Ask the Physicist!

QUESTION:
My name is James. I am a 12 year old student in grade 8, and once in math class we talked about a cube in the forth dimension and I was a little confused, so I want to ask you some questions. Why would a tesseract or a forth dimensional cube be like a big cube with a small cube inside, why wouldn't it be a cube that constantly exists I its time axis, because in fourth dimension you would add a time axis to the three dimension xyz axises so when a three dimensional cube has a side length of x then its volume would be x^3, I think that when this is applied to the fourth dimension the volume or some kind of time volume of that object will be x^4, so I think that similar to side lengths the object or tesseract will have some sort of time length and that tesseract will exist in a certain amount of time according to its time length. Is my theory correct? If its wrong, what is wrong about it? How is the big cube small cube one correct? what about a cube in the 5th dimension?

ANSWER:
This is not really physics, more mathematics. I am not a mathematician, so you should read the Wikepedia article on tesseracts to get a more rigorous treatment. A tesseract is not a picture of a four-dimensional cube, that is impossible to draw. But we can ask what are the rules we might follow to move up in dimensions in constructing cube-like shapes? In zero dimensions we have a point; to go to one dimension we use two points to define a straight line; to go to two dimensions, we can use four lines to create a square; to go to three dimensions, we can use six squares to create a cube. (See the first animation.) Get the pattern? 2⇒4⇒6 to go 1⇒2⇒3 dimension. So the analogy of a cube in four dimensions would be an object in four dimensional space which was composed of eight cubes. The tessearct has eight distinct ten sided volumes like a cube has (the big and little cubes and the six volumes connecting the big and little cubes), so it can be topologically manipulated into eight cubes as shown in the second animation. A five dimensional "cube" would be composed of ten tesseracts; try drawing that one!

QUESTION:
Is the speed of light really constant? Like, is it an enduring, eternal constant, or does it somehow have a limit?

ANSWER:
The speed of light in a vacuum is a universal constant. Although there is some speculation that the universal constants may change with time, but there is no evidence for that. To understand why light speed is a universal constant, you should read some earlier answers:

Why is the speed of light constant for all observers?
Why does the speed of light have the value 3x10⁸ m/s and not some other value?

QUESTION:
My husband and I are in a heated debate about the following: Two genetically 100% identical people, assuming everything about their lives are identical aside from the following: Each one is given their own identical atomic clock at the exact instance of their birth. One person is sent into travel approaching light speed with their atomic clock, the other person stays on earth with their atomic clock. Does the person approaching light speed "live longer" than the person on earth? If you are going by what their independent clocks say, then theoretically yes, because time slows at light speed (pardon the very layman's interpretation of the theory of relativity). So the clock with the person approaching light speed with be slower than that of the person on earth, and at the instance of their deaths, the light speed person's clock shows they lived "longer" than the earth person's clock shows for the person on earth. But say you have a perfect and totally independent clock in another fixed and remote place in the universe (it could be anywhere, as long as it's position is fixed). According to the independent and perfect clock, do the two people die at the exact same moment? I say yes, according to the independent clock, they die at the same time. My husband disagrees. Please help!

ANSWER:
You are playing fast and loose with time! The biggest takeaway of the theory of special relativity is that time and length are no longer universal quantities. Someone in one frame will not see his time passing at the same rate that he measures the rate of an identical clock which is in motion relative to him. Clocks in your frame always run just like you expect them to—when your clock says 80 years from your birth, you will be an old woman, regardless of whether you are the stay-behind or zoom-off person. But you cannot even meaningfully ask the question you want to ask because "at the same time" is meaningless when comparing two different frames. (A related result of relativity is that simultaneous events in one frame are not simultaneous in another frame.) You should carefully read my earlier answer regarding the twin paradox; if the traveler goes away and then comes back to the earth and stops, then the two can compare their clocks (the traveler will be younger). The easiest way, in my opinion, to understand this "paradox" is to introduce length contraction—moving lengths are shorter than when at rest. So, if the traveler is going toward a star 100 light years away with a speed of 99% the speed of light, he will see the distance to be only be 100x√(1-0.99²)=14.1 ly; the time it will take him to get there, by his own clock, is 14.1/0.99=14.3 yr. The stay-behind twin will say the time to get there is 100/0.99=101 years and he will likely be dead. And people on earth will have to wait 100+101=201 years to "see" him arrive because it takes 100 years for the light emitted at the arrival to reach earth.

QUESTION:
I would like to get some help with the angular quantities especially angular velocity md angular acceleration. So if we have a fixed mass on lets say a merry-go-round, will this mass have a greater angular velocity or angular acceleration if it�s closer to the center or when it is at the edge?

ANSWER:
If the merry-go-round is neither speeding up nor slowing down, the angular acceleration is the same, zero, for everyone on it. But, the angular velocity is the same for everyone also. Why? Because velocity tells you how far you go in a given time (e.g. miles per hour), whereas angular velocity tells how far you rotate in a given time (e.g. revolutions per minute). Now, if someone near the center goes around 3 times in a minute, someone near the edge also goes around 3 times in a minute. The outer guy has a higher velocity but the same angular velocity. Angular acceleration is the rate at which the angular velocity is changing. For example, if the merry-go-round starts out at rest (zero angular velocity) and ends up at 3 revolutions per minute angular velocity after 2 minutes, the angular acceleration was 3 (revolutions/minute)/(2 minutes)=1.5 rev/min/min, the same regardless where you are standing.

QUESTION:
I�m writing some Sci Fi and I�m wondering:

Do we have a space-time calendar which tells us time as well as our relative place in space-time continuum? Like a map + time... Since the sun and earth are themselves moving and the galaxy itself is moving ...
Is there an opposite of time dilation where the traveller would live years but only days go by on years?

ANSWER:
The answer to your first question is that the most important thing we learn from the theory of special relativity is that time and space are not absolute; they depend on the frame of reference (that's why it is called relativity). So what space-time "looks like" depends on your frame of reference. You and anybody in motion relative to you would have different "space-time map/calendar" if there were such a thing.

Your second question doesn't really make any sense to me. If you mean by "opposite" that you might see a moving clock running fast, then the answer is no.

QUESTION:
This electric pole along the highway near a village called 'Chordi' has become a tourist attraction. It appears bent towards the observer but appears straight when one sees it from the side. What can be the scientific explanation?

ANSWER:
This is not really physics, but since the questioner resubmitted the question four times when I did not initially answer, I got curious. As luck would have it, when I googled chordi pole there was one relevant hit, a video (three images above are from that video). There is really no mystery here. Each of the two identical poles has a bend in it. They are oriented so that the entirty of both poles is in a single plane. When the poles appear straight you are viewing them edge-on with the plane.

QUESTION:
How much weight and force would it take to tip something over that weighs 1130 lb, has a base of of 6'x6', and is extended 41' in the air?

ANSWER:
You want the force, not "weight and force"; the weight is given as 1130 lb. The force you would have to use depends on three things—where you apply it, its direction, and where the center of gravity (COG) is located. I will do a general solution assuming that the COG is somewhere on the vertical line above the center of the base. There are, as shown in the figure, three forces on the object, its own weight W which acts at the COG a height C above the ground, the force F which is applied a distance Y above the ground, a frictional force f which acts at ground level. I will assume that the frictional force is large enough that the object does not slide before it tips and that F is horizontal (the red vector). I will generalize to a force not horizontal (green) later. The object will try to rotate about the edge below where the force is applied, so the friction will exert no torque. The weight exerts a torque of magnitude Ww/2 which is trying to keep the object from tipping and the force exerts a torque of magnitude FY which is trying to tip the object. The object is just about to tip when Ww/2=FY, so F=Ww/(2Y). In your case, F=1130x3/(2Y)=1695/Y; so, for example, if Y=10 ft then F=169.5 lb. If F' is applied at some angle θ relative to the horizontal (green), a larger force would need to be applied: F=Ww/(2Ycosθ).

SLIGHTLY REVISED ANSWER:
I forgot about the normal force (dashed red vector N) from the floor up on the object. But it does not matter because, when the object is about to tip, N acts at the edge and therefore exerts no torque.

QUESTION:
If you were to drop a tablespoon of a neutron star from 6 inches would it penetrate and go through a persons hand or would the person's hand simply give way and the neutron star hits the ground without penetrating through the hand?

ANSWER:
First, be sure to note that it is not possible to have a tablespoon of neutron star. The thing which holds a neutron star together is gravity and there is not nearly enough in a small quantity of the neutron matter. Second, I have no idea how the skin of your hand (normal matter) might interact with the nuclear matter. So, for your purposes let's just assume we have a tablespoon of something with the same density as a neutron star which is totally inert. I reckon that such a thing would have a weight of about 4 billion tons. I will leave it to you to imagine what would happen if such a weight were to hit you in the middle of the palm of your hand. That is more physiology than physics. There is no reason, I believe, to think that it would penetrate your hand; smash it, maybe.

QUESTION:
If an object's acceleration is zero in one inertial reference frame then is its acceleration zero in all other inertial reference frames? And If an object's velocity is zero in one inertial reference frame then is its velocity zero in all other inertial reference frames?

ANSWER:
Absolutely not! You can prove it mathematically. However, it is easy to understand it intuitively. If you are driving in a car at a speed of 60 mph, the trees, which are at rest in their frame, move past you with a speed of 60 mph.

QUESTION:
Do magnetic fields repel vibrating charged particles?

ANSWER:
That depends on how you define "repel". Usually we define repel to mean that the particle experiences a force which has at least a component along the direction of the field. For example, a positive charge will experience a repulsive force (along the field) but a negative charge will experience an attractive force (opposite the field). So, let's look at how charged particles behave in magnetic fields.

If the particle is at rest, it experiences no force at all.
If the particle is moving in the direction of the field or opposite the direction of the field, it experiences no force at all.
If the particle has a velocity which is not zero and is not parallel or antiparallel to the field, it experiences a force which is perpendicular to the magnetic field. This is shown in the figure for a positive charge. The magnitude of the force is F=qvBsinθ; for a negative charge the force would have the same magnitude but opposite direction, still perpendicular to B.

Therefore, a moving charged particle never experiences a force along a magnetic field so, according to my definition of repulsion, never experiences a repulsive force.

QUESTION:
If a piece of aluminium foil is held under mercury and scratched with a sharp object, something unusual happens to it when it is removed. Almost immediately a white substance appears to extend from the scratch marks, soon reaching a centimetre in height. As this happens the foil becomes distinctly warm. What evidence is there that a chemical reaction has taken place?

ANSWER:
This sounds suspiciously like a homework problem, but it is so cool I thought I would share with my readers. See this youtube video. Anyhow, this is clearly more chemistry than physics, but everything is physics at its root! (Note that I have not answered the question in the last sentence which is where the homework would be.)

QUESTION:
Is the red shift of light the impetus for the expanding universe (big bang) and if so we know gravity can bend light. Maybe gravity can also slow down light which can cause the red shift which in turn would mean there is no expanding universe. It always was and will be.

ANSWER:
Your thoughts have merit. Indeed, gravity does cause a gravitational red shift; it does not, however, cause the light to slow down; light in a vacuum always has a constant speed. Gravitational red shift is a much smaller shift than Doppler red shift. However, gravitational red shift for star-sized objects is very small compared to the Doppler red shifts we observe. Also, since the shift is so well-known to increase for stars as they are farther away, gravity could not account for the data. A static universe would also be impossible because the attractive gravitational force felt would have the effect of trying to collapse the universe. Believe me, the universe is expanding. Perhaps we do not understand all the details, e.g. it has been shown that distant stars are actually speeding up, we certainly know that it is expanding .

FOLLOWUP QUESTION:
Don't mean to argue with you on gravity effecting light speed and causing a red shift but what proof or experiment was done to prove that light's speed in a vacuum is constant? Is that just an assumption that Einstein made that no one questions? It would seem to me that if gravity can cause light to bend then gravity has an effect on light and to say it doesn't effect its speed in a vacuum may not be correct. Since the speed of light I assume can be effected if it isn't in a vacuum then gravity can effect its speed. Light coming from a long distance may feel the effects of gravity. What I am really asking is how the speed of light is constant if in a vacuum was determined?

ANSWER:
First, some background. There are good physical reasons why the speed of light must be constant. Rather than rehash answers I already have written, I refer you to two of the FAQs:

Now, you might argue that this is dependent on Einstein's principle of relativity—the laws of physics are the same in all frames of reference. To my mind, that is the foundation of the theories of special relativity and general relativity. Now, in more than 100 years since these theories were proposed, there has been not one single instance of their predictions being incorrect; the most recent, of course, is the observation of gravitational waves. So, in some sense one does not have to measure the speed of light because if it were not a universal constant these theories would not correctly predict what we observe in nature.

But, in fact, lots of measurements of c have been made. Historically, the famous Michaelson-Morley experiment tried to observe variations in the speed of light observed at different times of year as the earth moved around its orbit, the observer obviously in motion relative to the source of light; it was an extraordinarily precise experiment and had absolutely a negative result. The speed of light from stars has been measured. The relative speeds of each star and the earth are different for for every star, but the speed of the received light is always the same.

Your statement that "the speed of light I assume can be effected if it isn't in a vacuum then gravity can effect its speed" is false. Indeed, the speed of light in a material is less than c but that is because Maxwell's equations are different than in a vacuum because of the influence of the electric and magnetic charge and current densities due to the the atoms inside the material. But it is faulty logic to suggest that this means that gravity could have similar effects because there are no charge and current densities in a gravitational field in a vacuum.

Finally, if you want to learn more about how gravity bends light, you can find links to previous answers in the FAQ page.

QUESTION:
If Newton's third law is a fact then wouldn't that mean that for this universe there is an equal and opposite universe parallel to the one we live in? And one for that universe, and so on and so forth?

ANSWER:
Newton's third law does not state that for every thing there is an equal and opposite thing. It states that for every force there is an equal and opposite force. More precisely, if object A exerts a force on object B, object B exerts an equal and opposite force on object A. That statement of the law turns out to not be literally true for all forces in nature; see an earlier answer for more detail.

FOLLOWUP QUESTION:
So is the possibility of an equal/opposite paralell universe possible?

ANSWER:
Some theories include parallel universes, but I have no idea what an "equal/opposite" universe would be. It certainly has nothing to do with Newton's third law.

QUESTION:
I am in a wooden country boat in a lake with a stone tied to a rope. If I tie the free end of the rope to the boat, can I propel the boat by throwing the stone 4 or 5 meters in the water or will stone curve back under the boat? I hope I have been able to clearly state the question.

ANSWER:
If the rope is long enough and the lake is shallow enough, the rock will drop to the bottom of the lake and you will be able to pull yourself forward by pulling on the rope. When you throw the rock, the boat will recoil backwards but the boat and contents are much heavier than the rock, so the recoil speed of the boat will be much smaller than that of the rock. So, assuming that the rock ends up on the bottom forward of the boat, your scheme will work.

QUESTION:
One of my Encore students would like to know if the Flask (assuming you mean the Flash) could make DaVinci's copter work. Seems like Superman beat him in a race and Supes travels up to Mach 4.

ANSWER:
Well, all the superpower stuff is simply not physics, simply not possible. Now, the Flash's super power is to move really fast, so there is really no point in asking if he could make a helicopter work—that is not his thing! Just googling around I find that his maximum speed is the speed of light (impossible), faster than that (impossible), or infinite (impossible). I seem to remember some Superman movie where he sped many times around the earth at such a speed that he could make time run backward so he could save Lois or somebody after but before they were killed by an earthquake or something (just plain crazy) and that would sure be a lot faster than Mach 4! Well, it's all fun but you can make up just about anything you fancy which is what all the creators of these superheros do anyway!

ADDED THOUGHT:
I found the around-the-earth video!

FOLLOWUP QUESTION:
But I think the question was more along the lines of "is there any speed at which a human-shaped-figure could run to create enough lift to make the DaVinci helicopter fly?"

ANSWER:
OK, so I did a little research on the DaVinci helicopter. Apparently his design was for four men turning the screw while running around the base which is not supposed to turn; so I guess you are asking if it would fly if, propelled by the Flash, it spun fast enough. There is a lengthy discussion which you can look at here. Certainly the screw will provide a lift and I would guess that, at some rotational speed, the lift would be equal to the weight of Flash plus the copter; since, as we saw above, Flash can run any speed he likes, he could achieve this critical lift to lift off (provided that materials strong enough for it to hold together were possible). (The "human-shaped figure" has nothing to do with it, though.) There is one real problem not anticipated by DaVinci, though—because of friction, as soon as it lifted off the base would start spinning too; that is the function of the tail rotor in modern helicopters, to counter this tendency. Also, the machine has no way to control the direction of flight. So the answer is that nothing will make this thing "fly" because it could not be controlled. But probably it could be made to provide a lift force greater than the weight so that it could momentarily lift off before it was totally out of control.

QUESTION:
So my question is related to drones and was a point of much debate with some work colleagues over dinner. If we assume I have a drone that is powerful enough to rise 20ft above the ground (max), I turn it on and it hovers 20 ft above ground. Then I get a big object below it (let�s assume a large 10 square foot box). Will the drone go up to 30 feet above the ground (20 feet above the box)? So I�m interested in the answer (yes or no) but more interested in the explanation behind it also.

ANSWER:
The brief answer no. Now, the long-winded part. How high a drone can fly does not depend on the power at all. Power is the rate at which energy is given to the drone, and if you can get to 20 ft, you can subsequently get to 40, then 60, and so on. Suppose the engine can deliver the energy to the drone to lift it to 20 feet in, say, 10 seconds, then the minimum power P necessary is Wx20/10=2W where W is the weight of the drone; if the weight is 20 lb, P=20x2=40 ft⋅lb/s=54 Watts=0.073 horsepower. So, it will just keep rising.

But surely it is not going to rise all the way to the moon if it does not run out of fuel! In order for the drone of weight W is to rise, it must experience a force of at least W upward and that force comes from the whirling rotors pushing down on the air and the air pushing back up on the rotors (Newton's third law). If, at sea level, the engine is running full power, it will rise at about a constant speed. Since the air for several hundred of feet up is nearly constant density, it will continue rising; but, eventually, the air gets thinner a few thousand feet up, the upward force gets smaller, and the drone stops rising.

QUESTION:
For an object travelling at the speed of light, is time dilation absolute? i.e does time cease to exist for an object travelling at the speed of light? If so, doesn't this mean that an object travelling at the speed of light exists in all it's possible locations until the point in time that it stops travelling at the speed of light? Assuming that "possible locations" would all have to be distinct from each other to avoid self annihilation, would this then extend to explain the results of the double slit experiment? Would the lack of existence of the time dimension not also explain the effect of quantum entanglement?

ANSWER:
No object can travel with the speed of light.

FOLLOWUP QUESTION:
That's not very helpful. Substitute the word particle for the word object and try again or don't bother. I thought the point of the email address was to help enthusiastic but not expert people interested in science.

ANSWER:
I am very sorry you take such umbrage with my answer. Indeed, the point of this web site is to address questions from "enthusiastic but not expert people interested in science". However, the site does have ground rules because, for one thing, I cannot address questions which make impossible assumptions and are therefore not "science" at all; for another, it is very difficult to give good answers to questions which are really many questions (time dilation, travelling at the speed of light, self annihilation, double slit experiment, quantum entanglement) without writing a rather lengthy essay. So, if you had taken the trouble to read the site ground rules you would have found:

I no longer answer questions which are based on premises like: "...if we could travel faster than the speed of light..." or "...ignoring the fact that nothing can go faster than the speed of light...", "...if I were traveling at the speed of light..., etc.
Please submit single, concise, well-focused questions.

Now, since I have gone this far, I will address your question as best as it can be addressed. The only "particle" which can travel at the speed of light is one which has zero mass. The only particle we know which has zero mass is a photon, so it can go with the speed of light. But, the photon is a stable particle (does not decay) so it carries no clock with it and it is therefore pointless to ask how a photon perceives time—a photon does not have a "point of view".

QUESTION:
Why moon does not fall on earth due to gravity?and why it does not its centrifugal force throw it away from ?

ANSWER:
See an earlier answer. Also, the centrifugal force is a fictitious force.

QUESTION:
A football is inflated in a warm room. It is used out of the door on a cold day . what happens to the ball and why?

ANSWER:
See an earlier answer.

QUESTION:
If I were to roll a ball on a flat surface, how could I figure out how much going up the 45-degree incline of a hill would affect the ball's speed? And what would happen when it goes back down the hill with the same angle and distance? This is assuming that the ball never leaves the ground, and there is no air resistance.

ANSWER:
You also need to assume there is no rolling friction and that the ball rolls without slipping. The ball will lose its speed linarly with time until it stops and comes back down; coming down it will have the same speed at each point that it had on the way up. I will do a general calculation. The sphere has a mass m and a radius R; at the bottom of the incline it has a speed v₀; the angle of the incline is θ. At the bottom its kinetic energy is ½mv₀²+½Iω₀²where I=2mR²/7 is the moment of inertia of a solid sphere about the center of mass and ω₀=v₀/R is the initial angular velocity about the center of mass. The total energy is thus E_bottom=7mv₀²/10. At the top there is no kinetic energy but the potential energy is E_top=mgh. If there are no frictional losses, the energy at the top and bottom are equal; setting E_bottom=E_top and solving, you find that h=7v₀²/(10g). Note that this does not depend at all R, m, or θ.

If you want to know what its speed is after it has risen to some height y, the appropriate energy conservation equation is 7mv₀²/10=mgy+7mv²/10, so

v=√[v₀²-(10gy/7)]

QUESTION:
If you have 3 hot metal balls that each are 1000 degrees, and you put them all in a bucket of water, would the water be 1000 degrees, or 3000 degrees?

ANSWER:
None of the above. First of all, water can only exist (at atmospheric pressure) up to 100⁰ C, so anything above that would be impossible. But secondly, particularly since you have not specified the amount of water in the bucket nor the size and composition of the balls, you should know that you cannot predict the final equilibrium temperature. For example, suppose the "bucket" was Lake Superior and the three "balls" were bbs—surely the temperature of the lake would increase by an absolutely unmeasurably small amount. You are envisioning, I think, a situation where the water does not boil before the balls and water reach the same temperature. It would be fairly simple to calculate the final temperature, somewhere between the initial temperature of the water and 100 degrees, if you knew the amount of water and the size and composition of the balls (assuming the bucket itself absorbs no heat). If there was too little water compared to the balls for the final temperature to be less than 100⁰, it gets more complicated because the boiling water will result in steam and it takes energy (heat) to evaporate water. In that case, the final temperature of the remaining water and the balls would be 100⁰unless all the water evaporated in which case there would be no water left to specify the temperature of.

QUESTION:
Could matter be described as a single object, as a single mass?

ANSWER:
Why would you even think of such a thing? For all of recorded history philosophers and scientists have been seeking to understand the structure of matter. Even the ancient Greeks had for "elements"—fire, water, air, earth—which they hypothesized all of matter. Maybe you are thinking of an idea like string theory where it is imagined that everything is composed of vibrating strings of some sort. The problem is that string theory makes no predictions so it can be neither proven nor disproven. Who knows, maybe someday there will be a theory where we find a single root of everything, but I doubt it because of the richness of the composition of the universe.

QUESTION:
If we make a hole vertically throughout the earth then how much time will a thing take to reach bottom of earth?

ANSWER:
In order to solve this kind of question, it is necessary to assume there is no air drag on the "thing". Technically, you need to know the mass distribution over the whole volume of the earth. Usually, for problems like yours, it is assumed that the density is constant everywhere. This is actually a quite poor approximation, the density is much larger in the center than near the surface. But, for your question, let's assume that the earth density is uniform throughout. In that case, it turns out that the period of oscillation (time from when you drop it until when it gets back to you) is exactly the same as the period of a near-earth orbit satellite, 128 minutes. I do not know what you mean by "bottom of earth". If you mean "center of earth", the time is ¼ of the period, 42 minutes; and if you mean "other end of the hole", the time is ½ of the period, 84 minutes.

QUESTION:
we have studied that the rate of change of displacement is known as Velocity and Displacement is the shortest distance between two point. Than how we could say that the electron revolves around the nucleus with some velocity But there is no another point on the orbit?

ANSWER:
Well, what you have described is average velocity, not velocity. So if something like an orbiting electron is moving in a circle of radius R and it takes it a time T to go all the way around once, you can ask what is the average velocity during the time that it goes from one point on the orbit to another. For example if you ask what the average velocity is if the electron goes halfway around, it would be 2R/T. If you ask what the average velocity is for one complete revolution, it would be zero; this makes sense because over the time you watched it, it went nowhere on average!

QUESTION:
I've recently been learning about QED and have started to become curious about some of the things involving charge. In specific, I am curious, how do particles know that their neighboring particle is charged? For example, how does a up quark, know that the other up quark nearby is also positive? I once read it was because of the exchange of virtual particles but couldn't find anything about that and was hoping you could clear the confusion for me.

ANSWER:
Electromagnetism is a field theory. When you try to include quantum theory into electromagnetism (or any field, for that matter), it is called quantization of the field and that is what QED is. When a field is quantized, one of the results is the emergence of a boson (spin 0, 1, 2, etc.) which may be thought of as the "messenger" of the effects of the field. The "messenger" for the electromagnetic field is the photon. So a simple explanation is often given that "exchange of photons" between the charges, sort of like two ice skaters throwing a ball back and forth, is responsible for the repulsive force; this is a very qualitative explanation and should be considered to be a "cartoon". It is useful mainly to emphasize that photons are the force messengers. QED is a pretty mathematical theory and a cartoon cannot be expected to convey all its complexities. The photons in the field are "virtual", they pop into and out of existence due to the uncertainty principle. If a photon suddenly appears, where did its energy E come from? The presence of a photon would imply a violation of energy conservation. But, that is OK as long as this violation lasts only a time t such that Et≤ℏ.

Nature has three other fundamental forces, the strong and weak nuclear forces, and gravity. The nuclear forces also have "force messengers": gluons for the strong force and Z⁰ and W^± bosons for the weak force. The gravitational field, however, has not been successfully quantized and this remains one of the "holy grails" of physics; it has been hypothesized that a boson called the graviton will be the messenger, but this is just hypothetical at this point.

QUESTION:
if a rocket is made up of magnet with certain dimensions and launched into space,what happens? can the old satellite pieces get attracted by this rocket or not?

ANSWER:
So, I guess you are wanting to put a magnet in orbit and use it to "vacuum up" space junk? First of all, you cannot imagine the vastness of the volume where orbiting junk is; the likelihood of encountering any piece of junk being close enough to be strongly enough attracted is misiscule. Second, I expect that a lot of the junk is not ferromagnetic (iron or some kinds of steel) to be attracted to a magnet with much strength.

QUESTION:
How much energy is used to pull a 15 pound suitcase on wheels versus lifting same poundage And lifting and carrying it the same distance Note: I had an accident and can�t lift over 10 pounds. However my mini suitcase weighs 15 pounds and it is easy to roll. I can do it with one finger. Does this 15 pound suitcase On wheels constitute listing or pulling 15 pounds?

ANSWER:
You don't want to ask "how much energy" because you are told not to exert a force more than 10 lbs. You want to ask "how much force"; and you have answered that by saying that you can pull it with one finger—almost nothing, maybe a few ounces. And it makes no difference how far you pull it, because your prescription only stipulates the maximum force you are allowed to exert, not how long you exert that force. Lifting it, however, is verboten. You can "lift" it, though, by dragging it up a ramp; I reckon that the steepest ramp you could use would be about 40⁰ and the force you would have to exert would be a little less than 10 lb.

FOLLOW-UP QUESTION:
Is there a physics formula used to determine the force used in the 15-pound suitcase moving it by lifting it or rolling it on a straight plane, flat had floor as if in an airport.

ANSWER:
If you lift it straight up, you must exert a force greater than or equal to the weight W, F_lift≥W. If you pull it horizontally, you must exert a horizontal force greater than or equal to the frictional force due to the rolling of your wheels; this frictional force f is f=μW where μ is the coefficient of friction, so F_pull≥μW; for example, for your 15 lb suitcase, if μ=0.005 then f=0.075 lb so F_pull≥0.075 lb=1.2 oz. If the force you exert is not horizontal, it is a little trickier to work out; if the angle your force makes with the horizontal is θ, the force you exert is F_pull≥μW/(cosθ-μsinθ). Again taking μ=0.005 and W=15 lb, I find F_pull≥0.107 lb=1.71 oz for θ=45⁰. In all these equations, if you replace ≥ by = the suitcase will move with constant speed; with bigger forces, the suitcase will accelerate.

QUESTION:
Why is potential energy negative?

ANSWER:
Because of the way potential energy is defined, it is arbitrary to within an additive constant. This ambiguity is usually handled by choosing where the potential energy will be defined to be zero (or simply choosing the potential energy to have some value at some position. For example, you probably know that the potential energy near the surface of the earth is V=mgy; but to know what to use, you need to specify where you choose V=0. If I choose V=0 to be at the ceiling and +y is up, then the potential energy at the floor is V=-mgh, a negative value, where h is the distance from the floor to the ceiling.

QUESTION:
Why do cavities that form in a wood ﬁre seem to glow brighter than the burning wood itself? Is the temperature in such cavities hotter than the surface temperature of the exposed burning wood?

ANSWER:
Here is what I think. A point on the surface is continually emitting radiation and all of that radiation escapes, taking energy away from the surface cooling it. In the cavity, much of this emitted radiation does not make it out but is absorbed by the walls of the cavity, so less energy is lost and the temperature is higher.

QUESTION:
I accidentally magnetized the tip of a steel rod about 0.5 cm in diameter using a neodymium magnet of unknown, but relatively strong field strength. The rod is installed in a larger device so that I only have access to the tip that I magnetized. I need to demagnetize it. I tried reapplying the opposite pole of the same magnet. That didn't work. I also tried spinning the magnet near the tip of the rod using a drill, but that also didn't work. Commonly available demagnetizers seem to produce magnetic flux densities in the range of hundreds of gauss, whereas neodymium magnets apparently have densities in the range of thousands of gauss, so I doubt they would be sufficient for the purpose. What would be the easiest way to demagnetize this rod without heating it to its curie temperature? (I can only safely heat the rod to about 90 degrees C.)

ANSWER:
Applying the opposite pole will just magnetize it in the opposite direction. Spinning the magnet might work but only if you spun it but only if you spun it about an axis passing through the center and parallel to the faces (first figure); this will cause a magnetic field which oscillates in direction. (Don't have the rotation axis pointing at the rod tip.) Now, move this spinning magnet slowly away taking advantage of hysteresis. But, the classic way to demagnitize is with a coil/solenoid which has an AC current which you slowly decrease to zero. There are also static demagnitizers (see second figure) but there seems to be some question about how effective they really are. Finally, I am sure that the magnetization you caused is much weaker than magnetization of the neodymium magnet you used, so do not assume that you need a field as large as that to demagnetize the rod. If you have access to an AC power supply, I would just wrap wire around the rod tip (a lot of turns), connect it to the power supply, turn it on at very low voltage, increase it until you have the current which the power supply is rated at, and then slowly turn it back down. You may have to go through this a few times.

QUESTION:
I've been thinking about the difference in time required to cook food from a cold vs a preheated oven. Simplistically assuming that the "food" is "cooked" once it has taken in q joules, how would you: a) find a function of time that gives the energy change of the food in the static temp oven based on the oven's temperature, and constants related to the object (specific heat capacity, mass, etc) and secondly: b) the same energy change function when the oven heats up along with the object when given dT/dt for the oven and the same constants for the object to go on to then finally work out the time the food would need to cook for in both scenarios.

ANSWER:
As a person who loves both cooking and physics, I would say that your question is purely academic. The success of a particular recipe you are cooking may very well depend on how the food absorbs the heat from the oven to be the tasty creation you were aiming for. And, it is simplistic to assume that "cooked" is synonymous with an absorption of a certain amount of energy. Some foods, a beef roast for example, require that the inside ends up at a significantly lower temperature than the outside. Other foods, for example pork ribs, require a long slow cook to become both tender and juicy. Many recipes require an initial very high temperature to sear the outside to seal in juices. In my many years of cooking, I found only one case which sometimes benefitted from putting it into the oven at the same time you turn on the oven—if a loaf of bread has not risen as much as you want, and you are in a hurry, placing it in the cold oven allows the bread to rise more before a yeast-killing temperature is reached. To analytically solve this problem would be very hard because heat absorption also depends on the shape of the item being cooked; think of a chicken whose wings become done must faster than the breasts.

QUESTION:
I'm somewhat baffled. Mr. Einstein stated in his special theory that the upper-bound velocity in the universe is c^1. Yet his energy - mass equivalence equation has c^2. In his General Theory of Relativity his field equations have c^4. How can this be reconciled. Thank you.

ANSWER:
The speed of light is just some constant which has the dimensions of length divided by time (L/T). Energy is a different thing from velocity and it may be expressed as a mass times the square of a velocity (ML²/T²). You seem to think it should be mc instead of mc², but then it would have dimensions of ML/T so it would not be energy. Perhaps it is simpler to think about a length L. Suppose we have a square and each side is of length L. What is the area of the square? Clearly L²! "How can this be reconciled"? It does not have to be reconciled because a length and an area are two completely different things, just as speed and energy are two completely different things.

QUESTION:
Why does anything that moves get from point A to point B at all if (mathematically speaking) you can cut the distance in half and in half again to infinity? Mathematically you will never reach your destination but in reality you do reach it. How do you reconcile the two?

ANSWER:
This is often called "Zeno's paradox". Suppose that the speed of the arrow is 100 m/s and the target is 100 m away. Then the time it takes to go the first half way is 1/2 s, the second half time takes 1/4 s, the third half time takes 1/8 s, etc. So the total time it takes is the infinite series (1/2+1/4+1/8+1/16+1/32+�). Although this is an infinite series it has a finite sum and that sum is, guess what, 1 s. So, when viewed this way, you take an infinite number of steps but in a finite amount of time. You can read more about the history and details of Zeno's paradox on Wikepedia.

QUESTION:
I found a stack of these pie pans in a recycle bin, and I want to use them but am worried that there's something I can't see (like radioactivity) wrong with them.

ANSWER:
I believe that you have no worry whatever about radioactivity. The only way I can think of that would create radioactivity is if the pans were near a reactor and exposed to a large neutron flux. Corningware is a ceramic-glass mixture and composed entirely of light elements. There is no way it could become radioactive with a half life more than a few hours. For example, glass is mainly silicon and oxygen. If the most abundant isotope of silicon were to capture a neutron it would become a heavier but stable silicon isotope; the most abundant oxygen isotope plus a neutron would become a heavier but stable oxygen isotope. Some components could become radioactive but their half lives would be very short. For example, aluminum plus a neutron makes a radioactive aluminum isotope which decays to a stable silicon isotope with a half life of 2.24 minutes. So the point is that in the unlikely case of these pans being exposed to something which would make them radioactive, they would have done all their radiating long before you found them in the trash.

In the event that you are interested, here are the reactions I alluded to:

²⁸Si+n→²⁹Si
¹⁶O+n→¹⁷O
²⁷Al+n→²⁸Al→²⁸Si+β^-

QUESTION:
What decides which electron would revolve in at which energy shells?

ANSWER:
Actually, in a quantum system like the atom, all the electrons are indistinguishable, you cannot label them by what orbital they occupy. In fact, each orbital is a superposition of all the electrons in the atom.

QUESTION:
does hydrogen explodes or implodes when it reacts with oxygen?

ANSWER:
When hydrogen combines with oxygen heat is generated. If the oxygen and hydrogen are in a closed vessel, the pressure will increase as the temperature increases so, if the container is not strong enough, it will explode.

QUESTION:
The Sun is a Massive star of our solar system. Its gravity is so strong that it holds the eight planets, comets, asteroids and many other celestial objects in the solar system. Now my question is, how the photons escape from the surface of the sun with such a strong gravity of the Sun?

ANSWER:
One of the ways to quantify the strength of the gravity is to calculate the escape velocity, the speed which an object needs to totally escape from the sun; that speed is about 6.2x10⁵m/s and anything going that fast or faster than that at the sun's surface will will keep going and never come back. A photon has a speed of 3x10⁸ m/s. The objects lose energy as they go farther away by slowing down but light does not. It does, however lose energy; as it gets farther away its wavelength gets longer (toward red) which is called the gravitational red shift. If a photon is close to a black hole, however, it cannot escape and just eventually disappears, all its energy captured by the black hole.

QUESTION:
In regards to the earths core magnet is the north pole the positive or the negative pole?

ANSWER:
Sometimes the north pole of a magnet is denoted as being positive. The end of a compass which points to what we call the earth's north pole is the magnetic north pole of the compass pointer. But a magnetic north pole is attracted to a magnetic south pole. Therefore what we call the north pole is actually a magnetic south pole.

QUESTION:
So my dad gave me a square piece of paper and told me to get through it I've thought a few seconds and came up with an idea to make a cut in the paper diagonally and turn it inside out As you can imagine, he broke out in laughter. Can you please explain me why this answer is wrong. I really feel like an idiot right now.

ANSWER:
This is not physics. But see the figure for the trick.

QUESTION:
When dropping a strong magnet into a copper tube, it slows down as a result of eddy braking. However what happens if you increase the speed at which the magnet enters the tube, will the magnet exit the tube at different speeds each time or at a particular speed relative to the strength of the magnet, thickness of the pipe? If the velocity is increased, according to Lenz's law the opposing force should also increase, would this mean the final velocity as the magnet exits the pipe will be the same for all the trials?

ANSWER:
Let's look at what is happening. When the magnet moves through the tube it experiences a force opposite to the direction of the velocity and which increases as the velocity increases.

If you drop it, it speeds up initially but eventually it is going fast enough that the force up is equal to the weight down and it falls with a constant speed. This speed is called the terminal velocity v_t. If this happens before it gets to the bottom, it will have speed v_t when it exits.
If you throw it down with a speed less than the terminal velocity, it will speed up until it reaches v_t. If this happens before it gets to the bottom, it will have speed v_t when it exits.
If you throw it down with a speed equal to v_t. it will move the whole length of the tube at this constant speed.
If you throw it down with a speed greater than v_t, it will slow down until it reaches v_t.If this happens before it gets to the bottom, it will have speed v_t when it exits.

So, the speed it exits at the bottom depends on the length of the tube as well as strengh of the magnet, its mass, etc. Normally v_tis achieved quite quickly and you will have that speed at the bottom for all situations except for extremely large initial speeds. See a nice demonstration.

Note, though that all the discussion above is an approximation because, in theory, the magnet will never reach the terminal velocity. If you are interested, there is an excellent analysis of the physics where it is shown that the force is proportional to the velocity, F=kv where k is a constant which depends on the particular setup—strength of magnet, size and mass of magnet, geometry of the tube, susceptibility of the tube, etc. The solution of this problem is v_t=mg/k and the speed as a function of time t is v=v_t[1-exp(-kt/m)]; so, strictly speaking, v=v_t only when t=∞. Technically, the discussion above should include something like "…will have speed within 1% of v_t when…", etc.

QUESTION:
Click here to see the original question.

ANSWER:
Wow, is this really a "single, concise, well-focused question" as required by the site ground rules? I should have simply thrown this question out, but it is kind of interesting so I will edit this absurdly long question and provide an answer:

A SINGLE, CONCISE, WELL-FOCUSED QUESTION:
Please convince me that it is possible for a rocket to achieve a velocity, starting from rest, which is four times larger than the velocity of the exhaust gas (relative to the rocket).

ANSWER:
Be sure to note in the following that the analyses are in empty space, ignoring gravity.

The easiest thing to do is to assume that the mass m of the fuel burned is much less than the mass M of the payload itself. In that case, if the thrust experienced by the rocket because of the exhaust is T, then the acceleration of the rocket is a=T/M and its speed at some time t is v=at; there is no limit to how big v can be, just wait long enough.

But, that is not really a good approximation because the mass of the fuel consumed is not small compared to the mass of the payload, particularly for very large final velocities. So this now becomes a much more difficult problem. I will not work out the whole problem here because you can easily find it by googling "rocket physics"; a quite easy-to-understand derivation can be found on real-world-physics-problems.com. The final result, equation (4), in my notation is v=uln[(M+m)/M] where u is the velocity of the exhaust gas relative to the rocket. This can be rewritten as [(M+m)/M]=e^v/u. For your case, v/u=4 and e⁴≈55, so you will find that m≈54M. So it takes a lot of fuel to achieve this speed. It is often virtually impossible to achieve a very high speed because the payload includes the fuel tanks, valves, nozzles, etc. associated with the propulsion itself. Often all this excess "payload" is discarded when its fuel is exhausted and a second stage is now used to further increase the speed.

QUESTION:
Is there any practical limits from the laws of physics in terms of how tall skyscrapers we could build? Like would a ten or thirty kilometer tall one be still possible to build?

ANSWER:
This is basically the same question as "how high can a mountain be?" If you Google this you will find several ways to get an estimate. One of the best, in my opinion, is Weisskopf's estimate. Generally speaking, the order of magnitude answer is ~10 km.

QUESTION:
Despite of the absence of the gravitational force in the space, how the rockets that are launched to the mars or the moon is returned to the earth again?

ANSWER:
I do not understand your question. First, gravity is not "absent" in space; halfway between earth and Mars the gravity is pretty small but not absent. In fact, the gravity from the sun is probably bigger than that from either earth or Mars. So, to return from Mars you need a rocket to speed you up to the "escape velocity" and then you can just coast the rest of the way; when you get close to earth again, you will speed up so you will have to use a rocket again to provide the "brakes".

QUESTION:
How does a rain drop comes down to the Earth,if drag force is cancelling the weight and as it's already in rest so inertia tend the body to continue it's state in rest? What is terminal velocity?

ANSWER:
Drag force depends on the speed the raindrop has, so if it is at rest in a cloud the weight will be a downward force which will accelerate it downward. The upward drag force F is approximately proportional the the square of the velocity v: F=Cv², where C is a constant which depends on the size and shape of the drop and the density of the air. So F starts at zero and increases as the drop speeds up; eventually F will equal the weight W and that will happen when v=√(W/C). This speed is called the terminal velocity and the drop will stop speeding up and fall with this constant speed.

QUESTION:
If a parachutist jumped off a building 10000 feet high and a parachutist jumped out of a plane at 10000 feet travelling at some speed. Let�s say 125 KTS, given exact same fundamentals such as air density, pressure, temp body mass Everything right....woulda the two parachutist hit the ground at the same time?

ANSWER:
If there were no air, they would indeed land simultaneously. But there is air and so both parachutist would eventually end up falling straight down with a constant speed (terminal velocity). For something like a parachute the terminal velocity is achieved quite quickly, so we have a situation where two objects move over most of their paths with equal speeds. Then the one who drops straight down will land first because his path is shorter.

QUESTION:
If I throw a baseball at a glass window, and the glass does not break, then I could understand that the force exerted by the window onto the ball is equal to the force of the ball on the window. However, if the glass breaks when the ball hits the window, that should mean that the overwhelming force of the ball exerted on the window is greater than the force of the window on the ball, hence shattering the glass. Now, isn't it disobeying the Newtons's third law of motion?

ANSWER:
Let's look at the forces involved here. The ball which does not break the window bounces back but the ball which did break the window continues but with a slower speed. From this we can conclude that during the time when the ball was in contact with the window (or pieces thereof), the non-breaking ball experienced a larger force. Newton's third law is not violated because, at any instant, the force which the window experiences due to its contact with the ball is equal and opposite the force the ball experiences due to its contact with the window.

QUESTION:
OK, I hear PHYSICISTS say that black holes are black because photons of light can't escape their gravity which has an impossible escape velocity greater than c (i.e. the "event horizon") however photons if I remember correctly are MASSLESS, how is gravity still able to interact with photons despite them having NO MASS?

ANSWER:
One of the most famous features of the theory of general relativity, our best theory of gravity, is that gravity does have an effect on light. For example, during a solar eclipse light from a star actually behind the sun can be observed because light which we would not have seen otherwise is bent by the gravity of the sun and observable. This was one of the first verifications of general relativity. For a black hole, if a mass is inside the Schwarzschild radius and moving away, it slows down until it turns around and falls back, no matter how fast it is moving. For a photon moving away from the black hole, it cannot slow down because it always moves with speed c; however, it still loses energy by being shifted to larger and larger wavelengths until it finally loses all its energy and disappears. The energy it had now contributes to the mass of the black hole.

QUESTION:
What is the size of a photon. If we arrange ten photomultiplier tubes around a source, and that source emits one photon, how many of the photomultiplier tubes register the emission? More specifically, is it only one? Or do they all? Or some number between one and ten?

ANSWER:
The question of size is hard to answer because there is not really one answer. However, the question about how many detectors the photon interacts with is not hard—the answer is one. Otherwise the detectors would interact with part of a photon which is not possible. It is possible for there to be a secondary interaction with another detector; for example, if the photon Compton scattered from one detector, the scattered photon (of different energy from the original) could interact with another detector.

QUESTION:
my question has foundations on two things: 1.) according to relativity , things in motion shrink down relative to some else that is not in motion. 2.) for every body ,there is a radius after or on which the body will become so dense that even light wont escape it, aka become a black hole. that radius , called as "SCHWARZSCHILD RADIUS". so CAN ANYONE GO THAT FAST , so RELATIVELY they shrink lower or equal to this radius and become a black hole ? is this possible? if yes , is gravity relative and not dependent on mass? if no, why isn't special relativity compatible in this situation?

ANSWER:
Whether something will become a black hole is determined by its size and mass in its own rest frame.

QUESTION:
I wonder... Velocity is all relative, but you can't go faster than 300Mm/s. What is that relative to? Couldn't you use that to test whether you are really moving or not? If you have two objects moving in opposite directions at 150Mm/s, to someone stationary, they are both moving at 150Mm/s, but to each of the objects, the other one is moving at the speed of light!

ANSWER:
Actually, velocity is not relative in the sense that you are trying to apply. You can tell that it is not relative because at least one velocity, the speed of light, is the same no matter what the motion of the observer is. One of your speeding objects sees the other as having a velocity different than an observer on the ground would see, but not 300 Mm/s. The velocity it sees is (150+150)/(1+(150/300)²)=240 Mm/s. See an earlier answer.

QUESTION:
Can you please tell my why physicists ever enter a value other than 0 for speed or velocity? Surely the act of measurement renders the object under observation immobile and in any single moment movement is arrested? Recognition of this would solve a number of physics paradoxes; the twin paradox, the andromeda paradox et al.

ANSWER:
Observing where a particle is at a particular time does not "render the object…immobile". You are a photographer, so surely you know this; if your shutter speed is too slow, you will get a blurry photo because it moves some distance while the shutter is open. The velocity of an object is the time rate of change of position, a quantity entirely different from the position.

QUESTION:
when i put up a swing and my bolts container says do not exceed 225 lbs is that for the whole thing or each bolt then add up the weight ex. 2 bolts in beam box says 225lb max, does it become 450lbs with 2 bolts and 675lbs with 3 and so on?

ANSWER:
To answer definitively, I would need to know exactly where the bolts were placed and how the load was being supported. If the bolts are the only parts that could fail under load and the bolts were placed such that all held up an equal share of the load, then the answer is that N bolts could hold up 225⋅N lb.

QUESTION:
Winds are blowing that can lift 90 lb objects. My 2 friends and I would like to go out side to play. Our weights are 85 lbs, 79 lbs, & 83 lbs. Individualy the wind would toss us around, if we held hands, would our combined weight keep us on the ground?

ANSWER:
The force which a wind exerts on something is determined mainly by three things—the speed of the wind, the mass of the object, and the geometry of the object. So the first sentence of your question is not really informative: if the 90 lb object was a giant sail, even a slight breeze would "toss it around" but if it were a 90 pint jug of water, it would take a hurricane to move it. Since the second part of the question talks about people, we can assume the wind is such that if one person stands upright facing the wind, the wind will blow her away if she has a weight less than 90 lb. So, if 3 people stand, all facing the wind and holding hands, they will be collectively blown away. But if the three stand hugging each other, their total area exposed to the wind will be probably just a little larger than the area of one of them alone and, since their total mass is much greater than 90 lb, they would not be blown away.

QUESTION:
When putting a golf ball I feel there is a point on it's roll where the ball "dies" or loses its momentum and rolls out. Would this point be visible on a graph of the balls deceleration or measurable in some way?

ANSWER:
This depends a lot on the properties of the green itself—its slope, how the slope varies with position, how fast the green is (how dry it is, the type of grass and its length, etc.) I will do the simplest case, motion on a level green where the ball is rolling and not slipping, as an example; I think this will provide the answer to your question. Motion if the green is not level is considerably more complicated and I will not do that here. It may be shown that the force f on a ball of mass m is of rolling friction is given by f=N(5/7)ρ_gwhere N is the normal force (the weight mg on level ground where g=9.8 m/s² is the acceleration due to gravity) and ρ_gis a dimensionless parameter explained in the web page referred to above; I would call the coefficient of rolling friction for the golf ball on the green μ_R=(5/7)ρ_g. Typical values of ρ_gare between 0.065 (fast) and 0.196 (slow); I will use an average value of 0.13 for my calculations, so μ_R=0.093. So, you see, the "deceleration" is constant and therefore its graph would not be informative. More interesting would be graphs of position and velocity.

So here is the analysis for a perfectly level green. Let the initial velocity be v₀, N=mg, a=f/m=μ_Rg. Then the equations of motion become
- v=v₀-0.91t which implies that v=0 at t=v₀/0.91, and
- x=v₀t-0.455t²which implies x=0.55v₀²when v=0.
- You could also write an expression for the velocity as a function of position, v=√(v₀²-1.8x)
- Here are some plots for three initial velocities, v₀=1 m/s, 1.5 m/s, 2 m/s.

The first graph shows that, regardless of the initial velocity, the rate at which the ball slows is the same for all; the faster you start, the longer it takes to stop.
The second graph shows the the distance traveled increases quadratically (proportional to t²) with initial speed.
The third graph shows more clearly the distances traveled by each ball: about 0.5 m, 1.25 m, and 2.25 m.
Finally, keep in mind that I have assumed that the ball rolls the whole way and that may not be the case always. Particularly if the ball is struck hard, it is likely to slide for a bit before it stops sliding and purely rolls.

If you want to do similar calculations for going straight up or down a sloped green, then the normal force would be reduced by an amount proportional to the cosine of the slope angle and there would be an additional force due to the weight pointing down the incline proportional to the sine of the slope angle.
If you want to examine the case of a putted ball on an incline and not straight up or down it becomes a two dimensional problem and much more difficult.
If you are really interested, there are lots of more detailed web sites addressing the physics of putting.

QUESTION:
Friction is independent of area but all factors which cause friction don't adhere to this statement i.e molecular bonds and hence adhesive force are directly propotional to no. Of molecules which further to surface molecular density� area of surface. Deformation of surface also favour area factor. So why it doesn't depend on area. Either the cause is wrong or mathematical equation ( which cannot be wrong) showing f= k�mg. Whats your opinion sir?

ANSWER:
The force of friction is proportional only to the force and the proportionality depends on the surfaces in contact, F=μN. Is that a law of physics? The answer is certainly not, it is an expression of the results of experiments which have measured the frictional force over many materials and situations. It is an approximation, not an absolute law, but for normal conditions it does a quite good job. Your arguments about molecular bonds or number of molecules is wrong because what is going to matter is the force per molecule and a larger area has more molecules but a proportionally decreased force per unit area. The finer details of more microscopic descriptions of friction, though, can get very complicated and there are whole institutes devoted to studying friction. An earlier answer discusses this issue for automobile tires.

QUESTION:
I just started exercising at a local high school. I found a rectangular, rubber "field weight" that lift while I walk on the beach. It is 9"x14". It's 7 lbs. I'm wondering how much mass I'm actually lifting when I hold it extended out horizontally and lift over my head. My arm is appx 26".

ANSWER:
The mass (and weight) does not change. No matter what the orientation of your arm, the force you need to apply with your hand to hold it steady is 7 lb. But it feels harder when horizontal than vertical, doesn't it? The reason is that your muscles must exert a torque about your shoulder and that torque is greatest when your arm is horizontal. That torque is 26x7=182 in⋅lb when horizontal and zero when vertical. When vertical the muscles which provide the torque are relaxed but other muscles must provide the 7 lb force. If your arm makes an angle θ with the horizontal the required torque is 182cosθ in⋅lb.

QUESTION:
Is it true that Energy can neither be created or distroyed?

ANSWER:
This is conservation of energy. It is only true for isolated systems. If an external force does work on a system, energy in the system changes.

