge; we would find experimentally that if T=F(D/d) the stick would not rotate. We would conclude that the net torque about the hinge is now zero and that this torque changes linearly with the distance. Of course, the torque exerted by F would have to be opposite that exerted by T. If we choose clockwise torques as positive, counterclockwise torques would be negative. Newton's first law for rotational equilibrium would be Στ=FD-Td=0.

This is not the most general case but I feel you wanted to be convinced that torque depends on where the force is applied relative to an axis about which you are computing torques, which was easy to see.

QUESTION:
Can someone please really answer the question about a 747 on a massive treadmill that is set up in a manner to match the speed of the wheels? This seems to be different then the Mythbuster's version and no one really just outright answers it from a physics position, and maybe because it is just simply it will take off. However, with the weight of the plane, the friction of tires and wheel bearings and the treadmill's speed changing I can only assume two scenarios, the plane remains in place because the engines cannot produce enough thrust to overcome the friction generated by the wheels spinning at a ridiculous speed (and the reality is tires would blow and bearings would seize) or the wheels and treadmill spool up to infinity. It just kinda sucks that no accredited organization has stepped up an answered this.

ANSWER:
I cannot, for the life of me, understand how this is apparently such a big mystery. I have answered this twice before, first and second times. In a nutshell, if there is not air moving over the wings with sufficient speed, there is not sufficient lift on the airplane for it to leave the ground. I agree that Mythbusters' version of this is not the same as I understand your question and previous questions where whatever the conveyer is doing, the plane is not moving forward through the air; the Mythbusters have the conveyer at rest relative to the plane but moving forward with the same speed as the plane, so the wheels are not spinning.

I guess it is possible that a propeller plane could take off from rest since the props send a wash of air over the wings, but probably not unless there were a pretty hefty headwind; you do see some planes, designed for flying into and out of tight places, which take off in a very short distance.

FOLLOWUP QUESTION:
Regarding the 747 Jet on a rapidly moving treadmill, I can only find your answer unfathomable. The four General Electric CF6-80s 44,700 lbs of thrust each or 178800 lbs forward thrust total would push the plane forward off the treadmill almost instantly. It's difficult to believe that the friction of the wheel bearings would/could be enough to hold it back. If that were the case, the 747 would be forever grounded and could never take off, which we know full well to be false. Furthermore, lift is not a factor because lift does nothing to negate the 178,800 lbs of thrust pushing the plane forward. Wheel bearing friction isn't going to stop this monster from going ahead. You'll need well over 100,000 lbs of counter-thrust to do that. I seriously doubt that you can spool up a treadmill fast enough to lock up the wheel bearings before the CFC-80s were brought to bear and force the plane off the treadmill, especially given the monstrous size and weight of the treadmill and mechanism that would be required to support a 747, loaded or empty. You seem like one of those physicists who hasn't had a lot of real world experience with tremendous forces and loads.

ANSWER:
My goodness, you do not need to get so upset that you have to be insulting! The problem here is that you and I are clearly looking at different aspects of this problem. You are more interested in the engineering aspects of whether it would be possible and/or practical to build such a treadmill and could the bearings in the wheels withstand the punishment they would have to take as the wheels spun faster and faster. I agree that it would be folly to try to build such a contraption. The issue which interests me is if this airplane does not move forward would it be possible for it to become airborne? As I suggested in all my answers, the treadmill is not necessary at all to discuss that question—just lock the brakes, give the engines full power and ask what happens. First, ask if you can get enough friction to hold the plane. The weight of a 747 is about 800,000 lb and the coefficient of static friction between rubber and dry asphalt is about 0.8, so the maximum frictional force you could get before the plane started skidding forward would be 0.8x800,000=640,000 lb, far greater than 178,800 lb of maximum thrust. So you could run at full thrust and remain at rest by just locking the brakes. So, now, your statement that "lift is not a factor" is nonsense because if the plane is to fly there must be a lift greater than its weight and I sure don't know where that lift will come from if there is not air passing over the wings. When you say that lift is not a factor, you are assuming that the plane has broken away and is roaring down the runway which then violates the premise of the whole situation that the treadmill gizmo will work and ensure that the plane does not move forward. One important thing which you have gotten wrong is that it is the friction in the wheel bearings which will hold the plane back. The friction which would hold the plane back would again be static friction between the tires and the treadmill; even though the tires are spinning and the treadmill is moving, they are still not slipping on each other and so it is still static friction which would keep the plane from moving forward if this thing were practicable to build. Also, if you were to try to do this crazy thing, you would not just open your throttle at the outset but gradually increase thrust while the treadmill adjusts to match the wheel rotation; could you imagine a treadmill which would keep the plane at rest if the thrust were just right that the plane would move forward at 20 mph if not on the treadmill? Although the wheels on this (impractical to build) treadmill would probably end up spinning faster than on normal takeoff, it would probably not be destructive to them until you stayed there too long.

Let's just call a truce on this whole thing. A 747 at rest for any reason is not going to take off. And practical problems with the fabrication and utilization of this stupid treadmill will ensure that it is never constructed or tested.

QUESTION:
In Star Wars there is the Executor class star dreadnought that is 19 kilometers long from bow to stern. And a few kilometers wide. So I was wondering how large would it appear in our sky if it was at the low orbit of 100km altitude? As big as the moon does? How about the Death Stars which were 120km and 160km in diameter?

ANSWER:
The principle you need here is that of subtended angle of an object with length L a distance d from where you are viewing it. You can see that the angle subtended by the moon is 31 arcminutes≈0.52⁰. The angle is expressed in radians as θ=L/d. To convert that to degrees, multiply by 180/π, so θ=(L/d)x57.3 in degrees. For example, your 19 km dreadnought 100 km away would have θ=10.9⁰, about 21 times larger than the moon.

QUESTION:
Why don't electrons get pulled into the nucleus? What forces prevent electrons from being pulled to the protons?

ANSWER:
See an earlier answer which answers the same question except for the earth-moon and sun-earth pairs. For your question it is the electrical force which holds the electron in its orbit. There is an important difference, though. The electron-proton system, if you look at in classical electromagnetic theory, should radiate energy (it is essentially a tiny antenna) and fall into the proton. This does not happen; the reason is that for such small systems classical physics does not work and this was one of the milestones of physics when Niels Bohr proposed his model of the hydrogen atom where certain orbits simply did not radiate. To understand the details of quantum mechanics, which eventually evolved from the Bohr model, would require a much longer explanation.

QUESTION:
Why don't the gigantic cruise ships at sea today simply flip over? It appears there is a lot more above the water than below.

ANSWER:
Because the heaviest things are at the bottom—engines, fuel tanks, drinking water tanks, sewage tanks, ballast tanks into which sea water may be pumped, etc. For a more detailed explanation, see this link.

QUESTION:
Why is angular momentum conserved for a charge in an electric field? I don't know, I just can't see how the relation between distance and velocity could justify that. It made sense in the gravitational field, since when a satellite gets closer it also gets faster. Now, if I have a stationary positive charge and a smaller positive charge in its field, the first charge will accelerate the other to repel it, so with the increasing distance of the second charge there's also an increase in velocity. positive charge and a smaller positive charge in its field, the first charge will accelerate the other to repel it, so with the increasing distance of the second charge there's also an increase in velocity.

ANSWER:
Angular momentum is not necessarily conserved in an electric field, but the problem you are describing (which is essentially Rutherford scattering) does result in the angular momentum of the moving charge q about an axis passing through the fixed charge is conserved. It is not hard to understand. What is the condition for which angular momentum is conserved? There must be no torques on q. The expression for torque is τ=rxF=qrxE where r is the position vector of the moving charge and E is the electric field due to the stationary charge. But, r and E are parallel so rxE=0 and there is therefore no torque on q and its angular momentum will remain constant.

QUESTION:
If a mass is accelerated it will create gravitational wave.The earth is always accelerating as it moves around the sun. Does always creating gravitational wave? If yes, then where is this energy coming from?

ANSWER:
This exact problem has been worked out on Wikepedia: "Gravitational waves carry energy away from their sources and, in the case of orbiting bodies, this is associated with an in-spiral or decrease in orbit. Imagine for example a simple system of two masses—such as the Earth�Sun system�moving slowly compared to the speed of light in circular orbits. Assume that these two masses orbit each other in a circular orbit in the x�y plane. To a good approximation, the masses follow simple Keplerian orbits. However, such an orbit represents a changing quadrupole moment. That is, the system will give off gravitational waves. In theory, the loss of energy through gravitational radiation could eventually drop the Earth into the Sun. However, the total energy of the Earth orbiting the Sun (kinetic energy + gravitational potential energy) is about 1.14�10³⁶ joules of which only 200 watts (joules per second) is lost through gravitational radiation, leading to a decay in the orbit by about 1�10⁻¹⁵ meters per day or roughly the diameter of a proton. At this rate, it would take the Earth approximately 1�10¹³ times more than the current age of the Universe to spiral onto the Sun. This estimate overlooks the decrease in r over time, but the majority of the time the bodies are far apart and only radiating slowly, so the difference is unimportant in this example." So the answer is that the energy comes from the orbiting earth which causes its orbit to get smaller as it loses energy; however, as the calculation shows, the loss of radius of the orbit is trivially small.

QUESTION:
Assume a stationary satelite in outer space at a distance of 299,792,458*3,600 meters from the earth. So basically the distance is the speed of light times 3600 seconds (or 1,079,252,848,800 meters) This satelite has a big lantern atached to itself that is pointing to the earth. The lantern is initially turned off, but can be turned on by sending a "turn on" signal from the earth, through radio waves. So, a "turn on" signal is sent from the earth. We assume that the "turn on" signal will reach the satelite in 1 hour and will turn on the flashlight instantly. So, when the signal reaches the satelite (after 1 hour) we send a rocket from earth that points to the direction of the light beam that is emitted from the flashlight. The speed of the rocket relative to the earth is 10,000 m/s. The question is: How much time will it take the rocket to see the light beam? Is it 1 hour? Or is it less? If it is less, does it mean that the light beam traveled faster than the speed of light relative to the rocket?

ANSWER:
So, using a clock on the earth, we will receive a light pulse exactly two hours after we sent the first pulse toward your satellite. You are thinking that the rocket will see that pulse a little bit earlier because it will have met the light before it gets to earth. That is not surprising, but that is from the perspective of a clock on the earth. But you have chosen a really tiny speed for the rocket�10,000 m/s=0.000033c where c is the speed of light. In order to see the effect which I think interests you, you need to choose a much larger much larger speed for the rocket; I will choose v=0.8c, 80% the speed of light. I will choose my unit of length to be one light hour and my unit of speed to be light hours per hour, so c=1, v=0.8, and L=1. First, look at what an earth-based observer will measure with his clock. If t is the time when the light arrives at the rocket, then vt=0.8t=x and ct=t=(L-x)=(1-x); solving, x=0.8/1.8=0.444 and t=x/0.8=0.556. Also, note L-x=(1-x)=0.556.

But, what you asked for is how long a clock on the rocket takes to see the light. The important thing is that the distance between the earth and the satellite is not 1 light hour any more, it is length contracted such that the distance seen for L by the rocket is L'=L√[1-(v/c)²]=0.6 rather than 1. Of course the same factor occurs for x'=0.6x=0.267 and the time this happens is t'=x'/0.8=0.333 because even though the distance to the satellite is shorter, it still zips past the rocket with a speed of 0.8c. Finally, c'=(L'-x')/t'=(0.6-0.267)/0.333=1! So, you see, c really is the speed seen by all observers.

QUESTION:
If I throw a bottle forward in a car as we are going 50 miles an hour what speed is the bottle going after it leaves my hand? Over 50 miles an hour or as fast as my hand can throw as if I wasn�t moving?

ANSWER:
So many times have I answered this question or its equivalent! You cannot ask what a velocity is unless you specify velocity relative to what. As a concrete example, suppose that you throw the bottle with a velocity of 20 mph relative to you (the speed you would throw it if you were at rest on the ground). But you are are at rest relative to the car, so you and anybody else in the car would see it moving forward with a speed of 20 mph. But someone by the roadside would see it moving forward with a speed of 70 mph. Somebody in a car going in the same direction as you with a speed of 70 mph would see the bottle drop straight down to the ground (no horizontal velocity). If you threw it backwards (relative to the direction of your car) with a speed of 50 mph, someone by the roadside would see it drop straight to the ground (see a Mythbusters episode).

QUESTION:
How would one calculate the buoyancy force/pressure of a static fluid acting on an object entering said fluid from the side?

ANSWER:
I see no reason why Archimedes' principle would not work, the buoyant force is equal to the weight of the displaced fluid.

QUESTION:
If u had a grain of sand with the density of the universe, next to ur average gravel driveway rock would the grain of sand have a greater mass?

ANSWER:
The density of the universe is about 10^-29 g/cm³, so it isn't even close.

QUESTION:
Would the black hole in the image released today have appeared to be a circle, rather than an ellipse, regardless of the position from which it was viewed? And if so, why?

