Recent content by Geronimo85

For a molecule of water, I understand that there are 6 degrees of freedom for each of the three atoms within it; 3 translational and 3 due to the potential energy of the bonds. Is this at all affected when the water goes from solid to liquid or gas?

whoops, missed the heading about posting homework questions... I'll mosey on out of here. Sorry!

I understand that for H20 in solid form, each atom within the molecule has 6 degrees of freedom; 3 translational and 3 due to potential energy from bonds. Does the number of degrees of freedom change when water changes phase to a vapor?

Ok, how does that help?

I have a 3x3 symmetrical matrix A: 1, 8^.5, 0 8^.5, 1, 8^.5 0, 8^.5, 1 that I need to find A^5 for. Is there a method aside from brute force multiplication to do so?

Nevermind, I've got it. I'm a bit slow on the uptake today

I need to prove that the uncertainty principle can be expressed in the form delta L * delta theta = hbar/2 where delta L is the uncertainty of the angular momentum and delta theta is the uncertainty in angular position. I know that L = m*v*r and I think I can express theta as x/r...

yes, that makes sense, but how is that any easier to integrate when multiplied by e^ax?

sorry about reposting it, I just feel like I'm going in circles

I'm supposed to integrate the following expression, and supposedly there is a very simple way to do so. Maple comes up with something rediculous, so I'd appreciate any input. Sorry about the short hand, don't know how to make everything pretty on here: Integral[(e^ax)cos^2(2bx)dx] where a and...

I'm supposed to integrate the following expression, and supposedly there is a very simple way to do so. Maple comes up with something rediculous, so I'd appreciate any input. Sorry about the short hand, don't know how to make everything pretty on here: Integral[(e^ax)cos^2(2bx)dx] where a...

I know that (a + b)² = a² + 2ab + b² and (a^b)^c = a^b*c so if: (e^ax)cos^2(2bx) = (e^ax) [(e^(i*2*b*x) - e^(-i*2*b*x))/2] because: cosx = (e^ix - e^-ix)/2 why when I square inside the brackets don't I get (e^ax)* [((e^(i*2*b*x)-e^(-i*2*b*x)/2)^2] ? I'm just really not...

but I have: (e^ax)cos^2(2bx) = (e^ax)* [((e^(i*2*b*x)-e^(-i*2*b*x)/2)^2] when I square [""] how do I not end up with x^2?

something really ugly that doesn't seem to cancel out: integral(dx(e^ax)cos^2(2bx)) =(e^ax)*(2e^(-4*b^2*x^2)-2e^(4*b^2*x^2))

because I'm an idiot... got it, thanks a bunch. Also, with the integral problem (I haven't done calc in a long long time), once I substitute cosx = (e^ix - e^-ix)/2 how do I solve?