QUESTION:
Which golf ball each struck at 100 mph swing speed would fly farther, 46.0 grams, 56.0 grams, or 65.0 grams?

ANSWER:
Simply stating the club speed does not adequately define the problem. What really matters is the initial speed of the balls. What matters is the time the ball is in contact with the club and the average force the club exerts on the ball during that time. I have worked out two possible scenarios below, all balls with the same initial speed and all balls with the same linear momentum.

My first approximation would be to assume that all start off at the same speed and angle relative to horizontal. The speed of a ball off the tee is about 180 mph=80 m/s. In that case, they would all go the same distance if there were no air drag. Since you describe them as golf balls, I would assume that they all have the same radius, about 21 mm. Since the air drag would be proportional to the cross sectional area of the ball, they would all experience the same drag force at a given velocity; but the force would have a bigger effect on the lighter balls, so the 65 gram would go farthest. The three graphs below show a calculations done numerically on a website I found; the distances are about 270 m, 300 m, and 320 m for a launch angle of 45⁰. The green curves are the trajectories with no air drag, the black include drag. See below for more details about how air drag is calculated. (I note that the 56 and 65 gram balls are illegal, the maximum allowable mass being about 46 grams.)

m=46 grams

m=56 grams

m=65 grams

My second approximation is that all three would experience the same average impulse from the club; impulse is I=Ft where F is the average force and t is the time of contact with the club. The impulse is equal to the change in the linear momentum mv, so all three would start off with the same linear momentum. This is a more likely scenario assuming identical strikings by the club because a heavier ball has more inertia and would leave the club with a smaller speed. So, if the 46 gram ball had an initial velocity of 80 m/s, the 56 and 65 gram balls would have speeds of 66 m/s and 57 m/s. With no air drag, the lighter ball would go the farthest because it started out the fastest. But air drag would take some or all of that advantage away, so we need to estimate the drag. For a smooth sphere, the drag force can be approximated (only in SI units) as ¼Av²=¼π(0.021)²v²=3.5x10^-4v^² N. But a golf ball is not smooth—it has dimples to reduce drag to about half of this value, about 1.7x10^-4v² N. The corresponding accelerations at launch would be a₄₆=F/m₄₆=24 m/s², a₅₆=13 m/s², a₆₅=8.5 m/s². The light ball loses speed, at the start, about twice as fast as the middle ball and about 3 times faster than the heavy ball. So which is more important, the rate at which the speed is removed or the speed when it starts out? The three diagrams below show the calculated trajectories.

m=46 grams, v₀=80 m/s

m=56 grams, v₀=66 m/s

m=65 grams, v₀=57 m/s

So you see that the best ball in my second approximation is the lightest one, the distances being about 270 m, 240 m, and 210 m for a launch angle of 45⁰.

QUESTION:
I was wondering if you could tell me how Einstein came up with the equation E=mc²

ANSWER:
You have to know a little about physics to understand this answer. I will mostly be qualitative, but you need to know what force, acceleration, velocity, time and energy are. Let's first talk about classical physics first. Newton's first law may be stated that the force F experienced by a particle is equal to the time rate of change of linear momentum p=mv, where m is the mass and v is the velocity, so

F=m(Δv/Δt)=m(Δv/Δx)(Δx/Δt)

assuming m is constant. Here Δv is the change in the velocity over the time Δt and Δx is the distance traveled in that time. You can rewrite this FΔx=mvΔv=ΔW. ΔW is called the amount of work done if F acts over the small distance Δx. The energy which a particle acquires when work is done on it is called kinetic energy K. Now, if m is at rest when t=0 and F is applied, the work done and hence the kinetic energy may be shown to be (using calculus) K=½mv².

It turns out that when the velocity of a particle is very large, comparable to the speed of light c, the classical definition of linear momentum is wrong. Instead,

p=mv/√[1-v²/c²)]=γmv.

Now proceed just as we did in the classical case but now we must solve the modified equation, γmvΔv=ΔW; this is a little trickier because γ depends on v. The final result is that the kinetic energy is K=γmc²-mc²; although this does not look at all like ½mv², it is pretty easy to show that for very small v (compared to c), this is almost exactly equal to the classical value. You can read the final answer as the kinetic energy is something (γmc²) minus a constant (mc²); this is now interpreted as something=the total energy E=γmc²and the constant=energy the object has by virtue of its mass E₀=mc². This makes sense because if v=0, γ=1 and so E₀=E=mc². So, just to be sure that you understand the iconic equation, it is the rest energy which equals mc², not the total energy which is γmc².

QUESTION:
Does it require more energy to accelerate an object from 201 kmh to 202 kmh than it does to accelerate the same car from 1 kmh to 2 kmh, neglecting aero drag and all other friction.

I say this because the formula for Kinetic Energy seems to suggest it DOES take more energy. Since KE goes up by velocity SQUARED, the different between the "V" components of the formula will be much greater at higher speeds than lower speeds. I have an Excel file illustrating this if that is easier.

KE=.5MV^2
Mass=4
KE1 = .5*4*1^2
KE2 = .5*4*2^2
KE3 = .5*4*201^2
KE4 = .5*4*202^2
(KE2-KE1)>(KE4-KE3).

Since conservation of energy means whatever KE is "taken out" of the system was at some point put INTO the system, I would imagine it takes more energy to accelerate an object from 201 to 202 kmh as it would from 1 to 2 kmh, right? I thought of this because a supercar maker would need more energy to get to 202 kmh than it would to get to 2 kmh even neglecting things like rotational intertia, aero friction and other friction. I find this interesting, something a supercar maker might not even consider much but maybe I am missing something.

ANSWER:
I have answered this question (although with different numbers). You should be able to read this and understand what you are "missing".

QUESTION:
As a particle moves closer and closer to the velocity of light, its mass grows continuously. Would this particle eventually gravitationally attract the whole universe leading to a universal collapse?

ANSWER:
First of all, have a look at an earlier answer to get some perspective on how small the mass of an accelerated proton is even if going 99.9999999% the speed of light. Now, keep in mind that the gravitational force falls off like the square of the distance away from the source; there are black holes out there which have masses billions of times the mass of the sun, and do you feel any force which would have us collapsing into them? And the energy being expended accelerating your particle will require the loss of mass from your energy source. The ultimate fate of the universe depends on all the conditions which exist right now—it will either keep expanding forever or will reverse and fall back into itself, depending on these conditions. A lot will depend also on dark matter and dark energy, both of which are poorly understood at present.

QUESTION:
The magnetic quantum number means that the electron generates a magnetic field?

ANSWER:
The magnetic quantum number represents the z-component of an angular momentum of a particle. The electron can have two different kinds of angular momenta, orbital and spin. The orbital angular momentum results, simplistically, from the electron orbiting around the nucleus; of course, this orbit is a tiny current loop and current loops have magnetic fields. It turns out that the electron also has an intrinsic angular momentum as if it were spinning on its own axis; since it is charged and a spinning charge is an electric current, it is not surprising that a free electron has a magnetic dipole moment. Any moving charge distribution creates a magnetic field. Historically, angular momenta of atoms and nuclei were most conveniently observed by measuring the resulting magnetic fields and that is how they acquired the name "magnetic"; a better name would probably be projection quantum number.

QUESTION:
I have a three floor home. Each floor has its own air conditioning unit and thermostat. I am wondering how to set the temperature on each level to maximize cooling and minimize AC use? Comment: what I have done is to set the top at 75, middle level at 73, lower at 71. My thought is since hot air rises, I will help the upper units, but I have no evidence that this model is correct. And my son suggested the opposite. Since cool air sinks, he suggested setting the top at 71, mid 73, bottom at 75. Furthermore, the way the house is designed, the air from the third floor can�t really work it�s way down to the floors below. And there is a wide open space starting at the bottom level where a column of hot air can rise to both upper levels. I wonder if an answer exists or if it really depends upon the exact layout of the house?

ANSWER:
I am afraid that my answer is more common sense than physics. What does a thermostat do? It turns the AC on or off to maintain the temperature you want to be comfortable. Set each thermostat to the comfort level you want on that floor of the house. It will automatically take care of whatever causes the temperature to change, in particular any influences from other floors. To save energy, set all as high as possible. If it were my house, I would set all at 76 or 78.

QUESTION:
When a glass falls will the impulse be greater if it lands on plush carpet than if it lands on a hard floor?

ANSWER:
The impulse is the change in linear momentum, and in both cases the momentum the glass hits the floor with momentum mv and ends with zero momentum, both the same change. But, the result will probably be that the glass breaks on the hard floor and not on the carpet. The reason is that, since impulse is force times the time the force lasts, the force the floor on the glass is impulse/time, much bigger for the hard floor since the time is much shorter.

QUESTION:
Newtons gravitation creates a dependency on the distance for the FORCE of gravity (mass)(acceleration). Why is it that an object dropped at 200K feet elevation measured in time for a total of a 50 foot interval, will take the same amount of time if dropped at 200 feet elevation for that same 50 foot interval? I am assuming (mass)(acceleration) means just that in the equation when referring to the "Force" of gravity.

ANSWER:
Where did you get the idea that the time over a 50 foot fall does not depend on altitude? The acceleration of gravity gets smaller as you get farther away from the surface of the earth where g=32 f/s². The acceleration of a dropping object a distance h above the earth is g(h)=MG/(R+h)² where R≈2x10⁷ ft is the radius of the earth, M is the mass of the earth, and G is the universal gravitational constant. Since g(h=0)=32 f/s², we can write that g(h)=32/(1+(h/R))² and can apply this to your question: if h=200 ft, g(200)=32/(1+(2x10²/2x10⁷))²=31.999 ft/s² and if h=200,000 ft=2x10⁵ ft, g(200)=32/(1+(2x10⁵/2x10⁷))²=31.369 ft/s². So it will take slightly longer to fall the 50 ft at the higher altitude, but not a big difference because even 200,000 ft is much smaller than the radius of the earth. The bottom line is that the FORCE of gravity is not the same at different altitudes. (I have neglected air drag.)

QUESTION:
If I understand the Higgs field correctly in that it provides mass to otherwise massless particles. Does it not follow the universal speed limit C would be better defined in relation to the strength of the Higgs field in any specific region of spacetime?

ANSWER:
First of all, the speed of light is a universal physical constant; something which is a universal constant does not have a definition, c is simply a speed which must be measured. I would have no idea how to relate the Higgs field to the speed of light.

QUESTION:
I was wondering if a cotton ball going at 1000 mph will kill you? If not , then at what speed will the cotton ball kill you?

ANSWER:
See a recent similar question. Since you specify "cotton ball", I am thinking the kind you buy a pack of for first aid type uses; googling, I find that the mass of a cotton ball is about 0.5 grams (g). 1000 mph is about 450 m/s. So if this cotton ball collides inelastically (meaning sticks to) with a person of mass 100 kg (about 220 lb) the recoil speed can be calculated from momentum conservation: 450x0.5x10^-3=(100+0.5x10^-3)v or v≈0.0023 m/s=2.3 mm/s, almost at rest. But what was the force? To get that we need to estimate the time it takes for the collision to take place. Suppose that the cotton ball compresses from a typical thickness of 1 in≈2.5 cm=0.025 m to a thickness of 1 mm=0.001 m. I estimate that the time to stop is about 10^-4 s which would correspond to an average force over the collision time of about 22.5 N≈5 lb; this would certainly not kill you or even break any ribs.

I think I will not flounder around trying to estimate the speed necessary to kill you. One reason is that air drag has a huge effect on a speedy cotton ball. If you could get a cotton ball going 1000 mph, it would almost immediately lose its high speed. I figure that the air drag force on the cotton ball is about 250,000 times its weight! As the graph shows, the speed of the cotton ball will fall from 450 m/s to 4 m/s in one second! This is easy to understand: how far do you think you could throw a cotton ball? Even a big league pitcher who could give it a 100 mph speed when he released it would be unable to throw it anywhere near the distance to the plate. (If you are interested in how I calculated this, I estimated the air drag force to be ¼Av² where A is the cross sectional area of the cotton ball, roughly π*0.0125², v is the speed, and the "exact method" given at the end of an earlier answer has been used.)

QUESTION:
Hypothetical question. If it were possible to create a copper tube going through the centre of the earth (I.e. North to South). Then drop a magnet through the tube; would the magnet keep falling between the opposite poles and generate infinite electricity?

ANSWER:
When a magnet moves through a conducting tube, it experiences a force due to Lenz's law; this force increases in magnitude as the velocity increases and points in the direction opposite the direction of the velocity. So if you have a vertical tube and drop a magnet in it, the magnet will start speeding up but the upward force will get larger and larger as it speeds up until the upward force is equal to the downward force of the magnet's weight. Once this happens the magnet falls with constant speed, called the terminal velocity; it is very similar to the motion of an object falling through air and experiencing drag. See a very good demonstration on YouTube. So your magnet will quickly be falling at some small speed. But, as it gets farther and farther toward the center, the gravitational force gets smaller and so the terminal velocity gets smaller and so the magnet will slow down more. So all the way to the center of the earth where the gravitational force is zero, the magnet gets slower and slower. Eventually it will end up at rest at the center of the earth. Even if it kept going all the way to the other pole, the "electricity" which its motion generates is in the form of currents running around the circumference of the tube and therefore not accessible to you.

QUESTION:
Hey I'm doing an Extended Response Task which requires me to design a rollercoaster. The first bit of it is to find an appropriate motor to pull the cart up the incline to the top of the hill. My cart was 2588kg and the incline was 40 degrees. It started from rest and accelerated to 3m/s on a flat plane then enters the incilne where is remains at a constant velocity. I tried using: KEi+GPEi+Wmotor=KEf+GPEf+Wfriction+Wdrag But when I converted the work to horsepower, it was 76hp. This can't be true. What am I doing wrong?

ANSWER:
For starters, hp is a unit of power, not work. Power=work per unit time. Secondly, surely I need to know how high it is to the top of the incline. Third, if you calculate the change in total energy you are calculating the total work done on the cart, which is the work done by the motor (positive)+friction (negative)+drag (negative). Unless you have a way of estimating or measuring the work done by the dissipative forces, you cannot know the work done by the motor. Otherwise, if you give me the height I can calculate the net work done on the cart; also, since I know its speed I can calculate the time to the top so I can get the power (which will include the power due the dissipative forces).

RESPONSE:
Hey, the height is 50m

ANSWER:
Keep in mind that the period of acceleration is not really relevant. The problem of interest begins with the cart at the bottom of the incline with a speed of 3 m/s. Also, the calculations I do below for the frictional forces are just rough approximations the main importance of which is to show that they are more or less negligible for your problem.

The kinetic energy up to incline is unchanged, so the change of total energy is the change in potential energy which is mgh=2558x9.8x50=1.25x10⁶ J. This is the total work done by the motor, rolling friction, and air drag (keeping in mind that work done by the latter two is negative). Now, since the length of the incline is s=50/sin40⁰=77.8 m, the time it takes to get there is 77.8/3=25.9 s so the net power is P=1.25x10⁶/25.9=4.83x10⁴ W=64.8 hp (which raises the question why you thought 76 hp was so off what you expected).

Now we should address the question of how important a role dissipative forces are since you are particularly interested in the power requirements of the motor. If there were no friction at all, the force required from the motor to pull the cart at constant speed up the incline is F=2558x9.8xsin40⁰=16,100 N. Air drag can be approximated only in SI units) by f_drag=¼Av² where A is the area presented to the onrushing wind and v is the velocity; so I imagine a cart 3 m wide and 3 m high with A=9 m². Therefore, f_drag=¼x9x3²=20.3 N, basically negligible compared to F. I googled and found that the coefficient of rolling friction for a roller coaster cart is about μ=0.01, so f_friction=μN=μmgcos40⁰=192 N; this is still much smaller (about 1.1%) than F, although you still might want to make a correction. To correct for dissipative forces, add them to F to get the required force from the motor: F_motor=16,100+20.3+192=16,300 N. So the work done by the motor is 16,300x77.8=1.27x10⁶ J and the power of the motor is P_motor=1.27x10⁶/25.9=49,000 W=65.7 hp.

QUESTION:
If one wheel is rated for, say, 300 lb, how much could a pair spaced three feet apart hold? Want to build a hand cart.

ANSWER:
If the load is evenly distributed between the two wheels, that is, the center of gravity is halfway between the wheels (d₁=d₂), each wheel would support up to 300 lbs and the load could be as large as 600 lb. However, if d₁≠d₂, one wheel would support more than half the load. The forces which the ground exerts on the wheels are N₁and N₂; the weight is W. The two equations which describe the equilibrium are W=N₁+N₂ and N₁d₁=N₂d₂. If you solve these you find N₁=W/(1+d₁/d₂) and N₂=W/(1+d₂/d₁). So, you see, if d₂=d₁, each wheel holds up half the total weight. But, in your case, if W=600 lb and d₁=1 ft and d₂=2 ft, N₁=400 lb and N₂=200 and the left wheel would fail. (Note that how far apart they are makes no difference.)

QUESTION:
I am told that atmospheric air pressure depends on altitude and the height of the column of air above me. But I'm also told that the vapor pressure around me does not depend on altitude and is solely a function of temperature. But the atmospheric pressure is also a combination of the partial pressures of all the component gases (oxygen, nitrogen, water vapor, etc.). Why is water vapor special in this instance? Why don't we consider the partial pressure of oxygen to be solely a function of temperature as well? Why does the partial pressure of oxygen depend on the atmospheric pressure but the pressure of water vapor does not? How would I go about calculating the partial vapor pressures of all component gases independent of altitude, or can this be done? If not, again, why is water vapor unique?

ANSWER:
I do not know. (I was dodging the question! Thermodynamics was never my strong suit and I really did not want to research it! On the other hand, if someone is persistent I will usually pursue the question further as long as it is not an "Off the Wall Hall of Fame" candidate!)

QUESTION:
Ok. Is my question legitimate in your mind (i.e. phrased properly)? Is there another resource to whom you think I should pose this question? Do plan to pursue this in any capacity going forward?

ANSWER:
OK, you win! The reason that water vapor is unique is that, at the temperatures and pressures we are talking about, water vapor is able to make a phase transition to either water or ice. You do not need to worry about O₂, for example, turning into a solid or liquid at usual atmospheric temperatures or pressures. You can probably find everything you need in the Wikepedia article on vapor pressure of water and links therein. (It is not simple!)

QUESTION:
Is the following analysis correct? For a photon of light traveling at the speed of light, time comes to a stop. Thus, in its own reference frame, the photon does not experience time. Therefore, from it's reference frame, the photon does not exist because a particle cannot exist for zero time.

ANSWER:
I have answered essentially this question several times. The crux is that a photon does not have a "point of view" and cannot carry a clock with it; therefore it is not meaningful to ask the rate at which a clock in that frame runs because you can never get a clock into that frame. See an earlier answer.

Here is a history of questions and answers processed by "Ask the Physicist!". If you like my answer, please consider making a donation to help support this service.

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

QUESTION:
do uniformly rotating body just like fan have linear accelration? in this regard every rotating body has linear accelration so a rotating body cant be in complete equilibrium however it can be in rotational equilibrium because linear or tangential velocity during circular motion changes at every instant however if a body is moving with uniform linear velocity it satisfies both conditions of equilibrium

ANSWER:
This question addresses whether a fan which is running is in equilibrium. It is true that if you take a point on one of the blades to focus your attention on, that point is not in equilibrium. Or, if you focus on the center of mass (COM) of one of the blades, that blade is not in equilibrium. But, if you sum the forces on the three COMs, they will add to zero. Therefore the net force on the fan from the rotating blades is zero. When solving an equilibrium problem for a system, the translational equation is that the sum of all forces must equal zero. For the fan pictured, the only forces on the fan are its own weight (acting at the COM of the whole fan) and the normal force from the table. All internal forces add to zero, do the normal force must have the same magnitude as the weight and act in the opposite direction. The normal force must also act through the COM in order to satisfy the second condition of equilibrium, all torques must add to zero.

QUESTION:
Light speed is constant in all inertial frames.This constant speed is what allows us to see. However, what happens to light speed in non-inertial (accelerating) frames? Would we be able to see in that case?

ANSWER:
What makes you think that the constancy of the speed of light has anything to do our ability to see? In noninertial frames, the speed of light is still a constant but the velocity may not be since light is bent in a gravitational field. This is predicted by the theory of general relativity and has been observed many times.

QUESTION:
Is it possible the rotation of the earth is what causes gravity?

ANSWER:
Certainly not! In fact, the rotation of the earth causes a centrifugal force which is trying to throw you off so it is more like an antigravity. But that depends on your latitude and there is no centrifugal force at the poles. Even at the equator where the centrifugal force is biggest, it is about 0.35% the gravitational force (your weight). The origin of gravity is well described by the theory of general relativity.

QUESTION:
why does a confined gas exerts pressure ?

ANSWER:
Because the molecules of the gas are moving very fast and when a molecule hits a wall of the container it exerts a force on the wall.

QUESTION:
What is the difference between linear and non linear sound?

ANSWER:
The mathematical equation for linear waves is aptly called the wave equation. This equation is a linear equation which means that if f₁ and f₂ are both solutions to the wave equation, so is f₁+f₂. The hallmark of the wave equation is the superposition principle where if more than one wave is traveling through a medium, the net wave is simply the sum of all the waves. Sound waves are normally linear, but if the amplitude of the waves is very large, the equation which describes them is not linear and therefore the superposition principle does not hold.

QUESTION:
Not a homework question. Just something I've been wondering about on and off for years. Question: (If) someone could travel at light speed or close to light speed and they traveled 10 light years away from earth and then took another 10 light years to come back to earth. For the people on earth how much time has passed? Is there a formula?

ANSWER:
First of all, nothing can travel the speed of light except light. So, as seen by someone on earth, the traveler will take 20 years for the round trip. For the traveler, however, the distance will shrink by a factor of √(1-β²) where β is the ratio of the speed divided by the speed of light, L'=L√(1-β²). For example, at 99% the speed of light, L'=10√(1-0.99²)=1.41 ly. So the time elapsed on the traveler's clock will be 2x1.41/0.99=2.85 years. You should read the earlier answer about the twin paradox.

QUESTION:
This is not really a question about my homework, but I have problems with the uncertainty analysis, which I need for my lab report. I've got 3 results for an oscillation period. Now I have to get an average value of those 3 results. All of these results have their own uncertainty (the same for all of them). How do I get the average with its uncertainty now without ignoring the uncertainty of every measurement. It would be great, if you could help me.

ANSWER:
If you make N measurements of x the uncertainty of each is Δx, then the uncertainty of the average, Δx_avg, is Δx_avg=Δx/√N.

QUESTION:
Hey I have a report due on in 3days and its an experimental investigation on the double slit experiment. The laser I used was 650+-10nm. If the experimental values I found were 656+-33nm and an extreme one such as 575+-150nm (this was from the max and min slope uncertainty) how would you calculate the percentage error? I dont understand becuase we dont know the *exact value of the wavelength.

ANSWER:
Independent sources of error should be added in quadrature. For example, if u₁=33 nm and u₂=10 nm, u=√(33²+10²)=34.5 nm.

QUESTION:
Why does the candle flame always burn in upward direction?

ANSWER:
Because "hot air rises". The hot gasses in and around the flame expand to a smaller density than the density of the cooler air around it. It is then like a hot air balloon which has a upward buoyant force on it. This would not happen if there were no gravity which is basically the reason there is a buoyant force.

QUESTION:
Are time and temperature proportional? If so, how?

ANSWER:
First, let's define "proportional". It means, in your case, that the temperature is a linear function of time; this means that the ratio of temperature and time is always a constant. Therefore, for example, if the temperature at 5:00 is 10⁰ and at 6:00 is 20⁰, then at 7:00 the temperature will be 30⁰ if temperature is proportional to time. Is the temperature always proportional to the time? Obviously not. For example, the temperature outdoors is sometimes increasing, sometimes decreasing, sometimes remaining constant. If you place a cold object in a warm room, the object will warm up but not linearly (see a recent answer as an example). Is it ever a linear dependence? Yes, you could arrange for it to be proportional. For example, if you add heat at a constant rate to an object which is well insulated so that no heat would escape, the temperature would increase at a constant rate, i.e. linearly.

QUESTION:
I race mountain bikes and currently have 27.5 inch diameter wheels on my bike. Currently many people are switching to 29 inch wheels which they claim are easier to maintain rotation when pedalling If the two different size wheels weigh the same, and all other factors are equal, how much easier would it be to keep the rotation going thru peddaling on the 29 inch wheel than the 29.5 inch wheel? Show me what the difference in force needed to maintain the same rpm on a 29 vs a 27.5 with all things about the wheels and tires being equal (EVERYTHING) weight, tire pressure, temperature of air, same tires being used, same surface etc. Example: if two wheels, one 27.5 and the other 29 inch, with same tires weight say, 800 grams, how much more force does it take to maintain rotation on the smaller wheel once both tires are rolling at the same rpm? If possible, could you give me an answer in percentage, as that would greatly help me in determining the truth about how much an advantage the larger wheels are for racing and if the price is worth the expense.

ANSWER:
You are really asking the wrong question, or at least one of the least important questions in comparing the efficacy of the different wheels. You are asking the force required to maintain the same rotational speed which would correspond to constant speed of the whole bike+rider. The figure shows all the forces on the bike+rider: the total mass m times the acceleration due to gravity (g) is the weight and acts at the center of gravity, the normal forces (N_i) on each of the wheels from the ground, the forces (f_i) due to the rolling friction of the wheels on the ground, and the air drag force f_d which depends on the speed (proportional to the the square of the speed, see below). The relation between the speed (v) of the bike and the rotational speed (ω) of a wheel of radius R is v=Rω. So, for constant ω the ratio of the speeds is v_27.5/v₂₉=27.5/29=0.95; since you specified constant angular speed and the drag is proportional to v², the air drag would be 0.95²=0.9 smaller (10% smaller) for the slower (27.5) bike. However, this does not seem to me to be the right thing to look at; rather you would want to know how much force you would need to apply to keep the 27.5 and 29 inch bikes going with the same constant speeds and that is what I will do now. The rolling friction forces are both proportional to the the corresponding normal forces and N₁+N₂=mg and the force (shown in green) you need to keep it going forward with constant speed is f_you=f₁+f₂+f_d=μmg+f_d where μ is just some constant (called the coefficient of friction) and f_d is the same for bikes going the same speed. Therefore, the only things which determine the force you need to apply to keep either bike going with some constant speed are the total weight and the speed. Note that f_you is the frictional force exerted forward by the rear (driven) wheel, not the force you exert on the pedals; the force on the pedal would depend on the length of the pedal crank and the gear you were in.

QUESTION:
What are the three places on a pinewood derby car where friction primarily occurs?

ANSWER:
There is an earlier question on pinewood derby which has a link to the best video I have seen on how to win. If you watch this you will find, I believe, the main friction losses are to wheels rubbing on axles, air drag from poor aerodynamics, and wheel rubbing on the rails.

QUESTION:
Can water exist as a vapour at 25 ⁰C and 1 atm while we know that the normal boiling point of water is 100 ⁰C?

ANSWER:
We certainly know that water can exist in a vapor state under those conditions in the air. But you cannot have pure water vapor under those conditions as you can see by the phase diagram. The maximum amount of water at a particular temperature and pressure which air can support is called the saturation fraction and the relative humidity is the fraction of that amount actually present.

QUESTION:
Hey, I had a thought that I am not qualified to answer, nonetheless still curious about. Just as positively charged objects have an out-going electric field and negatively charged ones have inward electric fields, Is it possible that our idea of inward gravity is a result of us living in a "negative" universe? Could there be "positive" universes where gravity would behave similarly to a positively charged object? Hope this isn't a poor question it's just been on my mind a lot.

ANSWER:
Negative and positive charges and inward and outward fields are arbitrary choices. In other words, a universe where all masses were "negative" and the fields all pointed out would be indistinguishable from our own. Although electrostatics and gravity appear to be very similar, in fact they are very dissimilar—there are two kinds of charge but only one kind of mass. This allows in their being both attraction and repulsion in electrostatics but only attraction in gravity.

QUESTION:
Would a green bowl weighing 2lb travel the same distance on the green as a bowl of 2lb 6oz if the same force were applied to both (the green is damp and heavy)

ANSWER:
Your statement about "the same force…to both" does not adequately constrain the problem. You need to specify that the balls start with the same speed (v), the same kinetic energy (½mv²), or the same linear momentum (mv). The most important force once the ball is released is the rolling friction due to the green. This force is proportional to the weight of the ball, so a 32 oz ball will experience a force slowing it down which is 32/38 (84%) as large as the force felt by the 38 oz ball. Suppose that a ball of mass m and starting speed v goes a distance d before stopping, and the friction force is F=km where k is a constant depending on the conditions of the green. Then it can be shown that d=Cv² where C is a constant which does not depend on the weight of the ball (it depends on k and what units you choose to use). Therefore, if the balls start with equal speeds, they will go the same distance. The reason the heavier ball, which experiences a larger retarding force, goes as far as the lighter ball is that it has more inertia. However, if you give the balls the same initial kinetic energy or linear momentum, the more massive ball will have a smaller initial speed and therefore not go not go as far as the less massive ball.

QUESTION:
Why are the walls of my vocal tutor's voice room covered in foam and carpet? Is it a wave thing she is trying to prevent?

ANSWER:
So that the sound will stay inside the room, the foam/carpet absorbing the the sound that hits it. People in nearby rooms probably do not want to hear you practice. Although you could not perceive an echo if the walls were not padded (because the time of the echo arrival is so small), there would be sound bouncing around in the room which would not be the best environment to practice. Did you ever notice the difference between a room with rugs, furniture, pictures on the wall, etc. and that same room empty? The empty room "sounds empty".

QUESTION:
What will happen if 100g cotton hits you with the speed of 2500km per hour?

ANSWER:
Strange question! Let's think think about recoil first. If the cotton ball and you are stuck together immediately after the collision, momentum conservation can be used to estimate the speed of the two of you: m₁v₁=(m₂+m₁)v₂. Using v₁=2500 km/hr≈70 m/s, m₁=0.1 kg, and m₂≈100 kg, I find that v₂≈0.07 m/s. 7 cm/s is pretty slow, not a very dramatic effect. But, how much does it hurt? In other words, what is the force you feel during the collision and how long does the collision last? Assuming constant decceleration during the collision and estimating that the cotton ball compresses by about 10 cm=0.1 m, I find that the collision time is about t≈0.003 s and the average force is on the order of F≈m₁v₁/t=70/.003=23,000 N=5200 lb. The force to break a rib is about 3300 N, so this is going to cause some serious bodily harm, almost certainly killing you if you are hit in the head or chest.

QUESTION:
Who coined the term 'mass' and when? I can't find it anywhere.

ANSWER:
The Online Etymology Dictionary has the following entry for mass:

"'lump, quantity, size,' late 14c., from Old French masse 'lump, heap, pile; crowd, large amount; ingot, bar' (11c.), and directly from Latin massa 'kneaded dough, lump, that which adheres together like dough,' probably from Greek maza 'barley cake, lump, mass, ball,' related to massein 'to knead," from PIE root *mag- 'to knead, fashion, fit.' Sense extended in English 1580s to 'a large quantity, amount, or number.' Strict sense in physics is from 1704."

I was unable to determine who first (1704) used the word for the current physics useage. It is difficult to determine because Newton's Principia was written in Latin and translations would incorporate current physics usage. I believe, but am not certain that in the first edition (1687) of Principia he used the Latin gravitas (heaviness) for both force and mass and that would be sorted out in the second edition (1713) and in subsequent translations. Because of linguistics twists and turns, it is probably a fool's errand trying to give credit to a single person, sort of like "…who invented TV?"

QUESTION:
Let your mass be m₁ and the mass of large box be m₂. Now imagine that you are in the box, now the combined mass would be m₁+m₂. Now start applying the force on the floor of the box from inside the box. Does the mass of the body vary or its just the weight that varies?

ANSWER:
The "body" is me plus the box. The mass of the body, m₁+m₂, does not change. The weight of the body, (m₁+m₂)g, does not change either because the weight is defined as the force which the earth exerts on the body; if the body is sitting on a scale, this is what the scale reads. If I push down on the floor with a force F, the floor pushes back on my hand with a force F (Newton's third law); since these are both forces on the body, they cancel out so the scale reads the same. (I have assumed that the body is not being accelerated.)

QUESTION:
I was trying to find out what would happen to Supercritical Water (3200psi/374C) if the pressure was suddenly reduced to 1 atm (14.7 psi). My guess is that it would immediately (and powerfully) flash to steam. But I've had many people tell me that "what happens" depends entirely on HOW the Supercritical Water was created. Was it heated in a boiler to supercritical state, compressed mechanically to supercritical state, found in moonrocks, etc. ?? Something about the "latent heat of vaporization", and if no heat was applied to CREATE the supercritical state, then the water would not expand to steam.

ANSWER:
Refer to the generic phase diagram above. (The green line qualitatively represents the line between solid and liquid; the green dashed line qualitatively represents the anomalous solid-liquid interface for water.) Since you specify "supercritical water" I assume that you are not at the critical point but in the region just above that point in the area labelled supercritical fluid in the phase diagram above. At any temperature above the critical temperature (T_cr) there is no distinguishing between vapor and liquid. Above the critical pressure the water is called a supercritical fluid, below it is called a supercritical gas.

Contrary to what "many people" have been telling you, the sample retains no memory of how it got where it is on the phase diagram; if you were exactly at the critical point, I guess that it might be argued that the fractions of liquid and vapor depend on how you got there, but you have specified that the water is supercritical, beyond that point. What does matter, though, is how you reduce the pressure. You can either reduce the temperature (keeping the volume constant), increase the volume (keeping the temperature constant), or some combination of the two. If you increase the volume while holding the temperature constant, as indicated by the orange arrow A, you would still be supercritical but you would perhaps call it "supercritical steam" instead of "supercritical water". If you then cooled it, it would become normal water vapor below T_cr. (I have also included an accurate phase diagram for water.)

QUESTION:
How much force is exerted on an upright motorcycle rider at highway speeds ??

ANSWER:
A fairly good approximation of air drag F (in Newtons) is F≈¼Av²where A (in m²) is the area presented to the wind and v (in m/s) is the speed. (This only works if you work in SI units; I will convert the answer to pounds in the end.) So, taking v=70 mph≈30 m/s and A≈½ m², F≈¼x½x30²≈113 N≈25 lb.

QUESTION:
When we say that a clock in moving frame runs slower than a clock in stationary frame. What does it mean?

ANSWER:
If you are moving relative to me, I observe your clock to be running more slowly. The easiest way to understand this, I believe, is the "light clock" which you can read about in an earlier answer. (Key to understanding this is the notion that the speed of light one measures is independent of how you are moving.) It would also help you to understand if you read about the "twin paradox".

QUESTION:
Could we extract the Helium-3 isotope from the solar wind?

ANSWER:
From data given in one publication I could find, I figure that the flux of ³He in the solar wind is less than 3000 atoms/cm²/s. That is not a very promising source of an isotope!

QUESTION:
I need the simplest description possible of the double slit experiment. Is this correct? A scientist takes a box and at one end puts a laser with the ability to shoot photons, the building blocks of light. The other end of the box is just a plain wall. In between, the scientist puts a piece of cardboard with two vertical slits. In step one of this experiment the scientist shoots photons through the slits to see what kind of pattern they'll create on the wall on the other side of the box. If light is made up of particles, then the photons will hit the wall as a series of dots, randomly placed. If light is a wave, the pattern on the wall will be uniform. Here's where it gets very strange. Whenever this experiment is carried out, if the scientist is observing it, the pattern on the wall will be uniform, indicating the light is acting like a wave. However, if the scientist doesn't observe the experiment, if he does it within a closed box and looks in when it's complete, the pattern will show the particles hit the wall in random fashion. In other words, the result of the experiment depends on whether someone is watching it or not.

ANSWER:
There are a couple of problems with your question. First, "If light is made up of particles, then the photons will hit the wall as a series of dots, randomly placed." What actually happens is that the photons make two stripes on the wall which are simply images of the two slits. Second, "If light is a wave, the pattern on the wall will be uniform." In fact there is a pattern of many stripes on the wall with diminishing intensity as you get farther from the central brightest stripe; this is called an interference (or diffraction) pattern. Your "very strange" part has problems too. The scientist cannot just "look in", she must do a particular kind of looking—she must do an observation to determine which slit the photon went through. If she does not make that observation, the photons will make an interference pattern meaning that they are waves. However, it she does observe which slit the photon goes through, there will be no interference pattern. A cute little video may be seen here (it is for electrons instead of photons, but the idea is the same).

QUESTION:
If you were driving 90 mile an hour and you hit a 600 pound horse and the horse stayed in the car with you how much wood it slow the car down our way 3200 pounds

ANSWER:
Use momentum conservation: 3200x90=3800V, so V=75.8 mph.

QUESTION:
What is a black body?

ANSWER:
It is an object which absorbs all electromagnetic radiation which falls on it. There is no such thing, but an excellent approximation can be made. The most common way to fabricate a near perfect black body is to take a hollow metal object and drill a tiny hole in it. Radiation hitting this hole will enter the body, get absorbed in the wall, and reradiate inside. But the reemitted radiation will have a very low probability of "finding" the hole and coming back out. A black body is also a perfect radiator. Using the hollow metal object, imagine that you heat the metal to some temperature and observe only radiation coming out of the hole (not from the outside of the metal object). This is called black-body radiation (see spectrum above). Around the beginning of the 20th century, the failure to understand the spectrum using electromagnetic theory led to the discovery of quantum theory in physics. Stars turn out to be very good approximations of black bodies and by observing the emission spectrum of a star astronomers can determine its temperature. Our sun has a temperature of around 5000 K.

QUESTION:
Does mono-atomic hydrogen (1 proton + 1 electron, in a bound state) actually exist in universe? Not di-atomic molecular hydrogen, not the ion, and not hydrogen bound in molecules (e.g. organic compounds), but a simple "atom of hydrogen"? If you believe so, what experimental evidence is there, and under what conditions does it exist?

ANSWER:
The Wikepedia article on hydrogen states that "Its monatomic form (H) is the most abundant chemical substance in the Universe, constituting roughly 75% of all baryonic mass." Most of this resides in the interstellar medium, the vast regions betweens stars and galaxies. Temperatures greater than 3000 K break the molecules into atoms.

QUESTION:
If I have two positive charges, that are repelling each other, I force them closer together, doing work on them, and then pin them in place. The work I did on them appears as energy in the electrostatic field between them. now, it is explained, that the "force" is mediated by the exchange of virtual photons, and these virtual photons "borrow" energy versus time. ( which is explained as a violation of energy conservation, but is allowed, briefly, in quantum mechanics) However, if the field has energy in it, why do the virtual photons need to borrow energy vs time? Wouldn't the virtual photons just "borrow" energy from the field, and when nothing happens (since they are pinned in place)just return the energy to the field?

ANSWER:
You should first read the answer to an earlier question. This answer emphasizes that this idea of photon exchange is a "cartoon" to help visualize the qualitative role of virtual photons in a quantized field and should not be taken too seriously or literally. Your statements regarding the photons "borrowing energy" is totally unnecessary. No energy is borrowed from anywhere. When a virtual photon appears, the energy of the system suddenly increases, violating energy conservation; it turns out that, because of the Heisenberg uncertainty principle, it is perfectly ok for this to happen as long as it doesn't happen for too long a time.

QUESTION:
I would like you to comment a thought that I have concerning the definition of weight. Almost every definition (from textbooks to physics sites and encyclopedias) of weight is to say that it measures the force of gravity on a object. I strongly disagree with these notion and I think it's quiet armful. By means of an example I'll illustrate "my case". Grab an object and suspended it from a dynamometer. Situation 1: If the dynamometer is at rest relative to earth (inertial frame of reference) it measures directly the object weight, and indirectly, the force of gravity on the object. So far so good. But... Situation 2: if we let go of the dynamometer you will clearly observe that it's reading is zero. So you find yourself in the situation where there is still the force of gravity (the dynamometer is falling...) but there is no weight (the scale reading is zero). This simply experiment shows two things. First, in a inertial frame of reference weight equals the force of gravity (situation one). In a non inertial frame of reference weight is different from the force of gravity (situation 2). In our daily lives most of the time we find ourselves in situation 1. But now and then (elevators, jumping from a higher place to the ground), we are in a situation where weight no longer is equal to the gravity pull.

ANSWER:
I am sorry, but you cannot disagree with a definition. A definition is a definition. You can say that you would not define the quantity that way, but you would be mostly alone. The most uniformly accepted definition of weight is the force which the local gravitational field exerts on a mass. And this is completely reasonable. In your situation 2, the scale reads zero but the weight is just what it was before you let go. Even in an accelerating elevator, your weight is just the same it is when the elevator is at rest. "Weightless" is misnomer for an orbiting astronaut. I once found myself teaching an introductory physics course from a textbook which had been chosen by a committee. To my horror I found that the weight was defined as "what a scale reads", so your weight, using that definition, increases when you are in an upward accelerating elevator. I have talked to scores of physicists since then and all have agreed with me. You also mentioned inertial/noninertial frames of reference. If you are thinking about what weight is in special or general relativity, you need not worry about it because force is not a useful concept in relativity theory. The reason is that acceleration is not an invariant quantity and therefore different frames of reference would see a given force as having different values; in relativity mass is the important concept for determining inertial properties.

QUESTION:
How much power would be necessary to create a forcefield that could hold back over a hundred million cubic feet of water at a depth of over 6,000 ft? I'm trying to come up with a way to create temporary force fields for ocean archaeology the idea being basically creating a site of 8 Points as markers for the site then produce a forcefield to empty the water out from within the designated site area. I'm thinking a magnetic force field would be the best solution due to the fact that electricity can be harnessed from the Earth's electromagnetic field.

ANSWER:
Water experiences zero net force from either a magnetic field or an electric field. In an electric field the water would become polarized (it is a dielectric) but no net force. There is absolutely no field that I can think of which would simply hold water out of a volume. To do what you want to do you would need to simply build a water proof box around the site and pump the water out. At 6000 feet down the pressure is about 2600 psi; if you had one atmosphere of pressure in the air inside, this would correspond to having about 176 atmospheres of pressure on the outside, so the box would have to be really strong. I am also curious how you will "harness…Earth's electromagnetic field". If you could extract energy from the air, surely it would have been done by now. I think there was something like that in Ayn Rand's novel Atlas Shrugged. but that was purely science fiction.

QUESTION:
I have four axle stands, well made but not certified to any particular weight. If a vehicle is supported on four stands will each stand need to support only 25% of the total weight?

ANSWER:
The answer is no, probably not. How much weight each stand will carry depends on where the center of gravity of the car is located. The singlest biggest weight of the car is the engine so, in a front-engine car, the center of gravity is going to likely be closer to the front axle than the rear; so the front axel stands will bear a larger fraction of the total than the rear. Usually the center of gravity is equidistant from the left and right sides of the car, so the left and right stands of either front or rear axles will bear the same weight. To make this more quantitative, suppose each stand in the front (rear) exerts an upward force of F (R) on the car as shown in my figure; the weight W acts at the center of gravity. Now, if the center of gravity is a distance d_F from the front axel and a distance d_R from the rear axel, it may be shown that F=½Wd_R/(d_R+d_F). So, if d_R=d_F, F=W/4 or if d_R=2d_F, F=W/3.

QUESTION:
We know that the speed of light is the same in every reference frame whether stationary or moving. Is there any physical reason behind this constant speed of light? Or is it from mere observation that we have come to know about it?

ANSWER:
See faq page. Two faqs of interest are here and here.

QUESTION:
If you have 800 lbs of steel falling 12 feet what is the force of the fall?

ANSWER:
I get this kind of question all the time. The question has no answer unless you know how long it takes the weight to stop (or how far it falls before stopping). I can calculate the time it takes to fall 12 ft (0.87 s) and the speed it has after having fallen 12 ft (28 f/s) but I cannot find the force it exerts on whatever it has fallen on unless I know how long it took to stop. For example, suppose it takes 0.1 s to stop; then it can be shown that the average force over that 0.1 s is about 7800 lb. If the time were instead 1.0 s, the force would have been about 1500 lb. See the faq page.

QUESTION:
Are the laws of physics always changing because of our planets forever changing position in the universe?

ANSWER:
One of the most important tenets of physics is that the laws of physics are the same everywhere in the universe. People have looked for evidence of the possibility that the laws change as time passes, for example by observing astronomical objects which we can see billions of light years away, but all observations have given no indication of changing laws.

QUESTION:
I recently received 3- rug runners from a company wrapped in plastic but to my surprise their was a round orange sticker on each of the 3 -runner rugs that said x-ray and under the word x-ray there was a small box with a check mark and next to the marked box it said tick after x-ray. WHAT DOES THIS ORANGE STICKER MEAN? Can you help me? Any answer would be much appreciated.

ANSWER:
I do not know for sure. My best guess is, particularly if they were sent air freight or came from overseas, that they were x-rayed to make sure there were no bombs or contraband items (drugs, weapons, etc.) and the stickers were to notify the recipient that this had been done. But do not worry—the x-rays are long gone and have done no damage nor have caused any lingering radiation.

QUESTION:
Why is it required that the lever of a weight balance should be horizontal to compare equal weights on the two sides of its pan? Is it possible to compare two equal weights on the two sides of the pan even if the lever is in an inclined position? Is it a misconception that to compare equal weights on the two sides of a weight balance the lever must assume a horizontal?

ANSWER:
I have explained the workings of a beam balance in an earlier answer. The key is to realize that the center of gravity of the balance itself is below the pivot point.

QUESTION:
While studying magnetism, I got curious why magnetic field lines can be visualized with iron filings. To my understanding, the field lines are not actually existing physical objects, but just man-made imaginary lines to help us understand the field intuitively and visually. One explanation that I thought of was that friction between the iron filings and the paper(as paper is typically used in iron filings experiment) prevent the filings from moving closer to the magnet at some point, but I am not entirely sure and it is incomplete.

ANSWER:
Certainly the "lines" are not really there; but the field itself is real. What happens is that the iron is magnetizable in a magnetic field. Each filing is like a tiny needle so it becomes a tiny bar magnet when placed in a magnetic field. Since it is so tiny, it thinks that it is in a uniform magnetic field, which on its scale is an excellent approximation. In a uniform field a bar magnet feels no net force but it does experience a net torque trying to align it with the field, the north pole of the bar magnet pointing along the field.

QUESTION:
Is there any relationship between a sine wave and the bell shaped curve used in statistics? They look similar. Is there a reason for this or is it just a ...

ANSWER:
Is this a joke? A test of my memory? These are the exact words of a question I answered many years ago.

QUESTION:
I was heading towards a bus terminal with my son during a steady rain. When we walked on a long ramp, we have observed that the water was not flowing continuously, but there were waves at roughly equal distances, rolling down pretty rapidly on the surface. The water supply was continuous on the elevated end, so the emergence of the waves was not apparent, but the resulting pattern was spectacularly regular. Why do these waves appear? What is their velocity? What is the time interval between waves?

ANSWER:
Looks like the 'c' on your keyboard is not working! Don't know what happened to the 'fl'. Anyhow, it is really difficult to give a definitive answer without more examination of the situation first hand. My best guess is that, although the supply was "continuous", there was a slight depression or obstruction where the water was being added which was periodically emptying and getting refilled.

QUESTION:
here is another question for you. The questions I sent are a deal me and my friends are doing for fun becuse we just learned about the mole number deal in chemster and we wanted to do our own little project together it is not a homework deal. here is the question. how many times would a mole of eiffel towers go to the VEGA star and back to earth?

ANSWER:
The height of the Eiffel tower (ET) is about 1000 ft/ET; the distance to Vega is 25 ly≈8x10¹⁷ ft; so the number of ET to Vega is about 8x10¹⁴ ET so a round trip is about 16x10¹⁴ ET. Avagadro's number is about 6x10²³, so the number of round trips is about 6x10²³/16x10¹⁴≈3.8x10⁸=380,000,000.

QUESTION:
Can the energy of a holon be harnessed and used? Can a split electron be put back together? Does an electron give off holon energy when split? Is a holons energy limitless? What is holon energy like, is it like electricity?

ANSWER:
The following quote is from Physics Stackexchange and more or less summarizes why your question is basically unanswerable: "Spinon, orbiton and holon are quasiparticles, they are not actual constituents of the electron. We know about particles by their interactions, and it appears that in condensed matter physics some phenomena are best described as interactions of new kind of particles which are actually arising from complex collective behavior." An electron is a particle without constituents. However, sometimes under special conditions where there are very many electrons all closely packed, the results of their interactions which involve a very large number of them can be well described by inventing new particles. These particles do not really exist, per se, but assuming that they do gives accurate descriptions of physical phenomena. As far as I could find, these collective behaviors are best observed in one dimensional systems. If you google holon physics you will find places where you can read about experiments.

QUESTION:
Why does to every action there is an equal and opposite reaction?

ANSWER:
Because that is the way nature works. This is a law of nature (Newton's third law) which comes from experimental observation.

QUESTION:
Why does energy can never be created nor be destroyed?

ANSWER:
It is partly due to the way energy is defined: if no forces outside of any system do work on the system, the total energy will not change because of the way energy is defined.

QUESTION:
Why does when an object fall on the sand it displace into the sand.why not it just stop as it will also get the same reaction force from the sand

ANSWER:
The reason is that the sand is not just one thing but many little things. When the object hits the sand it is really hitting hundreds of tiny grains of sand and each grain of sand experiences a force which moves it independently of the other grains. All these tiny forces add up to a larger reaction force on the object.

QUESTION:
How light is formed?

ANSWER:
Generally speaking, all electromagnetic radiation (of which visible light is just a tiny part) is generated when electric charges are accelerated. Much of the visible light you see comes from atoms. When an atom is excited to a higher energy state it jumps back down to a lower state and emits a little bit of light which carries off that energy.

QUESTION:
How tall are radio waves? (AM or FM) Absolutely nothing on the Internet addressing this question.

ANSWER:
The reason that you could not find anything is that the amplitude of of radio waves (electromagnetic waves) is not measured in length. A water wave (water moving up and down) or a wave on a string (string moving up and down) can be expressed in meters or inches, but not all waves are something waving in that way. What waves in a radio wave are electric and magnetic fields. An electric field, for example, is measured in Newtons per Coulomb, a force divided by an electric charge, so it is a meaningless question to ask how tall it is.

QUESTION:
Suppose there were two particle accelerators which accelerated particles to a high percentage of c. if the particles were then directed on a collision or near collision course, what would happen?

ANSWER:
I do not know what you are asking. Each particle would see the other approaching with a speed which is a higher percentage of c. What happens depends on what the particles are and on chance and on how fast they are actually going.

QUESTION:
I was wondering what would happen as the particles (with some mass) approached each other at greater than c? Also what would an observer see? Would space deform to limit the approach speed?

ANSWER:
I see now what your question is. So, if each particle has a speed of 0.9c, you are assuming that each would see the other as having a speed of 1.8c, right? You are thinking classically and, indeed, that is not the right way to add velocities when the speeds are not small compared to c. See an earlier answer. The correct way to calculate the relative speed if one particle has a speed u and the other is moving in the opposite direction with speed v is v'=(u+v)/[1+(uv/c²)]. So, for example, if u=v=0.9c, v'=1.8c/(1+0.9²)=0.9945c.

QUESTION:
My question has to do with Voltage, aka potential Voltage is always described as potential, as in potential energy. In most mechanical systems, potential energy can get turned into kinetic energy, and usually the kinetic energy is what does the energy transfer to other things. In electricity, the voltage drops (energy transfer) seem to be described as if the electron never acquires a change in kinetic energy, and the energy automatically comes from the "potential" the drift velocity of the electrons doesn't change very much, the energy transfer to the load, comes directly from the potential energy. why don't electrons first change the potential to kinetic then kinetic to the load? it seem that the energy goes directly from potential to the load?

ANSWER:
To keep this as simple as possible, I will work only with uniform electric fields; so a point charge experiences the same force in magnitude and direction no matter where it is. This allows us to focus on the principles without all the integral calculus necessary to do it more generally.

This is tricky stuff, and you have some confusion about electric potential which is common to almost all students when they first learn it. I prefer to talk about electric potential, leave the term "voltage" out of it; if you insist on using voltage, it is the same a the difference in electric potential between two points in space. Potential is not the same thing as potential energy, just the same as electric field is not the same thing as electric force.