ANSWER:
Read the following question/answer to get a detailed description of the experiment. The resolution achieved was, in fact, much better than what was expected and there is no way you could call this dark region anything other than a circle or close to it. Certainly, if you viewed this object edge on you would not have seen the dark area at all; if your viewpoint were somewhere between edge on and straight on, it would have appeared to be elliptical.

QUESTION:
About this new descovery I have several questions https://www.bbc.com/news/science-environment-47873592 Why this cannot be done by one single telescope and how the group of telescope functions in order to work (I cannot think why it needed so many to be focused in a single point). Also the earth is rotating so most of them are not aligned with the point of focus. Second question is why we only have an image? if we can take a photo, we can surely take more photos and create a video. In that way we can see the moving plasma or whatever is there.

ANSWER:
The first thing you need to understand to appreciate this problem is that if you were much closer to the black hole where you did not need a telescope; instead what you would see would be a very bright and large source of light which would hot have that black hole in its middle. So you cannot use optical telescopes at all. The actual telescopes in the experiment are all detecting photons in the millimeter range, on the order of 10^-3 m wavelength; this wavelength is expected to originate only from the accretion disk which forms around a black hole. Now a large optical telescope has a size on the order of 10 m in diameter and detects photons around 500 nm=5x10^-7 m. Increasing the size of a telescope gives it better resolution, so the size S for a millimeter telescope comparable to the best optical telescopes would be S=10^-3(10/5x10^-7)=20,000 m=10 km; but, looking at an object so far away (55 million light years) would be impossible for visible light, but since the earth has a diameter of about 13,000 km, we can get very much better resolution with a large millimeter telescope array by having the individual telescopes have separations comparable to the size of the earth. Although the size is very huge, the total area of all the radio telescopes is extremely small, so trying to make an image would be like blacking out 99.999% of the mirror in an optical telescope—you would still get an image but you would have to wait a very long time. So the data from all the telescopes in the array were gathered over many hours during a 10 day observation period. Then, all the data had to be put together using a computer to reconstruct the image; the data analysis took two years to complete, a very difficult problem.

Finally, you should ask what the dark circular region in the center is—it is not the black hole itself which has zero size. The outer edge is what is called the photon radius, the distance from the black hole where a photon will orbit. Somewhere inside of the photon radius is the event horizon, the sphere inside of which nothing, including light can escape.

A warning: as I clearly state on the site, I do not usually answer complex questions in astronomy/astrophysics/cosmology, so take what I say with a grain of salt!

QUESTION:
How has the Bristlecone Pine been able to survive the Earth's pole shifts if they occur every 3900 Years?

ANSWER:
The oldest bristlecone pines have lived more than 5000 years. However, the current polarity has been unchanged for about 780,000 years. The frequency of polarity changes are notoriously and unpredictably variable, typically from 80,000 to 780,000 years over the last 5 million years. I can still answer your question, I think, supposing that the period were within the lifetime of the plant. Most life is relatively insensitive to magnetic fields at all and would not be significantly affected by changes in magnitude or direction of ambient fields. I believe some birds navigate using the earth's magnetic field and they would get messed up and maybe fly north in wintertime were the field to flip. There has been some speculation about a lack of magnetic field resulting in higher radiation levels from cosmic rays, but no link has been found between polarity changes and extinctions.

QUESTION:
I am confused about the actual model of the atom. Some say that electrons move in circular paths about the nucleus.Other times its a wave and then I read about orbitals with different shapes. What is the actual motion of the electron about the nucleus. I am a high school student.

ANSWER:
You need to understand a little bit of history of atomic physics to fully appreciate the places of various models in the overall scheme of things. First came Niels Bohr who proposed a model which had electrons moving in circular orbits; but not just any orbit but only those with particular values of their angular momentum�the angular momentum L was quantized, L=nℏ where ℏ is the rationalized Planck constant. This was actually wrong, but it gave the right answer for the spectrum of light observed from a hydrogen atom and that was a great leap forward; but purely empirical. Improvements

and extensions of this model were made, but they were all equally empirical. Still, this first empirical model set everyone on the right track to achieve a more fundamental model. During this time, it was beginning to be understood that trying to imagine a model based on classical ideas, ideas like an electron orbiting in a circular orbit, were invalid at such tiny scales�that realization ultimately gave rise to the development of quantum mechanics. You can no longer even think of an electron as a little point charge when it is bound in an atom. Rather, you solve an equation, known as the Schrodinger equation, which gives you a mathematical function, called the wave function, which tells you the likelihood of finding the electron at any particular place in space. So the best way to think of an electron in an atom is akin to picturing a cloud over which the electron is smeared. The shape of this cloud is dependent on four quantum numbers which quantize the energy, the angular momentum, the projections of the angular momentum and the spin angular momentum on some axis. Some examples are shown in the figure. To visualize the "clouds", imagine rotating each around a horizontal axis through the center; so, e.g., the upper right cloud is donut-shaped.

QUESTION:
If I could travel away from a light source at the speed of light looking in the direction from which I came would the photons that hit my eye continue to be visible or bounce off and away from my eye at the speed of light in the opposite direction leaving nothing to see as the photons behind it would not reach me?

ANSWER:
If you had bothered to read the site ground rules, you would have seen that "I no longer answer questions which are based on premises like: '...if we could travel faster than the speed of light...' or '...ignoring the fact that nothing can go faster than the speed of light...', '...if I were traveling at the speed of light...', etc." Suppose that you were traveling at 99.99% the speed of light. The light which you would see if at rest would be visible, but in your moving frame the wavelength would be Doppler-shifted to a longer wavelength, into the region of far infrared which your eye cannot detect.

QUESTION:
I am currently working on a school science project for what is the equivalent of High School in Norway. It is about trying to measure different materials ability to repel radioactive materials. We do this by pointing a radioactive source toward a geiger counter, and sticking a materials to block between. All distances and thicknesses are fixed. Our results after some 100 different test, pose some problems. We were expecting to see that most Alpha radiation were completly blocked because our materials were 13mm. But the results show that alot of the materials, like cotton and cardboard, does not stop the radiation. In addition to this, the results form water showed that water, on all radiaion sources, were the worst at blocking radiation. We are having some trouble explaining these phenomena because they are very stable and clear. Do you have any answers to why the results are the way they are?

ANSWER:
There is no way the alpha radiation will get through 13 mm of either cotton or cardboard. I am quite sure that the reason the Geiger counter detects radiation in the anomalous cases in your experiments is that any alpha source will also have gamma radiation associated with it. For example, above I show the decay scheme of ²⁴¹Am. In addition to the four energies of alpha particles there are gamma rays of energies up to about 100 keV. I cannot explain why water is the least absorptive than any other material you tested.

QUESTION:
We say that a electron has +1/2 or -1/2 spin.What is this spin actually and how do we get the value?

ANSWER:
There are actually two answers on this page which answer this question. The first is here, the second is linked to from that answer.

QUESTION:
As a hot rod building enthusiast and garage mechanic I have a torque question that has never been adequately explained to me. Why does the definition of torque differ between a physicist and engineers\mathematicians? Engineers and mathematicians say a physicist's definition of torque is what to them is the definition of a torque moment. To the engineer and mathematician torque is defined as a measurement of "the amount of change" between torque moments and when there is no change measured, torque is at equilibrium and equals zero. Wikipedia acknowledges this difference by disciple on their torque page but doesn't discuss why. I've always accepted the reason the definitions differ has to do with the solutions needed for problems each discipline is trying to solve. If possible. I'd like to understand why the different definitions are needed with more substance than my assumption. What's never been adequately explained to me is what is it about the problems being solved by disciple that requires the different definitions of torque?

ANSWER:
It is mainly simply semantics. In physics, rotational Newtonian physics uses definitions which end up with physical laws which are wholly analogous to the laws in translational Newtonian physics. Newton's second law is, perhaps, the best example. In translational physics, force equals mass (inertia) times acceleration, F=ma. In rotational physics, moment of inertia I plays the role of mass and angular acceleration α plays the roll of a, and if torque (due to a force F applied a distance r from the axis about which the object rotates (or does not)) is defined as τ=rxF, then τ=Iα. I do not know much about engineering notation semantics but if what you say is correct, that torque means that which is zero if the object is not rotating at all (or with a constant angular velocity), then that is what would be called net torque in physics which seems to make more sense to me. According to Wikepedia, mechanical engineers (and also British physicists) refer to (physics) torque as moment of force, not torque moment as you state. Similarly, the definition by mathematicians is that torque is equal to the rate of change of angular momentum which is exactly the same as τ=Iα and so torque, again contrary to your statement, is the same as in physics. So, I read the Wikepedia article differently from you and, to me, there is no difference in anybody's definition of what I would call torque other than semantics. And I read nothing there which implies that the word torque means the sum of all moments of force to an engineer. So my final answer is that everybody's definitions of the physical quantities are the same, only the words used to label them differ.

QUESTION:
Why doesn't the moon fall to the earth?

QUESTION:
A six year old has just walked up to you and asked you why the Earth keeps spinning around the Sun but never goes out of orbit. After all, there's nothing that holds them together, right? What explanation do you give?

ANSWER:
There is something which holds them together—it's called gravity. I think the best way to understand why the earth does not fall into the sun, nor the moon fall into the earth, nor any artificial satellite fall into the earth is Newton's Mountain thought experiment. If you are on a very high mountain and shoot a cannon horizontally, it will eventually fall to the ground; as you give it more and more speed, it will go farther and farther; eventually, with enough speed, it will go all the way around, always falling but never hitting the ground. Click on the picture to see how this works.

QUESTION:
Two drops of water of the same size are falling through air with velocities of 10 cm/s. If the drops combine to form a single drop, how do you find the terminal velocity?

ANSWER:
The things you need to know:

the volume of a sphere of radius R is 4πR³/3;
the cross sectional area A of a sphere of radius R is πR²;
the air drag force on an object having speed v is proportional to Av²; and
the terminal velocity U of an object of mass m is proportional to √(m/A).

I will use subscripts 1 and 2 to denote the original and final drops, respectively; so m₂=2m₁ and R₂/R₁=2^1/3. Therefore, A₂/A₁=2^2/3. Finally, U₂/U₁=[(m₂/m₁)(A₁/A₂)]^1/2=[2⋅2^-2/3]^1/2=2^1/6. So, finally, U₂=10⋅2^1/6=11.22 cm/s.

QUESTION:
Am interested in calculating the B field a certain perpendicular distance from a large stream of moving like charged particles.

ANSWER:
The questioner had considerable confusion regarding how to do this, so I have not included most of his question. After a couple of communications with the questioner I believe that the question asks for the magnetic field for a long straight circular cross section wire of radius R with the current I distributed uniformly throughout the wire. The figure shows the field due to the wire, the vector B having a constant magnitude at a distance r from the axis of the cylinder defined by the wire and always tangential to the circle of radius r around the wire. Hence field lines, as shown are circles, the direction defined by the right-hand rule as shown. The magnitude of the field is determined by Ampere's law,

∫B⋅ds=μ₀I_enc

where the integration is all the way around the path defined by the circle and I_enc is the amount of current which passes through the area defined by that closed path. Given the cylindrical symmetry of this special case, the integral is simple because B is constant and B and ds are, everywhere on the integration path, parallel, ∫B⋅ds=2πrB=μ₀I so B_r>R=μ₀I/(2πr) for r>R. This is only correct outside the wire itself because the integration path to determine B inside the wire encloses less current. Inside, I_enc=I(r²/R²) so B_r<R=μ₀Ir/(2πR²) for r<R. Note that when r=R both results give the same value of B, B=μ₀I/(2πR). So, the field increases linearly with r inside the wire and decreases like 1/r outside.

The questioner also wanted to express this in terms of the velocity v of electrons with charge Q=-e and having a density of n electrons/m³. The fact that the electrons have negative charge means that, in the figure above, the direction of v is opposite of the direction of current flow. In a time Δt the electrons go a distance ΔL and so sweeps out a volume ΔV which contains a total charge ΔQ=-neΔV=-neΔL(πR²) and therefore I=ΔQ/Δt=-ne(ΔL/Δt)(πR²)=-nevπR².

FOLLOWUP QUESTION:
Thanks for the answer. Am not sure but am thinking you missed the fundamental part of my question, I am looking for the B field adjacent a large stream of like charges in free space, not in a wire. That I know. I also however think I figured out how to do it this morning using the formulas I provided earlier.

ANSWER:
My answer was exactly the same as if there were many uniformly distributed point charges moving with speed v in a cylindrical beam of radius R. The second part of my answer showed you how to relate I in the first part of the answer to the charge, speed, density of charges, and R. So the quantity you seek results from setting I=nQvπR² into B=μ₀I/(2πR) (assuming that "adjacent" means right on the edge of the moving charge distribution).

SECOND FOLLOWUP QUESTION:
I am trying to compare the "B" fields associated with two moving electrons in free space, both travelling in the same direction and at the same velocity.

