Recent content by zush

No, it's actually the same process. Whenever light interacts with a medium, (reflection or refraction) the interaction is an electron from the medium absorbing the photon and re-emitting it. There's a very good book on this by Richard Feynman called...

Actually, the photons are affected by the the atoms, specifically, they are absorbed and re-emitted by the electrons in the atom.

If you are referring to gravity as moving the nucleus then you may assume so. Seeing as the relative distance from the nucleus to the the first orbit is incredibly large, gravity would not have an effect on the nucleus.

Could this system work using both anti-neutrons, and p-bars?

If I understand your problem correctly, the answer is zero resistance, seeing as the more resistors which are added in parallel in a circuit, the lower the resistance. If there are an infinite number, than there is an equivalent resistance of zero.

For true specifics, you're going to have to take some four years of college and then graduate school to learn about quantum electrodynamics, or QED. There's a great book on it by Richard Feynman.

How did you get that velocity? since v=(2gh)^0.5, you get v=32.833 m/s. Also, what's the mass of the roller-coaster?

The thing is, it takes energy to oscillate the magnet which passes through the metal coil. This is the energy which becomes the work done by the electric field.

It's impossible. You need some type of charged object/physical particle to move it. It's like asking to move a stationary ball without touching it or using a gravitational attraction.

A good link to a pdf on quantum field theory is right here. http://damtp.cam.ac.uk/user/tong/qft/qft.pdf

Could you please be clearer? What exactly do you want to know?

This depends. Are the charges moving or are they stationary?

thank you I found what I was looking for

18. That is the correct explanation 22. P=IV=V^2/R. Because of this, assuming that the voltage is constant, the greater the resistance, the less power output. 23. The top bulb would be unchanged, because the same amount of current leaves it. Its hard to explain For bulb 2, but it has to do with...

looks about right