Recent content by mjc123

Remember to account for the different radius; g = GM/r2. So g x 1.19/1.062 = 10.39 m/s2

https://www.unz.com/PDF/PERIODICAL/Encounter-1983nov/97-98/

In the 1980s the British left-wing miners' union leader Arthur Scargill visited the Soviet Union and made a speech in which he referred to "Ronald Ray-gun" and Margaret Thatcher as "the plutonium blonde". The apparatchiks were bemused because of course the puns didn't translate into Russian...

"Evening of Clairvoyance: Cancelled due to unforeseen circumstances."

Don't try it! https://en.wikipedia.org/wiki/Cat_bin_woman_incident

They don't intrinsically. A sodium atom with a single electron in the 3rd shell is more stable than a sodium ion (with a closed shell configuration) and a free electron. Why then does sodium form ionic compounds containing the Na+ ion? You have to consider the total energetics of the compound...

Yes. I remember when the company was in the news for boring car-industry reasons.

To simplify, imagine X is drawn randomly with 50% probability from either of 2 distributions N(μ1,σ2) and N(μ2,σ2), with μ1 - μ2 >> σ. Do you imagine the variance of X will be equal to the mean of the two separate variances, i.e. σ2? It will obviously depend on the difference between the means...

Or in England.

Because quick brown foxes jump over it?

You misunderstand the question. 1 m is a measure of height, not steepness. The question is, over what horizontal distance does the height change by 1 m?

Could have called it a "lift".

Remember: a floating body displaces its own weight of fluid; A submerged body displaces its own volume of fluid. Whether it floats or sinks depends on its own density relative to the fluid.

But I don't accept your premise. You assert that the top plate's charge distribution doesn't change, therefore the electric field doesn't change. Why? I say that the electric field changes, depending on what's between the plates, and therefore the charge density will change too. The key point is...

I think you are right. The book solution implicitly assumes that the voltage at the bottom of slab 1 is constant across the width, which is not the case. You have to treat it as two series combinations in parallel.