In the first case the two electrons are very close together and are moving to the right at say 50 m/s, what would be the "B" field be 50 mm directly above, i.e. in the positive y axis above these electrons?

In the second case, same two electrons traveling to the right at 50 m/s but this time they are separated. Call the first electron, "e1"and the second "e2". The first electron e1 is located directly 1 mm below the point of interest where I want to calculate the combined "B" field. The second electron e2 is located directly 98mm below e1 or 99 mm below the point of interest where I want to determine the combined "B" field. In effect from the first case I have moved e1 closer to "B" by 49 mm and e2 farther away by 49 mm.

I was using the formula B=μ₀NqV/(4πR²) where μ₀ is permeability of free space, N is the number of charges, q is charge on an electron and V is the velocity of the electrons, R is the vertical distance between the charges and the point of interest for "B" the strength of the magnetic field.

ANSWER:
The equation you are using for the field, a special case of the Biot-Savart law, will work for the specific case you state. It is important that you understand that this gives only the magnitude of B and not its direction. Also the position vector R of the point where you calculate B must be perpendicular to V. The correct form of the Biot-Savart law is B=μ₀NqVxR/(4πR³); note that the magnitude of VxR is VRsinθ where θ is the angle between V and R. So, for θ=90⁰, your equation is correct. Be careful, though, because the direction matters when you are adding vectors from two sources; the case you are looking at has the point the same side of both electrons so the magnitude is the sum of the magnitudes, but if you look at a point between the electrons, the magnitude will be the magnitude of the difference. The form of Ampere's law which I used in the earlier answer is simply not applicable in this situation because making reference to a current I only makes sense in magnetostatics. Finally, I should note that all this is only approximately correct when you are in a nonrelativistic regime which you certainly are for V=50 m/s.

QUESTION:
Regarding the twin paradox. I read an example where the traveling twin aged 6 years while the earth bound twin aged 10 years. Since a year is 365 earth rotations, 10 years would represent 3650 rotations, while 6 years much fewer. But if they are both counting the days by watching the earth rotate, their counts must be the same when they meet, I would think. Will they both count 3650 or the 2190 days for 6 years? As a corollary, whichever count they get, they would also count the same number of birthdays during the trip.

ANSWER:
You have fallen for the trap of assuming that there exists some "real clock" which tells you the "real time", in your case the rotation of the earth. I hope you have already read my explanation of the twin paradox (I have attached here the figure from that post for your convenience) which shows how the traveling twin "sees" how the earth clock runs by having the earth-bound twin send him one message per year; this is different from your situation where essentially a daily rather than an annual message is sent, but the ideas are identical. When he arrives home, the traveling twin has "seen" the same number of years (days) on earth pass as the earth-bound twin has but, as you can see, a different number of years (days) have passed on his on-board clocks. "Watching" some clock in another frame of reference is not in any way relevant to time in your frame; the traveling twin would count his birthdays using his clock, not earth's clock. Note also that the traveling twin "sees" the earth spinning very slowly on the trip out but spinning very rapidly on the trip home.

QUESTION:
My 9 yo son come up with this idea that there is a limit at how hot can something get becase heat are particles in movement and nothing can move faster than the speed of light hence the limit to how hot can something get. Either if he is correct or not could you tell us why?

ANSWER:
This is a good question. In fact, even though there is a limit on the speed that a particle can have, there is no limit on the energy it can have. The temperature of something depends on the average kinetic energy of the particles, not their average speed. The kinetic energy in Newtonian physics is ½mv², so it is natural to think that the maximum possible kinetic energy would be ½mc². But Newtonian physics is not valid at speeds close to c, and in relativity the kinetic energy is ½mv²/√[1-(v²/c²)]. To learn more, see an earlier answer to the same question; you might also look at the answer following that one.

QUESTION:
How is quantum spin defined? and how did scientists get the idea of this?

ANSWER:
See an earlier answer. Spin is nothing more than a specification of the angular momentum of a system; e.g., the spin quantum number of an electron, proton, neutron, and other so-called fermions is s=½. The actual angular momentum is given by S=ℏ√(s(s+1)) where ℏ is the rationalized Planck's constant. Usually, we think of spin as the intrinsic angular momentum of an elementary particle or entire quantum system like an atom; particles may also have orbital angular momentum, for example an electron orbiting around a nucleus in an atom. But each angular momentum will be quantized (have a specific quantum number) and the total angular momentum J of any angular momentum number j will be J=ℏ√(j(j+1)). Orbital angular momentum quantum numbers are integers rather than half integers like electrons; so, an electron in a "p-orbit" in a Bohr atom has an orbital quantum number 1 and a spin quantum number �. As to where the idea came from, there were several things which suggested it. The experiment which showed that spin of an electron was the Stern-Gerlach experiment, but there were also hints from spectroscopy and, notably, that there were twice as many states in quantum systems (like atoms) as expected from the Pauli exclusion principle.

QUESTION:
What direction was the earth rotating during the ice age? And is the north pole now on a different axis?

ANSWER:
The earth's axis is pretty-well fixed relative to the earth itself. However this axis, in absolute space, precesses much as a spinning top does. The earth's axis is inclined by about 23.4⁰ relative to the normal of the plane of its orbit but that axis precesses around the perpendicular axis about once every 26,000 years. This precession is caused by torques exerted on the earth by the sun and the moon. Currently the axis is pointing almost exactly to the the north star, Polaris; however, in a few thousand years there will be some new star taking that role. The last ice age was about 2.4 million years ago and lasted much longer than 26,000 years, so the answer to your question is that it pointed many directions during the ice age. You are asking about the geographic north pole which is essentially constant in its location (although the land masses move over the surface because of plate techtonics). The gravitational north pole, though, is forever wandering around and has actually jumped to the southern hemisphere around Antactica and back many times.

QUESTION:
I�ve been having a discussion with a friend of mine, a professor of philosophy at the Autonomous University in Mexico, for the past two months about aesthetics. I had just finished reading Phillip Ball�s new book about quantum physics as well as Sabine Hossenfelder�s critique of "beauty" as a criterion for judging theories in physics. I told my friend that I thought "quantum mechanics is an art form". He asked me - as good philosophers do - why I thought so. And that has led me on a happy quest into philosophy as well as what physicists have to say about what they do. But I have not read a physicist saying that they thought they were creating "art". So I thought I�d ask a real physicist. If what you do can be both "beautiful" and "elegant", then do you believe you are creating "art"?

ANSWER:

My third book, PHYSICS IS… The Physicist Explores Attributes of Physics, has a chapter titled Physics Is Beautiful, devoted to beauty and elegance. However, I have never thought of it as art. A sunrise over the Sangre de Christo mountains is beautiful but you would certainly not call it art. I found that it is impossible to find an unambiguous or not vague definition of art; art is qualitative which sets it apart from science which is quantitative. And, consider this: There are many paintings of the of Chartre Cathedral, four of which I have copied here, and all are indisputedly art; there is no "correct" painting of this subject. The theory of electromagnetism, on the other hand, is beautiful and the only one of its kind; you might think up another electromagnetic theory, but it would be wrong even though it might be considered very imaginative, creative, even beautiful. There is no well-defined test which specifies which painting is the best; there is a test as to which is the theory of electromagnetism—it must quantitatively describe what really happens in nature and if it fails to do so, it is false. I have always thought of string theory as not a theory (and not beautiful) because it makes no attempt to quantitatively interface with the real world and is therefore not falsifiable.

QUESTION:
What is the distance to stop a truck weighing 7200 lbs. at 45 mph ?

ANSWER:
That depends on what the composition and condition of surface on which the truck is moving and on the composition of its wheels. I will choose a truck with rubber tires on level dry asphalt and that the truck has antilock brakes. The maximum force which can be exerted by the friction on the tires is approximately F=-μ_sN (negative because the force is opposite the velocity. where μ_s is the coefficient of static friction between rubber and dry asphalt which is about 0.5-0.8, and N is the normal force of the truck on the road which, on level ground, is the weight; I will choose μ_s=0.65, halfway between the extremes. At this point I will switch to SI units which scientests prefer: weight N=7,200 lb=32,027 N, mass M=32,027/9.8=3,268 kg, and speed v=45 mph=20.1 m/s. Now, using Newton's second law, F=Ma=-μ_sN; so acceleration a=-6.37 m/s². Now, you can compute the time to stop: v=0=v₀+at=20.1-6.37t, so t=3.16 s. Finally, the distance traveled is d=v₀t+½at²=20.1x3.16-½x6.37x3.16²=31.7 m=104 ft. It is interesting that this answer is independent of how heavy the truck is. The reason is that the acceleration is proportional to the weight/mass which is just 9.8 m/s².

QUESTION:
How a rubber ball and an iron ball with same mass hits on the same ground with same force but the rubber ball gets more bounce without violating Newton's third law of motion?

ANSWER:
What makes you think that they experience "the same force"? The force which the ground exerts on the ball is determined by what the mass of the ball is, how long the collision lasts, what the nature of the gound and ball are, and how fast the ball was moving when it first encountered the ground. But all that is beside the point if what your question is concerned with Newton's third law. Whatever upward force from the ground which a ball experiences during the collision, the ground is experiencing an equal and opposite force down from the ball. The iron ball may not bounce at all, but these two forces are always equal and opposite. The rubber ball might rebound with a speed half the speed it came in with, but, during the time the ball and the ground are in contact, Newton's third law is still obeyed.

QUESTION:
Would an object that weighs 4.5 ounces going at 2,500 feet per second kill a person? I'm asking because my friend asked how fast you have to throw a beanie baby at someone in order to kill them, and I'm slightly concerned for them but I want to know

ANSWER:
Scientists prefer to use SI units, so I will switch and then convert back when I have finished" 4.5 oz=0.128 kg, 2500 ft/s=762 m/s. What matters is, when it hits you how much force do your experience over the time that the collision was occuring. Suppose that the beanie baby flattened by about ½ inch or 1.27 cm=0.0127 m before it came to rest. If you assume that the object comes to rest by having a uniform deceleration over a half inch, you can show that time to stop is about 3x10^-5 s and the acceleration is about 2.3x10⁷ m/s². The corresponding force to stop the beanie baby is F=ma=.0128x2.3x10⁷≈3x10⁵ N≈67,000 lb. I think that if the beany baby hit you anywhere except possibly your limbs, you would be killed. Keep in mind that this is a very rough calculation, but certainly in the right ballpark.

QUESTION:
We know if V=4πr³/3 is the volume of a sphere and if it is increasing with a constant rate dV/dt=C, then C=4πr²dr/dt. Solving, dr/dt=C/(4πr²) which is velocity at which the surface is expanding. So the acceleration of the surface is (d²r/dt²)=-[2C/(4πr²)]dr/dt=-2C²/[(4π)²r⁵)] or, combining all constants, kr^-5. I just don't see if the radius of the sphere is expanding at a rate of Cr^-2, how it is accelerating at kr^-5?

ANSWER:
This is your question which I have edited to make it more clear what you have done. I see no errors in your differentiations or integrations. So, your constant k is k=-2C²/(4π)². I think that the problem you are having is that you have carried the minus sign into your constant. But that minus sign is important because it tells you (since everything else in k is constant) that the radius of the surface is increasing, but the rate at which it is increasing is decreasing as r changes. So, if the volume of a sphere is increasing with a constant rate, its radius is increasing like 1/r², but the rate at which the speed of the surface is decreasing goes like 1/r⁵.

QUESTION:
Can you tell me why/how this snow is holding together and curling backwards after sliding off a metal roof? Facebook pic mystery.

ANSWER:
This is not really so rare, at least the sliding off the roof part. You can see, in the other picture, my son at his home in Vermont with a similar (not curled) situation. The roof is warmed from inside and the snow slides down. If conditions are right, the wet snow on the bottom will refreeze and hold it from breaking off and also refreezing to halt the slide. I would guess that the curling would result from repeated cycles of sliding, pausing, sagging repeatedly. I cannot really see why the end of the slab would be being pushed back toward the house.

QUESTION:
When you let a lead ball and an aluminium ball of the same radii to free fall from a height of 100 metres which will ground first?

ANSWER:
Assuming that the balls are not hollow, the lead ball will have more mass. At a given speed, both balls will experience the same drag force. But that force will have a bigger effect on the ball with the smaller mass (Newton's second law). So the lead ball will reach the ground first. If you are interested in more detail, a recent answer is very similar to yours.

QUESTION:
I am a man of some girth, 490 lbs to be exact. I also enjoy "competing" in 5K race events. As I struggle to maintain a 3 mph pace, Cutler, skinnier people zip past me unphased and undaunted by hills, heat, etc. I have often pondered but do not have the requisite knowledge to figure the answers. My question is how much force is required to propel a body, weighing 390 lbs, at 3.1 mph. Compare that to someone at 130 and 180...or share the formula. Also, how can I calculate KCAL consumed at different external temps and humidity.