The force felt by a charge Q in an electric field E is F=QE. Since we are using a uniform electric field, I will choose a coordinate system such that the field points in the +x direction. (This very question was dealt with very recently except for gravitational forces and fields.) Now, the difference in potential energy (PE) between two points, x₁and x₂, is the negative of the work done by the electric field on a charge Q moving from x₁to x₂: ΔU=U_final-U_initial=U(x₂)-U(x₁)=-QE(x₂-x₁). Now, suppose we choose a coordinate system such that x₁=0, x₂=x, and U(x₁)=0; Then U(x)=-QEx. The minus sign says that as a positive charge moves in the direction of the field (+x), the PE gets smaller (just as in the case of a gravitational field); but as a negative charge moves in the direction of the field, its PE becomes larger. So, left to its own devices a positive charge will "fall" in the direction of the field but a negative charge will "fall" opposite the direction of the field. So, one reason one might want to introduce electric potential is to avoid this ambiguity. We follow the same procedure to define as was done to define electric field—divide out the charge: E=F/Q and, now, ΔV=ΔU/Q. So, you see, you have to be careful how you use the terms, force, field, potential, and potential energy. Force is N, field is N/C, potential energy is J, potential is J/C.

Now I will finally address your question which, if I understand correctly, is why the electrons in a conducting wire don't just keep accelerating because their potential energy should be decreasing all the time. In fact, the electrons are accelerating all the time but they are trying to make their way through a forest of atoms and they do not go very far at all before they bump into an atom and transfer their kinetic energy to the atom and have to start all over again. So it's go-stop, go-stop, go-stop, go-stop, etc. And, what happens to the electron energy losses from all those tiny collisions? It goes to heating up the wire. Don't you worry—energy is conserved!

QUESTION:
I suspect you may have been asked if anti gravity is possible and I understand that there is no current theory stating that anti Gravity can exist in they way it is displayed in fiction. However my question is this. Do you know of or every heard of a theory that could support shielding from the effect of gravity? Generally speaking it would cause a large mass to be manipulated like a boat floating on water. If I have gone beyond what you normally accept for questions I apologize. But searching on the internet on this subject has only lead me to questionable information at best. I thought perhaps you could lead me down a better path. I ask this because I armature writer of fiction. When possible I like to use concepts that can sound not impossibly implausible.

ANSWER:
Your hand is an antigravity machine—you can hold a mass up, overcoming gravity! My point is that you need only find a way to exert a force on a mass which is equal and opposite the gravitational force to overcome gravity. E.g., if you have have an object which has some electric charge Q and you turn on an upward electric field E, the object will float if EQ=W where W is the weight of the object. In the last 20 years it has been observed that the stars which we can observe near the outer edge of the universe are not only moving away from us (that has been known for nearly 100 years now) but they are actually speeding up. If gravity were the only force which those stars see, this would be impossible because gravity is an attractive force; therefore there must some repulsive force pushing out on those distant start and its origin has been called dark energy. Nobody really understands what dark energy is, but in your fiction you could imagine that somebody has managed to harness dark energy and if you had a dark energy power generator on your space ship, you could use it to push your ship away from a planet. So, dark energy could be thought of as antigravity.

QUESTION:
What is the difference between gravitational force and gravitational force per unit mass?

ANSWER:
Suppose you have a 1 kg mass and it experiences, due to the gravity of some large masses in the vicinity, a gravitational force of 3 N. If you now replace that mass with a 2 kg mass, you observe a force in the same direction as the 1 kg felt and which you measure to be 6 N. Now put a 10 kg mass where the 1 and 2 kg masses were; you find that the force it experiences is 30 N. You do this enough times to convince yourself that the force which a mass m experiences at that point in space is proportional to m. You could write this as an equation, F=mG where G is called the gravitational field. So you can solve for the field, G=F/m. F is a force measured in Newtons (N) and G is a field which is measured in Newtons per kilograms (N/kg). In the example I gave, G =3 N/kg and you would have to specify some direction to specify the field completely, but it will be in the same direction as the force F. Force tells you the force which a particular mass will feel, field tells you the force any mass will feel.

QUESTION:
With Boyle's law, what would happen to the pressure of a gas inside a sealed bottle, if the bottle was squeezed tightly, reducing the volume of the gas by half?

ANSWER:
It depends on how you do it. If the temperature remains constant (isothermal) while you squeeze, Boyle's law would tell you that PV=constant, so if V becomes half as big, the pressure becomes twice as big. But, you have to allow energy to flow out of the bottle in order for the temperature to remain constant; one way to do this is to squeeze very slowly so that the temperature of the gas can equilibrate with the ambient temperature. On the other hand, if you reduce the volume very quickly or thermally insulate the bottle very well, it is called an adiabatic compression. For this situation it may be shown that PV^γ=constant, where γ=C_P/C_V and C_Pand C_V are the specific heats at constant pressure and volume, respectively. For a monatomic gas γ=5/3 and air is pretty well described by γ=7/5. So, let's assume your bottle is filled with air. Then P₁V₁^7/5=P₂V₂^7/5 so P₂/P₁=2^7/5=2.64. You can also find the temperature because it may be shown that TV^γ-1=constant so T₁V₁^2/5=T₂V₂^2/5so T₂/T₁=2^2/5=1.32. (Temperatures must be absolute temperatures. Similarly, pressures must be absolute, not gauge, pressures.)

Q&A OF THE WEEK, 3/24/2018

QUESTION:
I am constructing a floating dock out of pressure-treated lumber and 55-gallon empty plastic barrels and want to ensure that I provide sufficient buoyancy. Although the dock comprises a ramp and a platform, they will be only loosely connected (with slack rope and eye bolts), and the buoyancy calculations for the platform are easy; it's the ramp that is giving me headaches. Specifically, one end of the ramp will be sitting on the shore and the other end will be supported by the plastic barrels. The ramp is T-shaped, with the lakeside end wider to accommodate the barrels. Please give me some guidance as to how I might analyze a T-shaped structure with the narrow end on shore and the wide end kept afloat by barrels, so that a weight of w placed at any point on the ramp will not submerge the barrels more than 45% (because, although I think I did a good job sealing off the barrel caps, I'd rather not have to find out). Although I've meant for this question to be somewhat general to allow for design modifications, I will mention that I've built part of the ramp already, with the walkway being 3' wide and 8' long, and the cross of the T being 5' wide and 3' long and covering two empty barrels; I'm willing to add another section with more barrels if needed. Any help in analyzing buoyancy of a T-shaped ramp with the narrow end resting on land would be greatly appreciated!

ANSWER:
I will not be able to do any quantitative calculations without knowing the weights of the ramp and dock. Also, will the level of the water remain fairly constant?

FOLLOWUP:

I must mention that I modified the design to make the walkway eight feet longer, so that it is now 16 feet long. Also, I had another empty 55-gallon barrel on hand, and placed it lengthwise under the walkway at the end where the walkway joins the cross of the tee. The salient data for the components are as follows:
1. T-shaped ramp (total weight 648 lb)
2. Walkway (463 lb) is 3 feet wide and 16 feet long; a 55-gallon drum is placed under the walkway on the lake end.
3. Cross of tee (185 lb): 5 feet wide and 3 feet long, covering two 55-gallon drums.
4. The dock platform weighs 515 lb and is 8'x8'. Four 55-gallon drums support it.
My dock will be located in a tributary of the Potomac River. As such, the water is tidal, varying in depth between 2 and 5 feet. (This is the reason I did not join the ramp and the platform rigidly, as I anticipate that the angle of surface of the ramp relative to that of the platform will fluctuate, given that one end of the ramp rests on the shore.)
In order to help visualize the situation, I am attaching three JPG images which I created in Sketchup. Please note that these images do not include the latest modification, whereby I lengthened the walkway and placed a barrel under its far end. Also, I recognize that you will probably make certain simplifications in order to expedite the analysis. That is fine with me; I only wish to obtain a rough idea of the limits of motion of the ramp and dock as I walk upon them.
At high tide the walkway and platform are approximately horizontal. At low tide the water level is about 2-5 feet down.

ANSWER:

Some preliminaries:

The volume of each barrel is 55 gallons=7.35 ft³;
the density of water is 64.2 lb/ft³;
each barrel is 45% submerged, so each barrel provides a buoyant force of 7.35x64.2x0.45=212 lb;
each barrel has a weight of 21.5 lb;

I will first do the platform which is easiest if you assume that the load is at the center.

The weight of the platform is 515 lb;
the weight of the four barrels is 21.5x4=86 lb;
the weight of the load will be denoted as W;
the buoyant force of the four barrels will be 4x212=848 lb.

Therefore, W=848-515-86=247 lb.

If the load is not in the center, the total buoyant force will still have to be 848 lb but the platform will tilt so that two of the barrels will be submerged more than 45% and the other two less than 45%. I made an estimate of how much the platform would tilt if the load were moved over to 1 ft from one side edge. Without going into details, the heavy side would go down by about 2.8 inches and the other side would go up by the same distance; the corresponding tilt would be about θ=4.50. This would make two of the barrels 65% submerged, beyond your desired limit.

Next I will look at the walkway.

the weight of the tee and two barrels under it as well as the net buoyant force of those barrels act at 17.5 ft from the shore;
the weight and buoyant force of the third barrel act at 14.5 ft from the shore
the maximum weight W acts at a distance x from the shore;
there is a force F exerted up by the shore which we will not need to know;
I have assumed that the ramp has no interaction with the platform, since reference is made to "slack ropes".

All this is shown in my diagram. If one now sums the torques about the point of shore contact and sets that sum equal to zero, the product Wx can be solved for.

0=212x14.5+424X17.5-228x17.5-21.5x14.5-Wx-463x8=2488-Wx.

Wx=2488 ft⋅lb. So, for x=19 ft, the end of the ramp, W=131 lb. I am guessing that this result does not make you happy!

I am not sure how rigidly coupled the platform and walkway are ("slack ropes"), but suppose that they are coupled as if, when horizontal, they were rigidly attached. So now the summed torque equation is 0=212x14.5+424X17.5-228x17.5-21.5x14.5-Wx-463x8+848x23-601x23=8169-Wx. So W=8169/x. So, for x=19, W=430 lb and for x=27, W=303 lb. It would seem that it is important that the coupling be designed such that the platform can help hold up the walkway. I figure that the walkway will only go down a maximum of about 15⁰ which should not alter my estimates for horizontal significantly. So you should allow a fairly rigid coupling like some kind of hinge.

QUESTION:
If you have two metal spheres of the exact same volume, however with differing masses, say one sphere at 1kg and one at 10kg, attached each to an identical parachute, will they fall at different speeds? We know that two spheres of the same volume with differing mass will fall at a nearly identical speed, as the drag is identical. I have had it put forward to me that somehow adding a parachute into the system dramatically affects the outcome.

ANSWER:
"We know that two spheres of the same volume with differing mass will fall at a nearly identical speed…" Sorry, that statement is wrong. It is true, as you state, that, since they have the same size and shape, the drag forces will be the same on both. But, that force will have a much bigger effect on the less massive sphere because it has much smaller inertia. Adding identical parachutes will still result in the drag forces being identical, but the more massive ball+parachute will still fall faster. (You can understand this intuitively. Imagine a bowling ball and a balloon the same size. Drop them from a height of 2 m and surely the bowling ball will hit the floor first.)

To be a bit more quantitative, the drag force is of the form f_D=Cv² where C is a constant determined by geometry and v is the speed the object if falling. As the object falls from rest, v gets bigger and bigger. Eventually f_D will equal the weight W and thereafter it will fall at constant speed because the two forces add to zero: W=Cv², so v_t=√(C/W) where v_t is called the terminal velocity.

QUESTION:
Hi! Just another curious soul here. How do subatomic particles (especially fundamental ones that are parts of the Standard Model) transfer energy to each other? Can it be achieved by means of physical proximity? (E.g. A magnetron pointed to a cathode ray tube and shoots it with electrons in a high-energy state).

ANSWER:
The best current description of interactions among elementary particles is quantum field theory. The mathematics of the theory are difficult but the basic ideas can be understood qualitatively. Each quantized field has particles which respond to that field. The three fundamental fields in the standard model are the electromagnetic field, the strong or nuclear field, and the weak field. The way that the particles interact with each other is by "exchanging" field quanta. The field quanta, respectively, are the photon, the gluon, and the Z⁰ and W^±. Any charged particle feels the electromagnetic field, any particle made up of quarks feels the strong field, and the particles called leptons (the electron being one) feel the weak field. Interestingly, the fourth fundamental field, gravity, has not been quantized and so, although a graviton has been hypothesized, there is no successful theory of quantum gravity.

QUESTION:
The question relates to a medical procedure. If you fill a balloon to a set pressure to use in a semirigid tube to dilate it, will the radial force exerted on the tube by the balloon be different if the balloon is filled with water as opposed to air? In otherwords, does the density of the medium filling the balloon change the radial force exerted on the walls of a hollow tube if pressure is kept constant? Is there a known law that describes this?

ANSWER:
I am afraid that your question is very ambiguous. I do not get the picture at all. What is the idea, to use the balloon to pressurize the tube or vice versa? When you say "to use in a semirigid tube to dilate it" what does "it" refer to? Can you give me a description of what this device does and how it is used? Would the tube be filled with water also? Is the patient horizontal or vertical? Is the balloon above or below the tube?

FOLLOWUP:
The tube is a hollow tube with muscular walls. The patient is laying flat. The balloon is within the tube at an area along the tube that is narrowed compared with the rest of the tube.

ANSWER:
Ah, one picture is worth a thousand words! Water has a density about 1000 times the density of air. Therefore, although you can ignore the pressure differences between places at different heights for air but possibly not for water. That is, the pressure will be higher at the bottom of the baloon than at the top by an amount ρgd where ρ is the density of the fluid, g≈10 m/s² is the acceleration due to gravity, and d is the diameter of your inflated cylindrical balloon. For example, if d=1 cm=0.01 m the pressure on the back side is 0.1 N/m²=1.5x10^-5 psi larger at the front for compared to 100 N/m²=0.015 psi for water. In other words, the back side of the esophagus (I assume that is what this is) must support the weight of the water creating a larger push by the balloon on the back than the front. The density of the fluid does make a significant difference; I have no way of judging whether the difference is enough to make a difference in efficacy of your device.

QUESTION OF THE WEEK 3/17/2018

QUESTION:
I have a 40' ladder that weighs 300 pounds. Standing it up with it's foot against a wall. And walking it up. It gets heaverier and heavier; at the half way point it feels the heaviest. How much does it weigh half way stood up?

ANSWER:
The weight of the ladder does not change if you go up it. It is always 300 lb. Why do you think it gets heavier?

FOLLOWUP QUESTION:
With the feet of the ladder against the wall..with ladder not extended...each section is like..21and a half feet long... Starting at the inter end of the ladder..standing it up you have to go ring by ring...it just feels like it weighs a ton before it is stood all the way up..feels heaviest at a 35-40 degree angle . it takes all you can do to stand it up

ANSWER:
There are four forces acting on the ladder: the weight W of the ladder, the force F which you apply, the force N which the floor exerts up, and the force f which the wall exerts to hold the ladder from sliding. I have assumed that F is applied perpendicular to the ladder. At any angle θ the force necessary to hold it can be determined by summing torques about the corner of the wall and floor: Στ=0=FD-½LWcosθ. So, the force you must exert to hold up the ladder is F=½W[(L/D)cosθ]. Now, this is not so simple because as the ladder gets higher, cosθ gets smaller but L/D gets bigger. So, we have to make a further assumption about how the ladder is lifted. Having done this, my recollection is that I start off lifting the end straight up over my head and then raise it by walking toward the wall lifting as I go. So F is always applied at a constant height h above the ground; then sinθ=h/D and cosθ=√[1-(h/D)²]. So, F=½W(L/D)√[1-(h/D)²]. So we should put in some numbers, W=300 lb, L=21', h=7': F=(3150/D)√[1-(7/D)²]. The graph illustrates how F varies: starting at D=21' the force you must apply increases from a little less than half the total weight until you get about halfway (10', just as you perceived qualitatively) where it is about 225 lb. For the rest of the way F decreases rapidly to zero when the ladder is upright. The angle for the maximum F is about 35⁰, also about what you observed.

ADDED COMMENT:
To determine in general the location of the maximum value of F, set its derivative equal to zero and solve for D. dF/dD=LW(2h²-D²)/[2D⁴√(1-(h/D)²)]=0. The root of interest to us is D=h√2, so for h=7', D=9.90'.

QUESTION:
I was in a vehicle accident and had some injuries to my neck. For the life of me, I could not figure out how. I was sitting still in a light vehicle called a Polaris slingshot. I had a full jaw helmet on as well. I was curious about the average head around 10 to 11 pounds, and maybe a 2 pound helmet(?), what was the weight of my head on impact. The truck weight appx. 18,000 pounds I think (single axle box truck) and hit us about 10 -12 mph. Maybe that will help me understand how a slight bump hurt my neck so bad because I just don't get it!

ANSWER:
I can make a rough estimate of the force which your head felt during the collision using momentum conservation. The weight of the Polaris is about 1800 lb and the weight of the truck is about 18,000 lb. I will assume a perfectly inelastic collision, the two are stuck together after the collision. In that case 18,000x10=(1800+18,000)v where v is the speed of the two after the collision. Solving, v≈9 mph. If the collision lasted some time t, your head experienced an accleration of about a=(9 mph)/t. To translate that into a force F, use Newton's second law, F=ma where m is the mass of your head. Suppose the collision lasted about 1/20^th of a second, t=0.05 s. Then, if you work that out, the average force your head felt during the collision was about 80 lb, quite enough to cause a neck injury, I would think.

QUESTION:
Hi there, I homeschool my son and we have come across a few questions in the syllabus that I'm having trouble explaining to him, I was wondering If you could please elaborate on the question/s and how he can work out similar problems in the future and what I can use as future reference material for him to brush up on?

A large box of mass M is moving on a horizontal surface at speed v₀. A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are �_s and �_k , respectively. Find an expression for the shortest distance d_min in which the large box can stop without the small box slipping.
A pickup truck with a steel bed is carrying a steel file cabinet. If the truck's speed is 15 m/s, what is the shortest distance in which it can stop without the fi�le cabinet sliding?

ANSWER:
These are homework-like questions which I usually refuse to do. However, I am happy to help this homeschooler dad get the basic idea. These problems are essentially the same question. I will do the first one (which is more general) and let you see if you can do the second. You will need to look up the coefficient of static friction of steel on steel.

The first thing to do is to choose a body upon which to focus your attention; ignore everything else at this point. I choose to look at m.
Next, draw a diagram and show all the forces acting on the body you have chosen. In this problem, there are three forces: the weight of the box, mg straight down; the normal force N which M exerts upward on m, unknown for now; and the frictional force f which M exerts on m in the negative direction, also unknown.
Also, choose a coordinate system and resolve the forces; in this simple example, all the forces are already trivially resolved: N_x=(mg)_x=f_y=0, N_y=N, (mg)_y=-mg. f_x=-f. The acceleration a is in the negative x-direction, a_x=-a and therefore f_x is in the negative x-direction. (If you are not sure of the direction of a force, choose it to be in the positive direction and if you find that the magnitude of the force is negative, you drew it in the wrong direction.)
Apply Newton's laws; the box is in equilibrium in the y-direction (Newton's first law) and has an acceleration in the x-direction (Newton's second law): ΣF_y=0=N-mg so N=mg, and ΣF_x=ma=f so a=f/m.
Now, since the box is not slipping, what we know about f is that f≤μ_sN=μ_smg. Therefore, a≤μ_sg so the largest the acceleration can be is a=μ_sg.
Now apply the kinematic equations of motion which in this case would be v=v₀-μ_sgt and d_min=v₀t-½μ_sgt²where t is the time for the box to stop, i.e. for v=0. Finally, t=v₀/(μ_sg) so d_min=½v₀²/(μ_sg).

Note that you did not ever need to know either M or μ_k. It is very instructive, while we are at it, to look at the bottom box now. Suppose the coefficient of kinetic friction between M and the ground is μ_k. Now, what are the forces on M? Clearly its own weight, Mg down, a frictional force I will call F, and a normal force N' up from the ground. But M is also experiencing forces from its contact with m. Newton's third law tells you that the forces which m exerts on M are equal and opposite to the forces M exerts on m; therefore, there are forces -f (magnitude μ_smg) and a force -N (magnitude mg) on M. So now, going through the same procedure as above, ΣF_y=0=N'-Mg-mg so N'=(M+m)g and, ΣF_x=Ma=-Mμ_sg=μ_smg-μ_k(M+m)g so μ_k=μ_s. It was a little surprising to me that the kinetic friction coefficeint between ground and M is the same as the static friction coefficient between M and m for maximum acceleration. It is worth going through this to illustrate Newton's third law.

Here is a little warning. It is tempting to say that the weight of m is a force down on M. But this is wrong because that is a force on m, not on M. If you had that mistake in this problem, you would have ended up with the right answer. However, there will be many problems (e.g., blocks on inclines) where making that mistake will cause the final answers to be incorrect.

QUESTION:
Please. Explain the results using classical electricity and magnetism of the test depicted in the figure on page 10 where results of the 2-path experiment match exactly the results of an earlier experiment in the same lecture (page 2) demonstrating the property of repeatability of Color from one Color box to another Color box. For reference, please see this link.

ANSWER:
This reference is about superposition in quantum mechanics and beyond the scope of this site. The first time the questioner asked, he was trying to understand the Stern-Gerlach experiment with all the "color box" stuff in the reference. I told him that I could explain the experiment with just classical E&M and a little QM. If he cares, I have attached the appendix from my second book, Atoms and Photons and Quanta, Oh My! which explains how the apparatus works and how its results are interpreted.

QUESTION:
Ok my question is somewhat a "what if" question but it pertains to basic laws of physics and questioning them in a way. Ok, say me and a friend each set a time on our watches to go off at a certain time and it's the same time for each of us. While the watches are counting down he goes to Saturn, and I stay on Earth. Now sunlight takes, I don't know how many minutes to get to Saturn but let's say 30 mintues. Ok when our timers go off on our watches I'm going to be zoomed in on him with the best telescope around , one where I can see him standing on Saturn. When the timers go off he will also start to waive at me. Now the question is , do I see him waiving right then, or 30 mins later?

ANSWER:
The shortest time of transit of light from Saturn to earth is about 70 minutes, so let's go with that. I will assume that both clocks read zero when you synchronize them and he leaves. What you see through your miraculous telescope depends on how fast he traveled to get there. If his speed was very small compared to the speed of light (3x10⁸ m/s), your clocks will, for all intents and purposes, still be synchronized and it will take 70 minutes for the light from his waving to reach you; you will not be surprised to see his clock reading 70 minutes earlier than yours when you see the waving. On the other hand, if he is traveling very rapidly, say 80% the speed of light, he will get there, according to your clock, 70/0.8=87.5 minutes. But his clock will not read 87.5 minutes because he sees (length contraction) the distance to be shorter, 70√(1-0.8²)=42 light-minutes, so you will see him wave at 87.5+70=147.5 minutes on your clock; you will be surprised, however, if your telescope is good enough, to see his clock reading 42 minutes.

QUESTION:
But for my work, I need to figure out one parameter. Usual passenger vehicle. We know door velocity (for instance driver door) and I need to convert it to Joule. I am suppose that also need a mass (27 kg trim assembled door), distance from hinge to outside handle (approximately 1000 mm) What also? Please share a proper equation.

ANSWER:
Joules measure energy. This would be kinetic energy of rotation. You do not tell me what you mean by "door velocity"; I assume it is rotating about the hinges with some angular velocity ω radians/second. Since it will be rotating about the hinges, you need to know its moment of inertial I about that axis; if you model the door as a uniform rectangle, I=ML²/3=27x1²/3=9 kg m². If it had an angular velocity of ω=1 rev/s=2π s^-1 then its rotation kinetic energy is K=½Iω²=178 J.

FOLLOWUP QUESTION:
I am working in vehicle assembly plant and at the end of a final line we measure all doors speed of closing (velocity). For this check, DURR 5000 is used. It devise install at the edge of the rear door when the front door is measured and at the rear body panel when rear door is measured. For this measurement, we have upper limit borders: for front/rear own. As I know, velocity 1.62 m/s it is about 5 J and 1.49 m/s is 4 J for our C-class model, at least I found these data in my documents. I am puzzled that the result 178 J it is not appropriate equation (after my poor explanation). Can you please figure out in Joules or provide equation then is known: door mass = 27 kg, hinge to handle distance = 1000 mm, velocity = 1.1 m/s.

ANSWER:
Everything I did above was correct, but since you did not give me a velocity in your original question, I had to make up a speed of 1 revolution per second for the angular velocity; now that you gave me some velocities, I can see that my guess was way too big. If the door is closing, then it is rotating about the hinges. So specifying a velocity (1.1 m/s) does not tell me anything because the speed depends on how far from the rotation axis you are measuring. For example, if this is the speed of the center of the door, the outer edge has a speed 2.2 m/s and the hinge edge has a speed of 0 m/s. If I take the outer edge (1 m) as where the speed is measured, I find that ω=v/R=(1.1 m/s)/(1 m)=1.1 s^-1 and therefore, K=½Iω²=½x(9 kg m²)x(1.1 s^-1)²=5.45 J. The equation you want is K=½Iω²=½(ML²/3)(v/R)² =M(vL/R)²/6 where L is the distance from the hinge to the outer edge of the door and R is the distance from the hinge to the point where v is measured. If L=R, then K=Mv²/6. (Keep in mind that this (I=ML²/3) assumes that the door's mass is uniformly distributed in the radial direction which will certainly not be exactly true. It would be most accurate if you could get a good measurement of the moment of inertia about its hinges.)

ADDED THOUGHT:
I looked up the use of this Durr 5000 device and saw it also applied to sliding doors in addition to the hinged doors I have considered in my answer. In the case of a sliding door the energy is K=½Mv².

QUESTION:
Is it possible to convert mass to energy and send it some where at the speed of light?

ANSWER:
Sure. A nuclear power plant uses nuclear fission to turn mass into thermal energy. This thermal energy is used to turn a turbine (mechanical energy) which turns a generator to create electrical energy. This energy is sent down a wire at approximately the speed of light to you. In fact, even an exothermal chemical reaction (like burning coal) converts mass to thermal energy, but the amount of mass is really tiny compared to nuclear reactions. A more "pure" example is when an electron meets its antiparticle (a positron); the mass of both particles is turned completely into the energy of two photons which, of course, move away at the speed of light.

QUESTION:
If I launch a very dense piece of tin foil and a same sized less dense piece which one would fly farther on a homemade catapult and why?

ANSWER:
There are two considerations for this question. The first is the fact that the launcher will impart a different initial velocity to each of the two. The second is that air drag will have different effects on each.

To illustrate the first consideration, I will model the catapult as a simple spring pushing the mass m (a sphere of aluminum foil) up an incline. The easiest way to solve this is to use conservation of energy. The spring is compressed by an amount s and so has a potential energy of ½ks² where k is the spring constant; the ball is at rest and at a position I define to be y=0, so it has no energy. At the instant of launch, the ball has a kinetic energy of ½mv² and a potential energy of mgh. Equating the initial and final energies and solving for v, I find v=√[(ks²/m)-2gh]. So, you can see, the bigger the mass m, the smaller the speed at launch. If you ignore all friction, it is clear that the lighter ball will go farther. However, it is quite likely that the air drag (friction due to the air) will play a role.

To illustrate the second consideration, look at the ball at the instant of launch. There are two forces on it, its own weight mg straight down and the air drag f which points in the direction opposite the velocity. Now, the magnitude may be approximated as f≈-¼Av² where A=πR² is the area presented to the onrushing wind and R is the radius of the ball. So the air drag causes an acceleration opposite the direction of v which is a=-¼πR²v²/m; the negative sign means the ball is slowing down because of air drag. If the balls had the same velocity, the one with smaller mass would lose speed faster and therefore go less far.

So, the first consideration gives a higher speed to the lighter ball but the second takes speed away faster from the lighter ball. My hunch is that the air drag will be more important and the more massive ball will go the farthest; my reason is that the catapult force mechanism ("spring" in my simplified model) also will have to accelerate the bucket and arm, so the difference in the velocities leaving the machine will be quite small. I am quite sure that the heavier ball will go farther.

QUESTION:
In Brownian movement, atoms and molecules in motion impart motion upon the particulate matter. But what imparts motion upon the atoms and molecules in question. Gravity, the wobble of the earth?

ANSWER:
Those molecules and atoms in motion have kinetic energy. It turns out that if you heat up the fluid you increase their speeds, so the kinetic energy is changed by adding heat. This property of fluids is what we quantify as temperature—temperature is proportional to the average kinetic energy per molecule.

QUESTION:
Came across this particular question on a forum with a multitude of different answers and their opinion of it, I would like to hear your input on it. Here is the question: How many days will it take a spaceship to acceleration to the speed of (3.0x10⁸ m/s) with the acceleration g ? how far will it travel during this interval and what fraction of a light year is the answer to the previous question?

ANSWER:
(This sounded like a homework problem to me.) What forum?

REPLY:
Reddit.

ANSWER:
Now I know it is homework because this section of AskReddit has a title cheatatmathhomework! And if you can read you see that AskThePhysicist.com does not do homework so I should just trash this question. However, this question was submitted to AskReddit a year ago so the present questioner is presumably interested in the answer since it is marked UNSOLVED. But, the problem is not unsolved because the one comment points the reader to a site where the one-dimensional, constant acceleration kinematics is discussed; this is the appropriate way to help someone do their homework, not working it out to the final answer. The solution to the problem is a simple application of these equations. So here is my answer:

You have not accurately conveyed the problem. Most important, the final speed is not 3x10⁸ m/s but rather 1% of that number. But that is a small enough fraction of the speed of light that you can do the problem classically, not relativistically (which would be much more difficult). The problem is not unsolved since the single answer points you to the appropriate equations to use, Specifically, x≈½gt² and v≈gt. Here, g=9.8 m/s², v=3x10⁶ m/s. Do the algebra to solve for the two unknowns, convert t from seconds to days and convert x from meters to light years.

ADDED NOTE:
I worked out the answers and found that the problem, as stated, does not have the correct answer for the distance traveled. The correct answer is 4.6x10¹¹ m=4.9x10^-5 ly.

QUESTION:
I know photons are exchanged between a proton and an electron. Are photons exchanged between 2 protons or between 2 electrons?

ANSWER:
Any two electric charges "exchange photons"; photons are the quanta of the electromagnetic field. Keep in mind, though, that if you try to apply this idea too literally you will not really be describing what happens. Photon exchange is a "cartoon" classical representation of a quantum field effect to convey a qualitative description of the origin of the electromagnetic force. The photons cannot be directly observed because they are "virtual photons".

QUESTION:
I am currently writing a short story that entails time dilation for a person in space. If the average life of a person on earth is 250 years and that said person were to live in space via a space station or ship for about 75%-95% of their life, what's the amount of additional time would this person add to their life?

ANSWER:
According to clocks which traveled with that (average) person, she would live 250 years. I will neglect gravitational time dilation, just do the calculation for the usual special relativity time dilation due to speed. The answer will depend on the speed v of the space ship relative to someone on the earth. The time t' according to her clock elapsed when she returns to earth after a time t elapsed on earth is t'=t√[1-(v/c)²] where c=3x10⁸ m/s is the speed of light. You mentioned a space station. The ISS has a speed of about 4.76 mi/s=7660 m/s, so if she is up there for 80% of 250 y, 200 y as measured on earth, her clock will read, upon returning to earth, t'=200√(1-(7.66x10³/3x10⁸)²)=199.9999999348 y, 0.652x10^-7y=2.06 s younger than she would have been if not traveling. On the other hand, if she were on a spaceship going half the speed of light, t'=200√0.75=173 y, 27 years younger.

QUESTION:
For a bit of background, i am an adult with post secondary education who adores science but did not become a scientist because calculus is my nemesis. I am a science fiction writer, but don't let that scare you. This has nothing to do with anything i am writing or plan to write. I'm just really curious. I was thinking about E equals mc squared in the shower today and wondered why we think about trying to push mass along to the speed of light in the first place. In theory, couldn't we just convert a particle into energy with it's anti-matter pair, send it on it's way and then reconvert it at the other end. I'm aware this would be extremely dangerous due to the matter anti-matter pair, i just want to know if it's possible in theory.

ANSWER:
What is that you are trying to do? I presume you know that no particle can go as fast as the speed of light. But, when you push on it you are giving it energy, so the faster it goes, the more energy it has. This is clear since the correct way to write the famous energy equation you refer to is E=m₀c²/√[1-(v/c)²] where v is the speed of the particle and m₀ is the mass it has when it is at rest. So, the faster you get it going, the more energy it will have. In your "scheme" you start out with a particle-antiparticle pair at rest, say an electron and a positron which have an energy of 2m₀c² where m₀is the rest mass of an electron. The energy they create is a pair of photons with no mass but they still have between them a total energy of 2m₀c² and, in fact each will have energy m₀c² because they must come off in opposite directions with equal energy to conserve momentum and energy. So getting them back together will be a trick which could probably achieved with mirrors. But, I know no way that these two photons could be made to coalesce into an electron/positron pair but if that could be achieved, they would still have the total energy of 2m₀c² and be at rest, so you have gained nothing. Why is it that you think this would be dangerous? The total rest energy which gets converted to photons is really tiny and while the photons are of X-ray energies, there are only two of them and one X-ray photon is not the least bit dangerous.

QUESTION:
when heat is supplied to a solid body the spring-like bond between them turns into a string-like bond when change of phase takes place and when the liquid changes into a gas the spring-like bonds between them break so when condensation takes place how are those strings formed again and how are those string get converted back to spring when solidification takes place?

ANSWER:
It all boils (pun intended) down to kinetic energy. When the temperature of a collection of molecules increases, the average kinetic energy per molecule increases. Molecules have attractive forces between them and, if the temperature is sufficiently low, these forces can be modeled as springs. But, as the kinetic energy of the molecules increases, eventually these forces stop behaving like classical springs where the force increases linearly with separation but stretch out (or you can think of them as breaking) so that each molecule is no longer required to remain in place relative to its nearest neighbors, but not enough to escape altogether (your string picture, I guess). But eventually, as you continue adding energy, they get enough energy to escape from all the others. Cooling is just the reverse, condensation being clusters of gas molecules forming but not with spring-like forces between them, and solidification occuring as the molecules have less kinetic energy and become bound to only their nearest neighbors.

QUESTION:
My question is about space ship propulsion. Let us assume that we have an electromagnetic propulsion system, similar to particle accelerator like the Large Hadron Collider. We need to propulse a 100-ton ship with the velocity of 10,000 km/s. Energy needed for that equals: 50,000 * 10,000,000 * 10,000,000 = 5*10^18 joules. My question is -- is this model physically possible, given enough energy and tech advancement? Or it's just a magical concept, like releasing total energy of matter without antimatter?

ANSWER:
I see that you have calculated the kinetic energy of the ship. I do not believe that there is a word "propulse" but what I think you mean by "…propulse a 100-ton ship with the velocity…" is "…propel a 100-ton ship to the velocity…" But I think you are asking the wrong question. You could easily supply the ship with 5 J/s if you did so over 10¹⁸ s≈3x10¹¹ yr. So you really need to say how long you want it to take to give the ship this energy, that is, what power do you need to supply? For example, there are about 3x10⁷ in a year, so if you want to achieve crusing speed in a year, the power required would be (5x10¹⁸ J)/(3x10⁷ s)=1.67x10¹¹ W=167 GW. The largest non-hydroelectric power plant in the world, the Kashiwazaki-Kariwa Nuclear Power Plant in Japan, generates about 8 GW, so the notion of this ship achieving the speed you want in a reasonable time is, for all intents and purposes, impossible.

QUESTION:
How many seconds do you need to endure 3.7 g forces going at 56mph to feel effects of 10g-force at 224 mph for a duration of 60 seconds?

ANSWER:
If you are experiencing a 3.7g force it will never feel like a 10g force regardless of how fast you are going or how long the force is applied. There is one simple example I can give where you are going at a constant speed (not velocity) but experiencing a force. Shown in the figure is a popular carnival ride where you (the guy in the striped shirt) is moving in a circle of radius R with a speed v because the whole thing is spinning like a wheel. You feel like there is a force F pushing you outward; this is called the centrifugal force but what is really happening is that the wheel is pushing you toward the center (centripetal force). The magnitude of your acceleration is a=v²/R. If you want to express the force in gs, the force is a/g where g=9.8 m/s² is the acceleration due to gravity. Suppose R=4 m. Then if a=3.7x9.8, v=√(3.7x9.8x4)=12 m/s and you feel a force of 3.7g. Similarly, if a=10x9.8, v=√(10x9.8x4)=19.8 m/s and you feel a force of 10g. (You can express v instead as revolutions per second, rps, by f=v/(2πR) so f_3.7=0.48 rps and f₁₀=0.79 rps.)

QUESTION:
Just wondering... if matter and energy are constant in the universe doesn't that make time travel impossible because the traveller would bring new matter and energy in and reduce the matter and energy from where/when he came? or.... is where/when he came from just another part of the universe which would mean no gain or loss to either the place he went or from where he came? or... is time an illusion and past , present, and future just an all-present "now"?

ANSWER:
Well, this is an interesting idea. However, it cannot be right because we already have demonstrated that time travel to the future happens: a radioactive particle with some half life will last much longer if traveling fast than if at rest. The laws of physics currently prohibit time travel to the past. I think the trick here is, as hinted at in your last sentence, that you need to think of the universe as four dimensional and it is just fine if you take energy away from some "now" as long as you add it to a later "now". Incidentally, I would not talk about "matter and energy"; rather just say the total energy is constant since mass is just a form of energy.

QUESTION:
Hello, I'm not a student or anything I was watching a TED Talk video and it inspired a curiosity question. He was talking about antimatter accelerators and transverse seeing the universe at various speeds of light, and he said that if the ship could approach 99.99 percent the speed of light then we could reach the edge of the viewable Universe in 30 years relative to the spacecraft. However 13.9 billion years would have passed here on Earth.

My question is—how would a radio signal behave if this ship pointed a transmitter at the earth when it started this journey and broadcast a constant radio signal as it grew closer and closer to light speed traveling further away?

ANSWER:
Well, you got your numbers wrong somehow. 99.99% of c is not really that big for such a long distance. If you go at that speed, the distance (as seen by the ship) would be 13.9x10⁹√(1-0.9999²)=2x10⁸ ly. Since the speed of the ship is 0.9999 ly/yr, the time necessary would be 2x10⁸/0.9999≈2x10⁸ yr, not 30. I have seen this kind of estimate before and the situation being described was probably that the ship has a constant acceleration (as seen by the ship) equal to the acceleration of gravity (about 10 m/s²) essentially forever; so the speed (as seen from earth) gets to be much larger than 0.9999c. You can see the calculation of the speed of accelerating objects as seen from an inertial frame in an earlier answer. If you want a really detailed discussion of getting to the edge of the universe in a few decades, see this post by Dave Goldberg; his number is more like 99.9999999999999999998% of c!

But your question does not depend on the details of how far you went and how long it took you to get there. The radio signals being received on earth will be Doppler shifted to lower frequency, the shift getting larger as the speed gets larger. This is the same idea as the Doppler shift of a siren going away from you getting changed to a lower pitch. The expression for the shift in frequency is f_observed=f_source√[(1-(v/c))/(1+(v/c))]=√[(1-0.9999/(1+0.9999)]=0.0007f_source, a much lower frequency.

QUESTION:
How does a pebble get the energy to bounce up and hit a windshield?

ANSWER:
There are many ways, but most commonly the pebble is thrown up by the car ahead of you. You say "hit the windshield" but usually it would be more accurately described as the windshield hitting the pebble. The pebble will usually have a relatively small speed relative to the ground whereas the car may be going 70 mph.

FOLLOWUP (DIFFERENT QUESTIONER):
Follow up to one a friend asked recently. The rock or pebble on the road which is run over by a vehicle and when it is run over, the rock becomes energized somehow and lifts off of the hard road surface and usually finds my windshield as it bounces down the road. What gives the initial energy to the rock to make it lift off of the surface of the road when the wheel crosses over it?

ANSWER:
Imagine that you and the car in front of you both have a speed of v. You look at a rear tire of the car in front of you and what you see is the tire not moving either toward or away from you but spinning with a rotational velocity where each point on the circumference is moving with a velocity v in the direction tangential to the tire. Now, if a stone becomes stuck in the tread of the tire but breaks loose at some point it will leave in a tangential direction as shown in the figure. As you can see, that trajectory is likely to be coming your way!

QUESTION:
what is the physics behind: 6balls r placed in the centerof the u shaped curved ramp. if 1 ball is released from a certain height, another one ball at the other end is displaced.why?

ANSWER:
The same principles as apply to Newton's Cradle.

QUESTION OF THE WEEK 4/1/2018

QUESTION:
I apologize if this question seems a bit simple. I'm setting up a small malthouse, and I have a plastic holding tank (not insulated) that holds the water for the malting process. I'm trying to determine if I need to heat the tank in the winter or not. The tank holds 18928 liters. The water comes into the tank at 9 degrees C, and the ambient temperature around the tank is 17 degrees C. My question is, how long will it take the water in the tank to reach equilibrium with the ambient temperature.

ANSWER:
To answer this question it is necessary to know the material from which the tank is fabricated and its thickness and its geometry (mainly the area exposed to the ambient temp).

FOLLOWUP:
Okay, the tank is a cylinder with 102" diameter, and 154" high. The walls are 1/4 thick, high density polyethylene.

ANSWER:
The equation for heat transfer through a conducting barrier is

dQ/dt=(kA/s)(T_high-T)=3.17x10³(290-T)

where k=0.47 W/(m·K) is the thermal conductivity of high density polyethylene, s=0.25"=0.00635 m is the thickness of the barrier, A=43.15 m² is the area of this tank, T is the temperature of the water as it warms, and T_high=290 K. dQ/dt is the rate at which energy (heat) enters the tank.

The equation involving the rate of increase of the temperature of water to which heat is being added at a rate dQ/dt is

dQ/dt=mC(dT/dt)=8.51x10⁷(dT/dt)

where m=20.5x10³ kg is the mass of the water and C=4.19x10³ J/(kg·K) is the specific heat of water.

Combining the two equations, I find [dT/(290-T)]=3.73x10^-5dt . Integrating and solving for T I find T=290-8·exp(-3.73x10^-5t)=290-8·exp(-0.134t); the first expression is for t in seconds, the second for t in hours, and the temperature is in K. The final result is shown in the graph converting the Kelvin temperatures to ⁰C. Note that the temperature never (theoretically) reaches 17⁰C, but a day to a day and a half would probably be fine for your purposes. (By the way, the volume which I calculated and used was 20.5 m³, a tad larger than the volume you quoted, 18.9 m³.)

FOLLOWUP (DIFFERENT QUESTIONER):
You recently worked out a heat transfer/tank problem where you came up with: the equation "Integrating and solving for T, I find T=290-8�exp(-3.73x10-5t)=290-8�exp(-0.134t)". Is this e raised to (values), or is it 290-8^(-3.73x10-5 t). It's not clear about the intermediate steps? Using Wolfram on-line integration, I get: -log(290-T) + constant = 3.73x10^-5t. Just looking for clarity.

ANSWER:
exp(x) is standard alternative notation for e^x. The reason that an exponential function appears in the answer is that Wolfram's log(x) is the natural log, ln(x) in more common notation, and x=e^ln(x). So, if you have an equation of the form ln(x)=y and you want to solve for x, x=e^y. Also, the integrator you used is for indefinite integrals and you need to do an integration here from 282 to T. You also need to remember the properties of logarithms, viz., ln(a)-ln(b)=ln(a/b) and ln(a)=-ln(1/a).

QUESTION:
The gravitational potential energy of an object will be greater if its taken higher above the ground.When the object is released this energy will convert into kinetic energy.Thus energy conserves.But if that object is taken to high above the ground where gravity won't effect the object,what would happen to the potential energy?

ANSWER:
There is no place where "…gravity won't effect (sic) the object…", gravity gets weaker the farther you go but never zero. So it has potential energy, relative to the surface of the earth equal to the work you did on it. Even if you lift it a million miles, when you release it it will start falling toward the earth and, when it finally comes to the surface it has kinetic energy equal to its potential energy way out there. (This ignores the effects of all other objects.)

QUESTION:
My question concerns wind power. If one creates artificial wind by using fans would it be possible to generate an endless supply of electricity? Since the fan will create wind, that will be enough to generate maximum electricity from wind power and some part of generated electricity will be used for fans and the rest for other purposes

ANSWER:
The energy produced by the windmills would be way less than the energy consumed by the fans.

QUESTION:
If a wheel is rolling without slipping on an inclined plane, friction force is the one that provides the torque, why does the torque provided by friction increase as the angle of the inclined plane is increased yet the friction force decreases as the angle of inclination is increased?

ANSWER:
You are wrong about how f changes with θ. You are probably thinking that f=μN=μmgcosθ, but this is static friction and the correct expression is that f≤μN, so in this problem f is whatever it has to be in order to satisfy Newton's laws, as long as it does not exceed μN, in which case the wheel will not roll without slipping. This problem has three unknowns, so you must generate three equations:

-f+mgsinθ=ma
N-mgcosθ=0
fR=Iα=Ia/R=mRa

To simplify the algebra I have assumed that the wheel is a hoop of radius R, so the moment of inertia is I=mR². So, from equations 1 and 3, f=ma=m(gsinθ-ma) or a=½gsinθ and f=½mgsinθ. You can see that f increases with θ. The third unknown is easily found, N=mgcosθ.

You can also calculate the largest angle for which the hoop can roll without slipping since the maximum static friction you can get is f_max=μN=μmgcosθ_max=½mgsinθ_max. Solving, θ_max=tan^-1(2μ). For example, if μ=0.5, θ_max=45⁰.

QUESTION:
Okay, you have two objects in space with the same gravitational pull. Those objects can not be moved. Those objects are close enough where right in the middle of them one gravitational ends the other starts. What would happen to let's say a human in the middle? Or another oject?

ANSWER:
It is incorrect to say that "…one gravitational ends the other starts". The gravitational force of each object extends all the way to infinity. However, there will be a spot where the gravitational field of one is equal and opposite the field of the other. At this point, an object will remain at rest but the equilibrium will be unstable—give it the slightest nudge and it will "fall" in that direction. Technically, this is only approximately true for an object very much smaller than the distance between the two objects. If a human were at that point he would feel forces from each trying to stretch him.

QUESTION 1:
A star's mass is almost as much as black hole but the star is not a blackhole. If it moves its mass would increase relative to the observer and if it moves fast enough its mass will be as much as a blackhole. So if it moves fast enough what will the observer see, a blackhole or a star?

QUESTION 2:
If a mass moves relative to an observer its mass increases relatively. But its real mass does not increase. So why can't something travel faster than light?

ANSWER:
Both these questions are about mass increasing as speed increases. However, as I have emphasized in earlier answers, modern physicists do not really ever think of the mass of a moving object, but rather of its linear momentum and its energy. When we talk about mass we are normally talking about the object's rest mass, the inertia it has when it is at rest. If you read earlier answers (here and here on the faq page) you will understand this better. Often the mass of a moving object is talked of as increasing with speed because mass is a measure of how much inertia (resistance to acceleration) an object has and it certainly gets harder and harder to accelerate an object as it gets closer and closer to the speed of light. But, questioner #2, there is no such thing as "real mass" because if you think of mass as inertia, it is something which we say is not an invariant quantity, it depends on the motion of the particle. And, questioner #1, the only thing that matters as to whether a star becomes a black hole is the mass it has in its own rest frame, in other words its rest mass.

QUESTION:
How does matter "transition" to energy and vice versa without consuming energy to complete the process?

ANSWER:
There are two kinds of reactions, exothermic and endothermic; the former releases energy when it occurs, the latter must have energy added to happen. What matters is that in an isolated system energy must be conserved. That is, the energy you start with must be equal to the energy you end up with plus the energy you added. If the energy you add is negative, the reaction is exothermic; if positive, endothermic. Nuclear fission of heavy nuclei is an example of an exothermic reaction. A uranium nucleus will spontaneously split into two lighter nuclei whose total mass is smaller than the mass of the original nucleus; the missing mass energy shows up in the kinetic energy of the fragments. But, lighter nuclei will not spontaniously fission: if you take an alpha particle (a helium nucleus), composed of two neutrons and two protons and break it apart into two deuterons (hydrogen-2 nuclei) each composed on one neutron and one proton, you will end up with more mass than you started with because you had to do work (add energy) to cause it to happen. So the answer to your question is that sometimes you must consume energy for a transition, other times you get energy out.

QUESTION:
My friends and I were messing around one night and posed the question "could a falcon break a human neck" and we tried to solve it but couldn't get anywhere with it (we are all theater majors not physicists lol) Could you solve that for me?

ANSWER:
A falcon weighs more than 1.5 lb and can fly at speeds exceeding 200 mph. I think no fancy physics is required—surely a 1.5 lb projectile hitting your head at 200 mph could easily break your neck. (Wouldn't be too pretty for the falcon either!) Just to make it a little more rigorous, I calculate the average force exerted on the falcon as he (and the head) move a distance d before stopping would be about 25,000/d lb if d is measured in inches; Newton's third law says that the head would experience an equal and opposite force. So the average force for the head moving 3 inches would be about 8,000 lb. Information I could find indicated that 1000 lb should be adequate to break a neck but that the actual force would depend on circumstances—the individual, where the force was applied and its direction, and whether the neck was initially bent or not. I think that 8,000 would do the trick for almost all situations.

QUESTION:
Do objects falling from the sky reorient themselves to reduce friction with air (even if just a tiny bit)?

ANSWER:
It depends on the object. For example, a parachute would not work if it reoriented itself to minimize drag. Consider a simple parachute having a rigid disc and the load hanging from the center. Orienting such that the disc were in a vertical plane would reduce the air drag but it does not happen. Some things, though, will reorient to minimize drag. For example, a cone will orient pointing down because if it is falling sideways there is a torque trying to orient it pointing down because the drag on the pointy end is smaller than on the blunt end.

QUESTION:
Flat earthers believe gravity doesn't exist and that the earth is accelerating up at 9.8m/s^2 and the earth is 6,000 years old. I stated that, at that rate the earth would be traveling at approximately 6,185 x the speed of light, which is impossible. They introduced a formula that suggested that I was wildly incorrect. I have a degree in electrical engineering but took a different career path so I have admittedly lost my edge in mathematics and projectile motion so I was a bit lost by their equation. Am I close to correct or wildly incorrect?

ANSWER:
I hate to do anything which endorses the views of flat earthers, and their views are absolute nonsense here, but it is possible to maintain a constant acceleration for an indefinite time (as observed from "their earth") without exceeding the speed of light. Of course you need to define a frame relative to which the velocity of the earth is measured; let's put you on an inertial frame where the earth was at rest 6000 years ago. The catch, of course, is that you must be constantly exerting some force on the earth to keep it accelerating, F=Mg where M is the mass of the earth; who is exerting that force? Your analysis assumes (I presume) that you will also see the earth accelerating away from you with a=g but, you will see the acceleration getting smaller and smaller as time goes on. The graph above shows the velocity and acceleration which you would observe. This comes from an earlier answer which, along with links to even earlier answers, fully describes how to handle acceleration in special relativity. In this graph, a₀=g. Using the result from the earlier answer, I find that the speed you would measure after 6000 years is given by v/c=0.999999974, very close to but not larger than c; the folks on the "earth" would all the time see a constant acceleration of g. If you are actually going to discuss this with a flat earther, ask her where the force providing the acceleration is going to come from? God?

QUESTION:

In need to pull a 25,000 lbs building across a field, all the weight will be on runners in each side 3"x130' what does this do to increase the pulling weight

ANSWER:
"Pulling weight" really has no meaning; I assume you mean the force necessary to move the building with constant speed. This, I would assume, you would want to minimize, not "increase". I assume that your runners are fixed on the ground and that the building will slide on them. How hard you have to pull would depend on the nature of the surfaces—steel on steel, for example would be slipperier than wood on wood. Better yet, grease the surface. Still better, put rollers between the runners and the building.

QUESTION:
What would be the rate of acceleration of a falling object subjected to the sun's gravity? Ex: I drop a two-pound diamond a mile up from the sun's surface. How long does it take to hit the ground?

ANSWER:
Well, the sun does not have a well-defined surface and its atmosphere is very turbulent. Even if the environment did not destroy your diamond, drag from the gases and plasma would be very important. But, I think you want an imaginary scenario where the sun is taken as a solid uniform sphere with a radius of about R≈7x10⁸ m and mass of about M≈2x10³⁰ kg and no atmosphere. In that case the gravitational acceleration would be g≈GM/R²≈272 m/s². Then t≈√(2h/g)≈3.4 s; I used h=1 mi≈1600 m and G=6.67x10^-11 N·m²/kg².

