Questions and Answers

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QUESTION:
the speed of light is constant,yet we know gravity pulls on light,wouldn't light aimed directly towards a high source of gravity,such as a black hole move faster ?

ANSWER:
As light falls into a black hole, it gains energy, but not by speeding up. The frequency increases as it falls meaning the energy of each photon increases, but the speed stays just the same.

QUESTION:

1. With increase in velocity during free fall is there increase in simple kinetic energy without gain in mass or there is increase in relativistic KE as per velocity?

2. If object becomes more massive due to gain in velocity and relativistic KE and velocity approaches c then would there be increase in gravitational attraction exponentially as mass increases exponentially?

3. What happens to total energy if the object hits the ground and KE is dissipated.

ANSWER:
This is three questions (I numbered them), not one as required by the site groundrules. In future, please abide by the rules.

1. The "free fall problem" has been earlier solved. There the problem is one of constant force F=mg where m is the rest mass. There it was shown that β=v/c=(Ft/(mc))/√[1+(Ft/(mc))²]. Since E=mc²/√[1-β²] and K=E-mc², it may be shown that K=mc²[√[1+(gt/c)²]-1]. For very large times, this becomes K≈mgct, kinetic energy increasing linearly with time while velocity approaches a constant (c). Hence, the answer to your first question is the energy increases where the velocity changes hardly at all; I have often said that "accelerators" should really be called "energizers" because once the particle gets near c, it no longer accelerates.{Note that it is problematical what is meant by free fall. Since relativistically you cannot have constant acceleration forever, you have to define what you mean. I define it to be constant rate of change of relativistic momentum as discussed in the earlier example.}

2. In general relativity, the gravitational field is more generally determined by the energy density of the source, not just its mass, so the gravitational field of the moving object increases as its speed increases.

3. Total energy remains the same, but now goes into sound, heat, tearing stuff apart, etc. See an earlier answer to get some idea of how much energy a relativistic object has.

QUESTION:
The definition of e.m.f in my book is given as, 'The e.m.f of a cell is defined as the energy spent per unit charge in taking a positive charge around the complete circuit of the cell'. The definition of terminal voltage is given as 'the terminal voltage of a cell is defined as the work done per unit charge in carrying a positive charge around the circuit connected across the terminals of the cell'. Can you please explain the difference between the two? How is the work done different in these cases?

ANSWER:
The EMF characterizes the potential difference between the terminals of a source of power when no current is flowing through the device. It is, for example, the voltage measured across the terminals of a battery with an ideal (infinite resistance) voltmeter. The terminal voltage is the voltage measured by an ideal voltmeter across the terminals of a battery if it is connected to a circuit through which current is flowing. These can be different if the battery is not ideal, that is, if it has some internal resistance.

QUESTION:
I am currently writing a science fiction novel and have been researching the effects of vacuum on the human body. According to the sources I looked up, a person will not explode or freeze to death if exposed directly to space. What I am curious to know is if being exposed to too much gravity would have the effect of making someone explode. If not, can you tell me what physical forces, even artificially, could exert enough internal pressure to make someone explode. This research is for use within my novel. It would help me to maintain scientific accuracy if you could give me, at least, a theoretical explanation for what could create the scenario I describe.

ANSWER:
Well, increased gravity will not cause a body to explode either. Imagine that your weight suddenly increased by a factor of 1000. Your puny legs would be incapable of holding you up. All your muscles would have the same problem. You would be crushed to the ground. If the gravity became strong enough, it would be like you were in a hydraulic press, liquified. I do not know of anything, short of a bomb inside, that would cause a body to explode.

QUESTION:
Light behaves like a particle and a wave(in that it collides with matter and it has no real mass). Right? SO, why does light travel from objects straight into our eyes (and everywhere else) why doesn't it collide with ALL of the other light in between said object and our eyes and then when the light reaches our eyes it would not have traveled directly from the object but from a multitude of angles, thus distorting the image of the object(plausibly too much to even see anything).

ANSWER:
Light does not always travel straight from an object to your eyes. The reason you see a blue sky and not a black sky is because some light from the sun strikes a molecule or a dust particle in the atmosphere, scatters from that point to your eye. Sometimes light from some source finds two paths to your eye and the light from those two paths can interfere with each other either being brighter or darker than you would expect; this is called diffraction but usually we do not see this because the wavelength of light is so short. When you are looking at a leaf outside, you are really looking at light which originally came from the sun.

QUESTION:
If you are doubling the distance what happens to the time and velocity.

ANSWER:
This question makes no sense because time and velocity are not coupled. If you double the distance you travel, the product vt doubles and there are an infinite number of ways that can happen:

• speed twice as fast, time the same;

• speed the same, time twice as long;

• speed ten times as fast, time 1/5 as long;

• speed half as fast, time four times as long;

• etc., etc., etc.

QUESTION:
I am a teacher in New Zealand, and am much more familiar with Chemistry than with Physics. I have been trying recently to understand magnetism better, in general, and as applied to Earth and Space Sciences. When a permanent magnet is created, what is actually lining up to create the magnetic field? I have heard it described as atoms, or pockets of charge, or micromagnets, but what does this chemically look like? Are they polar molecules within the solid? My understanding of bonding of metals doesn't include any polarity, they are positive nuclei in a 'sea of electrons', so how could a metal be aligned by charge? Is it an alteration of electron orbitals? Am I over-thinking this?

ANSWER:
An electron is, itself, like a tiny bar magnet, called an electric dipole moment. In most materials, the electrons point in random directions resulting in no bulk magnetism. In a ferromagnetic material like iron, there is a quantum mechanical effect which causes the electrons responsible for bonding to neighbors in the crystal to align their dipole moments. An ordinary piece of iron is normally not a magnet because the aligning of electron magnets happens only below a certain temperature called the Curie temperature and when the iron cools down the alignment occurs in local small volumes, each of which is pointing in a random direction; these are called domains and are hugely bigger than atoms but still microscopic. To magnetize the iron, you put it in a strong magnetic field and the domains already aligned with the field will grow at the expense of those not aligned with the field. You can also cool iron in a strong magnetic field and then most domains will form being aligned with the external field.

CONTINUED QUESTION:
Also, how does the movement of charged particles in the Earth's liquid outer core, or in the Sun's convective zone, actually produce a magnetic field? I have read about this effect many times, but I just can't picture chemically what this would look like. Is it that the general movemnet of ions within a fluid tends to align in the same direction over time, thus influencing other ions to do the same, and this spreading the force?

QUESTION:
We know there are EM waves. All other waves need a medium to propagate. How does the EM wave propagate? There must be an ether.

ANSWER:
Here is what we know for certain: the speed of light in a vacuum is independent of the velocity of the source or the velocity of the observer. There is therefore not a medium with respect to which it moves. It is no puzzle how light propogates, Maxwell's equations clearly predict electromagnetic waves which have a speed c in vacuum. If you want to believe in the æther, be my guest. But it is sort of like religion, you need faith and will not get proof. It is not needed in electromagnetic theory, and if it is not needed, why add it? Did you ever hear of Occum's razor?

QUESTION:
So if black holes have so much mass that light can't even escape and gravity has almost an infiniate range of influance, when our universe is at its end and black wholes have swallowed up everything will the black holes start to move toward each other and then come together? If so would this singularity be so heavy that it would rip space time and then "big bang" again?

ANSWER:
Again, I emphasize that I usually do not answer astronomy/astrophysics/cosmology questions. I believe, though, that the current best model of the universe has, because of dark energy, the most likely fate of the universe not recollapsing (big crunch) but ever expanding and black holes eventually all evaporating via Hawking radiation.

QUESTION:
Here is the setup:
You have a larger container with 2L of water in it, and above that you have a smaller container with an airtight piston/plunger inside it. A counterweight attached to the piston. The upper (smaller) container is attached to the lower (larger) container via a pipe. The question: How much would the counterweight need to weigh, to pull the piston up and fill the smaller container with water? Assumptions: - The piston weighs nothing and has no friction. - 1L of water weighs 1kg My best guess would be that the counterweight needs to weigh 1kg? Is this correct? (If not, what would it be, and why?) (It seems simple enough, but there might be something I don't know about, that I'm not factoring in?)

QUESTION:
A question about capactance, electrostatic. If a Van De Graaf or Wimshurst were connected to the ungrounded side of a parallel plate capacitor--the other side is grounded, would the capacitance be nearly unlimited, like an electrophorus? Else how could a plate with such little surface area hold such a high voltage?

ANSWER:
The capacitance of a capacitor is determined only by its geometry and the properties of any dielectric material which may be in or around it. Too high a voltage will simply cause it to break down.

QUESTION:
Is there a factor in Newtonian Gravity similar to the factor for precession as in GR(ie., the factor in the Schwarzschild equation for planet precession)?

ANSWER:
If you are wondering whether an orbit precession is possible in Newtonian gravity, the answer is yes. For example, if the sun were not perfectly spherical (which it is not), orbit precession is possible.

QUESTION:
Just read Ender's Game, and it got me thinking of relativistic travel. What would be the optimum speed (relative to Earth) that you could send a probe (say, 1 light-year), and return to Earth? Go too slow, and it takes a long time. Faster is better, but if you go too fast, then relativity starts to work against you.

ANSWER:
I am puzzled by your statement that "relativity starts to work against you." And, what are the criteria which determine the "optimal speed"? If you are on earth and want to get the probe back to earth as quickly as possible, the faster the better. If you are on the probe and want to get there and back as quickly as possible, the faster the better. How does relativity work against you?

FOLLOWUP QUESTION:
So, say there is an object 1 light-year away from the Earth. I want to send a robotic probe there, spend 1 day gathering a sample, and then return it to the Earth. If I send it at 50% c, (if I did my math correct) I calculate it would take (from my point of view on Earth) ~ 4.6 years (plus one day) to get the sample back (with only 4 years + 1 day elapsing relative to the probe). But, if I set the probe speed to .999 c, it would take me ~ 44.8 years ( + 1 day ) to get my sample, which is almost 10 times longer than .5c. This is what I mean by relativity working against you.

ANSWER:
Well, I do not know what you did, but your answers are certainly wrong. As I understand it, you stay on earth. So, using your clock you see a probe going 0.5c to take two years to go one light year and 2 years to get back: 4 yr + 1 day. If the speed is 0.999c, the time out (and back) is 1/0.999=1.001: 2.002 yr + 1 day. From your point of view, relativity is irrelevant if all you are interested in is measured by your clocks and your meter sticks.

You might ask how much time elapses for a clock attached to the probe. (See my discussion of the twin paradox.) Each probe will see the distance you see (1 ly) Lorentz contracted, the slower by √(1-0.5²)=0.866 and the speedier by √(1-0.999²)=0.045 and so the elapsed times will be 1.73 yr + 1 day and 0.09 yr + 1 day.

QUESTION:
I have been debating this for years and have read many online forums. My question regards solar heat and how it is conducted into water. I have a large swimming pool. My current solar cover is silvery with round bubbles underside. The silvery material supposed holds heat better and warms up the water better, as well as preventing evaporation, dirt in the pool etc... I now need to replace the cover which has worked reasonably well for 6 years, but I always thought it doesn't help heat the pool much. My thought is that diamond shaped air bubbles on the underside will have a greater surface area for transmitting heat into the water. But I cannot get a good answer on what color the material should be. These covers come in silvery, clear, dark blue and light blue. I thought dark blue would absorb more heat, but others say clear helps magnify the sun's rays and produce more heat. Still others say light blue is best, others say the silver. The benefits of these covers are multiple. They definitely do keep heat in overnight, stop evaporation, and keep pools cleaner. But I really want a cover that will help heat up the pool during the day if no one is swimming yet. So what is best? Diamond bubbles or circular? Clear or dark blue, or silver as I currently have?

ANSWER:
My knee-jerk reaction is to agree with you that the best choice would be a dark color. However, there are many variables and you could get tricked. To get heat into the water is the main concern for you, not necessarily how much heat gets absorbed in the cover itself. A little search resulted in some data which, if accurate, I think tells the story. Evidently, the most important thing which determines heat transmitted to the water (heating power) is how thick the cover is. This makes sense since the thicker the cover, the more absorbed energy it can hold and therefore the more energy is avaiable for heating the pool. So choose any color you like best, anything else would be fine tuning at best. I suspect bubble shape is pretty irrelevant too.

QUESTION:
Time dilation. 3 observers: earth, spaceship A, and spaceship B. Both ships are rushing by the earth at .9 of c, and experience a 2:1 ratio of time passage as predicted by relativity. If spaceship A observes clocks on spaceship B is time dilation double that as from earth? A simple yes or no would work, but a link describing this would be appreciated.

ANSWER:
Relativity does not predict a 2:1 ratio of clock rates of A and B relative to earth. The gamma factor for γ=1/√(1-β²)=2.29, where β=v/c, not 2; elapsed time on A or B as measured by earth is earth time divided by 2.29. To find the velocity of B realtive to A you need the relativistic velocity addition formula β_BA=(β_B+β_A)/(1+β_Aβ_B)=1.8/1.81=0.9945 and so γ_BA=9.53 and  the time dilation relative to that seen on earth is 9.53/2.29=4.16, not double. The answer to your question is a simple no. I guess I am your link since I gave you the description.

QUESTION:
If I listen to two radio stations in a travelling car that's moving away, why is it that I can still get the signal from the 10,000 watt station while the 1,000 watt signal soon fades. Both signals are EM waves, so why should either fade with increasing radial distance? Don't these signals continue to propagate @ light speed into the solar system and beyond?

ANSWER:
Once you are far from the broadcasting antenna compared to its size, it looks like a point source and the signal drops off like 1/D² where D is the distance from the antenna. The signal from the 1 kW station is smaller than that from the signal from the 10 kW station by a factor of 10. There is a threshold below which your radio cannot process accurately the information in the signal and the 1 kW station drops below that threshold first. They do continue propogating as the distance gets bigger but with ever-decreasing intensity.

The animation on the home page is a catylatic converter where a CO molecule comes to the platinum surface, rattles around for a bit, then picks up an O atom and departs as CO₂.

QUESTION:
I would like to ask another question, this one a variation on the twin paradox in relativity. In this case, instead of twins, let us suppose triplets, Alex, Ben, and Charlie. Alex gets in a spaceship and rockets away from the earth at near the speed of light; Ben gets in another spaceship at the same time (in Charlie's frame of reference) and follows him on the same trajectory at half that speed; Charlie remains on earth. After rocketing for some originally agreed upon distance, Alex turns around and rockets back to earth, again at near the speed of light. After rocketing half of Alex's distance (in Ben's frame), Ben also turns around and rockets back to earth, again at half Alex's speed. Now, as I understand it, in Charlie's frame, Alex is rocketing away and back at near the speed of light, and should therefore experience observable relativistic effects; in Alex's frame, by contrast, Charlie is the one moving; in Ben's frame, however, both Alex and Charlie are moving, but at slightly less than half the speed of light, at which there should be no (non-trivial) relativistic effects. So what happens? It seems that in Charlie's frame, upon the return of his brothers, he and Ben should still be the same age (or close enough as makes no difference), since Ben underwent no observable relativistic effects, but Alex should be younger now, since he did undergo observable time dilation. In Ben's frame, though, it seems like all three brothers should still be the same age, since none of them ever went past about half the speed of light. So how is this apparent paradox resolved?

ANSWER:
This is just the twin paradox doubled and there is still nothing paradoxical. First, read my original twin paradox answer to see what I am doing here. I will do it generally although you could do it for a specific number (e.g. the 8 ly trip at β=v/c=0.8 for the original answer). One problem with your question is that if Ben goes half the distance that he measures Alex going, they will not arrive home at the same (Charlie) time. You have to say that relative to Charlie the distances are L_Charlie for Alex and L_Charlie/2 for Ben and the relative speeds are β and β/2. Then, Charlie's clock when Alex and Ben arrive will be T_Charlie=2L_Charlie/β. Now, Alex and Ben both see the distance they have to travel to be length contracted, L_Alex=L_Charlie√(1-β²) and L_Ben=(L_Charlie/2)√(1-(β/2)²). So the clocks (when Alex and Ben arrive home) will read T_Alex=2L_Alex/β=(2L_Charlie/β)√(1-β²)=T_Charlie√(1-β²) and T_Ben=(2L_Ben/β)=(2L_Charlie/β)√(1-(β/2)²)=T_Charlie√(1-(β/2)²). That's it.

To do my original example where L_Charlie=8 ly and β=0.8, T_Charlie=20 y, T_Alex=12 y, and T_Ben=18.3 y. As you can see, the time dilation is smaller for Ben with β=0.4, but certainly not neglibible.

The triplet paradox is resolved!

QUESTION:
why can photons make a laser but electrons and protons can't make a laser like beam.

ANSWER:
Laser is the acronym for Light Amplification by Stimulated Emission of Radiation. So, by definition, a laser emits electromagnetic radiation. There are apparently some methods to create coherent electron beams, but you would not call them lasers. There is something called the free electron laser but it uses electrons to create electromagnetic radiation, not a coherent electron beam.

QUESTION:
I have been reading a bit about time dilation and am curious about the subject. The fact that a clock on the ISS and another on earth report slight differences in rate of time show that something is going on. However, I simply cannot imagine that if a spacecraft where to fly to away from earth traveling near the speed of light and then immediately return, that they would return to a future earth. It just does not seem logical. This brings me to my questions: Are there any other examples of solid proof that time dilates based on gravity and velocity or is time dilation still a theory based solely on the ISS vs earth clock example? Also, a more theoretical question, if time dilation is true, could stasis (hypersleep) be achieved by moving an object near the sleep of light backward and forward -- basically a extremely high frequency vibration?

ANSWER:
Sorry it does not "seem logical" but that is not surprising because it is certainly not intuitive. Intuition is the feeling we have about something based upon experience and our experience does not extend to very large velocities where time dilation becomes apparent. Maybe some earlier answers of mine would help you find it more believable: see my discussion of the light clock and the twin paradox. Both of these require that you accept that the speed of light is the same for all observers; if that does not "seem logical", other earlier answers might help. There are innumerable examples of time dilation in the real world. Have you ever used a GPS? The way it works, essentially, is that satellites buzzing around triangulate your position. Obviously, accurate timing of signals between you and the satellite are crucial and if corrections in the software were not made for time dilation in special relativity (speed effects) and general relativity (gravity effects), a GPS would simply not be accurate enough to be of any use. Another example is the lifetime of elementary particles, a simple clock. If some particle is measured to have a lifetime of 10^-6 s and has a speed of 2.4x10⁸ m/s (80% the speed of light), "logic" would dictate that it will travel a distance of 2.4x10²=240 m before disintegrating, right? In fact, it travels 400 m because of time dilation. Regarding your idea for hypersleep, imagine the acceleration the sleeper would have to endure―the resulting force she would experience would crush her beyond recognition.

QUESTION:
I read somewhere that photons carry electromagnetic force. Also charged particles exchange photons and the momentum of the exchanged photons appears as force. My question is, from the above description I can very well visualize repulsion between like charges but how can it explain attraction between unlike charges?

ANSWER:
The idea of photon exchange must not be taken too literally. It is meant as a "cartoon" picture to convey the qualitative ideas of field quantization. For example, you cannot just think of the recoil of the emitting particle which you normally would in classical physics and seems like an easy way to see a repulsive force. The reason is that these photons are necessarily what are called virtual photons, in some sense they do not exist. You cannot observe a virtual photon. It is easy to see that they cannot be real photons, because if the charged particle suddenly spit out a photon, where did the energy needed for the creation of that photon come from? Energy conservation has been violated! However, in quantum physics, it is ok to violate energy conservation by an amount ΔE and that violation can last as long as Δt as long as ΔE·Δt < h/(8π) where h is Planck's constant, about 6.6x10^-34 J·s; this is the famous Heisenberg uncertainty principle. Often in physics simple, qualitative models are good for getting insight but not necessarily good as a detailed theory. Another example is the Bohr model of the atom: we know that it describes a lot of data very nicely but we also know that the idea of electrons running around in little circular classical orbits is total nonsense in a rigorous model of the atom. Another example is the often cited "trampoline model" for gravity, that mass (a bowling ball, for example) attracts other masses (a marble, for example) because of the deformation it causes in space-time (represented by the deformed trampoline).

QUESTION:
what is coriolis effect
how does it influence earths atmosphere

ANSWER:
See an earlier answer.

QUESTION:
Is it possible to make carbon monoxide (carbon MONoxide, NOT dioxide) into a liquid?

QUESTION:
I remember being taught that if a photon encounters an atom, the atom will absorb the photon only if the photon's energy is exactly the right amount to raise the atom's electrons to one of its permissible states. If the photon's energy doesn't match, then there'll be no absorption and the photon will just pass by. How exact does the match have to be, really? It can't require a bang-on precise match, because then absorption would never happen -- the photon's energy would always be some tiny fraction off. And if it doesn't have to match exactly, say if the photon has a tiny amount too much energy, what happens to the excess -- conservation of energy says it has to go somewhere.

ANSWER:
There are several considerations here. First, a photon does not have a discrete amount of energy. Second, an excited atomic state does not have a discrete amount of energy. These are consequences of the Heisenberg uncertainty principle and discussed in an earlier answer. Therefore, to excite an atom there is a range of energies which may be added to achieve the excitation. A third consideration is that when the atom absorbs the photon it recoils; that is, part of the photon energy goes into the kinetic energy of the atom not into exciting it. (An interesting exception to this is the Mössbauer effect where the whole solid, not the individual atom, recoils resulting in virtually recoilless absorption). Finally, the atom may well be moving (in fact it will be moving) before the absorption occurs and so it sees the photon as Doppler-shifted, again allowing energies different (as seen by you) from the excitation energy to cause excitations.

QUESTION:
The hydraulic brake has brake fluid inside and it could be dangerous if there are air bubbles it it because a part of the pressure is used to compress it. I am baffled by this due to my assumption that the pressure inside a bubble is atmospheric pressure and i quote from wikipedia " the pressure inside 100 nm diameter droplets can reach several atmospheres " in the laplace pressure page. How could the bubbles 'pop' ?

ANSWER:
The bubbles of interest are on the order of a mm (10^-3 m), certainly not 100 nm (10^-7 m). The pressure is probably roughly atmospheric, but when you increase the pressure in the fluid it compresses the bubbles which results in less movement of the brakes at the other end of the line, and a “spongy” feeling of the brakes.

QUESTION:
This is a question about tractive effort of locomotives. I am a model railroader and we have been debating this situation for a while with no clear answer. If there is a locomotive on the track that has 4 driving wheels touching the rails, and the only change you make is to the number of wheels , now 6 driving wheels on the track, will tractive effort go up, stay the same or drop. The argument for staying the same is that each wheel now supports less weight, so tractive effort of each wheel would drop. The argument for tractive effort going up would be that there are more contact points on the rail, so it would pull more.