ANSWER:
To figure out the force is very difficult because running is a series of accelerations (when you push off with one foot) and decelerations (when the other foot hits the ground). So your speed is constantly speeding up and slowing down as you run and the forces which the ground exerts on you is constantly changing in both magnitude and direction. It is daunting to try to think about such a calculation. But, I believe what you really want is energy expended over a certain distance in a certain time. I found a handy little calculator which you can look at here. For example, for your current weight, a 5K race at 3.1 mph (which is about equal to 5 km/hr, so it takes one hour for you to complete), I find "You burned 1151 calories on your 5 kilometer run. At that pace, your calorie burn rate is 230 calories per mile and 1151 calories per hour." Sorry, I could not find any information on temperature and humidity.

QUESTION:
The total amount of energy is always same in the universe. But if there were two observer who were travelling at different velocity,would their measurement of total energy differ?

ANSWER:
Kinetic energy is not invariant, different observers see different values. Imagine a universe which is entirely at rest and you are at rest in that universe; the universe has no kinetic energy. Now imagine that you move through that universe with some velocity v; you now observe that universe to have a kinetic energy of ½Mv². where M is the mass of the universe except for your mass. And just a minute ago that energy was not there. Obviously, the total energy of that universe cannot be the same in all frames. But, if you stay in one frame and keep track of the total energy of the universe in your frame, it will not change.

QUESTION:
if a person dove off the back of a boat going forward at 35mph would that person actually be pulled backwards? I say no he would just de accelerate

ANSWER:
The answer would be easy if you dived from helicopter moving 35 mph—you would have a forward velocity of 35 mph and the water would be at rest, so you would experience a drag force opposite your motion, backwards, which would cause you to slow down (decelerate) in the horizontal direction. But, hydrodynamics can be very complicated, and it is quite possible that the water behind the boat is moving with the boat; I cannot imagine, though, that it would be moving faster than the boat (and you), so it would not pull you toward the boat.

QUESTION:
Is the pull of gravity the same for every part of Earth? (32 feet per second?)

ANSWER:
The number you are stating is not the "pull of gravity" but rather the acceleration due to gravity. And, the units are ft/s/s, not ft/s. But it is an indication of how hard gravity is pulling, so I can go ahead and answer your question. We (physicists) usually prefer to work in SI units, meters instead of feet; in SI units, g≈9.8 m/s². The acceleration due to gravitation does vary at various places on the earth. The animation shows you how it varies. The variation is small, no larger than 50 milligals; a gal is 1 cm/s²=0.01 m/s², so 50 milligal=0.005 m/s². So, compared to 9.8 m/s², the variations are only on the order of 0.05%. There are several reasons for gravitational variations, notably:

altitude (the farther you are from the center of the earth, the weaker the gravity);
local geology (large volumes of more dense earth will increase the gravity);
the earth is not a perfect sphere.

If you want a lot more detail, see the Wikepedia article on the earth's gravity.

QUESTION:
when a stone is thrown up with a velocity and it collides with a particle in mid air(the particle being at rest during collision) , in this situation , is vertical momentum conserved?

ANSWER:
Linear momentum is conserved only if there are no external forces. But there is an external force acting, the force of gravity, so the vertical component will not be conserved for a "real" collision by which I mean one which lasts for some nonzero time. But, you have stated that the struck particle remains at rest during the collision; the only way this can be true is if the collision is instantaneous. So the momentum will be conserved because the collision lasts for zero time. The reason this is not a physically possible collision is that it requires the struck particle to instantaneously acquire its velocity which requires an infinite force. Nevertheless, these kind of problems with an external force present are often approximated as having momentum conserved if the collision happens very quickly. Another example is shooting a bullet into a block of wood sitting on a horizontal surface which has friction; only if the collision happens very quickly will momentum be approximately conserved.

QUESTION:
Newton's third law of motion says that for every action, there is an equal and opposite reaction. If i throw a ball towards the wall with some Force it will not Bounce back to me with the same force. Why is that? I mean like if i have a bowling ball and I threw it Towards the ground it will not bounce back. This is driving me crazy

ANSWER:
You are thinking of Newton's third law incorrectly. It applies only to forces during the collisions, it does not tell you how fast the rebound will be. When the ball hits the wall it exerts a force on the wall, and the wall exerts an equal and opposite force on the ball. But knowing the force on the ball does not tell you how fast the ball will be going after the collision. What happens is that some energy is lost by the ball during its collision so it comes off the wall more slowly than it went into it. The bowling ball is an example of the same thing except it loses all its energy when it collides with the ground. These are called inelastic collisions, the bowling ball is a perfectly inelastic collision. If the ball were to leave the wall with exactly the same speed it came in with, it would be called an elastic collision.

QUESTION:
What the heck is spin? Quantum objects don�t literally have angular momentum, do they? Does the concept of spin have any relationship to angular momentum? Or is it just a name like �color� that isn�t descriptive of what�s happening?

ANSWER:
They do "literally have angular momentum". So, how can we observe experimentally whether an object has an angular momentum? One way is that if you can find a way to exert a torque on an object which has angular momentum, the torque will cause the angular momentum to change. For example, a spinning top has angular momentum and if the top does not have its symmetry axis vertical, there will be a torque because its own weight exerts a force which makes it want to fall down. But, because it has angular momentum, it does not fall but instead precesses about the vertical. Now, we know that an electron has an intrinsic angular momentum, a spin. Presumably because of this spin the electron also has a magnetic dipole moment—it looks like a tiny bar magnet. So when we put a magnetic dipole in a magnetic field, there is a torque exerted on it which tries to line it up with the magnetic field. This torque, because the electron has angular momentum, causes the dipole to precess just like the top did. Without going into detail, it is possible to infer this precession from measurements. What is not possible, though, is to understand this intrinsic angular momentum classically. For example, if you try to make some classical model imagining a little sphere with some reasonable radius and mass of an electron spinning on an axis, you will get nonsensical results like the surface of the sphere having a speed greater than the speed of light.

QUESTION:
If a space vehicle left earth at 18000 mph in a direct line to the sun, how fast would vehicle accelerate to taking gravitational pull into consideration?

ANSWER:
The escape velocity from the earth's surface is about 25,000 mph, so your vehicle would fall back to earth.

QUESTION:
If you built a perfect sphere in space that was 187,000 miles wide, the inside was a perfectly polished mirror, and you fired a Laser inside for 1 second, how long would that beam of light bounce around? Provided the sphere was dust free and a vacuum inside. This question has bothered me for 35 years.

ANSWER:
You have chosen the distance to be about the distance which the light would move in 1 second. I cannot imagine the reflectance of the sphere would be greater than 99%, so if you solve the equation 0.99^N=0.001, N would be the number of seconds it would take for the intensity of the light to drop to 0.1% of its original value. Using Wolfram Alpha, I find N≈687, a little more than 11 minutes.

QUESTION:
If two objects that weigh the same are attached by a string will they sink at the rate of combined mass or will they sink equally?

ANSWER:
Since you refer to "sink" I will assume that they are falling in water. There is no simple answer to this question because the drag force on a sinking object depends on its geometry. For example, suppose one object is a big flat sheet and the other is a long thin needle suspended below the sheet which behaves like a parachute; clearly, this will sink with a much smaller speed than if both objects were long thin needles. The most appropriate approximation of the magnitude of the drag force f in water is a quadratic function of the velocity v, f=½ρ_wC_DAv², where ρ_w=10³ kg/m³ is the density of water, A is the area presented to the onrushing water as the object falls, and C_D is a constant which depends on the geometry of the object, called the drag coefficient. For a sphere of radius R the drag coefficient is given by C_D=0.47. The upward force f is only one force on the object as it sinks; it also has a weight down W=mg and a buoyant force up B=Vρ_wg where V=4πR³/3 is the volume of the sphere. If the sphere is solid and has a density ρ=m/V, we can write B=mg(ρ_w/ρ). Also note that we can write R as R=³√[3m/(4πρ)]. Finally, we can write the total force on a sphere: F=[mg(ρ_w-ρ)/ρ]+½ρ_wC_DAv². Note that when a speed is reached where F=0, the sphere will fall with a constant velocity called the terminal velocity, v_t=√{[mg(ρ-ρ_w)/ρ]/[½ρ_wC_DA]}.

So, let's do a specific example to be able to answer your question for a particular situation. Suppose we have two 1 kg balls of solid iron for which ρ=7.87x10³ kg/m³. Then the radius of each ball will be R=0.0312 m. Doing the arithmetic, F=-8.56+0.72v². So each of two balls of mass 1 kg, tied together will accelerate until their speeds would be v_t=√(8.56/0.72)=3.45 m/s. On the other hand, if we have one sphere of mass 2 kg, its radius will be 0.0393 m. So, F=-17.1+1.14v². So one ball of mass 2 kg will accelerate until v_t=√(17.1/1.14)=3.87 m/s.

You could do more examples to come to the conclusion that the answer to your question is that, in general, two spheres tied together will not sink at the same rate as one sphere with their combined mass. I am sure, by carefully tailoring the shapes of the objects, you could force the rates be the same, but that is not you would find in general.

QUESTION:
If we go inside a mine and drop a 10 kg iron ball and 1kg aluminium ball from the top of a high platform. what will happen?

ANSWER:
I do not understand the point of going inside a mine. I will assume that the atmospheric pressure and acceleration due to gravity are the same as at sea level. In that case the air drag f can be approximated as f≈¼Av² where A is the area presented to the onrushing air and v is the speed. (You will note that Av² is not the same dimensionally as a force; this approximation only works if you work in SI units, A in m², v in m/s and f in N.) So the net force F on each ball (which I assume to be solid) is F=¼Av²-mg where m is its mass; and, by Newtons second law, the acceleration a is a=(¼Av²/m)-g; and there is a velocity U, called the terminal velocity, at which the drag equals the weight so the acceleration is zero and U=2√(mg/A).

Iron and aluminum have different densities, but it is pretty straightforward to compute the radii of each

ball and from that compute A for each. I find A_Fe=0.014 m² and A_Al=0.0064 m². The accelerations are a_Fe=-(9.8-3.5x10^-4v²) and a_Al=-(9.8-1.6x10^-3v²). At any given speed the magnitude of the acceleration for the iron ball is larger than for the aluminum ball. The terminal velocities would be U_Fe=170 m/s and U_Al=78 m/s. The iron ball always has a larger speed than the aluminum ball and will hit the ground first.

It is also enlightening to find the distance traveled as a function of time, s=(U²/g)ln(cosh(gt/U)). This may beyond your math ability (ln is a natural logarithm and cosh is a hyperbolic cosine), but the graph shows the behavior of each ball and clearly the iron ball is falling faster. Note that for the first few seconds (6 or 7 s) the balls pretty much keep pace with each other and after about 15 s both have nearly reached their respecitve terminal velocities because the curves are nearly linear.

QUESTION:
Would the weight of any object become less as it approached the center of the earth? Neglecting the obvious environmental difficulties of survival could i fall thru a tunnel bored completely thru the axis of the earth and slowly decelerate to the center of the earth. Possibly oscillating back and forth near the center and finally come to rest in the weightless middle?

ANSWER:
Do you want the Physics 101 answer or the answer based on our best understanding of the density distribution of the earth? I will give you both.

When this kind of problem is presented as an exercise in elementary physics, it is usually assumed that the density of the earth is uniform throughout—a cubic meter of earth from the surface has the same mass as a cubic meter at the center. In this case, it is easy to show that the acceleration due to gravity, g(r), is linear: g(r)=9.8(r/R) m/s² where r is the distance from the center and R is the radius of the earth. Since the acceleration is always toward the center, you speed up until you reach the center (not decelerate as you suggest) and then slow down until you reach the other end, neglecting any frictional forces. More realistically there will be at least air drag which will get bigger as you speed up until you eventually reach a constant speed; then at the center, you start slowing down until you stop somewhere before the other end. Then you speed up back to the center getting there with a smaller speed than the last time you were at the center, etc. eventually stopping at the center as you suggest. One thing you might note is that, for the frictionless case, your motion is exactly as if you were attached to a spring with spring constant k=9.8/R N/m.
But, the earth is not a uniform sphere. The core is much denser than the mantle and therefore the acceleration is not a nice linear function. The figure shows that the acceleration is almost constant until about one third of the way to the center. The above qualitative description of your motion remains the same except for the similarity to the simple spring.

ADDED THOUGHT:
I never answered your first question, regarding the weight of the object. In the uniform density earth the weight (mg) decreases linearly to zero at the center. In the realistic model, the weight remains constant for a while, then increases slightly, then falls almost linearly to zero at the center.

QUESTION:
I am a professional tennis coach and would like to explain the relationship between mass and spin of a tennis ball. So in my words, I experience hitting the ball with allot of spin makes the ball heavier. When a fast flat ball is hit, I feel it is faster but on contact lighter than the spin ball. Does this mean the increase in rpm, will increase the mass exerted on the inside of the ball, thus making it heavier? And is there a formula for this?

