Perhaps nice compact solutions simply don't exist if your expressions are too complicated. Do you really require an algebraic expression? If not, plug in your expressions for w and v, then use Solve/NSolve.
In no way does the classical theory imply that the statement is false. As said above, quantum and classical theory do not disagree on the definition of the word "intensity".
That's because you shouldn't be touching the charged comb. If you do, you'll act as an earth and the comb will be discharged.
As for the other materials, it's for you to try and find out =D
1) Can you find the displacement (with respect to the flagpole) of each runner at any given time? When they meet, what must their displacement be?
2) How long does it take for the train to stop with the given acceleration? Thus how far would it have travelled?
The purpose of the second comb is to find out the polarity of the charges! You have no way of knowing if you're using just a single comb. You shouldn't expect to find them charged in their natural state.
If the photon is absorbed, it is completely absorbed, since its energy will match the gap between the energy levels of the atom. When the atom falls back to the ground state, it might not do so in one step, but instead it might have several transition states. If so, at each transition, the...
When you rub against the comb, you don't know if it gains a positive or negative charge, right? Thus you rub one comb with, say, cotton, and the other with rubber. If the combs attract, they have opposite charges.
Indeed the energy densities are the same precisely because the intensities are the same! The original sentence is false. The third paragraph you quoted does not imply in any sense that the statement is true, does it?
As mentioned before, this question is more than just acid-base equlibria. Take into account that Mg(OH)2 has a low Ksp as well, so consider its solubility equilibria.
I'm not sure how they got that number, but you can imagine that it'll be huge. Any OH- produced will precipitate out, lowering...
1) Good point. Question should have stated "with respect to centre of the earth" to be more precise. Clearly in this case you've done what they wanted
2) Exactly the same thing as your other thread. Pay attention to the directions in your vector diagram! You have to swim against the flow to...
If the intensity is the same, and the beam sizes the same, the energy carried by each is the same...
Besides, even going by classical theory, how can you assume that energy densities of the two different beams are the same?
"According to this view, the intensities of radiation are equal...
2) The time taken to cross the river would be the width of the river divided by the velocity component in that direction, right? The velocity component parallel to the river doesn't matter here as it doesn't contribute to getting you across the river.
3) Eh, 3 choices are arrows, the 4th is a...
Which identity are you referring to? Whatever the identity it would work for any vector....but of course the cross product of 2 equal vectors is zero =)