ANSWER:
What drives your train is the static friction between the wheels and the track. The nature of static friction is that a maximum amount of force may be achieved before the surfaces slip on each other. This force F_max is determined, to an excellent approximation in many cases, by the nature of the surfaces (steel on steel for your case, I presume) and how hard they are pressed together (called the normal force N); F_max=μN where μ is the coefficient of static friction, a number determined by the materials. You will note that this force does not depend on the area of contact. Essentially, adding wheels adds surface area but, as you note, the force pressing the surfaces together (namely the normal force which is determined by the weight) is simply distributed over a larger area. So, the simplest first-order physics answer is that the same maximum force from the wheels without slipping will be achieved regardless of the number of wheels. Friction, though, can be a tricky business and first-order physics does not always work for all situations. For example, there are instances where a car tire can have greater "road hugging" (read increased friction) if the road-tire contact area is increased. I believe in your case, though, since the wheels and track are so little deformed by contact, that increasing the number of wheels will not increase friction provided that the total mass of the locomotive remains the same. Of course, if you are adding wheels to an existing locomotive, you are adding mass so you will be increasing F_max. There is a very easy way to test this. Put the locomotive (wheels not free to rotate) on a piece track which is attached to a horizontal board. Gradually increase the angle of the board until the train slides down. The tangent of the angle at which it starts to slide will be equal to the coefficient of static friction. If increasing the number of wheels does not increase the slip angle, there is no advantage.

QUESTION:
if we paint an object with any colour, are we painting its elementary particles-electrons, protons, neutrons?

ANSWER:
No. You are putting a layer over whatever you are painting which has the desired color.

FOLLOWUP QUESTION:
Then are the elementary particles of the layer of our desired colour? I actually want to know that how do we perceive colour since most of the spacd in atom is empty.

ANSWER:
You misunderstand what makes something appear a certain color. It happens at the molecular and atomic level. When white light falls on an object, some is absorbed, some is reflected, some is absorbed and then reemitted. The colors absorbed, reflected, and reemitted are determined by the atoms or molecules which are present in the material you are looking at. Electrons, protons, and neutrons do not have a color because you are not looking at them; they, themselves, do not emit light which you see. If a particular object looks red, for example, it is because it absorbs other colored light more than red light which strikes it. The origin of light is at the atomic level. When an atom is excited to a higher state, it decays back down to its normal state. This means that the atom has lost energy and that energy comes out as light.

QUESTION:
I’m not sure if you would consider this an “off the wall” question (it seems like a “deceptively simple” type to me), but here it goes: what would be the relative density (to water’s 1 (1000 kg/m^3) say) of something with a theoretical “complete density,” i.e. something with no empty space between mass components (atoms, quarks, etc.)—a celestial goo, if you will. Would this be a singularity material or would this be less than singularity? Some sources list a singularity as being infinite density, so, if that is the case, I’d guess relative comparisons might be pointless. Part of what I am trying to figure out is how deceptive the term "density" actually is; I am thinking it will end up being something like the differential between the Fahrenheit scale and Kelvin. I would also like to know the how much “more” surface area a 1 meter by 1 meter square of this material would have than the traditional one square meter (of water I guess) of any traditional material. In that case, I figure ewe have always assumed complete density with surface area calculations, but, for some calculations I need to do, I need to know what “true” surface area is.

ANSWER:
Well, you are pretty close to "off the wall" here! First, a singularity has infinite density and zero surface area. A black hole is thought to be such a thing. I could quit there since this is certainly the limit. However, maybe you want something somewhere between the density of normal matter and infinite density, something which does not depend on what the material is. If we think about all matter as being composed of electrons, neutrons, and protons, we can imagine somehow compressing them all together to make some "primal stuff". Since electrons have very small mass compared to neutrons and protons, the number you seek is the density of nuclear matter, that is, the mass of a typical nucleus divided by its volume; that number is about 2x10¹⁷ kg/m³; so, that is 2x10¹⁴ times more dense than water. There are actually, in nature, objects with this density. Some stars, late in their lifetimes, collapse under gravitational attraction to this density and the enormous pressure causes electrons to be "forced" into the protons to make neutrons; they are called neutron stars.

Regarding your question about area, it really has no meaning because you cannot have a square meter of something, it always has a volume as well. If you had a sphere of nuclear matter of radius 1 m, it would have a surface area of 12.6 m² and a mass of 8.4x10¹⁷ kg. I guess you could imagine a "sheet of neutrons" with this density and ask what its area was. Taking the radius of a neutron to be about 10^-15 m, the volume of a sheet of area 1 m² would be 2x10^-15 m³ so its mass would be 400 kg. Taking the radius of a water molecule to be about 10^-10 m, a sheet of water of mass 400 kg would have an area of about 4x10⁹ m².

QUESTION:
While we were studying about atomic structure, our teacher taught us something (I don't remember precisely how was it) about the reason that electron can't be in the nucleus of an atom. She made some calculations and showed that if electron were in the nucleus, it would have to move with a velocity which is greater than that of light. But what about beta decay? Doesn't an electron originate from the nucleus during beta decay? Does it travel faster than light when it originates?

ANSWER:
What your teacher was trying to convey is that an electron cannot be confined in a nucleus, that is, the structure of an atom cannot include electrons which spend most of their time inside the nucleus. There is nothing forbidding an electron from ever being in there. In fact, the wave function of any atomic electron has a nonzero value inside the nucleus so it spends some (tiny amount of) time inside. I am not familiar with her "faster than light" argument; usually the uncertainty principle is used which shows that the energy uncertainty of an electron confined in a box the size of a nucleus is too large to be consistent with measured atomic masses. So, you see, your beta decay example does not violate what she taught you since the electron does not remain inside the nucleus after its creation. (Be sure to realize that the beta decay electron did not exist before the decay happened, it was not hanging around waiting to pop out.)

QUESTION:
how much negative charge do I accumulate by touching the earth? Here is how I (tried) to answer this question: The Earth carries a negative electric charge of roughly 500 thousand Coulombs (according to different sources I've seen). If I touch the Earth I should therefore pick up some of this electric charge (through conduction) and become negative charged. Assuming the earth can modeled as a conducting sphere with radius 6371 km and me as a conducting sphere with radius 1 m, around how much negative charge would I accumulate? The reason I ask is because I'm trying to prove to myself that grounding does indeed render a charged object neutral (i.e. transfers all the object's charge to the Earth). Using the well known equation for two connected conducting spheres with different radii (see Example 3-13 on page 115 in David Cheng's "Field and Wave Electromagnetics, 2nd Ed."), I calculate 0.0785 C, which is way too big and must be wrong.

ANSWER:
There is an old joke in physics the punchline of which is "consider a spherical cow". I like the spirit of your question, but your approximations are killing you, I think. But, I cannot quarrel with the earth being approximated as a sphere and approximating yourself as a sphere is probably ok too. The problem is the assumption that you and the earth are conductors. If you take your number for net charge you will find that the surface charge density is about 1 nC/m². If your answer were anywhere near right, you would have to get all the charge from around 10⁸ m² in your vicinity. But it is not free to move around, either on the ground or in you. Although you are standing in an electric field, caused by the earth's net charge, on the order of 150 V/m, this field cannot cause any significant electric charge to flow onto you because you are not a good conductor.

QUESTION:
I've been studying Schrodinger's equation in one dimension as well as wave functions, and I came across the equation Ae^i(kx-wt) which supposedly gives you the wave function. I know what all the quantities in this equation except for the "A" represent. I've heard that A has to do with boundary conditions, but I've also heard that A must be a complex number. Could you please tell me how to calculate what this A is supposed to be? Nothing I find on the internet or in books tells me how to find A.

QUESTION:
What is fire? I ask this because I recently viewed an exchange on the Internet where someone was crucified for saying that the sun was a big ball of fire. The prominent theme was that the sun's light and heat was from plasma, not fire. That got me thinking: what is fire? Is there a scientific definition? Is the term fire limited to the traditional fires we see where there is combustion with oxygen?

ANSWER:
I am not sure there is a scientific definition of "fire". I would say that fire is the ionized, radiating gas resulting from a chemical reaction. I would tend to agree with the opinion that the sun is not a "ball of fire." Certainly chemistry is not the source of energy emanating from the sun.

QUESTION:
Will a non-magnetic metals interfier with a magnetic field when placed next to a super strong magnet?

ANSWER:
All materials have magnetic properties. How they will affect a magnetic field is a matter of degree.

QUESTION:
How are protons produced?

ANSWER:
Protons are primordial, the universe is composed of mostly hydrogen which is protons and electrons. Protons can be made in many ways, usually in beta decay where a neutron decays into a proton, an electron, and a neutrino.

QUESTION:
Last summer I hit a deer while riding my large motorcycle. The combined weight of cycle and load would be nearly 1,000 pounds, and I believe I was going about 65 mph. The deer weighed approx 150 pounds. Potentially how far could have the deer been thrown by the impact? Furthermore, evidence at the scene of the crash suggested that the deer actually rode on the motorcycle for a short time. How much would that have slowed the motorcycle down immiediatly after impact?

ANSWER:
It is not clear to me how the deer could both "stick to" ("…actually rode on…") the bike and then fly off. If it sticks, that is called a perfectly inelastic collision and you can easily find the speed right after the collision using momentum conservation. Momentum is the mass times the velocity and the momentum before the collision must be the same as before: M_bikeV_bike+M_deerV_deer=M_bothV_both=1000x65+150x0=1150xV_both, so V_both=65,000/1150=56.5 mph (assuming the deer had no velocity in your direction). It is much harder to predict what happened after the collision if they do not stick together. You could only guess, say maybe you gave the deer a speed of 30 mph going at a 45⁰ trajectory. The distance would then have been about 60 ft.

QUESTION:
Hydrogen can be metalized through high pressures to create metastable metallic hydrogen which has been theorized to be present on Jupiter. Is it possible to use a particle accelerator to cause a collision between two hydrogen atoms and replicate the pressures present on Jupiter and metalize the hydrogen?

ANSWER:
No, because an accelerator does not accelerate atoms. Protons are accelerated, not hydrogen atoms.

QUESTION:
If a plane, while mid-flight, had an explosion from the wing or rear area, would the debris maintain the exact same momentum as the plane and go forward alongside it, would the debris go in front of the plane, or would the debris go behind the plane when the accident first happens? Me and two friends got on the topic after watching a movie dealing with a plane crash. One friend thinks debris would go in front, the other thinks the debris would maintain precise momentum with the plane, and I'm thinking it would go behind the plane. We'd really love for a physicist to clear up this question for us, thank you so much!

ANSWER:
What anything does is determined by the forces on it. A plane flying in a straight horizontal line with constant speed has four main forces acting on it: gravity (its own weight); lift which is the force which counteracts the gravity to keep it flying level, drag caused by the air, and the forward force exerted by the engines which counteracts the drag to keep it from slowing down. If a piece of the plane suddenly separates from the plane, it no longer has any foward force and it no longer has any significant lift; so, it will start dropping vertically and slowing down horizontally, falling down from and behind the plane. The falling is generally more prounounced than the slowing down (gravity usually a greater force than drag) so there would be a tendency for the piece to appear to just drop straight down as seen from the plane. (Be sure to realize that the forward velocity of the piece is approximately maintained so that someone on the ground sees it moving forward with about the same speed as the plane.) This is often seen in bombs dropped from a plane. On the left photograph the drag on the bombs is small so they keep pace with the planes as they drop. On the right, the bombs have little parachutes to increase drag and so they fall behind the plane.

QUESTION:
Could a beam of light be used to move an object?

ANSWER:
Yes. See an earlier answer.

QUESTION:
If some object, at some distance far from the surface of the Earth, but much closer to the Earth than anything else (meaning everything else in the universe is negligible) is at rest initially but then begins accelerating towards the Earth due to gravity, how long will it take to get to the Earth? I get that the acceleration is a=GM/r^2. And then you can write a as the second derivative of r which would make it a non-linear differential equation but I have no idea how to solve those really. I have taken a differential equations course but it was 3 years ago so I don't remember if we even solved non-linear ODE's ever.

QUESTION:
When electron transitions between two energy levels it emits a light photon of energy E=h*v. But a photon is a light particle, i.e. it is localized in space. Such is wave packet is made up of many frequencies and not just one. Then how is E=h*v valid?

QUESTION:
I am a physician on an internet mailing list. Someone asked why it is easier to push fluid into thick skin using a thin 1cc tuberculin syringe (6 cm long for one cc of fluid) compared to pushing that one cc through the same size needle but from a wider bore 3 cc syringe (1cc is marked off along only 2 cm of distance along the hub). My theory was that one could develop more pounds per sq cm of pressure using the thinner bore plunger than the same pressure expanded over the wider area of the 3cc syringe. I thought the difference between the two was based on the square of the radius of the syringes. Others argued for something called Poiseuille's equation which described flow through a tube based on the fourth power of the radius.

ANSWER:
Poiseuille's equation gives the pressure drop ΔP along a tube of length L and radius R due to the viscosity μ of the fluid, ΔP=8μLQ/(πR⁴), where Q is the flow rate. Although I did not do a calculation, common sense tells me that, since the tubes are not long, the fluid is probably not very viscous, and the flow rate is very small, that the pressure drop for the 6 or 2 cm is negligibly small. Your suggestion is the correct one. If you exert equal forces on the two with your thumb, the resulting gauge pressure in the thinner syringe will be 9 times greater.

QUESTION:
i tried to find the time it would take for two charges to collide under electrostatic force,realizing simple kinematics won't cut it,tired to integrate but failed,how is the question done?and how does it differ from gravitational force??

QUERY:
You have to define what "collide" means. Since it is a 1/r² force, if you use point charges the velocity will be infinite when they collide but they will do so in a finite time. Also, what are the initial conditions (velocities, positions), masses, charges.

REPLY:
My initial question was the time it would take for a 1/r^2,to collide,for example two bodies lets say 1g,and a charge of 1micro coluomb,initial at rest attract each other,and collide,I can't use simple kinematics to solve this question,what shall I do?

QUERY:
How far apart are they?

REPLY:
ok,for simplicity 1m apart,is there a general formula than can help?

ANSWER:
Whew! I finally have everything I need. This is the Kepler problem, the same, as you suggest, as the solar system with gravity. You may want to look at an earlier answer similar to yours. It is very lengthy to work out the whole problem in detail so I will refer you to a very good lecture-note document from MIT; I will just give you some of the necessary results to calculate what you want. First, a brief overview of two of Kepler's laws:

• Kepler's laws refer to problems where the force is of the form F=K/r² where K is a constant and the force is attractive. So it could refer to either two masses or two opposite charges.

• The first law states that bound planets move in ellipses with the sun at one focus. This is really only true if the sun is infinitely massive but the generalization still leads to an elliptical orbit for each body, both of which move around the center of mass of the two. Still, the semimajor axis a of the ellipse (which we will later need) in the center of mass system can be found for any orbit from the simple equation a=-K/(2E) where E is the energy of the system.

• For your case, the particles move in a straight line toward each other and then turn around and return to their original positions. This is just the most elongated possible ellipse with an eccentricity of 1. Of course this would never really be possible in the real world since the particles would be going an infinite speed when they "collide". That means we really should do the problem relativistically which would greatly complicate the problem. Keep in mind that you are asking an unphysical question requiring point charges and infinite forces and velocities. But the answer below should be a good approximation of the time if they have some finite size small compared to their initial separation.

• The third law relates the period of the orbit T to a: T²=4πμa³/K where μ=m₁m₂/(m₁+m₂) is the reduced mass. In the gravitational problem, K=Gm₁m₂ and in the electrostatic problem, K=k_eq₁q₂ where k_e=9x10⁹ N·m²/C².

For your case, the energy is given since the charges are initially at rest and separated by some distance S, so E=V(S)=k_eq₁q₂/S and so a=k_eq₁q₂/(2k_eq₁q₂/S)=S/2=0.5 m; the reduced mass in your case is μ=m₁m₂/(m₁+m₂)=10^-3x10^-3/(2x10^-3)=0.5x10^-3 kg; and K=kq₁q₂=9x10⁹x10^-6x10^-6=9x10^-3 N·m². Finally, the time it takes for a complete "orbit" (which would correspond to the particles returning to their original positions) would be T=√[4πμa³/K]=√[4π(0.5x10^-3)(0.5)³/9x10^-3]=0.3 s. But, the time you want is just half a period, T/2=0.15 s.

To help you visualize the orbits, the figure below shows the orbits for the two charges when the eccentricity is just less than 1; imagine the orbits getting flatter yet, approaching two straight lines.

NOTE ADDED:
I got to wondering what the limits of doing this classically are, that is, how good an approximation my calculation above would be for some real system. This requires that I determine how close the two charges would approach each other before their speed v became comparable to the speed of light c. I will use the same notation as above and write things classically. If released a distance S apart, then when they reach a distance r apart energy conservation gives: k_eq₁q₂/S=k_eq₁q₂/r+½μv² which results in v=√[(2k_e|q₁q₂|/μ)(1/r-1/S)]. For the case in point, if I solve for r when v=c/10, a reasonable upper limit for a classical calculation, I get r=4x10^-8 m, about 100 times bigger than an atom. Alternatively, we could ask what the velocity would be for a 1 mm separation, r=S/1000: v=√[(2k_e|q₁q₂|/μ)(999/S)]=1.9x10⁵ m/s=0.00063 c. In either case, I think we can conclude that the time remaining to complete the half orbit will be extraordinarily small compared to 0.15 s.

QUESTION:
I read recently that Lene Vestergaard Hau stopped a beam of light in its tracks. Isn't this a violation Heisenberg's uncertainty principle as both the velocity and the position of the photon are known simultaneously?

ANSWER:
The light is not really so much as stopped as trapped and it is confined in a Bose-Einstein condensate which is non-zero in size. It certainly does not violate the uncertainty principle.

QUESTION:
If a ship is traveling toward the Earth at a speed of 0.25c from 4 light months out, and they are looking at the Earth and collecting optical images (which at the start would be 4 months old) what would the collection rate be? Would it still be 1 to 1? Or would they be collecting it fast since they are traveling toward the source of the light?

ANSWER:
It would certainly not be 1 to 1. I have worked this problem out in earlier answers. Yours is one of those questions which asks "how fast do moving clocks appear to run at high speeds?" Be sure to note that this is a different question from how fast clocks run. The speed I normally use for these situations is 0.8c because the arithmetic comes out easily and I am here to convey concepts, not arithmetic! A problem virtually identical to yours (but including also how the ship would observe things going away and how someone on earth would observe things on the ship) but for 0.8c has been previously answered; in that case, a clock on earth appears to run 3 times too fast. For 0.25c, the earth clock would still appear to run too fast but at a substantially slower rate. If you really need to get a numerical answer, you can get all the details in my discussion of the twin paradox.

QUESTION:
It's well established that a rocket propelled spaceship accelerating forward will gain mass as it goes faster and faster. Appreciably so as it approaches light speed. My question is, how does this mass gain manifest itself? That is to say, does it occur by the atomic particles themselves gaining mass (protons, electrons etc.) or do new particles simply pop into existance due to the energies involved? Also, by what mechanism does this occur? How does energy being expelled from the rear result in the addition of mass to the ship and it's contents?

ANSWER:
When you think of mass, you think of "amount of stuff". When I think of mass, I think of inertia. Mass measures how hard something is to accelerate, and that is inertia. In classical mechanics, inertia does not depend on on the velocity; a 1 N force will provide exactly the same acceleration on a mass regardless of how fast it is going. In special relativity, as an object with mass goes faster, it becomes increasingly more difficult to accelerate it. In my view, special relativity requires only the concept of rest mass, its inertia when at rest; beyond that we introduce new definitions of energy and momentum which then take care of this inertia issue.

QUESTION:
I recently read an article about the world's largest diesel engine. It is 44 feet tall, 90 feet long and weighs 2300 tons. This engine makes 109,000 horsepower at 102 rpm and is used to power a 1300 foot long cargo ship. Reading this brought a question to mind. It seems that in general, the larger a mechanical engine is the slower it rotates. If you consider a small RC airplane engine, it can rotate up to 20,000 rpm whereas my car’s engine can only rotate at about 6000 rpm. Is there a physical law, theory, or area of physics that studies and defines the scaled-strength of materials? More basically – why does this seem to be true?

ANSWER:
The acceleration of a mass M moving in a circle of radius R with speed V is V²/R; therefore the force necessary to keep it moving in that circle is MV²/R. Consider two machines, each with the same rotational velocity ω (ω is in radians per second, but measures the same thing as your RPM) and one is 10 times bigger than the other. Because V=Rω, F=MRω,  so the force required to keep a gram of some gear or wheel from flying off the big machine is 10 times bigger than for the small machine. I believe it is mainly a strength of materials issue although other reasons are certainly possible depending on what the machine is designed to do.

QUESTION:
With all the space junk circling our planet, I wonder how much mass do we have to lose before it affects our orbit around the sun?

ANSWER:
First of all, we have not lost that mass, it is still attached to the earth. Second, the orbit of an orbiting body is independent of the mass; a 1 gram earth, given the same orbit and speed, would take one year to go around. For example, there are hundreds of communications satellites in orbits high above the equator and all have a period of 24 hours, although they have many different masses.

QUESTION:
I just read in a book that gravitational force b/w two bodies is due to the exchange of particles called gravitons between them but isn't gravitational force action-at-a-distance force..so from where comes this concept of particles in relation to force,...and how the interactions of gravitons lead to the force gravity?

ANSWER:
Gravitons are a speculation, not a fact. Here is a little background: You would probably characterize electric forces as "action-at-a-distance" forces. And while, in some sense, that is true, it is not very satisfying because it is always nice to be able to understand the mechanism for any force. In the 1940s Richard Feynman and others developed quantum electrodynamics (QED) in which the electromagnetic field is quantized, thereby making electrodynamics a field theory compatible with quantum mechanics. In QED, the quantum of the field is the photon and may be thought of as the "messenger of the force". It has long been a holy grail of physics to do the same thing for gravity but nobody has succeded in developing a theory of quantum gravity; if someone were to succeed, it is widely accepted that the graviton would be the quantum of the field. There is already a very successful theory of gravity, general relativity, which explains the force mechanism as resulting from warping of spacetime. However, this theory is not compatible with quantum mechanics and, until a theory of quantum gravity is found, the graviton is purely sepculation.

QUESTION:
Convection, conduction and radiation with respect to heat loss in the human body. If 2 equally dressed, equally insulated and body surface area and BMI candidates are sitting in rooms with no air movement, at an ambient temperature of 40 degrees, with the difference that one subjects' room is at nearly 100 percent humidity and the other subject's room is at 35% humidity. Is the saturation of the air and loss of heat by the first subject a function of conductive heat loss? I understand that movement of heat in a liquid or gas state is convection, but would not the high humidity increase the conductive principle of heat loss (and a reason that cold and damp is very uncomfortable)? Or is that still a principle of convection, the gradiant differences in air and body causing a convective heat transfer.... I believe the humidity adds a conductive component... Am I misunderstanding?

ANSWER:
Heat loss by conduction of a gas is always very small compared to convection and radiation. However, here we are talking about organisms, not just blocks of something or other, and we have evolved to cool ourselves by evaporative cooling (sweating) which is really important. Since evaporation occurs at a lower rate when the air is humid, humidity plays an important role. Still, you are interested in whether conductivity of air (small though it may be) depends on humidity. The answer is yes, but the effect happens to be exactly opposite of what you expect―conductivity decreases as humidity increases as shown on the graph at the right.

