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16

Yes, in special relativity, the mass of a system is synonymous with the energy of the system in a frame where its momentum is zero. This, as you observe, would directly follow from the relation $E^2=p^2+m^2$. I will drop the factors of $c$ for convenience (or, in other words, I will use natural units and set $c=1$). Thus, in spirit, saying that the mass is a ...

5

A watt is a unit of power, not energy. Power is energy per unit time. Every second we use on average 14 trillion joules of energy. A joule per second is a watt. By the way, when you divide 14 trillion watts by the roughly 7 billion humans on Earth, each human uses about 2,000 watts as a continuous average. This is an easily grasped number. (For example, one ...

3

Well, they don't. An ideal wire would have no resistance so no charge carrier electron collisions with the wire lattice electrons. This would mean that no energy is lost, similar to assuming no friction on an ideal surface. It could mean that electrons have the speed of light as drift speed. This idea probably doesn't sound right! That's because the wires ...

3

You are right in saying that it's weird that this static system does work. The reason it does work is because it is not a static system, but rather electrons are moving through a resistive medium (the electromagnet). Though a normal electromagnet would in fact consume power, this is not necessarily the case. Because the system is not moving, and there are no ...

2

The most direct evidence for the acceleration of the expansion of the universe comes from the observation of type Ia supernovae. Specifically, their luminosity is measured to determine their distance from us, and their spectrum is measured to determine their redshift. The way that the distance is found to depend on the redshift is what would be expected if ...

1

You should think about the meaning of $x$. For instance, in the casse of NaCl dissociation it would consist of the coordinates of all Na$^+$ and Cl$^-$ particles, i.e. it would be a highdimensional vector with about $3\times6\times10^{23}$ components. Notice that usually also the momenta are also included, but i assume these were integrated out...

1

The answer already given is comprehensive and correct. Concerning your example problem however, using your approach you must put r=a/2 for L=mvr; so that squaring gives the factor of 1/4 on the top. Otherwise, for interacting particles, the point of the exercise is presumably that in quantising such a system of particles, one calculates I=mr^2 using r=a to ...

1

You can quantize it anyway you want, but some ways are better than others. Lets say you have two non-interacting particles. Then the position of each particle relative to another one should not matter. Therefore you expect the Hamiltonian of the whole system to commute with rotation operators of both particle 1 and 2. Lets us put it into maths. You have ...

1

Momentum is a Vector quantity, it is measured with both value and direction. Kinetic Energy is a Scalar quantity, measured only in value, but dependent on momentum. If momentum is constant this means that we are traveling at a constant speed in a constant direction, our Kinetic Energy is then guaranteed by the equation K=P^2/2m A constant kinetic energy ...

1

Momentum is a vector, but only its magnitude enters into kinetic energy. So kinetic energy can be constant while the momentum vector varies as long as its magnitude stays constant and only its direction changes (for example like in perfect circular motion).

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