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1. ### Energy Needed to Ionize a H-atom

Nevermind. After consulting my excellent, omniscient professor, I learned I have the right set up and he explained the electron volt stuff. Thanks!
2. ### Energy Needed to Ionize a H-atom

Homework Statement A H-atom when in its lowest energy state consiste of a proton nucelus of charge +e and an electron of charge -e and mass 9.11e-31. In the Bohr model of the atom, the electron moves around the nucleus in an aprrox. circular orbit of radius .51e-10m. The speed of the...

You wouldn't happen to know a dude with a yo-yo, would you? Because if you did, then yes, same class. Definitely the same book. That, I chalked up to book-error. I couldn't get any other answer and the book is known to be wrong occasionally.

Due to your confusion, I double checked and realized that I didn't need to use the v=omega*L constraint.

I got it I got it I got it. This unit check is going to suck big time, but I got the numbers. Pardon me while I pass out from mental exhaustion.

I replaced omega with V/L

Sorry, Latex was giving me some trouble when the equations started getting complicated and I had to clean it up.

Okay, here goes: m'vL=I\omega, I= \frac{m}{3}L^{2}+(m'+M)L^{2} m'vL=(\frac{m}{3}+m'+M)L^{2}\omega m'vL=(\frac{m}{3}+m'+M)L^{2}(\frac{V}{L}) m'vL= (\frac{m}{3}+m'+M)LV V=m'vL/((m/3)+m'+M)L .5((m/3)+m'+M)[/tex](L squared)(V/L) = mg(.5L)(1-cos(theta)) + (m' +M)gL(1-cos(theta))...

well, r = L = 1m, so none of the values would change, so the erroneously way I calculated it before would have yielded the right answer. But that's obviously not right.

It's defined L=Iw, but the bullet is traveling in a straight line (I'm so ignoring projectile trajectory when I say that) so it doesn't really have a rotational axis. It does once it hits the block, but then that wouldn't be pre-collision, so I don't see how it can. I can completely see how...

Okay, I've seen several solutions to similar problems where they set it up mvL=Iw...where does the L on the left side come from?

Angular velocity would be conserved?

Btw, thank you!

Okay, I double checked it solving for g instead of the height (I habitually check the constants) and it came out as 9.80002. So, either my book is scary wrong or I just set the problem up wrong to begin with...any brainwaves?

Yes, it is. The squared latex was messing with my square-root latex and it just got messed up. Hmm...if you're doing the delta-heights, wouldn't the *delta-height* be the same for both the masses and the rod? I can see the height measurements themselves being different, but in that case...

Why would I need to calculate center of mass when i'm completely bypassing the moment of inertia?

Homework Statement A 2.0kg (M) block hangs from the bottom of a 1kg (m), 1m (L) rod. The block and rod form a pendulum that swings on a frictionless pivot at the top end of the rod. A .01kg bullet (m') is fired into the block, where it sticks, causing the pendulum to swing out to a 30 degree...

\alpha=2(50Nm)/(250kg*.75m^{2})=.711rad/s^{2}...I guess parentheses might help. I'm sorry. Does this help?

If I was using the wrong Beta, then my answers for A, B, and C would be incorrect. However, they aren't, so I'm assuming Beta = 1/2 is correct....? Where does that leave D?

Calculations as requested: A) T(motor)=I*alpha. I-Beta*Mass*Radius squared, Beta =1/2. Alpha=2* Torque(motor)/Mass*radius squared= .711 rad/seconds squared. Omega final = alpha *time. omega final = 1200rpm * 1min/60sec * 2pi rad/1 rev=125.664 rad/s. t = omega/alpha= 176.743 B) KE(r)=...