QUESTION:
If I am driving my car at 70 mph down a straight highway road, and I drop a brick outside the driver's side window, it would leave the car, initially, at 70 mph relative to the unmoving ground below. Correct? However, if I dropped a brick outside my car at 70 mph on the windshield of another car headed in the exact opposite direction at 70 mph, would the collective force be the same as if I had dropped a brick on a still standing car at 140 mph since the other car is headed in the opposite direction? My knowledge of physics is quite elementary, I'm afraid. But, if I am thinking of Newton's formula F=M x V, does force change if the velocity is a relative velocity (ie two objects headed nearly straight at one another)? And if it does, is it a simple addition problem like I mentioned above (70mph+70mph= 140mph relatively), or does force get influenced by gravity/wind friction/absorbing quality of the windshield?

ANSWER:
If you look on my faq page, you will see that there is no way you can calculate the force if one thing, going speed v, hits something else going with any other speed. And your expression of "Newton's formula" is wrong, force is mass times acceleration, not mass times velocity. However, the effect of a 140 mph brick hitting a stationary windshield and the effect of a 70 mph hitting a windshield approaching it at 70 mph would be the same.

QUESTION:
I am a high school physics teacher, and am troubled somewhat by inverse beta decay. From what I gather, it can occur when a proton absorbs an electron antineutrino and an up quark flips to a down quark. It then emits a positron and becomes a neutron. What troubles me is where the energy comes from. Flipping a proton to a neutron requires an input of energy, it seems, because of the mass difference. And then there is the mass and kinetic energy of the positron. Can the electron neutrino possibly supply all of this energy? It seems unlikely.

ANSWER:
If you put in the rest mass energies you find that the neutrino must bring in at least 1.8 MeV of energy. This is not a large amount of energy for a neutrino. The main source of neutrinos in the universe is beta decay, so it is to be expected that they would have the requisite energies to induce the inverse processes. For example, neutrino spectra from nuclear reactors are shown in the figure above. As you can see, there are many neutrinos with energies greater than 1.8 MeV.

QUESTION:
I am really confused about what Heisenberg principle actually states...why cant the position and momentum can be measured together....cant we measure the position and momentum at any particular instant of time since physics can do anything???

ANSWER:
Physics can do anything? Unfortunately, nature has put limits on how precisely certain things can be determined. The underlying reason for the uncertainty principle is the fact that there is really no such thing as a particle. Anything which you might think of as a particle has a dual reality, it is also a wave. In what way, for example, is an electron like a wave? Think of a very long wave with some frequency f. It turns out that the momentum of the wave depends on f. The "position" of this wave train is uncertain, you will admit because it has some length where it exists. Now I said earlier that the wave has a frequency f, but in reality, it has a well defined frequency only if the wave train is infinitely long. If you want to write an expression for the finite wave, you find that it is really made up of an infinite superposition of frequencies; this is called a Fourier transform, decomposing some wave into simple sinesoidal functions. The most important contributor is the momentum corresponding to f, but there is a whole distribution of others. So, you see that the wave has uncertainty in both position and momentum. As I said above, infinite uncertainty in position will result in zero uncertainty in momentum. Similarly, if you try to pin down the position by making the wave train shorter and shorter, you find that the momentum distribution becomes larger and larger.

QUESTION:
Acc. to E=mc2 energy and mass are inter convertible....based on this principle atom bombs have been designed.....so following this equation, a rigid object, say a chair, which is made up of large no. of atoms can be converted into energy which will release a large amount of energy..so can a domestic thing like this be super dangerous!!!!!!!!!!!! if not then why??

ANSWER:
It is usually not easy to convert a large amount of mass into energy. See an earlier answer.

QUESTION:
does the terminal velocity in a parachutist's fall occurs two times? one before opening the parachute and one some time after opening the parachute Am i right?

ANSWER:
The terminal velocity is determined by (among other things) the geometry of the falling object. In the simplest approximation, what matters is the cross sectional area of the falling object. So, when the parachute is opened, the area gets much bigger and the terminal velocity gets much smaller. If the sky diver has achieved some larger terminal velocity, she will slow down to the new terminal velocity. However, this need not be the first time the terminal velocity changed. The sky diver can orient in a ball or like a down pointing arrow and have a relatively large terminal velocity or she can fall spread-eagle and slow down. So the terminal velocity probably changed several times before the parachute was opened.

QUESTION:
How do photons react to heat?

ANSWER:
Heat does not refer to some thing. The word heat refers to energy transfer, and to see more detail you should look on the faq page. So, heat is not really something a photon can "react to". (In fact, heat could refer to photons if radiation is transferring the energy.)

QUESTION:
I was just watching a video from IBM and how they were moving single atoms. I heard that atoms are mostly empty space between the nucleus and the electron(s). So when we're looking at a single atoms, what exactly are we looking at? The nucleus or the electrons orbit?

ANSWER:
I suspect the video referred to is "A Boy And His Atom: The World's Smallest Movie". First, you should appreciate that you are not really "looking" at atoms in the sense that your eyes are receiving light which comes from the atoms. Because atoms are much smaller than the wavelength of visible light so you cannot create an image using visible light. It is popular to refer to an atom as being mostly empty space, but that is not really true. Because of the quantum nature of atoms, it is inaccurate to think of the electrons in the same way we usually think of them when they are free, essentially point-like particles. Rather, the electrons "spread out" to form a cloud of charge and mass around the nucleus of an atom. It is true, though, that the mass density of most of the volume is enormously smaller than the mass density of the material because most of the mass is in the very tiny nuclei. When you are far from an atom, it has zero net charge so you do not see any electric field. But, as you get close to it, you see the negatively-charged electron cloud as closer to you than the nucleus and so you see a net electric field which can be measured and recorded. The way this is done is with a very tiny needle (atomic scale) which is the probe for an instrument called a scanning tunneling microscope. A computer then takes the measured electric field and makes a graph of the data and this is what you are looking at. So, you are looking at a representation of the electric fields close to the atoms. You can also watch a video of how this movie was made.

QUESTION:
My question relates to the Hubble constant. As I understand it the further away an object is in the universe from us the greater red shift it exhibits. By looking at how red shifted emission and absorption lines are we can apparently deduce how far away an object is. People always seem to assume that a photon's progress along it's path is loss less. Is it possible that part of the red shift is due to drag along the path the photon travels. Since a photon has no rest mass but has mass simply at lim v>>c it will simply continue at the speed of light but would exhibit a loss of energy that appears as a red shift. Or for that matter part of the red shift. I think my question involves both classical E and M and Quantum mechanics.

ANSWER:
Again, I emphasize that I usually do not answer questions in astronomy/astrophysics/cosmology. I think you are thinking about the photon in a classical sense, like it is a ball falling through molasses and losing energy as it goes. But, a photon is an all or nothing thing if it is traveling in a straight line. Photons can be either absorbed or scattered if they encounter some interstellar medium, there is no "drag" in the classical sense. If a photon is absorbed by an atom or molecule in a molecular cloud, its energy will eventually be reemitted but in random directions not toward us. If it encounters an atom or molecule, it may be scattered (losing some energy) but, again, out of our line of sight. Astronomers do need to worry about this loss of intensity when trying to determine intrinsic brightness of stars, but this light is lost to us, not red shifted.

QUESTION:
I learnt recently that the gravitational field strength which is represented in Einsteins space time fabric is the masses gravitational acceleration. Yet I know that the value of the gravitational acceleration of any mass grows exponentially. Why in the visual representations of the fabric only shows dents (if it were exponential, the visual representations should be an endless dent which approaches negative infinity).

ANSWER:
I am having trouble understanding what you are asking. When you say "grows exponentially" are you referrring to the 1/r² behavior of the gravitational force? Your "endless dent" would occur only for a black hole which has zero size. For any larger object, the magnitude of the force no longer increases like 1/r² once you are inside it. For example, the gravitational force is zero at the center of the earth.

QUESTION:
If I person could build wings for themselves like Icarus and fly at 99.999 percent the speed of light, all the way to the Andromeda galaxy, two and a half million light years distant, how much time would pass from their perspective? Would they experience said trip as taking two and a half million years, or would time pass more quickly for them? Would they even be alive upon reaching said galaxy? Do we even possess the mathematics to work a problem like that out?

QUESTION:
One of Maxwell's equations is simply a correction to Ampere's law, which states that a magnetic field is induced by a current. Maxwell corrected this to say that a magnetic field can be induced by a both a current and a changing electric field (or more accurately, a changing electric flux). Here is my concern. Since a current is a movement of electric charges wouldn't these moving charges create a changing electric flux? This electric flux would in turn induce a magnetic field, which becomes the equivalent of saying that the current induced the magnetic field. If the current and the changing electric flux are inducing the same magnetic field, it is in fact the same phenomenon, and Ampere's law doesn't need a correction. By this reasoning Maxwell's correction should instead be a generalization and would simply say that a magnetic field can only be induced by a changing electric flux (where in the case of moving charges the electric flux is caused by the current).

ANSWER:
Ampère's law applies to magnetostatics, all current densities being steady. And, if you think about currents in wires, there is no net electric charge and so there is no electric field at all, so even if the current is changing, there would (according to Ampère's law) be no electric field. You are apparently somewhat advanced to be asking this question, so I put a more detailed explanation here, this being primarily a site for layman questions.

QUESTION:
i am unable to understand a pulley problem

QUESTION:
If a rogue planet the same size as Earth was about to hit us, would there be a period of weightlessness here before impact from the gravity of the rogue planet?

FOLLOWUP QUESTION:
What if a rogue planet 2X the mass of Earth was about to collide, would the oceans be pulled from their beds before impact?

ANSWER:
I have added a curve for the near side case you ask about. As you can see, there is a time before impact (when r=2R) when the force goes negative. That means the net force is pointing toward the other planet, not earth, so anything on the earth's surface near the impact point, including the oceans, will "fall up".

QUESTION:
If all the galaxies are accelerating away then is nothing an intertial reference frame? If so, why do Newton's laws work the way they do?

ANSWER:
I am not, as I emphasize on the home page, a cosmologist, so take what I say with a grain of salt. I do believe, though, that there is no such thing as an inertial frame of reference and it really has nothing to do with acceleration. General relativity posits the equivalence of accelerating frames and frames in a gravitational field, so any frame where there is any gravitational field is not an inertial frame. Because there is no place in the universe where there is truly zero field, there are no inertial frames. Therefore, Newton's laws are false! However, there are many frames which are very close to inertial and for which they are almost not false. On earth, a patently noninertial frame, Newton's laws work remarkably well in many situations. For much of celestial mechanics (calculating orbits of satellites and planets, for example), Newton's laws work almost perfectly. Special relativity originally overthrew Newtonian mechanics which does not work, even in an inertial frame, at very high speeds without serious modification of definitions of things like linear momentum and energy.

QUESTION:
I think I understand that the speed of light is a constant and it is fast (ok non physicist language, but humour me). However, it is only a huge number because of the units we choose to represent it.

ANSWER:
The only reason we use units is to have universally agreed upon "yardsticks". If we had chosen a length standard (call it the baker) such that 1 baker=10⁸ m, the speed of light would have been c=3 bakers/s. A small number but still a big speed.

QUESTION:
I've been trying to write a action sci-fi screenplay, but there is one problem that I can't get my head around. During the climax, two characters fall from a 91 story building, with the first jumping off, and the second falling approximately 10 seconds later. The second character proceeds to catch up to the first, and they then brawl. My question is, how much time would pass before they would hit the ground?

QUESTION:
What is the difference between electric potential(infinity to a point) and potential difference(between two points)? Why do we need two terms - electric potential and potential difference? What's the use?

ANSWER:
Because of the definition of electric potential, it is always arbitrary within an additive constant. Therefore, the only meaningful quantity is the potential at one place relative to the potential at another, so only potential difference is really of significance. For example, if you call the potential at one place V₁ and that at another point V₂, you could just as well call the potentials V₁'= V₁+C and V₂'= V₂+C and there would be no change in any physics you do; note, though, that ΔV=ΔV' because the C subtracts out. However, if you clearly state where you have chosen the potential to be zero, then the potential difference between that point and somewhere else is unambiguously meaningful, and that is what is usually referred to as electric potential. The electric potential at a point in space is simply the potential difference between that point where you have chosen zero potential to be. For example, for any localized charge distribution (not extending to infinity), it is customary to choose V(r=∞)=0.

QUESTION:
For a story I'm writing I have a space station about 500 meters long with a diameter of about 100 meters. How does one figure out how fast the station has to rotate (correct term?) around it's long axis so that it has 1 Gravity around the circumference inside?

ANSWER:
If you move in a circle of radius R with speed v, there must be a force on you of F=mv²/R where m is your mass; this force is directed toward the center of the circle. On your space station, the force would be provided by the walls and you want that force to be equal to your weight on earth which is W=mg, g=9.8 m/s² being the acceleration due to gravity. So, that means you want v=√(gR)=√(50x9.8)=22.1 m/s. The circumference of the space station is 2πR=314.2 m, so the time to complete one revolution would be 314.2/22.1=14.2 s; so the rate of rotation would be 1/14.2=0.07 revolutions per second or 4.2 RPM.

QUESTION:
‘Electron can pass through two holes at the same time without splitting into two.’ I read it in a book, is it true? If so, explain it to me.. Please.

ANSWER:
This results from wave-particle duality. All particles are both waves and particles and, if you look for an "electron wave" (and you will certainly agree that a wave can pass through two holes at the same time), you will find one. See an earlier answer for more detail.

QUESTION:
Can you make your own Northern Lights (Aurora Borealis) in a Lab?

ANSWER:
The aurorae result from electric charges (mainly electrons) interacting with high-altitude atmosphere. The atoms become excited and then deexcite and emit light. The laboratory equivalent would be a gas-discharge tube.

QUESTION:
I have a question upon which I have been pondering for a while and it seems increasingly difficult to answer the more I look into it. I was wondering, what causes fundamental particles to possess a charge? Is it a similar mechanism to the property of mass whereby particles interact with a field of some sort, is it to do with spin, or perhaps something else entirely different?

ANSWER:
This is the kind of question―"what causes" the existence of something in nature―that science often does not or can not address. Electric charge is that property which allows matter both to create and experience electromagnetic fields. If particles could not interact via the electromagnetic force, we would have no atoms, no light. Fundamental particles have fundamental properties―mass, electric charge, spin, baryon number, isospin, charm, etc.―which essentially describe how they interact with other particles in nature. For the most part, these are simply empirical quantities for which we have no answer as to what "causes" them. That is not to say that we never find a "cause" as illustrated by the recent excitement over the Higgs boson; the Higgs field is thought to be the "cause" of mass. To my knowledge, there is no similar theory proposed for electric charge.

QUESTION:
Am failing to understand a logic of nuclear reactor. (The questioner refers to this link.) There have been 11 nuclear accidents till now. Considering they occurred during a cumulated total of 14000 reactor years of operation (which i dont understand), how would 15000 nuclear reactors cause an accident every month? Please help understand this.. Am getting so confused an restless here!

ANSWER:
The rate at which major accidents have occurred since reactors have been being built is R=11/1.4x10⁴=7.8x10^-4 accidents/reactor/year. If you have N reactors, then the rate at which accidents will occur for those N is NR=1.5x10⁴x7.8x10^-4=11.79 accidents/year≈1 accident/month.

QUESTION:
I'm wondering what causes the fluctuations that happen in perfect vacuum in quantum physics? Are they just an inherent part of the universe, or are there theories about what cause them?

QUESTION:
My question is about the doppler effect. It is understood that a moving objects sound waves are bunched up in front of it, and elongated behind it, which creates the sound effect that we hear. What is throwing me off is the bunching up of sound waves in front of the object. Why would the sound waves bunch up in front of the object, wouldn't they just blend together and harmonize? I keep thinking in my mind that the sound difference we here on a moving object is the harmonizing of the sound waves together as the waves get closer to your ear and not that they are just bunched up together.

ANSWER:
What does that mean, "blend together and harmonize"? Harmonize usually means when two or more sounds blend together. But the source of sound we are talking about here is a single sound, so harmonizing seems to be inappropriate. Although most sounds are complex, it is easiest to understand by first assuming that the sound you are listening to is a single wave with a single wavelength λ and frequency f which travels with a speed c in still air. Think of the source sending out the peak of the wave once every 1/f seconds. So, a peak goes out and then, 1/f seconds later, the next peak goes out. But, if the source has a speed v, it has moved a distance x=v/f while the first peak has moved a distance c/f=λ. Therefore, the distance between peaks, which is the new wavelength λ', is given by λ'=λ±x=c/f' where you choose -(+) sign if the source is moving toward (away from) you and f' is the frequency you hear. So, the waves either "bunch up" if the source is coming toward you and "spread out" if the source is going away from you. The Doppler effect is usually expressed in terms of the frequency, f'=f/[1±(v/c)]. [Incidentally, no generality is lost if you consider a more complex wave form than a simple single-frequency wave, because any complex wave form may be represented by a superposition (add up a bunch) of simple waves of single frequencies. Since the new frequency is proportional to the old frequency, you get exactly the same shape of complex wave but just "bunched up" or "spread out". This is called Fourier analysis.]

QUESTION:
Was killing time on a slow day at work, and involved a coworker in the site. Some time later he ask a question, which might be up the physicist alley. Assuming we had a person seal in a vacuum with a piece of paper and a baseball, if he threw the baseball and then crumpled up the piece of paper and threw it. Would the paper fly as far as the baseball having no friction to affect it, or does gravity rule supreme and bring it to a halt just as well as it does outside a vacuum?

ANSWER:
Let's assume that you are throwing both as hard as you can which means you exert equal forces on them. Also assume that you exert that force over the same distance for each case which means you do the same amount of work (force times distance) on each. That means that they have equal kinetic energies when you release them. But, kinetic energy is ½MV² where M is the mass and V is the speed. So, suppose the baseball has a mass 100 times the sheet of paper; then, if they have equal kinetic energies, the speed of the paper will be 10 times larger than the baseball and therefore go 10 farther. Your last sentence makes no sense because gravity does not "bring it to a halt", it is friction which does that.

QUESTION:
My question relates to electromagnetic radiation waves like light. I don't understand what the wave looks like in three dimensions. Usually I am asked to picture it like a sea wave or an undulating piece of paper but then how could a radio wave be received by many receivers at the same time in many locations out of the plane of the wave? The other question on waves is why there is a wave at all. What cause the peaks and troughs or the energy to change direction?

ANSWER:
If you have a point source on a water surface, you generate wavefronts that propogate like circles outward (2-dimensional waves). But, in three dimensions, the wavefronts are spheres and they propogate outward as shown in the little animation to the right. So really, 2D waves are not such a good representation of 3D waves. There is "a wave at all" because the radio antenna creates waves which propogate outward. Most antennas are directional, that is they do not send out spherical waves but wave fronts which are more intense horizontally so that energy is not wasted sending waves into the ground or into space; maybe the waves are more cylindrical than spherical. The animation to the left shows a cylindrical wave but for a radio antenna, the cylinder axis would be vertical instead of horizontal.

FOLLOWUP QUESTION:
I now understand the the wave is promulgated as a sphere (or maybe a cylinder) but I cannot get my head around how that translates into the generally depicted view of the light wave being like an undulating snake. With the sphere analogy surely I would just see arcs of a circle as an observer?

ANSWER:
The pictures above show only the wave fronts, which usually represent the maximum amplitude of the outgoing waves. But the spaces between the surfaces shown also have the "undulating" waves in them. If you were to look out along a line and measure the electric field at each point in space at one time, you would see your "undulating snake". The graph to the left shows what the results of your electric field measurements along a radial line would be. The high spots are where you would be crossing the wave-front surfaces in the 3D pictures above. As time goes on, the whole wave would move to the right. The reason that the amplitude decreases as you get farther away from the source is that the waves carry energy and, as you get farther from the source, that energy gets spread out over a larger area but the total energy of the wave has to remain constant. (The energy is proportional to the square of the electric field.)

QUESTION:
Suppose a bullet is fired parallel to the ground, due to perpendicular direction of work done with respect to gravity, no work is done against it. Then why does not the bullet fall immediately to the ground as it does when it is not in motion? I have speculated that motion in in dimension reduces the effects of forces of other dimensions. Is there really such a thing in physics? If not, then what is the correct explanation?

ANSWER:
First, do not worry about what was happening to the bullet while the gun was firing it; that is past history and has nothing to do with what happens to the bullet after it leaves the gun. Newton's first and second laws tell us that the only reason something will not move in a straight line with constant speed (or be at rest which is a constant speed of zero) is if there is a force acting on it. Further, the change in motion the object experiences is in the direction which the force points. Normally in elementary physics classes, we neglect the force of air drag (which is a pretty poor approximation for a bullet) such that the only force is that due to gravity which we call the weight of the object. So, all the bullet can do is change its motion in the vertical position because that is the direction that the weight points. So even though your bullet starts with no vertical motion, it will fall just as it would if you simply dropped it. In other words, it will take just the same time to drop to the ground whether you fire it horizontally or simply drop it. In an earlier answer, you can see a strobe photograph where a ball launched horizontally and one dropped fall vertically the same. In the real world, air drag is not negligible and is a force which always points in the direction opposite the velocity. This causes the bullet to slow down in the horizontal direction but to also speed up more slowly vertically. The figure shows an example where the conditions allow the whole range of scenerios to play out: because of the air drag, the horizontal motion is eventually stopped and, because of the interplay between weight and drag, the projectile ends up falling vertically with a constant speed called the terminal velocity.

QUESTION:
In simple artificial bombardment experiments with alpha particles, how do I determine the correct resulting isotope and if/which particles are emitted? For example, 27Al (a,N) 30P results in the release of a neutron. Some such alpha bombardments result in the release of a protons. How do I determine which will occur? Determining either the remaining particle or the resulting isotope is the issue.

ANSWER:
You cannot "determine which will occur." Almost anything will occur as long as the final products are bound. It is just that they will occur with different probabilities. A second consideration is, as you suggest, energetics must be considered. If you subtract all the mass energy after the reaction from the mass energy before the reaction, you get a number called the Q-value. If Q<0, the incident particle must bring in that much energy or else the reaction cannot happen. Here is what we will need: 1 AMU (atomic mass unit)=931.494 MeV/c² and a table of atomic masses (a handy one is here) which gives the atomic masses of isotopes in AMU.  I will do a some examples.

• Your example: ²⁷₁₃Al(⁴₂He,¹₀n)³⁰₁₅P; M_Alc²=26.981538578x931.494=25133.1413 MeV, M_Hec²=4.00260325413x931.494=3728.4009 MeV, M_n=1.00866491600x931.494=939.5653 MeV, M_Pc²=29.978313753x931.494=27924.6194 MeV. So, Q=27924.6194+939.5653-25133.1413-3728.4009=2.6425 MeV. So, this reaction can occur since there is an excess of of energy after the reaction. There is an additional detail which I will discuss below.