QUESTION:
why the earth rotate on its own axis? which source generated the torque and how?

ANSWER:
This angular momentum was not generated by a torque, it was already there as the earth formed. When the sun was young, there was a disk of matter all around it. The chunks of rock and dust and gas began clumping together to form the planets. But since all their velocities were basically in a plane (the disc), each capture brought angular momentum which was perpendicular to the plane so the earth ended up, when it finished adding this matter, with angular momentum perpendicular to the plane.

QUESTION:
A small body starts falling onto the Sun frame.the distance equal to radius of the earth's orbit. The initial velocity is zero in the heliocentric frame.find how long the body will be falling using Kepler laws??

ANSWER:
You will find the answer in the FAQ page. Most pertinent to you would be a particular earlier answer.

QUESTION:
Does g=9.8 or -9.8?

ANSWER:
Lower case g is usually used to denote the magnitude of the acceleration of gravity at the surface of the earth. Since the magnitude of a vector is a positive-definite number, g is then positive. If we call the vertical direction the y-axis, then the y-component of the acceleration of an object in free fall is a_y=-g if +y is up or a_y=g if +y is down.

Q&A OF THE WEEK, 4/21/2018

QUESTION:
Glad I found your site, my name is Ben, and this question is something I'm working on professionally, not homework. I use Autodesk Inventor's Dynamic Simulation to model collisions and some of the results for a specific simulation don't seem consistent. So, a colleague suggested we simplify the problem and work the numbers out by hand, but we can't figure it out. Here is a diagram:

A rod (blue) of length 2m and mass of 1kg can freely pivot about it's center (orange dot) which is connected to a frictionless track (purple) running in the Y direction. The rod is positioned at 45 degrees to the track, and a ball of mass 1kg travelling at 15 m/sec strikes the rod at 0.5m from the rod's center. Collision is perfectly elastic, friction is zero. How can I determine the final velocity of the rod along the track, and what is it's rotational speed? And more importantly, I need to know what effect the rod angle has on these two answers. Then I can solve and plot a chart for all angles from 0 to 90 and compare with my Inventor results. I'm really having trouble with the moment of inertia part, being that it's not struck in the center, or the end.

ANSWER:
There are important errors with the problem as you state it. The first obvious error in this problem is that the direction and magnitude of the final velocity of the ball is impossible. If the ball carries off all the energy it came in with, the rod must end up with no energy. And, conservation of linear momentum in the y direction (not conserved in the x direction) demands that the speed of the center of mass (COM) of the rod along the rail would be 15 m/s meaning it would not be rotating. And, the angular momentum relative to the COM of the incoming ball is equal and opposite that of the outgoing ball, so the rod would have to be rotating to conserve angular momentum. So I thought to simply redraw the picture but with the final ball velocity having unknown components (2 unknowns). Two other unknowns are the final speed of the COM and the angular velocity of the rod about the COM. However, I only can see three equations: conservation of energy, linear momentum (only y), and angular momentum. I am still pondering the question, but there is too little information or I am missing something.

CONTINUED ANSWER:
Since the questioner is ultimately interested in a more general solution than the special case, I will set it up generally from the start and apply that solution to his initial question; I will, however, set m₁=m₂=m as soon as I have written the most general solutions to simplify the algebra after setting up the problem. As noted above, the problem begins with three conservation equations:

conservation of linear momentum in the y-direction:
m₁v₁=m₁v_2y+m₂u₂
conservation of energy:
½m₁v₁²=½m₁v₂²+½m₂u₂²+½Iω²
conservation of angular momentum:
m₁v₁Ssinθ=Iω+m₁v_2ySsinθ-m₁v_2xScosθ

The moment of inertia of a thin rod about its COM of mass m and length L is I=mL²/12. Therefore, if m₁=m₂=m, the three conservation equations are:

v₁=v_2y+u₂
v₁²=v₂²+u₂²+L²ω²/12
v₁Ssinθ=L²ω/12+v_2ySsinθ-v_2xScosθ

There are now three equations and four unknowns, v_2x, v_2y, u₂, and ω. To generate a fourth equation, consider the impulse J delivered by the rod to the ball: J=Δp=m(v₂-v₁)or J_x=mv_2x and J_y=mv_2y-mv₁. But, the impulse must be normal to the surface of the rod, so tanθ=|J_y/J_x|=(v₁-v_2y)/v_2x. The fourth equation is

v_2x=(v₁-v_2y)cotθ.

I first solved these four equations for θ=45⁰, L=2 m, S=0.5 m, and v₁=15 m/s, the conditions specified by the questioner; note that, because it is a quadratic equation we are solving, we get two solutions:

v_2y=(15, 8.33) m/s
v_2x=u₂=(0, 6.67) m/s
ω=(0, 14.15) s^-1=(0, 2.25) rev/s.

The meaning of the first (trivial) solution is that there was no interaction between the rod and the ball (you missed!). Another situation for which we can check the solution intuitively is θ=90⁰, for which v_2x must be zero:

v_2y=(15, 4.1) m/s
v_2x=(0, 0) m/s
u₂=(0, 10.9) m/s
ω=(0, 16.35) s^-1=(0, 2.6) rev/s.

Finally, the general solutions are:

v_2y=v₁(A-1)/(A+1) where A=cot²θ+1+[12S²/(L²sin²θ)]
v_2x=(v₁-v_2y)cotθ=2v₁cotθ/(A+1)
u₂=(v₁-v_2y)=2v₁/(A+1)
ω=[12(v₁-v_2y)S/(L²sinθ)]=[24v₁S/((L²sinθ)(A+1))].

The plots requested by the writer are shown to the right.

ADDED NOTE:
An alternative way to specify the vector v₂ is to specify its magnitude, v₂, and the angle φ it makes with the +x axis:

ACKNOLEDGEMENT:
Thanks to "haruspex" at Physics Forums for helping me realize how to get the desperately sought fourth equation!

QUESTION:
We know that mass is relative.So why can't energy be relative?

ANSWER:
Energy is relative, even classically. An object with mass m and speed v has a kinetic energy classically of K=½mv²; but if you are running with speed v along with it, it has zero kinetic energy. In special relativity, the kinetic energy relative to you is K=m₀c²[(1/√(1-v²/c²)-1] where v is its speed relative to you; if you start moving, the kinetic energy of the object changes relative to you.

QUESTION:
Because the earth is rotating, is there a centrifugal force that is acting against gravity. If the earth stopped rotating, would a mass on the earth's surface weigh more?

ANSWER:
Yes, there is a centrifugal force (except at the poles). A scale you are standing on would indeed read more if the earth stopped rotating, but the increase would be too small to notice. (It is customary to call "weight" the force which the earth's gravity exerts on something, so in that context you would not weigh more even though the scale would read more.)

QUESTION:
what is force of attraction and repulsion between two charges when placed in oil

ANSWER:
The magnitude of the force between point charges q₁ and q₂ separated by a distance r in a medium with permittivity ε=ε_rε₀ is F=q₁q₂/(4πεr²). The answer would depend on what kind of oil it is.

QUESTION:
What would the kinetic energy of a 1kg mass traveling at 50% the speed of light? Or a 5 kg mass? Need to know for a book I'm writing

ANSWER:
The expression for kinetic energy of an object with speed v and rest mass m0 is K=m₀c²[(1/√(1-v²/c²)-1] where c=3x10⁸ m/s is the speed of light. For v/c=½ and m₀=1 kg, K=1.39x10¹⁶ J. K for a 5 kg would just be five times as large.

QUESTION:
I understand the practical aspect of buoyancy of a helium balloon and the Archimedes principle. But I don't understand why the balloon goes up. I understand that if it were in water, the the force on the bottom of of a 12 inch balloon would be approx 0.5 psi more than the top of the balloon and there would be a force upward. But in the atmosphere, the pressure differential would be very small. So why does it go up?

ANSWER:
It is exactly the same as in water, but the change in pressure from the bottom to the top of the balloon is much less. A balloon filled with air will have a buoyant force less than the weight of the baloon so it will fall rather than rise. Filled with hydrogen or helium, though, the balloon will rise if the weight of the balloon plus contents is smaller than the weight of the air it displaces; of course a lead balloon will not rise even if filled with helium. A hot-air balloon rises because if you heat air it expands and becomes less dense.

QUESTION:
i am using 12.5HP pumping motor. How much time is required to to fill the water in 1 Hectare pond, with average depth of 1m, its rectangular in shape?

ANSWER:
There is no way to calculate this unless you specify where the water is being pumped from. And even if you did this, a calculation would be an approximation because you would probably not account for viscosity of the water or efficiency of the pump.

QUESTION:
I think this question applies to the theory of relativity well the motion part of it anyways, if an object is sitting completely still on earth would its speed by measured by the revolution of the earth on its axis or would it just be considered to have 0 speed?

ANSWER:
As I have probably said in more than 20 previous answers, specifying a velocity without specifying the frame in which it is measured is meaningless. The object is at rest with respect to the surface of the earth. The object moves in a circle with a period of 24 hours relative to the axis of the earth. The object has a velocity 60 mph east relative to a car going 60 mph west. Etc., etc.

QUESTION:
A passenger is seated in a bus at rest. If the bus starts to move without his knowledge, why is he pushed backwards?

ANSWER:
He is not pushed backwards, it just feels like he is. What actually happens is that the back of his seat applies a forward force on his back which results in his accelerating with the bus.

QUESTION:
Me and my brother are having a debate about running on a treadmill vs running on a track. Not counting mental stuff like learning how to pace i feel that the only real difference is wind resistance. He seems to think inertia comes into play differently for each. I feel like once you are on the treadmill and it starts moving that the treadmill track is your inertial frame of reference so any change in inertia would require just as much energy as it would on an outdoor track. He seems to think that since you don't let the treadmill ever move you actually that you never have to work against inertia of rest while you're accelerating or decelerating on a treadmill. I mean if the treadmill was already moving at 10mph and then you just jumped on it i could see his point but not if you start off standing on the treadmill while its at rest and accelerate along with it.

ANSWER:
In terms of simple introductory physics, you are correct. However, there are often subtle differences between simple physics and the real world when applied to very complicated systems like the human body. Your brother is also wrong because he is just trying to explain any differences in terms of simple physics also. In fact, there are many differences between running on a track and running on a treadmill due to biomechanics. A good article to read is a post on the RunnersConnect blog. Or just google treadmill vs. track.

QUESTION:
It takes 375 Joules of energy to crack a bone. Moreover, a baseball player hits a ball. The speed of the ball off the bat is 90 mph. The ball the player hit has a weight of 5 oz. If this ball were to hit someone in the head would it crack their skull? What other injuries could be expected?

ANSWER:
This question makes no sense to me. Surely it is harder to crack a femur than a skull. Besides, you would specify a force needed to break the bone, not an energy. The kinetic energy of a 5 oz baseball at 90 mph is about 145 J.

QUESTION:
One of the ways to define horsepower is 550 lb-ft/sec. My understanding is that means 1 hp is required to maintain a velocity of 1 ft/sec straight up with a weight of 550 lbs. However wouldn't it take more than 1 hp to lift a 550 lb weight that is sitting on the floor motionless 1 foot in 1 second? How would you calculate the hp needed to lift 550 lbs from a dead stop to a height of 1 foot in 1 second?

ANSWER:
"Maintain a velocity" means just that—the object begins, ends, and always has that velocity. So your worrying about how it got that velocity in the first place is really irrelevant to this example.

Let's get straight what all these units are.

A pound (lb) is a unit of force. A net force on an object accelerates it. For example, the weight of an object is a force on it downward and if you exert an equal force upward the object will move upward (or downward) with a constant velocity, any velocity.
A foot (ft) is a unit of length.
A foot-pound (ft-lb) is a unit of energy. If one lb is exerted over a distance of one ft, the energy delivered is 1 ft-lb.
A second (s) is a unit of time.
1 ft-lb/s is a unit of power and it measures the rate at which energy is being delivered. Other units of power are the horsepower (hp) and the Watt (W). 1 hp is 745.7 W.

Your question about how much hp it would to take to move an object up 1 ft starting and ending at rest is meaningless because there are an infinite number of ways to do that but you could not do it with a constant power. For example you could deliver a lot of power for 0.1 s, ending up with a speed (at 0.1 s) which would be just right to get the object up to 1 ft in an additional 0.9 s before falling back down; or you could do the same thing with a smaller power to 0.2 s, etc. Regardless of how you achieve this for a 550 lb object, the total energy required is 550 ft-lb. So, if it takes 1 s, the average power is 550 hp.

QUESTION:
How does a credit card reader work? Specifically, when you slide your card in one?

ANSWER:
See an earlier answer.

QUESTION:
If two dice were floating in space about a centimeter apart, what would eventually happen to them and why?

ANSWER:
They would be attracted to each other and, after a time, stick together. The details are given in an earlier answer to a question very similar to yours except the dice were separated by 10 cm there. If you want to calculate the time your dice would take to come together, you can follow the calculation given there changing the numbers. I find the time to be surprisingly small.

QUESTION:
This question concerns angular momentum as related to motorcycles. If a motorcycle rests atop a trailer, but sits on a roller system, the bike will stand upright without any supports or tethers if the wheels are rolling (throttle is locked in the "on" position and wheels are spinning). If the trailer itself travels forward in a straight line, the bike should remain upright. But if the trailer takes a sharp turn, what happens? Does the bike fall, or does it instead do what a rider does to achieve a turn-countersteer and remain upright?

ANSWER:
I have waited a long time to answer this question because I am bothered by the way the problem is stated. First of all, if the throttle is locked on, only the rear wheel will be spinning, so we can discuss the problem by looking only at the wheel. It is certainly correct that if the truck goes straight the wheel will continue running upright (assuming that the center of gravity of the bike is in the vertical plane passing through the center of gravity of the wheel). Imagine the bike and rollers to be mounted on a big "lazy Susan" the base of which is bolted to the truck bed. So, if a north-bound truck turns to the west, the angular momentum, experiencing no torque, will remain constant and continue pointing in the same direction (originally either east or west). Viewed from inside the truck it will appear that the whole bike rotated through 90⁰ relative to the truck.

Now, if the rollers are attached to the truck bed, when the truck turns the rollers turn and the wheel, trying to not turn, will come off the rollers at some point. We first need to understand the physics relationship between the torque and the angular momentum of the wheel. The rotational form of Newton's second law is τ=ΔL/Δt, torque equals the time rate of change of the angular momentum. The first figure shows the wheel, as seen from above, turning through some small angle which results in a change of angular momentum from L₁ to L₂ and ΔL=L₂-L₁. So the torque which you must apply to make it turn this way is in the direction of ΔL.

If you think you can just steer it as if it were not rotating, you would fail. The second figure shows what would happen if you try to steer it like your intuition would have you do it by exerting a force like F in the figure. The torque points up and so the wheel would not turn but lean in the opposite direction from the way you would lean on the bike if you were riding it and making a turn. You can find some videos showing this by googling gyroscope in a suitcase video.

If you want the wheel to turn with the truck, you need to have a torque which causes that. One way I thought of to achieve this was to have strings attached from the axles on each side and the truck bed below. These need to have no tension on them when the truck is going straight. When the truck turns as indicated, the tension in the string on the right-side string will be bigger than the other side and there will therefore be a net force N on the wheel. This results in a torque in the horizontal plane which will cause a change in angular momentum in the direction consistent with the wheel turning with the truck. Two strings are needed because the truck might turn either left or right.

QUESTION:
I wonder, how step by step, is parallax method used to find distance of a star? When you find the parallax angle, then it's easy to find distance using tangent formula or other trigonometric formule. But how is that parallax angle exactly determined? In some online articles, like in Wikipedia or Sky and Telescope they show such pictures. But in this picture, only distance between A to B is known. Distance between A to Eros or B to Eros is unknown. Angle of A and B are unknown. So we can't know angle of p with so little information. What we see only is a star on the sky do change its position (like you see right side of the picture) slightly (some millimeters) every 6 months. How do astronomers determine from that motion the parallax angle of the star?

ANSWER:
The angles A and B are not unknown. These are measured by the direction in which the telescopes at those locations are pointed. Once you know those angles you know the angle p; and if you know p and the distance AB you can also calculate the distances from A and B to Eros.

QUESTION:
Can you explain the differences between circular motion & rotational motion?

ANSWER:
Rotation normally refers to an object spinning about some fixed axis; e.g., the earth rotates about the axis which passes through the north and south poles with a period of 24 hours. Circular motion refers to the motion of an object which moves in a circle; e.g., a race car on a circular track is in circular motion. If the circular motion is with constant speed it is called uniform circular motion; e.g. a point on the equator moves in a circle whose radius is the radius of the earth with constant speed. More general than circular motion is revolution; e.g., the earth revolves around the sun in an elliptical path once a year.

QUESTION:
Is there a formula for calculating the side-ways deflection wind has on a lawn bowl(over and above the bias deflection ) running at 12 s, the time a bowl takes from delivery to stop over a 26 m distance over bowling green grass?

ANSWER:
Once again, doing Ask the Physicist has led me to learn something new. I never really knew anything about lawn bowls other than it is done on grass and rolling balls are involved. For the benefit of others who are ignorant of the game, let me summarize by describing the ball. (A good article on the physics of lawn bowls balls can be found here.) The ball is not a sphere but rather an oblate spheroid which makes it sort of like a door knob but not so extremely flattened; but it is slightly more flattened on one side of the ball than on the other which results in a center of gravity being displaced to one side of the equatorial plane as shown in figure (a). This results in a tendency for the ball to curve left if it is rolling the angular velocity shown in the figures; this motion is the "bias" referred to by the questioner which I am to ignore. When rolling in the x direction (figure (b)), there is a frictional drag force called, rolling friction D, which opposes the motion (v) and eventually brings the rolling to a halt. If there is a wind, there is a force W due to the wind which tries to make the ball roll to the right (figure (a)) but if it does roll, there will also be rolling friction trying to keep it from rolling. In order for the wind to have any effect at all, it is clear that we must have W>D; if this is not the case, there will only be static friction in the y direction which will be equal and opposite to W. A lawn bowls ball has a mass of about m=1.5 kg and a radius of about R=6 cm=0.06 m.

To get the equations of motion for the x and y motions, we first need expressions for D and W. The rolling friction may be expressed as D=-μmg where μ is the coefficient of rolling friction and mg is the weight of the ball. The force due to the wind may be approximated as W≈¼AV² where A=πR² is the cross sectional area of the ball and V is the speed of the wind; this approximation is only correct if SI units are used. The equations of motion in the x-direction are

D=-μmg=ma_x ⇒ a_x=-μg
v_x=v₀-μgt
x=v₀t-½μgt².

Here t is the time and v₀is the speed of the ball at t=0. If the ball is rolling in the y-direction because of the wind, the equations of motion are:

W=¼AV²-μmg ⇒ a_y=(¼AV²/m)-μg
v_y=[(¼AV²/m)-μg]t
y=½[(¼AV²/m)-μg]t².

It should be noted that if (¼AV²/m)<μg, these equations imply that the ball will accelerate opposite the direction of the wind, obviously not correct; hence the wind will have no effect on the ball if V<√(4μmg/A). In that case, a_y=v_y=y=0.

So, having found the general solutions, let us now apply the solutions to the specific case from the questioner. We are told that when t=12 s, v_x=0 and x=26 m. With that information you can solve the x-equations to get v₀=4.32 m/s and μ=0.037, reasonable values compared to numbers in the article I read. The area is 3.14x0.06²=0.0113 m². The first question we should ask is what is the minimum speed of the wind to have any effect at all: V_min=√(4x0.037x1.5x9.8/0.0113)=13.9 m/s=31 mph=50 km/hr; this is a pretty stiff wind, so the wind probably has no effect on bowling under normal conditions. So, just to complete the problem, consider V=15 m/s=34 mph=54 km/hr.

v_x=4.32-0.363t v_y=0.0612t
x=4.32t-0.182t² y=0.0306t²

The trajectory during the 12 seconds is shown in the graph below; after 12 seconds the ball will continue accelerating in the y direction.

So the bottom line is that unless you are playing in a gale-force wind, the wind has no effect on the ball if the wind has no component along the original direction of the ball (which I have called the x-axis). You can tell if wind makes a difference by simply setting the ball on the ground--unless the wind blows the ball away, you need not worry about its effect. If the wind is blowing in the +x or -x direction, that is a whole different thing, but the questioner asked for the sideways deflection.

ADDED THOUGHTS:
This question continues to intrigue me and I have carried my investigation further. The question originally stipulated "over and above the bias deflection" so my whole discussion totally ignored the fact that the ball, owing to its off-center center of mass, will curve. At the very end of my answer I noted that if the wind is not perpendicular to the path of the ball, it would be a different story; indeed for a spherically symmetric ball I showed that, except for very strong winds, a wind perpendicular to the path has no effect at all. However, for an actual lawn bowls ball, the path curves to where a wind in the y-direction might have a significant component along the path. I have calculated (graphed below) the x and y positions of a realistic path with no wind using equations (10) and (11) of the article referred to above. To do these I used all the numbers used above (R, m, A, v₀, μ); I used the moment of inertia for a solid sphere ( I₀=I_cm+mR²=(7/5mR²)) and chose the COM off-center distance to be d=1 mm. As you can see, the curving is substantial, carrying the ball about 4 m from its original direction. You can see that now a wind of any magnitude can have an effect on the trajectory. The angle φ which the tangent to the trajectory is given in the article as φ=(2/p)ln(v₀/(v₀-μgt)) where p is a constant also given in the article. As can be seen, once the trajectory leaves the x-axis the wind contributes with the component of its force along the trajectory; this has the effect of reducing the effect of the frictional force causing the ball to slow down less rapidly. However, this is now like having a time dependent force of friction which, I believe, will lead to equations of motion which will not have an analytical solution but would have to be solved numerically.

QUESTION:
How can rockets move in space. Because they have nothing to push against won't trust be useless. Like for example the reason the rocket gets out of the atmosphere is because of the thrust pushing against the earth then on the air in the atmosphere and if space has no air how can rocket move in space.

ANSWER:
The reason is that your statement that "the thrust pushing against the earth then on the air" is the reason the rocket accelerates is totally wrong. The reason a rocket accelerates is that hot gas is expelled out the back at a high velocity. The reason this propels the rocket can be illustrated by the following example: an astronaut is floating in empty space and happens to be holding a bowling ball; she throws the ball and the result is the balls moves away and she recoils so she is moving also. This is called conservation of linear momentum—if the ball has 1/20^th as much mass as she does, she recoils with 1/20^th the speed the ball has. The rocket is not "pushing against" anything.

QUESTION:
Earth moves around sun at 67000mph. A rocketmoving in same direction as earth at 25000mph isnt it actually moving at 25000 plus 67000 mph?

ANSWER:
There is no such thing as "actually moving". The only way a stated velocity has any meaning is if it is relative to something else. If the rocket has a velocity of 25,000 mph relative to the earth, then its velocity relative to the sun is 92,000 mph if the velocity of the earth relative to the sun is 67,000 mph.

QUESTION:
This is something I remember discovering when I was younger. If i would take a small cylindrical object (like a AA battery) , set it on a flat hard surface, then using my fingers I would apply downward pressure on the edge of the battery. This would create backspin on the battery but also shoot the battery across the floor. When the battery's finally stopped moving forward it would spin rapidly on its axis until stopping completely. Could you explain the physics behind movement of the battery?

ANSWER:
Problems like this are standard in intermediate-level classical mechanics. Round objects can move translationally on a horizontal surface by rolling without slipping or by sliding. The two classic extremes of the slipping scenario are a skid where the object is initially not spinning at all (e.g. a bowling ball begins by approximately not rotating but may end up rolling without slipping), or is at rest but spinning (like "peeling out"); in both cases, the problem is usually to find the time elapsed or distance traveled before rolling sets in. In your case, the object is initially both translating horizontally and rotating with a backspin. What will happen depends on the initial conditions—the initial speed and the initial angular velocity; also the properties of the object will also matter—its mass m, radius R and shape, and the coefficient of kinetic friction μ. Three possibilities are that

it will reverse directions and come back still spinning and slipping, eventually stop slipping and roll (what I think you are remembering),
it will stop slipping and continue rolling in the initial direction, or
it will stop dead.

In the figure there are three forces on the object: its weight mg, the normal force N up from the floor, and the frictional force f which the floor exerts on the sliding object; the object begins with velocity v₀ and angular velocity ω₀. The mass of the object is m and its radius is R. The friction slows down both the velocity and the angular velocity. Sliding ceases when v=-Rω (see footnote*). If v₀ is very small and ω₀ is very large, possibility #1 will happen; if v₀ is very large and ω₀ is very small, possibility #2 will happen; if v₀ and ω₀ are just right, possibility #3 will happen. I think that this gives you a good qualitative overview of the physics of the problem.

For those interested in the quantitative solution, I will give it here for a uniform solid cylinder. For the translational motion the two equations (taking +x as to the right, +y up) are -f=ma and N-mg=0; since f=μN=μmg we can write a=-μg. For rotational motion about the center of mass (choosing the direction it is initially spinning as positive), only the frictional force exerts a torque, so -fR=-μmgR=Iα where I is the moment of inertial about the center of mass and α is the angular acceleration. For a cylinder I=½mR², so α=-2μg/R. So, we can write the equations for the velocity and angular velocity as functions of time t: v=v₀+at=v₀-μgt and ω=ω₀+αt=ω₀-2μgt/R. Now, the time when slipping ceases will be when v=-Rω; solving, t=(v₀+Rω₀)/(3μg). Finally, put this into the equation for v to get v(t)=(v₀-Rω₀)/3. So if Rω₀>v₀, possibility #1 will happen; if Rω₀<v₀, possibility #2 will happen; and if Rω₀=v₀, possibility #3 will happen.

*The condition v=-Rω is a little tricky. The negative sign is because of my choice of the positive direction for ω which is opposite that which would be the case for no slipping.

QUESTION:
Do cars need friction to move?

ANSWER:
If they are not moving, they need friction to move. If they are moving, they need friction to stop. (I have assumed that you mean that you control things from inside the car. Somebody outside the car who was anchored to the ground an held a rope attached to the car, for example, could move the car. Also, if the car had a rocket engine it could move.)

QUESTION:
Sometimes gravitational force is proportional to r^2 and sometimes to 1/r^2. What's the reason?

ANSWER:
The gravitational force depends on the way that mass is distributed. For example, if you have two spherically symmetric spheres of mass, like the earth and the moon, the force between them is proportional to 1/r². If you are inside a sphere which has its mass uniformly distributed inside, the force you would feel is proportional to r. If you are inside a sphere which has its mass distributed so that the density varies linearly (e.g., the density at the surface is twice as large as the density halfway to the center), the force you would feel is proportional to r².

QUESTION:
Can you please explain why putting water into a bottle AFTER putting in baby formula does not give the correct exact amount of water necessary for proper amounts. E.g,the instructions on the formula tin state 1 scoop of formula for 30mls of water

ANSWER:
Because the formula occupies volume also. So when you put the formula into the bottle and then fill the bottle up to the 30 ml level, you will have less than 30 ml of water. If you measure the 30 ml of water before adding it to the bottle, it should make no difference. (Although, sometimes if you are mixing something dry with something wet, how easily the dry dissolves depends on the order of mixing.)

QUESTION:
If a weight of 7.5 kg is falling 0.6 m and land on the head of somebody, I understand that the energy at impact will be 44.1 J. But how can I translate this in g as it is the value reported by accelerometers? In other words, if the person has an helmet with an accelerometer, which value in g the instrument will register?

ANSWER:
When you measure an acceleration you are measuring a force, not an energy. It is true that a 7.5 kg mass has a kinetic energy of about 44.1 J if dropped from a height of 0.6 m (mgy=7.5x0.6x9.8), but that does not tell you the average force exerted by the mass as it came (accelerated) to a stop. The speed which the mass has when it hits is v=√(2gy)=3.43 m/s; if it stops in a time t, then the average force over that time is F=mv/t=25.7 N. For example, if t=0.1 s, F=257 N. To convert this to g-force, the weight of the object is 7.5x9.8=71.3 N so the force in gs is F=257/71.3=3.6 gs.

QUESTION:
How would you convince a relative who had not studied physics that the international space station is really a falling body?

ANSWER:
My favorite example is Newton's mountain. The idea is that you imagine a cannon on top of a very high mountain. You fire the cannon in a horizontal direction and the cannonball falls to the ground eventually. The more speed you give to the cannonball, the farther it goes until it eventually goes all the way around the world, falling all the time but never hitting the ground. The illustration here is the original drawing by Newton which was published in his classic book, Principia. There is a nice animation you can play with here.

QUESTION:
I observed the following: Placing a photo flashgun very close to a butterfly produced a wisp of what looked like steam from the butterfly's wings. It seems that the intense burst of light is vaporising some dust on its wings? The creature does not seem to notice and flies off after three flashes all of which caused the steam or dust to rise? Was I seeing dust or vaporised dust?

ANSWER:
It is well known that there is "dust" on butterfly and moth wings. In fact, this is not dust but tiny scales on the surface of the wing. It comes off rather easily as you have demonstrated. Although I have seen explanations of similar results of flashguns which attribute the effects of the flash to the momentum transfer of the photons in the flash (for example, the "singing cymbal" demonstration where a cymbal is made to sound with a flashgun), that has been shown to be incorrect; what is happening is that there is a thermal pulse caused by the light pulse which is responsible for, in your case, dislodging some of the scales.

QUESTION:
What is weight and why there is no weight while falling body (pulling) by earth..?

ANSWER:
Several years ago I had to teach an introductory physics course using a textbook in which the author defined weight as what a scale reads. I do not know a single physicist who likes this definition. Historically and almost universally weight means the force which the earth (or other massive object) exerts. When something is dropped it accelerates downward because the weight is a force pointing vertically down. Therefore, your question is wrong because "there is no weight while falling" is incorrect—the force the earth exerts on you does not disappear when you are falling. Similarly, astronauts in orbit are not "weightless" as is usually said because there is still a force of attraction to the earth.

Nevertheless, it is clear that the astronauts seem to have zero net force on them. The reason is that they are in "free fall" when they are in orbit and are therefore accelerating. There is a legend about Albert Einstein related to this. The legend is that Einstein saw a painter fall from a ladder and he said to himself "there is no experiment which he could do which would allow him to say that he is not at rest in empty space". This is called the equivalence principle and forms one of the keystones of the theory of general relativity.

QUESTION:
What is weight and why there is no weight while falling body (pulling) by earth..?

ANSWER:
Several years ago I had to teach an introductory physics course using a textbook in which the author defined weight as what a scale reads. I do not know a single physicist who like this definition. Historically and almost universally weight means the force which the earth (or other massive object) exerts. When something is dropped it accelerates downward because the weight is a force pointing vertically down. Therefore, your question is wrong because "there is no weight while falling" is incorrect—the force the earth exerts on you does not disappear when you are falling. Similarly, astronauts in orbit are not "weightless" as is usually said because there is still a force of attraction to the earth.

QUESTION:

Imagine a cart moving on a frictionless surface with a cannon aimed opposite the direction of travel. At 40 km/hr the cannon fires a projectile accelerating the cart-cannon to 50 km/hr. A second projectile (identical to the first) is then fired accelerating the cart to 60 km/hr. The impulse (and thus recoil) against the cart should be the same no matter what the moving velocity of the cart was. After all, the cart is only moving within my frame of reference. If I selected a frame of reference in which the cart was stationary, an impulse acceleration of 10 km/hr would be the same irregardless of whether in another frame of reference the cart was initially moving at either 40 or 50 km/hr. It's true that the second firing of the cannon would be acting on slightly less mass, after the first cannon ball was no longer part of the system—but that would seem a trivial difference. I also understand that the momentum of the total system is the same before and after the cannon is fired.
Does it take more energy to accelerate an object from 40 km/hr to 50 km/hr than it takes the same object to accelerate from 50 km/hr to 60 km/hr ? I know that the Kinetic Energy at 60 km/hr is 60/50 squared or 1.44 times greater than at 50 km/hr so the answer is undoubtedly, yes. What am I missing about Kinetic Energy - does it depend on your frame of reference?

ANSWER:

I have rearranged your question so we can talk about one question at a time. Your first question, not really a question, basically says that, although the momentum of any object is dependent on your frame of reference, that the change of momentum if you do something to it (impulse) is not frame dependent. Why is that? The reason is that Newton's second law may be written as F=dp/dt and the rate of change of momentum must be frame independent because we know that the force is frame independent.
Now, when you change the momentum you also change the kinetic energy because either the speed or the mass is changing. But does everyone see the same change in kinetic energy? The answer is no. And your example shows very clearly that the change in kinetic energy does depend on the frame of reference. A more general example, calculating the work done by a constant force as seen in different frames, can be found in an earlier answer.

Q&A OF THE WEEK, 4/7/2018

QUESTION:
For a snow plow that is very heavy, is there an advantage to having the connection point of the winch line up high so that there is less weight to pull or is there no difference? I can send a picture for clarification.

ANSWER:
Yes send me a picture. You are talking about a winch which is used for what? Pulling a stuck vehicle? Lifting and lowering the plow? What?

FOLLOWUP QUESTION:
Yes, lowering and lifting a plow. The problem is that the plow is out very far out and that my winch is mounted pretty low on my machine. So instead of the winch simply lifting the plow up, right now it is mostly pulling backwards and then that is making the plow come up. I attached a picture of what the manufacturer recommends but I haven't had too good of luck with them in the past. The last two pictures are of my machine. You can see how far out the plow is and how low my winch is. Would that pulley and cable system helped at all?

ANSWER:
You may not want to get the full physics explanation here, so I will first give you a qualitative explanation. The tension T in the strap is what is lifting plow and any part of it which is horizontal (T_H) is wasted. Your gut feeling is right, "…it is mostly pulling backwards…" Anything which you can do to increase the vertical part (T_V) will make the lift easier, and moving the winch up is a good way to do this but it might be easier to have a pulley higher up which then brings the strap back down to the winch.

The figure shows all the forces on the plow assembly: the weight W which acts at the center of gravity (yellow X), the tension T in the strap (where I have shown vertical (T_V) and horizontal pieces (T_H)), and the force the truck exerts on the support which I have represented as its vertical and horizontal parts, V and H respectively. (Note that the various forces are not drawn to scale since T has to be much larger than W to lift the plow.) Suppose T is just right that the plow is just about to lift. Then the sum of all the forces must add to zero, or V+T_V-W=0 and H-T_H=0. The sum or torques must also add to zero; summing torques about the point of attachment to the truck (light blue X), WD+T_Hs-T_Vd=0. Note that the T_V is trying to lift the plow but T_H is trying to push it down. Now, in order to get a final answer for the unknowns (which are T, V, and H) we note that T_V=Tsinθ and T_H=Tcosθ where θ is the angle which the strap to the winch makes with the horizontal. The final answers I get are:

T=DW/(dsinθ-scosθ)
V=W-Tsinθ
H=Tcosθ

I put in some reasonable numbers just to get an idea of the answers, W=500 lb, θ=20⁰, D=2m, d=1.8 m, and s=0.1 m. Then T=1920 lb, H=1800 lb, and V=-157 lb. The negative value for V means that V is down, not up. In this scenario, the pulling force has to be nearly four times greater than than the weight being lifted.

Now we need to look at whether the manufacturer's suggestion will be better than a straight shot to the winch. Now there are two forces pulling up, the tension T from the pull point to the winch and the tension P from the pull point to some anchor higher up. Of course the magnitudes of these two tensions are the same, P=T. The picture shows only the pulling forces, the rest are the same as in the picture above. There are still three unknowns, T, V, and H. I will call the angle that P makes with the horizontal φ. I will not show the details, just give the final results:

T=DW/[d(sinθ+sinφ)-s(cosθ+cosφ)]
H=T(cosθ+cosφ)
V=W-T(sinθ+sinφ)

As a numerical example, I will use the same numbers as above and add φ=40⁰. Then T=624 lb, H=1070 lb, and V=-115 lb. It is definitely advantageous to use the manufacturer's suggestion here which, in my numerical example, reduced the force the winch needed to exert by a factor of about 3.

QUESTION:
Is it possible that if one person lives in another planet (say Mars) can we chat with them live? (like on a mobile phone) or is it impossible? also will that person age differently compared to me?

ANSWER:
Certainly not using your mobile phone; the current cell phone network depends on earthbound transmitters and receivers. But communication is certainly possible or else we would not be able to communicate with the rovers and probes already there. Trying to carry on a conversation would be very frustrating, though, because of the transit times of the signals. Depending on where earth and Mars are in their orbits, transit time is 3-13 minutes. So, if you ask a question, you would have to wait 6-26 minutes for an answer.

QUESTION:
We are a seventh grade science class, and we are discussing friction. We understand that life as we know it would end if there were no friction, but we were wondering what, exactly, would happen to your body without friction.

ANSWER:
It would have been impossible for your body to form in the first place. When two surfaces are in contact with each other they exert two forces on each other. As an example, think about a block which is sitting at rest on a slight incline. The part of the force which the incline exerts on the block which keeps the block from breaking through the incline and that is called the normal force. The part of the force which the incline exerts on the block which keeps the block from sliding down the incline and that is called the frictional force. With no frictional force, the block would slide down the incline. Now imagine a single bone cell on the surface of your leg bone. It has a weight which makes it want to slide down your leg to the ground; friction between it and its neighbors with which it is in contact keeps that from happening. With no friction, all the cells in you body would slide down to the floor and spread across the floor.

QUESTION:
When operating a forklift and wearing a hip style seat belt, what is the g force the body is subjected to during a head-on collision with a immovable object at 2-3 mph?

ANSWER:
This question has been answered, although for different numbers, before. Here is the answer I gave there but substituting your numbers. You should definitely read that answer before proceeding here. Let's say that the forklift stops dead and that it plus the pole deform by 1 cm so that the distance traveled during the collision time is d≈1 cm. Since these are very rough calculations, I will let the mass of the forklift be m_f≈5000 kg and the initial speed be v=3 mph≈1.3 m/s. If the forklift stops uniformly, it will stop in a time t=2d/v=0.015 s. The force F experienced (by both the pole and the forklift) will be F=m_fv/t≈1.4x10⁶ N=315,000 lb. To get the g-force you divide by the weight of the forklift, mg=5000x9.8 N=49,000 N; so, g-force=6.4. That is the force experienced by the forklift and would be the same for anything which moves the same. The driver would experience that force where the lap belt was restraining him but his upper body would experience a smaller force since it would take longer to stop. As emphasized in the earlier answer, this is a very rough approximation.

QUESTION:
What is the ratio between the height H of a mountain and depth h of a mine,if a pendulum swings with the same period at the top of the mountain and at the bottom of the mine?

ANSWER:
This must be a homework question where you assume that the density of the earth is uniform.

QUESTION:
No it's not. I am a student who is trying to crack NEET. While I was solving questions, I saw this one but I couldn't get the explained answer. The answer I got was 1,but the answer given here was 1/2. I just wanted to know was my answer correct.

ANSWER:
There is no way to solve this problem without an assumption regarding the density of the earth. The standard assumption in introductory physics is to assume that it is uniform (which is not a good approximation); I will do that. Also, I will assume that it is a simple pendulum with a small angle amplitude so that the period is T≈2π√(L/g) where L is the length and g=MG/R² is the acceleration due to gravity. At a distance H above the earth's surface g_H=MG/(R+H)²; at a distance h below the surface of a uniform-density earth, g_h=MG(R-h)/R³ (see footnote*). Now, for the periods to be equal it is necessary that g_h=g_H; a little algebra shows that equivalently (1-(h/R))=1/(1+(H/R))²=g_H(h)/g.

It is instructive to look at h/R as a function of H/R: h/R=1-1/(1+(H/R))²shown in the first graph above. When h=H=0 you are at the surface of the earth; when h/R=1, H/R=∞ and g_h=g_H=0.

Next, look at the plots of g_H(h)/g as a function of H/R (black) and of h/R (Red). There are only two locations where g_h=g_H, at the surface where h/R=H/R=0 corresponding to g_h=g_H=g and near h/R=H/R=0.6. The third plot shows a closer look around 0.6 showing h/R=H/R=0.618 corresponding to g_h=g_H=0.382g.

This was a pretty long-winded answer, but the upshot is that you were right and the answer key was wrong: h/H=1.

*M is the mass of the earth, R is the radius of the earth, and G is the universal gravitational constant.

ADDED NOTE:
Actually, I did not need to solve this problem graphically, I could have solved it analytically. If you assume (guess) that h/R=H/R≡x, then the equation to determine x is 1-x=1/(1+x)² or (1-x)(1+x)²-1=0=(1-x²)(1+x)-1= x³+x²-x=x(x²+x-1)=0. One solution is x=0 (the surface) and the positive solution to the quadratic equation is x=½(√(5)-1)=0.618. I guess I shied away from this because I know I am not very good at solving cubic equations, but I can handle this one!

QUESTION:
If you could spin a person on the moon how many revolutions per minute would it take to release them to escape the moons gravity?

ANSWER:
The moon's escape velocity is about 2400 m/s. The angular frequency for an object moving in a circle with radius R and speed v is ω=v/R. Taking R≈1 m, ω=2400 Hz=23,000 rpm. The velocity required for a low altitude orbit is about 1700 m/s, so that would require about 16,000 rpm.

QUESTION:
With Star Wars being in fashion again with the new movies coming out, I thought to ask this.

So Emperor Palpatine is known for his ability to shoot lightning out of his fingertips quite effectively. And here's a GIF of him doing it: So how much energy, current and charge generally speaking would producing that much electricity to penetrate that much air require from a human being, and how does it relate to the amount of energy a human body actually can produce, and finally assuming a human could produce the electric power needed to do that, how would a real life Palpatine shoot lightning out of his hands with real life physics?

ANSWER:
For the air to break down and create a spark, about 10,000 V/cm is required. It appears that these sparks are about 3 m long, so about 300x10,000=3x10⁶ Volts=3 Megavolts of potential difference is required. Of course, this is fiction so I will not go any farther with this answer.

QUESTION:
How the f*** can the velocity be negative? And why isn't the acceleration 0 when the velocity is 0?

ANSWER:
Gee, don't get so upset! Velocity is a vector—to specify the velocity you must specify both magnitude (called speed) and direction. To specify the direction, it is convenient to define one direction, say north, to be positive and the other, south, to be negative; so a car having a speed 50 mph on a north-south road has a velocity of +50 mph if going north, -50 mph is going south. (Speed is never negative.) Acceleration is the rate at which the velocity is changing. Suppose you throw a ball vertically; at the very top the ball is at rest, albeit only for an instant, but at rest. Now, if at the top the acceleration were zero, that would mean that its velocity is not changing and it would therefore just stay there. It is possible for the acceleration and velocity to both be zero; a book sitting at rest on the table would be one example. It is just not necessary that both be zero.

QUESTION:
Suppose the Sun and the Earth were each given an equal amount of charge of the same sign. Just sufficient to cancel their gravitational attraction. How many times the charge on an electron would that charge be? Is this number a large fraction of the number of charges of either sign in the earth?

ANSWER:
So we want kq²/R²=Gm_em_s/R² or q=√(Gm_em_s/k)≈3x10¹⁷ C. The number of electrons this would be is q/e=1.9x10³⁶. If we roughly estimate that the earth has is 50% protons and 50% neutrons, then the approximate number of protons would be (m_e/2)/m_p=1.8x10⁵¹, far larger than 1.9x10³⁶. If you want to check my arithmetic, look up all the constants I have used.

QUESTION:
Theoretically, if I look through a telescope at a mirror 2 light years away, the light bouncing back would give me a instant view of my location from 4 years ago, right? With that, if I was in a ship that safely accelerated to the speed of light, say 1 g acceleration, so that I would reach the speed of light after approximately 1 year, traveling 1/2 of a light year, then turning around and decelerating at 1 g for one year to come to a complete stop, then doing the same to get back to my original location, thereby traveling 4 years total, why would I change 4 years in age, but my original location (and final destination) be more than 4 years older?

ANSWER:
Yes, the mirror would give you the view four years earlier. However, your understanding of the accelerating spaceship is very wrong. It is a pretty difficult thing to understand, so I suggest that you read an earlier answer as well as the earlier answer it links to. When you have done that, you should be able to follow my answer to your question.

First a qualitative discussion. You decide that you want your spaceship to have a constant acceleration of g as measured by you, captain of the ship. You therefore arrange to have your engines always exert a force on the ship of F=mg where m is the mass of the ship. So you will always see your ship having the desired acceleration. But, as I have emphasized in many earlier answers, acceleration you measure is not the same as acceleration someone on earth measures. In Newtonian physics, everyone measures the same acceleration. This is really a glorified twin paradox problem and I am not going to calculate how long the round trip will be for you (but I will calculate how long it takes in earth time). What I can tell you is that you will take longer than 4 years to return to earth and the time elapsed on earth will be longer than your time. Your desire to accelerate "to the speed of light" is impossible, as you will see, so forget that; with your scenario the fastest you will ever go is about 86% of c.

What I want to focus on is what someone on earth will measure regarding your motion. This is mostly worked out in the earlier answers I have referred you to, in particular v/c=(gt/c)√[1+(gt/c)²] and a/g=[1+(gt/c)²]^-3/2; these are plotted in the graph above with the red and blue curves. Things to notice are that after 1 year you have only acquired a speed of about 0.7c and your acceleration has dropped to about 0.35g. What was not derived earlier was your position as a function of time; integrating the velocity equation, I find gx/c²=√[1+(gt/c)²]-1, shown by the black curve. Things to notice are that it starts out looking like a parbola as it would if you had a constant acceleration of g, but eventually becomes a straight line as you approach the speed of light which you never reach. Reading off the graph, the time it takes you to get halfway, 1 ly, is about 1.7 years (as measured from earth); your speed will be about 0.86c and your acceleration (again as measured from earth) will be about 0.13g. This will be ¼ of your total trip, 6.8 years. The time you measure will be somewhere between 4 and 6.8 years.

Incidentally, the axes may be shown to be numerically equal to years and light years if you approximate g≈10 m/s².

ADDED COMMENT:
I see that I have misread the question. For some reason the questioner reverses his thrust at ½ ly out so he only goes to 1 ly before turning the ship back to earth. From the graph you can see that the time to get to this position is about 1.1 years and the speed there is about 0.75c. Thus the total elapsed time on earth before your return would be 4.4 years in this scenario. Light would take 2 years to make this trip. Again, I have not calculated the time on your clock but it would be between 2 and 4.4 years.

QUESTION:
Why the projectile motion of a body in air is not three dimensional?

ANSWER:
In the real world the motion is three dimensional. In an elementary physics course, the situation of the body having no interaction with the air is first considered for simplicity. In this case the only force on the object is its own weight which is vertical and therefore the object moves in a plane defined by the initial velocity and the vertical. This is a good approximation for many everyday applications of projectile motion. However, the object will interact with the air. If the air is still and the object is a point mass, it will experience a force opposite the direction of its velocity and therefore also move in a plane; the path will not be a parabola. If there is a wind whose velocity does not lie in the plane defined by the initial velocity and vertical, the projectile will move out of the plane—true three-dimensional motion. If the object experiences a greater air force on one side or the other, for example for a spinning ball, the projectile will curve out of the original plane.

QUESTION:
For a book pushed horizontally against a vertical wall, we know that the friction force is equal to the weight of the book. We also know that the friction force is equal to the normal force multiplied by the coefficient of friction. So, technically speaking, the harder you push against the book, the greater the normal force. Therefore, the friction should get bigger the harder the you push, but we know that isn't true since friction is equal to the weight of the book, which doesn't change when you press the book harder to the wall. Thus the only way to appease the equation: (mu)(normal force) = (weight of book), with an increase in normal force (caused by an increase in applied force to the book), that would mean the (mu) or coefficient of friction has to change. So my question is, why does the coefficient of friction change when we apply more force to the book on the wall? Its still the same two surfaces!

ANSWER:
You should have stopped after the first sentence which is correct! You misunderstand static friction. Although it is true that f=μ_kN for kinetic friction, for static friction the equation f=μ_sN is not true. The correct formula is an inequality rather than an equality: f ≤ μ_sN. For one single force you may write an equation, f_max=μ_sN which is the largest frictional force you can get for a given N. If you do not push your book with a force equal to or larger than N=mg/μ_s it will fall to the floor. (The Q&A right below yours is closely related to your question.)

QUESTION:
How two objects of the same material with different masses when placed on a sloped surface the force of gravity overcomes the force of friction at the same angle for both objects, this would bring the question why would they have the same angle if one object has more inertia than the other.

ANSWER:
An object on an inclined plane which makes an angle θ to the horizontal is at rest and in equilibrium. There are three forces acting on it, the friction f, the normal force N, and the weight W=mg. Then, using Newton's first law, f-Wsinθ=0 and N-Wcosθ=0. Solving, f=Wsinθ and N=Wcosθ. If the block is just about to start moving f=μN where μ is the coefficient of static friction, so μ=(Wsinθ/Wcosθ)=tanθ. As you can see, the angle is independent of the weight; this is because both N and f are proportional to W, so W cancels out when you calculate their ratio.

QUESTION:
In what distance will a car skid to a stop on a dry concrete road if its brakes are locked when it is moving at 89 km/h?

ANSWER:
The coefficient of kinetic friction for rubber on dry concrete is μ≈0.7. The speed of the car is v=89 km/hr=24.7 m/s. The change in kinetic energy must equal the work done, (K₂-K₁)=-μmgd=(0-½mv²); the mass m cancels out, so -½x24.7²=-0.7x9.8xd and therefore d=44.5 m.

QUESTION:
can length contraction of the crew of a spaceship reach the point where it affects the biological processes? I am not talking about time dilation effects. Is thinking changed? Can one be squeezed down to the point of injury? At very near the speed of light could the density of the spaceship be so great that it becomes a black hole? What about charge density in an electron? Can fundamental constants like Planck's constant be affected?

ANSWER:
You misunderstand length contraction. In the frame of reference of the spaceship there is no length contraction. Lengths on the spaceship as measured by someone at rest are shorter, but everyone on the spaceship sees everything just the same as if it were at rest.

QUESTION:
We have been talking in science class about air pressure and vacuums,and how a straw even in perfect vacuum will only suck water up to ~10m, my question is why does the diameter of the straw not effect the height to which the water is pulled as a greater diameter would mean more mass.

ANSWER:
The mass m of the water in the tube is the density of water (ρ=10³ kg/m³) times the volume of water (V=Ah); therefore the weight of the water is mg=ρgAh≈10⁴Ah (taking g≈10 m/s²). The pressure at the bottom of the tube is atmospheric pressure, P_bottom≈10⁵ N/m² and at the top the pressure is zero; therefore there is a force F up on the tube which is F=(P_bottom-P_top)A≈10⁵A N. But F is holding up the weight, so F=mg or 10⁵A=10⁴Ah or h≈10 m. A cancels out!

Q&A OF THE WEEK, 4/14/2018

QUESTION:
hi, so, there is a fairly recent video going around the internet of nasa ISS astronaut Randy Bresnik spinning a fidget spinner, in space, in a quite low gravity environment and then apparently grabbing hold of the spinner then video cuts and we, presumably some short time later, see the entire body of the astronaut holding the spinner spinning fairly rapidly in space. Perhaps it is a quite elementary question, but, I wonder if that video might have been faked, wondering if the what must be a relatively small amount of energy in the spinner could cause the entire body of mass of astronaut plus the spinner to spin in the fashion observed in the video. That is to say: wouldn't the energy from the spinner, say, be absorbed by the astronauts body or the muscles in his arm, or some such, upon grabbing the spinner; or, if there was movement, wouldn't it be far far less than what is demonstrated in the video?

ANSWER:
You are right, I think the astronauts are messing with you here! You have looked at it from the perspective of energy conservation; however, energy would not be conserved here and energy would actually be lost, not gained as it appears. So, your reasoning was sound—where did the energy come from? What is conserved is the angular momentum. If the moments of inertia are I_toy and I_man and the original angular velocity of the toy is ω₁, then the initial angular momentum is L₁=I_toyω₁ and the final angular momentum is L₂=(I_toy+I_man)ω₂ where ω₂ is the angular velocity of the man+toy. Conserving angular momentum and solving for ω₂, ω₂=[I_toy/(I_toy+I_man)]ω₁<<ω₁. Now, I actually found the moment of inertia of a fidget spinner, I_toy≈7x10^-5 kg·m² and I estimate I_man≈70 kg·m². This means that ω₂=ω₁/1,000,000! The astronauts have angular velocity of about 1 revolution per second and you know perfectly well that the fidget spinner did not have a speed of a million revolutions per second.