ANSWER:
I am surprised, if you are a professional coach, that you do not tell me what kind of spin the ball has. The two most common spins are top spin and back spin (the reversed direction of top spin); there is also apparently something called slice spin.

First, though, let me address your reference to the ball being "heavier" or "lighter" depending on the spin on the ball. This is totally incorrect as the ball always has the same weight, a force which points vertically downward. The way that the ball moves results from the forces which the ball experiences after it leaves your racquet; there is the weight straight down, the force the ground exerts on it when it hits, and the forces which the air exerts on it. The figure above shows a ball with top spin; the top of the ball is spinning in the same direction as the ball is moving to the right. (If you are confused, the diagram is shown in the frame of the ball, so the air is shown like a wind blowing in the opposite direction as the ball is moving.) Here the ball is experiencing the force of its weight straight down (not shown), a drag force to the left slowing it down (not shown), and an aerodynamic force called the magnus force F straight down. If the ball were not spinning, there would be no magnus force; if the ball had back spin, the magnus force would be straight up; if the ball had slice spin, the magnus force would act horizontally. (These are all also important for a pitched baseball.) I am guessing that the ball you judged as "heavier" is the top spin case because, due to F, the ball will drop faster than a nonspinning ball. I am guessing you call the nonspinning ball "flat" and, indeed, it will not fall as fast as the ball with top spin. A ball with back spin would fall even less fast than the nonspinning ball, so you would call it even lighter. (In fact, I believe that the magnus force on a ping pong ball with sufficient backspin, can actually be larger than the weight of the ball.) But in all cases the ball has exactly the same weight. (I also note that an actually heavier ball, in the total absense of air, would move identically to a lighter ball because the acceleration due to gravity is the same for all masses.)

For completeness I would like to talk about what happens when the ball hits the ground because these are strategically important in tennis. Shown in the second figure are the forces (in red) which the ground exerts on the ball in top and back spin cases. For clarity, I have not shown the forces due to the weight or the air. In each case there is a normal force N upwards which is what lifts the ball back in the air; but there is also a frictional force f which slows down the spin in both cases. In the the top spin case, the friction both slows down the spin and speeds up the ball so it comes up faster and lower than a nonspinning ball would; in the the back spin case, the friction both slows down the spin and slows down the ball so it comes up slower and higher than a nonspinning ball would. If the spin were large enough, the back spin ball could actually bounce straight up or even backwards!

QUESTION:
When an electron absorbs photon does its mass increase?

ANSWER:
I guess you are thinking about the photoelectric effect. A photon strikes an atom and disappears. An electron is ejected from the atom. You are thinking that the photon is totally absorbed somehow by the electron. But what really happens is that the energy, hf, which the photon had is given to the electron; the electron ends up with a kinetic energy equal to hf minus the binding energy it had in the atom. The rest mass of the electron remains unchanged.

QUESTION:
"If a ladder you are standing on begins to fall...hold on... it will reach the ground slower than a falling body." Is the above advise true?

ANSWER:
One could work out an analytic solution to this problem, but that is not necessary if you just want to be convinced that the advice is good. All the forces on you are your own weight, straight down, and the force the ladder exerts on you. The vertical force which the ladder exerts is upward, so the net force in the vertical direction is smaller than if you were just falling.

QUESTION:
Is there anything that can block the effects or decrease the effect of gravity? Any experiments that tested it? Magnetism can be blocked and why not gravity?

ANSWER:
Both magnetic and electric fields can be manipulated to be zero, but that is much more difficult for gravitational fields. The main reason is that there are two kinds of electric charge but only one kind of mass. That is not to say it is impossible. For example, there is a point between the earth and the moon where the field is zero, where the attraction to the earth is equal and opposite the attraction to the moon. But, in order to have any effect on the field near the earth, you need to manipulate giangantic masses. As far as we know, there is no such thing as "antigravity".

QUESTION:
When I carry my cell phone (iPhone 7) in my left pocket and it vibrates, I feel the vibe on my right hand, while my left hand feels nothing. It happens quite often and I even check to see inside my right pocket (or in my tote bag on my right shoulder) I fear of having misplaced my cell there. What does the trick?

ANSWER:
This is not really so surprising. The brain is divided into two halves and there is some crossing of nerve impulses sometimes. I used to hear and feel a thumping in my right ear when I tapped my left big toe. It seems to not do that anymore.

QUESTION:
If you take out the spaces between all the atoms of the nuclei in your body, would you end up half the size of a flea but still weigh the same?

ANSWER:
Technically, there are no spaces between atoms in your body. I think what you are asking is that if we compressed all the matter in your body (as happens in a neutron star) to nuclear density, how big would it be? Suppose your body has a mass of 100 kg. The density of nuclear matter is about 2x10¹⁷ kg/m³, so your volume would be about 10²/2x10¹⁷=5x10^-16 m³. So, if you are a cube, your dimensions would be ³√5x10^-16=8x10^-6 m≈10 microns. A flea has a size of about 1 mm=1000 microns, one hundred times bigger than the compressed you. And, yes, you would still have a mass of 100 kg.

QUESTION:
If we are looking back in time through telescopes now at stars and galaxies, would it be possible to calculate the distance required away from earth in order to look back and see a moment in history here on earth, theoretically ?

ANSWER:
Of course, but you could never get there in time to see it. An alien who was 5000 light years from earth would see, if he had a good enough telescope, the earth as it was 5000 years before he was looking.

QUESTION:
Since even light cannot escape the gravitational pull of a black hole, does this mean that the theoretical particle, known as the graviton, have the potential to move faster than the photon?

ANSWER:
Sorry this took so long, but I had to consult with an astrophysicist. The best answer he could come up with is at this link.

QUESTION:
Attached is a picture culled from the internet on an object rolling down a plane. I understand everything except how to explain and draw the forces to my AP physics students. Should this look like a box sliding down the plane? with mgsintheta down the plane and mgcostheta as the normal force and mg down and friction up? Or are some of these properly represented as torques? Could you please draw how this should be properly represented?

ANSWER:
I have not included your picture because there is too much stuff on it (the velocity, the R, the Rω, etc. all distract from the crux of the problem, in my opinion). I have redrawn the problem showing some axially symmetric object of radius R, mass m rolling down an incline of angle θ without slipping. There are only three forces acting on the object, the frictional force f and the normal force N, both of which act at the point or line of contact, and the weight mg which acts at the center of mass of the object. There are three unknowns, N, f, and a, the acceleration of the center of mass. To solve for these unknowns, we must apply Newton's laws, both translational and rotational (ΣF=ma and Στ=Iα); here I is the moment of inertial about the axis about which torque is calculated and α is the angular acceleration about that axis. You are a teacher, so I will skip a lot of the details which you can fill in easily. From ΣF=ma I find N=mgcosθ and ma=mgsinθ -f. The first equation nails N, one of our unknowns, the second is one equation for the two remaining unknowns, so we are not done. So the only place to get a second equation is the rotational form of Newton's second law; I think this is the trickiest part of this problem. I find many students have an inclination to choose the symmetry axis to sum torques; they have been told, when they learned statics of non point objects, that you can choose any axis you like, but this is not a statics problem and there is a very important constraint—Newton's laws are not valid in an accelerating system, and the center of the rolling object is accelerating. (See correction below!) So Στ=Iα yields Iα=Ia/R=mgRsinθ, or a=(mR²/I)gsinθ; knowing a, then f=m(gsinθ-a) But, what is I? It isn't what you look up in a book because usually what is listed is the moment of inertia about an axis passing through the center of mass (I_cm). Here the axis is a distance R from the center of mass, so using the parallel axis theorem, I=I_cm+mR².

I will do one example, a solid sphere where I_cm=2mR²/5, so I=7mR²/5, so a=5gsinθ/7 and f=2mgsinθ/7. If you sum torques about the center of mass, you get the wrong answer.

CORRECTION:
OK, I slipped up here, albeit in a fairly minor way. While it is true that Newton's laws are not correct in an accelerating frame of reference, there is one exception: the sum of the torques about an axis passing through the center of mass is indeed I_cmα. A little detail I had forgotten about classical mechanics. Let me work that out for my example: I_cmα=2maR/5=fR, so, using the result above, f=2ma/5=m(gsinθ-a) or a=5gsinθ/7. Then it easily follows that f=2mgsinθ/7. But you should try to sum torques about an axis halfway between the point of contact and the center of mass—you will not get the right answer.

Nevertheless, it is important that your students be careful in choosing the axis. I have edited the original answer somewhat.

QUESTION:
How massive is the largest black hole ever discovered?

ANSWER:
See the Wikepedia article List of Most Massive Black Holes. This is the most up-to-date information I could find.

QUESTION:
Does the core of the earth spin? If is does how does it spin faster than the mantel with the viscosity of magma and the 4+ billions years old the earth is? I just don't understand what I am being told.

ANSWER:
The core of the earth is composed mainly of iron and is comprised of an inner core which is solid and an outer core which liquid. Apparently, from what I understand from an article I found, recent research indicates that the outer core rotates with the same speed as the mantle (once a day) but the inner core's rotational speed is slightly faster by about 0.3-0.5 degrees per year.

QUESTION:
If I dropped a circular, triangular, and square parachute from the same height which will land first assuming the parachuted object's weight is the same and the surface area is the same for all three parachutes?

ANSWER:
I am assuming that your parachutes are flat, horizontal plates and that drag forces due to the load and lines attaching the load to the parachute are negligible or at least the same. The drag coefficients C_D for the three shapes, determined experimentally, are shown in the figure. The drag force is given by F_D=½ρv²C_DA where ρ is the density of the air, v is the speed of descent in still air, and A is the area. Therefore the drag force is least on the circular parachute so it reaches the ground first. (A proviso is that the values of the drag coefficient depend on a quantity called the Reynolds number which depends on the velocity. However, the values given are for low Reynolds numbers which are suitable for parachutes.)

QUESTION:
How did Einstein get the idea that the speed of light must be constant for all observer?

ANSWER:
Einstein's original paper on special relativity, On the Electrodynamics of Moving Bodies, states two main problems in electromagnetic theory at the end of the 19th century:

The first paragraph states, essentially, that Maxwell's equations will not have the same form in a frame of reference moving with constant velocity relative to a frame where they are the laws governing electromagnetism if you transform them using Galilean transformations.*
The beginning of the second paragraph acknowledges "…the unsuccessful attempts to discover any motion of the earth relatively to the 'light medium' [luminiferous æther, e.g. the Michaelson-Morley experiment]…"

Einstein had a strongly-held philosophical belief that, for something to be a true law of physics, it must be the same in all reference frames—regardless of where you are or how you are moving, the laws of physics for you are the same as for anybody else in the universe; this is the principle of relativity. Since he believed Maxwell's equations to be laws of physics, he concluded that Galilean relativity was not correct and this led to his postulate that the speed of light was independent of the motion of the source or the observer. To my own mind, I believe that the constancy of the speed of light is not a necessary postulate, rather a consequence of the principle of relativity applied to Maxwell's equations.

*A Galilean transformation is, for example, observing a car moving 60 mph east relative to the ground to be moving 100 mph east if you are in a car moving 40 mph west relative to the ground.

QUESTION:
What type of impact will cause the metal in the car to "splash" (eliminating all hope of a simple conservation of momentum calculation) should the New Horizons interplanetary space probe hit your car?

ANSWER:
The only thing which prevents linear momentum from being conserved is an external force. If there are no forces other than those exerted by the car and the probe (or pieces thereof), linear momentum is conserved. For example, if the collision happens in empty space with the probe of mass M moving with velocity V in a frame where the car is at rest, the total momentum of all the pieces will be MV after the collision or any time during the collision. What will not be conserved is the kinetic energy ½MV².

QUESTION:
In a recent answer, you mention the Cosmic Microwave Background and say that it "provides a frame of reference for the whole universe". It was my understanding that one of the underpinnings of the basic theory of relativity is that there CANNOT be such a universal frame of reference (UFOR). I assume you're not claiming Relativity has been disproven (well, shown to be a special case of something else), so can you explain the difference between the frame of reference meant for relativity and the CMB as you described it?

ANSWER:
Good question! Special relativity rests on the assumption that the laws of physics must be the same in all inertial frames (the principle of relativity); no frames are ruled out as long as they are inertial. In other words, no one frame is better than all the rest. But suppose that we find a frame of reference which contains a "gas of photons" which are all moving in random directions (average velocity of all these photons is zero) and this "gas" pervades the whole universe. Then, if our frame happens to be moving with some velocity v through this gas, we will see that gas having an average velocity -v which would be determined by looking at the doppler shift of the photons; this is analogous to feeling a wind on your face when you run in still air. Having found such a frame allows us to specify an "absolute velocity" which, before now was not possible. What makes this frame special is the fact that, because it is a remnant of the big bang, it does pervade the whole universe.

QUESTION:
I was curiouse at which height would a 6kg turkey have to fall for it to be perfectly cooked when it hits the ground?

