Recent content by JoshuaR

Let's say you put a block on a flat scale. It measures weight of mg. You exert force exactly parallel to the scale, does the reading increase? How about it you put a second bloc on top of the first? Now the scale reads (M+m)g. How about if you slide one block but not the other, using only a...

Don't think in equations, think in terms of the picture you see. You grabbed the range equation, but don't just try to plug things into it. The rock strikes the ground at 45 degrees. That means it's final velocity has x and y components that are related to each other how? You know x velocity...

For part B: You know the meteor's initial KE. What is that of the earth's? You know mass of earth, and velocity of earth-meteor, so you should be able to find KE of earth. There you can find the fraction. An additional question: Does this answer make sense? If yes, why, if not, why? What type...

Where did you get these units? Is this a final answer of yours, a constant, or from something else?

Do you know any relativity? What might be useful about a particle accelerator, in regards to things like time dilation, length contraction, and how might that be more useful than a normal lab setting? That's just one idea. Good luck.

Right, borgwal, I think that was the problem. Thanks again!

I see. It all makes so much more sense now, lol. Just take away that darn factor... New question on this. I figured I should use minimum uncertainty for \delta x since we were confining it to a box of size 5E-15m. Do you suppose my problem here was assuming the box was \delta x? After all...

Gah! Where's the math mistake? -.511MeV + sqrt[.511^2 + (6.58E-22*3E8/(2*5E-15))^2)

Mmkay, I think I've tried that. I have \delta p\delta x = \hbar/2. Thus, when I use the KE form from before: KE = sqrt[(mc^2)^2 + (\hbar c/\delta x*2)^2] - mc^2 I must be doing incorrect math... sigh.

What have you tried thus far?

Not sure what your question is exactly. 75 feet? Doesn't your problem say 70m? BTW, have you learned energy yet? If so, try equating potential and kinetic energy to find some useful information. That will help you with some of the parts. For (c), total distance, if you know the other...

That was indeed my guess on the second approach. If I take just that: KE = sqrt[(mc^2)^2 + (pc)^2] - mc^2 , I still don't get there. Using the formula p = h/\lambda If I do as I did on the "second approach" then setting energies equal, I also have trouble. I'm a bit puzzled because I am sure...

Homework Statement Consider an electron confined in a region of nuclear dimensions (about 5fm). Find its minimum possible KE in MeV. Treat as one-dimensional. Use relativistic relation between E and p. Homework Equations KE = p2/(2a) = \hbar2/(2ma2) p = h/\lambda E = hf E2 = (mc2)2 +...

Um, I don't know if it's terrible to reply to a thread from so long ago, but I ran into the exact same problem and a different derivation (incorrect). Idea as follows, in the distance through the magnetic field, l, the electron moves a height h=ltan\theta. This same distance can be related...

Someone suggests that force P may be only applied to beam BEF, not on beam DCE. Would that make it work?