Finally, if you are interested, you can show that energy is not conserved. The expression for kinetic energy is E=½Iω²so E₁=½I_toyω₁²and E₂=½[I_toy/(I_toy+I_man)]ω₁²≠E₁.

QUESTION:
I have an idea to help enhance mankind's interplanetary manifest destiny. The idea hinges on pragmatics and so I want to know if it is worth considering further. Without giving away too much of the idea, and without more ado, here is my question. Is the escape velocity the same at sea level as it is atop a mountain? The only equations I can find show escape surface velocity. As an added bonus question: If my initial velocity is extremely large but I have no propulsion, how would this scenario affect escape velocity?

ANSWER:
First of all, let's be sure we know what escape velocity from earth is: it is the minimum speed an object must have at a distance R from the center of the earth to totally escape (to infinity) the gravitational field of the object; the expression is v_escape=√(2GM/R) (which assumes that the mass of the escaping object is very small compared to the mass of the earth, M). That is the initial speed with no additional propulsion. Obviously you could escape any gravity by simply pushing with a force greater than the force of gravity you experience; if you had a means of propulsion which kept you moving at 1 mph, you could completely escape the earth given enough time. Therefore your "bonus question" essentially has no meaning. Your first question is whether the escape velocity atop a mountain is different from sea level; the answer is yes because R is bigger atop the mountain, but since the height of even the highest mountain is tiny compared to the radius of the earth, the difference is neglegible.

QUESTION:
Why do raindrops fall down at a pronounced angle with the vertical when seen from the window of a speeding train? Is this angle necessarily the same as that of the path of a water drop sliding down the outside surface of the window?

ANSWER:
For the same reason that stationary objects outside the train appear to be moving backwards with the same speed as the speed of the train. For example, if the drops are falling vertically relative to the ground with velocity v, the negative of the velocity (horizontal) of the train u must be added to v; the velocity w seen by an observer in the train will be this vector sum.

This has nothing to do with how drops adhering to the window move. If the train were still a drop would drip straight down the window with some speed which would be much smaller than the speed of a falling rain drop. If the train is moving the drop would see a wind of speed u which would push it backwards with some speed considerably less than the speed of the train. Again, the vector sum of the two velocities would be seen as the drop velocity by an observer in the train.

QUESTION:
Why is it when you sit on a swing and swing really goes smooth but you get to a certain height and it gets bumpey

ANSWER:
It is pretty complicated to discuss in detail. It has to do with the "pumping" the swinger performs. Look at the little animation above. Notice that the chains do not stay in a straight line. If someone were pushing you and you sat perfectly straight, I believe that the "bumpey (sic)" motion would not happen until you were swinging above the bar. When pumping, you eventually reach a point where the chain between your hands and the seat goes slack which you can see in the photograph of the swinging girl if you look carefully (see my yellow line). If the girl is at the very top of her path, she is at rest and there is no force from the chain on the swing itself. So when she starts to fall back she will be in a sort of free fall until the chain becomes taught; at that instant there will be a jerk.

DISCLAIMER:
I was unable to find any discussion of this anywhere and I have invented this answer using what I find to be reasonable arguments. It might not be the right answer!

QUESTION:
So, you have a train that is not moving and inside the train you have a drone that is hovering. No one is controlling the drone it is just hovering in one place. If the train begins to move, would the drone move with the train or would it run into the back wall of the train?

ANSWER:
As viewed by an observer outside the train, for the drone to accelerate with the train it would have to experience a force in the direction of that acceleration. But there is no such force so the drone would stand still; viewed from inside the train, the drone would appear to accelerate toward the back wall of the train.

QUESTION:
With the recent discovery of "seeing" gravitational waves emitted by two merging neutron stars, how does one "see" gravitational waves.

ANSWER:
There is a good description of the detectors here. What is particularly exciting is that light from the merging neutron stars could be seen; earlier observations, of merging black holes could not. Also, the abundance of heavy elements like gold is much larger than could be understood using standard models of stellar evolution; it had been hypothesized some years ago that perhaps the excess could be produced in neutron star collisions, and the current observation seems, at first glance of the spectroscopy of the light, to support this hypothesis.

QUESTION:
What would a chart of an 8g impact look like for an 8g impact with a 100 ms duration? I can send a jpg of what I am told fits the requirement. But I seriously doubt that it does fit the requirement. But then again I am barely understanding the test. I see spikes to 20-30g. I am having a hard time finding anyone that can answer this question. (I asked for more information.)

FOLLOWUP QUESTION:
These are the charts provide by the AF to me when they tested a pallet that we built.

The requirement is: Ultimate load. When uniformly loaded to 10,000 pounds, the load being restrained to the pallet by chains, the pallet installed between restraining rails locked to the rails by 2 locks through each rail and engaging 2 lock notches on each side of the pallet, and resting on 4 rows of conveyor as specified (see 3.4.5.1), the loaded pallet shall withstand a dynamic load of 3 times the force of gravity (g's) for a period of 0.1 second. The pallet shall be serviceable after undergoing the test. In addition, the pallet shall withstand a dynamic load of 8 g's for a period not less than 0.1 second. The pallet need not be serviceable after undergoing such a load; however, the pallet shall remain in one piece.

These are the result of those tests. These charts just do not look right to me in their conclusions. I am not a math major, but I do understand some stuff and this does not look correct. ie. "Area under the curve." or correct averaging. If you can set me straight or put me on another path it would be greatly appreciated.

ANSWER:
So, I must admit that I do not understand all the details of the requirements (locks, chains, notches, etc.) What I can do, though, is estimate the area under the curve which seems to be one of your main concerns. I drew in a rough fit (green) to the data (red) and calculated the area under the green curves (keeping in mind that the area of the triangular segment below the axis is negative). As you can see, the area I got as a rough estimate is 0.37 compared to the actual area of 0.35. The average would simply be the area/time=3.5. My guess is that your pallets did not fulfil the requirements since the precipitous drop before the 0.1 s had passed would seem to indicate a collapse of the pallet. Both tests show this behavior, with the 8g test having the "collapse" occur earlier as would be expected. However, since I do not understand the details of the test well enough, I would not take my guess as gospel.

QUESTION:
How would i explain bus or a truck(high CG)flip over, when driving in a curve with high speed, from a inertial reference frame? If centrifugal force is fictitious and bound to non-inertial reference frame, what torque causes truck to flip over observed from an inertial ref. frame?

ANSWER:
I have attached figure from the earlier answer where I did essentially this problem in the noninertial frame; this saves me having to draw the whole picture again. For your problem, the vector labelled C is zero. Part of what makes your problem harder than in the accelerating frame is that you must sum torques about the center of mass (green x). For equilibrium, the sum of torques must be zero; summing torques gives 0=H(f₁+f₂)+LN₁-LN₂. (Translational equations are f₁+f₂=mv²/R and N₁+N₁-W=0.) Now, when the truck is about to tip over the left wheel is about to leave the ground so N₁=f₁=0, f₂=mv²/R, and N₁=W; therefore Hmv²/R-LW=0. So, if v>√[LWR/(Hm)] the truck will flip over.

QUESTION:
If you jumped vertically upwards, strictly speaking and according to Newton's 3rd Law, what would the Earth do?

ANSWER:
The earth would recoil but with a velocity much less than you do. Suppose that you have a mass m and initial speed v and the earth has a mass M and initial speed V; then conservation of linear momentum (essentially Newton's 3^rdlaw) says mv+MV=0 or V=-(m/M)v; the recoil velocity of the earth is unmeasurably smaller than your velocity.

QUESTION:
Suppose I am in space, in a 100 meter barrel strapped to one end. The barrel and I have mass 100 kg (I'm pretty light). I throw a 100 kg ball down the center of the barrel towards the other end, accelerating it to 1m/s velocity. Presumably, it takes the ball 100 seconds to reach the other end and it sticks because it is covered in velcro. I presume the barrel started in motion in space when I threw the ball from the reaction from the force I applied to the ball and it stopped when the ball hit the other wall. Could I keep moving the barrel by throwing more balls? How is this possible without violating Newtons laws? Surely this is not a case of propellantless thrust?

ANSWER:
These 100 kg balls just materialize from nothing, do they? So you must have a supply of them which means that the total mass of the entire barrel and its contents is much bigger than 100 kg which means you go much less far with each throw. Eventually you will run out of balls, but you will have gone some distance. What propelled the barrel? You did.

FOLLOWUP QUESTION:
Thank you so much for you previous answer. My question about throwing balls at the back of a barrel in space was inspired by what I had read about some experts criticizing the "EM Drive" for violating Newton's Laws. Basically the writers quote experts as saying that you can't have forward thrust without throwing mass out the back " no such thing as propellantless thrust". I suppose these experts are simplifying their response for the lay press because qualifying their answer would take up to much column space. In retrospect, I should have known that some net movement was possible as I believe NASA uses large gyroscopes to adjust the pointing of some space telescopes. This did get me to thinking though. Suppose I attach a laser to one end of the barrel and fire a single photon towards the other end. Presumably, momentum is imparted to the wall attached to the laser and the barrel moves through space as the photon moves to the opposite wall where it is absorbed. Now I wonder why I can't keep firing photons and moving the barrel. Is the hitch in my "propellantless thrust" scheme come from the absorption process? Is the heat generated through absorption dissipated out of the barrel preferentially so as to counteract my momentum drive?

ANSWER:
The situation is similar to your first question since, as you apparently know, photons have energy and momentum. But, to operate your laser you have to supply energy somehow. Suppose that you use a fusion reactor to generate that energy. Then as you shoot more and more photons, the whole ship gets less and less massive. But, energy has to be conserved (mass is a form of energy, mc²) and when the photons get absorbed, the heat generated will "retrieve" the lost mass. Eventually you will run out of fuel for your reactor and you will be right back to being at rest. (I am assuming that the rules of the game forbid letting the heat at the back radiate into space.)

QUESTION:
Can you calculate the negative g when a 38sq metr vehicle hits sea level atmosphere at 5,000kph?

ANSWER:
I presume that you want to know the acceleration of the vehicle due to the air drag. There is an approximate expression (only valid for SI units) for the drag for an area A and speed v, F=-¼Av². The problem is that I do not really know how accurate this is at such a high speed as 5000 km/hr≈1400 m/s. So, view this as a very rough calculation. F≈-2x10⁷ N. Now, to get the acceleration you need to know the mass m (in kg). Using Newton's second law, F=ma, a≈2x10⁷/m m/s². To express this in gs, divide by 9.8 m/s². For example, if the mass of the vehicle is 10,000 kg, a≈2000 m/s²≈200 gs.

FOLLOWUP QUESTION:
What would be the approximate negative g on a rocket having a 7sqr metre frontal area when exiting from an evacuated tube doing 5,000kph at sea level? This is a further question related to impact of air on a vehicle.

ANSWER:
The only thing you changed is the area A. Since the drag force is proportional to A, all estimates given above need only be multiplied by 7/38=0.18.

QUESTION:
My question is using the flight time equation of (1/2)x g(flight time/2)^2 or the more simplified (g x (flight time)^2)/8 i have the answer and know it is m but i was just wondering what the origins of the equation is, such as what is it called or what is the rule. Or if it is derived and what from. this is just for myself to have a deeper understanding of what i have done and worked out. for me the the equations was (1/2) x 9.81(0.0557/2)^2=0.38m or 38cm. i know how to use the equation just want to know where it comes from.

ANSWER:
The equations of motion for the vertical component of projectile motion are v=v₀-gt and y=y₀+v₀t-½gt²; note that this is only for the vertical component of the motion, the horizontal motion does not interest us for your question. Here v is (the vertical component of) the speed at time t, v₀ is (the vertical component of) the speed at time t=0, y is the altitude at time t, and y₀ is the altitude at time t=0 (which I will choose to be 0). You are interested in the height to which the projectile rises, and at that position v=0; therefore v₀=gt and so y=h=(gt)t-½gt²=½gt². But my time here is only half the "flight time" of the projectile, so h=½g(t_flight/2)². You do not need to memorize "the flight time equation". Everything you need to know is contained in the equations of motion which you should memorize.

QUESTION:
A Newton's Cradle is some times used to demonstrate the Law of Conservation of Momentum but, I noticed anomaly. Specifically, whenever the steel balls collide, I hear a clicking sound and that requires energy. This tells me some of the kinetic energy of a moving ball transforms into thermal and accoustic. How do physicists explain that momentum is conserved when a ball leaving a collision is moving slightly slower? If there is less kinetic energy but no change in the total energy, it seems logical to think a certain quantity of motion has vanished after any collision. Both momentum and kinetic energy use the same variables. I also understand that when I pull one ball back, after a collision one ball emerges. If two balls are pulled back and released, two balls pop out.

ANSWER:
You are right, energy is not conserved in the collisions. And, you can tell that it is not conserved because the machine eventually stops. But momentum is always conserved in an isolated system (no external forces on the colliding bodies). I commend you for thinking about this which seems paradoxical. To illustrate, let's choose the simplest situation, the collision of two balls of equal mass m, one at rest the other with speed v, in one dimension. As you know, if the collision is perfectly elastic the first ball ends up at rest, the second moving with speed v, so the momentum is conserved because it is mv before and after the collision. Next you say that if the second ball emerges with speed u=v-δ that momentum has been lost. But you have assumed that the first ball is at rest, but it cannot be if momentum is conserved: mv=mu'+m(v-δ) and so the first ball has a speed u'=δ after the collision. In the video I have inserted above, look closely at the first demo, with one ball striking four and one popping out; you will clearly see that the next to last ball is not at rest after the collision(s).

QUESTION:
If I take two solenoids with dimensions of 6 inches in length, and 2 inches in radius, each producing .5 Tesla and I combined the opposite poles so they attract each other, will the combined solenoids be a 1 Tesla solenoid?

ANSWER:
No. The argument goes like this: Suppose you have an infinitely long solenoid; it has a perfectly uniform field 0.5 T aligned with the axis of the solenoid throughout its whole volume. Now cut out two 6" sections of that solenoid; each will still have a nearly uniform field 0.5 T near its center but the field will become weaker as you approach the ends. Now, put those two together. It is just the same as if you had cut out a single 12" section to start with which would have an approximately uniform field of 0.5 T except near its ends.

Here is a history of questions and answers processed by "Ask the Physicist!". If you like my answer, please consider making a donation to help support this service.

QUESTION:
As I understand it, a helium balloon can float upward because of the difference in pressure between the helium in the balloon and the atmosphere. Suppose the balloon had a strong hard surface and the inside of the balloon was a partial vacuum, Would the balloon float upward or at least weigh less? In this case there would also be a difference in pressures like helium.

ANSWER:
Your understanding is incorrect. The reason anything floats up in the air is that the atmospheric pressure up on the bottom of the balloon is greater than the atmospheric pressure down on the top resulting in a net upward force F. If F>W where W is the weight of the balloon plus its contents, there is a net force up causing the balloon to accelerate upwards; this force is called the buoyant force and also explains why things float in water. The reason you use helium is that it is lighter than air. Then the answer to vacuum question is clear: taking away the helium but keeping the balloon the same size would make it lighter and therefore the net upward force would be larger.

QUESTION:
What would be the difference between a bowling ball size meteorite hitting the ground at an average meteor speed compared to a meteor that has been propelled past lightspeed? I'm just wondering how much more damage that would do and if we would even see it coming. Sorry just a random thought.

ANSWER:
First, it is impossible for anything to be "propelled past lightspeed"; no material object can go faster than (or as fast as) the speed of light. I will take the meteor mass to be 300 kg and its "average speed" to be 5x10⁴ m/s. Then the energy would be E=½mv²=4x10¹¹ J. For the super fast meteor, let the speed be 99% the speed of light. Then the energy would be E=mc²/√(1-.99²)=2x10²⁰ J. For comparison, the energy of the Nagasaki atomic bomb was about 10¹⁴ J.

QUESTION:
If I threw a ball straight up into the air and it stayed there for 6hrs would it drop back to excactly the same spot?

ANSWER:
Any object thrown vertically will not fall back exactly to where it started. To understand why, see a recent answer. Your question has a slightly different twist: when this type of problem is calculated it is usually assumed that g, the acceleration due to gravity, is constant; if your ball stays in the air for six hours, that will not be the case. Nevertheless, the ball would always have a Coriolis force on it so it would be deflected westward by some amount. If you had said something like the ball goes up 1000 m, you could have done a pretty good calculation since 1000 m is very small compared to the radius of the earth and it would be a reasonable approximation to say that g=9.8 m/s² the whole time (which would be about 28 s). It can be shown that, for a latitude of λ, on the way up the Coriolis deflection is [(4ω/3)√(8h³/g)]cosλ west and on the way down it is [(ω/3)√(8h³/g)]cosλ east, so the net deflection is x=ω√(8h³/g)cosλ west; the angular velocity of the earth is ω=7.27x10^-5 s^-1, so x=2.1 m west at the equator (λ=0).

Finally we should note that at the poles (cosλ=0) there is no deflection due to the earth's rotation, but since the earth revolves around the sun you are still in a frame where there would be a Coriolis force but it would be much smaller; it would be smaller yet because the angular velocity of the motion around the sun is almost parallel (about 23⁰away) to the vertical at the poles. The deflection for the example above (h=1000 m)would be about 2 mm compared to 2 m; the direction would be opposite the direction the earth is moving in its orbit.

All of the above ignores air drag.

QUESTION:
Is the graviitational force on Earth the same everywhere or is the force different near the south or North Pole?

ANSWER:
If the earth were a perfect sphere with a spherically symmetric mass distribution, the gravitational force would be everywhere the same. Since it is not, there are small variations as you move around the surface. Even if it were, it would appear to vary as you moved from the poles to the equator because of the earth's rotation. A scale would read your weight (gravitational force on you) correctly at the poles but would, because of the centrifugal force (a "fictitious" force), read slightly less at the equator.

QUESTION:
I can get a neon lamp to ignite (light up) at less than ignition voltage two ways. One is at a voltage a little lower than ignition voltage to shine a UV lamp on the neon bulb which causes it to light. This can be explained by the photoelectric effect. The other way, in the same situation as above, is to generate a spark near the neon lamp. One way is to use a piezo electric barbecue igniter which generates about 15KV. This produces the same effect as the UV light - the neon lamp comes on at a lower voltage than its ignition voltage. There is also a YouTube video of a fluorescent that will no longer come on starting when a spark is generated near it. I know a spark produces RF radiation but it is said to not have enough energy to produce a photo electric effect. I have be unable to find any explanation of what is happening here.

ANSWER:
Well, you can be sure that the spark does not generate only RF because you can see it! Seems to me that if visible light is being produced by the spark, there is probably also UV radiation being produced.

QUESTION:
If time is relative does that mean space would be too/ or to put it another way is space actually fixed or can it be and has it been streatched or bent like time in the famous high altitude clock experiments??

ANSWER:
Yes, space is also relative to frame of reference. The most obvious example is length contraction where lengths along the direction of relative velocity are shortened by the factor √(1-(v/c)²) where v is the velocity and c is the speed of light. The most elegant way to think of this is to not think of space and time as being two things, rather think of space-time.

QUESTION:
Why does a satellite shot straight up from the equator deflect to the west?

ANSWER:
There are two ways to look at this.

If you view it from outside the earth, the satellite will continue moving directly away from the center of the earth. But, viewing the earth from above the north pole, the earth rotates counterclockwise, west to east. So the earth is actually rotating under it which gives it the appearance of deflecting to the west.
The second is a little fancier and harder to understand. If you view Newton's laws from a rotating coordinate system (the rotating earth with you on it) you find that they are wrong; these "laws" of physics only work in inertial frames of reference. However you can force Newton's laws to be correct if you invent just the right fictitious forces. I will not go into all the complexity here, but one of the fictitious forces is called the Coriolis force and can be written as F=-2mωxv . Since the vector ω points along the earth's axis and out through the north pole and the velocity vector v points radially outward, the negative of their cross product points west. Hence, it will appear that the satellite is pushed west when viewed from the ground.

QUESTION:
I am a safety coordinator in a warehouse operation. I am attempting to impress the need for safety with my co-workers. To that end, I am trying to answer the following: A 10,636 lb. forklift (4825 kg) traveling at a speed of 8 mph (13 kph) strikes a fixed object (metal pole). How much force is transferred during this collision? . Secondly, same forklift, same traveling speed strikes a 150 lbs. (68 kgs) person. How much force is transferred to the person in this collision?

ANSWER:
There is no accurate way to calculate the forces. They depend on the details of the collisions. In particular, how long do the actual collisions last? Also, how do the forklift and the person move after the collision has occurred (assuming the metal pole stays stationary)? I could make some rough estimates to get an idea:

For the first problem, let's say that the forklift stops dead and that it plus the pole deform by 1 cm so that the distance traveled during the collision time is d≈1 cm. Since these are very rough calculations, I will let the mass of the forklift be m_f≈5000 kg and the initial speed be v=8 mph≈3.6 m/s. If the forklift stops uniformly, it will stop in a time t=2d/v=0.0056 s. The force F experienced (by both the pole and the forklift) will be F=m_fv/t≈3.2x10⁶ N=720,000 lb.
For the man-forklift collision, assume the forklift keeps right on moving with the same speed but with the man stuck to the front. Suppose that the man's body compresses 3 cm during the collision, so the time of collision will be t=0.06/3.6=0.017 s. The force now is F=m_mv/t=68x3.6/0.017=1.44x10⁴ N=3200 lb.

QUESTION:

do the proton and the neutron have exactly the same mass?
how do masses of the proton and neutron compared to the mass of the electron? w
which particles make the greatest contribution to the mass of an atom?
which particles make the greatest contribution to the chemical properties of an atom?

ANSWER:
I usually do not answer multiple questions, but these are pretty much one question about properties of atoms.

The neutron is about 0.14% heavier than the proton.
The electron is about 1837 times lighter than a proton.
All stable atoms but ¹H have neutrons as having to largest fraction of the weight. All stable nuclei except hydrogen have more or an equal number of neutrons than protons.
The electrons, which orbit the nucleus comprised of protons and neutrons, determine the chemical properties of the atom

QUESTION:
CAN A STRUCK GOLF BALL SLICE, IF HIT IN A VACCUM, LIKE THE MOON ?

ANSWER:
Because a slice is curving due to a spinning ball moving through air, it will not occur in a vacuum. For a more detailed discussion of golf on the moon, see an earlier answer.

QUESTION:
After searching through so many questions on here relating to the speed of light (each one being fobbed off by saying 'it just can't get faster') my question is why ? If a craft was able to reach light speed and had a torch pointing forwards, the light emitted from that torch will be travelling at twice light speed (craft speed (of the source of light) and the actual light projected from the source) ? Don't say it just can't.

ANSWER:
Well, all I can say is that you must not have done a very good search of my site because both your questions have been answered multiple times. The first thing you should always do in such a situation is go to the site's FAQ page if there is one—there is one on AskThePhysicist.com! For the question regarding why the speed of light is the highest possible, see this link. Your question/statement that the speed of light would be the speed of light plus the speed of the source is simply wrong. To understand why, see this link.

QUESTION:
In your book From Newton to Einstein, page 2-15, you show the resulting solution to variable gravity effect on increasing mass. Is there a web site where the step by step solution (rather than the end solution) can be found?

ANSWER:
In the original Q&A there is a link to a more detailed derivation although I do not think it will make this simpler for you. Rewrite the first equation as

(g/c)dt=√(1-β²)·d[β/√(1-β²)]=dβ/(1-β²).

There was a little tricky differential calculus here to get the the last step:

d[β/√(1-β²)]=(1-β²)^-1/2dβ+β(-½)(-2β)(1-β²)^-3/2dβ=dβ/(1-β²)^3/2.

If, like me, your integration skills are rusty, I suggest the Mathematica on-line integral calculator which gets you to the solution for gt/c; don't forget that the difference of logs is the log of the ratio. Then, to get the final answer, exponentiate 2gt/c to get rid of the natural log and algebraically solve for β. Of course, the starting point is recalling m=m₀/√(1-β²).

QUESTION:
I am interested in the temporal phase relationship of electric field excitation and magnetic field excitation for a circularly polarized electromagnetic wave. I was taught that linearly polarized waves had temporal fields in sync, while circularly polarized waves had an orthogonally delayed temporal phase difference. An Electrical Engineer with 50 odd years in antenna design tells me that this is a common misunderstanding of circularly polarized waves - it is his opinion is that ultimately ALL EMWs have temporally synchronised pahase relationships, even circularly polarized EMW!

ANSWER:
I believe your friend is correct. An electromagnetic wave, regardless of its polarization has a magnetic field which at every point and time is proportional to the electric field. Therefore they are in phase. Quoting the article in Wikepedia on circular polarization, "Since this is an electromagnetic wave each electric field vector has a corresponding, but not illustrated, magnetic field vector that is at a right angle to the electric field vector and proportional in magnitude to it. As a result, the magnetic field vectors would trace out a second helix if displayed." The electric field, which has a constant magnitude, is illustrated in the animation; the helix traced out by the magnetic field would be π/2 out of phase with the electric field which may be the source of confusion.

QUESTION:
Another question relating to gravity is what does 9.8 meters per second squared actually mean or at least the "squared" part of it? When I square 9.8 I get 96.04 then when I square the resulting answer the number escalates to over 9,223 and falling objects clearly don't reach 9,223 m/s in just 2 seconds!

ANSWER:
Oh my, this is so fundamental to understanding physics. One of the first things you learn in an introductory physics class is kinematics, the specification of position, rate of change of position (velocity), and rate of change of velocity (acceleration). Position is specified by a length, so many meters (m) from somewhere. If the position is changing at some constant rate, called velocity, is the change in position divided by the elapsed time (m/s). If the velocity is changing at some constant rate, called uniform acceleration, is the change in velocity divided by the elapsed time ((m/s)/s). So, if a car goes from 0 to 60 in 6 seconds, its acceleration is (60 mi/hr)/6 s)=10 mi/hr/s. If you drop a stone and measure its speed after it has fallen one second, you will find that the velocity has speed up to 9.8 m/s; therefore its acceleration is (9.8 m/s)/(1 s)=9.8 m/s/s=9.8 m/s². Students usually find the s² as strange—what in the world is a square second? So, m/s² is just shorthand for m/s/s. That makes sense, right: 1/4/4=1/4²=1/16.

QUESTION:
How does lightning conductor works?

ANSWER:
See an earlier answer.

QUESTION:
If a person ran around in a circle at the speed of light, would he cause something like a cyclone?

ANSWER:
No material object can travel the speed of light or faster. But let's imagine someone who could run a million miles per hour (way smaller than the speed of light). In order to cause "something like a cyclone", she would have to be running in air. But way before she even got to top speed she would burn up due to the air friction.

QUESTION:
what is the terminal velocity of a soccer ball sized piece of hail?

ANSWER:
I don't think there is hail that large! But, the calculation is straightforward. I usually estimate the air drag of something of cross section A having a speed v to be ¼Av²=0.038v². When this is equal to the weight of the object, mg=(4πR^3/3)ρg=402 N, the object will have the terminal velocity; so v=√(402/0.038)=103 m/s=230 mph. I have used R=0.22 m, ρ=920 kg/m³, and g=9.8 m/s².

QUESTION:
If you were traveling at a speed of 60mph, what would happen if you ran into a 180mph wall of wind.

ANSWER:
Assuming that you did not change the accelerator setting, the speed of the air hitting your car would change from 60 mph to 240 mph. Since the drag force is proportional to the square of the speed, the drag force on the car would become (240/60)²=16 times larger. This would be like brakes and probably eventually stop the car. Depending on the details of the car, it would possibly eventually reverse the direction of the car.

QUESTION:
I need to know when a pingpong ball is under 10' of water and when ball is released how much work is the ball capable?

ANSWER:
This is a peculiar question. To find out the work you can get out of the ball you need to calculate the work you have to do to get it there. Imagine pushing the ball down very slowly so that you can neglect the drag forces; also neglect the work you need to do to initially submerge it since that will certainly be neglegible compared to the work the rest of the way down. The data for the ball are a mass of m=2.7x10^-3 kg and a radius of R=0.02 m; the volume is V=4πR^3/3=3.35x10^-5 m³. The forces are its weight down mg=9.8x2.7x10^-3=0.0265 N and the buoyant force B up which is the weight of the displaced water (the density of water is 1000 kg/m³), B=ρgV=1000x9.8x3.35x10^-5=0.328 N; the net force on the submerged ball is therefore F=0.328-0.0265=0.301 N up. So, since s=10 ft=3.05 m, the work done to move place it down there is W=Fs=0.301x3.05=0.918 J=2.55x10^-7 kW·hr=0.677 ft·lb. Keep in mind that if simply released the ball will quickly reach terminal velocity and move up with a constant speed, most of the work being against the frictional drag force.

QUESTION:
A mathematician while he was a kid was punished by teacher to add 1 to 100. But he gave the answer in less than 10 minutes? How is it possible with out modern devices in 1700 NC

ANSWER:
It is actually pretty simple: 1+99=100, 2+98=100, … 49+51=100, so that is 4900. But we still have 50 and 100, so the answer is 5050. (This is not physics, though!)

QUESTION:
Recently in my university, we had to do an experiment on beta spectroscopy. They gave us a sample of radioactive strontium-90 and told us that it decays beta particles at energies of 0.546MeV and 2.274 MeV and below. After finishing the experiment and submitting the report, the lab instructors told me that i had understood the concept wrongly. I had thought that at the energies of 0.546MeV and 2.274 MeV i would find a high number of electrons emitted. They told me that this was not the case and that at these energies you would expect to find no electrons emitted. They told me that electrons were not emitted at this energies because the neutons would absorb all the energy and leave the electrons in the sample thus no electrons would be detected. However, my understanding is that at these supposed max energies shouldn't the eletrons be emitted and leave the atom behind, allowing us to detect it? Why does the electron not leave the atom when it has high energies? Also during the experiment if there is supposed to be no electrons detected why did i detect a lot of electrons at these energies that is orders of magnitudes higher than that of the background radiation?

ANSWER:
Let's first talk about what happens in β^- decay: a nucleus with too many neutrons to be stable changes a neutron to a proton (which stays in the nucleus), an electron, and a neutrino. The energy released by this decay is therefore shared between the electron and the neutrino (assuming that the nucleus, being enormously heavier than the other two particles, will have almost no recoil energy. The electron, as you demonstrate in your lab assignment, is pretty easy to observe. The neutrino, however, is extremely difficult to detect—the sun produces copious amounts of them and nearly all pass completely through the earth without interacting at all. When β^- decay was first discovered, it was a huge mystery because the electrons, the only thing observed, were not emitted with a single energy but with a broad spectrum of energies. A typical electron spectrum is shown in the graph above; in that graph you see a maximum near 0.1 MeV electron energy but, in this example, the total energy should be about 1.15 MeV. There seemed to be only two explanations: either energy is not conserved or else there is a third particle which was not observed and carries off some of the energy. Everyone believed the second explanation and the existence of neutrinos was totally accepted by all physicists. They were so hard to observe that it was 26 years after Wolfgang Pauli proposed them (1930) and they were observed by Cowan and Reines (1956).

In your experiment, there were two decays going on simultaneously: ⁹⁰Sr decaying to ⁹⁰Y and ⁹⁰Y decaying to ⁹⁰Zr. Both have a maximum but they are far smaller than the total energies of the decays. Nearly nothing after the first sentence starting with "They told me…" makes little or no sense but that is probably because you did not understand what they were trying to tell you.

QUESTION:
The centre of mass of a uniform ring lies at its geometrical centre i.e. outside the body but as we generally define centre of mass that if we apply a force at that unique point the whole body will move in a same way of whole of the mass is assumed to be concentrated there. So my question is how can we apply a force to a point not on body and still see the force's effect??

ANSWER:
If the force can be represented as a field, put the ring in a uniform field. For example, near the surface of the earth there is a nearly uniform gravitational force. If you drop the ring it will not rotate regardless of how you orient it at the time you drop it. The effect is exactly the same as if all the weight were acting at the center. Another way would be to put a very thin membrane across the ring whose mass was negligibly small compared to the mass of the ring; then apply a force at the center of the ring.

QUESTION:
We know that electric field lines do not intersect because they shows two direction but it is a vector quantity so we can find the resultant of the two direction but this not happen. Why?

ANSWER:
You may observe the electric field at any point in space and map the field. You will never find that that field crosses itself. Of course, if you have more than one source of field and you look at each separately, those fields will cross each other just about everywhere. However, the net field is the vector sum of all the fields everywhere. The electric field obeys superposition.

QUESTION:
We know that if a body rotates about an axis that is not the axis of symmetry then the angular momentum about a point on the axis does not point along the axis. If we rotate such body about such axis at a constant angular speed then a contradiction arises: since angular acceleration is zero, the torque about that axis has to be zero but since the angular mometum vector is changing (since the axis is not the axis of symmetry) a net torque is required. Where am I getting wrong? Please help.

ANSWER:
Your question is a little ambiguous since you do not clearly specify which axis you are talking about after the first sentence of your question; if you just say axis, I do not know if you mean the symmetry axis or the rotation axis. In the laboratory frame, though, the angular momentum points along the rotation axis and does not change. As you state, there are no torques on the system about the rotation axis so the angular momentum is not expected to change. But, you are probably looking at the problem from the body frame and see that the angular momentum is changing in that frame; but the body frame is not an inertial frame and therefore Newton's laws are not applicable (in particular, torque is not necessarily equal to the rate of change of angular momentum). That is probably what you are getting wrong.

QUESTION:
If I take my foot off the gas pedal on a flat road will my car eventually stop or do I have to hit the breaks? I tried to test it myself but I felt unsafe towards the end each time.

ANSWER:
That depends on the car. Some cars, when in gear, will move forward when idling unless brakes are applied. Certainly a car with manual transmission will not stop unless it is in neutral but will probably eventually stall if in a high gear at sufficiently low speeds; if it stalls, it will certainly stop soon after.

QUESTION:
can a body have a charge of 1.8�10^-19 C?

ANSWER:
The electric charge on an object is due to the excess or deficiency of electrons. The charge of an electron is 1.6x10^-19 C, so the net charge must be an integral multiple of this. So the answer to your question is no.

QUESTION:
What is the velocity of the electron when it orbits around nucleus? I mean the distance taken by electron on the perimeter of the orbit divided by time not its angular speed

ANSWER:
See this video for a derivation. For a hydrogen atom in its ground state, v=2.16x10⁶ m/s.

QUESTION:
How many Suns would it take, laid side by side, to reach a star 9.3 light-years away?

ANSWER:
The diameter of the sun is about 865,000 miles which is about 1.47x10^-7 ly. Therefore 9.3 ly is about 9.3/1.47x10^-7=6.33x10⁷ sun diameters (63.3 million).

QUESTION:
If the space inside an atom is totally empty, therefore a vacuum, why doesn't the atom collapse?

ANSWER:
If the space inside the solar system is totally empty, therefore a vacuum, why doesn't the solar system collapse? Because there is a force, gravity, pointing from the object to the center of the sun which keeps all the moving objects in the solar system moving in elliptical orbits. The same is true of the Bohr model of the atom except the force is the electrostatic force rather than gravity. You should be aware, however, that the inside of an atom is not a vacuum. The electrons are not really in simple orbits as the Bohr model would have it, but they have wave functions (distributions of the probabilities of their being at any place in space) which extend all the way into the nucleus of the atom.

QUESTION:
I've been thinking about this one, and cannot find a good way to reason it out. if you are in an elevator that is going down and you jump in there while its going down would your head hit the elevator's ceiling??

ANSWER:
If the elevator is moving with constant speed, it is exactly as if it were standing still—if you give yourself an adequately large velocity you will hit the ceiling. If d is the distance from your head to the ceiling at the instant that your feet leave the floor, then you will hit the ceiling if the speed v you launch with is greater than √(2gd) where g is the acceleration due to gravity. If the elevator has an acceleration a, the critical speed will be √(2(g-a)d). For example, if a=g you are in free fall and even the tiniest velocity will cause you to hit the ceiling (assuming that the elevator does not get to the ground before you get to the ceiling).

QUESTION:
I understand that the motion of a light source doesn�t add to or subtract from the speed, or change the direction, of the light that it emits. So, if I was a passenger on a spaceship, and I shone a flashlight from one side of the ship to the other, perpendicular to the direction of travel, the light beam wouldn�t strike the wall exactly where the flashlight was pointed, but rather a little towards the rear, as the opposite wall would have moved forward some small distance in the time it took the photons to cross the ship. And if the ship was traveling at, say, half the speed of light, the light beam would leave the flashlight at a noticeable angle. Is this correct?

ANSWER:
If the spaceship were moving with constant velocity, the light beam would go straight across the ship. As viewed from outside, the light would have a component of its velocity along the direction of motion equal to the speed of the ship, so it would still hit exactly opposite of your position which also moves with speed v). The speed of the light would still be c. If the width of the ship is W, you would see the light take a time W/c but the outside observer would see it take a time (W/c)/√[1-(v/c)²]. If the spaceship were accelerating you would see the light hit somewhere backward (forward) of where you aimed it depending if you were speeding up (slowing down).

FOLLOWUP QUESTION:
Your answer confirmed what I thought was the case for three-dimensional objects propelled mechanically through space. For example, an arrow shot from a bow across the spaceship would hit the bullseye or miss, according to whether the ship's velocity was constant or accelerating/decelerating. However, I thought electromagnetic radiation, such as light, was a different animal. What led to my original question was reading that the speed of light is a constant which can't be exceeded, so I assumed it couldn't take on additional velocity from the motion of its frame of reference. When you say "the light would have a component of its velocity along the direction of motion equal to the speed of the ship" it sounds like you're saying the light is leaving the flashlight at speed c, and then the ship's speed of, say, 10 km/sec is being added to make the effective speed c + 10 km/sec (viewed from outside) which I thought was impossible. My mental image of light radiating is of energy being "conducted" (analogous to electricity conducted by a copper wire) at a constant speed through an intrinsically motionless medium (space), not of a physical object being propelled.

ANSWER:
First a minor point. The arrow, speed u across the ship which moves with speed v, will have a speed u'=√(u²+v²), not u+v as you suggest. However, as you correctly surmise, this cannot be true for light since, in your example, you could not have the light with speed √(c²+10²) as seen by an outside observer. The speed must still be c. So, if the component along the direction of the ship is c_x=v, you must have c_y²+c_x²=c²=c_y²+v² or c_y=c√[1-(v/c)²]. Regarding what you would see in the ship, the principle of relativity states that the laws of physics are the same in all inertial frames of reference; if your ship is at rest, we could agree that anything aimed straight acoss will go straight across. In other words, there is no such thing as absolute rest—any inertial frame with constant velocity relative to another inertial frame would have the same laws of physics.

QUESTION:
How did the eclipse of the sun move from east to west across the earth.

ANSWER:
The shadow the moon casts on the earth (which is the eclipse) moves west to east, not east to west. A clear explanation was provided by NASA: "Because the Moon moves to the east in its orbit at about 3,400 km/hour. Earth rotates to the east at 1,670 km/hr at the equator, so the lunar shadow moves to the east at 3,400 � 1,670 = 1,730 km/hr near the equator. You cannot keep up with the shadow of the eclipse unless you traveled at Mach 1.5."

QUESTION:
I teach middle school science, however I've spent more time advancing my understanding in several of the other science disciplines I teach so my ability to work through this problem is above my current skill set. However I would really like to be able to walk my students through this scenario, in part because I'm not good with subtlety. I would like to know what other variables I need to measure/know in order to calculate the force a hypothetical student's head would experience at the moment it hit the floor after they tipped their chair back too far (despite numerous warnings not to). I imagine I'll need the Mass of the hypothetical student + the mass of the chair. The Time from when the hypothetical student's center of gravity passes over the point of rotation for the chair, the Height of the hypothetical student's head is above the floor. With Gravity being constant, I "should" have all the necessary information to calculate this (no chance at all will I let my students try to measure the force in real life) but I still have the niggling sense that I'm missing some variable.

ANSWER:
Let's first do the simplest estimate: imagine dropping the head from about h=1 m above the floor. The speed the head would hit the floor could be found from energy conservation, ½mv²=mgh or v=√(2gh)=√(2x9.8x1)=4.4 m/s. Now, you either need to know the distance the head stops in or the time it takes to stop. There is not much give in either the head or the floor, so let's just guess that the distance it takes to stop is 3 mm. If you do the kinematics, the acceleration of something moving 4.4 m/s and stopping in 3 mm is about 3000 m/s², about 300 gs! If you take the mass of the head to be about 4.5 kg, the force is about F=ma=13,500 N=3000 lb. The time over which this force acts is very short, about 4.4/3000=0.0015 s.

To be fancier, you would have to do the rotational problem. The student plus the chair have a certain moment of inertia I and when they rotate about the rear chair legs they acquire a rotational velocity ω when the center of mass has fallen a distance h. In that case ½Iω²=mgh and so ω=√(2mgh/I); now you can get the speed the head hits the ground by writing v=Lω where L is the distance of the head from the rear legs of the chair. But to calculate the moment of inertia of the student/chair would be very complicated and, in the end, the speed of the head would not be all that different from the simple calculation above. And you still have to make approximations over how far it would travel to stop. Just use the simple head dropping calculation to convince the students that the order of magnitude of the force would be a few thousand pounds.

QUESTION:
A swing is suspended by nylon ropes and connected to a limb that is 30 feet in the air. The problem is that the kids cannot build any momentum in order to swing. They pull and kick their legs but the swing will not swing. Is there a way to fix this swing?

ANSWER:
30 ft is a very long pendulum. The period T of a pendulum (the time it takes for one swing back to where it started) is approximately T≈2π√(L/g) where L is the length and g=32 ft/s² is the acceleration due to gravity. In your case, L=30 ft, T≈6 s. Now, if you watch a kid swing a much shorter swing, you will see that they "pump" once per cycle; this is called a driven oscillator and, driving with the same frequency as the natural frequency of the swing is called resonance and each pump will increase the amplitude. No doubt the kids, being used to a much shorter swing, are pumping with a period much shorter than 6 s, far off resonance, therefore not having the effect of increasing the amplitude. I suggest that you start them with a good push and tell them to pump each time they reach the bottom of the arc and moving forward.

QUESTION:
When a man going from pointA to pointB which is at 1km and return via same path according to physics the work done is nothing but man still feel tired why?

ANSWER:
First, what force moves the man along the path? It is the frictional force between the ground and his feet. Since friction is not a conservative force, the work done over a closed path is not zero. But walking is more complicated than that, way too complicated to discuss here. But, it is pretty easy to understand why we get tired—any time the muscles exert a force they are doing work whether or not they are actually moving something (see earlier answers).

QUESTION:
With respect to special relativity, is it correct (or not) to say that time dilation & length contraction are "real" because Nature has them in place to protect the upper speed limit "c"? Is there a better way to say this?

ANSWER:
First of all, what does it mean for something to be "real"? I would argue that, for example, length contraction is real because experimental measurements confirm its reality. Second, I find anthropomorphizing "Nature" to be fatuous; "Nature" does not "place" or "protect". Finally, you have it exactly backwards: time dilation and length contraction are the result of the fact that the speed of light is a universal constant, not vice versa. I would also point out that these would also be true if the speed of light were different from what it is; as illustrated by George Gamow's book Mr. Thompkins in Wonderland, they would be much more evident if the speed of light were very small.

QUESTION:
I am working on a science fiction book and have a bombardment of a planet happening. I would like to get an idea of the force of impact of 2 kinetic kill vehicles
1 is a 10 pound Depleted Uranium ball.
2 is a 100 pound Depleted Uranium dart.
They are traveling at 1/10th light speed.
They impact an Earth like planet.
I only need a ballpark value. If either or both are some how destroyed interacting with the atmosphere or cause some other effect, that would be good to know too.

ANSWER:
I am not sure what the material being depleted uranium has to do with anything. Also, "force of impact" cannot be calculated unless you know the details of the collision, in particular how long the collision lasted. You could estimate the kinetic energy classically for your projectiles because the speed is much less than the speed of light. 1 lb is about 4.5 kg, so K≈½mv²=½x4.5x(3x10⁷)²=2x10¹⁵ J=2000 TJ (terajoules); the Hiroshima bomb had an energy of about 63 TJ. If the collision lasted 10 s, this would correspond to a power of 200 TW (terawatts). To give you a feeling for the magnitude of this power, the total output of all power sources on earth is about 15 TW. The 100 lb projectiles would have 10 times the energy and power as the 10 lb projectiles. So those things would have quite a punch. Here is the problem that most sci-fi writers never think about, though. Whoever is firing these projectiles has to give them this energy—where are you going to get 200 TW in the middle of empty space?

Now, the second part of your question. These things have a speed (3x10⁴ km/s) which is much faster than the fastest meteor (72 km/s) and you know what happens to them—they burn up and break up. The recent (2013) Chelyabinsk meteor exploded at an altitude of about 20 mi and most of its energy (about 1500 TJ) was absorbed by the atmosphere. It had a much smaller speed than your projectiles (about 20 km/s compared to about 30,000 km/s) but a much larger mass (about 10⁷ kg compared to 4.5 kg), so the energies were quite comparable (1500 TJ compared to 2000 TJ). So I would guess that your projectiles would not do much damage to the objects on the surface of the planet.

QUESTION:
i'm traveling at the speed of sound and a gunshot is fired at the exact moment I am passing the gun, what is the resulting sound that I hear and for how long?

ANSWER:
The figure shows the plane at three times:

just after the gun has been shot;
at the time when the sound reaches where the plane had been at time 1;
at a time twice as long as 2.

Also shown are the corresponding spherical wave fronts of the sound from the shot. As you can see, the wave fronts never catch up with the plane. You will never hear the gun.

QUESTION:
I am a nurse at a long term care facility. My back hurts every time I get finished pushing a medication cart for the 8 hour shift. My question is...If the medication cart weighs 220 pounds when assembled, how much weight am I pushing given the fact it is on wheels?

ANSWER:
You may assume that the wheels almost remove the frictional force of moving forward; in other words, it takes very little force to keep it moving once it is up to speed. The times you need to exert a significant force on the cart would be when it speeds up or slows down. The more quickly you bring it up to speed or bring it to rest, the larger force you need to exert. So plan ahead and speed it up or slow it down gradually. Here is an example: if you sped the cart up to a speed of about 6 ft/s in 1 s, you would need to exert a force of about 40 lb, whereas if you took 2 s, the needed force would be about 20 lb. Also, when you turn a corner, go slowly since it takes a force to turn the cart also and the faster you take the corner, the greater the required force.

QUESTION:
I have a small (15' diameter) swimming pool. I have a round leaf net, about 17' in diameter, which I suspend over the pool as follows: I took 5 10' lenghts of 1" PVC electrical conduit, and joined them end-end to form a hoop/circle ca. 16' in diameter. The round leaf net has a drawstring, so I placed the 17' diameter leaf net on the ground, then placed the 16' diameter on top of the net. Then I folded the excess leaf net up above the conduit hoop and tightned the drawstring. So far so good -- net below hoop, edges wrap around hoop, drawstring holds the net on the hoop. I suspend this hoop-supported net from a single point in the middle of the circle formed by the hoop/net. From this central point, I have 10 "spokes" of cord (parachute cord) going to 10 points along the hoop. The pieces of cord forming these "spokes" were measured carefully to all be the same length. Here's my question: When I suspend this hoop/net from this central point and the 10 lengths of cord (spokes), the hoop/net does not want to stay in a plane. Instead, 2 opposite sides (e.g. at 12 and 6 on a clock face) lift

up, and the other opposite sides (e.g. at 3 and 9 in the clock face) drop down. I know in German, when a bycicle wheel gets stressed too much and deforms, this is referred to as an "eight" -- I suppose because from the edge it may resemble an eight. I suspect there is something similar going on with my hoop/net, but don't really understand enough to know for sure. As a followup question, I would be grateful for any thoughts as to how I might get this hoop/net to remain in a plane. Many people suggest just tightening up (shortening) some of the "spokes" but I have tried this and it does not help.

ANSWER:
The surface defined by your deformed loop is called a saddle point. Because the pvc is pretty flexible (you were able to easily bend it into a hoop), unless the weight is distributed very symmetrically, it is quite possible for this kind of warping to take place. To minimize this uneven distribution of weight, your spokes should be connected to the points where you have used straight through connectors to connect the pvc pipes and the points opposite as shown in blue in my drawing. But I suspect that there will still be an asymmetric distribution of mass and the circle will deform again. If so, then you will need to add crossbars as shown in red; this should keep the hoop pretty resistant to deformation. You will probably only need 3 spokes now and if it does not hang so the plane of the hoop is horizontal, you can just adjust the lengths. It would also be good to have a cord coming straight up from the center, maybe another piece of pvc to which the spokes could be attached or just another cord. You could get some idea of the balance by just hanging it from the center.

QUESTION:
Even though radio waves can't raise electrons to higher energy levels due to their extremely low energies do they still interact with orbiting electrons in some way? Just curious, how exactly do radio waves and microwaves interact with orbiting electrons within atoms when their energies are far too low to excite electrons to higher energy levels?

ANSWER:
You know that a radio wave will not travel unaffected through a large mountain, so something is interacting with it. The electromagnetic wave is electric and magnetic fields, and electric and magnetic fields interact with electric charges. The wave sees mainly electrons. In a conductor, the conduction electrons are essentially free-floating and respond to the fields approximately like a free electron would. In a nonconductor, there is still an interaction with the crystal as a whole, but again via the atomic electrons. These interactions result in the properties of refraction and attenuation of the incident wave. The details of how you model these interactions are beyond the purposes of this site.

QUESTION:
I was fishing and the water was rough. Waves crashing all over. Then I look over and see a perfect about 20M circle of calm water while the whole lake is thrashing about. It lasted for about 5mins. The water was about 20feet deep and there was not large rock anything different where the circle appeared. The calm circle did not appear to be like a vortex or whatever of water. It was calm and not rotating. Can you explain the physics behind this or at least give me a guess of what you think caused this. (It almost looked like an object that was not visible to the naked eye caused it but I know thats not true)

ANSWER:
I was just about to reply that I have no idea but just googled calm water in a storm. It turns out that there is an old sailor's trick to calm water in choppy conditions: just pour a little oil on the surface. There seems also to be physics understanding of this phenomenon. For a good overview, see the discussion on Physics Stack Exchange. Perhaps there was some oil in your calm area.

QUESTION:
If a bullet of mass m moving at v strikes a stationary rod of mass M at its center and passes through exiting with u, there is simple conservation of momentum if the rod is free to move/slide: m(v-u)=MU. What happens if the bullet strikes and passes through the rod at its end (same v and u), causing the rod to move forward while at the same time start to spin?

ANSWER:
Your question is essentially the same as an earlier question except that in that question the collision was stipulated to be elastic. In your case, there is no conservation of energy equation and so there are only two equations—conservation of linear momentum and angular momentum. But, there are only two unknowns, U and ω, because you stipulate u to be known. (Also, d=L/2 for your question.)

ADDED THOUGHTS:
I thought it might be interesting to look at the energy change for these collisions. It is straightforward algebra to show that U=m(v-u)/M and ω=6U/L. The energy before the collision is E₁=½mv². Referring to the earlier answer, the energy after the collision is ½mu²+½MU²+½Iω². For the simple case of hitting the center (ω=0) I find E₂=½m(u²+(m/M)(v-u)²), and for the case of hitting the end I find E₂'=½m(u²+4(m/M)(v-u)²). Since E₂'>E₂, less energy will be lost in the case where the collision happens off center (assuming u and v are the same in each case). For example, if m=M/4 and u=½v, I find ΔE=-(11/32)mv² and ΔE'=-¼mv². In both cases, energy is lost in the collision.

QUESTION:
If the farther out in space we look, we are looking back through time how is it we do not see ourselves?

ANSWER:
You are looking at how things which are far away looked when they were at an earlier time. But none of those things are at the position where you were in the past so you would not see yourself.

QUESTION:
How could I measure the viscosity of pizza sauce (and other materials) using "at home" equipment? I want to determine the viscosity of a sauce, then take the pizza sauce and place it on a turntable whose speed can be controlled and see what speed is required to make the sauce flow from the center to the edge.

Then, I want to alter the sauce and make it more and/or less watery (changing its viscosity) and measure that new sauces viscosity.

Then, I'll take the new sauce and re-measure the turntable speed necessary to make the new sauce flow from the center to the edge.