ANSWER:
It is very difficult to do a quantitative calculation because the air drag (which would cause heating) changes with altitude as the air gets less and less dense. I doubt that there is any height where this could work because, giving it a high enough speed to compensate for the short time (keep in mind that it takes several hours to roast a turkey at 350⁰F) of fall would result in the outside being burnt and the inside being raw. It might be fun to do a couple of very rough calculations just to get an order of magnitude feeling for what you are interested in.

I will approximate the turkey to have the properties of water, in particular that its specific heat is C=4.186 J/gram⋅⁰C; so the energy which will raise the temperature of a M=6000 gm turkey is E=MCΔT=6000x4.186x56=1.41x10⁶ J. (70⁰F→170⁰F is 21⁰C→77⁰C, hence ΔT=56⁰C.) The change in potential energy in falling from a height h is V=Mgh=6x9.8h=59h, so if the turkey falls with constant velocity the potential energy lost goes into heat. If the heat is 1.41x10⁶ J, the height is h=1.41x10⁶/59=24,000 m. The atmospheric pressure at this altitude is only about 3% what it is at sea level; so it is clear that, if your turkey is given any velocity at this altitude, it will not continue moving with constant velocity. (Whereas, if the pressure were atmospheric and you gave the turkey a speed equal to the terminal velocity it would continue moving with that speed as it fell.) If you dropped it from very far away (essentially infinitely far), it would arrive at the upper atmosphere going nearly the escape velocity, 11,200 m/s. At this speed, which is faster than the reentry speed of the shuttle which needed special ceramic tiles to shield against burning up, the turkey would be burned to a crisp, probably would not reach ground before totally burnt. Although there would be some speed of entry to the atmosphere where the turkey would absorb 1.41x10⁶ J, it would almost certainly not be "perfectly cooked".

QUESTION:
In a nuclear fusion reaction hydrogen atoms combine to create a helium atom. But in a hydrogen atom there are no neutrons,then where does the neutrons in helium nucleus come from ?

ANSWER:
There is more than one answer to this question. We can look first about how hydrogen (with no neutrons, just one proton) becomes helium (two protons, two neutrons) in a star like our sun. As shown by the figure, it begins with pairs of protons which interact and fuse in a reaction which results in one deuteron (heavy hydrogen, one proton and one neutron), one positron (a positively charged electron), and one neutrino. The deuterons thus created now fuse with another proton to create a

light isotope of helium, He (one neutron, two protons), and also a gamma ray. Two of these ³He now fuse, throwing off two protons in the process and one alpha particle (⁴He, two protons and two neutrons). The two thrown-off protons are now free to continue the process, so this chain, called the proton-proton cycle, is self-sustaining until the star runs out of hydrogen which ends the life of the star.

For more earth-bound hydrogen fusion, like fusion reactors or hydrogen bombs, single-step reactions are sought. The most frequently used is the deuterium-tritium reaction, involving the fusion of one deuteron and one triton (³H, one proton and two neutrons) shown in the other figure.

QUESTION:
If time ticks at 1 second per second in a 1g gravity at the earths surface. Then my question is would a clock tick twice as fast 2 second per second if the earth surface was 0.5g.

ANSWER:

Most certainly not! The expression for the time t elapsed in the gravitational field of a nonrotating spherically symmetric mass M is t=t₀/√(1-δ) where δ=2GM/(Rc²)=2aR/c²; here, R is the distance from the center, t₀ is the time elapsed when R=∞, G is the universal gravitational constant, M is the mass of the sphere, and c is the speed of light, and a is the local acceleration due to gravity. If a=g and R=6.4x10⁶ m (earth radius), it is easy to calculate that δ=1.39x10^-9. Now, t_g/t_½g=[√(1-½δ)]/[√(1-δ)]; using the binomial expansion (1+δ)^x≈1+xδ+… if δ<<1, we finally get t_½g≈t_g(1+¼δ)=t_g(1+3.5x10^-10), quite a bit different from t_½g=2t_g!

QUESTION:
Why does steam rise from the cooking pot when I turn off the heat after the water starts boiling?

ANSWER:
Water evaporates from the surface and the higher the temperature, the faster it evaporates. The higher the rate of evaporation, the more likely it is that individual evaporated molecules will coalesce with others to form droplets of water large enough to be seen. That is what steam is and what clouds are. But steam can be observed even at low temperatures under the right conditions, for example the morning mist on the surface of a lake or fog. The water in the cooking pot, if you observe it carefully, will begin to steam before it starts boiling. So it should be no surprise to find that it still steams after you turn off the heat.

QUESTION:
what makes my radiometer occasionally accelerate wildly when bumped? I can't duplicate the action but have seen it a few times. It has happened around a desk lamp with an LED bulb. Could it be part of an electrical emission around the radiometer? I was allays taught that the photons are what push the vanes

ANSWER:
Well, as it turns out, photon pressure is not what makes the radiometer turn. You can learn about radiation pressure by reading an old answer and the links it contains. I do not know what "accelerate wildly" means in this context but, since you mentioned "bumped", mechanical bumping probably caused unexpected behavior of the radiometer.

QUESTION:
I wanted to ask a question regarding airfoils. When looking at videos online about them, the instructors say that airfoils work because the top of the wing turns air down due to the coanda effect and the downturned air also turns the air under the wing down to create lift. I think I understand the law applied here to be with every action there is an equal and opposite reaction however, with the explanation they gave, it seemed like they were saying that the air on the top that was turned down was the only thing that is turning the air under the wing down. I that true? I would assume that, even though that does happen, the lift generated from the bottom of the actual wing deflecting the air down was more and more important than the coanda effect on the top. The video also said that disrupting the flow on top of the wing through too high of angle of attack would cause it to stall, but wouldn't the actual reason being that the bottom of the wing converts too much lift into drag by changing the opposing force that is lifting it from mostly up to mostly back? There was another source that said that there was a vortex at the back of the wing that pushed the air at the bottom of the wing so that the net force would cancel out a bit, slowing the air down and making a pressure difference to lift the wing. What is that about?

ANSWER:
This question violates the site groundrule for "single, concise, well-focused questions". I can tell you that what is learned in elementary physics courses that airplanes get lift from the Bernouli effect of lower pressure due to higher velocity of air on the top of a wing is a relatively small part of how an airplane is able to fly. As you seem to refer to in your question, the main thing is that air leaves the wing with a downward component of its velocity; this means that the wing exerted a force down on that air which means, via Newton's third law, that the air exerted a force up on the airplane. To quote Wolfgang Langewiesche, author of Stick and Rudder, the private pilot's bible on the art of flying, "The main fact of all heavier-than-air flight is this: the wing keeps the airplane up by pushing the air down."

QUESTION:
How many light-years or parsecs away from 2018 is 1961? The solar system, the earth and the galaxy would have been in a different part of the universe. Do we know the direction of where our galaxy, solar system used to be before we arrived here in the present moment? How many miles has the Earth traveled in space in fives minutes? Including the movement of the solar system and galaxy moving through the vacuum? Millions of parsecs? Billions?

ANSWER:
It never makes sense to talk about velocity of something without specifying the frame of reference relative to which the velocity is measured. And when you are talking about the earth, the motion is very complicated because it rotates about its center (about 0.5 km/s), revolves around the sun (about 30 km/s), and the solar system revolves around the center of the Milky Way galaxy (about 220 km/s). What about the motion of the center of the galaxy? Until recently, you could only measure that speed relative to another object outside our galaxy, e.g., another galaxy or the center of our local cluster of galaxies. However, the discovery of the cosmic microwave background (CMB) has now given us a reference frame which seems to define the reference frame of the whole universe. Very careful measurements and data analyses from the satellite COBE have shown that our galaxy moves with a speed of about 580 km/s relative to the CMB.

So, what can we do to answer your question which is, essentially, how far does the earth move in 57 yr=1.8x10⁹ s? (Your 5 minute question is essentially impossible to answer since it depends on the time of year. Over years the orbital motion averages out to zero.) Relative to the center of the galaxy, the distance is D=220x1.8x10⁹ km= 4x10¹¹ km=0.013 pc. The center of the galaxy in that time will move 580x1.8x10⁹ km=10¹² km=0.032 pc. But I do not know the relative directions of these two velocities, the sum of which would be the average velocity of the solar system over these 57 years; this speed would be anything between 360 and 800 km/s. However, in my researching this question I found that the speed of sun relative to the CMB is about 370 km/s, so the final answer is that the earth moved about 370x1.8x10¹¹=6.7x10¹¹ km=0.022 pc. It is interesting to me that, at this time, the sun and the center of the galaxy are moving in nearly opposite directions (370 km/s compared to 360 km/s); since the time for the sun to go all the way around the galaxy is about 225 million years, in about 110 million years our speed relative to the CMB will be about 800 km/s.

Keep in mind that I am not an astronomer/astrophysicist/cosmologist! I have gathered these data as best I can.

QUESTION:
I have a potentially stupid question about pressure in a vacuum. So I just found out that in a vacuum chamber, microns can be used for measuring vacuum pressure because it's representative of a DISTANCE- the distance being the displacement of columns of Mercury. In understanding Leak Up Rate tests in a vacuum furnace, the Leak Up Rate is a function of Difference of Vacuum Levels/Elapsed time. My question is, does the displacement of Mercury happen faster with more time? IE, does the displacement of Mercury gain momentum with extended periods of time under vacuum? Or, IE, is it exponential instead of linear?

ANSWER:
Very likely there is not a column of mercury being used to measure pressure in your chamber. There is probably some mechanical device using strain gauges which is calibrated in mm or μm of Hg. So, you do not have to worry about what a column of mercury is going to do as the pressure changes. 1 μm (micron of Hg)=0.133322 Pa (N/m²)=1.93368x10^-5 PSI=1.31579x10^-6 atm.

QUESTION:
After a math class yesterday (homeschool) , my son and I were discussing an article I read a few days earlier regarding a exploded star that went super nova about 150 million years ago. The event bathed the earth in a stream of high energy cosmic rays for a couple of months and possibly caused an extinction level event that effected ocean life of the day.

As the conversation progressed, we considered the implications and began to wonder if it was possible to shield against cosmic radiation? To the best of my knowledge, I explained that no substance I know of can block cosmic radiation and that the Earth's magnetic field would have little effect either. The only effects I know of are when they impact atoms in the atmosphere or the earth. If I understand this, if it doesn't hit something fairly dense, it just keeps going, understandable given the space between atoms.

I have read about the gravitational lensing effects on light by massive objects like neutron stars and galaxies that bend light around them. I have also read about the possibility of warping space as a method space travel by changing the shape of space in front of and behind a craft so that it "falls" in a particular direction.

While we were discussing these effects, he looked at me and said something to the effect, "Maybe we could bend space and make them go around?". To me, this was a pretty profound leap in my book. So I decided to ask physicist if he could be right.

Please forgive the length of this post before asking our question. I wanted to provide sufficient context before asking. So, here it is. Is it possible to make cosmic rays go around something, either by gravitational or microwave warping of space?

B.T.W. I really really enjoy your web site. It is awesome!

ANSWER:
I am not a paleontologist, so I had to do a little research on mass extinctions. I discovered that there is no known mass extinction 150 million years ago. In addition, no known extinction is associated with a nearby supernova. (The timeline here is clickable for an enlarged picture.) I would conclude that the article which you read is a "fringe theory" of mass extinctions, not a mainstream fact.

But, I can speak to your main question and comment on some of your statements. Cosmic radiation refers to a broad range of radiations but mostly protons and light nuclei; but also it could refer to neutrinos, electrons, positrons, antiprotons, gamma rays, x-rays, and others. What is not the case, though, is that "no substance…block cosmic radiation"; just about any cosmic radiation can be blocked by a dense solid or liquid. Neutrinos are the exception and most pass all the way through the earth without any interactions. You may know many neutrino experiments are performed in deep abandoned mines to block out all other radiation. The atmosphere also is very effective since a single energetic proton will strike an atom and many other particles, all with less energy than the original proton, will form a "shower" of particles. And magnetic fields are not very effective deflecting the very highest energy charged particles but are known to be an effective shield for lower energy particles.

Finally, let's address the question of warping of space as a way to deflect the rays. The only way to warp space that I know of is to have an enormous mass nearby. The earth already warps the space around it which would cause a particle passing close by, but not on a collision course with earth, to be deflected a tiny bit (viz. gravity). But putting a star sized object near enough the earth to "shield" us would obviously be a catastrophic thing to do! The whole solar system would be disrupted. I have no idea what "microwave warping of space" means.

I commend your son for his hypothesizing and you for your tutoring and encouragement of his curiosity.

QUESTION:
I understand that LED light won't refract a full spectrum of colors through a prism. Is this true? I want to combine a grouping of different wave length diodes to deliver a white light that will separate from indigo to red. Can this be done?