• Suppose instead we do the (α,p) reaction: ²⁷₁₃Al(⁴₂He,¹₁H)³⁰₁₄Si; M_Sic²=29.973770136x931.494=27920.3870 MeV, M_Hc²=1.00794x931.494=938.8901 MeV. So, Q=27920.3870+938.8901-3728.4009-25133.1413=-2.2651 MeV. So, for this reaction you must bring in an excess of at least 2.2651 MeV of energy for it to proceed.

• Finally, a reaction ((p,3n) which bombards with protons and three neutrons come out) to illustrate my first point above that the final product must be bound: ⁶₃Li(¹₁H,3¹₀n)⁴₄Be. ⁴₄Be would be a nucleus with 4 protons and 0 neutrons and, when you look up its mass you will not find it because it does not exist; 4 protons will not bind together to make a nucleus, ⁶₄Be is the lightest known berylium isotope.

Here is the additional detail mentioned above. The first example, with a positive Q-value might seem to imply that you can just have a bunch of Al and He nuclei at rest and they will react to give you a bunch of neutrons and P nuclei plus a bunch of energy. But, both nuclei are positively charged and they therefore repel each other. Therefore, you must give the alpha particles at least enough energy so the two nuclei can "touch" and interact. I calculate that energy to be roughly 10 MeV for alphas on ²⁷Al, so your reaction will go for the kinetic energy of the α-particles of greater than 10 MeV. For the (α,p) reaction, the kinetic energy must be at least 12.27 MeV.

ADDED DETAIL:
Here is the detail of how I estimated 10 MeV. The potential energy U of two point charges Z₁e and Z₂e separated by a distance R is U=kZ₁ Z₂e²/R where k=9x10⁹ J·m²/C² and e=1.6x10^-19 C. The projectile, therefore, must start with its kinetic energy equal to U in order to come from far away to R. I took the size of the ²⁷Al nucleus to be 1.2xA^1/3x10^-15 m=1.2x27^1/3x10^-15 m=3.6x10^-15 m and treated the alpha as a point charge. So, for Z₂=13, Z₁=2,  U=26x9x10⁹x(1.6x10^-19)²/3.6x10^-15=1.7x10^-12 J=10.4 MeV. I have used 1 eV=1.6x10^-19 J. This assumes that the mass of the Al is infinite compared to the mass of the alpha-particle, but this is just meant as an order-of-magnitude estimate.

QUESTION:
which takes more fuel- a voyage from earth to moon or from moon to earth?

QUESTION:
If electric fields and magnetic fields both result from the same force (electromagnetic), then why do they interact differently with charged particles?

ANSWER:
Because the electromagnetic field is not a vector field, it is a tensor field. OK, that is kind of technical, so let me try to clarify. An electric charge at rest will interact differently with the x- and y-components of an electric field (the E_x will cause an acceleration in the x-direction, E_y in the y-direction). The full field may be roughly thought of as having six components and an electric charge will respond differently to all six. And, they are all mixed up depending on your motion. If you have a pure electric field but observe it from a frame moving by it, magnetic fields emerge like magic.

QUESTION:
is there some limit where magnets will no longer cause another to move? If I set a magnet near another and it repels or attracts the other, can I do that an unlimited number of times? If so where does the energy to move that mass come from? Is it infinite?

ANSWER:
Of course you can do it an unlimited number of times. Where does the energy come from? From you since you do work every time you put the two magnets close together.

QUESTION:
We know small amount of matter can be converted into large amount of energy but is the reverse of this possible? Can a large amount of energy be converted into matter and if it is possible has it been observed?

ANSWER:
I often get this question. See an earlier answer.

QUESTION:
Astronomy:Astrology :: Physics:Nucleonics ?

ANSWER:
It took me a minute to realize that this is one of those SAT-style analogy questions, "astronomy is to astrology as physics is to nucleonics?" The answer is that this is false because nucleonics (variably the study of nuclear energy, or of nucleons or nuclei, or nuclear phenomena) is a real science and astrology is certainly not. A better analogy would be astronomy:astrology::physics:psychics.

QUESTION:
When you create an electric field or magnetic field is the field established everywhere in space the moment you activate the source? Or does the field build outward from the source? example: It is my understanding that when you turn on an electric magnet that the field it creates is created everywhere at once. So something x distance away that reacts to the field will feel it's effects as soon as the magnet is activated and something x + x distance away also feels the fields effects at the exact same moment. If this is true have there been experiments to prove it and can you site any famous ones?

ANSWER:
An electromagnetic field propogates at the speed of light through a vacuum. This is well known from electromagnetic theory which is one of the best-understood theories of physics. Also, if an electric field were to propogate instantaneously, you could send messages anywhere in the universe instantaneously which is forbidden by laws of physics which are known to forbid the transmission of information at a speed faster than the speed of light. So, your understanding of an electromagnet is wrong. But it might as well be right if you are doing an experiment in a laboratory where the field appears very close to instantaneously across the room because the speed of light is so large.

QUESTION:
Tritium (1 proton and 2 neutrons) is reasonably stable, with a reported half-life of 12.3 years. Why are there no stable isotopes of hydrogen with more than 2 neutrons?

ANSWER:
It is generally not easy to say in a simple way why any particular isotope is not stable. One of the most important features of the nuclear force is the pairing interaction. Protons like to pair up with protons and neutrons with neutrons. So, tritium has a pair of neutrons which increases the binding energy enough to bind it. A third neutron would find nobody to pair up with and, in this case, the nucleus would be unbound. You might think, taking this argument a step farther, that adding a fourth neutron would result in stability. However, for light nuclei there is a tendency for there to be equal numbers of neutrons and protons, so the farther you get fron N=Z, the less stable you get. If you add a proton to tritium you get an α-particle (⁴He), one of the most tightly bound nuclei.

QUESTION:
an object is thrown vertically upward on earth with the speed of light . will it leave earth ? My teacher said that it will travel vertically with decreasing c ( gravity ) and reach zero at one point and fall back earth with increasing speed until it reaches earth with the initial speed c , is he right ? Nd what will be the object's route after it starts moving, straight or performing elliptical orbit ?

ANSWER:
First of all, site groundrules forbid questions assuming an object can move with the speed of light. But, since your question is a little more substantive, I will modify it slightly so that your object may go any speed less than the speed light, say 99.999999% the speed of light. The fact is that any object with a speed greater than or equal to the escape velocity will never fall back down. The escape velocity from the surface of the earth is only about 11 km/s≈7 mi/s. Your teacher is wrong. The shape of its trajectory if its initial speed is greater than the excape velocity depends on the direction it is fired: a straight line if straight up, a hyperbola if shot nonvertically.

QUESTION:
Is a brick considered to have a high specific heat capacity?

ANSWER:
It is about average for a nonmetalic solid, around 0.2 cal/g·K. Most metals are around 0.03-0.1 cal/g·K, water is 1.0 cal/g·K.

QUESTION:
We've all been reading about dark matter and dark energy for some years now and I believe you've said that you (among many others) are not yet persuaded that dark energy and dark matter exist. If matter and energy, as traditionally and conventionally understood, comprise only a very small part of the substance of the Universe, does it follow that classical mechanics, thermodynamics, relativity and quantum theory, etc., correspondingly apply only to that (seemingly) tiny aspect of the world? Is there any reason to think that the laws and theories of physics that humanity has discerned to date would apply also to dark matter and dark energy?

ANSWER:
First, a disclaimer: as I state on the site, I am not an expert in astrophysics, astronomy, or cosmology, so you can take my opinion with at least a grain of salt or ignore it altogether! I would not say "many others"! Most astrophysicists, astronomers, and cosmologists talk about dark matter as if it is surely there but just not directly observed yet. My own point of view is that I need to see some direct evidence before I accept that such a thing really is there; there is lots of indirect evidence of dark matter―the dynamics of galaxies, the time when galaxies first began to form, to name a couple―but it is altogether possible that we do not understand gravity as well as we assume that we do. The best theory of gravity, general relativity, makes many assumptions which are not necessarily true over really large distances. If this were the case, maybe dark matter is the 21^st century equivalent of the lumeniferous æther (see the following answer) and is something we are looking for in vain because there is no such thing. There are lots of good ideas about what dark matter might be (including WIMPS, for which some evidence has recently been observed in the observed excess of high-energy positrons) and I will be happy to accept experimental evidence when it happens. Dark energy is a different matter in that it has not caused a search for some "stuff". There is already a place for dark energy in general relativity, known as the "cosmological constant"; in other words, many cosmologists do have the point of view that dark energy does result from an incomplete theory of gravity.

QUESTION:
I believe that Einstein at age 16 did a thought experiment that demonstrated that if one were traveling at the speed of light in a capsule (inertial system) with a light bulb behind you, it would be dark because the light would not travel from the light bulb to you. You would be going as fast as the light leaving the bulb. Thus, you would know how fast you were traveling, and that would be a paradox since the Newtonian model says one cannot know the speed of an inertial system from within. Did physicists at that time know that the speed of light is constant and not effected by the speed of the source? If so, could you briefly tell me how they had determined that?

ANSWER:
There are innumerable anecdotes about the young Einstien's ruminations; I will not attempt to comment on this aspect of your question although the more mature Einstein would have surely not suggested that he could go the speed of light relative to anything else. Regarding what was known about the speed of light, it was certainly a matter of concern. Although electromagnetic theory predicted the observed speed of light in vacuum, light was unique in that it could travel in a vacuum at all. Basically, when comparing with everything else we knew about waves, most people could not believe that there was not some medium permeating all of space through which light traveled and called it the lumeniferous æther. This is what the now famous Michelson-Morley experiment was all about―observing the æther; the results of that experiment were negative. In later years, Einstein said that he did not remember whether he had even heard about that experiment when he formulated special relativity. His arguments for the constancy of the speed of light were based on electromagnetic theory and the principle of relativity which states that the laws of physics are the same in all inertial frames of reference and electromagnetic theory is a law of physics.

QUESTION:
If there was an simple molecule in a precise position and its bond energy was X Joules. And you aimed a laser beam at it that produced (1/2)X Joules. Then aimed another laser at the atom that also had (1/2)X Joules. Would the two lasers energies add to overcome the bond energy and split the molecule?

ANSWER:
This is called multiphoton excitation and can happen. When you talk about the energy of the laser beam, it should be the energy of the individual photons of that beam. So if the sum of the energies of two photons equals the excitation energy of a process (which would usually be excitation of an atom or molecule instead of dissociation), the excitation can proceed. However, the timing is crucial and to have a reasonable possibility to excite the system with two photons you need very intense beams to enhance the probability of two interacting "simultaneously".

QUESTION:
I am in my powerful rocket. If my orbital speed around the sun was zero and I was the same distance from the sun as the Earth, and my rocket was firing at the exact velocity to hold my position to keep the rocket from from falling towards the sun but no greater, what would my 200 lb Earth body weigh on the rocket ship scales? I am guessing a bit less than 800 lbs. What do you say?

QUESTION TO QUESTIONER:
What does that mean, "my orbital speed around the sun was zero"? If your speed is zero you are not orbiting. If you are at rest a distance of the earth's orbit from the sun, I would call that hovering. Then you have to calculate the sun's gravitational force on you to know your "weight".

REPLY:
You are right, I did wish to know my weight when hovering,

ANSWER:
The force W which the sun exerts on you (aka your "weight") is given by W=GMm/R² where G=6.67x10^-11 Nm²/kg² is the universal constant of gravitation, M=2x10³⁰ kg is the mass of the sun, R=1.5x10¹¹ m is the distance to the sun, and m=91 kg is the mass of a 200 lb weight. If you do the arithmetic, the result is W=0.54 N=0.12 lb. Quite a bit less than 800 lb! You might be interested in a similar question where the questioner wanted to know the time it would take the earth to fall into the sun if it had no orbital speed.

QUESTION:
why can nothing escape event horizon when all it needs to escape is just to exceed the gravitational force and not to reach the escape velocity-speed of light in this case- the way rockets on earth do as far as I know they do not cross the escape velocity

QUESTION:
I have a 1 ton heavy weight suspended with a rope connected to an electricity generator - I let the heavy weight fall 10 meters - how much electricity can I generate? I was told it was 10 kW but I do not know how to check.

ANSWER:
The question "how much electricity can I generate" does not have any physics meaning. You have to either ask "how much energy can I get" or "how much power can I get". And does 1 ton mean 1 metric ton? 1 metric ton is usually written 1 tonne, abbreviated 1 t and is 1000 kg. If 1000 kg falls 10 m at a constant speed without losing any energy to friction, the energy delivered will be E=mgh=1000x9.8x10=98,000 Joules of energy. The power generated would be P=E/t where t is the time it takes to fall; power is the rate at which energy is delivered. For example, if t=20 s, P=98,000/20=4900 W=4.9 kW.

QUESTION:
Are electrons and protons respectively attracted preferentially to north or south magnetic fields-?

ANSWER:
There are many possible layers to this question:

• You are probably just asking: "If I put an electric charge near a magnet, what will happen?" The answer is nothing because an electric charge only experiences a force in a magnetic field if it is moving. Either a positive or a negative charge could be either attracted or repelled from either a north or a south pole depending on how it was moving.

• However, electrons and protons are not just charges but also have magnetic moments, that is, they are like tiny bar magnets. So, treating them as infinitesimally small, they will experience a torque in the magnetic field which will cause them to align with the field so that the N (S) pole of the electron or proton will point toward the S (N) pole of the bar magnet. Still, there will be no force.

• If we allow that the electron or proton magnetic dipoles are not of zero size, then the nonuniformity of the fields (the fields get weaker as you get farther away from the bar magnet) will cause the proton or electron will be attracted to whichever pole it is closest to.

QUESTION:
As you arrive closer to the sun, it would be logical that it would be warmer. But would the cold vacuum of space be warmer? Wouldn't there have to be some type of surface for the heat from the sun to warm for the area around the sun to be warmer? If the area around the sun is warmer, that would lead me to believe that there is something within the vastness of space that has surface for the radiation of the sun to adhere. And if there are things in the vastness, wouldn't they be detectable by the type of radiation that they absorb?

ANSWER:
We usually think of temperature as somehow measuring the energy of the atoms and moledules. But, if there are no atoms or molecules, there can still be energy in the form of radiation (photons) and so we can still assign a temperature to a vacuum as being proportional to the energy of photons in a unit volume. So, clearly, there will be a higher such energy density closer to the sun than farther away. To read a lengthier answer to a similar question, link here.

QUESTION:
How much time would pass on earth if you were travelling 99% light speed for 7 days in your spacecraft?

ANSWER:
I presume you mean that your clock, in the space ship reads 7 days. Then, an observer on earth who observes your clock running slow by a factor of √(1-0.992)=0.14 would observe his clock to read 7/0.14=50 days. Of course, the time elapsed on earth, as measured by you, would be 0.14x7=0.98 days; each observer measures the other's clock as running slow. However, if you come back home (twin paradox), you will both agree that the spacecraft clock was the slower for your round trip.

QUESTION:
E=mc squared or so we are told. Can you give an example of energy being converted into matter since presumably energy and matter are interchangeable (at the speed of light)?

ANSWER:
The best-known example of energy into mass is pair production. Another example can be found in nuclear physics. For example, take a ¹²C nucleus with 6 protons and 6 neutrons bound together. Because they are bound together, it will take energy to take the nucleus apart, you must do work. The mass of 6 protons and 6 neutrons is measurably larger than the mass of a ¹²C nucleus because E=mc² and you increase E by pulling it apart. Your statement "at the speed of light" is meaningless; no matter can travel that fast.

QUESTION:
I have a part (think like a round thick plastic cylinder- it actually is a safety device for heavy trucks that if the air shock system fails the bed of the truck rests on this cylinder). Similar plastics are used in the part (a toughened with rubber nylon actually) on two exact tests with slight differences that my customer is saying should be NO DIFFERENCE- but I get a failure in one and not the other.
1rst test- 200 lb load at 4.00 feet drop- parts passed
2nd test - 220 lb load at 3.66 feet drop- parts failed
Now I know the "Force" is ~ the same...but could it be failure due to Kinetic Energy change (definition more dependent on mass?)?

FOLLOWUP QUESTION:
I did made a mistake when I say "force": the customer has the 200 pound weight dropped from a 4 foot height with the end result of 800 foot lbs of energy in one test and in the other test it is 220 lb weight x 3.66 feet in the other test- 800 foot lbs of energy (805 actually). Are you saying using this kind of "math" for failure analysis is flawed?

QUESTION:
Are the protons and or electrons that make up a platinum atom identically the same as the protons in a gold atom with the only difference being the gold atom has one more proton than a platinum atom? This question has intrigued me since I learned that elements that are heavier than hydrogen and helium are products of fusion in supernova explosions. I guess the biggest explosions yield the heaviest elements. In the fusion process, are the individual protons and electrons changed or are they the same?

ANSWER:
First of all, only elements heavier than iron require a supernova to create them. But, the answer to your main question is that any isolated proton or electron is indistinguishable from any other isolated proton or electron. The protons and electrons which were used to make platinum are just the same as those used to make gold or lead or tungsten or carbon. when the protons are bound together in the nucleus of an atom, though, they become part of a new system and are changed ever so slightly. This is evident because if you measure the mass of, for example, a carbon nucleus which is made of 6 protons and 6 neutrons will not have the same mass as 6 protons and 6 neutrons. This is because of E=mc²; since the nucleus is bound together, you must add energy to take it apart and that causes it to get heavier. To read more about fission and fusion, see an earlier answer.

QUESTION:
So, in today's science, there is this "Law of Conservation of Energy". And all motion is supposedly kinetic energy and comes from potential energy. But I noticed one day that in Newton's three laws of motion, motion is created not by energy or potential energy, but by an "unbalanced force". And I remember that fridge magnets can give off "endless" force apparently without any electrical energy input. So, wouldn't it be theoretically be possible to break the "Law of Conservation of Energy" using this "force-energy gap"? 

ANSWER:
A force alone cannot create energy. The force must act over a distance, thereby doing work. Let's not use a magnet as an example because it is sort of complicated because a magnet never does work (see an earlier answer). Just consider an object in a gravitational field. If you drop a stone, you would say that its energy is not conserved which is true; but conservation of energy applies only to isolated systems, systems on which no external forces are doing work and clearly the gravity is doing work on the stone, thereby changing its energy. The other way of looking at this system is that we can incorporate the work done by the gravity into the problem; this is called introducing potential energy. Now, you have made the gravitational field part of the system and the total energy of this system will not change.

QUESTION:
I understand that air has molecules in constant motion and are constantly in collisions with other molecules but what are they traveling through and how much free space is there? I think a better way of posing my question is that say you vacuumed out all of the matter out of a chamber, then introduced some specific small volume of gas. Clearly there are gas molecules in the chamber now but there's got to be empty space through which they can travel, right? Is that a space a vacuum? Assuming my assumptions are true how much of this "free vacuum space" is there in ambient air?

ANSWER:
The density of air is about ρ≈1.2 kg/m³. The mass of a typical air molecule (N₂ or O₂) is about m≈5x10^-26 kg, so there are about N≈ρ/m≈2.4x10²⁵ molecules/m^-3. The approximate radius of an air molecule is about R≈1.5x10^-10 m, so the volume of a molecule would be about V≈4πR³/3≈1.4x10^-29 m³. The volume occupied by all the molecules in one cubic meter would therefore be NV≈3.4x10^-4 m³ or about 0.03%; the rest, 99.97%, is, as you surmise, vacuum.

QUESTION:
So I understand that protons and neutrons are made of quarks. A proton is made of two up quarks and one down quark and a neutron is made from two down quarks and one up quark. I know that much. But there are so many other particles in the standard model. Are there any other large particles we can (or cannot) observe besides protons and neutrons that are composed of quarks or bosons or whatever? I have been wondering this for a while.

ANSWER:
There is, of course, no concise answer to your question. The standard model has numerous particles and some are made of quarks (hadrons) and some are not (leptons and field quanta). The hadrons which are bosons (integral spin) are made of two quarks, hadrons which are fermions (half odd integral spins) are made of three quarks. There are also many "particles" called resonances which are excited states of the more fundamental hadrons. The figure to the right which I stole from Wikepedia gives a pretty good overview, I think. For more detail, read the Wikepedia article on the standard model.

QUESTION:
i have to know that why the mass of bob in a pendulum does not affect the time of oscillation.....i think it should as the more massive bob should have more potential energy when string is pulled for 10degrree but at the same time when its moving more air resistance would also act on it. My book of physics says that the number of oscilllations does not affect the final answer of time period per each oscillation. as we know that a pendulum should automatically stop moving after some time. This means that if 100 oscillations are taken then the time per one oscillation should be decreased. I know im wrong as my physics book of GCSE says so but i want to know why im wrong and what part of my staement does not make sense

P.S : please use simple language or formulas(if u wanna use) as i wont be able to understand complex ones.

ANSWER:
Let's get straight what we are talking about. You refer to air resistance and you note that the pendulum will eventually stop. Of course, the main reason it stops is air resistance. Any air resistance is assumed to be negligible in a treatment of the pendulum at your level (9th grade), so imagine that there is none. Second, your book should tell you that the pendulum only behaves in a simple way if the angle you start it at is small; but even 30⁰-40⁰ is sufficiently small for the standard treatment of the simple pendulum to be quite accurate. However, your expectation that the air resistance will affect your measurement of the period is wrong because, to an excellent approximation, the period does not depend on the amplitude and so one period when you start and the amplitude is, say, 20⁰, will be almost exactly the same as one period when the amplitude has decayed down to 5⁰. Given your level of school, I cannot give a very rigorous explanation for why the mass does not matter. But, if you can understand resolving a vector into components and know Newton's second law, that F=ma, then you should follow my explanation. In the picture to the right you see the forces, mg and T, which act on m. But we are interested in the motion of m along its arc and neither T nor the component of mg perpendicular to the arc can affect that motion, so all that matters is the component of mg along the arc which is mgsinθ and this is the force which causes the mass to accelerate along the arc. So, applying Newton's second law for motion along the arc, mgsinθ=ma; you will notice that m cancels out so that a=gsinθ. Clearly the acceleration will determine the motion along the arc and therefore the period will not depend on m. But, of course, a real scientest tests his hypothesis. Go to a park where there are swings and find a little boy and his father both swinging; you will see that they both have the same period even though they have very different masses. (Your argument about potential energy is not valid because, although the potential energy at the top is, indeed, proportional to m (U=mgh), the kinetic energy at the bottom is also proportional to m (K=½mv²). So, again, mass cancels out.) Finally, I should tell you what the period does depend on: only on the length of the string and the acceleration due to gravity, T=√(L/g)]/(2π). A pendulum clock will not keep good time on the moon.

QUESTION:
If a rope is threaded through 3 adjacent pulleys and attached to a load at the end what is the mechanical advantage of the system? The rope would go over one pulley under another and over the third at which point it would drop down to the load? I believe the M.A is 1 because the load is only supported by one line. Everyone else contends that it is three since there are three pulleys. What is the right answer?

ANSWER:
The tension in the rope is the same everywhere; call that T. The bottom pulley (to which the load is attached) has three ropes pulling up on it, so if you pull with a force T you can lift a load 3T. Everyone else is right.

