Recent content by rissa_rue13

Just one more quick thing, I'm not sure how I would factor L out of the left hand side without cancelling out the L on the right hand side.

Z2=R2+(wL-(1/wC))2 Z2-R2=(wL-1/wC)2 sqrt(Z2-R2)=wL-1/wC sqrt(Z2-R2) + 1/wC = wL [sqrt(Z2-R2) + 1/wC] /w = L substituting in C: [sqrt(Z2-R2) + 1/w(1/w02L)]/w = L I think this would simplify to: [sqrt(Z2-R2 + w02L/w]/w = L

Okay, so making that adjustment, it becomes: (sqrt Z2+R2) (1/L(2pi*f0)2) = L To factor out: sqrt Z2+R2 = (2pi*f0)2L2 Then: (sqrt Z2+R2)/(2pi*f0)2 = L2 And then take the square root of the left side to get: 7.58 x 10-4 H = L Am I supposed to be using F0 or Fv for this equation?

So what I've done, and I have no idea if this is correct: f0=1/(2pi x sqrt(LC)) C = sqrt(1/f02piL Z = sqrt(R2 + (XL-XC)2) simplifies to: sqrt(Z2 + R2) x C = L substituting C: [sqrt (Z2 + R2) / f0 x 2piL] = L I ended up getting 0.147 H = L Is that heading in the right direction?

I'm sorry, but what you just wrote makes absolutely no sense to me. I'm not trying to find impedence and I don't understand how the inductive reactance= -capacitive reactance reveals the value of a resistor.

Since Z=R, then XL=XC Using an equation I found in the book, this would mean that: 2pif0L=1/(2pif0C) This gets me back to the equation: f0=1/(2pi x sqrt(LC))

Homework Statement A series RLC circuit has a resonant frequency of 6.00 kHz and a resistance of 575 ohms. When connected to an AC power supply that varies the voltage at 8.00 kHz, its impedence is 1.00 kOhms. What are the capacitance and inductance of the circuit? Homework Equations...

I think I made a mistake, I think it should be: Vnet= (kQ/(L/2)) + (k2Q/(L/2)) - (k4Q/(L/cos 30)) This can then be simplified: Vnet=kQ[(2/L)+(4/L)-(4cos30/L)] Am I on the right track?

Homework Statement There is an equilateral triangle with one point charge at each vertex. The point charges have charges of +Q, +2Q, and -4Q. The length of one side of the triangle is L. Determine an expression in simplest form for the voltage at a point halfway between the +Q and +2Q point...

So fa/fb should still equal 2, right? It was just the other part of the equation that was wrong? 2 = (v+vs)/(v-vs) 2v-2vs=v+vs v=3vs v/3=vs assuming v=343 m/s (speed of sound in air): vs= 114 m/s Therefore, vs/v = (114 m/s)/(343 m/s) = 0.333 right?

The only number I'm given is that wa=1/2wb. So would I substitute that in by doing: (2v/w)/(v/w) which would reduce to 2. And then put that in to: 2=[(v-vs)/(v+vs)] ? Wouldn't that make v=vs ?

So, fa/fb=[fs(v/v+vs)]/[fs(v/v-vs)] right? Then, wouldn't that reduce to: fa/fb=[v(v-vs)]/[v(v+vs)] ?

fa=fs[v/(v+vs)] -> v/(1/2w)=fs[v/(v+vs)] fb=fs[v/(v-vs)] -> v/w=fs[v/(v+vs)]

fa=v(1/2w) fb=vw so wouldn't that make fs different for Abigail and Bertha?

But fa and fb are different because the wavelength Abigail observes is 1/2 that of the wavelength Bertha observes.