Recent content by CFede

The equation you have there (xf=xi+vi t+0.5 ai t^2) is only true for constant acceleration. You should workout the instantaneous velocity by integrating the acceleration with respect to time. THen do the same thing to get the position.

As for (1) and (3) I'm not sure where your questions points at. As for your second question, a weightless balloon filled with water is just water being limited to a confined space so, asking how much the balloon weights in water is equivalent to asking how much does water weights in water. I...

Yes to both, you are headed in the right direction

What you say is usually ok, but not for this paticular problem. All the current will flow only trough the resistor at the top of the circuit. As for the bottom part, all the current will flow trough the resistance-free cable.

There is a little trick here. Actually the circuit is quite simple (in fact, as simple as a circuit can be). Hint: If you have two paths in a circuit and one of them has o resistance, all charges will flow trough that path.

Yes, in that equation, the v is the inicial velocity of the body, usually named v0.

There are only 2 things you don't know here, the acceleration and the final speed. All other things you do know. As the problem says the body starts moving from rest, so there's v0. Aside from that, you now it travels a distance of .25 miles, which all you need to now, simply define x0=0 which...

You have a problem with the units of measure. Thinking only of units what you have is \left[\frac{n^2\pi^2\hbar^2}{2mL^2}\right]=\frac{(eV)^2s^2}{kg\cdot m^2}=\frac{eV^2}{J} which is obvioulsy not in eV. You should work with h bar in joules and, at the end of your calculation, go to eV.

That's true in the region between the infinite wall and the delta potential, however, for x>0 you can assume that there are no waves propagating from the right since there's no potential to the right of the delta function were they could be reflected.

There is quite a common trick for this kinds of things, you can always express the dot product of two angular momentums as J_1\cdot\J_2=\frac{1}{2}((J_1+J_2)^2-J_1^2-J_2^2) The operators on the right hand side are easy to handle. Hope this helps.

You should use chain rule I think, so \frac{\partial }{\partial x}=\frac{\partial u}{\partial x}\frac{\partial}{\partial u} If I understood your question, this is what you are looking for.

What did you do to determine the constants in the case where its open on either side? I'm guessing you can follow a similar procedure here for the open side. That plus normalization should be enough.

The probability is 2/3. There is no need to square anything if your performed the braket.

It's correct. You are not assuming that every state is equally likely, you know it from the fact that all are multiplied by the same constant (in this case 1) in the non normalized wave function.

This has no problem with the fact that acceleration is the derivative of the velocity. As it turns out, the velocity of a free falling body is v(t)=g*t, then: a=dv/dt=g which meas the acceleration is contant and equal to g. On the other hand, if you evaluate the velocity a t=0, then...