Recent content by myanmar

I'm going through the 2007 Physics Olympiad F=ma test tonight (first 27 questions). It's available http://www.aapt.org/Contests/upload/olympiad_2007_fnet_ma.pdf" , if you're wondering. I'll post my attempts here as I get to them. Tell me what I'm doing right or wrong. I don't really know this...

LURCH, this is exactly the kind of answer I'm looking for. However, wouldn't adding more cups (or making them bigger) increase the rotational force and make this irrelvant? If not, explain.

While this is certainly true, we should be able to prove this without relying on the conservation of energy. That is, there is somewhere in the system where energy is lost or something I have ignored.

That almost sounds like saying "It won't work because of the conservation of energy". which I understand. I just don't see where the flaw is that makes it stop working.

perhaps you can help me. on http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm fundamentally, what is the reason that the picture under the words "buoyant wheels and belts" doesn't work? (Except remove the movement of water between two buckets. replace that with compressing and...

I'm really having trouble with these three problems. I'd post my attempts but most of it is in graph from. 3. David subjects a cylindrical can to a certain transformation. During this transformation the radius and height vary continuously with time. The radius is increasing at 4 in/min...

Oh, sorry. I meant \frac{-(x)(12x^2-26x-27)}{(2x-3)^{4}(3x+2)^{2/3}} if {x} > \frac{3}{2}or {x} \leq \frac{-2}{3} \frac{(x)(12x^2-26x-27)}{(2x-3)^{4}(3x+2)^{2/3}} if \frac{3}{2} < {x} < \frac{-2}{3}

Based on Halls of Ivy / Dick 's help, I get this for 1 \frac{-(x)(12x^2-26x-27)}{(2x-3)^{-4}(3x+2)^{-2/3}} if {x} > \frac{3}{2}or {x} \leq \frac{-2}{3} \frac{-(x)(12x^2-26x-27)}{(2x-3)^{-4}(3x+2)^{-2/3}} if \frac{3}{2} < {x} < \frac{-2}{3} is this right?

Thanks for the help so far, trying 13 i get a. u=-x^2 + 2x du = -2(x-1)dx int((x-1)/(-2(x-1)) e^u) su int( -1/2 e^u du) = -1/2 e^u u' + c = (x-1)(e^(-x^2+2x)) + c b. so, I have int(1/x e^(-2lnx/ln3)) dx u=lnx du = 1/x dx int (e^(-2u/ln3)) du = -2/ln3 e^(-2u/ln3) + c = -2/ln3...

I'm having trouble on these two problems. Can anyone give me a step by step explanation on how to do them (or one of them)? 7. Find f'(x), if f(x) = \left|\frac {x^2\left((3x + 2)^{\frac {1}{3}}\right)}{(2x - 3)^3}\right| 13. Perform the integration: (a) \int(x-1)e^{-x^2+2x}dx (b)...

Reducing it to three questions, because I'm pretty confident on the others. 9. Find the volume of the solid generated by revolving about the line x = -1, the region bounded by the curves y = -x^2 + 4x - 3 and y = 0. --- I graphed everything, and then translated the graph 1 to the right making it...

Since I can't seem to find the edit button, I'll post only what I need here. Ignore other problems So, I'd appreciate some help on the following. Even one problem would be great. In order of preference: 5, 10, 11, 9. (6&7 if possible) 5. (I figured out how to do b, but if #5a is wrong...

Last few I can't seem to do, but on the others, can someone check my answers?. The last few I need help getting through. 1. Find the volume of the solid formed when the region bounded by the curves y = x^3 + 1, x = 1 and y = 0 is rotated about the x-axis. My answer: int(pi(x^3+1)^2,x,-1,1)...

So I've got the first one now, it's quite easy due to the help I've recieved. However, on the second one. I have my x values for the extrema, but I can't figure out the y values because I'm not sure how to take the integral of \frac{t^2-4}{1 + cos^{2}t}. I just need to take the def. integral...

1. Find \frac{dy}{dx} and \frac{d^{2}y}{dx^{2}} if \int^{3x}_{1} \frac{1}{t^{2}+t+1}\,dt I expect that I'd make u=3x, then du=3dx. I think when I differentiate, I'd end up with \frac{dy}{dx}=\frac{1}{3t^{2}+3t+3}. I think that \frac{d^{2}y}{dx^{2}} would just be the derivative of \frac{dy}{dx}...