Recent content by Jaccobtw

What if I phrase it like this: We want to find the disturbance at ##(x, f(x))## at time ##t## when the wave travels at velocity ##v##. Subtraction shifts the graph to the right. ##x - vt## is smaller than ##x## because it gives the coordinate where we want to find the disturbance in the moving...

Thank you, I think I’ve got it down. (x, f(x)) is where we want to find the disturbance. x-vt is the coordinate in the moving reference frame. It says, what is the disturbance at (x, f(x)) when the wave has moved a distance vt.

f(x) = 10 for t = 0 For t = 1, and v = 1 10-(1)(1) =9 We began at 10 and now we’re at 9, so hasn’t the wave moved to the left?

The only way I can see x-vt being a rightward moving wave is if x-vt = some position the wave had initially. As t increases, x-vt gets smaller, despite the fact that it is a rightward moving wave. For example, if x = 10 m and v = 1 m/s, as t increases, x-vt describes a position the wave had in...

So after Newtonian mechanics and electricity/magnetism, what do physics majors study? Also, is there a textbook you can reference me to that applies to that?

1.) So first I differentiate and set it equal to 0 and get: $$\frac{A}{r^2} -\frac{Bn}{r^{n-1}} = 0$$ 2.) When solving for r, I'm not quite sure how to take away the exponent so I get up to the second to last step: $$r^{n-3} = \frac{Bn}{A}$$ Would it be: $$r = \sqrt[n-3]{\frac{Bn}{A}}$$ ...

Thanks for showing me a much simpler way. I think it's because they would cancel each other out. For example, taking the integral you'd end up with some constants times ##z^4 - z^4## which would give you zero. I think.

Hi I wanted to revisit this and finish the problem to see if my reasoning skills have improved since. So to start I think it's important to define ##q_{enc}##. We are given ##\rho## as a function of z. ##\rho## has the constant in it with units of C/m^5 with z^2 over it which has two...

Thank you. I got ##\frac{2 E_o \epsilon_o}{r}##

I know we're supposed to attempt a solution but I'm honestly super confused here. I think the second an third terms of the del equation can be cancelled out because there is only an E field in the r hat direction, so no e field in the theta and phi directions. That leaves us with ##\nabla \cdot...

Thank you @kuruman and @cnh1995 I finally figured it out

You're right. I explained that poorly. The gain is the voltage across the resistor over the total voltage which is 0.5 $$\frac{V_R}{V_t} = 0.5$$ $$V_R = iR$$ $$R = 8$$ Rearranging gives, $$\frac{V_t}{i} = \frac{8}{0.5} = 16$$

My reasoning is that because half of the voltage goes across the resistor which is 8 ohms the other half must go across the inductor and capacitor making the total impedance 16. The current is constant throughout an entire series circuit

I’d imagine at high frequencies, X_L - X_C is true and X_C -X_L at low frequencies is true

Not sure what I'm doing wrong. When I set up a system of equations for L and C, I don't see how having two frequencies makes a difference. Am I on the right track at least or no? $$\sqrt{8^{2} + (X_L - X_C)^{2}} = 16$$ $$X_L - X_C = \sqrt{192}$$ $$\omega L - \frac{1}{\omega C} = \sqrt{192}$$...