Recent content by Coop

Thanks for following up :). Yes, I was confused at where the integral c=∫vf(v)dv was coming from (I don't like using random equations which I don't know where they came from) but I figured out it was pretty much just a definition.

Yes sorry, I see that my wording was ambiguous. I am having trouble seeing why integrating v*f(v) gives you an average.

Hi, I am having trouble with this concept... "A mean speed (c) is calculated by multiplying each speed by the fraction of molecules that have that speed, and then adding all the products together. When the speed varies over a continuous range, the sum is replaced by an integral. To employ...

Thanks for both your help, guys. P.S. I found a free download for Maple from my university.

Ray, I do! Thanks :) Is Maple a free program I can use to check my answers?

Homework Statement Find (\frac{dV}{dp})_{n,T} for the Van de Waals gas law Homework Equations Van de Waals gas law: (\frac{p+an^2}{V^2})(V-nb)=nRT The Attempt at a Solution I just started doing problems like these so I would like to know if I am doing them right... What I did was I took...

So only the parallel component is responsible for doing the work, correct?

@Nathanael Ah, I think I was just making a stupid mistake. I was letting F = the force of gravity on the weight. Thanks for pointing that out.

Homework Statement A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin60 = 0.87; cos60 = 0.5. Ignore friction and the weights of the pulleys.) A) 50 J B) 100 J C) 174 J D) 200 J...

Homework Statement Two adjacent allowed energies of an electron in a one-dimensional box are 2.0 eV and 4.5 eV. What is the length of the box? Homework Equations E_n=\frac{h^2n^2}{8mL^2} The Attempt at a Solution My question is, since E_n and n^2 are both on separate sides of the equation...

Thanks a lot everyone :) I appreciate the input.

Thanks guys, So is absorption actually taking place here or is it a different type of interaction? I am getting mixed answers.

Hi, My textbook describes Compton scattering the following way: "...an x-ray photon...has a collision with a slow moving electron...the photon transfers energy and momentum to the electron [and recoils]..." Is it not true that by definition when a photon collides with a particle it...

Thanks :)

So I can't use I^2R, because by definition the current would also be a different value for a new R? But we know the voltage will stay the same for a new R, so we can use V^2/R?