Recent content by klile82

Your explanation of H bonding is right on the money but I think you are missing an important concept. Hydrogen bonding occures between + and - poles of polar molecules (this is analogous to the attraction between + and - poles on a magnet). Hydrogen bonding is an important force in water because...

Snell's Law for the general case says that n1sin(theta1) = n2sin(theta2) and since the angle into the second interface is the same as the angle out of the first interface, n2sin(theta2) = n3sin(theta3). So n1sin(theta1) = n3sin(theta3). So the index of refraction of the middle medium doesn't...

A linear combination of x and y is a*x + b*y. So you want to know if you can find some a and b such that a*first column + b*second column is equal to third column. Can you?

Why the 1? The 1 isn't a hack, it's there because of an identity used in the derivation. Basically, as you go through the derivation you wind up with a term that looks like the sum from n=0 to infinity of exp(-nh\nu / kT) . There's an identity that says that the sum from n=0 to inifity of...

I'm not sure what's hiding in your k1's and P's (is k the same as k1?), but there's a really good treatment of this in McKelvey's Solid State Physics (section 8.3 in my version).

Here's my guess at a starting point: Lasers work because of the resonant cavity. For a cavity with mirrors at either end, a resonant wavelength is one such that after making a round trip in the cavity, there's some multiple of 2pi phase shift. (That's where your \lambda = 2L/n comes from...

Catastrophe? So, Rayleigh-Jeans says that the energy density is U(\nu) = \frac{8 \pi k T \nu^4}{(c^4)} and Planck says U(\nu)d\nu = \frac{4 \hbar \nu^3}{c^3 (\exp(\hbar\nu / k T)-1)}. When \hbar \nu >> kT, exp(\hbar \nu / kT) >> 1, so you can treat it as U(\nu)d\nu = \frac{4 \hbar...

Here's what I would do: Take the partial of \phi_{u} with respect to u. Forget about the intermediate step in the solution, I think that required some extra simplification. You care about the end result. Start with the chain rule Substitute in for x and y Pull out of each partial...

Since Joe has to increase his velocity by 22.8% to match speed with Tom, you want 1.228vJoe = vTom, not .228vJoe=vTom.

You're using Snell's Law correctly, but I think this is a two-part problem. So first, you need to apply it to find the angle of refraction into the glass at point A. This is the angle you solved for. You need to use that angle to apply Snell's law again at point B, to find the angle at which the...

A) Can you show your work for A? I think you're on the right track with setting their kinetic energy equal, (and then using what you know about Joe's velocity in terms of Tom's velocity) but I'm not sure how your method didn't work.

Also, since x<0, when you multiply both sides of 1/x < eps by x, the inequality flips. Does that make sense? You know that eps is positive and x is negative, so you can't say that 1/eps < x. Sorry for my lack of tex. I'm working on that.

Again, your limit is as x approaches neg inf, so x is negative. So the absolute value |1/x -0| = -1/x is positive.

You can take the log of both sides

Can you use what you know about x<0? So from your next to last line, say |x| >\sqrt[4]{\frac{2}{\epsilon}-1} -|x|<\sqrt[4]{\frac{2}{\epsilon}-1}< |x|, but you know that x=-|x| Okay, I obviously need to work on my tex communication skills, but hopefully you can make sense of this.