What is said in the video is true, but... remember that the atomic theory is just that: a theory. The theory itself predicts that perturbations will have a really big influence on the results. Take ...

The definition of speed (please, let me call it velocity hereinafter) is not random at all. It seems you understand that it must depend on the distance $d$ and the time $t$, so I'll skip to the next ...

This is a very interesting phenomenon. Roughly speaking, the thing is that pressure affects the "effective length" of the tube. Let me explain, tubes are not as easy as strings. A string has a ...

The confusion comes because you think that something can be seen if it is facing the reflective surface. But that's not true. Check iamge 1. You think (if I understood right) that the rays from the ...

Schrödinger's equation makes a microscopic-level description, while temperature is a macroscopic variable.

The charge of $1C$ was derived from the definition of Ampere. If you look at the SI units, you'll check that, surprisingly, intensity of current is a basic unit, whereas charge is a derived quantity. ...

The circle is just one way to see those oscillations (the x component of a circular movement is a sinusoidal function). But the actual reason is that sines and cosines are the simplest solutions to ...

I guess you're asking for an intuitive explanation, because it seems to be mathematically clear. Well, imagine a set of three columns of raindrops, each row containing three drops. Let's call columns ...

This question is never addressed by teachers, altough students start asking it more and more every year (surprisingly). Here are two possible arguments. A particle is meant to have 0 volume. Maybe ...

Okay, I'll try with a poor but "intuitive" explanation. According to relativity theories, "it is impossible to tell if you're at rest or moving with constant velocity". We know that a charge at rest ...

You can compute the mean value of the velocity vector. However, it turns out to be useless sometimes. A trivial example is a circular motion. The mean velocity of a full loop in a circular motion is ...

You need to add surnames to just "work." W-E theorem is $$W_\text{total}=\Delta K_e$$ For conservative forces $$W_\text{cons}=-\Delta P_e$$ Consequently, if you combine both formulae, since ...

You ask for "why". Well the reason is that the rotation operator in matrix form does that. For a spinor $\left( \begin{array}{} a \\ b \end{array} \right)$, you find that the rotation operator about ...

For small oscillations, you find that $$T=2\pi\sqrt{\frac{L}{g}}$$, where $L$ is the string lenght, $g$ the gravity, and $T$ the period. You see that this doesn't depend on the mass. But the key is ...

Seeing that it is dimensionally correct should be easy. Just plug the units and check that they match. $F=kx$, so $k$ must be newtons/meters. Remember that newtons are $kg\ m/s^2$. So, check that $$... View answer 6 votes An electron in an orbital is described by QM, so it doesn't make sense to talk about position vector or momentum vector. You can, however, calculate the mean values of the coordinates. But, if you're ... View answer Accepted answer 6 votes By default we usually suppose that the electric field vanishes in infinity, since for a point charge it is KQ/r². We usually stablish V=0 at infinity in order to cancel one of the terms of the ... View answer 5 votes The definitions of "second" and "metre" have not changed. One second is defined in terms of frequency. Frequency is measured in hertz (1\ \rm Hz=1\ s^{-1} We take an atom of \rm ... View answer 5 votes Okay, I'll try to answer. First of all, there's one idea that some people forget at the beginning: commutators are operators That is, they act on a wavefunction, like all operators. A commutator is ... View answer 5 votes There are no dumb questions, just bad teachers. In your case, both capacitos are in series. IT doesn't matter if they are upside or in the red boxes, it's series. How can you tell? This is a rule of ... View answer 5 votes The other answers are correct. Just in case you don't see it very clearly, here's my attempt. There are few ideas. 1.- The flux of a field through a surface is defined as \iint \vec{E}\cdot d\vec{S}... View answer Accepted answer 5 votes The main factor is geometry. The Sun doesn't change, and it emits a constant "radiant intensity", defined as "luminous flux per solid angle" I=\frac{dF}{d\Omega} And the illumination depends on it.... View answer Accepted answer 4 votes Both L and S are quantum angular momenta. Angular momentum is different in quantum mechanics (QM). Almost everything is different in QM. In QM, angular momentum is complete if you give two ... View answer Accepted answer 4 votes That is not the momentum vector, it is the modulus. If you recall the definition of kinetic energy,$$K=\frac{1}{2} m v^2 = \frac{1}{2} \frac{m^2}{m}v^2 = \frac{1}{2 m} (mv)^2=\frac{p^2}{2m}$$So ... View answer 4 votes The surface is not a real existing surface, it is an arbitrary stroke that you draw. In other words, the space is filled with electric field lines, and then you draw a circle on that space; but that ... View answer 4 votes It might depend on your definition of "simple". For easy cases (low numbers, direct steps), yes, it is simple. However, it gets complicated too fast. What we do is: calculate only the easy ones, and ... View answer 4 votes Short answer: No, you can't. This is one of the few things you can think of classically. If you have an arrow pointing upwwards, can you expect that the arrow reduces its lenght (to 0 in this case) by ... View answer Accepted answer 4 votes It can either be linear, or elliptical... or any other! There are not just 2 possibilities. The terms "linear" or "elliptical" refer to the basis that we have chosen. If we choose a linear ... View answer 4 votes Let \vec{v} be te velocity vector, and \vec{r} the position vector. Indeed, you are right, it is true that$$ \vec{v}=\frac{d\vec{r}}{dt} But, you should be aware of one thing: all ...