Last, I want to replicate the tests enough times so I can create an equation that would allow me know what turntable speed would be necessary to correctly flow the sauce based upon the viscosity of the substance.

ANSWER:
Why not just experiment with sauces until you find the right thickness to achieve what you want? Measuring the viscosity (not an easy task) is just an unnecessary step in the process.

FOLLOWUP QUESTION:
The reason is that I want to be able to alter the sauce, test the viscosity, then know how much to alter the speed of the rotating table consistently. Knowing, for example, a 1% increase in viscosity requires a 5% increase in rotation speed allows me to continuously alter the sauces and know with certainty the speed the table must rotate.

ANSWER:
This must be a science fair project or something because it will certainly not be of any help in making pizzas in the real world. In a pizza you want to spread the sauce uniformly over the whole area, right? What causes the sauce to move out on the rotating turntable? There is a (fictitious) force F, the centrifugal force, which pushes the sauce out, F=mv²/R for a mass m with speed v when at a distance R from the center. But, the speed depends on the distance from the center, v=Rω where ω is the angular velocity; so F=mRω²and an ounce of sauce experiences a bigger force as it moves out. In other words the sauce will tend to all be pushed out the the rim of the pizza regardless of its viscosity. If the viscosity of the sauce too large, the centrifugal force might be too small to move the sauce at all, so there would be the tendency for the sauce to stay in the center. In any case, I cannot imagine that it is possible to use rotation to get a uniform spread of sauce.

If you still want to pursue this, I found the description of a straightforward experiment to measure viscosity μ. You get a tall cylindrical container and fill to a depth d with the fluid (density ρ_f). Drop a sphere (density ρ_s radius R) into the fluid and measure the time t it takes to reach the bottom; then the velocity of the falling sphere was v=d/t. The viscosity is then μ=(4R²g( ρ_s-ρ_f))/(9v). It will be a little tricky since the sauce is not transparent. Also, it is important that the sauce be homogeneous, no chuncks in it.

QUESTION:
Lately I've been dealing with the Work, Force and Energy thing. I can't deny the fact that understanding some of the equations included in most of the textbooks has been one hell of a challenge to me. I'm currently trying the understand something here: we know that the work done (W) is equal to the change of kinetic energy. Is it also the same when it comes to potential energy? Does it amount to the done work as well? I never see the work of the gravity force included in the equations. I think I know the answer but I just want to make sure. There's something more bugging though--when we write the conservation of energy equations for a certain setting, do we include the work there or is it already included? For example, if there's a body on the top of an inclined plane and we take friction into account, should the equation be "potential energy at the top + kinetic energy at the top = potential energy at the bottom (0=0) + kinetic energy at the bottom + the work done by the friction"? Getting an answer would mean a lot to me as an engineer-to-be. :)

ANSWER:
This question verges on a violation of a "concise, well-focused" question as required by site groundrules. I will do a couple of examples to try to clarify work-energy for you. The "guiding principle" which is always true is that the the total work W_total done by external forces on an object of mass m is equal to the change in kinetic energy, ΔK=½mv₂²-½mv₁². The picture is a mass m on an incline θ being pushed by a constant force F up the incline and moving a distance s; for simplicity, I choose m to be at rest when you start pushing. Other forces on m are its own weight W=mg, vertically down, and a possible frictional force f pointing down the plane. For the time being I will assume f=0. Only the component of W along the plane W_s=-mgsinθ will do work. The net work w (sorry if it is confusing to have big W and little w) is w=Fs-W_ss=Fs-mgssinθ=ΔK=½mv² and so v=√[2s(F-mgsinθ)/m].

Rather than derive the gravitational potential energy, U=mgy, I will assume that you already know that. I will now do the problem over but using potential energy. w=Fs=E₂-E₁=(K₂-K₁)+(U₂-U₁)=½mv²+mgh=½mv²+mgssinθ=Fs. Solving this, you find exactly the same answer: v=√[2s(F-mgsinθ)/m]. So here is how I look at it: potential energy is a very clever bookkeeping device to keep track of the work done by a force which is always present like gravity. It just makes life a lot easier to not always have to calculate the work done by some force that you know is always there. So, here is the so-called work-energy theorem: the change in total energy, kinetic energy plus any potential energy you have included, is equal to the work done by all external forces, E₂-E₁=W_ext; here an external force is any force for which you have not introduced a potential energy function. I like to rearrange the work-energy theorem as E₂=E₁+W_ext—what you end up is what you started with plus what you added (or subtracted if W_ext<0).

Finally, let's include the frictional force, f=μW_N=μmgcosθ. (μ is the coefficient of kinetic friction.) The work which the friction does is negative because it is opposite the direction of s, w_f=-μmgscosθ. Therefore, W_ext=Fs-μmgscosθ=½mv²+mgssinθ; solving, v=√[2s(F-mg(sinθ+μcosθ))/m]. You might wonder why we did not introduce a potential energy function for f. The reason is that there are two kinds of forces in nature, conservative forces and nonconservative forces and the latter kind cannot have a meaningful potential energy function; friction is a nonconservative force. But this answer has rambled on long enough and that is a topic for another day!

ADDED NOTE:
I see that I did not answer your question ("…if there's a body on the top of an inclined plane and we take friction into account, should the equation be "potential energy at the top + kinetic energy at the top = potential energy at the bottom (==0) + kinetic energy at the bottom + the work done by the friction"?") explicitly. You have it wrong. It should be that the total energy at the bottom equals the total energy at the top plus the work done by friction; don't forget that, as in my example going up the plane, the work done by the friction is negative. Some books like to write it your way replacing the friction part by the work done against the friction; I actually do not like that one bit.

QUESTION:
Why does dust move behind fast moving cars?

ANSWER:
Because there is a pocket of turbulent air behind the car which moves with the vehicle.

QUESTION:
If you were on the moon and used an electric motor - would the armature spark as it does here in an atmosphere like ours?

ANSWER:
It is not possible for a visible spark to form in a vacuum, so the answer is no.

QUESTION:
As a physicist do you think it's theoretically possible for energetic photons of x-rays and gamma rays (keV-MeV) to be converted to visible light photons (1.77-3.1 eV) with the compton scattering effect despite the HUGE energy difference?

ANSWER:
It is pretty easy to figure this out. The change in wavelength in compton scattering is λ'-λ=(h/mc)(1-cosθ) where h is Planck's constant, m is the mass of an electron, c is the speed of light, and θ is the angle of the scattered photon; (h/mc)=2.4x10^-12 m=2.4x10^-3 nm and is called the Compton wavelength of the electron. The largest shift is for the largest angle, θ=180⁰ where cos180⁰=-1 so the largest possible shift is (Δλ)_max=4.8x10^-3 nm. Visible light has wavelengths around 1000 nm, soft x-rays (the closest to visible light) around 1 nm: so that would require (Δλ)_max≈1000 nm>>4.8x10^-3 nm. So the answer to your question is no.

QUESTION:
You can't pressurize a liquid, correct? So why is it when you turn your hose off at the faucet bib, the water still is under pressure when you squeeze the sprayer trigger? It still shoots our with force.

ANSWER:
Certainly you can pressurize a liquid. Most liquids are almost imcompressible, so increasing pressure does not significantly decrease volume.

FOLLOWUP QUESTION:
BUT, the pressure is from a giant water tank, correct. So , when you shut off the faucet, you've isolated the water tank. Therefore, the pressure should be gone.

ANSWER:
Why do you think that isolating the water makes the pressure go away? The water deep in your giant tank has to support the weight of all the water above it, so the pressure gets bigger as you go deeper. And the pressure at the top is not zero because atmospheric pressure is pushing down on the surface. I suggest you put on a bathing suit and swim down to the bottom of the tank and then decide whether there is pressure in the water down there.

QUESTION:
I am writing a novel and am trying to understand how my interstellar spacecraft will behave. It is going to Iota Persei. 34.38 ly away. I was thinking it would travel at .6c (somewhat arbitrary but...). If I accelerated the space craft at a consistant 9.8m/s, it would take ~212.44 days at a distance of ~.1744 ly. Presumably, it would take about the same time and distance to slow the craft as it approached Iota Persei, roughly speaking. While I've no idea of my craft's mass, it would be large, like 40,000 people large, and be propelled by an anti-matter drive. Would the energy output of the antimatter drive need to increase due to the increased mass as speed increased? Is it feasible to suggest that an antimatter drive would be able to produce enough energy to accomplish the above scenario?

ANSWER:
I have answered variations on this question several times in the past. The most useful for you to read and understand compares acceleration of an object as measured from both a rest frame and from the frame of the object; I recommend you read both this and all the links in that answer. [The derivations of the equations below can be found in the earlier answers.] From the perspective of your space ship the velocity v as a function of time t is v=(a₀t)/√[1+(a₀t/c)²] where a₀ is the acceleration as measured by the ship (g in your case). In order to achieve this acceleration, you must exert a force F=ma₀ where m is the mass of the ship. The velocity is plotted in red in the graph; reading from the graph, the time when v/c=0.6 is when a₀t/c≈0.8=9.8t/3x10⁸. Solving, t=2.45x10⁷ s=0.78 yr=284 days. The position as a function of t is x=(mc²/F)(√[1+(Ft/(mc))²]-1)=(c²/g)(√[1+(gt/c)²]-1); for gt/c=0.8, x=2.58x10¹⁵ m=0.273 ly. At this point you turn off your engines and coast at v/c=0.6. The distance you need to go to get halfway to the star is 16.9 ly and it will take you 16.9/0.6=28.2 years. The second half of the journey will take the same time, so the time for the whole trip is T=2(28.2+0.78)=58 years.

Your answers (212 days, 0.17 ly) are different from mine (284 days, 0.273 ly) because you have assumed that the acceleration, viewed from the earth, is the same as the acceleration (9.8 m/s²) seen from the ship. The black curve in the graph shows the ratio of the acceleration seen from the earth compared to the acceleration seen from the ship; reading from the graph, at 0.6c the ratio is about 0.5, 4.9 m/s². Be sure to read the discussions of acceleration in the earlier answers if you want to understand this issue with acceleration in special relativity.

Since there is no such thing as an antimatter drive engine, I can make only a rough assessment of how feasible this is. The force you need to apply during the 0.78 year accelerations is F=mg. What is m? The space station has a mass of about 400,000 kg and you have 40,000 passengers. Let's say each passenger needs one space station of mass to support his/her needs, so the required force would be 1.6x10⁹x9.8=1.57x10¹⁰ N. The energy you have to supply to the ship to v/c=0.6 I figure to be about 36x10²⁴ J and the amount of mass converted with 100% efficiency into energy would be about 4x10⁸ kg and you would need that much again to stop it. So your ship of mass 1.6x10⁹ kg would have to carry 8x10⁸ kg of fuel, one third of the total weight! Obviously, I have not included this mass change in my estimates of time. And, of course, there is the issue of how you are going to store 4x10⁸ kg of antimatter. The whole scenario certainly does not look feasible to me!

Finally, note that all the distances and times are as measured by the earth bound observer. In other words, the passengers will not be 58 years older when they arrive at their destination. Because of length contraction, they will have a shorter distance to travel. Ignoring the accelerating periods, I can estimate the time of the trip by just assuming they travel 0.6c the whole trip. In that case, T'≈T√(1-.6²)=58x0.8=46.4 yr.

QUESTION:
If energy cannot be destroyed or made, only transferred from one form to another, how does the level of matter and energy in the universe stay at a constant, if black holes can consume both matter and radiation in huge quantities?

ANSWER:
If a black hole consumes an amount of energy E, its mass increases by E/c². If it consumes an amount of mass M, its mass increases by M.

QUESTION:
I have 1,000 kilowatt hours of energy. Could you help me understand how much weight I would need to lift an object up 100 feet in the air to use that 1,000 kWh of energy up completely? This is not homework. I have tried researching online, but every answer I have seen ends up being different than the last. Trying to understand for a meeting I have with some friends at work.

ANSWER:
I will work in SI units because a Watt is a Joule/second (and because scientists prefer SI units): 100 ft=30.5 m. The energy to lift a mass M to a height h=30.5 m is E=Mgh=299M where g=9.8 m/s² is the acceleration due to gravity; the answer will have units of Joules if M is in kg. Now the energy available is 10³ kW⋅hr(3600 s/hr)(1000 W/1 kW)=3.6x10⁹ J. So, 299M=3.6x10⁹ so M=1.204x10⁷ kg=2.65x10⁷ lb.

QUESTION:
If a fly is in a car that is travelling, say, 50 mph... and the fly is not resting in the car (i.e. he is flying, not attached to any surface), does that mean he has to fly at 50 mph as well? Or would he be pressed to the back window of the car?

ANSWER:
The fly flies relative to the air in the car. But the air in the car is at rest relative to the car. Therefore the fly flies exactly the same as she would if the car were at rest.

QUESTION:
How is this possible. My brother while playing baseball hit the ball in such a way that when it broke the glass window in the living room it left a perfect circle. What are the chances of that happening?

QUESTION:
If an asteroid was in space but not moving, and you were within reach of it, and you pushed on the asteroid to get to the spaceship, would you lose force because the asteroid isn't moving and you were acting upon the asteroid? (The asteroid wasn't pushing against you) Wouldn't the asteroid go faster than you?

ANSWER:
You have misunderstood this problem on many levels. First, there is really no such thing as moving or not moving in an absolute sense. If you are on some other asteroid and see this one moving with a speed of 500 mph it is not at rest; but if you get in a space ship and go 500 mph to just keep up with the asteroid, it is at rest. If you are in a frame where the asteroid is at rest, its lack of motion has nothing to do with how it would react to a force you might exert on it. And, Newton's third law says that if you exert a force on the asteroid, it always exerts an equal and opposite force on you. And, whether you or the asteroid goes faster after the push depends only on your relative masses. An asteroid is most likely much more massive than you and therefore you will be moving faster after the push. For example, if the asteroid is 1000 times more massive than yours, its speed after the push will be 1000 times smaller than yours.

QUESTION:
What will be the work done by a person who is moving forward and carrying the bucket in his hand?

ANSWER:
The person does zero work on the bucket assuming that the bucket moves horizontally with constant speed. However, the person does do work; see an earlier answer.

QUESTION:
If you could please explain the experiments that supposedly confirmed quantum entanglement and "spooky action at a distance" I would appreciate it. The experiments I read about all involved a pair of entangled particles, that are then moved "far apart". Since they are entangled, one has an "up" spin and the other a "down" spin. However, we don't know which is which until they are separated and one is observed. The it is found that by observing one particle to have, let's say, the "up" spin, miraculously the other particle "far away"has the down spin, with the information between them being transmitted instantaneously even though they are far apart. What I don't get is, how is this different than having two cards; and ace and a king. Both face down. You don't know which is which until you observe them. You place them "far apart". Then you flip one over, and see it is the ace. Instantaneously, the other card becomes the king. This does not seem like "spooky action at a distance". You observed one of two possible particle states, and therefore, the other particle must have the other state. Obviously, I am missing something, but can never find the explanation.

ANSWER:
If entanglement were as you currently understand it, one up and one down, this would not be entanglement at all and there would be nothing mysterious about the second one being, instantaneously, the opposite of the first once you had revealed it. But that is not how it works because quantum mechanics, which is the operative physics in the realm of particles, describes the state of a particle. Let us first start by looking at a single particle which has two possible states which I will denote as |+> for up and |-> for down. A particle does not have to be one or the other, it can be a mixture of the two, for example 25% |+> and 75% |->; as long as the two percentages add up to 100%, this is a possible state. Now, when you observe the particle by measuring whether it is up or down, you will find it in one or the other (three times more likely to find it in |->) and now the measurement has put it into a state which is purely up or down; this is called "collapsing the wave function". In a typical entanglement experiment the two particles are initially in a state 50% |+> and 50% |->, half up and half down and therefore the total spin of the two must be zero. When you observe one to be up, the other is instantaneously observed to be down no matter how far apart the two are at the time of observation—that is the spookiness! The trick is to devise an experiment which clearly demonstrates that this is what is happening, not what you described as one being up and the other down from the beginning. It is pretty technical and tricky to understand this, but Roger Pennrose in his book The Emperor's New Mind has a pretty good explanation.

QUESTION:
If I were to coil 500' of 1/2" OD rope in a single layer, how large would the circle be?

ANSWER:
No homework.

FOLLOWUP QUESTION:
I am a 49 yo woman making a rug. Even my kids are too old for that kind of homework.

ANSWER:
If you lay the rope out straight, the total area will be 500x12x½=3000 in². If you coil it up, the area will be the same and the area of a circle is πR². Therefore the radius of your rug would be R=√(3000/π)=30.9 in.

QUESTION:
Frankly I wasn't sure if this was a crank question or deeply profound, I'll let you be the judge. I've been an electronics hobbyist all my 50 years, and I've used resistors, capacitors, and less often inductors. I started to wonder how many different types of passive electronic components were possible. Were there more than just these? I searched for and found something called a memristor which apparently IS a brand new kind of component that behaves differently than any of those. Wikipedia has a diagram. I'm curious, there are components on all the outside edges, what about the diagonals, what do these represent? They have different equations so they MUST behave differently, what ARE these components?

ANSWER:
As in many situations where you are trying to count something, it becomes problematic how to define what you are counting. What about diodes? Are Zener diodes different from diodes in your count? You should really think of this kind of question as qualitative and not quantitative; the answer is not of any importance. Your questions about the device itself would be better directed to an electrical engineer. The diagram you refer to is not a circuit, it is merely a representation of how these four devices relate to each other and to the quantities voltage, current, charge, and flux. The outer edges show how devices manipulate the quantities, e.g. changing the current through a resistor changes the voltage across it (dv=Rdi). The diagonal lines indicate relations between the corner quantities, e.g. current is the motion of charges (dq=idt).

QUESTION:
What happens to the atmosphere inside a cupping vessel when a flame is introduced and causes a partial vacuum (negative pressure) - which allows the therapeutic vessel to adhere to the skin surface? What are the physics that explain this happening?

ANSWER:
Cupping vessels have been in use for thousands of years. Ancient Greek and Roman physicians used them to assist in blood letting. Chinese medicine uses them to treat a variety of maladies; cupping therapy is generally considered pseudoscience by modern medicine. The idea is to provide suction on the surface of the skin and is achieved by first heating the cup and air inside and then placing it on the skin. As the air inside cools, the pressure decreases. The physics of this pressure decrease can be understood by examining the ideal gas equation which relates pressure P, temperature T, volume V, and amount of gas N: PV=NRT where R is a constant which depends on the units you use. So, as you can see, keeping the volume and amount of gas constant, if the temperature decreases the pressure must decrease. The fact that P∝T is sometimes called Gay-Lussac's law.

A similar thing happens when canning food in glass jars. The canning is done with the contents very hot and a lid which is slightly domed is affixed. As the contents cool, the pressure decreases causing the dome to pop inward toward the contents, signaling that a good seal has been achieved.

QUESTION:
I am a hot air balloon pilot. I am trying to develop a mathematical formula, for calculating where my Scoring baggy will land, when I participate in a Ballooning event. I am given a location in a Pilots briefing, as to where the Scoring X's are located. I attempt to fly to that particular location, drop my Scoring Baggy on the X, and earn points depending on how close I am to the center of the X. (Each Leg of the X is approximately 100' to 300' long, depending on the Ballooning event.)

The Scoring Baggy (6 ounces) has a constant weight.
My Balloon's Speed, as I approach the Target, is a variable. the Scoring Baggy will share that speed once I release it, Correct ?
And, the height I release the Scoring Baggy from, at the time I release it, is also a variable, correct ?

My question is, is there a Mathematical formula that will calculate, how many feet from where I release the Scoring Baggy, it will land ? And how long it will take ?

ANSWER:
This is not a simple question. Although it would be simple if air drag were neglected, I suspect that it is not negligible for the bag weighing only 6 oz. I will do the calculation without air drag here. I will not include the details, just the final results. The time t in seconds it takes for the bag to hit the ground is t=√(2h/g)=¼√h where h is the height in feet from which you drop it and g=32 ft/s² is the acceleration due to gravity. For example if you drop it from h=144 ft, t=3 s. The distance x in feet it will travel horizontally in this time is x=vt where v_x is the horizontal speed of the balloon in ft/s when you drop it; so if v_x=20 ft/s (about 13.6 mph) and you drop it from 144 ft, x=60 ft. The equation for x can be written as x=¼v_x√h which is handy if you do not care about the time. Note that the weight of the bag does not come into this at all.

Warning: this is pretty mathematical and probably not a computation you would want to do in the heat of a competition! For my own interest, I want to estimate how much error is introduced by neglecting air drag. It is much more complicated if you include air drag. For this case, I need to do the calculation in SI units rather than English units. The reason I need to use SI units is that I will estimate the air drag force as F_d≈¼(v_y²/A) where v_y is the vertical velocity of the bag and A is the area it presents to the onrushing air; this estimate is correct only for SI units because it has things like the density of the air at sea level built into it. Now, it is my understanding that the speed of a hot air balloon moving horizontally is the same as the speed of the wind; in other words, from the perspective of a person riding in the balloon, he is in perfectly still air. This greatly simplifies the problem because the bag will drop straight down as seen from the balloon, i.e. it should strike the ground directly below the balloon. So, the bag sees two forces, its own weight mg down and air drag up ; Newton's second law becomes ma=m(dv/dt)=-mg+¼(v_y²/A). This is a first-order differential equation has a solution v_y=-[√(g/c)]tanh([√(gc)]t) where c=A/(4m). This solution is also a differential equation since v_y=dy/dt where y is the distance above the ground; solving this differential equation, y=h-(1/c)ln(cosh([√(gc)]t). A graph of this function compared to the case above for dropping from 144 ft (note that 144 ft=44 m) is shown. (I approximated the area of the bag to be 0.01 m²≈16 in²=4x4 inches.) The time for the bag to hit the ground is now 3.33 s rather than 3 s, approximately 10% longer meaning that you should drop it when you are 66 ft from the target rather than 60 ft. If you are lower the correction is smaller, if you are higher the correction is larger. Given the circumstances under which you must act, I would expect the inclusion of air drag to be unnecessary unless you are at a very high altitude and that you should just drop it when you are about x=¼v_x√h from the target.

ADDED COMMENT:
It occurs to me that I have assumed that the wind speed and direction are the same at all altitudes. This will not be true in the real world and I am told that taking advantage of this is how hot air balloons can get some control over direction of travel. Obviously, trying to do a calculation including this would be impossible other than for a particular set of wind velocities as a function of altitude.

QUESTION:
The ideal gas law says that pv=nrt. If I have a balloon filled with an inert gas and put heat into the balloon while holding the volume constant the pressure will increase on the left side of the equation and the temperature will increase on the right side of the equation. If I remove the heat source and allow the balloon to expand the volume will increase and due to the gas laws the pressure will decrease by the inverse of the volume keeping the value for pv constant. However, on the right side of the equation with a constant number of molecules the temperature will decrease. How can pv remain equal to nrt with the temperature decreasing and pv remaining a constant value?

ANSWER:
What makes you think that PV will remain constant? This is called an adiabatic expansion, a process where no heat enters or leaves the system. What remains constant is PV^γwhere γ is a constant which depends on the gas. For example, for a monotonic gas γ=5/3 and for a diatomic gas γ=7/5. Once you know what the new P and V are you can get the new T: T=PV/NR.

QUESTION:
Work done by ship's engine = KE of ship + Work done to overcome frictional forces.
Work done by ship's engine - work done to overcome frictional forces = KE of ship
Eventually when the ship travels at constant speed, and all work done by engine is used to overcome frictional forces, then mathematically,
Work done by engine - work done to overcome frictional forces = 0
But KE of the ship is not zero. So where is the source of the ship's KE coming from?

ANSWER:
While the ship is accelerating, the engine is both overcoming drag and accelerating the ship. The drag is dependent on the speed, gets larger as the speed gets larger. Eventually, the drag will become equal the force the engine applies so the net force will be zero so the ship will move at constant speed. But I do not understand your question since the work done by engine was clearly adding kinetic energy to the ship during the acceleration time. Your error was to assume that the work to overcome friction was always the same which is clearly not the case.

QUESTION:
If the Sun is comprised of largely unignited fuel, and enough so that it can last for billions of years, how can it appear to be a fiery ball that is entirely aflame, even to the core? Why is the Sun not a hot, dark, ball?

ANSWER:
It is a basic physics fact that the amount of radiation from an object with temperature T is proportional to T⁴; also, the wavelength λ of the most intense radiation is inversely proportional to T. So, it is not possible for an object to be both hot and dark.

QUESTION:
I just wanted to know if a ball that is thrown downward would have an acceleration of 9.8m/s² or it would depend upon how much force the ball is thrown with. Is the acceleration of a stone thrown upward the same as that of a stone thrown downward?

ANSWER:
The acceleration of any object depends, according to Newton's second law, only on the sum of all forces acting on it. Ignoring air drag, an object going either upward or downward has only one force on it, its own weight, so its acceleration is g=9.8 m/s² downward; this results in the object speeding up if moving downward and slowing down if moving upward but both have the same downward acceleration of magnitude g. During the time you are throwing it down (up) there is a larger (smaller) acceleration because of the force your hand is applying, but that disappears the instant the object leaves your hand. If you take air drag into account, there is an additional force which always points in the direction opposite the velocity.

QUESTION:
Is inertia a form of energy? If not then why does the blades of a fan move even after the supply of energy is stopped because energy cannot be created nor be destroyed?

ANSWER:
When a fan is spinning at full speed, the only reason it needs power from the motor is that friction is constantly taking energy away—friction in the motor and bearings and air drag. With no friction, once you got up to speed you could turn the motor off. In the real world when you turn the motor off the fan slowly stops as friction takes its kinetic energy away and turns it into heat; always a "balance" as you expect.

QUESTION:
If light was trapped in a dome of mirrors and the light source was cut off by another mirror and there were no holes to have the light leak out would the light go on forever.

ANSWER:
Not in the real world. There is no such thing as a perfect mirror. See an earlier answer.

QUESTION:
I understand that adding energy to an object increases the motion of its molecules, and that the movement of molecules creates heat. But what I don't understand is WHY the movement of molecules creates heat. Why does an increase (or decrease) in molecular motion change temperature?

COMMENT:
First of all, you need to learn the vocabulary used in thermodynamics—see the faq page. You will then see that your question should have been written as:

I understand that adding energy to an object increases the kinetic energy of its molecules, and that the kinetic energy of molecules determines the temperature. But what I don't understand is WHY the kinetic energy of molecules determines temperature. Why does an increase (or decrease) in molecular kinetic energy change temperature?

ANSWER:
The simplest answer is rather arcane: Absolute temperature is defined to be proportional to the average kinetic energy per molecule. In what sense, then, can we understand why changing the average kinentic energy of the molecules causes you to feel different. It is easiest to talk about an ideal gas for which T=2<KE>/(3k) where <KE> is the average kinetic energy per molecule and k=1.38x10^-23 J/K is Boltzmann's constant. (Most everyday gases are pretty well described as ideal gases.) Molecules are continually striking your body and transferring momentum as they bounce off. This causes the molecules in your skin to gain energy, i.e. get hotter. Now, your skin is hotter than the interior part of your body so heat starts flowing, causing the inside part of your body to become hotter. Meanwhile, your body is doing what it does to keep your body temperature at 96.8⁰F. As the temperature of the air gets hotter, your body has more and more trouble cancelling out the flow of heat from the air and eventually fails and you get heat stroke or at least start to feel sick.

QUESTION:
So while I was a little kid due to numerous headaches I had, I was scanned in an MRI machine. I was feeling a little anxious when the machine was put on and my mom came to me to ease my anxiety. She had her wallet with her and pretty much all her cards went dead. What exactly causes magnetism to destroy payment/membership cards?

ANSWER:
The magnetic strip is just like magnetic recording tape. There is a layer of very fine particles which are magnetizable. Data is written on the tape by using an electromagnet called the recording head; when the magnet is turned on the particles become magnetized. So the data is written in stripes in a code, sort of like the UBS labels used to scan products at the cash register. In a magnetic strip, the card moves by a tiny coil in which a current is caused to flow when the magnetic stripe goes past it. Since magnetic fields are used to create the magnetized particles, magnetic fields can be used to destroy them. Even a relatively weak field, if present for a long enough time, can mess up the data on a magnetic strip. An MRI machine has a huge field and it would easily demagnetize the strip.

QUESTION:
If you spun a mirror-smooth metal ball at just about the maximum possible RPM... would the angles of light reflection change any due to the speed (as opposed to from the distortion of shape from centrifugal forces)? If so, would it be a visible difference, or something you'd detect with a precision instrument?

ANSWER:
There would be an observable effect but not due changes in direction of reflected rays. Wavelengths of light reflected from parts of the sphere moving toward (away) from the observer because of the rotation would be shifted to shorter (longer) wavelengths; this is called a blue (red) shift. This effect would be very small for speeds which would be achievable.

QUESTION:
If a fishbowl (full of water and a fish) was propelled forward (any direction really) at 2g. Would the fish feel the effects of the 2g acceleration in the water in the same way that a person feels this effect in air and therefore cause the fish to be pushed against the side of the bowl if it did not swim to remain in the centre of the body of water or would the fish be held in stasis in the body of water (Water neutralising the effect of the acceleration)?

ANSWER:
I have answered variations of this question before. You should read an earlier answer to see the complications and how you can understand your question. In that answer I addressed two possibilities applicable to your fish—if the fish has a density greater than that of water, it will move backward until it hits the back side of the bowl and if it has a smaller density it will move forward until it hits the front side of the bowl. But your question is different because the fish has the same density as water. Therefore it will remain in the center of the bowl if it started there. But the fish is accelerating, so he has to feel a net force; the water behind him will exert a force which is larger than the water in front of him does. So, the effect of the acceleration is not "neutralised", he feels it. It is analogous to your sitting in an accelerating car and the back of the seat exerts a forward force on you even though you remain seated in the seat.

QUESTION:
As I'm sure you know, LIGO has detected 3 black hole mergers in the last few months. This means that there have been and likely are a huge number of black holes in the universe. In parallel, the math says that 95% of the mass of the universe is not visible to us. Could it be that there are millions or billions of black holes floating around a few light-years from the nearest thing and that they would be virtually invisible - we could only detect them if they happened to line up with further distant light sources such that we might detect gravitational lensing. Could they account for the 95%? I know you don't do astrophysics but thought that the answer might be so simple that you might be happy to address it.

ANSWER:
Keep in mind that I have no real expertise in cosmology. Two things, I think, will shoot down your idea. The 95% number you quote is like 20% dark matter and 75% dark energy. The dark energy component cannot be thought of as "mass-like" because it makes itself known in the accelerating expansion of the universe, so it would appear to be some kind of repulsive force. The dark matter which everyone is trying to observe cannot be in black holes because to cause the things it is thought to cause (like the anomalous rotation of spiral galaxies) it needs to be spread out over huge volumes of space which, of course, black holes are not. My take on dark matter (absolutely not mainstream) is that perhaps dark matter is the luminiferous æther of the 21st century, i.e. it is not some invisible "stuff" but rather its purported effects are indicative that we do not really understand gravity as well as we think we do. Certainly general relativity is an incomplete theory since no one has successfully quantized it. You may know, but dark energy, at least the evidence for it, has an interesting history. When Einstein formulated general relativity, the expansion of the universe was not known and the universe was thought to be static; but with just gravity this is not possible—it will expand forever or expand until it collapses, depending on the initial conditions. Einstein therefore empirically inserted something he called the cosmolgical constant. Later when Hubble discovered that the universe is expanding, Einstein removed the cosmological constant and dubbed it his "biggest blunder"; now it is coming back into play, but still just empirically. There are doubtless millions or billions of black holes as you speculate, because all stars greater than a few solar masses end up that way. It is known that nearly all galaxies have a supermassive black hole at their centers. I would guess these came from early-universe stars coalescing. And early-universe stars were huge, thousands of times more massive than the sun because only hydrogen was available to make early stars. Still, I have learned that the size of the universe is just about impossible to comprehend and at its relatively young age still, I would guess these remnants of earlier stars are still a pretty small fraction of the mass of the entire universe.

QUESTION:
how can a photon have a frequency since moving at the speed of light would prohibit any internal vibration since time has stopped inside the photon. Textbooks show a tiny " football " shaped photon with an oscillating wave inside the " football " photon ?

ANSWER:
You are right, a photon does not have a frequency but does have an energy which is directly related to a frequency. Electromagnetic radiation is an example of particle-wave duality. Light is not a particle or a wave, it is a particle and a wave. If you design an experiment to "prove" that light is a wave (particle) you will "find" that it is a wave (particle). If you have light waves with some frequency f, that light can equally correctly be thought of as a collection of particles (photons) which have an energy E=hf where h is Planck's constant. The oscillating wave depected inside your "football" is a cartoon representation and really has no quantitative meaning. Incidentally, you can never have a precise knowledge of the energy of a photon because of the uncertainty principle; therefore you cannot have a precise knowledge of the frequency of the wave. Your "football" is called the envelope of the wave function and is representative of the uncertainty of the position of the photon.

QUESTION:
If one were calculating the design load for an hvac, would a cooler operating/"burning" 13 watt LED bulb be expected to add less demand on the system than a 13 watt halogen bulb which operates at a higher temperature and gives off less light? For the hypothetical, please assume the bulbs are installed in identical windowless closets. Does the additional light from the LED bulb convert to heat when it is absorbed by the closet walls? Would the temperatures of each closet be the same after a considerable period, or would the closet with the halogen bulb be warmer?

ANSWER:
To answer this question, you have to make some approximations. Before turning on the lights, the walls and room air are in equilibrium with each other, at the same temperature. I will assume that no heat escapes from the wall/room system and that the walls are ideal blackbodies. So the walls radiate a blackbody spectrum continuously and absorb energy from the air at the same rate. Now turn on the lamps, each radiating 13 Joules of energy per second, but it is not blackbody radiation because there are different amounts of heat (infrared) and visible radiation for the lamps. Nevertheless, it is my suspicion that eventually the two rooms will both come to where the walls and air are at the same temperature but are both increasing because of the constant input of 13 W of power. When that happens, I would think both would have the same temperature. This is an idealized situation; in the real world, heat would also flow out of the closet into the environment but, if everything is identical, both rooms should eventually be the same—energy is energy. (Thermodynamics has never been my forté, so I welcome comments here!)

QUESTION:
I am an Electrician and my Fitter friends have been arguing about the answer to a seemingly simple question. If I have a solar panel and it is rated to withstand a 1kg steel ball from a height of 1 metre, what size hail stone would it take to break it?

ANSWER:
This is a kind of tricky question because you have to estimate how fast the hail stone will be going when it hits and that speed will depend on the size of the hail stone. I will assume that the speed that it hits is the terminal velocity of a stone of a given mass; the terminal velocity v_t is the maximum speed something achieves when dropped from a very high height and may be shown, for of a sphere of mass m and radius R, to be approximately v_t≈√(9.7m/R²) at sea level. The density of ice is about 10³ kg/m³, so we can write m=4x10³πR³/3=4.2x10³R³. So, the speed of a hail stone of radius R may be approximated as v_t=√(4.1x10⁴R); for example a golf-ball (R≈0.021 m) sized hail stone will have a speed of about 29 m/s and a mass of about 0.039 kg.

Now, we need to make an estimate of the average force the steel ball exerts on the panel when it is stopping. The force is the linear momentum mv of the ball divided by the time the collision lasts; I will assume that any hard ball will stop in about the same time, so what matters is the momentum. It is easy to show that the speed of the steel ball when it hits is v=4.5 m/s so mv=4.5 kg·m/s; so we need to find the size of a hail stone with a momentum of 4.5 kg·m/s. But, we know the speed and mass from above, so 4.5=(4.2x10³R³)√(4.1x10⁴R)=√(7.2x10¹¹R⁷); solving, R=0.031 m, m=0.13 kg, v_t=36 m/s. So my best estimate would say that golf-ball sized hail would not break your panel but hail larger than a golf ball by 50% or more would. Keep in mind that all these calculations are estimates, not precise calculations.

QUESTION:
referring to this link: https://www.youtube.com/watch?v=Dt0JyXMjbNs At 18:48, why the partial differentiation notation can be "taken out" of the bracket?

ANSWER:
The equality of interest here is shown in the figure taken from the youtube video. It works simply because of the product rule for differentiation:

(∂/∂x){ψ*(∂ψ/∂x)-(∂ψ*/∂x)ψ}={(∂ψ*/∂x)(∂ψ/∂x)+ψ*(∂²ψ/∂²x)-(∂ψ*/∂x)(∂ψ/∂x)-(∂²ψ*/∂²x)ψ}=ψ*(∂²ψ/∂²x)-(∂²ψ/∂²x)ψ.

QUESTION:
I would like to perform a calculation of a man descending a tower using cords and subjected to the action of the winds. I have been trying to find some equations but it is somewhat difficult due to the drag force. My main objective is to calculate the maximum horizontal distance x that the man could reach due to the wind action against the technician descending the tower using cords.
Tower height: 78 m (please consider up tower as a zero reference).
wind speed: 20 m/s
Mass of man: 70 kg
Man descending with constant speed and slowly.
So, please what is the maximum distance when the man is at 66 m from the top of the tower ?

ANSWER:
As shown in the figure, there are three forces on the man, his weight mg, the tension in the cords T, and the force of the wind F. The equations of equilibrium are F-Tsinθ=0 and mg-Tcosθ=0. Solving, tanθ=x/y=F/(mg), so x=ytanθ=Fy/(mg). Now, how can we get F? There is a very good approximation to the drag force by air at sea level moving with speed v: F≈¼Av² where A is the area of the object presents to the onrushing air (and which works only for SI units). Finally, x=Av²y/(4mg). If I approximate g≈10 m/s² and A≈1 m² and use your numbers, x≈9 m.

ADDED COMMENTS:
I should have emphasized that air drag calculations are only rough calculations, probably accurate to maybe ±20%. Also, for your situation, the general approximation for x as a function of y is x≈y/7. Also, if the horizontal displacement seems too large, keep in mind that 20 m/s is a very strong wind, about 45 mph which is gale force.

QUESTION:
why a pen in a cup of water appears to bend away from the normal since the refraction of light entering a higher-index medium bends towards the normal?

ANSWER:
Because the light which allows you to see the submerged part of the pen is going from water to air, entering a lower-index medium.

QUESTION:
I understand why, if the speed of light is the constant from any viewpoint, the variable between two observers has to be their clocks but I don't understand why one is the reference clock and the other is the clock that, in the example below, is the one that is slower. On a train going a meaningful percentage of the speed of light past an observer on earth where both an observer on a train and the observer on the earth can see a light source on the train, both will measure the speed of light at the same rate so their clocks have to be calculating at different rates. My question is, how does one determine which observer is called stationary and which observer is moving? As a corollary question, if a rocket ship is moving away from the earth what is the difference between the rocket moving away and the earth moving away from the rocket?

ANSWER:
Actually, any inertial frame may be chosen as the rest frame. A's clock runs slow as B measures it and vice versa.

FOLLOWUP QUESTION:
Thank you for your reply. In that example neither person would actually age at a different rate though each would think the others clock was slow?

ANSWER:
You are thinking classically, but there is no such thing as absolute time. The rate at which any clock which moves relative to you (that's why they call it relativity) runs more slowly than yours, period. It is not a case of those clocks "seeming" or "appearing" to run slowly, they do run slowly. Your statement "…each would think the others clock was slow…" was wrong—they were both slow as observed by the other. I think you would understand the whole thing better if you read my earlier answer on the twin paradox.

QUESTION:
If wired glass takes 50 lbs of pressure to break and you throw a 1/4 pound orange at the window from 35 feet away how fast would you have to throw the orange to break the window?

ANSWER:
There is no way to calculate this with the information given.

FOLLOWUP QUESTION:
Alright. Well the orange is ripened so it is firm. Like fresh from the grocery store naval orange. I am not sure how to quantify that. The orange is the size of a baseball as well. So presuming that the orange is roughly the size and shape of a standard baseball but not quite as soft, we could use a baseball instead of an orange. And i did mean 50 psi not lbs. Would that help take out some of the confounding variables?

ANSWER:
I can only do a very rough estimate. And I certainly cannot replace the orange with a baseball since the orange is much softer, but I will use the size of the baseball, about 7.5 cm=0.075 m in diameter. So, I will assume that the orange compresses by an amount about a quarter of its diameter, say d≈2 cm=0.02 m; for simplicity, assume the orange stops with constant deceleration. If an orange of mass m=¼ lb=0.11 kg and speed v hits the wall and stops in a time t, the force it experiences is F=mv/t and, because of Newton's third law, this will also be the force the orange exerts on the window. During the collision, there will be some average area of contact between the orange and the window; I will estimate that area to be a circle of radius R=0.01 m so that the area will be about A=πR²≈3x10^-4m². The pressure will then be P=mv/(At); converting to SI units, P=50 psi=3.45x10⁵ N/m². For the assumptions I have made (uniform acceleration) it is easily shown that t=2d/v, so P=mv²/(2Ad)=3.45x10⁵=0.11v²/(2x3x10^-4x0.02); solving, v=6.1 m/s=13 mph and t=0.0066 s. Keep in mind that this is a very rough estimate of the speed, but it does indicate that it is not surprising if the glass broke since almost anybody can throw an orange with a speed more than 13 mph.

QUESTION:
If i was travelling towards a light pulse in opposite direction to its velocity wouldnt length contraction and time dilation make its velocity even bigger than c, for my length measuring rods will become shorter so i will measure more distance travelled by the light and my clocks will run slower so more time for light to travel ? I get it when the distance between the pulse and observer is increasing but not the other way.

ANSWER:
The very fact that the speed of light is the same for all observers is the very reason for time dilation and length contraction! So, the answer is no; the speed of light is the same for all observers.

QUESTION:
Hi. I'm creating an aluminum can crusher and I need to know the weight required in order to crush the can from a set height knowing the amount of Pa required to crush the can. I'm crushing a regular sized coke (375ml) can and need to get it to about 4mm long once crushed. On a pneumatic crusher I only needed 600000 Pa or 6 Bar to achieve this result. However I am trying to create another method of crushing the can using a weight (unknown) that drops 0.5m on top of the can, theoretically crushing it to the required length. My question is: what weight in Kg is required to crush a can when dropped 0.5m to reach a crushed can length of 4mm, or a conversion of 6 bar? And maybe the equation used to work it out?

ANSWER:
I looked up the geometry of the can: the height is about 128 mm and the radius is about 30 mm, so the crush distance is s=128-4=124 mm=0.124 m and the area of the top is about π(0.03)²=2.8x10^-3 m². The force which the press exerted on the can was F=PA=6x10⁵x2.8x10^-3=1680 N. This means if you put a mass 1680/9.8=170 kg it will crush the can. Your idea, I presume, is to drop a smaller mass from some height (0.5 m) above can. The way I will attack this problem is to first calculate the work done by the hydraulic press W₁ and equate it to the work done by the dropped mass W₂; I can then solve for the mass for any height. I will do it in general so you can estimate the mass from any height. W₁=Fs=1680x0.124=208 J; v=√(2gh) where g=9.8 m/s² and h is the height (0.5 m in your case); W₂=½mv²-mgs=mgh-mgs=208 J. Solving, m=208/[g(h-s)]. In your case, h=0.5 m and s=0.124 m, m=56 kg. Keep in mind that this is just an estimate. I would be curious to know if it was close. There is an earlier answer which might be of interest to you; it might give you an idea of how you might improve your crusher.

QUESTION:
I am a high school math teacher and have a student who asked a question I wasn't prepared to try to answer: Looking at the twins paradox in relation to special relativity, someone who left their twin behind on Earth while travelling near the speed of light would return to a much older twin while seemingly to have not aged as much themselves. Is there an equation or way to calculate this relationship in terms of how much one person would age in relation to another given a specific velocity in relation to each other?

ANSWER:
To understand the twin paradox intuitively, I recommend your student read my earlier answer on this topic. The equation you can use to compare elapsed time is T_away=T_home√[1-(v²/c²)] where T_home is the elapsed time for the earth-bound twin, T_away is the elapsed time for the traveling twin, v is the speed of the ship, and c is the speed of light. The easy way to understand this is that the traveling twin sees the distance to the star shorter by a factor of √[1-(v²/c²)].

QUESTION:
I'm reading the book "Aurora" and a question occurred to me: If a spaceship is traveling at 0.1C and is in communication with Earth (via radio waves), and C is relative to the observer, wouldn't the communications be distorted or "out of phase" between a "fixed" point (Earth) and the ship? Compare that to "Ender's Game", where as people are traveling close to C, they can not communicate with fixed objects.

ANSWER:
I do not know what you mean by "…C is relative to the observer." The speed of light is the same for all observers. To answer your question, you should read an earlier answer where your question is answered for a speed of 0.8c. Notice that your speculated "distortion" is experienced by both observers, not just the traveler. It also depends on the direction of travel—toward or away from the earth. For v=0.1c, the effect would be much smaller but qualitatively the same—for the traveler moving away, both observers find the communication slowed down, for the traveler moving toward the earth, both observers find the communication speeded up. Your statement that "…they can not communicate with fixed objects…" is false; it is just slowed down or speeded up.

QUESTION:
If a ball impacts a rod off center from the rods center of mass how does the rotational momentum and energy gained by the rod affect the amount of gained translational momentum and energy? For example, if the ball impacted at the rods center of mass the equations for conservation of momentum and kinetic energy could be used to determine the velocities of the objects after the collision, but with energy and momentum being transferred to the rotation of the rod would the ball loss more or less velocity or would the energy transferred remain the same? Would the rotational energy and momentum for the objects after the collision be solved for separately from the translational transfer?

ANSWER:
No homework.

REPLY:
Thank you for the response. I'm unsure if this could be a homework question for someone else, but I've posted in forums with no response or had the post deleted. This is an attempt to gain an understanding of basic physics concepts and I was given this site as a possibility to find answers from a local university. Would it be possible for me to ask the question again in a couple of months to certify that this is not a homework question or does "no homework" amount to no homework "type" questions that could require tutoring or classes?

ANSWER:
Well, you seem sincere about wanting to learn the physics, so I am happy to answer your question. The algebra to get the final solution is very tedious, but it seems you are most interested in getting the physics right, so I will set the problem up and if you really need all the final answers for a particular situation, I will leave that to you. Since you refer to conservation of energy, I assume the problem of interest is an elastic collision. The system is shown before and after the collision in the figure. A point mass m with velocity v approaches a uniform thin rod of mass M and length L; v is normal to the rod and the collision occurs at a point a distance d from the center of mass of the rod. After the collision, the center of mass of the rod has a velocity U, the point mass has a velocity u, and the rod has an angular velocity ω about the center mass. Because the rod is uniform, its moment of inertia about the center of mass is I=ML²/12; the angular momentum and kinetic energy of a rigid body are L=Iω and K=½Iω² respectively. There are no external forces or torques on this system, so angular momentum and linear momentum are both conserved. We now have three equations:

energy conservation: ½mv²=½mu²+½MU²+ML²ω²/24
linear momentum conservation: mv=mu+MU
angular momentum conservation: mvd=mud+ML²ω/12

Take stock of what we have: three equations and three unknowns—u, U, and ω. The physics is done, only algebra remains.

QUESTION:
I would like clarification on force experienced by a skydiver with air resistance factored in (not talking about during opening shock of a parachute) Someone at my work stated that you will experience a number of negative G's while simply skydiving and with only air resistance causing the negative G Force. Does this seem to be accurate? My understanding is that air resistance could not cause more than 1G of negative force or else you would technically be going upwards due to gravity only exerting 1 G of force downward while skydiving.

ANSWER:
The air drag can be well approximated as F_d=cv² where c is a constant (but which depends on the geometry of the falling object) and v is the speed; this is a force acting in the opposite direction from the velocity. When you jump from the plane, you originally have a horizontal velocity but the horizontal component of the drag will take away the horizontal velocity and you will end up falling vertically; I will assume the skydiver is falling vertically (or just jump from a hovering helicopter). In addition to drag, the only other force acting on the skydiver is his weight W=mg, a force vertically down. So, writing Newton's second law, ma=cv²-mg, where a is the acceleration. You start out with an acceleration g down, but as v increases a gets smaller and smaller until eventually v=v_t=√(W/c); this is called the terminal velocity. Now the acceleration is zero so you fall with a constant speed v_t. At no time was the drag bigger than the weight. But, can the drag be greater than the weight? If you were propelled downward from the helicopter with a speed greater than the terminal velocity, you would slow down until your reached the terminal velocity. In that case, the drag is greater than the weight; however, I cannot imagine this scenario in skydiving. By the way, you can control the terminal velocity—spread eagle will give a slower final speed than curled up in a ball.

QUESTION:
A ball dropped from the top of the mast of a moving ship falls straight down parallel to the mast. But with the movement of the ship, it also falls down on an inclined LONGER path. In other words, it travels over two different distances at the same time. Which is the actual 'real' distance traveled?

ANSWER:
There is no such thing as "the actual 'real' distance traveled". The distance traveled in one frame of reference will be different than for another frame. For your question, if the ball is dropped from a mast of height H, it will travel a distance H in a time t where t=√(2H/g) where g=9.8 m/s² is the acceleration due to gravity. Now, look at the path of the ball from a dock the boat is passing; if the boat moves with a speed v, the distance the boat travels before the ball hits the deck is vt=v√(2H/g). The distance between the starting and ending points of its path (blue line) is √[(2v²H/g)+H²]. For example, if H=4 m and v=5 m/s, then t=0.9 s and the distance is 6.03 m. The path followed by the ball in the dock frame is a parabola (red line), so its length is the distance the ball has traveled; it is rather complicated mathematically to compute this distance.

ADDED THOUGHT:
I got curious and calculated the distance the ball actually traveled in the dock frame. You need to integrate ds=√[(dx)²+(dy)²]=[√(1+(dy/dx)²)]dx=[√(1+4x²)]dx from x=0 to x=vt. The result is s=¼[2vt√(1+4v²t²)+sinh^-1(2vt)]; for the example above, s=9.19 m.

FOLLOWUP QUESTION:
Still confused, I appreciate your reply. The ball in the first frame of reference hits the deck at the same time as in the second, where it travels the longer parabolic path or distance. These are two observable facts. To travel the longer distance in the same time in the second frame of reference, the ball must either accelerate faster than 9.8 m/sec² - or time itself must slow down, which is both unlikely. Your "complicated mathematical computation" supposedly addresses the problem. But can a mathematical computation correct two conflicting observable facts? I am thinking of Albert Einstein's skeptical quote, "So far as the theories of mathematics are about reality, they are not certain; so far as they are certain, they are not about reality."

ANSWER:
The photograph above is a time-lapse photo of two balls falling. The red ball is simply dropped. The white ball is given a horizontal velocity. The only force on each ball is its own weight which points vertically down and is responsible, of course, for the downward acceleration of the balls. Since there is no force in the horizontal direction, the white ball's velocity in the horizontal direction remains constant. The thing to note is that both balls, having equal vertical accelerations, fall at exactly equal distances in equal times. This is exactly what is happening for your question. Your error was the false assumption that "…the ball must…accelerate faster than 9.8 m/sec²…"

QUESTION:
I question if there is an antimatter equivalent to the graviton which for now I will call a repelon. The repelon would be a massless, repelling force carrying particle. The antigravity force of antimatter would act the same but opposite of gravity. If that were the case, then at the creation of matter and antimatter an equal amount of gravity and antigravity would exist. Following subsequent loss of most of the antimatter, the repelon wave would still be prorogating through the universe, as indicated by the detection of gravity waves. If so, could this energy wave account for some or all of the dark energy? If there is a repelling force associated with antimatter, could the interaction of the gravity and antigravity forces also contribute to the loss of homogeneity in the soup at the big bang?

ANSWER:
A graviton (hypothetical particle which has never been observed and there is no good theoretical description of) would be a boson and bosons do not have antiparticles. Since there is no satisfactory theory of quantum gravity, I do not want to poo-poo your theory, but something is causing a repulsive force in the universe and if that field were quantized there would have to be a quantum of the field, and maybe a repelon would be a good name for it. I should remind you here that it is stated on the site that I do not usually do astronomy/astrophysics/cosmology.

QUESTION:
Why does the sun does not heat up the air as it heat up Earth and Water?

ANSWER:
The air does heat up, but because it is a gas most of the radiation which strikes it passes through without leaving much energy to heat it.

QUESTION:
when we stand on the surface of earth, every square centimetre of our body experiences atmospheric pressure that is equal to mass of 100 kg. This pressure is sufficient to grind our body, but nothing happens to us. Why?