FOLLOWUP QUESTION:
The Pulleys are side by side with the line going over the first pulley, under the second, and over and down the third to the load after the third pulley.

ANSWER:
Sorry, I misunderstood. You are correct since now just one rope is pulling up on the load.

QUESTION:
Suppose I built a tubesat (personal hobby sat) and sent it to 193 miles above the earth into orbit. Inside this sat would be a single shot co2 based "air gun" which would fire a single microchip based package approximately the size of a stick of gum into deeper space, hopefully propelling it forever outward. My estimates of package size and air gun capabilities lead me to believe I can shoot the package at as much as 1400 fps or so, thus around 1000mph at launch... My question is - will it work? Can I "shoot" a tiny package out into space from near earth orbit in this fashion?

ANSWER:
The speed of the orbiting satellite is about 7860 m/s and the speed of your package relative to the satellite is about 427 m/s, so the maximum speed you could achieve relative to the earth is about 8287 m/s. Escape velocity is about 11,200 m/s. So the answer to your question is no. [This is just a rough estimate since the maximum speed of the package would be tangential to the earth's surface which would require an even higher escape velocity.]

QUESTION:
Why are the ripples formed on the surface of water alwalys circular no matter which type of source has produced it .. i mean if v drop a square or a triangular stone then too v end up getting only circular ripples ??

ANSWER:
They are not always circular. Drop an 8' 2x4 in the water and nice straight waves will come from the long side. But on the corners the waves will come out rounded where the corners were because of diffraction. Any feature which is not large compared to the  distance between water waves will exhibit diffraction. So, if you drop a square pebble in the water it will have approximately circular waves almost immediately because it is small compared to the distance between water waves. The feature of diffraction can be understood using Huygen's principle which says the next wave may be generated by assuming each point on the previous wave behaved like a point source (making perfect little circles) and then drawing the envelope of all the little circles. The picture below illustrates that for the the 2x4. I have only drawn a few of the Huygen's wavelets at the left end to illustrate how the corners get rounded. As the wave (red) spreads farther the whole rectangle gets rounder and rounder until, finally when you are much farther than 8' from the 2x4, it becomes almost perfectly round

QUESTION:
Why is it that when you place a metal sphere on a flat magnet, it will spin real easily in one direction (i.e x-axis)? You can forceably spin it in other directions, but along that single plane it just spins and spins?

ANSWER:
The sphere becomes magnetized, so it looks approximately like a bar magnet whose south pole sits on the north pole of the original magnet (see figure at the right). To turn the ball about the x-axis requires no torque but to turn it about the y-axis does require a torque.

QUESTION:
Yesterday, while traveling on the train, I observed a fly come trough the door, it hovered in the centre of the ile and did not touch the inner cabin of the train. The train began to move and the fly continued to hover exactly where it was, it did not seem to struggle with maintaining the same momentum as the train, it just simply hovered. Can anyone explain to me why the fly was able to maintain the same speed as the train without having direct contact with the cabin?

ANSWER:
The fly is hovering in the air. The air moves forward with the train. However, when the train accelerates forward, you feel like you are being pushed backwards and have to adjust slightly to keep where you are. The fly also will have to make a little adjustment to keep from drifting backward.

QUESTION:
Am I right in assuming that any thing travelling north on a rail line for example, and then wishes to go south on that same rail line, must stop before going in the opposite direction? If this is the case how is it that a train weigh several hundred tons going 90mph heading north on our rail line meets a bumble bee weighing in at a few grams going south just doing 4mph. They collide What I want to know DOES THE BUMBLE BEE STOP before heading in the opposite direction and if the Bee stops does the train stop?????????

ANSWER:
Well, we have the problem that the bee will be smushed upon contact so various parts of him will have different motions. But I think we can simplify the situation by assuming the bee to be a point mass. If he stops instantainously when the train hits him, his acceleration would be infinite so the force on the bee would be infinite so the force felt by the train would be infinite. I think we can agree that the bee does not stop instantaneously. It takes some time for the bee to stop which would mean, assuming it is a point mass, that the train deforms a tiny bit. So, let's do a quick example: the mass of a bumble bee is about 1/3 of a gram, say about 3x10^-4 kg, and the mass of a freight train is about 10⁷ kg. The bee's speed relative to the train is 94 mph≈42 m/s. Suppose the bee stops (relative to the train) in one millisecond, 10^-3 s; then his average acceleration is about 42,000 m/s². So, the average force on him is ma=3x10^-4x42x10³=12.6 N. Because of Newton's third law, this is the same force experienced by the train, so the train's acceleration would be about a=F/m=-12.6x10^-7 m/s² so the train would slow down by about 12.6x10^-10 m/s≈28x10^-10 mph. After this the bee would have accelerate up to 90 mph so he would feel a continuing force until he got there, but you get the idea.

QUESTION:
I am trying to estmate the energy of a subatomic particle like a proton,by making the v=c or extremely close to c,but i am having a problem that the momentum is becoming Zero when calculating momentum at relativistc speed ,I my completely making a mistake,or am i missng somethin,

ANSWER:
E²=(pc)²+m²c⁴ where m is the rest mass. Also, p may be written as p=mv/√[1-(v/c)²]. As v gets closer and closer to c, p gets bigger and bigger until eventually the m²c⁴ term becomes negligibly small and E≈pc. So, I guess I do not see where your problem is.

QUESTION:
Pertaining to the asteroid that recently exploded over Russia, a television "news" station claimed the asteroid exploded with the same force of 20 atomic bombs... now I understand this space rock did quite some damage, but was it's demise really equivalent to the combined exploding force of 20 nuclear warheads?

ANSWER:
Well, its mass was about m≈1.1x10⁷ kg and its speed was about v≈1.8x10⁴ m/s, so its kinetic energy was about E≈½mv²≈18x10¹⁴ J. The energy of the Nagasaki atomic bomb was about 10¹⁴ J, so 20 is about right. However, its energy was not all released at once and it is estimated that the energy of the main explosion was about 4x10¹⁴ J, about 4 Nagasaki bombs.

QUESTION:
If force = mass x acceleration, how does that factor into something moving at a constant velocity? If it hit something, it would still have force, even though the acceleration is technically zero since there is no change in velocity.

ANSWER:
When it hits something it comes to a stop very quickly, therefore has a very large acceleration, therefore experiences a very large force stopping it. Because of Newton's third law, if it experiences a force from what it hit, it exerts an equal and opposite force on what it hit. It is totally untrue that "…acceleration is technically zero since there is no change in velocity."

QUESTION:
as earth behaves as a big magnet and even the sun has its own magnetic field, their magnetic fields must have an attraction or repulsion force that can overcome the gravitational force between them as electrostatic force is much stronger than gravitational force.But earth seems to be effected only by the gravity between it and the sun and not on the magnetic force.Why is it?

ANSWER:
It is true that the gravitational force is a weaker force than the electromagnetic force. This, however, does not mean that any gravitational force is always smaller than any magnetic force. The earth and sun each have an enormous amount of mass and therefore quite large gravitational fields. The force on the earth of the sun's gravitational field is huge compared to the the magnetic force the earth feels from the sun. Imagine you are holding a bar magnet in your hand―can you move the earth with it?

QUESTION:
we know that rotation off the earth is 1038miles an hour. Now consider that if I fly a helium ballon or any RC helicopter in space for about 15mins, it must go away from us as if I fly an RC helicopter inside a moving train.....But this doesn't seems to happen as I tried it many times in both the circumstances.Why is that? If your ans is because the atmosphere of the earth is also rotating about the same speed and it makes the helicopter to stay in the space in the same direction without moving away then my question is can we get the expected situation in the moon where there is no atmosphere?

ANSWER:
My answer is that both balloons and helicopters fly relative to the air. In a train, the air moves with the train and the hovering helicopter stays with the train. On the moon, neither will fly without air. But if you have a rocket hoverer, like the LEM, launch it vertically and it will stay over the same point above the surface because it started with (and therefore kept) the same tangential velocity as the surface; you could make a similar argument for launching a helicopter from the ground on earth.

QUESTION:
Gravitational forces affect everything that has mass? Light has mass? Yes we know this because it is affected by gravity (black holes, Einsteins test of relativity etc) So if light has mass, and traveling at the speed of light gives anything with mass infinite mass how is it that light is not more affected by gravity? Why doesn't light itself have a black hole gravitational effect if it should have infinite mass? I

ANSWER:
Gravity also affects everything without mass also. Particles without mass must travel at the speed of light, but if they have no mass they will still have no mass when going that speed. The reason that gravity bends light is that gravity warps space time so a "straight line" is not really straight if there is a gravitational field. To read about light being bent, the equivalence principle, and general relativity, see my FAQ page. I should add that gravity is caused not just by the presence of mass but by any energy density. So, a beam of light will indeed give rise to a gravitational field although quite weak for any realizable intensity.

QUESTION:
I would like to know, at what level of gravity (expressed either as G-Forces or acceleration) would an earthlike atmosphere flatten against the surface. I do understand that Earth's gravity does flatten the atmosphere against it, and that increasing the gravity would progressively flatten and thicken it more--all things exist on a gradient, after all. But I am wondering, at what gravity would there be a vacuum, or near-vacuum, at perhaps only six feet/two meters above the surface?

QUESTION:
Even though mars weighs one ninth of earth why does is have 38 percent of earth's gravity?

ANSWER:
For a spherically symmetric mass, the gravity at the surface is proportional to the mass of the planet and inversely proportional to the square of its radius. For Mars the radius is R_M=0.533R_E and the mass is M_M=0.107M_E, so M_M/R_M²=(0.107/0.533²)M_E/R_E²=0.38M_E/R_E². There is your 38%.

QUESTION:
This was part of an answer you gave to a question regarding matter "You understand wrong: matter can be created or destroyed. There is no such law as conservation of matter." I am afraid that I have to disagree with you there. Matter has always existed and is all around us in some form, seen and unseen. The point I am trying to make is, if you can create matter then you must have had something to start with, you cannot create something from nothing. Neither can you destroy it so that it does not exist, you can only change it. Take a pencil for instance, break it down to its smallest atom or smaller than that, it is still none the less a pencil in another form.

ANSWER:
The point is that matter is just a form of energy and energy without mass can create mass. So matter can be created when there is no matter. The best known example is pair production where an energetic photon can interact with an electric field and presto! an electron and a positron, each with one electron mass, appear. Similarly, mass can be destroyed and release energy. To read more about that, read the answer just below yours.

QUESTION:
I know Einstein said E=MC² and basically all mater can be equated to some quantity of energy; then why do we go to the gas station to fill our cars? Why can't we use garbage, which is mass and has energy, to power our cars? How can we convert matter to energy? I know we can burn gasoline to use perhaps 1/4 the heat content in the form of expanding gas to apply pressure to the piston in the engine. Has anyone invented a converter that changes matter to energy yet? We eat food and basically run on sugar which fuels a chemical based process. Any other matter converters?

ANSWER:
Most of the energy mankind uses comes from chemistry. Burn coal or gasoline, for example. When you eat and metabolize food, chemistry is going on. The energy which is extracted comes from―guess what―mass! For example, when you burn coal the main thing which is happening is that carbon is combining with oxygen to form carbon dioxide. One carbon dioxide molecule has a smaller mass than one carbon atom plus one oxygen molecule. So, chemistry is the best known example of your "matter converter". The problem is that an extremely tiny fraction of the mass is converted to energy, something like 0.00000001%, so chemistry is a very inefficient source of energy. Now, to get more efficiency we have work not with atoms but with nuclei of atoms. If a heavy atomic nucleus can be induced to split (fission), the mass of the fragments is smaller than the mass of the initial nucleus by an amount much bigger than with chemistry, something like 0.1% which is a huge improvement over chemistry; this is how nuclear reactors work. Also from nuclear physics, you can take very light nuclei and make them combine (fusion) and get like 1% of the mass converted into energy; this is how stars work and so, you see, solar energy comes from "matter converters" too and so does wind energy since the sun is the energy which causes winds to blow. If you want to get 100% efficient you have to go to particle-antiparticle interactions in particle physics. When an electron and its antiparticle the positron meet, their mass completely disappears and all the energy comes out as photons. Did you ever see the Back to the Future movies? Doc came back from the future where they had invented a small appliance called "Mr. Fusion" to do what you want, to convert garbage into the huge amount of energy needed to power the time machine.

QUESTION:
What is the closing speed of protons before they collide inside the LHC?

QUESTION:
Why any atom having magic numbers(i.e 2,8,20,50 etc.) of nuclides has special stability?(sry for 2 in 1 but i am very curious). How can we determine magic numbers? Is there any specific method to calculate magic numbers?

ANSWER:
Nuclei with particular numbers of protons or neutrons are particularly stable. The origin of these so called magic numbers is very similar to the noble gases of atomic physics where one has a shell model of the structure and a closed shell is particularly stable. In the case of nuclei, it took a lot of work to come up with an explanation for the shell structure. The shell model for nuclei won a Nobel Prize for Eugene Wigner, Maria Mayer, and Hans Jensen. The trick was to recognize the importance of the spin-orbit interaction for nucleons; in atomic physics this effect is also there but very small. The figure to the right shows how the magic numbers appear. On the left are the energy levels predicted for a simple potential well without the spin-orbit splitting. On the right are shown how, if very strong spin-orbit splitting is introduced, new shells emerge. Note that the first three magic numbers, up to 20, are explained without the splitting but not for higher numbers.

QUESTION:
If we travel through space at such a great speed on earth why do we not fall over? If earth stopped moving would everything fall down? Also (sorry for the 3 for 1 but I've always been curious) if I were on a train travelling at the speed of light and were wearing a special suit which enabled me to walk from the front to the back of the train, would I be travelling faster than the speed of light?

ANSWER:
If you are on an airplane going 600 mph, why do you not fall down? Because velocity has no meaning except with respect to… This is called the principle of relativity, that the laws of physics are the same all inertial frame of reference; once you find the laws of physics (which say, for example that you will not spontaneously fall over if you are standing on level ground), they are true in any other frame which moves with constant velocity relative to yours (you do not topple over in any frame). If the earth were to suddenly stop spinning on its axis, you would fall over but that is because during the time of acceleration, when it was slowing down, the earth is not an inertial frame; if you are standing in a bus which suddenly slammed on the brakes, you would fall down. (Keep in mind that your question is worded wrong because "stopped moving" has no meaning as I explained above.) Your third question (shame on you, violating site groundrules) you can find on my FAQ page; also note that no train can go the speed of light, but there is no limit as to how close you could get 99.999999999999999%. [One small catch: the earth is not really an inertial frame because it is always accelerating because it spins on its axis and revolves around the sun. These effects may be seen, but they are very tiny and it is a good approximation to say the earth's surface is an approximately an inertial frame.]

QUESTION:
Wonder if you could tell me what size of spring I would need to fire a ball of mass 200 grams up a slope of 30 degrees for 20 metres? What would I need to fire a 1KG ball the same distance?

ANSWER:
There is no single answer to this because the force a spring exerts depends on how much it is compressed (or extended). A spring's stiffness is characterized by its spring constant k where F=-kx. where F is the force a spring exerts if it is stretched or compressed by an amound x. The easiest way to address your problem is to use energy conservation. The energy which is stored in a compressed spring is E₁=½kx² and the energy which a mass acquires by being lifted a distance y is E₂=mgy. Ignoring any energy losses (to friction, for example), energy is conserved, E₁=E₂; ½kx²=mgy or k=2mgy/x². In your first question, m=0.2 kg, y=20sin30⁰=10 m, and suppose you want to compress the spring by 5 cm=0.05 m, then k=2x0.2x10/(0.05)²=1600 N/m. For your second question, m is 5 times bigger and since k is proportional to m, k=8000 N/m. Keep in mind that these are ideal-case situations and in the real world friction is usually not negligible, so you should build in a safety factor of probably 2.

QUESTION:
This is a question regarding fluid dynamics. When a ping pong ball is supported in a tilted airstream it seems to never be in the centre of the airstream rather slightly below it. Some people have tried to explain this using Bernoulli's principle but when I think about it the faster flowing air is at the top so the high pressure at the bottom will produce a force until it reaches equilibrium in the centre.

QUESTION:
if two bulbs one of 60w and the other of 100w are connected in series,then which of the two bulbs will brighter?

ANSWER:
I will assume that the resistance of the filaments is independent of temperature. Light bulb power ratings are made at a voltage of 120 V. The power is given by P=IV, so I₁₀₀=100/120=0.83 A and I₆₀=60/120=0.5 A. Ohm's law, V=IR, may be used to determine resistance: so R₁₀₀=120/0.83=144 Ω and R₆₀=120/0.5=240 Ω. Now put the two in series and get a net resistance of 384 Ω which means that the current through them will be I=120/384=0.31 A. Hence, the voltage across each is V₁₀₀=0.31x144=45 V and V₆₀=0.31x240=75 V. The power consumed by each is P₁₀₀=0.31x45=14 W and P₆₀=0.31x75=23 W. So it looks like the 60 W bulb will be brighter.

QUESTION:
Suppose I built a tubesat (personal hobby sat) and sent it to 193 miles above the earth into orbit. Inside this sat would be a single shot co2 based "air gun" which would fire a single microchip based package approximately the size of a stick of gum into deeper space, hopefully propelling it forever outward. My estimates of package size and air gun capabilities lead me to believe I can shoot the package at as much as 1400 fps or so, thus around 1000mph at launch... My question is - will it work? Can I "shoot" a tiny package out into space from near earth orbit in this fashion?

ANSWER:
The escape velocity from a near-earth orbit is approximately 25,000 mph. The speed of a satellite in near-earth orbit is about 18,000 mph. So, the maximum speed of your package is about 19,000 mph. Looks like it will not escape.

QUESTION:
I want to ask question about newtons law of gravitation.i want to ask about this exception. when we consider 2 like poles of magnet or two bodies having same charge then they do not obey Newtons law of gravitation.because there is a force of repulsion not attraction.then it can become exception to Newtons law

ANSWER:
Gravity and electromagnetism are completely different things. The existence of one kind of force does not make any other kind of force wrong. The magnets have mass and so they attract each other gravitationally but, since gravity is such a weak force, it is not noticed. The electric charges have mass and so they attract each other gravitationally but, since gravity is such a weak force, it is not noticed. Here is one more example: suppose a book sits on a table. The table exerts an upward force (repulsive) on the book. Does that mean there is no gravity?

QUESTION:
How do objects (specifically spacecrafts) accelerate in the near vacuum of space? What do they push off of in order to accelerate?

ANSWER:
That is not how a rocket works, it does not "push against" something. Imagine yourself standing on very slippery ice holding a very heavy rock. If you throw the rock, you recoil in the opposite direction. You do not recoil because the rocks pushes on the air; the same thing would happen if there were no air.

QUESTION:
I have question about thermodynamics. If you keep the pressure constant, and increase the volume, the temperature should increase or decrease? My thinking is that if you increase a volume, the temperature should decrease, as molecules would collide with walls less, so the average kinetic energy would decrease. But the formula PV=nRT, where nR is constant, if the volume should increase, the temperature should increase. Here I am confused, please help

ANSWER:
The ideal gas law, PV=nRT, has four variables―the pressure (P), the volume (V), the temperature (T), and the amount of gas (n). So if you change one of these, you can only find how another changes if you know how the remaining two change. For your specific question, we can probably assume that n remains constant (no gas added or removed) but we need to know how the pressure changes. If P is held constant (called an isobaric process), then T increases if V increases. You cannot assume that increasing V will decrease T because the collisions of the molecules with the walls determines P, not T. Another important kind of expansion is one where the expansion is done in such a way that no heat can enter or escape the gas; this is called an adiabatic expansion which is done either in a very insulated system or very quickly. Without going into any details, TV^X=constant where X is a number between 0 and 2/3, depending on what the gas is. So, for an adiabatic expansion the temperature does decrease. For example, if you double the volume of a diatomic gas like oxygen (X=½), T₁V^½=T₂(2V)^½, so T₂/T₁=1/√2=0.707. Keep in mind that T must be absolute temperature, so if T₁ is room temperature, about 300 K, T₂=212 K≈-61⁰C.

QUESTION:
With respect to the speed of sound travelling thru air is 343 m/sec. If the sound wave is coming from a car speeding along at 27 m/sec, then relative to a stationary observer along the highway, is the speed still 343 or 370 (343+27=370). If the speed of light is constant regardless of the source and is a wave, then doesn't it follow that sound (also a wave) obeys the same principle? If the speed of light is constant regardless of the source and is a  wave, then doesn't it follow that sound (also a wave) obeys the same principle?

ANSWER:
Sound waves travel with respect to their medium, air. So if the air is still and you are still, waves from a source propogate toward you with speed 343 m/s. However, if you are moving (say 27 m/s) toward a source of sound through still air, then you would measure a speed of sound as 370 m/s. If you are moving away from the source, the speed would be 316 m/s. If you and the source were at rest and there was a 27 m/s wind blowing from the source to you, you would measure a speed of 370 m/s. Light is completely different. The reason that its speed is constant to all observers is that there is no medium through which it travels, it can travel through a perfect vacuum. You might look at my FAQ page to learn more about the constancy of c.

QUESTION:
I have been in a debate with a friend on Facebook about a physics problem. He thinks that in the following scenario, the outcome will be equal to free fall acceleration time. I claim that the Law of Conservation of Momentum will slow it significantly. There are 11 iron cubes, each 1cm per side. The cubes are held 1 meter above the one below by a pair of magnets on strings, so when the cube begins to fall, the magnets drop away as not to interfere with the other cubes. The magnetic force holding each cube in place is exactly equal to its weight, so the slightest touch will make it fall. The first cube is one meter off the ground, the 11th cube at 11 meters. The magnets holding the 11th cube are pulled away and the 11th cube drops from 1 meter above the 10th cube. What percentage of free fall time will it take for the 11th cube to reach 1 decimeter above the ground? Does it make any difference if the collisions are perfectly elastic or perfectly inelastic?

ANSWER:
Whoa! This is too convoluted. We do not need to worry about magnets etc. to answer this question. Each block sits still until it is hit and then is free. And, it makes a big difference whether we are talking about elastic or inelastic collisions. And why 11 cubes? It gets a little tedious and I am not clever enough to get a general solution for N cubes, so I just chose to use 4 which can be extended to any number you like. And why ask how long before the last cube is some distance (1 decimeter?!) above the ground? Why not just ask when the cube closest to the ground hits the ground? That is what I will do. Also, let's treat them as point masses, not 1 cm cubes. Because it is a little involved, I will first outline the basic equations and then simply summarize the results I found after a couple of pages of algebra.

