Ask the Physicist!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ANSWER:
The things you need to know:

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

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

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

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

∫B⋅ds=μ₀I_enc

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ANSWER:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

QUESTION:
What the heck is spin? Quantum objects don’t literally have angular momentum, do they? Does the concept of spin have any relationship to angular momentum? Or is it just a name like “color” that isn’t descriptive of what’s happening?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ANSWER:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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