QUESTION:
As I understand the situation (perhaps incorrectly), if you were to throw a charged baseball, the baseball would radiate owing to the moving charge. The radiation would transport energy and momentum from the charged object into the field. Robbed of energy, it would eventually stop. Here, stop means that it would revert to a radial electric field and display no magnetic field at all. Is that true? If it were true, our physics seems to be very off. This is probably a beginner question, but it is not for some homework. What's the truth here?

ANSWER:
An accelerating charge radiates, one moving with constant velocity does not. If there is no gravity and no air drag, the ball will not radiate and not lose energy. But, here is something really interesting: if you are in a frame which is accelerating relative to the ball, the ball will radiate in that frame. See an earlier answer.

QUESTION:
I have attached some images to show the question. In one, a heavy package is affixed to the drone firmly. In the other it hangs by a rope that is affixed on the same level as the engines. The weight of the drone is 5 lb and the weight of the load is 7 lb. Where has the center of gravity moved?

ANSWER:
In the diagram the yellow Xs are the centers of gravity (COG) for the unladen drone and the load; the load COG is a distance L below the drone COG. The red X is the COG for the combination. Suppose the weight of the drone is W_drone and the weight of the load is W_load. Then the distance d by which the new COG is below the COG of the drone is d=LW_load/(W_drone+W_load). In your example, d=(7/12)L. Clearly, when L is increased d is increased. And, contrary to your expectation, the COG goes farther down when you hang the load from a rope increasing L. What might be a problem for stability, though, is that if the drone is not horizontal the rope will still be vertical and so the COG will no longer be on the center line of the drone.

QUESTION:
I practice medicine in Houston Texas. I am trying to workout a math formula that you are probably very familiar with. It is called the Schrodinger Equation. Below is one calculation based on Time dependent Theory. This would be a calculation for a single PHOTON PARTICLE, whereas this equation is for a wave of Photons? At any rate, Its all greek to me!!! I was hoping you help me in calculating the "pattern " of x rays (ionized Photon particles) that pass thru a slit and it calculated projected bands density, shape and size of the bands. (See Ex A below for what I am trying to calculate) Could you tell me which formula I would most likely use for the best predictable result? So, If took the Photonic energy wave (lets say...a wave length at 10 nm this is the conversion: As you can See, I could use Joules, Volts, Hertz ,etc...

Would you mind walking me through this one conversion example using Schrodinger Equation.? I am not a physicist and this Schrodinger Equation formula and values are Greek to me. But if you could set up the equation with this example at 10 nm wavelength, then I can simply cut and paste and copy and the do all the quantum math I want after that.!!! Thank you for considering my request.!

EX-A) Bands formed after wave passes thru Slit(s) Theoretical Photonic wave

ANSWER:
Something tells me that you are making this way harder than you need to. It looks to me like all you are really interested in is the pattern, called a single-slit diffraction pattern. You apparently found a reference for someone who did this using the Schrödinger equation, but you really do not have to go through all that in order to get an analytical expression for the intensity pattern on the screen; all you need to do is to use classical physical optics, a standard first-year physics calculation. I presume that the first figure, where you have used some computer code to get the photon energy of radiation of a particular wavelength is not a mystery to you; but you do not need this at all for the intensity, you just need the wavelength λ and the slit width a. As shown in the figure, you may specify a point on the screen by specifying either the angle θ or the distance y; the results are I=I₀sin²[πasinθ/λ]/[πasinθ/λ]² or I=I₀sin²[πay/(λD)]/[πay/(λD)]² where D is the distance from the slit to the screen; the quantity I₀ is some constant which would be determined by the intensity of the radiation on the slit and the exposure time. The details are shown on the Hyperphysics web site. Let's do an example, λ=5 nm, D=10 cm, a=32 nm; then I/I₀=sin²[2y]/[2y]² where y is in cm. The result is plotted in the graph of I/I₀vs. y.

QUESTION:
A bird is just started to fly in a open cage in spring balance system when it just started to fly no change in reading why, the reading should increase, if not, then why?

ANSWER:
This question is different from the on I usually get which is a bird in a closed box. A good explanation of this kind (closed box) of problem can be found in an old answer. In your question, the answer depends on how the bird takes off. If she propels herself upward by just using her wings, the reading of the scale will decrease by the weight of the bird; it's the same as if you simply lifted her off the perch. If the bird takes off by jumping, that is pushing off using her legs, the reading will increase and then, when she is airborne, it will return to the weight of the cage without the weight of the bird.

QUESTION:
The way I understand accelerating to the speed of light, is that mass increases, so that at the speed of light we would have infinite mass. What happens if we slow to an absolute stop – we stop walking, the earth stops spinning, the earth stops rotating around the sun, the sun stops rotating around the galaxy and the galaxy stops moving through the universe – Would our mass drop to zero and would we cease to exist? Or in other words, do we exist because we are moving?

ANSWER:
In relativity, you can always imagine yourself in a frame in which the object you are observing is at rest. The mass of something has meaning only relative to some frame of reference. If the object is at rest in some frame, it does not have zero mass, it has what is called the rest mass, m₀ this is the mass referred to in Newtonian mechanics and this is what the mass seems to be unless the velocities are comparable to the speed of light, speeds we never see in everyday life. In general the mass m depends on its speed v, m=m₀/√[1-(v/c)²] where c is the speed of light.

QUESTION:
If you were to heat up a substance and heat it up to the point where is protons and neutron split into quarks and cool it down,will you get the same elements the substance consisted of or will you get the elements created during the big bang(hydrogen,helium,lithium and beryllium)?

ANSWER:
For starters, you cannot ever have a free quark. As you heat it up, you cannot know for sure what will happen in detail; it will depend on how many atoms you have and the volume in which they are confined. You can be sure that the first things that happen will be at the atomic level—molecules dissociate and atoms ionize. But as long as the energies of the atoms and electrons are not great enough to cause nuclear reactions, the whole system will be dynamic, constantly changing as atoms capture free electrons and other atoms get more ionized. As all this happens, many photons will be generated and absorbed. As more energy is added, collisions begin to excite and dissociate nuclei and add more photons to the mix. As more energy is added, exotic and unstable elementary particles are created and decay. If you now start taking energy away, can you really imagine that it could possibly go back to its original state? You would eventually get back to where you had only neutral atoms, but it is impossible to know what atoms you would end up with. It would be imperative to know all the details of how you added and then subtracted the energy, and even then you could not predict everything that would happen.

QUESTION:
If you are encased in water in a jet and go into a high g turn would you feel any g force?

ANSWER:
I think you will find the answer to this in a recent answer; the main difference is that in that example the acceleration was due speeding up rather than due to turning. Essentially, you are accelerating because the jet is turning and no matter what, there must be a net force on you which provides that acceleration.

QUESTION:
A steel ball is released just below the surface of thick oil in a cylinder. How would you describe, qualitatively, the motion of the ball?

ANSWER:
There are three forces on the ball. Its weight down, a buoyant force up, and the frictional drag up. The weight and buoyant force are constant. The drag depends on the speed of the ball, probably approximately proportional the speed, but certainly increasing with speed. The ball starts at rest and starts falling down. Therefore, the drag begins increasing which means that the net force down is decreasing, so the magnitude of the acceleration is decreasing. As the speed continues to increase (but at a decreasing rate), eventually the sum of the buoyant and drag forces becomes equal to the weight and the ball will fall with constant speed.

QUESTION:
We know that when a magnet is taken near to iron it gets attracted. It is because of induction of magnetic field of opposite nature in iron. Is there no any phenomena that field of some nature is induced in iron so that the iron gets repelled?

ANSWER:
There is no other field which will have the opposite effect on iron. However, there are some materials where they are repelled by a magnet. These materials are called diamagnetic and the magnitude of the effect, compared the attraction for iron, is very small. Some elements which are diamagnetic are bismuth, copper, and lead. Actually, all atoms are diamagnetic but in most cases they are smaller than ferromagnetic or paramagnetic forces.

QUESTION:
I have to understand the classic formula for finding acceleration due to gravity's force. G x (m x M)/d Squared = m x (9.8 m/s/s) dividing the mass of the object out . . . G x M/d Squared = 9.8 m/s/s Finding acceleration from the mass x the universal constant and the inverse square of the distance to the center is so incredible and useful but still perplexing. My main question is: How are the units of Meters per second every second; m/s/s obtained when you are dividing Kilograms of mass? maybe the units of big "G" drives the conversion to m/s/s through the units m³ kg^-1 s² ? This classical formula is very interesting but very challenging to visualize. The numbers all work but an explanation of the resulting units are difficult to find. Please if anybody is able to visualize the logic behind 9.8 meters a second squared and can share, it would be GREATLY appreciated.

ANSWER:
Let's start from the beginning: Newton's law of universal gravitation states that the force between two point masses m and M separated by a distance d is proportional to the product of the masses and inversely proportional to the square of their separation, F∝Mm/d²; this is also true for two spheres if d is the separation of their centers. To convert this into an equation, we introduce a proportionality constant G, F=GMm/d². For a given M, m, and d we must measure F to determine G: G=Fd²/(Mm). Now, what are the units of G? N⋅m²/kg². This probably answers your question, but let's finish it off. If M>>m, m will accelerate toward M with an acceleration a=F/m=GM/d², à la Newton's second law. Since 1 N=1 kg⋅m/s², the units of a are [(N⋅m²/kg²)(kg)/m²]=[(kg⋅m/s²)(m²/kg²)(kg)/m²]=m/s².

QUESTION:
A photon of blue light has more energy than a photon of red light. Correct? Yet, when Herschel divided the sun's light into its colors with a prism and measured the resulting temperature produced by each color in thermometers, he found that red light increased a thermometer's temperature more than blue light. Blue light has more energy; so, why doesn't blue light make a thermometer warmer than red light? Also, apparently the thermometers were painted black. If that makes a difference.

ANSWER:
The answer is rather prosaic. As shown in the figure, since the red light is refracted less than the blue light, the region spanned by the thermometer at the blue end of the spectrum contains a smaller fraction of the total light than at the red end.

QUESTION:
I take my chocolate Lab to the river to play. The ball-flinger (Chuck-it) imparts backspin to the ball, and water flied off. F=MA - the ball is losing mass; is it accelerating?

ANSWER:
The mass is not disappearing, so the total mass of the system is not changing. You should actually think of Newton's second law as F=dp/dt where p=mv is the linear momentum, so if there is no net force the momentum does not change. Then, if you watch the water coming off one drop at a time you can calculate the new speed of the ball due to the expulsion of the drop; therefore the ball does accelerate in that sense but not because the mass is changing but that the momentum stays the same. As a simple example, let the mass of the ball plus one drop be M and the mass of the drop be m; the velocity u of the drop is in the same direction as the original velocity V of M and the velocity of the ball after the drop has left (now of mass M-m) is U. The momentum before the drop leaves is p₁=MV and after the drop leaves is p₂=mu+(M-m)U. Since p₁=p₂, the new speed of the ball is U=(MV-mu)/(M-m). Notice that if m is very small compared to M, U≈V.

QUESTION:
I've been watching Bill Nye's new show on Netflix, and, while I love Bill Nye, I think he just said something that's totally incorrect. He said "glass blocks infrared light." Sure, glass will hold in a gas or fluid, but he said that it actually blocks infrared light. That seems impossible to me because it would mean that infrared cameras couldn't have glass lenses, and also because sunlight through a window feels warm on my face. Does glass block infrared light?

ANSWER:
Actually, what he said is completely true. The graph shows the transmission of radiation as a function of wavelength through several thicknesses of a particular type of glass. Visible light has about 0.4-0.75 μm wavelengths and is about 90% transmitted through the glass. There is a precipitous drop in transmission around 2.5 μm and the glass becomes nearly opaque to wavelengths above 5 μm. The region 0.75-2.5 is referred to as near infrared (IR), so the glass does transmit most of that IR, but the IR radiation which is radiated by objects with temperatures between the freezing and boiling points of water is on the order of 10-12 μm. You are right, glass lenses are useless for infrared optics. Lenses have to be made from a material which transmits IR; common materials which are used are calcium flouride and germanium. This is part of the reason that IR optical devices are expensive. This property of glass is also how greenhouses work. Visible light enters gets absorbed and heats the inside. But as the inside warms up, it radiates IR but that radiation cannot get back out. The heating you feel from light coming through a window results from absorption of visible and near IR which does get through.

QUESTION:
Assuming a relativistic rocket travelling at 0.95 times the speed of light (c), what would be the drag force on the cross-section area (π500^2) of the ship facing the direction of travel assuming here that drag coefficient is 0.25. The equation for force drag in classical mechanics is: F_D=½ρv²C_DA where ρ is density of interstellar space; which should be about 2 million protons per m³. However I am not sure if in relativistic mechanics it is as such: F_D=ρv²γ² where γ is the lorentz factor (gamma factor). Both of these equations will give different results. Furthermore, what is the kinetic energy when matter within the vacuum of space impacts on cross-section area (π500^2) of the ship facing the direction of travel? The equation is as follows: KE=(γ−1)p c^2 However the results from these equations are about 26.6 N, 0.0028 N and 6.6e−4 J respectively. The latter is not much! However in many articles on the web and as per certain experts, the KE should have been greater and as explosive as nuclear bombs?! Where am I wrong?

ANSWER:
Calculations of air drag are always an approximation. And the equations you are trying to use are certainly not going to be valid under such extreme situations—huge speed and tiny density. Furthermore, when these protons collide with the ship, they will very likely penetrate into the metal unlike the conditions for which equations like you are using are applicable where the gas molecules just bounce off the object. This will cause radiation damage to the ship's hull. Furthermore, at such high speeds interaction with photons will become important, in particular the cosmic microwave background, causing additional drag. Your relativistic drag equation (which I do not where it came from) cannot be dimensionally correct.)

QUESTION:
I am trying to figure out the actual force on a RV or Tinyhouse when traveling against a head wind. From what I recall it is not as simple as taking the vehicle speed plus head wind. For example, lets assume you are towing a tiny house at 60 mph against a head wind of 40 mph - one would think the total wind speed/force on the tiny house to be 100 mph. However, I recall reading that the wind force increases as the speeds of both the trailer (tiny house in this case) and the head wind increase. E.g. 60 mph + 40 mph = 105 mph force but 60 mph + 50 mph head wind could have a force of 120+ mph, not 110. I could not find info online for this. Is this true and if so is there a formula to work this out? In the end I am trying to figure out if tiny houses should be built for hurricane force winds. If in fact, the above is correct and towing a tiny house in a head wind can be likened to a tiny house sustaining hurricane winds.

ANSWER:
Calculating air drag forces is an approximate exercise at best. For the speeds you are talking about, the approximation that the force is proportional to the square of the speed of the air is probably accurate enough for your purposes, F∝v². There is no such thing as "mph force". There is no difference, as far as air drag is concerned, between having a ground speed (GS) of 60 mph and a head wind (HW) of 40 mph; and being at rest with a HW of 100 mph; and having a GS of 100 mph in still air. I think what you are getting at is that the force does not increase linearly with air speed; so if the speed increases from 100 mph to 110 mph (an increase by a factor of 1.1), the force increases by a factor of (110/100)²=1.21. That said, I can now give you a way to estimate the force (very approximately) on the tiny house. Let A be the area which the house presents to the wind; then F≈¼Av², which only works for SI units (meters, kilograms, seconds). So, for example, if v=100 mph=44.7 m/s and A=(4 m)²=16 m², F≈¼x44.7²x16≈8000 N≈1800 lb. This would correspond to a pressure of about 0.07 PSI. (A complicating factor is the presence of the towing vehicle. To some degree, the tiny house would be shielded from the oncoming wind by the vehicle. I see no way to estimate this effect because it would be dependent on the vehicle.)

QUESTION:
What is the weight of impact (kgs) when a 0.5kg handsaw falls 11 floors at a speed on impact of 57mph?

ANSWER:
There is no such thing as "weight of impact". If you mean what is the force which the saw exerts on whatever it lands on, you need to know the time it takes it to stop, t. In that case, the average force F during that time, it is given by F=mv/t. We need to convert mph to m/s, 57 mph=25.5 m/s. For example, if t=0.1 s, F=0.5x25.5/0.1=127.5 N; the mass of a 127.5 weight is 127.5 N/9.8 m/s²=13 kg.

QUESTION:
Please explain me that why if a thin layer of water is spilled on a rough surface like plastered floor and we place our finger in it then why water move away from the point of contact of our finger on that surface and it appear to be dried around.

ANSWER:
This is probably akin to the similar phenomenon of a foot pressing down on wet sand and the area close to the foot is visibly dried. I found an explanation on Physics Forums which seems to be correct: "The phenomenon being described is called 'Dilatancy' and was discovered by Reynolds about 100 years ago. It works only when you have well compacted sand that contains just enough water to cover all the individual grains. When you stand on the sand you create a stress / force which causes the sand to move. In order for the sand to move / flow individual grains have to be able to move past one another. Imagine a bunch of oranges stacked as you might see them at a grocers. The first layer has them all tightly arranged and then the second layer sits down into the gaps between the oranges on the first and third layer. Now imagine trying to move one of the oranges in the second layer. In order to move it the oranges on the first and third layer mut move down and up respectively to enable the orange to move. Effectively the volume of the pile of oranges or grains of sand increases with bigger gaps in between. So when you put your foot down on the sand it shoves sand out the way but in doing so the volume in between grains has to increase temporarily to allow the grains to move relative to one another. Consequently all the fluid at the surface is sucked by surface tension into the extra gaps made by the rearrangement of the sand. Since there is no longer any fluid at the surface the grains of sand are now dry." If you go to the original Physics Forums question, ignore all the early answers which are wrong.

QUESTION:
In magnetism, how can the magnetic field be used to trap charges inside a magnetic bottle? as this requires deceleration of the moving charge at either ends of the bottle and changing its direction of movement, this will need a work done by the magnetic force although the force will always be perpendicular to the velocity vector of the moving charge from Biot-Savart Law. Please, if there is an image for clarification of directions of magnetic force and velocity vectors for more explanation.

ANSWER:
In the diagram, note that the forces on the charge are always toward the center of the bottle. A full explanation can be found at Physics Stack Exchange.

QUESTION:
If the law of conservation of mass states that matter can't be created or destroyed, then how is it possible in the deep future, after black holes die etc., the universe can be left with literal nothingness?

ANSWER:
There is no such thing as the law of conservation of mass. Mass can be created or destroyed. In chemistry, conservation of mass is assumed in chemical reactions because the mass changes are so small as to be almost unobservable.

QUESTION:
Some people like to think about what would happen if the universe's constants were changed (see fine-tuning argument for the existence of God). But what if the laws of nature were themselves altered instead? What if, instead of E = mc^2, we had E = mc^3 or E = mc^4? Or what if the law of conservation of mass-energy or the equations governing electromagnetism were altered? Or what if instead of F = ma, we had F = m/a or F = 2ma?

ANSWER:
Your first example, changing the power to which c is raised in the famous equation E=mc², is impossible because for any power other than 2 the equation is not dimensionally correct—e.g., mc³ does not have the units of energy; it would be like saying "the speed of my car is 55 pounds/foot". Your second example has two errors. F=m/a could be written as a=m/F which would mean that the harder you push on something, the less it would speed up, in violation of what happens. F=2ma would be perfectly ok but would imply that you have defined what you mean by force differently. Newton's second law is actually that the acceleration is proportional to the applied force and inversely proportional to the mass, a∝F/m or a=kF/m, where k is a proportionality constant. Assuming you have already defined what mass (kg), length (m), and time (s) are, your choice of k determines what you mean by force. The standard choice is k=1 which means that the unit of force (which we call 1 Newton) is that force which causes a 1 kg object to have an acceleration of 1 m/s². If you were to choose k=½ (your F=2ma), one unit of force would be that force which caused a 1 kg object to have an acceleration of ½ m/s².

You need to keep in mind that laws of physics are simply expressions of how the universe works. For example, Coulomb's law simply states that the force between two charges is inversely proportional to the square of the distance between them, F∝1/r² because this is what we find doing the most accurate experiments that we can. But, we can not really be sure that the "true" law is not F∝1/r^2.000000001 until we are able to perform an experiment which rules it out.

QUESTION:
I have two magnets, North poles facing each other (repulsive) Magnet A moves in the positive X direction, Magnet B moves in the negative X direction so that total momentum of system is zero. The magnets come to a stop do to the repulsion of the same N poles. When they come to a stop, I "pin" them in place so they cant move. Total momentum was zero to begin with, both magnets stopped, total momentum is still zero. The kinetic energy of both magnets rushing towards each other, now stopped, the energy is now in the field between the magnets, if I was to weigh this system, with a sensitive enough scale, I would find the total mass of the system to be: Magnet A + Magnet B + Field so far so good? Ok, I "unpin" the magnets, they fly apart with equal but opposite momentum, their kinetic energy coming from the energy stored in the field as mass. Is it wrong to say that the energy in the field is transferred directly to the magnets? or, is it more correct to say that "virtual photons" mediated the transfer of energy from the field to the magnets?

ANSWER:
This is a very difficult question even though it looks simple. First, I think it is a mistake to say that the field has mass. The field has an energy density U=B²/(2μ₀)+E²(ε₀/2); yes, there is an electric field because the magnetic field is changing in your example. As the two magnets approach each other the fields change and therefore the energy content of the fields changes also. To avoid the electrodynamics implied by your question (time varying fields) and the resulting radiation (carrying energy away) which would occur, I think we can get to the crux of your question by starting the two magnets at rest and bringing them together by doing a certain amount of work W and having them end at rest. If you measure the mass of the whole system before moving the magnets to be M, its mass after moving will be M+W/c². Again, thinking of the field as having mass is not the right way to think about it, you need to just think about the mass of the system. For example, if a nucleus has N neutrons and Z protons, its mass is less than Nm_n+Zm_p; you would not want to say that the field holding the nucleus together had negative mass, would you? Your second question, I think you can see, is not simple either. Because the magnets are moving, there will be an induced electric field. Because the magnets are accelerated, there will be electromagnetic radiation carrying energy out of the system. I think it is best to be thinking always about the whole system and not where the energy "resides".

QUESTION:
I need to calculate the effort to hold up ONE end of an 80 pound beam to a 40 degree angle. What is the formula?

ANSWER:
I am not sure what you mean by "…the effort to hold up…" I will assume you mean the force F (see figure) you need to exert to hold the beam at rest. In general, it depends on where the center of gravity of the beam is, how you exert the force, and how long the beam is. The smallest force you would need to exert is a force straight up, so I will assume that is the case. If you analyze the equilibrium problem for the beam in the figure, you find that F=Wd/L. Note that it does not depend on the angle at all. For your case, if the beam is uniform, i.e. d=L/2, then F=W/2=40 lb.

FOLLOWUP QUESTION:
Re: latest answer. The force required to hold up the end of the beam must involve a cosine of the angle.

ANSWER:
Details: Sum of vertical forces: ∑F_y=0=F+N-W; sum of torques about point touching the ground: ∑τ=0=Wdcosθ-FLcosθ=Wd-FL. Solutions: F=Wd/L and N=W(1-(d/L)). This results from choosing to have F be vertical. I could have chosen to have F perpendicular to the beam in which case the answer would depend on θ. (I now see that my statement in the original answer that a vertical force would be the smallest is wrong, although my answer for a vertical F was correct and independent of θ.) The new torque equation would be ∑τ=0=Wdcosθ-FL so F=Wdcosθ/L and the 80 lb uniform beam at 40⁰ would require a force of F=½(80)(0.77)=30.6 lb. It would require no force to hold the beam vertical since cos90⁰=0.

QUESTION:
If gravity has the same effect as acceleration (according to the theory of relativity ) gravity curves light, does acceleration also cause a curve ? And if it why do light particles move in a straight line ?

ANSWER:
The operative principle here is the equivalence principle: there is no experiment you can do to distinguish between being in a uniform gravitational field with associated acceleration g and having an acceleration g in zero gravitational field. Suppose that you are in an accelerating elevator in empty space which has a small hole in the side. Someone shines a beam of light into the hole. That beam will follow a parabolic path as seen by you inside your elevator. Therefore, light will also be bent by a gravitational field. There are two answers to your last question. First, if you watch the photons from your Euclidian frame of reference, the photons do not follow a straight line in that space. On the other hand, what mass does is warp the space around it so a photon will follow the shortest path between two points which you would call a straight line in that space which is, itself, curved.

QUESTION:
So, I am learning about the basics of waves in class and I was bored so I started messing around with some equations. λ=h/p p=mv, in the case of light, p=mc, c=λf E=hf c/f = λ c/f = h/p c/f = h/mc h=E/f c/f = E/mcf c=E/mc E=mc^2 Is this a correct derivation of E=mc² ??

ANSWER:
Well, "messing around" with equations you do not really understand seldom leads to good results! First, E=mc² does not apply to light (photons) because light has no mass and you would therefore you would conclude that light carries no energy which would be nonsense; for more detail, see the faq page. Your major error, though, is right at the start writing p=mv which cannot be true for a photon since it has momentum but does not have mass. In fact, this is not even true for particles with mass since, in relativity, the momentum is given by p=mv/√[1-(v²/c²)]. Incidentally, the energy of a photon is E=hf and the momentum is p=h/c.

QUESTION:
a light photon moving at the speed of light, can it have spin? and if so can the spin be added to its speed giving it a faster than light speed breaking the laws of physics? I have always wondered this but never seem to find out if they have spin....

ANSWER:
Every photon has a spin of 1; this means that it has an intrinsic spin angular momentum of L=√[1(1+1)]ℏ. So, I am guessing that you are visualizing the photon as a little spinning ball and think that you can conclude (see my figure) that the "equator" on the near side of the ball is moving forward with a speed of c+v. However, spin in quantum mechanics cannot be visualized using such a simple classical model, even though we often do think of spin this way to get a qualitative feel for spin. For example, if you visualize an electron as a spinning uniform sphere with a reasonable radius, you find that the surface must have a speed greater than the speed of light.

QUESTION:
Say I fill an airtight barrel with water and have a valve at the bottom and a feeder hose at the top. If this barrel is uphill and I have the feeder hose down lower say in a pond will the draining of the barrel through the lower valve create enough vacuum to pull the water uphill creating a siphon?

ANSWER:
First of all, I would call what you are proposing a pump, not a siphon. You are trying to "suck" water uphill using a vacuum. The first thing that comes to mind is that there will be a limit on how high the hill is above the water level below. Even if you have a perfect vacuum, the highest you can lift water this way is 10.3 m=33.9 ft. But you start off with a hose full of air, so you will never get a vacuum, so you will be limited further in the height to which you can pull the water from below. For example, if the volume of the air in the hose were 1/10 of the volume of the barrel, you could only lift the water 9.3 m. Or, if the volume of the air in the hose were equal to the volume of the barrel, you could only lift the water 5.1 m. So there is no simple answer to your question, but this is probably not a very workable way to lift water.

QUESTION:
If a person were in a closed container filled with water and the container was accelerated at high speeds, would the person in the container feel the g-forces the same as if they were not in the container?

ANSWER:
You would move backward until you hit the back wall and then the back wall would exert a force forward on you. If there were no water, the back wall would exert a force F₁=maon you where m is your mass. If there were water, the water would exert a force F₂ toward the back on you and the wall would exert a force F₃ forward on you. So now, F₃-F₂=ma so the force on you from the wall would be bigger (F₂+ma) than if there were no water. Plus the water would be trying to crush you.

QUESTION ABOUT THIS ANSWER:
You recently answered a question regarding the effect of acceleration on a person in a closed container of water. You suggested the result would be movement to the back of the container and an increase in the force experienced by the person. Buoyancy is dependent on relative densities, so a person will float with the same percentage immersed regardless of the local gravity/acceleration. This implies that the victim would, at least at moderate levels of acceleration, be forced to the front of the container. Initially I thought that if the container was not full, it would be quite a comfortable experience since the accelerating force would be evenly distributed. However, humans are not of uniform density, so the persons chest with its air-filled lungs would be forced to the front and his bony, less buoyant extremities would be dragged the the back. Unfortunately, the questioner filled the container completely so the person would be pressed uncomfortably against the front wall. At higher acceleration the body would be compressed sufficiently that he would become denser than the water, only then would he move to the back of the container to be further crushed by the water column.

ANSWER:
You are right, the motion depends on the density of the astronaut relative to the density of the water. If the ship is in empty space (no gravity) and not accelerating, there would be no buoyant force in any direction and the astronaut would float either at rest or at constant velocity until he hit something. The equivalence principle says that there is no experiment you can do to distinguish between being in a uniform gravitational field with associated acceleration g and having an acceleration g in zero gravitational field. If the ship had an acceleration a forward it would be the same as being in a gravitational field pointing backward in the ship. Therefore, there would be a buoyant force which would cause objects with smaller density than water to move forward ("float") and objects with larger density than water to move backward ("sink"); you correctly point that out and in my original answer I was assuming an astronaut whose overall density is larger (quite possible if he were wearing a heavy space suit, for example). In the back wall case, my original answer was correct. In the front wall case, the water would be pushing you forward and the wall backward, so F₂-F₃=ma or F₂=F₃+ma; now the water pushes on you with a force greater than the wall pushes back.

QUESTION:
Most physicist say that an antimatter engine is impossible because there is no way to store the antimatter but if you were able to get raw positrons and electrons and store them in seperate electrically charged tanks (each with the same charge as the particle they are holding). would you be able to sucessfully store the matter and antimatter and decide when to open it and close it. So basically is the atraction between matter and antimatter stronger than the electroweak force?

ANSWER:
If you have a hollow conducting tank and charge it up with electrical charge, the field inside is zero, so your plan to store charge inside will definitely not work; the only force felt by the charges inside will be the forces between those charges which will have the effect of pushing them all out to the tank. All the positrons in the positron tank would annihilate with electrons in the tank; all the electrons in the electron tank would end up on the outer surface of the tank. And, by the way, the only force between an electron and a positron is electrostatic.

QUESTION:
In the part of electromotive force, when we connect a wire to the terminals of the source of EMF like for example a battery, the potential difference between the two terminals create an electric field inside the conductor. Haw fast this field been established, is it simultaneous just at the time of closing the circuit or it takes time even if it is extremely small amount of time?

ANSWER:
In a vacuum, an electric field propogates at the speed of light. In the wire it will be a bit slower, but still close to light speed. The electric field causes conduction electrons to move, hence creating the current.

QUESTION:
I have been studying atoms and strong and the electroweak force for about a week and i came across something that caught my attention. When a radioactive isotope undergoes beta decay one of its neutrons ejects an electron and an antineutrino and leaves behind a proton. my question is if it leaves behind a proton and jettisions an electron is there any chance that the nucleus with one extra proton than electrons in orbit around the nucleus could catch the loose electron?

ANSWER:
Why not just ask if you can combine an electron and a proton to make a neutron? The answer is yes, and it is called inverse beta decay. If a proton absorbs an electron, an electron neutrino will be ejected: p⁺+e^-→n⁰+ν_e. This most commonly happens when an atomic electron is captured into the nucleus, a process called electron capture. This process also results in the emission of an x-ray because the hole in the K-shell is filled by a higher-orbital electron. It is also the main way neutron stars are formed.

Q&A OF THE WEEK,

3/19-25/2017

QUESTION:
Having a bit of a debate about whether this tennis ball would've landed in with a tennis player and we have a $100 bet on it. The ball machine fed the ball from the other side of the court at the baseline the player that hit the ball is a top ranked junior player... the ball hit the ball machine edge 5 inches off the ground at the top of the wheel base and the player claims that it would've landed on the line if it had not hit the ball machine of which the picture demonstrates the point of impact is 5 inches above ground at the back edge of the line and there are no external elements such as wind as we are playing indoor. the ball was traveling at approximately 30 mph at time of impact. We have attached an image for reference. We appreciate any clarity you could provide :-) [Note that the angle relative to the horizontal is specified to be 60⁰-70⁰ in the photograph attached by the questioner.]

ANSWER:
My first reaction was to say that, of course, it would not hit the line. That was because, as physicists often do, I was thinking of the ball as a point and ignoring its size. You can see from the figure that there could easily be a combination of h and v₀ where the ball would strike the line had the obstruction not been there. As best as I could tell, some part of the ball must touch the line so if we calculate where the bottom-most point of the ball strikes the ground (y=0) the ball will be in if x>0 in my coordinate system. My guess is that if we simply assumed that the bottom point went in a straight line in the direction of its initial velocity we would get the right answer; but the ball is actually a projectile and moves in a parabolic path so, since this is a $100 bet, I better do it right! The equations of motion are

x(t)=x₀+v_0xt

y(t)=y₀+v_0yt-½gt²

where t is the time it takes to hit the ground and g=9.8 m/s² is the acceleration due to gravity. Being a scientist, I prefer to work in SI units, so I will do that. OK, let's summarize everything we know. I will assume θ=70⁰ since that gives the best chance of hitting.

x₀=R=2.7"=0.0686 m

y₀=h-R=5"-2.7"=2.3"=0.0584 m

v_0x=-v₀cosθ=-30cos70⁰ mph=-4.583 m/s

v_0y=-v₀sinθ=-12.59 m/s

y(t)=0.

So now the task is to put these into the equations above, solve the y equation for t and put that value of t into the x equation to find x. I find that t=0.00463 s and x=0.047 m=1.9". Because x is positive, the ball will hit the line. Going through the same procedure for θ=60⁰, I find t=0.00502 s and x=1.4", again hitting the line. When the bet is settled, don't forget to reward The Physicist!

ADDED THOUGHT:
The curvature of the parabolic path is, as I had speculated, miniscule. For θ=70⁰ above I found x=0.0474 m and if you just assume the ball went in a straight line to the ground you find that x=0.0473 m.

FOLLOWUP QUESTION:
For clarity sake I am sending you two more pictures from the exact set up of which we have not moved. Previously you only were sent the Sideview so I have included below sent a top view showing that the ball machine is actually out and the court view from the approximate angle that the ball was struck showing the approximate ballpark at the ball was hit from approximately contact was made 3 feet off of the ground. I still can't wrap my head around if an apparatus is actually out and a ball strikes it from 5 inches above the court going in the opposite direction that it can still be in unless air resistance is involved of which there are no air elements. Just want to make sure that you still think the ball Woodland in from more data provided before I pay $100?

ANSWER:
The top view is helpful in verifying what I had assumed in the original answer—the surface which is hit is aligned with the outer edge of the line. The best way that I can convince you that it is possible that my first answer is correct is to show two figures, one showing a trajectory of the ball which lands it in bounds and one out:

Each figure shows the ball at the instant of impact with the machine and the instant of impact with the floor had the machine not been there. The dashed red line shows the trajectory of the center of the ball and the solid red line shows the trajectory of the bottom of the ball. The diameter of a tennis ball is 5.4". The ball with the steeper trajectory (which would include your 60⁰-70⁰ trajectory) is clearly in bounds by standard tennis rules.

FOLLOWUP QUESTION:
We did not know how to define the arc so we just said 60 70 as an ignorant guesstimate but we are ok with the actual real representation of the arc that we sent you in the last image...we just want to understand based on the ACTUAL real flight path arc that the ball really took when struck and its possibility of landing on the line of which the image with the actual art provided is a more accurate visual representation... and we are not sure of how to define the angle of approach based on that visual. Our original explanation of the ball angle degree of approach to the baseline could have been lost in translation somewhat we are not certain... that's why we contacted you and that's why I sent a VISUAL representation because we really aren't sure of how to define that...we reckon the more information we provide you the more accurately u can clarify our understanding. In saying that.... what approximate degree is the arc approaching the baseline from the image we provided this morning... assuming the ball was struck at about 3 feet off the ground traveling approximately 3 4 feet over the net? The distance from the net to the baseline is 39 feet and the player was approx 3 feet behind the baseline at time of contact so that would be approx 42 feet from the net at the time the contact was made. Does this change whether or not the ball lands on the line or is it irrelevant?

ANSWER:
This Q&A is turning into quite a tome! With the information you sent, I can calculate a pretty good estimate of the trajectory of the ball ignoring air drag. It turns out that your guess of 60⁰-70⁰ for the trajectory angle was way off. The algebra is tedious, so I will just give the final results. To check that I made no algebra errors, I plotted the trajectory, shown in the figure. I have worked in meters. The ball is hit at x=0 from a height of 1 m, passes 1 m above the 1 m high net at x=13 m, and then hits the machine at x=25 m just above the ground (y=0 m). You can see that my result describes the path you specified quite well. Do not be deceived by the picture, though, because the scales of the two axes are very different, only 2.5 m for vertical motion compared to horizontal motion of 25 m, a factor of 10. The inset shows the trajectory as it would look to the eye. As you can see, the angle is much smaller than you estimated. The analytical solution is that θ=15.6⁰and t=1.1 s; your estimate of the initial speed was pretty good, 23.5 m/s=52.6 mph but the final speed (although it is not important) is 23.9 m/s=53.5 mph not 30 mph.

Now I am compelled to recalculate with the best possible numbers whether the ball hits in bounds or not. However, given all that I learned above, I can make it brief: there will be a critical angle θ_c=tan^-1(y₀/x₀)=tan^-1(2.3/2.7)=40.4⁰; any angle smaller than θ_cwill be out of bounds. For 15.6⁰, x=2.7-(2.3/tan15.6)=-5.5", 5.5 inches out of bounds.

Finally, just for completeness, I would like to estimate the effect of air drag. Using the approximation in an earlier answer I find that t=1.13 s (0.03 s longer) and v=22.1 m/s (1.8 m/s slower). This would correspond to the angle being a bit bigger, about 17⁰, but not nearly enough to cause the ball to drop in bounds.

Q&A OF THE WEEK,

3/12-18/2017

QUESTION:
How much lateral force is needed to damage bearings in the hub motors of a skateboard when carving? So...if you roll in a straight line, on a skateboard, you exert a radial load on the eight bearings contained in the four wheels. But skating is more fun when you ride in wavy lines - carving! So the bearings start getting a lateral or axial load. How "hard" would the carve have to be (let's assume a rider of 100kg) to break the weakest of the bearings? The rotor is supported by two bearings, one that has an radial maximum force of 3.5 kN and the other of 7 kN. So the maximum axial forces would half those.

(The questioner and I had several exchanges. The bearings are cylindrical and he was only guessing that the axial force was half the radial force based on data for similar bearings. I have edited his several emails to get the gist of things in the question above.)

ANSWER:
The way I see the problem is shown to the right. The forces on the skater plus skateboard are his weight mg vertically down, the normal forces on the inner (N₂) and outer (N₁) wheels, and the corresponding frictional forces f₁ and f₂. Whatever the rated axial (along the wheel axis) maximum force is, that is what we want to use as the to find the limiting conditions for the "carve"; the determining factors will be the mass m of the skater plus board, the speed v he is going, and the radius R of the path. For the lean, d and h are the distances, respectively, of the center of mass horizontally and vertically from the front wheels; note tanθ=d/h. The wheel base is s. The normal forces and frictional forces shown each represent the forces on two wheels. Note that the inner wheels carry the most force. The easiest way to do this problem, since it is an accelerating system, is to introduce a fititious centrifugal force C=mv²/R pointing outward. Newton's equations are N₁+N₂-mg=0 (vertical forces), f₁+f₂-C=0 (horizontal forces), and N₁s+mgd-Ch=0 (torque about inner wheels). Now, from the torque equation, there will be a maximum speed you can go for a given m and R before the outer wheels leave the ground. At this time N₁=0=(Ch-mgd)/s or v²/(gR)=d/h=tanθ. Also, f₁=0, so f₂=C=mv²/R. Now, the inner two wheels each are experiencing the axial force of F=½mv²/R. Just to do an example, let v=10 mph=4.47 m/s, R=5 m, and m=100 kg; then F=0.2 kN. If your guess that F_max≈1.75 kN is correct, you should be ok. For this example, θ=tan^-1[v²/(gR)]=22⁰. If you execute the turn with a smaller lean angle, all four wheels will share part of the load. To do the general solution where both wheels have N≠0 would not be two difficult but would be quite a bit messier algebraically and probably not all that useful.

FOLLOWUP QUESTION:
Thanks for the explanation on your website. It wasn't really what I was looking for because it doesn't really give the answer something layman can understand. Also, you compare your estimated lateral force of 1.96 kN (0.2 kN, see below) to 1.75 kN, although they would be the breaking limit for only one bearing. You assumed that just the 2 inner wheels receive the load, so isn't that a total of 4 bearings... shouldn't the breaking limit be 4*1.75?

ANSWER:
I am glad you asked this question because it got me looking more carefully at what I had done late last night and I found an extraneous factor of g had crept into the final stage of my calculation, ("…so f₂=C=mgv²/R…") so my final answer was too big by a factor of 9.8 meaning that the lateral force per wheel is F=0.2 kN; this has been corrected in the original answer. So you should have no problem after all. I was in the process of writing an email trying to "verbalize" my answer somewhat to make it more accessible when I discovered this. Here is what I wrote:

You asked me "How 'hard' would the carve have to be (let's assume a rider of 100kg) to break the weakest of the bearings?" That is what I gave you. I assumed that the weaker of two bearings in a wheel can break even if the stronger does not. (I assume they are coaxial, one inside the other.) Sorry if the explanation was too technical, but that is as basic as I can get to convey the details. A few comments to try to clarify:

What I call C, the centrifugal force, determines how "hard" the carve is. C=mv²/R. For the example I did, C=100x4.47²/5=400 N.
When carving, the harder you carve the greater the load will be carried by the inner wheels on each axel. Eventually, the outer wheels will just lift off the ground which necessarily results in the inner wheels taking all the load, both lateral and radial. Since you asked me for how to calculate the maximum lateral force any wheel would experience for a given m, v, and R, that is what I gave you.
I do not understand your last question but I do know that if two wheels experience a force of 0.4 N then each experiences a force of 0.2 N (assuming they equally share the load).
ADDED THOUGHT:
I guess I have not really fully answered your question "How 'hard' would the carve have to be…to break the weakest of the bearings?" I suggest that the appropriate equation would be 1.75x10³=½mv²/R. Here are a couple of examples:
- What is the fastest a 100 kg rider could go in a curve of radius 10 m? v=√(2x10x1.75x10³/100)=18.7 m/s=67 km/hr=42 mph.
- What is the tightest curve a 100 kg rider could turn at 30 mph=48 km/hr=13.4 m/s? R=½x13.4²x100/1.75x10³=5.1 m.
These, of course, assume 1.75 kN is the strength of the weakest bearing.

QUESTION:
In a black hole what becomes of the system's angular momentum?

ANSWER:
Black holes have angular momentum. Any object which has angular momentum, will add its angular momentum to the black hole's when it is captured, conserving angular momentum.

QUESTION:
If the earth is spinning at 1040mph to the EAST, why do trips from LA to NY, take the same amount of time in either direction, give or take 30min? If the plane is traveling at 500mph, wouldn't flight times be significantly different from LA to NY vs NY to LA?

ANSWER:
The atmosphere rotates with the earth, to first approximation. The airplane flies relative to the air and therefore takes the same time in either direction because it is flying in still air. In reality, at the altitudes where commercial airliners fly, there is a strong air current called the jet stream which moves, relative to the ground, in a west to east direction. Therefore the trip from NY to LA is usually longer.

QUESTION:
Since a perfect black body would absorb all electromagnetic radiation would it mean that it would also have a perceived temperature of absolute zero and wouldn't it also mean that eventually the temperature of whatever the material the black body itself is made up of would eventually reach the planck temp since it's always absorbing radiation and never releasing it?

ANSWER:
A black body is also a radiator. If it is in a radiation field it will absorb all radiation striking it and, as it absorbs this energy, it will increase in temperature. As it gets hotter it will increase its own radiation. Eventually it will come into thermal equilibrium with its environment, absorbing and radiating at the same rates.

QUESTION:
Good Day ! Thank you for being there and dedicating the time to answer our questions. I have many... Can I start with the "Inverse Square Law" as it pertains to light. E = I/r2 , Using 100 as "S", S/4pi2 = I ( intensity at surface of sphere ) as the source strength, the farthest we would be able to "see" would not even get us to PC. it is 4.2 light years away, 5,878,499,817 x 4.2 = 24,689,699,231.40 miles... Assuming 100% at "S". Reaching the surface of the earth, using a nominal, 8,000 mile diameter, that works out to be about 1 (one) photon for every 1,205. square miles.... And, that is purported to be our closest star. How can this be ?

ANSWER:
I generally do not try to find errors in questioners' calculations, I just do the calculation myself. The sun is a pretty average star and its photon rate is about Φ≈10⁴⁵ photons per second. Using R=4.2 ly=4x10¹⁶ m, I=Φ/(4πR²)=10⁴⁵/(64πx10³²)≈5x10¹⁰ photons/s/m². That would be the photon intensity 4.2 ly from a sun-like star.

QUESTION:
If you would like to send a space probe to an asteroid that is 7 AU away, how many years would the journey take using the minimum energy direct trajectory?

ANSWER:
You seem to think that energy must be expended to keep it moving. In fact, once you give it a velocity greater than the escape velocity, it will coast the whole way. The minimum velocity you must give it is just slightly less than the escape velocity, but for the purposes of this answer, let's just say that we give it the escape velocity which is about 10⁴m/s. I find that the speed at r=7 AU would be about 28 m/s and the time to get there would be about 1.5x10¹² s≈48,000 years. So the energy required would be that amount needed to give your probe the escape velocity from earth's surface. Keep in mind that this estimate ignores everything except the earth. It would be a better question to ask, for example, how much energy would be needed for the probe to arrive in 10 years.

QUESTION:
I'm hoping this is not considered off the wall but our moon does not have an atmosphere. And our moon and earth are spinning (quite fast) and also our solar system is spinning while traveling through space. so my question would be how those men weren't blown clear off the moon when they landed on the moon?

ANSWER:
Where did you get the idea that the moon is "spinning quite fast"? The moon spins once on its axis approximately every 28 days so that it always keeps the same side pointed toward the earth. The earth, of course, rotates once every 24 hours. You can easily calculate the centrifugal force on something on the equator of the moon or earth, that force which I presume you are assuming "would blow [us] clear off". On earth that force is less than ½% or your weight; on the moon that force is 2.5x10^-16% of your moon weight! So, you see, in no way do you have to worry about being "blown clear off"!

QUESTION:
I am a Middle school science teacher and we do an unit on cell phones as part of our NGSS "waves" unit. The performance task is to build a device that blocks a cell phone signal. The kids quickly figure out (from the unit lessons and the internet) that aluminum foil is the best solution. My Question(s): What exactly is happening with the aluminum foil? everything I read says the waves are being absorbed, are they creating a Faraday cage, or just creating a barrier that blocks the waves? I have had students create "Faraday Cages" (with copper wire, following internet plans) and it did not work. What exactly is a Faraday Cage and how does it work? My teaching partner noticed a correlation between when the phone is touching the box, and when it is insulated from the box (with a cell phone case). Is there something to this? This is the second year of this project and I am in awe at the opportunities for learning for myself and the kids, thanks for your time.

ANSWER:
Do your students know what an electromagnetic wave is composed of? There are oscillating electric and magnetic fields as shown in the figure. The important thing is that there are electric fields and electric fields cannot penetrate into a hollow conductor (the Faraday cage). To understand how the Faraday cage works, look at the animation. When an electric field is applied the electrons experience a force opposite the field and migrate to one side of the cage while the other side has a net positive charge because of the missing electrons. Amazingly, the electrons arrange themselves in just the right way that their electric field is exactly opposite what the external electric field inside would be, giving a net field of zero inside. The waves are not so much absorbed as they are cancelled out. A Faraday cage does not have to be completely closed if you want to keep radio waves out as long as the size of the holes is very small compared to the wavelength of the radiation; it could be made of chicken wire, for example, as long as the wavelength is much greater than a couple of inches. From what I can find out the frequencies are in the not too far from f=1 GHz=10⁹ s^-1; since the wavelength of a wave is the speed divided by the frequency, λ=c/f=3x10⁸/10⁹=0.3 m. So, I would have thought that a cage you made would have gaps much less than 30 cm. Were all the wires in good contact with all others? Often copper wires have a varnish-like coating on them to serve as insulation. I don't know what you mean by "correlation", so I cannot answer that part of your question. (By the way, the aluminum foil is a Farady box if it completely surrounds the phone.)

QUESTION:
How much heat would it take to heat 1 gallon of water to 600 deg F in a pressurized system, from 70 deg F to 600 F in 1 hour. Not counting the ss vessel. Also since the water is not allowed to change states are the calculations just the Sensible heat cals or are there special calculations needed. This is part of a R&D Application.

ANSWER:
For my approximation to be fairly accurate, the water must remain liquid at a constant volume. I will work in SI units so 70⁰F=21⁰C and 600⁰F=315⁰C; I will convert back to Imperial units for the final answer. I looked up data for the specific heat of water which turns out to have a significant temperature dependence as shown in the figure (black). I did a quadratic fit (green) to these data and integrated over the temperature range to get E=1356 kJ/kg. The mass of a gallon of water is about 3.8 kg so the total heat is Q=1356x3.8=5.2x10³ kJ=1.44 kW⋅hr=1240 kilocalories. Keep in mind that the pressure will be very large at 600⁰F, about 1800 psi.

QUESTION:
I have been reading a lot lately about the prospects of nuclear fusion reactors; especially regarding electrical power plants. My question is this: if the hot plasma from the reaction is shielded from the equipment to protect it from the heat by a magnetic bottle, how then do you access the heat to make electrical power via steam for turbine generators or direct conversion from heat to electricity.

ANSWER:
The purpose of the magnetic bottle is to keep the hot plasma isolated from the physical containment; just touching the containment vessel would present two problems—it would cause instabilities in the plasma and it would damage the vessel. But the plasma is hot and anything hot will radiate heat energy; in other words, the magnet bottle contains the plasma, not the heat. This radiant heat would rapidly heat up the vessel which would heat up some coolant mechanism (imagine having water tubes coiled around the outside of the vessel) which would carry away the heat to drive the turbines.

QUESTION:
We could induce artificial gravity through centripetal acceleration. For example a ring-like structure in a spaceship could rotate about 1.34 rpm if the radius to the centre is 500 meters. This will give 1 g at the edges of the ring. However we can also induce artificial gravity in the spaceship through its propulsive power and hence a constant acceleration at 1 g of the spaceship is required. But what if the spaceship is accelerated at more than 1 g in order to achieve 75% the speed of light, even though the spaceship consists of the 500m radius ring-like structure that rotates at 1.34 rpm. What type of artificial gravity will the occupants feel? Will the excessive acceleration cancel out the centripetal effects in the ring? Is there an equation or formula that combine both linear acceleration of an object while the objects is also rotating?

ANSWER:
You seem to think that you need an acceleration greater than g to get to 0.75c, but any acceleration will do. But, since you seem interested in a greater than g acceleration, I will choose a₀=1.5g. So now a person of mass m in the ship sees two fictitious forces, mg which is radially out and 1.5mg which is toward the rear of the ship. A person in the ship would experience a net force of about 1.8mg pointing out and back. This would not be a comfortable situation. If you are just going to accelerate constantly, I would recommend not messing with the rotation at all, set

your acceleration to g, and make all the "floors" in the ship perpendicular to the acceleration. If you are interested in how long it would take to reach 0.75c with a₀=g, you can read off the graph (which I took from an earlier answer) that gt/c=1.8, so t=5.5x10⁷ s=1.7 yr.

QUESTION:
A worker was pushing a pallet forklift and when he put the metal against the metal at the bottom of a glass door - the glass shattered to pieces. He just barely touched the metal on the bottom of the door. What could have happened to make it shatter like that?

ANSWER:
Glass sometimes breaks under circumstances where it does not seem that enough force has been applied. As you probably know, glass is formed in very high temperatures and allowed to cool. Sometimes the cooling does not occur uniformly over the whole volume of the glass and the finally cooled object can have places where there are very large internal stresses. Just a small force at such a location can then cause the glass to break. I cannot verify that this was the case for your situation because it is a little hard to imagine a forklift having "just barely touched" anything!

QUESTION:
Is it true that the electric energy of one electron that is moved through a potential difference of 1 volt is called electron volt?

ANSWER:
You are on the right track, but your terminology is a little shaky (it is not clear what "electric energy" means). There are two ways you might want to define the electron volt (eV). First, if an electron at rest is allowed to accelerate across a potential difference of 1 V, it acquires a kinetic energy of 1 eV. Second, if you push an electron across a potential difference of 1 V (opposite the direction it wants to go), you will do 1 eV of work. 1 eV=1.6x10^-19 J. You need to realize that if the potential difference is greater than a few hundred volts, the kinetic energy will not be exactly ½mv² because of relativistic effects.