ANSWER:
Although a simple LED will not be monochromatic like a laser, it will tend to have a fairly narrow range of wavelengths emitted, not a full spectrum. The most common to make a white LED is to take a blue or near-UV LED and coat it with a material called a phosphor. The phosphor will absorb some of the light from the LED and reemit that energy in a broader spectrum creating white light. The details of the spectrum depend on the phosphor used; varying the phosphor used, for example, can create the whites described as cool white or warm white or bright white, two of these are shown in the figure. I do not understand what "separate from indigo to red" means.

QUESTION:
As I understand it, electricity and magnetism are the same force; this is where "electromagnetism" comes from. I also understand that radio waves, light, and non-particle radiation like x-rays are all the same thing; waves in the magnetic field. However, they're carried by different particles: electrons for electricity, photons for magnetism (radio, light, etc). Furthermore, there is a significant difference between electrons and photons; electrons have mass, photons do not. Wouldn't this mean that electricity and magnetism are actually separate forces which are so tightly coupled (have a very strong influence on eachother), that it's just easier or simpler to think of them as the same force most of the time? One example of this coupling in my hypothesis is how electrons gain energy when a photon collides with and is absorbed by an electron, and the reverse, when an electron emits a photon and looses energy.

ANSWER:
You have some serious misconceptions here. First, the waves you refer to are not "waves in the magnetic field", they are waves which are composed of magnetic and electric fields—they are electromagnetic waves. Second, both electric and magnetic forces are "carried by" photons—electrons never play that role. Indeed, there is only one field, the electromagnetic field; for example, if you have a charged particle at rest you will only see an electric field, but if you give the charge some velocity you will now also see a magnetic field and you will find that the electric field is somewhat smaller which you could interpret as having "morphed" into the magnetic field.

QUESTION:
I want to reach 90% c in 30 days. How many Gs would I have to accelerate at to reach that velocity in the given time? (in a vacuum)

ANSWER:
You should read an earlier answer to understand the answer here. The graph here is from that earlier answer and shows (in red) v/c as a function of a₀t/c where a₀is the acceleration measured in the frame of the object which is accelerating and t is the time since the object was at rest. Reading off the graph, you can see that v/c =0.9 when a₀t/c=2. Now, t=30 days=2.6x10⁶ s, so a₀t/c=a₀(2.6x10⁶/3x10⁸)=8.64x10^-3a₀=2; solving, a₀=232 m/s²=23.6 Gs. I would not suggest having anybody aboard this object!

QUESTION:
Everything being equal,including speed and mass, Like a man walking up a slope, does the amount of energy expended on say a 10% inclined slope double with a 20% inclined slope?, if traveling at the same speed and everything else the same? In other words, is energy expended related arithmetically and constant according to slope of incline. and resistance, or is it proportional to the slope by some other factor, geometric increase.?

ANSWER:
In the figure the man of mass m is walking up a hill with incline θ with a constant speed of v. His potential energy is E=mgy where y is his height above the bottom of the hill and g=9.8 m/s² is the acceleration due to gravity. The rate at which his energy is changing is dE/dt=P=mg(dy/dt)=mgv_y=mgvsinθ; so this rate (called power) depends only on the vertical, not the horizontal component of his velocity. Now, you have made things difficult for me because you have not specified the angle, but rather the inclination which is actually the tangent of the angle (rise/run) times 100%. I will call this inclination the grade G=100tanθ. For example, let's take a man with mass 100 kg walking with a speed of 2 m/s up the 10% and 20% inclines you specify. The angles would be θ_10%=tan^-1(10/100)=5.71⁰ and θ_20%=tan^-1(20/100)=11.31⁰. So the power values are P_10%=100x9.8x2xsin(5.71)=195.0 Watts and P_20%=100x9.8x2xsin(11.31)=384.4 Watts. The ratio of these two is 1.97, close to, but not equal to, 2.0; for small angles the sine and tangent are approximately linear functions but as the angle (hence inclination or grade) increases this is no longer a good approximation. The graph shows the power for grades up to 200% (about 63.4⁰); as you can see, the power increases approximately linearly only up to about a grade of about 30% which is about 17⁰.

QUESTION:
A vehicle , unknown speed of travel, strikes a 14lb object and displaces it 29ft. Could you please tell me what the approximate speed of travel is and how you come to that conclusion. I ask because a vehicle deliberately veered to the opposite side of the street in order to run over my cat. The driver was successful in killing my sweet lad and I am trying to collect as many facts as I can before I request help in prosecuting the individual under Arizona's Animal cruelty laws.

ANSWER:
I am afraid that you cannot deduce the speed of the vehicle from the information you gave me. You need to know how elastic the collision was and what the trajectory of the cat was.

FOLLOWUP QUESTION:
Thank you for responding to my question but I am at a loss as to what is meant by "elastic". As far as trajectory my cat was launched a distance of 29 ft from point of impact with the next point where he died on the street...� no evidence, ie; of blood in between with there being blood evidence only at the point of impact and final resting location. If it would help I can shoot a video of the street and the location of where Liam was struck and where he landed in relation to the surroundings.

ANSWER:
Elastic refers to the energy lost in the collision which gives you information on how fast the cat was moving immediately after the collision. The extremes are a perfectly inelastic collision where the cat would be stuck to the vehicle after the collision and perfectly elastic which would have the cat moving with approximately twice the speed of the car after the collision. Clearly, neither of these would describe a car hitting a cat and I cannot imagine how you could approximate it.

When I say "trajectory" I refer to the path from collision point to the ground. The horizontal distance gone from the point of collision to the point of landing depends on the angle at which the cat started; if launched at 45⁰ relative to the horizontal he will go much farther than if launched at 20⁰. Again, unless you witnessed the accident, there is no way to know the angle.

QUESTION:
If you are traveling on a spaceship close to the speed of light (pick a number e.g. 0.95c), With or without constant acceleration whichever works for the question. Time will slow down relative to earth and you will be able to travel a greater distance than you would be able to without relativistic effects. You would also be able to return to earth with significant time passed relative to your experience. My question is, what would it feel like on the spaceship...I know the standard answer is you wouldn't feel anything, but logically, if you were traveling through galaxy after galaxy in a relativistic time of human life then intuition says everything would look like it whizzing past. Basically, what is the human experience of being able to travel interstellar distances?

ANSWER:
Of course everything is whizzing past compared with what it would look like if you were at rest relative to the earth. What you would also see is that the distance to an object you were moving toward is shorter by a factor of √(1-0.95²)=0.31; so if you went to a star 1 light year from earth, it would only take you 0.31/0.95=0.33 years to get there but the earth-bound clock would say it took you 1/0.95=1.05 years. Also where the stars appear to be relative to where they would be if you were not moving would be different; see an earlier answer. In the case where v/c=0.95, the graph shows how the angle at which you see something depends on where it would be if you were at rest; for example, a star actually at θ=120⁰ (30⁰ behind you) would be seen as being about θ'≈30⁰ in front of you!

QUESTION:
how much times earth to be compressed to become black hole?

ANSWER:
A black hole is a singularity, of zero size, so compression is infinite. If the earth were a black hole, its Schwartzchild radius, R=2GM/c², would be about 1 cm. R is the radius within which nothing can escape. So, if you could get the earth down to that size, it would then collapse the rest of the way to a black hole, I presume.

QUESTION:
If momentum is applied not on the center of mass on an object, what would its momentum be at the center of mass and is there an angular momentum at the point of impact?

ANSWER:
Linear momentum is not something which is applied, force is what is applied. And, what happens also depends on the details of how the force is applied. I will work out one simple example which will illustrate the principles involved when the force is applied in a simple collision. A mass m approaches the end of a thin uniform rod of mass M and length 2L with velocity v perpendicular to the rod; when it collides with the end of the rod it sticks. The center of mass of the rod is at its geometrical center a distance L from where m exerts the force. Linear momentum and angular momentum are both conserved but, as we shall see, energy is not conserved. After the collision the center of mass has a velocity u; conserving linear momentum, (M+m)u=mv or u=v[m/(M+m)]. Before the collision m brings in an angular momentum mvL. After the collision, the rod plus mass has angular momentum (I+mL²)ω where I=M(2L)²/12=ML²/3 is the moment of inertia of a thin rod about its center of mass. Conserving angular momentum I find ω[m+(M/3)]L²=mvL or ω=mv/{[m+(M/3)]L}.

Just for fun I put in some numbers to illustrate what would happen in a specific situation: m=1 kg, M=3 kg, L=1 m, v=4 m/s. With these I find: u=1 m/s, ω=2 radians/s=19.1 rpm. You can also calculate the energy before and after the collision: E_before=½mv²=8 J, E_after=½(M+m)u²+½([m+(M/3)]L²ω²=6 J. So, energy was not conserved in this collision, 2 Joules were lost.

CORRECTION:
I realized that an error had been made in my analysis. After the collision the center of mass is no longer at the center of the rod but displaced a distance x=mL/(M+m) toward where m is stuck. It is this center of mass which moves forward with speed u, and u is still v[m/(M+m)]. The rotation is about this center of mass but now the moment of inertia is different; the net (rod plus mass) moment of inertia is ML²/3+Mx²+m(L-x)², so the corrected value of the angular velocity is ω=mvL/[ML²/3+Mx²+m(L-x)²]. In the example I did, x=1/4 m, so ω=16/7 radians/s=21.8 rpm and E_after=46/7=6.57 J (1.43 J lost).

QUESTION:
I have been trying to figure this out, but having some trouble doing it: In general, let's say I had a vehicle that could accelerate me (according to my perception) at 1G for 1 year (again according to my perception), then turn around and decelerate (by my perception again) for another year to return home. How far would I have travelled and how old would my twin at home be when I got back?

ANSWER:
The reason you are having trouble is that relativistic kinematics involving acceleration, even if the acceleration in one frame is uniform, is not easy; the main reason is that, unlike Newtonian mechanics, acceleration in one frame is not the same as acceleration in a different frame. As a prelude to reading my answer, you should read an earlier answer and the earlier answers referred to in that answer. The graph here shows the results derived or cited in that earlier answer. The three graphs all show what is observed of your motion as seen by the twin on earth. The red curve is your velocity v relative to c as seen by the earth-bound twin, v/c=(gt/c)/√[1+(gt/c)²]; the blue curve is your acceleration a relative to g, a/g=[1+(gt/c)²]^-3/2; the black curve is your position expressed (approximately) in light years, gx/c²=√[1+(gt/c)²]-1. Now, since I have set this up as seen from earth, let me specify the time you travel as the time of your trip out as observed from earth: earth-bound twin observes you to arrive at you destination in two years. Then, as you can read off the graph, gx/c²=√[1+(2)²]-1=1.24 ly; the return trip will be the same length of time, so the earth-bound twin's clock will have advanced 4 years. Light would take 2.48 yr to travel that distance, so your clock will have advanced a time somewhere between 2.48 and 4 years. Your maximum speed would have been about v/c=(2)/√[1+(2)²]=0.89. We can get a rough estimate of your clock reading by approximating your average relative speed as 0.89/2=0.445; so you would have seen, on average, the distance contracted by √[1-(0.445)²]=0.9 and your time would have been 0.9x4=3.6 years.

QUESTION:
What happend to accelaration due to gravity if the earth stops rotating?

ANSWER:
Because of the centrifugal force, g appears to be somewhat smaller than it really is, but the amount is very tiny. This effect depends on your latitude and is zero at the poles and greatest at the equator. Also because of this effect, the acceleration is not directly toward the center of the earth except at the poles and equator.

QUESTION:
Hello, is the mass of a person on a spaceship (to provide a suitable upward force) important in designing a spaceship? My teacher says otherwise but upon researching I found out that additional fuel is required for every kg of payload that will be sent out into space.

ANSWER:
If the ship and all the fuel are much bigger than the mass of the person, then you could neglect the mass of the person in any calculations you know; in other words, the rocket launch would be, as close as you could hope to be measure, identical whether or not an astronaut was on board. However you are also right that the amount of fuel you need depends on the payload you want to deliver, and that would include everything which is not fuel. You might find the an earlier answer illuminating.

QUESTION:
How exactly do electrons in electric discharges generate radio waves? Whenever someone uses an AM radio while there are electric sparks like lightning strikes nearby the radio waves they emit can be detected.

ANSWER:
Any time an electric charge is accelerated it emits electromagnetic radiation. There are many electrons in the spark which accelerate under the influence of the electric fields they experience.

QUESTION:
I just read about the kilogram being defined by measuring the current of an elector-magnet to balance a scale. How can various electromagnets have exactly the same force to current ratio? Would variations in the purity of the conductor, variations in the exact coil shape, etc. cause variations in this ratio? What am I missing about this new designation?

ANSWER:
You cannot use any old electromagnet. There is an exquisitely accurate balance called the Kibble balance which is used. Also see Wikipedia for more detail

QUESTION:
Is the following statement correct? Net radiated energy is made up of emitted and absorbed energy.

I stated that it was incorrect based on the fact that "made up of" usually means putting two or more things together (aka addition). Net radiated energy is the difference between emitted energy and absorbed energy. I have yet to see my exam paper but I believe that the mark I lost was due to this question on basis that �made up of� means that it is a result of a mathematical relation between the two values and not necessarily addition (in this case subtraction). What is the correct answer in this case and why?

