Recent content by mattyc33

Would the longer vector justify that angle...? i.e. would the 469 (E37.5S) vector cause the displacement to be in that direction (E37.5S) again?

Right I am confused on what the bearing would be

Oh wow I did draw it out wrong. So it is a right angle triangle with hypotenuse of 475.634m... I just have to give him the angle or degrees or something from the origin which I don't know how to do. This should make it more clear to you hopefully. Right angle triangle due to trigonometry one...

Ummm I'll try to visualize it for you. I'm new to the site so I'm not sure if there is a function to draw... Basically I have an origin I have a line (displacement of 74.896m 37.5 degrees North of East from the origin which is diagonally up and right (This comes from the velocity). I have...

I can... I have triangle here with two distances of equal angles compared to the +x line of the origin. I'm actually solving this for my little brother so this is really outdated for me. But to get that distance shouldn't it just be a straight line... AKA the subtracted number multiplied by cos...

Homework Statement "An electron moves at 24.8m/s (N37.5E) and undergoes a constant 103m/s2 (E37.5S) acceleration for 3.02s what is the displacement? Homework Equations 0.5at^2 = d d= vt The Attempt at a Solution I calculated the displacements to be 74.896m (N37.5E) and 469.7006m (E37.5S)...

I am just having problems factoring the constants out of the integral: ∫Fsinwt dt where F and w are constants. I can obviously take F right out of the integral but I forget how to take the w out of the sin, and I can't seem to find anywhere on the internet that will tell me how to do so. If...

I actually got help from a TA but thank you so much for your help until now.

sorry 2b*

So if I complete the integral: ∫(bsinθdx/dθ - acosθdy/dθ)dθ From 0 to pi = -bcosθ - asinθ from 0 to pi = b-1 Would b-1 be my final answer for work? That doesn't seem right to me, did I misplace a number that I can substitute in for b?

Sorry I must have mistyped the equation in the first place. It should have been: (ρ(naught)a^2q/18ε(naught)) (1-3((r/a)^2) + 2((r/a)^3) Hopefully that makes it more clear. The units for U should be Joules I believe (could be mistaken)

Thank you for clearing things up. At this point I'm just having trouble proving that my answer will be in Newtons (correct units). Which means I probably derived wrong.

Then y=b at theta = pi/2. y=0 at theta = 0 as well x=a at theta = 0 x=-a at theta = pi Therefore the values that I am looking for are theta = 0 and pi. I don't understand what the limits for this integral would be though. As well as what I am doing with these values of theta.

Homework Statement In a certain region, a charge distribution exists that is spherically symmetric but non-uniform. When a positive point charge q is located at (r,θ,φ) near this charge distribution, there is a resulting electric potential energy for the system given by: U(r,θ,φ) =...

Well for sinθ y=0 at 0, ∏ and 2∏ Therefore would "a" be pi/2 and theta be 0, 180 or 360? for cosθ y=0 at ∏/2 and 3∏/2 so b would be pi/2 or 3pi/2 and theta be 90 or 270... I am just really confused at this point as I feel like I need to review basic geometry however this question is one of...