Recent content by Hypercubes

Yes, sorry, I was confused.

Hello, I'm trying to solve for the allowed energies with the WKB approximation of the Schrödinger equation, using the Morse potential. So I have (as per equation 35 at http://hitoshi.berkeley.edu/221a/WKB.pdf), \int_a^b \sqrt{2m(E-V(x))}dx=\left(n+\frac{1}{2}\right)\pi\hbar However, how do I...

Ah, I see. Thanks you both!

Well, the exact question was "explain why the equation e^{-x} = x has exactly one solution", and the answer goes on to define f(x) = x-e^{-x} and says "Since f'(x) = 1 + e^{-x}, and 1 + e^{-x} > 0 for all x, the function f is increasing..." So they set it to 0 to get the derivative. However...

Homework Statement Differentiate x=e^{-x} The attempt at a solution \ln{x}=\ln{e^{-x}} \ln{x}=-x \frac{d}{dx}\left(\ln{x}\right)=\frac{d}{dx}\left(-x\right) \frac{1}{x}=-1 The correct answer is 1+e^{-x}. I know how to solve it that way; however, why is the above method wrong? I know there are...

Oh, I see. That makes a lot of sense, thanks.

I don't think that's the only factor. What originally prompted this question was playing with a toy car. I held it by the wheel in the middle (making it the fulcrum), and it tipped towards the heavier side. Changing the weight on either side changed the balance accordingly. It did not go to a...

But this occurs on a stationary pivot as well. Try balancing a pencil on your finger where one side is heavier than the other; it reaches an equilibrium.

Ah, I understand now. I think my mistake was not considering the mass of the board for the first case, which is why the answer that it would go to 90° does not work in non-ideal real-life situations. Thank you very much, greatly appreciated.

Okay, I guess that wasn't the best example. What about this one?

A trivial example would be something like this: The force exerted as "effort" is less than that of the load, but it remains stable, although it is not at 90° (never mind that it is not possible here due to the ground).

Thank you for the answer, I will remember not to cancel haphazardly. However, I don't understand how this could be correct. This suggests that θ=90° no matter the balance of forces, but this is not the case; everyday experience tells us otherwise.

Homework Statement This is not a homework question, I was just curious. I'm trying to find the angle θ that an unbalanced arm produces in the following case: Homework Equations τ=rF The Attempt at a Solution τ1=τ2 rF=rF r(98 N)(cos θ)=r(49 N)(cos θ) However, the cos θ...

Thank you, I think I understand now.

Shouldn't there be no potential difference between point I and the negative terminal of the battery since there is nothing to provide resistance or potential? In a real circuit there would be some resistance in the wire. If there is no potential difference between these two points, how can...