Recent content by exi

Done. What better way to spend $1.67 a month?

Will probably use this to subscribe; used to read that mag all the time years ago. Great to hear.

Thanks for taking a peek - turned out to be correct.

Homework Statement In this drawing: http://img245.imageshack.us/img245/6559/physgv8.png Index of refraction for glass: 1.52 Index of refraction for surrounding carbon disulfide: 1.63 Incident angle at point A: 43.0° At what angle does the ray leave the glass at point B? Homework Equations...

Thanks for the explanation. :smile:

I thought so too - but it was incorrect and my last submission on that question. Key won't be available until tomorrow am. Wonder where I went wrong...

If I do what I think is the RHR for this question (with my thumb pointing upwards, towards the ceiling, for the particle moving along the z-axis, and my fingers extending in the direction of the x-y plane magnetic field at 39.4007° to the right), and if the force is in the direction indicated by...

Thanks, berkeman/mps. Makes sense now. o:)

Oh. :biggrin:

Ahh, without keeping your thumb at your index finger's side and just left out naturally? Pointing to the right.

If I do that, my thumb's pointing upward, and the particle's still apparently curving perpendicular to that, so...

Really not sure what I'm not understanding here. The Lorentz article does make mention of the right hand rule, which - if I'm doing it correctly - tells me that the net force on a positively-charged particle would be directed straight downwards. Is there any significance to this? edit: WP says...

No problem - and sure, it happens :smile:

Whoa, for the first part? No, I did it the good ol' fashioned way, with |B| = \sqrt{0.056^2 + 0.046^2}. And yeah, that does look familiar, albeit in already-crossed F_B = qvBsin\theta form. I think I'm having a bit of a hard time with the angle bit, since I have an angle from the x- and y-...

I think that's what he's got. Like this: a = \frac{(v_f - v_{i})}{t} at = v_f - v_i v_f = v_i + at