Recent content by Steve4Physics

Why did the girl-eigenvector and the boy-eigenvector split up? They found that they had very different eigenvalues. And they both needed more Hilbert space.

No. The shape of a stress-strain graph doesn't depnd on the wire's diameter. So the breaking stress and strain are 'constants' for the material (in the same way that a material's density doesn't depend on its size/shape). E.g. wire-1 has cross-sectional area A and wire-2 has cross-sectional...

Here's one way to look at it... For a given material, there is a unique maximum value of strain, ##\epsilon_{max}##, at which fracture occurs.* This correponds to the unique end-point on a stress-strain graph. For a uniform wire of unstretched length ##L## and maximum possible extension...

Well (from what I've read), it depends. E.g. "A bluff charge is intended to intimidate, not to attack. A bluffing bear may run towards you, but it will likely stop short or veer off to the side. It may also stomp its feet, huff, or make other loud noises. While it’s a terrifying experience...

I ___________________________ @catedral3000, it’s worth addressing some problems with your answers to part a). From the first graph it appears that you are using cosines. On this basis... For convenience take ##\omega =1## and ##z=0##, then the question effectively specifies that ##E_H = \cos...

The non-slipping condition is simply required to enable the cone to rotate around the vertical axis. Friction provides the centripetal force. On a frictionless surface, the centre of mass would only move in a straight line.

That's a pity - I thought it [my Post #10 formulation] was quite good! How could it be improved? Yes - then the solution is nearly complete (as already hinted-at by @haruspex). The question does not involve a non-slippery condition. Or maybe you are referring to @Lnewqban's additional (and...

Possibly something like this... A solid right circular cone of uniform density has mass m, height h and a base of radius r. ‘V’ is the cone's vertex. If required, use the fact the cone’s centre of mass is located on the cone’s internal axis, a distance 3h/4 from V. The cone rolls, on its side...

A line can't be perpendicular to a point. You probably meant to ask: "Aren't gravitational field lines always straight and directed towards a body's centre of mass?". To add to what’s already been said, even single objects can have curved gravitational field lines. E.g. the shape of the earth...

I'll explain! In Post #24 you wrote: I looked at the link. The first (remarkable) sentence of the abstract is this: "The foremost grail of this academic indagation is to delineate a mathematical expression of the normalised charge density over a flat disk." Had to look-up 'indagation' and...

My wording was very poor. It was not the intended meaning. (But I hope the intended meaning was clear from the latter part of the post.) I’ve updated the wording. Caution needed. Net zero charge is not guaranteed. As already noted by @SredniVashtar (Post #14) and @haruspex (Post #16) it's...

I may misunderstand the issue you are indigating. But to try to answer your question... Take three conducting plates (equal sizes, aligned, parallel, negligible gap compared to linear dimensions). One outer plate is grounded and the other two plates have charges of, say, +5Q and +2Q. I believe...

Well, I'm not sure if this answers your question but here’s a qualitative (definitely not rigorous) attempt... Thinking visually, a field line starts on a positive charge and ends on a negative charge. Consider two (finite) parallel conducting plates arranged as shown below. The upper plate is...

@tellmesomething, this might resolve specific issues you raised…. ‘Ground’ and ‘infinity’ are not necessarily equivalent. I believe that we can only say that no work is done bringing a test charge from ground to the right side (or indeed to the left side) of plate 3. Minor (pedantic) point -...

Hi @tellmesomething. I understand what you mean - sorry for not picking it up sooner. I’ve edited what I said in Post #8 as it may be incorrect/misleading. I too had assumed the charges were distributed as you show in your Post #11 diagram. Maybe this is the problem. Or maybe there’s no...