ANSWER:
I do not believe that there is a hard and fast definition of what "net radiated energy" is. It is a matter of semantics. However, radiated energy usually means energy emitted and absorbed energy means energy absorbed; therefore, I would say that this is an incorrect statement because net radiated energy would mean the sum of all radiated energy, absorbed energy having nothing to do with it. Your reason is quite shaky because "made up of" is even more semantically ambiguous than "net radiated"; it would simply imply that "net" means total energy flux.

QUESTION:
How is momentum conserved for single or double slit diffraction? in other words:I have a photon gun that I can dial the intensity down to fire one photon at a time. I fire my photon gun, the gun recoils in the -X direction, the photon flies off in the +X direction. The photon then passes thru two closely spaced slits (double slit experiment)instead of appearing on the screen directly behind the slits, the photon ends up to the side (after many, many photons the interference pattern appears) however, for the single event of a single photon, how is momentum conserved? dos the single event photon exchange momentum with the walls of the slit? Since the photon ended up to the side, it was no longer travelling in the purely X direction, therefore shouldn't the photon have exchanged momentum with the slit somehow? If the photon interacts with the slit, that counts as �which way� information, and the interference pattern vanishes. If the interference pattern remains, then how do we account for the photons change in momentum? This is for a single event, one photon, not expectation value of many, many photons.

ANSWER:
You are forgetting that the photon goes through both slits, is being forced (if you like) to assume its wave-like identity. It behaves like a wave until something makes it act like a particle, so it is a wave spreading out in both directions, so its net momentum in the direction parallel to the screen is zero. When the screen is encountered it is a "measurement" which collapses the wave function at some point.

QUESTION:
I can imagine 2 similar objects rotating around a sphere at the same velocity but at different orbits. One object would rotate around the sphere faster than the other, right? What is puzzling to me is that if I think of both objects going now in a straight line one could not say that one object was going faster than the other. Can someone help me through this? Is there a subtlety here?

ANSWER:
You want to say same speed, not same velocity, but I get the idea. Say one orbit is twice the radius as the other. Then the smaller orbit has a circumference half that of the larger. So each time the object in the larger orbit goes around once, the one in the smaller orbit goes around twice. But they still have the same speed. But the angular speed of the object in the smaller orbit is twice that of the object in the larger orbit even though their linear speeds are the same. You cannot do this with planets or moons, though, because the speed of a satellite in a circular orbit depends on the radius.

QUESTION:
If space is curved, doesn't it need another dimension to curve into? If a 2d object is curved it curves into the 3rd dimension. So if 3d space is also curved it must curve into 4d.

ANSWER:
This is really mathematics, not physics; I am not a mathematician. But the definition of an N-dimensional space is, I believe, that N quantities are required to specify the location of a point in that space. Curvature of a space is defined as a space in which the geometry is not Euclidean. The physical existence of an N-dimensional space does not imply the physical existence of an N+1 dimensional space. You cannot make a generalization that curvature is necessarily "curving into" a higher dimension. For example, a planar sheet can be distorted without leaving the 2D. You also might "curve into" a higher dimension than N+1; for example, a coil spring is a one dimensional space which exists in a 3D space. You are trying to make generalizations based on the only spatial dimensions which we know to exist; but higher dimensions are defined by certain rules which in no way prove that they exist physically.

While researching this question I came upon a very interesting little essay about general relativity, curvature of spacetime, gravity. Many nonscientists (and scientists as well) are familiar with and accept as gospel that general relativity demonstrates that the distortion of spacetime by the presence of mass is what gravity is; the "cartoon" which illustrates this is the "bowling ball on a trampoline example in which a mass causes the distortion of the two-dimensional space defined by the trampoline surface. The implication is that gravity is just geometry. On the other hand, many theorists are struggling to find a successful theory of quantum gravity and to do that one must treat gravity as a force. The take-away from this essay, though, is that curved spacetime is not a unique representation of what the mathematics which are the framework of the very successful theory of general relativity mean. The author, Rodney Brooks, notes that "…you can view gravity as a force field that, like the other force fields in QFT [quantum field theory], exists in three-dimensional space and evolves in time according to the field equations." He also goes so far as to say "�shame on those who try to foist and force the four-dimensional concept onto the public as essential to the understanding of relativity theory."

QUESTION:
A rear wheel drive car accelerates quickly from rest. The driver observes that the car noses up. Why does it do that? Would a front wheel drive car behave differently?

ANSWER:
The car with rear-wheel drive has an acceleration a and a mass M. The forces on the car are its own weight W (W=Mg) acting at the center of mass a distance d from the real axle; the normal and frictional forces on the rear wheels, N₁ and f₁; and the normal and frictional forces on the front wheels, N₂ and f₂; the radius of the wheels is R and the distance between front and rear axels is D. I will ignore the rotational motion of the wheels, that is, I will approximate their masses as zero. If the front wheels are not lifting up from the ground, the sum of torques about the rear axel is (f₁-f₂)R+N₂D-Mgd=0; Newton's first law for the horizontal and vertical directions are (f₁-f₂)=Ma and N₁+N_₂-Mg=0. From these equations it is easy to show that N₂=M(gd-aR)/D. This is interesting because it shows that N₂=0 if a=gd/R; for larger accelerations, the front wheels will lift off the ground. (Note that as N₂ approaches zero, f₂ also approaches zero.) I have ignored the suspension of a real car, so this lift will not be abrupt; it will happen gradually as the front springs decompress and the rear springs compress; so overall the nose goes up but the rear end goes down.

If it is a front-wheel drive, the directions of the frictional forces reverse. The corresponding torque equation (this time about the front axle) is (f₂-f₁)R-N₁D+Mg(D-d)=0; Newton's first law for the horizontal and vertical directions are (f₂-f₁)=Ma and N₁+N_₂-Mg=0. From these equations it is easy to show that N₁=M(aR+g(D-d))/D. As a increases in magnitude, N₁ gets bigger and bigger until N₁=Mg which necessarily means that N₂ becomes zero. It is easy to show that this happens when a=gd/R,, exactly the same result as for the rear wheel drive.

Finally, it might appear that there is no difference between front- and rear-wheel drive. That is not the case in the real world at all. Keep in mind that the frictional forces are static friction and the larger the corresponding normal force is, the more friction you can get. Since in the rear-wheel situation the normal force gets bigger, the amount of friction you can get without the wheels slipping can keep increasing so that eventually (with a powerful enough engine) you could lift the whole front end far off the ground. However, when you have front-wheel drive, the normal force on the front wheels gets smaller as you increase a, so most likely the wheels will slip well before N₂=0 when f₂ must be zero. For very large accelerations, like required for drag racers, rear-wheel drive is imperative.

QUESTION:
Gravity is what makes objects orbit around other objects,and gravity is a reflection of an object�s mass.So why doesn� the mass of the objects appear in kepler�s third law??

ANSWER:
For the same reason why all objects have the same acceleration in the gravitational field, i.e. all objects near the surface of the earth have an acceleration of 9.8 m/s². The acceleration (which is what describes the orbit) is F/m but the force is MmG/r², so the acceleration is MG/r².

QUESTION:
If kinematics can give you data on a dropped object, would it be valid to use kinematics to calculate the acceleration experienced by the object upon impact with the ground? (the time from when it hits the ground to when it actually stops)

ANSWER:
When the object hits the ground until its velocity reaches zero is some time, call it Δt. Its momentum drops from mv (where m is the mass and v is the speed when it hits) to zero because an impulse I was delivered to it by the ground. The impulse may be written as I=F_avgΔt. where F_avg is the average force. Therefore you can find the average acceleration a_avg during the collision, but not the acceleration as a function of time: a_avg=F_avg/m=I/(mΔt)=v/Δt. So you cannot get detailed kinematics about the collision without measuring how the speed decreases with time.

QUESTION:
We know that air density decreases as you ascend into space, but what about looking out to the horizon? Is there a atmospheric lensing that occurs? Is there a formula for its calculation?

ANSWER:
Certainly there is refraction due to this density profile of the air. There is no simple formula because the actual density at any point is dependent on the pressure and temperature of the air as well as altitude, and these vary with time. One could calculate based on average density profiles, but I would suppose it is a numerical calculation, not an analytic formula. There is a nice description of this phenomenon here.

QUESTION:
I've been interested in physics for years but have never managed to get my head around the concept of space-time. I understand time and i understand space but somehow i could never get this. However today i think it finally clicked when i wasn't even trying. Can you tell me if my understanding of it is correct or not? I thought of space-time as like if i were on a roller-coaster or train tracks. When i am on the long flats, i travel slower, and when i am on the declines, i travel faster. Is space-time like this analogy? In that, when you are on Earth for example, you experience time moving faster than if you were floating in space millions of miles away. This is because the presence of Earth "warps" space time, much like a bowling ball on a trampoline. This "dent" in space-time causes a gravitational pull, pulling time into it, much like if i were rolling down a hill. I know that might sound very convoluted but it's the idea i have in my head and it has bothered me for YEARS, but i think i am on to an understanding here.

ANSWER:
Sometimes we make things more difficult than they need to be. The main takeaway from the theory of relativity is that time and space are not separate and universal things. In Newtonian times, it was assumed that any clock in the universe would run at exactly the same rate; now we know that if a clock moves with respect to us that it runs more slowly than ours—doesn't appear to run slower, it runs slower. Similarly, space depends on the observer; a meter stick moving by you is actually shorter than 1 m. What these facts about space and time tell us is that time and they are not different things, they are really entangled with each other. You cannot talk about one without talking about the other. They are two pieces of the same thing. So the term space-time is simply a way of acknowledging that. I do not find your roller coaster analogy very helpful. The bowling ball on a trampoline is a classic but keep in mind that it is really a cartoon and not rigorous. Generally trying to picture something in four dimensions futile!

QUESTION:
What does a wind speed (a gust is okay)in miles per hour, need to be to launch a baseball at a speed of 80 miles per hour into an object like a window. Given: Ball is 60 feet 6 inches from the window that it hits traveling at 80 mph Baseball is 9.25 inches in circumference and 2.94 inches in diameter and weighs 5.25 ounces My name is Karin and I'm testing to see if I use a baseball pitching machine from 60 feet 6 inches from a window protected with a protective panel, and set the machine to pitch at 80 mph, what would the hurricane wind speed need to be to get that baseball to fly into the window at 80 mph. I hope this is clear enough.

ANSWER:
If I understand your question, you are essentially asking what wind speed would be required for a baseball to reach a speed, starting from rest, of 80 mph in a distance of 60.5 ft due to the force of the wind on the ball. This is a problem which is very tedious to do in full analytic detail and not very illuminating. Also, all problems involving air drag are approximate, so it would be appropriate for me to find an approximate way to solve this problem. I am going to work in SI units so I will rephrase your question: what wind speed would be required for a baseball to reach a speed, starting from rest, of 35.8 m/s in a distance of 18.4 m due to the force of the wind on the ball? In SI units you may approximate the force at sea level which the wind exerts on the ball as F=¼A(u-v)² where u is the speed of the wind, v is the speed of the ball, and A is the cross sectional area of the ball. Applying Newton's second law, ¼A(u-v)²=ma=mdv/dt . This differential equation may be shown to have the solution t=cv/[u(u-v)] where c=4m/A and t is the time it takes the ball to reach the speed v if the wind speed is u. I find that c=137 s²/m. Note that after a very long time the speed will be constant at the wind speed; we are only interested in the time it takes to get to 35.8 m/s (80 mph). I have plotted t for three values of u, 70, 80, and 90 m/s.

Now comes the hard part. We are interested in finding the value of u for which v will be 35.8 m/s when the ball has gone a distance x=18.4 m. This means that we need to write the equation for t as t=c{dx/dt}/[u(u-{dx/dt})] and solve that differential equation for t as a function of x. That is actually doable, but the solution is so messy that I looked for a simpler way to do it. This took some trial and error, but I got a pretty good estimate, I believe. Look at the black line in the graph; it is not a straight line, but it is pretty close, so I am going to approximate the acceleration as being constant over this time interval of 1 s, a≈(35.8 m/s)/(1 s)=35.8 m/s². Now, an object with constant acceleration starting from rest goes a distance d=½at² in a time t, so if u=90 m/s=201 mph the ball will have a speed of 35.8 m/s when d=35.8/2=17.9 m=59 ft. As far as I am concerned, I have adequately solved this problem—it takes a wind of approximately 200 mph to accelerate a baseball from rest to the speed of 80 mph in a distance of about 60 ft.

QUESTION:
Why is light speed the fastest speed and why would people see a train going backwards if it is traveling at a velocity greater than light speed?

ANSWER:
I will not answer the second question because the site groundrules state that I will not answer questions of this sort. Your first question is equivalent to the question "why can nothing with mass travel as fast or faster than light?" which is included in the FAQ page.