Recent content by Munky

As my daughter is fond of saying... "Waaaahooo!" Thanks for your time. Now, according to Murphy, since I spent so much time on that, I won't need to know it tomorrow... lol.

I'm getting ready to crunch the numbers now, but I think I've found it... I knew it was some trig thing... it's literally been almost twenty years since I've done trig... lol. What I found is that in a right triangle, the sin of an angle is equal to the cosine of it's compliment. So it...

here's the image. all i can gather from the right angle formed by the norms is that the internal angles also add up to 90deg. The only way the external angles relate to the internal angles, though, is by n... which i don't know...

Snell's Law -- figuring out index of refraction Homework Statement A light ray enters a rectangular block of plastic at an angle of θ1 = 45.0° and emerges at an angle of θ2 = 79.0°, as shown in Figure P35.71. (the image is of light entering the block from the west, and exiting from the...

Thanks for the responses. I can see where the formula came from now (physics one was so long ago!), but the 4/3 is buggin me. The B field is perpendicular to the rod and the rails, pointed down. Would the fact that the rod is rolling as opposed to sliding have anything to do with it...

I am studying for an upcoming test and, while looking through my notes, came upon this formula: v= sqrt((4/3)(IAB/M)) It solved my problem, but I don't know where I got it. The problem is accelerating a rod with a current through it on rails across a magnetic field. Can someone...

Thanks for the response, cryptic as it seemed at first, it helped. I don't know if I was clear, but I was supposed to solve for v and r, and wasn't sure how momentum fit in. Your response was enough to make me look a little further, knowing that I had what I needed. I found that angular...

Homework Statement An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 1 mT. The angular momentum of the electron about the center of the circle is 4.00E-25 kg m^2/s. m= 9.109E-31 kg q= 1.602E-19 C B= 0.001 T I also know that the answer...