• First the totally inelastic case. Here when a collision occurs between a mass M with speed U and a mass m at rest, the stuck-together mass (M+m) will have a speed u=MV/(M+m) because of momentum conservation. Since the falling masses slow down with each collision, it should be obvious to you that the final blob of masses will strike the floor later than if the top block had simply free fallen to the floor, so basically, you are right (but, as you will see, only for inelastic collisions). But let's do the details. If an object starts with speed v₀ and falls a distance h, its final speed v will be v=√(2gh+v₀²) and the time it takes to go the distance h will be t=(v-v₀)/g where g is the acceleration due to gravity. So, the first block starts with v₀=0, falls a distance h=1 m and acquires the velocity v=√(2g)=4.43 m/s in a time t=4.43/9.8=0.45 s; it then collides with and sticks to the next block and the two depart with speed ½(4.43)=2.22 m/s which becomes v₀ for the next leg of the trip. Continuing this, I find that the next three times are 0.28 s, 0.23 s, and 0.19 s making the total time to drop the 4 m to the floor be 1.15 s. With no collisions, the top block would have acquired a final velocity at the floor of v=√(2x4g)=8.85 m/s making the time to free fall t=8.85/9.8=0.90 s.

• The elastic case is actually easier because you are interested in when the lowest ball hits the floor. Each time one of the balls strikes another, the struck ball continues down with the speed which the falling ball had and the falling ball stops momentarily and again falls from rest. Therefore the the first ball to reach the floor arrives exactly at the same time that a ball falling without colliding would. In this case, your friend is correct.

NOTE ADDED:
It turns out that the questioner was trying, by this question, to prove that there was a conspiracy involved in the 9/11 World Trade Center terrorist attack. His reasoning is totally faulty and I disavow any attempts using my calculations to prove his ideas.

QUESTION:
This is a question that I always wanted to know the answer to, but never learned the math to figure out on my own. If I am standing at, or near, the base of something spinning or swinging (like a fan, pendulum etc.) would I be moving slower that the far edge of it? How does the speed change at any given length?

ANSWER:
Speed is the distance from the axis of rotation times the angular speed in radians per second. For example, something spinning at 60 rpm has an angular speed (60 rev/min)(1 min/60 s)(2π rad/1 rev)=6.28 rad/s; so a point 2 m from the axis has a speed of 12.57 m/s, a point 3 m has a speed 18.84 m/s, and so on.

QUESTION:
I will be filling a pressure vessel (a soda keg) with old, flat tennis balls, then pressuring the vessel to ~30+ psi. The goal is to re-pressurize the old balls so they bounce "like new" (I'm told new unopened tennis ball cans are pressurized to ~17 psi). I believe this will be possible because the rubber membranes of the balls under their felt covers are permeable (to air). The vessel has a gauge to allow monitoring of the internal pressure. My question is ... as air molecules are forced across the rubber membranes of the balls, would you expect the vessel's pressure reading to remain constant, or would you expect it to drop?

ANSWER:
Suppose the pressure is not so large that the balls are crushed, they maintain their size. When you first pressurize the volume around the balls, the pressure outside the balls is greater than inside which, as you suggest, will cause the air to seep into the balls until the pressure is equalized. The ideal gas equation is PV=CNT where P is pressure, V is volume, C is some constant, N is the amount of gas, and T is the temperature. The volume of the gas outside the balls (which is where you measure pressure) does not change and I will assume that the temperature does not change. But, the amount of gas in that volume gets smaller. Therefore the pressure must get smaller. Incidentally, you will not really be getting the equivalent of "virgin" tennis balls. The "fuzz", probably all worn off, plays an important role in the aerodynamics of a tennis ball just as dimples on a golf ball do.

QUESTION:
My questions pertains to the gravitational constant. I'm a theoretical physicist, so I don't need an obvious answer, and I know the gravitational force equation works well enough to use but you have more experience than me so... Shouldn't a universally correct gravity equation not have a gravitational constant derived from earth? Are we missing something universally basic about gravity? Even Einstein's equations use an earth based gravitational constant as part of the definition. This seems very flawed to me... perhaps you could help me here? 

ANSWER:
If there is something which we believe to be a universal constant, we measure it; we are here so we measure it here. But the question you ask could be asked of any constant. How do we know that Planck's constant is the same everywhere (and "everywhen") in the universe? Once we have the value measured here, and we can measure most accurately here, we look elsewhere for evidence that physical constants may vary over distance or time. As far as I know, no evidence has been found for inconstancy of fundamental constants, but it has been looked for.

QUESTION:
Why do (two dimensional pictures of real) snowflakes ALWAYS have six main points and six sides. General answers on the web refer to the bonding angle of water molecules - 104.5 (doesnt correspond to 60, so dont understand why that would be the determining factor). I presume it has to do with locking in the hydrogen bond, but not sure why the six sided structure would be preferred.

ANSWER:
The best answer I can give you is that solid ice has a hexagonal symmetry as seen in the picture at the right. Here the red spheres are the oxygen and the grey rods represent the hydrogen. So the angle does not stay at 104.5⁰ when the water solidifies, rather opens to 120⁰. Looking at the picture, the crystal tends to grow by adding molecules to the outermost oxygen atoms with only two existing bonds and you can imagine how the snowflake grows. Also, it is more likely to grow in the direction of the planes of the hexagons than perpendicular to them, resulting in flat snowflakes.

QUESTION:
I was wondering wether a planets diameter relates to it's gravitational strength and why

ANSWER:
There are essentially two things which determine how strong gravity is at the surface of a planet―what its mass is and what radius is. The gravitational force is proportional to the mass so that if two planets have the same radius and one has a mass twice as big as the other, the more massive planet has twice the gravitational force at its surface. The gravitational force is inversely proportional to the square of the radius. So if two planets have equal masses but one has twice the radius of the other, the smaller planet will have 4 times the gravitational force at its surface.

QUESTION:
Is the acceleration due to gravity always -9.8m/s2 ? I mean when i solve a problem, could i possibly choose the downward direction as the positive direction to make the acceleration turn into +9.8m/s2 ?

ANSWER:
You are always free to choose any coordinate system you like. Indeed, if you choose a y-axis pointing down, the acceleration is a_y=-g where g=9.8 m/s². [I generally encourage my students to usually choose +y to be up because later when you study energy, your textbook will say that the potential energy of a mass m at a position y relative to the origin is mgy. It turns out that this is true only for +y up.]

QUESTION:
If a cargo plane allowed 10,000 pounds of cargo and a pilot had to transport 20,000 pounds of birds, would he/she meet the weight limit if there was a way to keep 1/2 the birds flying during the flight?

ANSWER:
Let's just suppose we have one bird in a box. That will illustrate the principle. I always tell my students to "choose a body" upon which to focus. I choose the hovering bird first. There are two forces on him, his weight down and a force which the air exerts up on him up which must be equal to his weight since he is in equilibrium. Next I choose the air. There are three forces on the air, its weight, the force of the bottom of the box up, and the force of the bird on the air. Since Newton's third law says the force of the bird on the air is equal and opposite the force of the air on the bird, we can conclude that the box exerts a force up on the air which is equal to the weight of the bird plus the air. Finally, supposing the box is sitting on a scale, choose the box as the body. The forces on it will be its own weight down, the force of the air down on it, and the force of the scale up on it, so the force the scale exerts up on the box is the weight of the box plus bird plus air. Therefore the box exerts a force down on the scale which does, indeed, include the bird's weight. Your 20,000 lb is what the load in the plane would be whether the birds were in the air or not. For a much more detailed discussion of a similar problem, see an earlier answer.

QUESTION:
Solve an argument for us please: My friend says that Sound is not part of the Electromagnetic Spectrum and I say it is. I CAN NOT find an answer anywhere on the Internet. What I find is that anything below radio waves in the frequency spectrum is just ignored. Why doesn't the EM Spectrum start at 0hz? If the audio spectrum is ~20hz-20khz then WHEN/WHERE does the EM spectrum start? Wikipedia "Radio Waves" says that radio frequencies can get as low as 3khz. That is an audio frequency, right? So, I'm just looking for a "break point" at which audio becomes radio. And at what point do audio waves become part of the EM spectrum? Is it a definition problem?

ANSWER:
Sorry to be the bearer of bad news, but sound is not an electromagnetic wave. What the wave is is determined not by frequency, but by "what is waving". Sound is a wave which travels in some medium, for example air. If there is no medium between you and a source of sound, you will not hear it. Electromagnetic waves are composed of electric and magnetic fields which are "waving" and can travel through a vacuum unlike any other wave. See an earlier answer. Another difference is that sound is a longitudinal wave (the air vibrates parallel to the direction the sound is going) and EM waves are transverse (the electric and magnetic fields vibrate perpendicular to the direction the wave is going).

QUESTION:
Is this a valid question??? With a valid answer??? It was in a teacher resource boook. I have my doubts. "It takes 20 million years for the light of a star to reach Earth. What is the distance between the star and Earth in light years?? answer: 20 million years".

ANSWER:
A light year is defined to be the distance which light travels, in vacuum, in one year. Therefore, one light year is a measure of distance, not time as you might guess. Given the way the question is written and answered, the answer should be either 20 million or 20 million light years, but not 20 million years.

QUESTION:
Im trying to figure out what speed a sphere weighing 4,506 lbs will attain on a 3 degree grade from complete stop to 95 ft. [After communicating, it is a uniform sphere of radius 4', rolls, does not slip.]

ANSWER:
The easiest way to solve this is conservation of energy. Let's work in SI units, m=4506 lb=2044 kg, d=95'=29 m, and R=4'=1.22 m. The energy at the top of the ramp is E₁=mgh where g=9.8m/s², and h is the height above the bottom which is h=dsin3⁰=0.052d. The energy at the bottom where the speed is v is E₂=½mv²+½Iω² where I=2mR²/5 is the moment of inertia and ω=v/R is the angular velocity. Putting this all together and solving for v, v=√(0.058gd)=4.07 m/s=13.4 ft/s=9.1 mi/hr. So, you see, the answer to your question does not depend at all on either the mass or the radius of the sphere, R and m cancel out as you can show for yourself.

QUESTION:
When an object falls to the ground, shouldn't the ground exert a greater force on the object than the weight of the object to stop its motion altogether? Because if the two forces were equal and opposite, they would only cancel the forces resulting in 0 acceleration, but the object should still continue to move at the velocity it had the instant before the force cancellation. (I know it sounds silly for the ground, since it's a solid, rigid body. Maybe it'd be better if you replace the ground for a trampoline and explain it for that).

ANSWER:
Yes, the force up by the ground must be greater than the force down of the weight in order for the object to do what we know it does, stop. How big the force is depends on how long it takes to stop. If it is an object hitting a hard surface, it stops very quickly which means that it requires a large force. If the surface is soft, say a big fluffy pillow, it takes much longer and the average force is much smaller.

QUESTION:
Figure skating - if you are doing a spin on ice and you leave your arms spread out it creates resistance, if you pull the in towards the body you create less resistance so you spin more easily. How do I express this with mathematical equation in physics? [The questioner is a 13 year-old high school student with no physics training.]

ANSWER:
While it is true that there is somewhat more air resistance if your arms are outstretched, you are barking up the wrong tree if you think this is the reason your spin speeds up when you pull your arms in. The principle involved here is conservation of angular momentum which is a topic usually covered late in an introductory physics course, so I have to give you a little background here. For translational physics (objects moving without spinning), Newton's second law is perhaps paramount: the rate at which linear momentum changes is equal to the force on the object. Linear momentum p is defined to be the mass m of the object times its velocity v, p=mv. If an object has no forces on it, the rate of change of linear momentum must be zero which means that it never changes; this is called conservation of linear momentum. For a single object, this means that it continues moving in a straight line with constant speed because the mass will not spontaneously change. In rotational physics, Newton's second law is analogous to translational physics: the rate at which angular momentum changes is equal to the torque on the object. Let's briefly discuss what the quantities are in translational and rotational physics:

• Force is what causes an object to accelerate, change its velocity. Torque is what causes an object to spin faster (called angular acceleration). Note that it is possible to exert a force on something without causing a torque, for example, pushing on a door where it is hinged does not open the door.

• Linear momentum is p=mv. Angular momentum is L=Iω, the moment of inertia (see below) times the angular velocity (for example, revolutions per minute).

• Mass is the resistance a body has to being accelerated; a one pound force exerted one second will cause a much bigger acceleration for a baseball than for a locomotive. Moment of inertia I is the resistance of a body has to being angularly accelerated (increased spinning speed). But, monent of inertia does not just depend on the mass of the body, it also depends on how that mass is distributed. In the case in point, the figure skater has a larger moment of inertia with her arms outstretched than not; the same torque applied to each will result in a smaller spin rate for the outstretched arms.

• If you exert no force on an object, its linear momentum must remain constant. If you exert no torque on the object which is spinning, its angular momentum must remain constant.

Are there any torques on the skater who is spinning? Yes, there is the friction of the skates on the ice and the air drag on her body, and if she just does nothing she will eventually stop spinning. However, those torques are small (it would take a pretty long time for her to stop), so we can say that angular momentum is approximately conserved. Now, suppose that her moment of inertia with arms outstretched is I_out and with her arms pulled in is I_in=I_out/3, and that she is initially spinning with ω_out=2 revolutions per second. Then apply momentum conservation:

• I_inω_in=I_outω_out

• 2I_in=I_inω_out/3

• ω_out=6 revolutions per second.

That's the whole story. Because angular momentum is conserved, reducing the moment of inertia results in increasing the angular velocity.

QUESTION:
If an object moves further away from the Earth it gains gravitational potential energy, this means that the energy used to lift the object above the ground is equal to the energy it contains. However there must surely be a point at which the gravitational pull of the Earth is negligible or is outweighed by a different, larger object in space that the object becomes close to. In this case what happens to the gravitational potential energy it had originally from the Earth? It cannot disappear as energy is not created or destroyed but surely it is not still there if the Earth has no pull on it or if a different object such as a planet or star is giving it a different amount of gravitational potential energy?

ANSWER:
Let's restrict the discussion to an "earth-only universe" because other objects will just complicate the situation and cloud the basic principles needed to understand. Once you understand the two-body problem, you can simply add other objects, it just makes the potential energy function more complicated but all the potential energies can be added up algebraically. As you lift the mass farther and farther away from the earth, the force gets smaller and smaller and so you add less and less energy with each meter you lift it and, eventually you might say that you are adding so little energy by lifting it further that you decide any additional energy will be negligible. But even if you only add 10^-10 J of energy for the rest of your life, you still are adding energy. But, the force never completely disappears because the universe is not infinitely large. So, no matter how far away you are, when you let go of that mass it will accelerate back to earth and all that potential energy you gave it will eventually (maybe in millions of years) be converted back into kinetic energy. No energy ever "disappears".

QUESTION:
I was working on a ship and was down in one of the holds cleaning out the bilges. The hatch cover was closed. No sunlight was entering the hold except for that coming through eight 1.5 inch diameter holes spaced out on the hatch cover. The eight holes cast eight bright circles of light on the deck 70 feet below the hatch cover, creating eight perfectly focused images of the clouds passing overhead. I get that part - pinhole cameras. However, after a couple hours when my eyes were well adjusted to the dark, I noticed another optical effect while sitting down there taking a break. About 25 feet above the hatch cover was parked the boom of a crane, running diagonally across the cover. When my eyes adjusted, I saw a perfect shadow of this crane on the deck of hold, a dimmer version of the shadow that would be seen if the hatch cover were to be opened. In no way was there a regular lay-out of the eight holes in relation to the crane boom. There were not eight different shadows - only one that somehow must have been a composite. Is there a name for this effect?

ANSWER:
Here is what I think is happening. Presumably, since you talk about shadows, it was a sunny day. The eight bright spots you see on the floor were eight separate images of the sun formed by your eight "pinholes"; the clouds you saw were clouds passing directly in front of the sun. As a check, I would expect that the diameter of these images would be about 7" since the angle subtended by the sun in the sky is about 0.5⁰, so d≈70'(12"/1')x0.5⁰x(π/180⁰)=7.3". But, the sun is only a tiny amount of the whole image formed by each pinhole; the whole rest of the sky is there but at a much lower intensity. The crane is part of that bigger image for each pinhole and I suspect there are really eight images which are pretty much overlapping because the spacing between pinholes is probably small compared to the size of either the crane or its shadow. I also think we should call the shadow an image instead.

QUESTION:
wanted to ask that when we say that for e.g i weigh 50kg is that my mass or my weight There is a difference because on earth mass X10=my weight and if 50kg is my mass then why people say that i weigh 80kg or 70kg or50kg or so..........? why dont they say that this is my mass and i weigh 500newtons please explain thi to me!

ANSWER:
In physics, you are completely right, a mass of 50 kg has a weight of 500 N (for g≈10 m/s²). In everyday life in most of the world, you buy 1 kg of fish, not 10 N of fish. That is just the convention and it works fine as long as you are at the surface of the earth because the weight is proportional to the acceleration of gravity. In countries where weights are expressed in lb, you are actually buying 1 lb of fish; but in those countries there is confusion as to what the mass of the fish is. The mass is W/g=1 lb/(32 ft/s²)=0.031 lb·s²/ft≡0.031 slug.

QUESTION:
What would happen if we were to compress a gas's atoms so tightly that they could no longer flow freely to be a gas would it change forms into a solid?

ANSWER:
It depends on what the gas is. Each material has its own phase diagram which shows what happens as pressure and temperature are changed. As an example, the figure to the right shows the phase diagram for water. You can see that at temperatures above about 0⁰C increasing the pressure on a gas at constant temperature (which is what you would do to "compress a gas's atoms"), you create a liquid (called a phase change) If you have temperatures below 0⁰C and very low pressures, you can have water in a vapor state and, if you increase the pressure, you will cause a phase change to solid water (ice).

QUESTION:
Im a young student of class 8th here's what i want to ask If two objects, one with bigger mass(but same surface area,which means air resistance will be same) and one with smaller mass are dropped from air which one will fall first? if the same objects(with same material and size, which means friction will be same) are slided from a ramp which one will reach down first? if same force is applied to both the objects (friction and air resistance is same) which one will speed up more? In my book it is written "in fact, the mass of an object does not affect how an object speeds up under the pull of gravity" i dont think its right(book is Heinamann Science 3) Do you agree? i dont. Please explain this to me as im very serious and anxious about it and use simple language and formulas and no complex formulas which i wont understand as im only 14.

ANSWER:
When you say "air resistance will be the same", you are neglecting that the air resistance will depend on velocity, not just surface area. An often excellent approximation to the air resistance force F on an object of cross section A moving with a speed of v is F=¼Av². When an object is falling, there are two forces on it, its own weight mg down and the air resistance F up. If you think about it, the object will stop speeding up when these two forces have equal magnitudes, mg=¼Av_t² where v_t is called the terminal velocity, the fastest that the object ever goes when free falling. So, v_t=2√(mg/A) and you can see that the bigger m is, the faster a free-falling mass can go; therefore, the more massive object will win a race with a less massive object. When your science book says that "the mass of an object does not affect how an object speeds up" it means that there is negligible air resistance. Similarly, for the race on the incline, the more massive should win the race because its terminal velocity is larger. You could have found many earlier answers on these subjects on my FAQ page, see 1 and 2.

QUESTION:
I teach a self defense course. I am trying to explain to my students that a 230 grain bullet moving at 850 ft per sec not stop a 200 lbs person running at them at full speed. If i am correct or not could ypu please put it into terms that i could use.

ANSWER:
The concept you want to use here is momentum conservation. Momentum is the product of mass times velocity and the total momentum of a colliding system must be the same before and after the collision. The one subtlety here is that if  the momentum of a 2 lb mass moving 300 ft/s north is 600 lb·ft/s, then a 2 lb mass moving 300 ft/s south  has a momentum of -600 lb·ft/s. I will use lb to measure mass and ft/s to measure velocity, so you have to be sure you get all the units the same to make any sense: 230 grain=0.033 lb. You say "full speed" so I will use 30 ft/s to be the person's velocity, which would correspond to a 10 second 100 yard dash. So, the momentum before the bullet strikes (and lodges in) the man is 200x30-0.033x850=5972 lb·ft/s. (Notice the - sign on the momentum of the bullet since it and the man are going in opposite directions.) After the bullet lodges in the man the momentum is 200.033xV where V is the velocity of the man (plus bullet) after he is hit. Setting the momenta before and after equal and solving for V, V=5972/200.033=29.9 ft/s. So, you see, the man loses almost no velocity at all due to the bullet. If you hit his head or his heart he might stop running but it is certainly not the impact with the bullet which stops him. A qualitative way to put this is that while the bullet has a speed about 30 times greater than the man, it has a mass almost 7000 times smaller. Related to that, when you see a movie in which a character is sent flying backward by the impact of a bullet, it is total nonsense and will not happen in real life.

QUESTION:
This is not a homework question. It's for an op-ed I'm writing about the pounding offensive football players take from ever-larger defensive players. What force would a 325 lb. object going 18 mph exert on a 215 lb. object going 20 mph? It would seem that if the 215 lb. object was stationary, the force would be 5,909 lbs. (F = MxA), but that seems way off.

ANSWER:
I have answered questions like this many times but every now and then it is worthwhile to answer it again. There is absolutely no way you can compute a force from the information you have given. When one object collides with another, Newton's third law requires that the force one exerts on the other is of the same magnitude (and opposite direction) as the force the other exerts on the one. So both players will experience the same force. That is not the same as saying that both incur the same damage which would depend on the details of the collision. To determine what the forces during collision are, you must know how long the collision lasts. One simplification is that football tackles are usually inelastic, that is, the two are stuck together after the collision. The units you use (English) are much harder to do calculations with than SI units, so I will convert units first and then convert back in the end: 325 lb=147 kg, 215 lb=98 kg, 18 mph=8 m/s, 20 mph=9 m/s. When they collide, momentum (which is mass times velocity) must be conserved: 147x8-98x9=(147+98)v, v=1.2 m/s, where v is the velocity after the collision. I assume that you want them to be going in opposite directions when they collide which is the origin of the negative sign for the momentum of the 98 kg running back. Now, you want to know what the force on the 98 kg running back is. An alternative way to write Newton's second law is F=rate of change of momentum (which is the same as F=ma and the way Newton actually wrote it). How much did the momentum (which I will call P) change during the collision? Momentum change=C=P_final-P_initial=98x1.2-(-98x9)=1000 kg·m/s. Now, the force is F=C/T where T is the time the collision lasted (which was the main piece of information missing in your question). For example, if the collision lasted 0.3 s, F=1000/0.3=3333 N=749 lb where "N" is Newtons, the SI unit of force. Note that the force is positive which means it is in the direction opposite of the running back's initial velocity, as you would expect. If you were to do the same calculation for the 325 lb player, you would find that F=-749 lb. The way to reduce the pain (force) is to lengthen the time of collision which is the function of padding; when you fall on a mattress it takes much longer to stop than when you fall on concrete. [If you want to do some calculations yourself, you might find the free unit conversion program CONVERT helpful.]

QUESTION:
does the amount of force(i.e the amt. of RECOIL that a gun experience) require to hold a operating gun in position ,depends on the mass of it? consider the following example - there are two persons A and B.A is having a 10 kg gun and B is having a 5 kg gun. Both guns are shooting the same type and number of bullets per second. Now ,which of the two persons would require to exert a greater force to hold the gun in position? If the force require to hold both the guns in position is the same(which I think it is ),then is it not counterintuitive?? I mean ,imagine to exert the same force to the 50 kg gun and a 5 kg gun, to hold them in position??

