Basically as you increase the slit width your interference pattern will become more "blurry", i.e. it will not stay as sharp as it would for a very small slit and therefore the intensity will not just simply scale quadratically with slit width. Here, specifically I'm refering to intensity of the interference fringes.

It's hard to give an accurate explanation without intensive mathematical analysis. However, I'll try to give some argument in favour of why it shouldn't be proportional to $\omega^2$ . Imagine a point for which the condition of constructive interference is satisfied in the limit of infinitesimal slit width (for a double slit situation). When you increase the slit width you obviously let more light through but you also slightly break the condition of constructive interference as now the phase difference is not exactly the multiple of $2\pi$ for every "pair of points" of two slits.

@lpz Unfortunately I don't understand french. Automatic translation to english allowed me to understand the approach but it's not super clear as I imagine a lot has been lost/misguided in translation.

@lpz Yes, that's what I meant

@MariusLadegårdMeyer Hmm, I think I see your point. In the meantime I found this paper which I think implies (if I didn't grossly misinterpret something, which could be the case) that relation is indeed a simple contraction factor : iopscience.iop.org/article/10.1088/0305-4470/12/9/011/pdf But obviously, I think my line of thinking about this was wrong.

What happens is that force of gravity acts on every small segment of your rod so total torque would be the sum of torques produced on every segment of your rod. That's hard to calculate though (well it isn't that hard if you know integrals), but luckily you can prove mathematically that the system (i.e. rod) behaves as if the force of gravity only acts on its center of mass which makes the problem much easier to solve. For example, try to understand Claudio's proof in the answers where he demonstrates that you can take that gravity acts only on body's center of mass (if g is uniform in space).

If they were to align then the resulting magnetostatic energy (stored in magnetic field produced by such a ferromagnetic sample) would be maximal. In order to reduce that energy (by supressing the magnetic field outside of ferromagnet) it's favourable for domains to form, although formations of domain walls also cost extra energy (because neighbouring spins in that region slowly turn from one direction to another). I guess different parts of the sample don't feel each other that much because exchange interaction is limited to 1st neighbours mostly, and long range dipole-dipole int. is weak.

In QM you use Hamiltonian formalism so you don't really need to deal with the forces as they are encapsulated in potential term. If I had to guess, you could maybe define force as expectation value of $-\frac{\partial V}{\partial x}$, but I'm not sure if there is straightforward connection of force with momentum in QM.

@Phys Correct, although multi-strand wire would have lower resistance only if its total cross-section is larger than for normal wire (assuming we use same materials). One wire with cross-section $S$ has the same resistance as 5-strand wire each with cross-section $S/5$ (you can verify that yourself). More complications arise if you use high frequency voltage source in which case resistance of multi-strand wire would be higher due to skin effect (basically for high frequencies current through a conductor is concentrated on its thin surface layer, lowering the effective cross-section).

Sorry for spamming comments, but there's limit to characters. Anyways, you should check out this paper (if you have access) : researchgate.net/publication/… EDIT : P.S. It's not my paper :D