Recent content by Sarial

Homework Statement An object is placed in front of a concave mirror, 15.0 cm from the mirror's focal point. The image formed by the mirror is five times farther away from the focal point. Calculate the focal length of the mirror. There are actually two possible image distances di that...

Yes! Thank you. I knew it was something small like that. .267c was the right answer, and using gamma sorted it all out! Case closed :P

Ohhh. That could be it. I'll try that soon, thank you. You should use the relativistic formula for .10c+?

I believe the +/- discrepancy, but the .314 apparently isn't right either :\

I tried +.314 and still no luck. I must be making a tiny mistake somewhere, and I just can't find it.

Bump. Due at 10 PM EST tonight

Bump? :( I don't know where else to go with this

1. Astronomers on Earth observe a feature at 442 nm in the spectrum of a distant quasar. However, it is known that this feature corresponds to 581 nm in the rest frame of the quasar. Calculate the speed of the quasar. Enter your answer as a fraction of the speed of light, positive if the quasar...

Homework Statement Simple question, but no matter how much I think about it I can't see what I'm doing wrong. http://spock.physast.uga.edu/res/uga/PhysicsLib/Matter_and_Interactions/Ch10/figs/mom_and_angmom_stick.png Homework Equations N/A The Attempt at a Solution Basically you...

Alright. Thanks a lot.

Awesome! That worked, thanks a lot. My one question is--where did the Erecoil = Egamma^2/(2Mc^2) come from? As in, what equation is it derived from?

I tried something else that I think is along the right lines from what my teacher was saying, but he wasn't going too in depth at the time: My problem is 67.4 keV and a Nickel-61 nucleus Knucleus + Egamma = 67.4 keV pnucleus = -pgamma pnucleus + pgamma = 0 mv + E/c = 0 Knucleus...

I'm sorry. I did read your post properly. 93.3 keV isn't the energy of the photon though, it's the recoil energy of the nucleus and the gamma ray. Egamma + Knucleus = 93.3 keV Is that not right? Also, how do you get the 400 odd m/s? If 1/2Mv^2 = 93.3 keV 9.33e4 = .5(67(9.315e8))v^2...

I understand that the recoil energy of the nucleus is described by (1/2)mv^2, and that the momentum of the photon and the nucleus are equal, I think my problem is not knowing the energy of the photon.

Bump. I'm having trouble with the same problem. It would be easy if it gave the energy of the photon, but it gives the change of energy between the nucleus and the photon. My line of thinking: 0=mv+E/c for momentum (delta)E=(1/2)mv^2 + pc for the energy I can't see where to go from...