QUESTION:
Let's say that I have mallet attached to a rotating axle, in such a way that the mallet is perpendicular to the axle, and the base of the mallet's handle is attached to the axle. Let's also say that the mass of the mallet is known (the entire mallet, not just the head), and the speed at which the mallet head is traveling is known. With that in mind, how do I calculate the kinetic energy involved if this silly rotating mallet contraption were to strike something with the mallet head? I'm aware that the equation for kinetic energy is normally k=(1/2)mv^2, but I don't know if that equation would apply in this situation. In short, I seek an equation to calculate the kinetic energy involved in this situation.

ANSWER:
The kinetic energy K of your rotating mallet is not ½mv², because the speed the whole mass m is not v. The velocity which is important is the angular velocity, ω, which is the number of radians per second the object is rotating; since there are 2π radians in a full circle, ω is related to the frequency f, the number of revolutions per second, by ω=2πf. Then the velocity of any point on a rigid body which is a distance r from the axis of rotation is v=rω. So, an atom, say, near the axis of rotation contributes less to the kinetic energy than one far from the axis. So your mallet's kinetic energy depends on the geometry and how the mass is distributed over that geometry. This is taken care of by introducing what is called the moment of inertia, I, which plays the role in rotational physics that mass plays in translational (in-a-straight-line) physics, such that kinetic energy may be written as ½Iω². I can give you an approximate moment of inertia for your mallet. If the size of the head is much smaller than the handle is long and the length of the handle is much larger than its thickness, you can approximate the head as a point mass and the handle as a thin stick. I will call the mass of the handle m, the length of the handle L, and the mass of the head M. In that case, I=[M+(m/3)]L². So, the energy would be K≈½[M+(m/3)]v² where v is the speed of the head.

QUESTION:
I am 15 years old and want to become a physicist, and I have a question about Sir Isaac Newton in relation to gravity and the moon. Newton asked the question "If the apple falls, does the moon also fall?" He then proved that the moon indeed falls, and is constantly falling around the Earth. But since every free-falling object falls at approximately the rate of 10 meters per second squared, then why does the moon orbit the Earth at a constant speed?

ANSWER:
Acceleration is the rate of change of velocity but velocity can change in two ways, either by changing its magnitude (speed) which is what you are thinking of, or by changing its direction. An object moving with constant speed v in a circle of radius R has an acceleration of v²/R.

QUESTION:
My Question is, if you were in a spaceship in space with no windows, what experiment could you perform in order to prove whether you were moving or not moving?

ANSWER:
There would be no such experiment, because "moving" only has meaning only relative to something else. If you were not burning your engines (that is there were no forces acting on your ship), Newton's laws of motion would be correct laws of physics in your ship. However if you were burning your engines but you did not know that for sure, there would be no experiment you could perform to determine whether you were accelerating or whether you were in a gravitational field.

QUESTION:
We know that space and time are not absolute in our universe. My question is: what causes space and time to be relative in our particular universe? In a multiverse scenario can we think of a universe where space and time may not be relative?

ANSWER:
The reason that time and space are not independent is that there is something in our universe (light) which, no matter who measures its speed the same value is found. Since speed is the ratio of a length (space) to a time, there is no way this could be true in a universe where time and space were not linked. I do not speculate about multiuniverse scenarios!

QUESTION:
A planet was discovered, Gliese 581g, that is 20 light years from Earth. This planet is the right size and distance from its' sun to have liquid water on the surface and therefore life. How long would it take to reach this planet traveling at 18,000 mph, the approximate speed at which our rockets can can currently travel?

ANSWER:
20 ly is about 1.2x10¹⁴ mi, so t=1.2x10¹⁴/1.8x10⁴=6.7x10⁹ hr=760,000 yr.

QUESTION:
Is there known upper limit of the acceleration? and if it exists, what is its formula as a function of physics basic constants?

ANSWER:
Acceleration is change in velocity over elapsed time, a=Δv/Δt. There is an upper limit on how much velocity can change, since the velocity of any object with mass must be less than the speed of light c, Δv<3x10⁸ m/s. There is, however, no known limit to how small Δt can be, so there is no limit to how large a can be. It has been suggested that perhaps there is a minimum time interval called the Planck time which is about 3.4x10^-44 s, so the maximum possible acceleration would be about 10⁵² m/s².

QUESTION:
How is the weight of a planet (e.g. Earth's) measured?

QUESTION::
Newtons Law states that to every action there is an opposite and equal reaction. If a cue ball is stuck hard enough and strikes another ball, the other ball moves off with the same speed at which it is struck, while the cue ball stops dead. This is providing the cue and cue ball and struck ball are in a dead straight line. Does this mean that the cue ball has transfered all of its energy to the ball that moves off therefore not being able to follow on? Have just found your site and find it very interesting and so give to you my thanks.

ANSWER:
This is an "action/reaction" example (Newton's third law) only in an indirect sense: because the balls, when colliding, exert equal and opposite forces on each other, their total momentum must not change (be conserved). Billiard balls are very elastic and therefore, to a good approximation, energy is conserved also. For the one-dimensional case where one ball is initially at rest which you cite ("dead straight line"), there are two equations to solve:

where m₁ is the mass of the incoming ball, m₂ is the mass of the other (at rest) ball, v₁ is the incoming speed, and u₁ and u₂ are the outgoing speeds. If you solve these, you get the speeds after the collision:

1. u₁=v₁(m₁-m₂)/(m₁+m₂)

2. u₂=2m₁v₁/(m₁+m₂).

So, if m₁=m₂ (as for billiard balls), u₁=0, u₂=v₁, as you note. And, certainly, as you also note, the first ball has transferred all its energy (and momentum) to the second ball. Since I have gone to the trouble of solving the problem generally, we might as well look at a couple of other special cases. Suppose that m₁<<m₂; this would be like colliding your cue ball with a granite wall. In that case u₁≈-v₁ and u₂≈0, where ≈ means approximately equal; this is just what you would expect, the ball bounces back (that is what the - sign means) with the speed it went in and the wall stands still. Similarly suppose that m₂<<m₁; this would be like colliding your cue ball with a BB. In that case u₁≈ v₁ and u₂≈2v₁; so the cue ball never knew anything happened and the BB ends up going twice as fast as the cue ball.

Incidentally, for completeness, there is a second solution when you solve the two conservation equations, sometimes referred to as the "trivial" solution. You should expect two solutions because one of the equations is quadratic. The second solution is

1. u₁=v₁

2. u₂=0.

This is called trivial because it would correspond to missing totally and you would certainly expect the energy and momentum to be conserved if the collision never happened.

QUESTION:
would it be possible to have a particle accelerator as an engine for a spaceship? Where by rerouting particles that are moving near the speed of light and making them collide with the back of the spacecraft would give the spacecraft a push in the back?

ANSWER:
You are essentially asking "if I, standing in my spaceship, push on the front wall, will it go faster?" Whatever impulse you gain by pushing forward with your hands will be cancelled by the frictional force your feet exert backwards. Similarly, the impulse you get from the push from the particles forward would be negated by the force necessary to "reroute" them or to accelerate them in the first place. Anything you do inside a closed system cannot change the momentum of that system. If you were to fire the particles out the back like a rocket you could propel the spaceship; this is the idea of ion-drive thrusters.

QUESTION:
if we see an object because of light from the source hitting the surface of the object make its electrons to excite to higher orbits and comming back to same orbit emiting a photon which strikes our eyes, then according to this, at one INSTANT of TIME, an excited electron can emit only one photon and similarly if the object contained (for example) ten electrons on its surface, only ten photons can come form that object at that instant of time and all those photons will have 3 degree of freedom to move in any direction.IF this is the case, we coudnt see a complete object at a given instant of time as all the photons emitted from that object woudnt have striken to our eyes and similarly, if 3 people are watching the same object, at a given instant of time, each person would have seen a part of that object as one electron emitts only one photon at that instant of time.... but this is not the case in real world....how is it?

QUESTION:
if a radioactive atom ejected a high speed (beta particle)from its nuclues, would the remainder of the nucleus recoil with the same speed?

ANSWER:
The total energy and total momentum of the system must remain constant. If the nucleus before decay is at rest, then it has zero momentum. After the decay, the β particle and nucleus must move in opposite directions to conserve momentum. If you think of momentum as mv, the nucleus, having a much larger mass, must have a much smaller speed to have the same magnitude of momentum as the β particle. To really do this quantitatively, you would not use mv as the momentum since the β particle is a relativistic particle (moves at a speed not small compared to the speed of light), and there is a third particle (a neutrino) which also carries energy and momentum; but a classical argument like mine above gets at the crux of the matter that the lighter particle must move faster.

QUESTION:
For my Science Fair project about balloon-powered Hover craft, I did an experiment to find whether air volume in a balloon (using different sized balloons) is proportional to the hover time of the hover craft. I found out that larger the air volume of the balloon, longer is the hover time. But it is not proportional. On an average, a 5" balloon hovered for 31.9 seconds, a 9" balloon took 2 min and 45.3 sec and a 12" balloon stayed for 4 min 17.5 sec. If it were directly proportional, 9" balloon should have taken 57.42 sec and 12" should have taken 1 min. 27.6 seconds. Why is the hover time not proportional to the air volume, in spite of keeping all other variables constant? Please help me understand.

ANSWER:
You make a common mistake―the volume of a sphere is not proportional to its radius, it is proportional to its radius cubed. Therefore, if you plot time as a function of the cube of whatever 5, 9, and 12 measure (diameter, radius, it really doesn't matter), you are finding how the time depends on the volume which you know is what should matter. Or you may want to be very fancy and plot time as a function of volume, volume being 4πR³/3 where R is the radius of the balloon.

QUESTION:
My figures are not exact, just my estimate. The circ. of the earth is close to 25,000 miles at the equator. From pole to pole it is less. I read that this is due to the earths rotation of about 1000 mph. at a radius of 12,500 miles. I believe that this is due to centrifugal force caused by the rotation at that distance.. Does that mean that a person would weigh less at the poles then at the equator due to centrifugal force? Also if the earth would stop rotating, would a persons weight increase at the equator?

ANSWER:
First, I have to give my stock response that the weight is the gravitational force exerted on something and that has nothing at all to do with rotation. What you really want to know is what is apparent weight, the weight you would read with a scale you were standing on. Think about it for a minute: if the earth spun faster and faster, eventually the gravity would not be able to keep you from going off the surface. So the centrifugal force would cause your apparent weight to be smaller at the equator than at the poles. It is a very small effect, though, only about 0.3%.

QUESTION:
Do objects colder than a person would normally experience radiate its thermal energy in frequencies longer than Infrared (ie Radio)? I know that as temperature increases an object will radiate in shorter frequencies, and it seems that initial radiation in Infrared would be arbitrary. If I have the wrong idea and the majority of radiation up to a certain temperature would be in Infrared I would be interested to find out why.

ANSWER:
All objects radiate at all frequencies. The most intense frequency does, indeed, increase with temperature and the colder an object is, the longer the wavelength of its most intense radiation is. Also, the total amount of radiation varies like T⁴, so if you want to look for radio frequencies there might be more at a higher temperature even though it was not the biggest.

QUESTION:
I understand why we use magnetism to smash particles against each other to find out what they are made from. Yet I don't understand why we don't find a way to use the strong force to bring those sub light speed particles up to and over the speed of light.You know like when the starship Enterprise goes back in time using gravity assit? At the last billionth of a second we might be able to suspend a stripped down atom to get to its strong force and it it again with colliding protons that would bend slightly by the strong force to jump the particles closer to the speed of light? So why not?

ANSWER:
The issue is not how strong the force is, rather how much energy can it deliver. For an object to exceed the speed of light, an infinite amount of energy is required. No finite force can do that in a finite amount of time.

QUESTION:
I have recently learnt about torque, but i have a confusion regarding its concept: Suppose you have two identical doors with the only difference that the hinges of one are well lubricated and the other's hinges are worn out, and cause more friction than the first one. To open the doors at the same speed, it is clear that i have to apply a greater force on the second one. As torque is the product of of force and perpendicular distance, the numerical value of the torque in the second door comes out to be greater. But in the sense that torque is that how much a body turns, both of them have the same torque. Which of the two ways is correct?

ANSWER:
You have to include all the torques on the door, not just the one you apply. For both doors there is a torque due to friction of the hinges, the second one being larger. Since the frictional torque acts opposite your torque (it tends to slow the door down, while yours tends to speed it up), it should be subtracted from your torque. So, to open the doors at the same rate, your torque on the second door must be larger.

QUESTION:
Is it true that a Wave Train travels at 1/2 the speed of its constituent waves? If so, why? I mean, say a storm in the Gulf of Alaska stirs up some seas. We might see, some distance away, that individual swell waves are moving at 30 knots, but from a bird's eye view, we would see that the area of disturbance expands toward Hawaii at only 15 knots. If this is true, is it only true of transverse waves. Is there such a thing as a longitudinal wave train?

ANSWER:
What you are talking about is that wave packets or wave trains are composed of more than one wavelength and there is dispersion, i.e. different wavelengths travel with different velocities in the medium; this is known as dispersion and is the reason that a prism breaks white light into a spectrum of colors. There are, as you note, two kinds of velocity. The velocity of the waves themselves is called the phase velocity and the velocity of the wave packets is called group velocity. For water waves the result is that the group velocity is half the phase velocity. I did not know this; it is apparently true for water surface waves in deep water. The little animation above is for just this situation; the green dots travel with the packets (group velocity) and the red dots travel twice as fast with the waves (phase velocity). Everything I have said applies just as well to longitudinal waves.

QUESTION:
We've learned recently about the strong and weak nuclear forces. We also learned that when an isotope has too few neutrons, the nucleus stabilizes itself by having an electron from one of its orbitals crash down into a proton and turn it into a neutron. What I don't understand is how does the nucleus as a whole know it is not stable, if it just an inanimate collection of matter? In other words, how does a proton know it needs to turn into a neutron, how does it know that there isn't that extra neutron on the other side of the structure, how does it know that it needs to sacrifice if it is inanimate, if it can't see or talk or anything? And furthermore, when this happens, how do the electrons know they have to crash into the nucleus? How do they know where to go, and where do they get the momentum to do this? Who pushes them? And to further my question, who's to say that an isotope with two few neutrons doesn't have all of its protons simultaneously trying to stabilize, causing the whole structure to be a bundle of neutrons?

ANSWER:
This is not exactly a single, well-focused question as stipulated by the the groundrules of this site, but I will answer some of your questions with a general description of beta decay which is what you are talking about. In some sense, that inanimate collection of matter as you call it, does know what to do because it is a quantum mechanical system subject to rules. The most important rule, perhaps, is that a nucleus will seek the lowest energy state. Consider a nucleus which has an atomic weight A, composed of N neutrons and Z protons such that N+Z=A. Nuclei of a given A are called isobars. So, N could be anything from 0 to A with Z=A-N. In practice, nuclei which are bound at all must have both protons and neutrons. Take a specific example: Plotted to the right is the binding energy of the known isobars of A=73 plotted as a function of Z. Binding energy is essentially the energy you would have to supply to totally disassemble a nucleus and the more tightly bound a nucleus is, the lower the energy state. So, if nature provides a way to keep A constant but change Z and N, a nucleus will take advantage of it. So, Zn and Ga have too few protons (or too many neutrons, however you choose to look at it), and they will decay by turning a neutron into a proton plus an electron plus a third particle called a neutrino (which has practically no mass and no charge). Because the energy of the nucleus decreases, the leftover energy is used to shoot out the electron and neutrino. This is called β^- decay. But, when these two get down to Ge they will go no farther because energy would have to be added to get to As and there would be none left for the electron and neutrino. On the other side, As and Se have too many protons (too few neutrons) compared to Ge and so they have to hope that nature provides a way to turn protons into neutrons. Indeed, a proton can decay into a neutron plus a positron (a positively charged electron) plus a neutrino. This is called β⁺ decay. Finally, there is a process which can compete with β⁺ decay which is called electron capture which is what you refer to in your question. There is a swarm of electrons around the nucleus and if a proton could combine with one of them it would do so. Indeed, this is allowed if a neutrino is ejected. The electrons do not have to suddenly "crash into the nucleus", they are already there. The electrons in an atom do not really run around in little circular orbits, they are smeared everywhere, even inside the nucleus; there is a small but nonzero probability of looking for an electron inside the nucleus and finding it there. I think if you read through my answer carefully you will find I answered all your questions. Nature is "smarter" than you give her credit for!

QUESTION:
how can the laws of relativity relate to quantum physics...since these are the only laws we know of , doesnt that invalidate using them for quantum physics, since quantum mechanics is so very different than the world we exist in?

ANSWER:
Relativity is certainly applicable on the quantum level. In fact, that is where we get most of the hard evidence to support the theory of special relativity because only for very small objects is it practical to achieve velocities comparable to the speed of light. Nuclear physics provides the most convincing evidence for E=mc² as evidenced by nuclear power and weaponry. The usual formalism of quantum mechanics (e.g. Schrödinger's equation) is not a relativistic theory in itself, but relativistic corrections can be made to calculations. There is a theory of quantum mechanics which is relativistically correct version of quantum mechanics (using either the Dirac equation for fermions or the Klein-Gordan equation for bosons). So, there is no incompatibility between quantum mechanics and relativity.

QUESTION:
What happens when two magnetic fields collide. Both positive into positive (or negative to negative) and positive to negative.

QUESTION:
Hey, so muons falling down towards the Earths surface survive decaying before they reach the surface because of length contraction as a result of Special Relativity. But I read SR doesn't apply when the object is accelerating. Aren't the muons accelerating due to Earths gravity? What have I understood wrong?

ANSWER:
First of all, special relativity can be applied to accelerating objects. Acceleration does not play a central role like it does in Galilean relativity because it is not invariant; what this means is that if two observers in two different inertial frames measure the acceleration of some object, they get the same answer in Galilean relativity but different answers in special relativity. Nevertheless, at any instant the object (muon in your case) has some velocity and special relativity may be applied at that instant. But, while that is important to understand, it is essentially irrelevant in the case of muons approaching the earth. A muon lives on the order of 10^-6 s and has a relativistic speed, say 2x10⁸ m/s; with an acceleration of about 10 m/s², its velocity changes by about 10^-5 m/s over its lifetime. Hardly accelerating, is it?

QUESTION:
I'm looking for a generator to produce 100% of my current electrical usage and my usage rate is 1,400 kw/h im seeking to find what volt generator i need to buy . would a 50 kw generator be too much or to little?

ANSWER:
Well, you have your units of electricity muddled, so let's get that straight first. A watt (W) is a measure of the rate you use energy, one joule (J) of energy per second. For example, a 1000 W dryer uses 1000 J per second or 1 kW. The most common measure of energy used is not the joule but the kilowatt hour (kW·h), not kW/h as you write. When you say your "usage rate is 1,400 kw/h", you must mean that you use 1400 kW·h per month. Now, there are 24x30=720 hours per month, so your average power consumption is 1400 kW·h/720 h, about 2 kW. So, a 50 kW generator is probably overkill. On the other hand, probably at least half of every day you use next to nothing, so a better average is probably like 5 kW, still 10 times less than 50 kW. But, what you  really need to think about is your peak usage. Energy hogs are typically anything which creates heat by getting a coil hot—clothes dryers, electric ovens and stove tops, toasters, water heaters, etc.—which are typically 1-3 kW each. Air conditioners also use a lot of power, often 3-4 kW, so your usage is probably seasonal. All things considered, something in the range of 10-20 kW should do it.

QUESTION:
if you could release ALL the energy in matter, that holding the quarks together, etc, completely, can you give me an example in ballpark terms as to what it would be like. for example if you released all the energy of a gram (or even a single molecule if it would be recognizable) of some heavy element, what would you see?

ANSWER:
This is pretty straightforward and does not depend on what that gram is made of. E=mc²=10^-3 kg (3x10⁸ m/s)²≈10¹⁴ J. The bomb dropped on Nagasaki in WWII had an energy of about 10¹⁴ J.

QUESTION:
how many thermonuclear bombs would you have to explode per sec to much the solar energy intersepted by the earth every sec.

ANSWER:
A typical H-bomb is about one megaton of TNT which is about 4x10¹⁵ J. The solar power striking the earth is about 170x10¹⁵ J/s. So you would need about 40 bombs per second.

QUESTION:
We've been learning about nuclear decay and radiation lately in AP Chemistry( I am junior in high school), and this question crossed my mind: If protons can absorb electrons, what if a neutron absorbed a positron? I was imagining that if we had a neutron emitter pointed to the same place a positron emitter was pointed, positrons might merge into some neutrons and create protons. Then i wondered, now what if we were to isolate this new proton into a separate chamber of this weird apparatus I cannot fully explain, in which we emitted electrons, and thus making the proton into a neutron again, except it is now one electron and one positron heavier. Now what if we continued this cycle? Would the neutron become unstable as do heavy elements? Are current-day neutrons nothing but the development of the said process to the most stable form, as the amount of positrons and electrons must be just right? If we managed to force more positrons and electrons into the neutron than naturally allowed, would it be like nuclear fission, but make an explosion that is thousands of times greater than a hydrogen(fusion) bomb? Or will there be such a tiny mass defect when the neutron gets too large that it only explodes the apparatus and nothing more?

QUESTION:
The concept of teamwork can be illustrated by the two-horse rule. If one horse can pull 700 pounds and another horse can pull 800 pounds, how much weight will they pull yoked together? The answer may surprise you. The two-horse team will pull their own weight plus the weight of their interaction. Therefore, yoked together, the horses can pull 3000 pounds! My coworkers and I were wondering if something like this was true, could two horses pull more than the sum of their individual maximum pulls, and if so why?

ANSWER:
This is perhaps the goofiest thing I have heard recently. When you say that a horse can pull 700 lb, you are specifying the maximum force he can exert in a horizontal direction, not how much weight he can pull forward. If that horse could just drag a 700 lb box sitting on the ground forward, he could probably be able to pull 7000 lb if mounted on wheels. The weight of a horse is a force which is vertically downward and any horizontal force would have no effect on that force. Those two horses could exert a maximum force of 1500 lb.

QUESTION:
I was reading Brian Greene's The Fabric Of The Cosmos and I was wondering if you could explain something to me. Einstein holds that Gravity & Acceleration are equal to one another. So Let's say I throw myself off a very tall building and experience the gravitational pull downward. Einstein would say that I am accelerating upward. Can you please explain to me how this is possible? Does Gravity & Acceleration zero one another out. In other words do they cancel each other? is this why they remain equal?I

ANSWER:
It is not quite right to say that "gravity and acceleration are equal to one another". So, let's review the equivalence principle which says that, if you are in a box with no windows, there is no experiment which can be performed which can distinguish whether you are in a gravitational field with gravitational acceleration g or that your box is in empty space and has an acceleration g. Einstein would not say that if you are accelerating down in a gravitational field that you were accelerating upward; he would say that there would be no experiment you could perform which could distinguish your current state (accelerating down in a gravitational field) from not accelerating at all in the absense of a gravitational field. An amusing anecdote (perhaps not true): Einstein, employed as a patent clerk in Switzerland, was able to discharge his duties quickly and spend the rest of the day daydreaming about physics, supposedly the environment which allowed his brilliant discoveries in 1905. One day when he was daydreaming and watching a housepainter across the square he witnessed the painter fall from his ladder and had a eureka moment vis à vis the equivalence principle.

