Recent content by macaholic

Thanks for the tag team help everyone :). I feel slightly silly now.

No idea. It just says "2m higher". I didn't actually plug in the numbers though... H_a = 5 meters H_b = 7 meters. WELP. Mystery solved. It's listed in italics... so I didn't think it was a unit. Thanks again haha.

Thanks guys, I stink at algebra apparently haha. So is this stuff about the "2m" nonsense problaby just some sort of weird mistake?

@Tsny, I will try to be more explicit, let's see... For B: KE_{trans} + KE_{rot} = GPE \frac{mv^2}{2} + \frac{1}{2} \frac{2mR^2\omega^2}{5} = mgH_b v^2(\frac{1}{2} + \frac{1}{5}) = gH_b H_b = \frac{7v^2}{10g}

Except that both balls don't always have total energy equal to \frac{mv^2}{2}. They're said to be rolling withoutt slipping initially, which means that they have both translattional and rotational kinetic energy, i.e. KE_{tot} = \frac{mv^2}{2} + \frac{I\omega^2}{2}. Ball a is still rolling with...

This is from an old course I took. I'm not sure what I'm doing incorrectly. Homework Statement Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is V = 9.9 m/s. Ball a encounters a frictionless ramp, reaching a maximum height...

Homework Statement Suppose there is a block attached to a spring with spring constant k. The block is pushed so that it compresses the spring a distance x_1. The block is released and slides without friction up a ramp, coming to a maximum height h_1 above the ground. Suppose we compressed the...

*facepalm*. Thank you! I feel very silly now.

I was looking over my old physics course problems, and I can't figure out how I'm doing this one wrong. Homework Statement Two identical stars, each having mass and radius M=2*10^29 kg and R = 7 *10^8 m are initially at rest in outer space. Their initial separation (between centers) is the...

'doh, thanks a bunch! I should have been able to try that on my own. While I'm here, would you mind helping with the other proof I'm stuck on? It wanted me to use: (\text{im} A)^\perp = \text{ker}(A^T) to prove \text{rank}(A)=\text{rank}(A^T) im is just the column space, and ker is...

I do? I guess that what I'm missing, I can't currently see how that arises from the equalities I'm given. It would make it trivial from there though, since then I would just say that \text{rank}(A A^T) = \text{rank}(A) = \text{rank}(A^T A) I keep wanting to "sub in" A A^T to one of the...

Funny you should mention that... That was the problem BEFORE the previous problem, which I still haven't figured out quite yet. In any case, I suppose I can use that fact. But I'm not quite sure how it applies since (A^T A)^T = A^T A So I guess I have these equalities...

Homework Statement Does the equation \text{rank}(A^T A = \text{rank}(A A^T) hold for all nxm matrices A? Hint: the previous exercise is useful.Homework Equations \text{ker}(A) = \text{ker}(A^T A) \text{dim}(\text{ker}(A) + \text{rank}(A) = m The Attempt at a Solution The previous exercise...

OH wait I think I may have something... I know that V\Sigma V^T = \Sigma (I can't figure how to show this though...) So that makes my inequality x^T(\Sigma x) \geq 0 Was that at least one of the right steps? Though I can't justify the first part...

Well I don't know how to show this, but I have a feeling that the product of an orthogonal matrix and a positive semidefinite matrix is positive semi definite... But alas I can't think of a way to reason this out. I'm no good at these matrix manipulation proofs it seems.