Recent content by Kant Destroyer

Homework Statement Homework Equations V = IR P = IV The Attempt at a Solution The attempt is much too long to type out, so here is a link to a picture of my work. I'm mostly concerned about part C, because I've seen some similar problems and my answer has been different by a magnitude of 10...

Thank you very much for taking the time to help me.

so cos2(θ)sin2(θ) = 1/4sin2(2θ) ?

Ifinal/cos2(θ)sin2(θ) = I0 EDIT: Misread your response. Ifinal = I0cos2(θ)sin2(θ)

I'm not even sure how sin(2θ) plays into the mix in this problem seeing as they are both cos2 functions, but no I do not know those. If I remember correctly the cos(90-θ) is actually just sin(θ) but I'm not sure about that.

The only trig formulas I remember are sin2+cos2 = 1 and cos2(θ) = 1/2 + 1/2cos(2θ).

I think you misunderstand my original post because I failed to use another variable for the angle between the two polarizing sheets and instead used (90-θ). Theta is meant to represent the angle of orientation of the first polarizing sheet, and per the relevant equation it is used to find I1.

As described in the question, theta is the angle of orientation of the first sheet relative to the polarization direction of the light entering the sheet. Per the relevant equation, this means that theta is used to find the intensity of the light after it has traveled through the first...

The relationship between I0 and I1 is: I1 = I0cos2(θ). Are you suggesting that I substitute this in for I1 in my equation and then solve for θ? I have tried this and I am not sure how to solve for theta in this instance.

The intensity of the light that is between the first and second polarizing sheets can be seen in my original post as I1.

Homework Statement A beam of polarized light of intensity 43.0 W/m2 is sent through a system of two polarizing sheets. Relative to the polarization direction of that incident light, the polarizing directions of the sheets are at angles θ for the first sheet and 90 degrees for the second...

I think I'm just confused about the angles. I'm not sure how to set up the right triangle and solve for it's different vector parts with this axis system because I can't seem to figure out what the triangle would look like. Is it a 30-60-90 triangle with the long leg being the y component and...

Thank you. I'm working it out right now. It's a 45-45-90 triangle, so would the Tension be equal to the normal force in magnitude?

I'm not sure exactly what you mean. I had a feeling that I was doing it wrong because of the different orientation of x and y, but I just went with what I'm used to. Are you suggesting that I just set N equal to the y component of the weight?

1. Homework Statement [/b] A block with mass M = 5kg sits at rest on a frictionless incline. The mass is connected to the wall by a string with linear density μ = 5.0 g/m. The incline is fristionless, with angle Θ = 30°. Let the positive x-direction point up along the incline, and let the...