QUESTION:
I was delivering a 5FT by 4FT glass picture frame. It was in my empty box truck, strapped up against the wall. I use standed tie down straps to secure the frame. May I note, that the front of the frame where the glass was covered with a cardbroad that covered the dimensions of the picture frame. I live in New York City where there are many potholes. While transporting the picture frame i heard a loud pop, then a few moments later I heard glass breaking. I would like to know if there was something i could have done to prevent this from happening or are there other forces involved?

QUESTION:
.5*mass*V2= KE but what do you do with kinetic energy? how would i turn that number in to say ft/lbs or PSI?

ANSWER:
You are asking me, I guess, how to convert units. I always recommend a nice little piece of free software called CONVERT. You evidently want to convert in English units, so the mass and velocity in ½mv² would be slugs and ft/s, respectively. A slug is a mass whose weight is 32 lb; the mass of an object which weighs W lb is m=W/32 lb·s²/ft. For example, suppose you have a 10 lb weight which has a speed of 5 m/s. Then KE=½x10x5² (lb·s²/ft)·(ft/s)²=125 ft·lb. So, you see that you should have asked me what the kinetic energy was in ft·lb, not ft/lb. Energy (and torque and work) is measured in ft·lb in English units. PSI, lb/in², is not a measure of energy, it is a measure of pressure. [I know this is tedious, but it shows you why scientists prefer to work in SI (metric) units because the kilogram is the unit of mass and the above calulations would not have required that I convert slugs into units like lb·s²/ft. If I had said that I had a 10 kg mass with a speed of 5 m/s, its energy would have been 125 kg·m²/s²=125 J where J is the Joule, the unit of energy in SI units. A Joule is also a Watt·second.]

QUESTION:
in the formula for HP (HP=TQ*RPM/5252) do you know where the constant 5252 comes from and what it represents?

ANSWER:
Horsepower is a measure of power, energy per unit time. The units of torque (I assume TQ is torque) are force times distance (e.g. ft·lb or N·m) which are also the units of energy and RPM is revolutions/minute, which has the dimensions of 1/time, so your formula for horsepower indeed has the units of power, essentially energy/time. The trick is how to get it into the proper units. I am guessing that torque is to be measured in ft·lb. Now, 1 hp=550 ft·lb/s and TQ*RPM=ft·lb·rev·min^-1·(2π/1 rev)·(1 min/60 s)=0.10472 ft·lb/s, so P=TQ*RPM/(550/0.10472)=TQ*RPM/5252.1 where P is power in hp, TQ is torque in ft·lb, and RPM is angular velocity in rev/min. So my guess that TQ is in ft·lb was right. So what the 5252 "represents" is what you must divide by to get the wrong units (ft·lb·rev/min) into the right units (hp). [The purpose of the (2π/rev) factor is that there are 2π radians in one revolution.]

FOLLOWUP QUESTION:
So if i'm understanding it right HP shows energy over time. What i was trying to understand is why HP matters. In my experiance with cars at the race track given as close to identical as posable cars, the one with more HP wins. Sometimes even if the other car had a signifigant torque advantage. With what you said this makes sense because in a given amount of time the car with more HP is inputting more energy in to the driveline than the car with less hp. Or did i not understand that at all?

ANSWER:
You have it exactly right. What matters is the rate at which energy is delivered to the car. You might also be interested in an earlier answer in which the power is part of the answer.

QUESTION:
if im traveling 40 mph on a scooter and the temp. is 32 degrees what will the temp feel like with the wind chill facter

ANSWER:
The chart to the left gives selected values of speed and temperature. Also given is the equation used to calculate it. In your case, Wind Chill=15.8⁰F. You might look also at an earlier answer for more detail.

QUESTION:
If someone sent a telegraph in say, 1862, could it radiate outward to the universe.

ANSWER:
Sure, but its intensity would be so low that you could never decode it or probably even recognize it as made by an intelligent being. Since the message in 1862 would have been over a wire and not by radio (the first wireless telegraphy was in 1899), the radiated signals would be extraordinarily weak.

QUESTION:
I am really facing great difficulty in understanding Planck's Theory.What is meant by energy being 'continuous' or being in 'packets'? What will energy 'look like' if it was continuous and what is meant by ''only discrete values of energy can exist" ? Please give some examples.

ANSWER:
Continuous is what we are used to in Newtonian physics. Consider a mass m on a spring with spring constant k. What energy can it have? The energy E of a simple harmonic oscillator is E=½kA² where A is the amplitude of the oscillation; the frequency of the oscillator is f=[√(k/m)]/(2π). What restrictions are there on what A can be? In classical physics there are no restrictions, A, and therefore E, can be anything you like. This is called a continuous spectrum of energy. But is this exactly true or simply because we cannot measure A carefully enough to be sure that it really can be anything. Suppose that only certain values A were allowed in nature and that they differed by 10^-30 m; this would be called a discrete spectrum but there would never be an experiment you could perform which could observe it. But, maybe if you went to values of m and k you do not usually experience in everyday life you might see a discrete spectrum. What Planck hypothesized was that the spectrum of a simple harmonic oscillator was discrete, not continuous, and the allowed values of energy were E_n=nhf where n=0,1,2,3,… and h=6.63×10^-34 m²kg/s. Suppose that we had an oscillator with f=1 s^-1, e.g. m=1 kg, k=4π² N/m; then the spacing between allowed energies would be 6.6x10^-34 J, clearly unobservable. But, an electron in an atom might very well have a frequency of 10¹⁶ s^-1 and the energy spacings would be on the order of  10^-17 J.  This sounds small, but you could easily measure such energy differences on the atomic level. Actually, Planck was wrong and the correct spectrum is E_n=(n+½)hf, but it was close enough to cause a revolution in physics. It turns out that any system which is bound has a discrete energy spectrum and any system which is not bound has a continuous energy spectrum. In everyday life we do not see discrete spectra because our senses are not sensitive enough to detect the tiny differences between energies.

QUESTION:
I understand that speed is a scalar quantity, but I don't get why. Since speed is written as d/t, and those are two separate measurements, what stops it from being a vector quantity?

ANSWER:
There is nothing esoteric here, it is simply a matter of semantics. The velocity vector is defined to be the rate of change of the position vector of something. Speed is the word we use to specify the magnitude of the velocity vector and the magnitude of a vector is always a scalar. So the reason that speed is a scalar quantity is simply that it is defined to be. For example, my speed is 60 mph but my velocity is 60 mph north.

QUESTION:
I'm just confused because according to newton's third law of motion, " any action has an opposite but equal reaction", hence forces in nature are balanced, and balanced forces means bodies are at rest, so how come in reality, we can observe motion?

ANSWER:
See my FAQ page.

QUESTION:
could centrifugal force force actually be used to simulate gravity like in so many sci-fi stories? one of my favorite sci-fi stories is the Ringworld by Larry Niven. The Ringworld of the title is a giant ring shaped structure the size of earth's orbit. It's centered on a star and has a habitable inside edge, gravity on this inside edge is simulated by the structure spinning fast enough to make objects feel as heavy as they would at 99.2% earth gravity. If I were stood on a real structure like the Ringworld, and I jumped up in the air, would I fall back down or fly off into space?

ANSWER:
Yes, as long as the radius of the ring is large compared to the size of the objects. In such a scenario, the centripetal acceleration should be set equal to g, so g=Rω² where ω is the angular velocity in radians per second. So, the picture to the right shows the ring as viewed from outside. You now jump straight up with a speed v. However, note that you also have a tangential velocity of Rω so your actual velocity is √(v²+R²ω²). So you see what will happen is that you will go in a straight line with constant speed (because there are no forces on you) along that velocity until you again strike the ring; it will seem that you jumped and came back down. You can calculate the time you were in the air and how high you went by doing some pretty straightforward geometry/trigonometry.

• I find the two angles labeled θ above are the same so sinθ=v/√(v²+R²ω²) and cosθ=Rω/√(v²+R²ω²).

• From these you can find the length of the chord (the length of your flight) C=2Rsinθ=2Rv/√(v²+R²ω²),

• the height you go above the surface h=R(1-cosθ)=R(1-Rω/√(v²+R²ω²)), and

• the time you are in the air T=C/√(v²+R²ω²)=2Rv/(v²+R²ω²).

Now, let's compare these with the time and height on earth. Remember that ω=√(g/R) and I will take R to be very large compared to compared to v²/g. So, now I find that

• h=R(1-Rω/√(v²+R²ω²))=R(1-1/√(1+v²/(R²ω²))=R[1-(1+v²/(Rg))^-½]≈R[1-(1-½v²/(Rg)+…]≈v²/(2g) and

• T=2Rv/(v²+R²ω²)=2R/(v(1+R²ω²/v²))=2R/(v(1+Rg/v²))≈2v/g.

These two approximate results are just the same as for a projectile launched straight up on earth. Keep in mind, though, that R must be large. Also, you do not want to jump with a velocity which has a component parallel to the axis of rotation. If you are too close to the edge, you will miss the ring when you "fall back".

QUESTION:
when someone pushes a box in frictional environment so that the speed of the box is constant, it is usually said in textbooks that the net work done is ZERO, because Fd=-F (frictional)*d. And this seems OK , because of the fact that the box does not accelerate: no energy input. But the box gets hotter and it must be connected with the force applied to the box. So how can net work be zero, shouldn't it be positive?

QUESTION:
Hello, I have an idea and would like to know if it is possible. If the north poles of two rectangular magnets are secured together and the sides are covered in magnetic shielding so that only the two south poles are exposed, will the geomagnetic field exert a net force on them? This was the premise behind a propulsion scheme that I devised (even though the force produced would be small). This kind of energy-free propulsion should not be possible, but I don't see any reason that would prevent it from working.

ANSWER:
You are not going to be able to shield your gizmo and make it look like a pure north pole. But, you do not have to. If you put a bar magnet so that its north pole is directly above the earth's magnetic N pole (which is its geographic S pole) and its S pole is farther away, it will experience a force away from the earth. The reason is that the field from a dipole (like the earth) gets weaker as you get farther away, so the attractive force on your S pole is smaller than the repulsive force on the N pole. I think if you work out the numbers, though, you will find it a pretty miniscule force!

QUESTION:
How can a photon or any massless particle bounce on things? or how could massless particles like photons be effected by forces of any kind? if it is massless shouldn't it go straight through bodies and not bounce? and what does it even bounce on anyways?

ANSWER:
A photon has momentum so it can collide with another particle and exert forces on the other particle as well as have forces exerted on it. (That is really a sort of classical way to phrase it, but it gets to the heart of momentum transfer happening.) Look up Compton scattering, for example, where photons scatter from electrons. A photon carries electric and magnetic fields and so it is very interactive with most things in nature which are composed of electrically charged objects, either bouncing or getting absorbed.

QUESTION:
If you shoot a bullet straight up into the air, its velocity at the very top of the trajectory is zero, even if only for an instant, as its upward velocity slows to nothing before becoming downward velocity. Downward vertical velocity then increases in the earthward direction . Would the velocity ever become dangerous if it landed on a living person? Is the weight of the bullet important? Does the atmosphere restrict the downward velocity?

ANSWER:
A falling bullet experiences a downward force of its own weight and an upward force of air drag. The result of the air drag, which increases with speed, is to have the falling object eventually reach a maximum velocity called the terminal velocity which is determined by its weight and its geometry (which is why you can jump out of an airplane with a parachute). A .30 caliber bullet weighing about 10 grams has a terminal velocity of about 90 m/s (about 200 mph) and a .50 caliber bullet weighing about 42 grams has a terminal velocity of about 150 m/s (about 335 mph). A bullet traveling 60 m/s (about 130 mph) can penetrate the skull so, yes, a falling bullet is dangerous. Dozens of people are killed every year by celebratory gunfire.

QUESTION:
I'm trying to explain to student drivers how to avoid going into a skid during bad weather. A common rule of thumb is "skids happen when there is a change in speed or a change in direction." Ok, so you are going up a hill at a steady 35 mph. When you reach that part of the incline when your automatic would normally downshift to get your rpm's up - you skid. I know it has something to do with gravity's effect and you are actually accelerating in order to go same speed but somehow I can't seem to explain it right.

ANSWER:
The crucial point really is that the coefficient of static friction is larger than the coefficient of sliding friction. That may be too technical for your purposes, so let me explain. Static friction is the frictional force which acts between two objects which in contact but not sliding; for example, when a car is parked on a hill, the thing which keeps it there is the frictional force between the tires and the road and that is static friction. But, as the hill gets steeper, the friction needed to keep the car from sliding gets bigger and, eventually, there will not be enough friction and the car will slide down the hill. But, what happens when the car just starts to slide? Does the car start creeping slowly down the hill? No, as soon as it "breaks away" it accelerates down the hill. The reason is that the sliding friction is less than the static friction. The same is true when the car is moving because the wheels are not slipping and therefore static friction is operative. If you are moving with constant speed on a straight, level road, almost no friction is needed to keep going. But, when you accelerate you need more static friction from the tires (just like you need more static friction on a steeper hill) and if you try to accelerate too much your wheels will spin (just like the car will slide down the hill). Or, if you brake you need more static friction from the tires (just like you need more static friction on a steeper hill) and if you try to brake too rapidly your wheels will skid (just like the car will slide down the hill). You probably tell your students that they can stop quicker if they do not slam on the brakes and skid; again, the smaller sliding friction is why and that is why anti-lock brakes are such a good safety feature. Turning a curve also requires static friction because a turning car is actually accelerating even if the speed is constant because the dirction of the velocity is changing; on a very icy road where you can get almost no static friction, when you steer into a curve nothing will happen and you will just skid straight forward off the road. Your example about going up a slippery slope and downshifting is more or the less the same as braking gently, trying to get more static friction. Essentially, trying to keep a car going up the hill at a constant speed is exactly the same as parking on the hill—the static friction required is the same and if the hill is slippery and steep enough, you will be able to neither park nor go up with constant speed. So, the downshifting has the same effect as accelerating, trying to get more static friction. Even if you do not downshift, to keep going up with constant speed would require that you give the car more gas than on a level road.

QUESTION:
Balancing a glass on knives how does it work? The video.

ANSWER:
It seems to me that they show you how to do it in the video.

FOLLOWUP QUESTION:
i'M sorry I mean why does this happen? How come the three knives hold the glass?

ANSWER:
Examine each knife. Each has a force A up on its handle from a glass; each has a force B up at the end of its blade from one of the other three knives; each has a force C down from the third knife; each knife has a force of its own weight W down at the center of mass of the knife (where you could balance it without any forces but the weight acting on it). On that knife, the sum of the forces is zero, A+B-C-W=0. On that knife, the sum of the torques about any axis is zero; for example, about the glass end the where A is applied, B(d₁+d₂)-Cd₁-Wd₃=0. These conditions are satisfied and the knives are all in equilibrium. There is another interesting thing you can note: the force C on this knife is the force one of its neighbors exerts down on it with the end of its blade; but, by Newton's third law, the forces forces these two knives exert on each other are equal and opposite. It follows, therefore, that C=B since the knives are symmetrically arranged and all three must have the same forces on them. You can also see that the three glasses must hold up the total weight of the three knives, so 3A=3W, A=W. If you solve for these forces you will find A=W  and C=B=W(d₂/d₃). Putting the beer glass on top simply increases the forces B, C, and A but the knives remain in equilibrium. Is this what you were looking for?

QUESTION:
With all this future talk about harvesting asteroids, I have an interesting thought experiment... how would the force of gravity shared by the sun and the earth (aka our orbit) be affected by an increase in the earths mass. Or, would the amount of asteroid mass be insignificant in that equation?

ANSWER:
The earth's orbit, both shape and period, are independent of its mass (unless the mass of the earth became comparable to the mass of the sun which will certainly not happen). But, your last sentence certainly hits the nail on the head that the added mass would be insignificant compared to the mass of the earth. If the mass of the earth increased signficantly, though, the period of the moon would change.

QUESTION:
I have always wondered why the "speed of light" is regarded as a constant when in a vacuum. If light has a finite speed in a vacuum, then the following questions puzzle me: 1. Does light accelerate? 2. what is its rate of acceleration? 3. If so, why is there a finite speed in a vacuum?

ANSWER:
To answer your first question, why is the speed of light a constant in vacuum, see my FAQ page. The answers to your questions:

1. Light can accelerate only by changing its direction, not its speed.

2. The rate of acceleration bending light depends on the gravitational field in the vicinity; this is called gravitational lensing.

3. Again, see the FAQ page. If you want to know why it has the value it has, 3x108 m/s, see another question on my FAQ page.

Note that it is important that you understand that "accelerate" does not have to mean change of speed, it can also mean change of direction.

QUESTION::
why is that during daytime when you turn on the headlights if your car the object ahead of you doesnt appearr to be brigther?

ANSWER:
Imagine you are sitting in a quiet room and whisper to the person next to you. She can hear you. Now imagine that a jet airplane is roaring right over your house. If you whisper, you will not be heard. The headlights are a whisper compared to the intensity of sunlight.

QUESTION:
I'm very confused about work. Let's say that I'm lifting a 10 kg backpack up 50 meters. That means that the work that I just did is 10 times 9.8 (acceleration of gravity) times 50 since W=Fd and F=ma. So if I did that work, then that means that I applied 10 times 9.8 Newtons of force. That is equal to the force that gravity applies to the backpack. Therefore, there would be no acceleration. But then, another definition of work as you know is the change of kinetic energy. But if there is no acceleration, then there is no kinetic energy. Which leads to the conclusion of no work. What am I doing wrong here? I'm very confused.

ANSWER:
You are not doing anything wrong, it is just that you have specify who does the work. When you carry the pack up the mountain, you do 4900 J of work. During that same time, the earth is doing  -4900 J of work, negative because the displacement (up) is opposite the force (weight, down). You and the earth together do zero net work, so the kinetic energy does not change. When you equate change in kinetic energy to work done, you must include the total work done.

QUESTION:
"A tube in the shape of a rectangle with rounded corners is placed in a vertical plane. You introduce two ball bearings at the upper right-hand corner. One travels by path AB and the other travels by path CD. Which will arrive first at the lower left-corner?" My problem is with the velocity of the ball at the end of traveling down the first side, whether it be A or C. The problem states 'rounded corners'. Surely the ball does not come to a stop -i.e. zero velocity, as it turns the corner. So I would guess that there is a significant velocity after 'rounding' the corner and hence must be accounted for.

ANSWER:
This is a problem out of Halliday and Resnick, Chapter 3 on kinematics. I have been assured that it not homework, rather someone brushing up on introductory physics. I must admit that I am a little puzzled by it because, although I can solve the problem analytically as a function of the sides A and C and the angle θ which C makes with the vertical, none of these is given; this means to me that there must be some obvious qualitative reasoning for this picture which does not require any details. In this picture, the ball going down C will have a much bigger acceleration than the ball going down A because it is steeper; since C is shorter, that ball will arrive at its turn earlier than the other ball. Also, because the first corner is lower along C than along A, the C ball will be going faster when it gets to its corner than the A ball will be. So, the C ball turns the corner earlier and with a higher speed. The C ball wins. That's the best I can do with qualitative reasoning. (I would be glad to have anybody write in with a better explanation.) However, my curiosity led me to solve the problem analytically so I could examine all the possibilities. I will not give the anaytical solutions for both of the two paths since they are a little messy, but they depend on A, C, and θ; click here to see these solutions. Plotted below are the times as a function of θ for 0⁰-90⁰. The case most closely resembling the given problem is where A=2C with θ≈30⁰; The C ball is the clear winner as speculated above. One interesting thing to note is that when θ=45⁰, the balls take the same time regardless of the lengths of the sides. This is easy to understand because both balls have identical accelerations of 0.707g the whole time. And, the ball starting on the steeper slope always wins the race, regardless of the lengths or the angle θ. Note also that as θ―>0⁰ the time for A ―>∞, and as θ―>90⁰ the time for C ―>∞. This is also easy to understand because a ball in a horizontal tube will not move at all.

QUESTION:
I'm hoping you can settle a bet between my father and myself. We are both movie buffs, and work together. While working we discussed the Ah-nuld movie "Eraser". I'm going to make the assumption you have never seen the movie which revolves heavily around the manufacture of man-portable super-railguns that fire aluminum rounds at 'close to the speed of light'. Being firearms enthusists as well, we discussed the flaws in their idea (beyond the fact no technology in the near future would allow such a weapon as the one depicted to be made, let alone one man-portable.) Where we came to a disagreement was why for other reasons it would be useless. While we agree that the mooks Ah-nuld tears through as he does in every movie would be out of luck. In practical situations the weapon would be useless but we disagree why. My father believes that a 5.56mm / 0.00401753242 kg bullet (our assumptions) traveling 80% of c would immediately flatten out and quickly lose all of its kinetic energy due to its low mass and aluminin's relatively soft nature upon striking anything solid such as a door, or brick. I say a round with the same dimensions would explode on contact with something solid (again like a brick) as the work-heating would vaporize a piece of aluminimum with that mass. If it could even get there without being melted by atmospheric friction. As I've submitted questions to you before we decided we would --drum roll--- ASK THE PHYSICIST!!! Are either of us correct? The loser has to buy the next Pizza we order.

ANSWER:
Neither of you are addressing the real issue here—a 4 gram bullet traveling at 80% the speed of light has an almost incomprensible amount of energy. So, saying that this weapon would "be useless" is way off the mark. Saying that it is impossible to make such a weapon is a different issue which I will address after talking about what such a speedy bullet would do. You are probably not interested in the details, so I will just give you the kinetic energy such a bullet would have—about 5x10¹⁴ Joules. To put that in perspective, if you took that energy and delivered it to the power grid over a period of a day, this would be the equivalent of a 6 gigawatt (6x10⁹ watts=6 billion Joules per second) power station. The largest power station in the US has a power output of about 4 gigawatts. Or, the energy of the Nagasaki atomic bomb was about 10¹⁴ Joules, 1/5 of the energy of your 4 gram bullet. So to argue whether the bullet's flattening or exploding is the reason it would not do much damage sort of misses the point, don't you think?! If the bullet takes 10 seconds to deliver its energy to whatever can take it, you are still talking about the energy of 5 WWII-era atomic bombs being delivered. I would not want to be within 50 miles of that. Finally, it means that the "gun" has to deliver all that energy in an unimaginably short amount of time (I presume that since, if it is "man-portable", the gun must be no more than a few feet long). I figure that a force on the bullet of more than 2 million pounds would be required during the time it was in the gun; do you think an aluminum bullet could withstand such a force? And, Newton's third law says that, if the gun exerts that force on the bullet, the bullet would exert that force on the gun. Imagine the recoil! This is such a ridiculous scenario, I don't think we even need to beat a dead horse and talk about what air friction would cause to happen as it flew to the target.