QUESTION:
why are sparks more likely to occur between two charged particles closer together rather than far.

ANSWER:
For air to become a conductor, there must be a sufficiently strong electric field to ionize the air molecules. There must therefore be a potential difference (voltage) between the two electrodes. The air has a dielectric breakdown strength of about 30 kV/cm which means that 30,000 V are needed for a gap of 1 cm but only 3000 V is needed for a gap of 1mm.

QUESTION:
If I drove my car home with the boot lid open, why would the air resistance be higher than usual?

ANSWER:
It probably would be higher, but not necessarily. If it were higher, it would be due to the fact that the area presented to the onrushing air was larger and the drag is proportional to the area. On the other hand, aerodynamics can be very nonintuitive. The spoiler on some cars is designed to break up the smooth flow of air over the car which actually results in lower drag at high speeds. The dimples on golf balls and the hairs on tennis balls have the same purpose, to break up smooth air flow.

Here is an anectdote which illustrates that your intuition is not always right regarding drag. Some years ago somebody called in to Car Talk on NPR and asked about these nets you can buy to replace the tailgate in a pickup truck to reduce air drag. Makes sense, right? The tailgate is like a wall in the wind and to get rid of it will reduce drag and increase your mileage. Click and Clack said that they thought these things were a great idea for reducing drag and increasing fuel efficiency. During the intervening week before the next show an engineer from GM called in and told them that removing the tailgate in fact greatly increases the overall drag on the truck. The reason is that the tailgate traps a bubble of air which rides along with the truck and the headwind slips over it. There was a lot of crow-eating that week at Car Talk Plaza!

QUESTION:
Does temperature affect a magnet's magnetism?

ANSWER:
Magnetism is a vast field and the temperature dependence depends on the material. I will just address a simplified description of a simple ferromagnet, a common permanent magnet. The basic way that these work is that electrons interact with their neighbors in nearby atoms such that their magnetic moments align with each other giving the bulk material a net magnetization. The most important temperature dependence is that, for any ferromagnetic material, there is a critical temperature called the Curie temperature where the magnetization disappears. So, if you want to destroy a permanent magnet, heat it up above the Curie temperature; unless you cool it in an exteranal magnetic field, it will not be a very good magnet when cooled. The behavior below the Curie temperature is likely to be complicated, but there will be generally a decrease in magnetization as it heats up.

QUESTION:
Photons don't have mass, but they have momentum. So this means that if a photon hit a mirror, for example, it pushes the mirror a little bit forward. Thus, the photon transferred momentum to the mirror. So this would mean that the photon lost a fraction of its speed or its mass decreased because p=m*v! But how could this be if photons are massless and only can travel at the speed of light?

ANSWER:
The linear momentum of a photon is not p=mv (since it could not have any momentum if m=0), but rather p=hf/c where h is Planck's constant, c is the speed of light, and f is the frequency the corresponding light wave. In everyday life, the light which reflects from a mirror is the same color of light that went in; this means that the mirror did not recoil at all, essentially has infinite mass. If you take into account the recoil of the mirror, some of the photon's momentum would be transferred to the mirror which would mean the photon would have to lose some momentum. But h and c are both constants of nature, so f would have to decrease; that would mean that the actual color of the photon would be slightly shifted toward the red end of the spectrum. You could never do this experiment, the shift is too tiny. But, if you "reflect" a photon from an electron (shoot gamma rays or x-rays at a solid which has many electrons in it) you can easily measure the change in wavelength of the radiation; this is called Compton scattering and was one of the pivotal experiments in the development of quantum physics.

QUESTION:
is the terminal velocity of a matchstick falling from 1000 metres the same as a 3000lb car falling from the same height.

ANSWER:
The terminal velocity is the largest speed something will acquire if it could fall forever. So it has nothing to do with how far it falls; a feather may acquire 99.999% of its terminal velocity after falling one inch but a bowling ball might have to fall a mile before it reached 99.999% of its terminal velocity. This problem you can do just using your common sense (although I will do a rough estimate below). Something as light as a matchstick will not fall far at all before it is moving with nearly its terminal velocity; if you had a terminal velocity as low as a match's, you could jump off the Empire State Building and not get killed. The car would go much faster.

This may be all that you want, but for the interested I will roughly calculate some numbers. The terminal velocity v of something with mass m which presents an area A to the onrushing air can be roughly approximated as v=2√(mg/A) where g≈10 m/s² is the acceleration due to gravity. The match I estimate to have m≈0.1 grams=10^-4 kg and area A≈(5 cm)x(1 mm)=5x10^-5 m²; then v≈9 m/s. The car I estimate to have m≈3000 lb≈1400 kg and area A≈(2 m)x(3 m)=6 m²; then v≈97 m/s.

QUESTION:
I was playing with my nephew and his toy cars and came to notice that when pushed, after a short distance, the cars begin to veer off course. Some simple "testing" determined this to be a (seemingly) random effect. This was not really new to me, but curious nephew eyes then began to ask me "why this is so". I assume it to be a mix of many effects arising due to the (lacking) production quality of the cars, but I didn't find a really satisfying answer. Perhaps you could give a "best guess" as to which effect is the main contributor?

ANSWER:
Suppose there is some small force which pushes the car left or right at a rate of 0.4 ft/s/s and there is some larger force which causes it to slow down at the rate of 3 ft/s/s; now push the car so it has an initial speed 10 ft/s. The path followed by the car until it stops will resemble the path shown in the figure. That is my best guess!

QUESTION:
In space like sci fi dogfights (gundam or macross) is it possible to accelerate from a stationary (from the pilots perspective like a ship) object straight forward (thinking in 4 axis) then if accelerate directly down from your original vector then accelerate again in a new vector does the orignal speed change or do only the amount of gravitational energy change? Also what would it do to the person inside the craft? If your already doing 500mph and you change directions like I said down at another 500mph then again at 500mph does it add up or is it like launching from a craft already doing 500mph where the energy felt going foward is 0 from their perspective? Like when a astronaut inside the ISS goes out for a space walk? Do they feel the pressures when they wear those jetpacks (sorry for childish name) and change directions? Again what type of math would I study to learn the answer to these types of questions? and hopefully solve them some day because space has space for everyone and I want people to one day go to the stars ALIVE!

ANSWER:
I am afraid that your question is quite muddled, but I think it shows that you do not really understand what acceleration is. Average acceleration is defined to be the change in your velocity (the difference between the final and initial velocities) divided by the time to affect the change. If I understand your example you are originally moving with speed v_initial=500 mph in one direction and later are moving with speed v_final=500 mph in a direction perpendicular to your original direction of travel; it is important that you realize that velocity is a vector so that it changes if the direction changes but the speed stays the same. The difference between the two velocities is shown in the figure, Δv=v_final-v_initial. So the average acceleration is a=Δv/Δt where Δt is the time to make the maneuver. The average force F you will experience is in the direction of a and proportional to your mass m, F=ma. For example, suppose your weight is 160 lb (then your mass is m=160/32=5 lb⋅s²/ft), Δv=500√2=707 mph=1037 ft/s, and Δt=1 s. The average force you will feel is F=5x1037/1=5185 lb! You do not want to change your velocity too quickly!

QUESTION:
I'm just curious: is there any geometric structure that could be built-up infinitely without collapsing under it's own gravity? A tetrahedral mesh, like diamond maybe?

ANSWER:
Let's refrain from talking about building up "infinitely" since there is not an infinite amount of energy in the entire universe. The gravitational field g of a point mass m is g=-mG1_r/r² where G is the universal gravitational constant, 1_r is a unit vector pointing radially out and r is the distance away from the point. To get the gravitational field for an object not a point you need to divide it up into infinitesmal pieces, each having a field dg=-dmG1_r'/r'² where r' is the vector from dm to the place where you wish to calculate the field; you then integrate over the entire object, usually not a trivial exercise. This is one of the main reasons that Newton had to invent the calculus, so that he could prove that he could treat the sun and the planets (spheres) as point masses. I understand that the difficulty of proving that the field of a spherically symmetric mass distribution outside the object is identical to the field there if all the mass were in a point at the center caused a delay of something like 20 years in the publication of his theory of gravity. I will only talk about spherically symmetric masses here because, the point mass field being spherically symmetric, all large bodies will tend toward a spherical shape. This is why the stars and planets are spheres; see an earlier answer on cylindrical masses. Calculation inside the object is trickier: a uniform sphere of radius R and mass M has zero field at the center which increases linearly until the surface where it has magnitude g=MG/R²; a hollow sphere of radius R and mass M has zero field everywhere inside and then jumps discontinuously to g=MG/R² at the surface. Since we know that a star with a large enough mass will collapse to a neutron star and/or a black hole, nothing will stop that from happening, certainly no "mesh" of any kind. On the other hand, if you want to create a hollow sphere with a thin outer shell, I believe that there would be no limit to how large you could make it without its collapsing. Suppose you have a hollow sphere of radius R₁ and and mass M₁ and it does not collapse; the field pushing in at the surface is g₁=GM₁/R₁². Now suppose you have a hollow sphere of radius R₂ and and mass M₂=2M₁; it is easy to show that if you keep the surface density of the shell σ=M/(4πR²) constant then R₂=(√2)R₁ and so g₂=GM₂/R₂²=2GM₁/(2R₁²)=g₁. The field remains the same and so the field per unit area actually gets smaller as the shell gets larger, making it even less prone to collapse.

QUESTION:
What's the physics of orbiting? Why don't satellites lose their velocity over time and fall straight to the earth's surface as they're orbiting (falling) around the planet?

ANSWER:
Orbital mechanics is just applications of Newton's laws of motion: the motion is determined by all the forces on an object. Start with the simplest assumptions: there are only to bodies and the mass of one body is hugely greater than the other and the only force is gravity. This is what is called the Kepler problem. See an earlier answer for the several possibilities of orbits but the important thing is that a stable orbit is the solution to Newton's equations and it simply continues forever until some other force changes it. Next you take into account the masses of both objects and find that the solutions are essentially the same except that the two orbit around their center of mass and the reduced mass μ=Mm/(M+m) replaces the mass M in your equations. Now you can add other non-ideal forces to the problem. For example, although earth satellites are above most of the earth's atmosphere, they are not above all of it and the air drag with the very thin atmosphere gradually takes energy away from the satellite and its orbit eventually decays to where it hits the ground. The farther out the satellite is, the slower this decay is. Other things can cause the orbit to change also. For example, the moon causes tides on the earth (in both the oceans and the solid earth) which causes some of the energy lost by the earth to be gained by the moon so it is gradually moving farther away (by like about 4 cm a year). And of course any third body perturbs the nature of the orbit. For example, the planet Neptune was discovered in 1846 because the observed orbit of Uranus was not exactly what its Kepler orbit should have been because of interaction with another body, Neptune; calculations predicted the necessary orbit of the unknown planet and it was subsequently observed where it was predicted to be. Imagine doing such a complicated calculation long before computers.

QUESTION:
Would a stove-top pot heat liquid faster if the bottom surface was concave instead of flat? My theory is the increased surface area on the liquid in the pot would cause it to heat faster, and heat would travel 'up' the concave 'cone' area so it would not lose any heat (or proportianally not a lot) compared with if that area was flat and closer to the flame.

ANSWER:
Such a pot would definitely not be best for an electric range. Here you want the element to be in contact with the metal pot bottom so that conduction is the main way of getting heat into the pot; with your pot convection would be the mechanism and result is smaller heat flow. For a gas range, it is possible that some advantage could result as you speculate, but I doubt it. The hottest point of a flame is in the visible part of the flame that you see, and you want this to be close to the pot.

QUESTION:
If traveling on an airplane from North Pole to Brazil, the pilot cannot fly on a straight line or he will miss his target; the pilot must actually fly on curved path to reach his desired destination. What is the reason for this?

ANSWER:
This is probably a homework question, forbidden here, but I am going to answer it because of its ambiguity. The problem with this question is that it does not define what is meant by a curved path. Of course, since we live on the surface of a sphere, no path between any two points on the surface of the sphere is a straight line in the three dimensional space from which we might view that path. Only by digging a tunnel between the two points is the path a straight line. But if you travel through this straight tunnel and your path is observed from outside the earth you follow a curved path because the earth is rotating. How do we usually define a straight line? It is the shortest path between two points. So if you observe things from our point of view on earth and define a straight line on this two-dimensional surface to be the shortest distance between two points, you can connect any two points on earth by a straight line. If the pilot flies due south on the longitude 43.2⁰ W, he will end up in Rio de Janeiro. The pilot does not have to keep steering his plane to account for the earth's rotation (which is what is hinted at here, I think); he flies relative to the air and the air rotates with the earth. If this is a homework question, it is a pretty stupid one.

Q&A OF THE WEEK, 2/18-25/2017

QUESTION:
Where can I get a graph (or other information) about the increasing "relativistic resistance" to the acceleration of a particle (an electron, hopefully) as its velocity is increased to near-relativistic speeds? If such a graph is not available, then how can I calcuate this increasing force that resists a particle being accelerated when it is already traveling at some relatively high percentage of the speed of light, like maybe 70%?

ANSWER:
It would definitely behoove you to read an earlier answer first which has lots of details and discussion of acceleration. There are two ways to approach this problem:

Suppose the observer in the inertial frame is doing the pushing and wants to know how hard to push the particle to achieve a particular acceleration. Resistance to acceleration is usually called inertial mass and the inertial mass m of a particle with rest mass m₀ and speed v is m=m₀/√[1-(v/c)²]. The first figure above plots m/m₀ as a function of v/c. So, to achieve an instanteous acceleration of a, a force of F=ma=m₀a/√[1-(v/c)²]; so for your example of v/c=0.7, you can read off the graph that you would need to exert a force 1.4 times larger than you would if the particle were moving slowly. This is probably what you want if you are interested in electrons accelerating since you would be accelerating them.
Suppose the observer was on the particle and the particle was a rocket ship. You adjust your engines so that the force F which they exert on the ship causes a constant acceleration of a₀=F/m where m is the rest mass of the ship. What acceleration a does another observer in an inertial frame (on earth maybe) see when the rocket has a speed v? Starting with the velocity derived in the earlier answer, v=(a₀t)/√[1+(a₀t/c)²], we can calculate the acceleration by differentiating with respect to t, [dv/dt]/a₀=a/a₀=(1+(a₀t/c)²)^-3/2. Now, reading off the second graph, when v/c=0.7, a=0.36a₀; the stationary observer will only see 36% of the acceleration seen on the ship.
ADDED NOTE:
It is important to note that the acceleration observed depends on the inertial frame you are in. In way#2 above, two different inertial frames, say with v/c=0.7 and 0.9, will observe the ship having different accelerations. This is why acceleration, and therefore also force, in special relativity do not play an important role as they do in Newtonian physics where all inertial observers see the same acceleration. This is discussed in the earlier answer.

Q&A OF THE WEEK (2/12-18/2017)

QUESTION:
If I am moving 55 MPH East (or West) at the equator how much weight would I gain (or lose) due to the Eötvös Effect. Thank You in advance. I am 73 years old and too dumb to figure this out myself.

ANSWER:
First of all, weight is the force the earth exerts on you so you never gain or lose weight when you are moving; you might want to say "apparent weight" which is the force which would be measured by a scale you were standing on. You experience two real forces, your weight W down and the normal force N (a scale, for example) up. One way to solve this problem is to note that an object with mass m with speed v moving in a circle of radius R has an acceleration a=v²/R which points toward the center of the circle; then apply Newton's second law, F=ma=mv²/R=W-N and solve for N to get your apparent weight of W-mv²/R, smaller than your actual weight. This is the "Eötvös effect". But there is another way to approach the problem. Rather than solving the problem from the outside the earth, we might want to solve it here on the earth. But Newton's laws are not valid in an accelerating reference frame (accelerating because it is rotating). You can force Newton's second law to work, though, by inserting a fictitious force which I will call E for Eötvös but it is more commonly known as the centrifugal force; E=mv²/R pointing radially out. Newton's first law now applies, N+E-W=0, so, again, N=W-mv²/R=W[1-v²/(gR)] where g=32 ft/s². In the figure above you have a velocity v=v_Earth+v_man. If you are at rest, v=v_Earth=1040 mph=1525 ft/s and R=3959 mi=2.09x10⁷ ft; so N=W(1-0.00348) and a scale will read 0.348% smaller than your actual weight. If you move with a speed of 55 mph=81 ft/s in an east direction, v=1525+81=1606 ft/s and N=W(1-0.00386), 0.386% smaller than your actual weight. If you move with a speed of 55 mph=81 ft/s in a west direction, v=1525-81=1444 ft/s and N=W(1-0.00312), 0.312% smaller than your actual weight.

FOLLOWUP QUESTION:
I made a donation, but the way I read your answer you have both directions being SMALLER, if I read it right. I know that West is an increase and East is a decrease, just thought you would like to know.

ANSWER:
Thanks for your support! Very generous, particularly because you say that I am wrong! I must stand by my calculations, though. It is certainly not unheard-of that I make an error, but this answer is right. Since I define weight to be what a scale would read if the earth were not rotating (or at north or south poles), you will see that the apparent weight (what the scale reads) increases if you go west and decreases if you go east, just the same as what you "know"! All apparent weights are smaller than the actual weight; the only exception is if you go west with speed of v_Earth in which case the actual and apparent weight will be the same. If you want to compare to your apparent weight at rest, it is 0.386-0.348=0.038% lighter going east, 0.348-0.312=0.036% heavier going west.

QUESTION:
How much "work" (physics definition) is actually accomplished in a gym workout? I'm currently using F x D x reps = actual work done. The upward lift ("D") x the amount of weight lifted ("F") x the number or repetitions... to get actual work done. Am I in making some mistake, here? Thanks, in advance, for your help.

ANSWER:
By "upward lift" I assume you mean the distance lifted. So, lifting a weight F over a distance D you would do W=FD units of work on the weight. For example, the weight of a 2 kg mass is about 19.6 N and the work to lift it 1 m is 19.6 J. But, and here is the catch, you use more energy than 19.6 J to lift that weight because your body is not a simple machine like a lever or a pulley. To understand why, see the faq page. In a nutshell, the reason is that to just hold up a 2 kg mass, not move it up at all, requires input of energy—you get tired trying to hold up a weight at arm's length, right? And, what about lowering the weight back down? The work done on the weight is negative which implies that energy is being put back into you but know that it also takes energy for you to lower the weight at a constant speed. A biological system is considerably more complex than systems we talk about in elementary physics classes. I think that it is of little use to try to analyze a workout in terms of elementary physics.

QUESTION:
Speed of light again becomes 3x10⁸ m/s when it emerges out in air from denser material without the loss of energy. Why?

ANSWER:
Just because it speeds up does not mean that it gains energy. For light, the energy is determined by the frequency, not the speed. When the light enters a dense medium its speed v decreases but its frequency f stays the same. Since v=λf, where λ is the wavelength, the wavelength decreases. Another way to look at it is to think of the light as a swarm of photons. The energy of a photon is hf where h is Planck's constant, so the energy of a photon depends only on frequency.

QUESTION:
If I'm driving and hit the gas and turn left would the angle between the velocity vector and acceleration vector be less than, greater than, or equal to 90 degrees. I would think it's greater.

ANSWER:
You would think wrong! The velocity vector v points straight ahead. The tangential acceleration a_t points parallel to the velocity vector because you are speeding up. The centripetal acceleration vector a_c points toward the center of the circle you are turning. As you can see in the figure, the total acceleration vector a makes an acute (less than 90⁰) angle with v.

QUESTION:
If the earth is curved how is it you can get a laser to hit a target at same height at sea level more then 8 km away? How is it that it's bent around the earth?

ANSWER:

First of all, light is not bent around the earth; it travels in a perfectly straight line and therefore, because the earth is curved, there is a maximum distance away for a target at the same altitude. What that distance is depends on the altitude of the laser. You say that the laser is exactly at sea level by which I presume you mean the surface of the earth; at this altitude you could not hit any target also at sea level. In the figure I have drawn the earth, radius R, a point a distance h above the earth's surface (laser location), and another point a distance h above the earth's surface (target location). The distance between them is 2d. Focus your attention on one of the triangles with hypotenuse (R+h). From the Pythagorean theorem, d=√[(R+h)²-R²]=√[2Rh+h²]; if h<<R, d≈√(2Rh). For example, if h=10 km, about the height a commercial jet flies, 2d≈714 km is the most distant target at the same altitude which you could hit.

QUESTION:
if gravity can hold back all the seas and heavy objects to earth how can a fly or leafs move threw the air does gravity not have same force on everything I don't get how it can hold back everything but same time let small things move so easy ?

ANSWER:
For starters, the force of gravity (often called the weight) is proportional to the mass of the object; the weight is ten times bigger for a 10 kg object than for a 1 kg object. Second, Isaac Newton taught us more than 300 years ago that to understand how an object moves (or doesn't) you need to consider all forces on the object. For example, for a leaf being blown upward, the force of the air up on the leaf is greater than the force of gravity pulling down.

QUESTION:
I have an old fashion balance scale, center fulcrum and two dishes on either side of equal weight. If I place two weights on either side and the weights are nearly the same, the heavier side dips slightly, if the difference between the weights are large, the heavier side dips much more. I do not understand why this is so. Logic says that if there is any difference at all, the heavier side should continue to drop until it reaches a barrier to the fall no matter what the difference in the weight. I asked a physics instructor and he did not have the answer either.

ANSWER:
The center of mass ⊗ of the scale itself must be below the fulcrum (suspension point). Then, as shown above, if the beam is off horizontal for the empty scale or equal weights in each pan, there will be a restoring torque to force a balance only for the horizontal beam.

FOLLOWUP QUESTION:
I don't think the answer given expained the phenomena described in the question. It only explained why when the weights are equal and there is a force on one side that it will bounce up again and continue to rock back and forth until it eventually evens out again. The problem I posed was a different issue, as to how the scale behaves when different weights are put on the heavier side.

ANSWER:
It does answer your question, but just not explicitly because I did not include examples of unequal weights. So let me do that now. Given the explanation in the original answer, I can model the scale as a massless T with all the scale mass M located a distance d below the pivot at the bottom of the T. As shown in the figure, when one side is loaded with a mass m the scale rotates to an angle θ with the horizontal and reaches equalibrium. The sum of the torques about the fulcrum is mgLcosθ-Mgdsinθ=0, so θ=tan^-1[mL/(Md)]. As m gets larger, θ gets larger. For small angles, θ≈mL/(Md) in radians.

QUESTION:
While i am reading Einstein book 'evolution of physics', i have encountered a very confusing reasoning process which i can not understand it. The topic is about how we can deduce that the gravitational mass and inertial mass are equal just by knowing that the acceleration due to gravity for all objects at the same height is constant. And here is the text i can not understand how he deduced that: "Now the earth attracts a stone with the force of gravity and knows nothing about its inertial mass. The "calling" force of the earth depends on the gravitational mass. The "answering" motion of the stone depends on the inertial mass. Since the "answering" motion is always the same —all bodies dropped from the same height fall in the same way—it must be deduced that gravitational mass and inertial mass are equal. More pedantically a physicist formulates the same conclusion: the acceleration of a falling body increases in proportion to its gravitational mass and decreases in proportion to its inertial mass. Since all falling bodies have the same constant acceleration, the two masses must be equal."

ANSWER:
Newton's universal law of gravity says that the force is proportional to the gravitational mass of the object, F=k₁m_g where k₁ is a constant. Newton's second law of motion says that the acceleration is inversely proportional to the inertial mass and directly proportional to the applied force, F=k₂m_ia; in SI units, k₂=1. Therefore, a=k₁m_g/m_i. (This is Einstein's statement "…the acceleration of a falling body increases in proportion to its gravitational mass and decreases in proportion to its inertial mass.") Now, since it is an experimental fact that a is the same regardless of the mass, m_g/m_i is a constant. If we choose to measure both m_g and m_i in kilograms, and since k₁ expresses the strength of the gravitational field, m_g/m_i=1. (For the field near the earth's surface, k₁=M_earthG/R_earthwhere G is the universal gravitational constant, M_earth is the gravitational mass of the earth, and R_earth is the radius of the earth.)

QUESTION:
well, i saw a theory that if you rotate the ISS, which weighs 450 tons at 10 rotations per second, it would generate its own gravity. would it be possible to spin something of proportion size at a proportional speed to generate artifical gravity like the ISS?

ANSWER:
You should read earlier answers regarding "artificial gravity" in rotating space stations. The idea is that if you are in an accelerating frame of reference you feel like there is a force on you; for example, when you drive fast around a curve you feel like you are being pulled toward the outside of the curve. It is not a "theory", it is elementary physics. The mathematics you need is that your acceleration if you are going in a circle of radius R with speed v your acceleration is a=v²/R. For a given R, if you choose v such that a=g=9.8 m/s² where g is the acceleration due to gravity, you will feel like you are at the surface of the earth. In your case, v can be determined from the frequency 10 rotations/second: each rotation has a distance of 2πR so v=20πR/1. This would imply that the acceleration would be a=9.8=(20πR)²/R=400π²R; solving, R=.0025 m=2.5 mm. This is obviously nonsense, so you must have remembered the the frequency incorrectly. Looking at the figure, the main cabins appear to have a diameter of about 5 m, R≈2.5 m, so if they spun about their central axis the required speed would be about v≈√(9.8x2.5)=5 m/s and so the frequency would be f=5/(5π)=0.32 rotations per second=19 revolutions/minute. But this would not work at all for the ISS because an astronaut's head would feel almost no "gravity" because it would be very close to the axis of rotation; if she were to lay on the floor/wall/ceiling it would be close to what it would be like on earth. The ISS is just too small.

QUESTION:
I've been told by many that the fastest thing is vacuumed light. since light travels at different wave lengths does some light travel than other. in other words does a ultraviolet rays travel the same linear speed from A to B the same as say an x-ray. If it does than would that mean we don't have a solid speed limit of light. if they do not do we just use the longest light wave to measure the max speed, OOORRRR do we throw light like a laser... but wouldn't that still have a wave of some form.

ANSWER:
Every-day language usually interprets "light" as that which we see with our eyes. When a physicist says "the speed of light", she means "the speed of electromagnetic radiation". All electromagnetic radiation travels with the same speed in a vacuum.

QUESTION:
I made a statement to somebody that a plane hitting a building was the same as if the building hit the plane at exactly the same speed,the plane now stationary. The results would be the same. In other words, if a man with large hands slapped my hand at 50 mph, it would be same as me slapping his hand at 50 mph.....its interchangable.....the other person said, no, the mass of the building and hand would have different results...

ANSWER:
Either you or your friend could be right depending on what you mean by "different results". Let me try to set up a simple example to demonstrate why.

Imagine we have a 2 lb ball of putty moving with a speed of 5 mph striking and sticking to a 18 lb bowling ball at rest; the time it takes to collide is 0.1 s. After the collision, the two move together with a speed of v₁. To find v₁, use momentum conservation: 2x5=(18+2)v₁, v₁=0.5 mph.
Next, imagine we have a 18 lb bowling ball moving with a speed of 5 mph striking and sticking to a 2 lb ball of putty at rest; the time it takes to collide is 0.1 s. After the collision, the two move together with a speed of v₂. To find v₂, use momentum conservation: 18x5=(18+2)v₂, v₂=4.5 mph.

So, you see that the two scenarios have different speeds after the colliision. But, suppose that you were the putty ball. During the collision you feel a force and the force is what is going to hurt you. Do you get hurt as badly, not as badly, or equally as badly during the collision? What determines the force you feel is the acceleration you experience during the collision, how quickly your velocity changes, which is your final velocity minus your initial velocity divided by the time of the collision.

For the putty ball moving initially, (v_final-v_initial)/t=(0.5-5)/0.1=-45 mph/s.
For the bowling ball moving initially, (v_final-v_initial)/t=(0-4.5)/0.1=-45 mph/s.

You could go through through the exact same process to find that the bowling ball experienced exactly the same force regardless of who moved initially. A physicist would say that you were right, but the ambiguity of your statement means that the other guy could split hairs. As far as physics is concerned, the only thing which matters is the relative velocities of the two before the collision. If the putty ball were moving 105 mph and the bowling ball were moving 100 mph in the same direction, the result of the collision which matters (the force) would be the same.

QUESTION:
The atmosphere is heavy. If the weight of a column of air above your desk is about the same weight as the bus you rode to school in, why doesn’t air pressure crush your desk?

ANSWER:
Because the atmospheric pressure also acts up under your desk.

QUESTION:
I have always wondered how much energy do you do with if you let a kettle at 1800 W be running for two minutes? What is the approximate cost for this? this is not a homework question.. just a question i wonder =)

ANSWER:
A Watt is one Joule per second, 1 W=1 J/s. Energy consumed by an 1800 W kettle in 2 min is 1800x120=216,000 J. But, we are more used to measuring electrical energy in kilowatt hours, (1 kW·hr)(1000 W/1 kW)(3600 s/1 hr)=360,000 J. So the energy used by the kettle is (216,000 J)/(360,000 J/kW·hr)=0.6 kW·hr. A kW·hr costs on the order of 5¢-15¢, so the cost would be between 3¢ and 9¢.

QUESTION:
Why do unstable elements always give off alpha particles with 2 protons and 2 neutrons. Essentially a Helium nuclei. Why not a hydrogen nuclei or a heavier nucleus

ANSWER:
Alpha-decay is only prevalent in very heavy unstable elements. Most unstable nuclei decay by beta decay, the ejection of an electron or positron along with a neutrino. The reason that alpha-decay happens is that the alpha particle is an extremely tightly bound particle and therefore there is a fairly high probability that it will spontaneously form inside a very heavy nucleus where there are a lot of neutrons and protons to contribute. For a more extensive discussion, see an earlier answer.

QUESTION:
Every day my wife reads the latest fake news about planet 9 and sits weeping in fear the whole time . As a scientist ,surely you can tell me one thing that just shuts this whole fiasco down . Please. Tell me that undeniable fact that will convince her that we are in fact safe from a collision or near miss from this nonexistent space oddity and it 's cohorts . It s affecting her health . And I love her . So it' i m feeling her pain as well .

ANSWER:
First of all, "Planet 9" is a serious astronomical topic. Minor irregularities in the orbits of some of the distant planets suggest the presence of another planet farther out. Serious efforts are underway to try to observe it. But its anticipated period is more than 10,000 years, so if it is far away now, it is not likely to be a problem for us for far longer than the lives of any of us. All reputable astronomers have declared that (if it actually exists) it is absolutely no danger to earth. Google "planet 9" to get lots of good information. But, even if there were a planet much closer, like the "fake news" "planet x", there is an amazingly low likelihood of its colliding with earth. Most lay folk like your wife have no comprehension of the vastness of space; the probability of two particular objects in the solar system colliding is, for all practical purposes, zero. It is often posited that a "rogue planet" passing through the asteroid belt would "shake loose" a "storm" of asteroids toward the earth. I recently answered a question addressing this possibility which you might find useful.

QUESTION:
Can tension ever be negative?

ANSWER:
This is an ambiguous question. If you mean can the magnitude of the tension force exerted by a string be negative, the answer is no; a string can only pull, it can never push. But, if you mean can the tension ever have a component which is negative, the answer is yes; it simply depends on how you have chosen your coordinate system. But, if you have drawn the tension vector T such that the string is pulling on something and you solve the problem and the magnitude T<0, you have made a mistake somewher.

Q&A OF THE WEEK (1/21-28/2017)

QUESTION:
My question is about the maximum tension experienced by a bow string. I'm specifically concerned with a traditional or recurve bow NOT a compound bow with pullies. I want to know the max tension compared to the draw weight so I have an idea how strong to make my strings. So here's the scenario, what's the maximum tension in the string for a recurve with a 70 lbs draw weight and a physical weight of 25 oz? I'm assuming the maximum tension is when it's at brace (not full draw), right after the arrow leaves. I think this because not only does the string have to oppose the restoring force of the bow limbs but it also has to stop the momentum of the limbs that isn't transferred to the arrow.

ANSWER:
To compute the tension I would need to know the geometry of the bow. I can tell you that the tension will be at the maximum draw for a simple bow, not where the arrow leaves the string.

REPLY:
The bow is 64 inches long and the string is about 4-5 inches shorter than the bow. It Is braced at 6 inches and has tips that are 3 inches recurved behind the handle. Its draw length is 28 inches and has an elipitcal/circular tiller shape at full draw.

ANSWER:
(An incorrect answer was posted earlier. This is a reposting, correct now, I hope!) When researching the physics of archery I discovered that this can be a very complicated problem requiring very sophisticated numerical calculations on computers if you want precision descriptions of all the details. You, however, require only a rough calculation for estimating the strength of the string. I can do that and it is more appropriate for the spirit of this site—to solve problems with simple physics concepts. This problem requires facility with trigonometry, understanding of Hooke's law, and application of Newton's first law. The simple model I will use was one used before the advent of computers; the bow is modeled as two straight rods (purple) the ends of which move on a circle as the string (red) is drawn. With this simple model, most of the details of your bow are not necessary. When the string is braced (undrawn) there is a certain tension in the string and this tension will increase as the bow is drawn. So, the maximum will be at the maximum draw. The figure shows, roughly to scale, the situation. Using simple trigonometry (law of cosines), I find β=59.6⁰. The point where the draw weight W is being applied must be in equilibrium, W-2Tcosβ=0; solving, T=69.2 lb. I think that your concern about the string having to "stop the momentum of the limbs" is misplaced because bows tend to be quite elastic so that nearly all the energy imparted to the bow by drawing it is imparted to the arrow and the limbs will end nearly at rest. Not so if you draw and release without an arrow, though; what I have read is that in that situation you are more likely to break your bow than the string. I am working on a general solution which I will later add here but thought I would post the part of the solution which answers your question regarding the tension in the string.

GENERALIZED SOLUTION:
To get a better understanding of this problem it is worthwhile to find an analytical solution for the tension as a function of draw distance. My research showed me that a traditional or recurve bow behaves, to an excellent approximation, like a simple spring (Hooke's law), the draw weight being proportional to the draw distance, i.e. W≈kx where x is the distance the string is drawn and k is the spring constant. In this case, since W=70 lb when x=28 in, k=2.5 lb/in. Using the law of cosines, cosβ=(L²+(x+d)²-R²)/(2L(x+d)). Again, the point where the force W is applied is in equilibrium so W-2Tcosβ=0 or T(x)=kx/(2cosβ). Now, note that in the limit where x → 0, β → 90⁰ and cosβ → x/L. Therefore T(0)=kL/2. Using your numbers, T(0)=37.5 lb and the angle and tension for all points are plotted below. At full draw, β=56⁰ and T=64 lb. It was interesting to me that in order for the calculated values to be correct at zero draw, very precise relative values of R and L had to be used because otherwise the expression for cosβ would not be 0 exactly when x=0. The value was R=30.59411708 inches for L=30.

QUESTION:
My question is related to photons. We have coherent light IE laser emission which over distance spread out much more slowly. Our sun emits incoherent light, which is the same for all observant stars. My query is why do we still see the stars as pinpoints of light no matter if they are near or very distant.

ANSWER:
The reason all the stars look like pinpoints is that they are very far away, not because they have tightly collimated beams like a laser; if the sun were very far away it would look that way too. If the light you see from a star were a tightly collimated beam, it would be pointed directly at you and if you stepped to the side it would disappear; you would see almost no stars at night because nearly all of them would be pointing elsewhere.

QUESTION:
when i kick the ball ,i exert a force on it and it exert the same force on me on the opposite direction according to 3rd law of newton ,bet what is the effect of the ball on me .the effect of the force that i exert on the ball make it move ,but my leg do not move .

ANSWER:
Do you feel it when you kick the ball? Of course you do and what you are feeling is the force the ball exerts on your foot. But your foot moves in response to all the forces on it, not just that one. During the kick your leg is exerting a force on your foot much larger than the ball's force on you. Your foot will be moving a little bit more slowly after the kick than if you had not kicked the ball.

QUESTION:
With the understanding that "gravity" is a fictitious force created or experienced by Earths acceleration what I don't understand is what is the source of this acceleration? How is it that the Earth is continuously accelerating us upwards to produce weight?

ANSWER:
Your understanding is wrong. Gravity is not a fictitious force caused by acceleration. Maybe it would help you to read some of my earlier answers regarding gravity and general relativity listed on the faq page.

QUESTION:
This is odd, but My family has just moved into a huge house with little outdoor space. We live in a climate that is cold in the winter, and I want my children to get some exercise on a daily basis. We own a trampoline, and have space for it indoors on the Second floor of our house. The ceilings are 12 feet high, so there would be no problem with the kids hitting their heads on the ceiling. My question is whether or not the house would stand up to the force generated by the trampoline. The walls of the house are made of concrete (you can't nail into it.) I am assuming the floors are quite solid as well, as they must support the weight of the house. They are concrete as well. My Youngest child is quite large (6 ft, 260 lbs)--he is only twelve. We need the activity.

ANSWER:
First, a disclaimer: I can give you an idea of how much force the floor will experience. I cannot predict whether this will cause your floor to fail because I have no information about your floor other than that it might be concrete. I have watched some videos and it seems that the jumper never goes as high as h=2 m and the trampoline never goes down as far as s=1 m. So I will just do my calculations with those to get an upper limit on what force might be expected. Your son's mass is about m=120 kg. An object falling from h=2 m will hit the trampoline with a speed of about v=√(2gh)≈√(2x10x2)=6.3 m/s. I will treat the trampoline as a simple spring so that I can write ½mv²=½ks²-mgswhere k is the spring constant. Putting in m, v, and s and solving for k I find k=7200 N/m; since the force exerted by a spring is F=ks, the largest force the trampoline exerts on your son is about 7200 N=1600 lb; Newton's third law tells you that this is also the force your son exerts down on the trampoline. Therefore, the trampoline exerts a force down on the floor of 1600+W where W is the weight of the trampoline. This is a little more than the weight of a grand piano. Keep in mind that this is the greatest force and just for an instant; the average force over the collision time would be half this. This is a little more than the weight of a grand piano.

QUESTION:
I recently watched with great interest a PBS program which described in layman's terms how Uranium 238 transforms into the different chained elements, to include U235. It also explained the basics of the chain reaction caused by splitting U235 using E=MC^2 as the basis for energy release. This is where I was a little unclear. The split was described as one U325 nucleus splitting into two separate nuclei with some individual particles released (can't remember if they are protons or neutrons) Those particles then collide with other U235 atoms in proximity triggering subsequent splits and particle releases as part of a chain reaction. My question is that the mass doesn't appear to be transforming into energy (E=MC^2). Rather it appears that it is simply splitting and being cast off, so what causes the energy release? This assumes that the number of particles in the remaining two nuclei + the particles independently released still equal 235. There was a general reference in the program to how the Strong Force reduces the size of the resultant smaller nuclei, but it didn't say if matter within each was converted to energy or if the number of particles are additionally reduced through such a conversion. Thanks for any clarification you can provide.

ANSWER:
Suppose that you weigh one ²³⁵U and one neutron. Now, when you add the neutron to the ²³⁵U it fissions and, after all is said and done you have two lighter atoms and a few neutrons; if you weigh all these byproducts, you will find that approximately 1% of the original weight is missing. Where did it go? Most of it went into kinetic energy of the byproducts, that is they are all moving faster. Kinetic energy of atoms is essentially what thermal energy is—the reactor (or bomb) has gotten hotter. For a bomb it all gets enormously hotter resulting in the explosion. For a reactor, the rate of fissioning is controlled and the heat is extracted to drive turbines to create electricity. More detail can be found in an earlier answer.

FOLLOWUP QUESTION:
I'm assuming (correct me if I'm wrong) that such motion was some percent of the speed of light which would account for the transformation of about 0.1% of its matter to energy.

ANSWER:
First things first—there was an error in my original answer, now corrected throughout: the amount of mass converted to energy is about 0.1%; nuclear fusion is about 1%. No, the motion of the atoms is nowhere near the speed of light. It is simply classical ½mv² type of kinetic energy. The "transformation" is simply that—mass energy transformed into kinetic energy. To understand, see a recent answer which explains why a bound system has less mass than if it is pulled apart and the mass measured. It turns out that heavy nuclei like uranium are less tightly bound than nuclei with roughly half their mass; therefore when they split the products are less massive. That is why fission works as an energy source.

QUESTION:
If a capacitor is made of oppositely charged plates, why do they look like cylinders inside computers, remote control cars and other electronics

ANSWER:
The easiest form of capacitor to understand and analyze in an introductory physics class is the parallel plate capacitor. But any two conductors insulated from each other is a capacitor. One possible capacitor is a wire along the axis of a hollow cylinder, but that is not what the common capacitor you are referring to is. Rather, it is a parallel plate capacitor! The analysis of the capacitance of two parallel plates shows that the capacitance is proportional to the area of the plates and inversely proportional to the distance between them. So, take a two ribbons of foil as long as a football field to make the area big and separate them by sandwiching a ribbon of mylar between them to make the separation small; then just roll it up so you can fit it in your device!

QUESTION:
Suppose a block is moving with constant velocity towards right on a frictionless surface and during its motion another block of slightly smaller mass lands on top of it from a negligible height. I argue that the lower block will eventually start moving to the left and upper block will end up moving towards right provided that there is friction between the blocks but not between lower block and ground . My friends can't accept my reasoning. Am I wrong? Please help!

ANSWER:
I hate to tell you, but you are wrong. This is actually a simple momentum conservation problem. Call the masses of the upper and lower blocks m and M, respectively. Before they come together the momentum is Mv where v is the incoming speed of M. When the masses come in contact they will slide on each other but, because there is friction, they will eventually stop sliding and both will move with a velocity u; the linear momentum will now be (M+m)u. Conserving momentum, u=[M/(M+m)]v. They both end up going with speed u and move to the right.

You can also determine the time it takes for the sliding to stop. m will feel a frictional force to the right of magnitude f=μmg and M will feel a frictional force to the left of magnitude f=μmg (Newton's third law). So, choosing +x to the right, the acceleration of m is a=μg and the acceleration of M is A=-μg(m/M). The velocities as a function of t are v_m=μgt and v_M=v-μgt(m/M); we are interested in the time when v_m=v_M, so solving for t, t=Mv/[μg(M+m)]. If you substitute this back into v_m or v_M, you will find the same value we found for u above: v_m=v_M=u=[M/(M+m)]v.

QUESTION:
What would it take for a falling body to travel 25' horizontally from a 350' height? I'm sorry, I didn't explain well. This is an actual situation. There is a guard booth at the base of the bridge, 25' away. If someone was to jump off the bridge at a height of approx 300+ feet, would it be possible that they would be able to strike the booth?

ANSWER:
One possibilty is if there is a steady wind blowing in the right direction. You should know that calculation of air drag, the force of the wind in this case, is always a rough estimate, not something you can predict with precision. I prefer to work in metric units, but I will switch back to ft and mph at the end. A rough estimate of the wind force is F≈¼Aw²where A is the area presented to the wind and w is the speed of the wind (this approximation only works for SI units). I will assume that the horizontal speed acquired by the jumper is small compared to the wind speed so that w is a constant during the fall. So, approximating A≈1 m², the acceleration horizontally is a_x=F/m=w²/(4m) where m is the mass of the jumper. The time to fall 350 ft is about t=4.7 s and the horizontal distance is then x=½a_xt²=w²t^²/(8m). Now, x=25 ft=7.6 m and I will take m=150 lb=68 kg. Solving for w I find w=13.7 mps=31 mph. A steady wind of 31 mph could cause the jumper to move 25 ft horizontally.

You have not told me where the the booth is. If it is under the bridge, then the jumper could not propel himself in that direction. But if the booth is 25 ft out from under the bridge, the jumper could jump out as well as drop down. The speed v_x he would have to give himself horizontally can be easily calculated: v_x=x/t=25/4.7=5.3 ft/s=3.6 mph.

QUESTION:
In electromagnetism we compute the intensity of a wave by taking the square of its amplitude. Why do we not do exactly the same thing with quantum mechanical waves?

ANSWER:
Actually, you could say that is exactly what we do. You just have to ask what intensity means for the wave function. In electromagnetism, intensity is just the energy density flux, the power per unit area, measured in W/m². The square of the wave function is the probability density and so is a measure of the likelihood of finding the particle in one small volume in space. If you add up the square of the wave function at all points in space (integrate), you must get the answer of 1 because the probability of finding the particle somewhere in space must be 1 for this interpretation to make sense; this is called normalization.

QUESTION:
My dad told me about your website, very interesting reading. My question deals with molecules. When a molecule emits a photon, the mass of the molecule decreases to account for the energy in the photon. So, the mass of the molecule as a whole decreases, but this mass does not come from the "parts" of the molecule. In other words, the mass of the constituent electrons does not decrease, the mass of the protons does not decrease, so the energy must come from the electric field between the electrons and protons. But the electric field has energy, not mass. Now mass is a form of energy, but I don't think you can say that the field has mass? But yet, it is said that the mass of the molecule decreases. The electric field contributes to the mass of the molecule, but yet it is incorrect to say that the field has mass?

ANSWER:
Actually, this is not as complicated as you are trying to make it. It all boils down to the fact that mass is a form of energy and must be factored into any energy conservation that occurs in an isolated system. You say that the masses of the protons and electrons do not change, but that is not right. Look at the simplest case, a hydrogen atom. If you measure the mass of this atom it will be less than if you measure the masses of a free electron and a free proton. Here is how you can see that: if you pull the electron away from the proton, that is you ionize the atom, do you have to do any work? Of course you do because the electron and proton are bound together. So, you have added energy to the system (p+e) and that energy shows up as mass. In a system as complicated as a molecule you cannot say which particle or particles changed their masses, but you can say for sure that the total mass of the molecule changed by exactly the energy of the emitted photon divided by c².

QUESTION:
This is my 2nd question related to the issue of time dilation - this one being related to the issue of motion [the other being based on gravity]. Since time dilation occurs for all moving objects, and considering further that the Earth has been revolving around the sun at 30 km per second for the last 4 billion years -- and further that our solar system is moving at roughly 45K mph through space, can't it be said that, compared to other objects in the universe, time dilation has occurred to a significant degree for our planet over those 4 billions years? And that really every object in the universe likewise has its own unique time dilation associated with it? Can't it also be said that every consolidated arrangement of matter in the universe is moving along at different "rates of time?" Wouldn't, over the course of several billion years, these "pockets" of different time spans become more and more "incompatible" with each other?

ANSWER:
The two speeds you quote are about the same (45,000 mph≈2x10⁴ m/s and 30x10³ km/s=3x10⁴ m/s). So let's just choose the larger one and see how much time dilation there is. Relative to the sun, an elapsed time T=4x10⁹ y would correspond to T'=ΥT wwhere Υ=1/√(1-(v/c)²)=1/√(1-(3x10⁴/3x10⁸)²)≈(1+0.5x10^-8). Therefore T'≈(T+20), a difference of 20 years. This may sound like a pretty long time to you, but relative to 4 billion it is less than 10^-6 %. That is not "significant" to my mind. You are right that every object has its own clock which, relative to other clocks, is not necessarily the same; every object also has its own meter stick, not necessarily the same as other meter sticks in the universe. The important thing is that you always must talk about velocity with respect to what.

QUESTION:
The movie "Interstellar" did a nice job of explaining how time dilation works in a massive gravity field. My question relates to how we on Earth measure the age of the universe to be 14.3B years. If I could make that measurement on the planet orbiting the "Interstellar" singularity [and since, theoretically, if I could view life on the singularity planet from Earth, it would all look to be in super slow motion], what would my measurement of the universe's age be from the my perspective on that planet? If time moves more slowly on the singularity planet, wouldn't my estimate of the universe's age be much less?

ANSWER:
As I say on the site, I do not usually answer questions on astronomy/astrophysics/cosmology. Maybe I can make a stab at this. The microwave background radiation which pervades the universe is generally considered the best source for information about the big bang and measurements are probably the best determinations of the age of the universe. In your high-gravity position, you would see the same microwave radiation I do. Likely, you would have to make corrections to your observations due to the intense gravity, but you would still conclude the same age as I did.

QUESTION:
hi, can and are earthquakes be caused by celestial alignments ie planets?

ANSWER:
Let's take a simple example. As seen from earth, Mars and Jupiter are aligned. I estimated the force on a 1 kg object which is sitting, let's say, on the San Andreas fault: F=3x10^-11 N; the weight of that 1 kg object is about 10 N. I would say that putting a 1 kg object on the ground is a great deal more likely to cause an earthquake than those planets, wouldn't you?

QUESTION:
So there's a powerline outside my bedroom window, and I thought, huh. Turns out I'm sleeping with my head in a 6mG AC magnetic field (according to two meters). Help me use physics to stop caring. How do I estimate/calculate which puts more force on the charged particles (calcium, potassium, sodium) in my brain: a) An aqueous solution at 98.6 degrees Fahrenheit or b) a magnetic field acting on charged particles moving at some estimated speed in said aqueous solution. My hope here is that the force of (b) is like an order of magnitude or two, or something, below the "noise floor" of (a) and then I can stop caring forever.

ANSWER:
How about this: the earth's magnetic field is about 0.6 G, two orders of magnitude bigger than the field due to the power line, and you are exposed to it 24 hours a day. It is also possible that there is some other source of field closer by than the power line which, though a much smaller current, would produce a much bigger field. For example, if there were a wire in the wall carrying a typical household current of 1 A, the field 2 m away would be 1 mG. There is no good scientific evidence that any magnetic fields you are likely to encounter have any effect, good or bad, on the functioning of your body.

QUESTION:
I do not want a theoretical answer, but has any experimentalist ever put a very sensitive weight balance below a vacuum chamber before and after vacuating it? Does it get lighter or... heavier? I do not have a sensitive balance nor a vacuum chamber. The reason I ask is that it would say something about the density, or absence of density, of the vacuum. If I understand pressure correctly, the scale would read a smaller weight value, due to less gas being in the column of air directly above it, but there might also be new physics there, if it is not the case. I simply do not know.

ANSWER:
You do not want a "theoretical answer" but you clearly do not understand the physics so I am obliged to give you one anyway. Let us assume the simplest possible "weight balance" so that I do not have to worry that it might operate differently in a vacuum. Envision just a simple string with a tiny butcher's scale which will measure the tension in the string and then hang an unknown weight of mass M and volume V from the string. Besides the string, there are two forces on the object being weighed, its weight Mg and the buoyant force B=ρVg where ρ is the density of air (about 1 kg/m³ at atmospheric pressure) and g=9.8 m/s² is the acceleration due to gravity. The scale will read W=Mg-B, an incorrect measure of the weight. Putting the whole device in a vacuum will change B to zero because the air is gone, so W=Mg, the correct weight. To get an idea of how important this is, consider weighing a solid block of iron whose mass is 1 kg. The density of iron is ρ_iron=7870 kg=M/V, so the volume of the block is V=1/7870=1.27x10^-4 m³. So, the true weight is W=9.8 N and the measured weight in air is (9.8-1.27x10^-4) N=9.7999 N, an error of 0.0013%. However, there are certainly examples where the effect of buoyancy would be very important. For example, consider an air-filled balloon. I did a rough calculation and estimated that the volume of an inflated balloon is about 5x10^-3 m³ so it contains about 5x10^-3kg of air; the mass of an uninflated balloon is about 5 gm=5x10^-3 kg, so the total weight of the inflated balloon is about 9.8x10^-2 N. But if you weighed it in air you would only measure half that amount. Your question, has anybody ever actually observed this, is a no brainer: since the existence of a buoyant force has been known and understood for well over 2000 years (Archimedes' principle), anyone wanting to make an extremely accurate measurement of a mass would either correct for it or eliminate it.

QUESTION:
Several years ago, I was caught in a massive windstorm in a skyscraper. I was on the 54th floor (approx. 756 feet from street level, full building height is 909 feet) , pulling cable, and I stopped for a break. I left a cable pulling string hanging from the ceiling (48 inches free hanging length) in the office, with a 1/4 lb weight attached, and when the storm hit, the weight began swinging like a pendulum. The arc was 16 inches (eyeballing it), and traversing the length of the arc took about 1 second. How can I calculate how far (full arc) the skyscraper was moving by observing what the pendulum in the building was doing?

ANSWER:
A 48" pendulum has a period of about 2.2 s, the time to swing over the arc and back. Since you were estimating, the pendulum was swinging with about the period it would if the building were not moving at all. I would conclude that either the pendulum got swinging somehow and the building was not perceptibly moving or that the period of the building's motion was about the same. If the building was swinging with a period significantly different from 2 s, the pendulum would be swinging with that same period; that is called a driven oscillator.