Recent content by thaer_dude

It works, ty

Homework Statement Show that: \sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{π^4}{90} Hint: Use Parseval's theorem Homework Equations Parseval's theorem: \frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2dx = \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2) The Attempt at a Solution I've been trying to solve...

Thanks for the method confirmation, I tried it again, very carefully, and it worked! :)

Homework Statement Find the centroid x,y,z of the region R cut out of the region 0<=z<=5sqrt(x2+y2) by the cylinder x2+y2=2x. Homework Equations x2+y2 = r2 x= rcosθ y= rsinθ The Attempt at a Solution Centroid x being Mx/m I'm guessing I've been working on this problem...

Yeah, but say I do Vr = V * Ur I would get Vr = (2t i + 2 j) * (cosθ i + sinθ j) Vr = 2tcosθ i + 2sinθ j Is that right? It strikes me as odd that the Vr I found has both θ and t in it.

cos(x) is equal to cos(-x) Those answers are equivalent. Btw, if you're confused as to what Wolfram Alpha is doing, press the "Show steps" button. Helps out sometimes.

Homework Statement At time t, a comet has the position R = (t2-1)i + 2tj At t = 2, find the radial and circumferential components of velocity and acceleration Homework Equations Vr = V * Ur Vθ = V * Uθ ar = a * Ur aθ = a * Uθ Ur = cosθ i + sinθ j Uθ = -sinθ i + cosθ j...

You have the initial horizontal velocity, and the distance traveled horizontally. Using your equations, you should be able to find the time it takes for the ball to hit the ground. You have all the information you need now. For the vertical "part" of this problem, you know what distance must...

A part of the problem is missing. A diagram most likely.

oh well :p

k=0.5mv2 you should know that formula

I'm kind of sleepy so i couldn't pinpoint the problem earlier but I'm pretty sure you just used the wrong value of r, that's why I recommended to draw a diagram. its the middle of a square with 10 cm sides so if you do pythagoras you have to use .05m and not .1m like you did

yeah, I had just done it and I got -2.55J as the answer right before I posted my reply. here: W=(kQq)/r get it into W=(k)*(Q)/(r)*(sum of qs) form W=(9*10^9)*(-5microC)/(0.0707m) *(.6microC + 2.2microC - 3.6microC + 4.8microC) W=-636396*(4microC) W=-2.55J

your method is fine, you must have done a calculation error somewhere. just redo it on a piece of paper.

I'm not too sure what you did in your calculations but.. First step is to do a diagram with the 4 charges each at a corner of the square. Then you have to find the potential energy at the center of the square (which is acquired by adding the V=kq/r for each charge). Now that you have the...