Recent content by Peter5897

Wow, I'm a little embarrassed that I forgot to distribute that constant but at least I wasn't totally lost with what I was doing. I haven't redone my work yet since I'm sick and I want to get some sleep but I'm sure it will all work out fine now. Thanks!

Thanks for the reply, I'm still unsure of what relation I'm looking for. Here are my steps for solving the partial derivatives. - \frac{h^2}{4\pi m}\frac{\partial^2\Psi(x,t)}{\partial x^2}+U(x)\Psi(x,t)=i\frac{h}{2\pi}\frac{\partial\Psi(x,t)}{\partial t} - \frac{h^2}{4\pi...

Homework Statement Show that \Psi(x,t) = Ae^{i(kx-\omega t)} is a solution to the time-dependent Schrodinger equation for a free particle [ U(x) = U_0 = constant ] but that \Psi(x,t) = Acos(kx-\omega t) and \Psi(x,t) = Asin(kx-\omega t) are not. Homework Equations - \frac{h^2}{4\pi...

Thank you, I always got stuck at step 2. In fact at one point I was sitting there with sqrt( ( 2t^2 + 1)^2 / t^2 ) ) and was baffled... Thanks again, the world makes sense now.

Ack, I wish that was my problem but I actually just mistyped it up there. I still don't see how sqrt( 4t^2 + 4 + ( 1/t^2 ) ) gets reduced to (2t^2 + 1)/t and I'm sure it's some simple step that I'm missing. Sorry...

Couldn’t decide if I should put this in the calculus or general math forums but... I’m studying for a final that’s coming up this Wednesday and I’ve been looking at some past quizzes with the steps to finding the solutions that my instructor has posted online. Given the problem: 1. Compute...

I understand that now. Am I correct in assuming that the magnitude will be the same and that only change will be direction?

Looking in my book it says The rate of change of momentum of an object is equal to the net force applied to it. My understanding was that momentum didn't change but from your question I'm now assuming momentum is a vector not a quantity... which would make sense because it is a force...

I'm a little confused about how you're using the formulas but you came out with the right answer. I'll show you how I did it. Our formula for height is going to be: Y(t) = -5t^2 + 4.9t + 0 I rounded gravity (9.8) to 10 and 1/2 of that is 5. I usually deal with gravity as being a negative...

One of those answers is what I got. Show me how you're setting up your formulas.

I think something is wrong with your first answer. How are you setting up your formulas? I got 7 m/s for the horizontal component though.

You're going to use the same equations as the last question you asked. Remember: V(t) = At + Vo X(t) = (1/2)At^2 + Vot + Xo Vo = Initial Velocity Xo = Initial Position These are good for any axis. You should try going through the problem real quick and looking at how your past questions...

I could help you out with 2-D probably but I haven't delt with 3-D yet. I actually go to DigiPen Institute of Technology and I'm learning game programming. I'm assuming everything would be the same except you would just add another axis into the equation but I'm unsure of how to find the...

What forces are involved? Gravity? Air resistance? Are we in space? Will Gravitation be invovled? Is this going to be in 3-D? 2-D?

Hi, My class has just started going into learning about momentum and I’ve found that I’ve been having quite a time putting all the formulas together to figure out homework problems… in fact I haven’t been able to answer one correctly yet. My first problem is: “A 4200-kg rocket is...