Recent content by Maxim Zh

For the gas in the capacitor you should add to the Gibbs free energy a term describing the electric field: dG' = dG + \frac{1}{4\pi}V_{\text{cap}}\vec{E}d\vec{D} This additional term is proportional to dn and makes a contribution to the chemical potential.

Using the relativistic dispersion formula: E^2 = m^2 c^4 + p^2 c^2 you can prove that the conservation laws for energy and momentum can not be satisfied simultaneously if Ep'=0.

Yes, V_A is the speed of rocket A.

Use Gibbs distribution.

You should add the kinetic energy of the support as a separate term: T = T_{\text{pend}} + T_{\text{sup}}

Yes, it is. The corresponding formulas are v^A_x = \frac{v^0_x - V_A}{1-(v^0_x V_A)/c^2} v^A_y = \frac{v^0_y\sqrt{1-V_A^2/c^2}}{1-(v^0_x V_A)/c^2} The "0" superscript means the origin's reference frame (RF) and "A" means A rocket's RF. Remember that emission direction in RF A...

If the balls are conductive the charges can redistribute inside them. This effect is considerable for short distances.

All the acceptor states will be occupied by electrons come from donors. So the semiconductor will behave like n-type one with n'_{\text{donor}} = n_{\text{donor}} - n_{\text{acceptor}}

The first way is incorrect. The observer should be at A position, not at the origin. The second way is correct, but you will have to recalculate the emission direction. I think the simplest way is to use Lorentz transformation to find photon's momentum-energy four-vector: p^i = \left(...

Try this software: http://www.abinit.org/

The general approach is to solve Schrödinger equation for 3 regions: x < 0; 0 < x < L; x > L; and then build the solution for whole space, using wave function and it's derivative continuity. This allows to answer 5). For large L this procedure gives...

The He+ ion behaves like a hydrogen atom with double-charged nucleus (Z=2). Hence the corresponding expression for hydrogen energy levels can be used. Relevant equations: E_{|n,l,m>} = -\frac{m_e Z^2 e^4}{2\hbar^2} \frac{1}{n^2} \hat{L}_\pm = \hat{L}_x \pm i\hat{L}_y

Yes, it's correct. When two operators commute their eigenvectors coincide. In quantum mechanics it means that the corresponding physical magnitudes can be measured simultaneously.

Yes, you would. \Delta E = m(e_2 - e_1). By the way, the solution in the first post (after the correction) will give you the work done by the piston on the steam. But if I see it right the question is about the opposite value.

e1 and e2 are specific energies. Multiply them by the steam mass. You can not sum [kJ] and [kJ/kg]. Watching the dimensions in formulas is a great way to verify your solution.