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That's one of my favorite problems. I think the answer is about 24,000mi. Don't forget to subtract the radius of the earth. I always forget to do that!

Me and my fellow class mates were able to find 32 as well for a. Nice work! You're a janitor that does physics in his spare time? Are you like Will Hunting? :D

Okay so I also missed: 1=3e, 2=2e, 3=e, 4=e Did we get them all this time? :cry: :rofl:

Yeah you got me there. I should specified identical particles. So say all are electrons. I ask because Prof. Webb specified "indistinguishable Fermions". So that would indicate were not talking about electrons and protons, in which case the Pauli Exclusion Principle doesn't apply, Protons...

Ahh I missed: 1=e, 2=e, 3=e, 4=3e So corrected answers: a) omega = 16 (I permute here by multiplying by 4 right?) b) omega = 4 c) omega = 1 Thanks.

Problem: Consider a non-interacting system of 4 particles with each particle having single-particle states with energies equal to 0, e, 2e and 3e. Given that the total energy of the system is 6e, find the number of microstates of the system (and identify the microstates) if the particles are...

Yi(x,t) = A*Sin(kix - {&omega}it), where i = 1, 2, 3 Phase velocity for two wave can be given by v = (&omega + [&omega]')/(k + k') and group velocity u = (&Omega - &omega')/(k - k') but what about three waves? I'm not sure what to do for three waves. I've looked all over my text and...

Navigation is one use off the top of my head. Our calander is based on astronomical patterns (loosely). Future uses. The cosmos is a place where we can conduct observational expirements to learn about physics that we can't replicate here on earth, black holes for example. Einstein's...

A surface current equal to Js is flowing on the surface of a perfect conductor in the x-z-plane traveling in the positive x direction. At a distance y = L along the y-axis lies the central axis of a cylindrical conductor with radius “a” and having a volumetric current distribution Jv= Jo*r*ex...

Thanks.

In my EM textbook it uses a equal sign with a triangle on top to define the intrinsic impedence of a lossless medium. I think it means definition but I'm not certain. The text is 2004 edition of author Ulaby titled Fundametals of Applied Electromagnetics, pg. 265. Thanks

Nevermind I fiquered it out. Not sure why it's right, but I discovered how to find the correct answer using, a different system of matrices, the correct system. I didn't need to include a diffraction for the scratched surface (apparently). If someone could explain to me why not, I would be...

Here is the problem: A glass sphere with a diameter of 5cm has a scratch on its surface. When the scratch is viewed through the glass from a position directly opposite, where is the virtual image of the scratch, and its magnification? The glass has an index of refraction n=1.50. Explain the...

I just found this: N-1 exists only if: det(NN-1 != 0 I'm a little rusty on my linear algebra, plus I got a concusion yesterday.

N= [i,1;-1,i] I used this theorem: N N-1 = In Thus: [i,1;-1,i]*[a,b:c,d]=[1,1;1,1] I then found: ia+c=1 ib+d=1 -a+ic=1 -b+id=1 Can I conclude an inverse does not exist. If so, how? If not, what do I do? Thanks, Frank

The a in your equation (the acceleration of gravity at the Earth's surface) should be negative, if you want the direction down towards the center of the Earth to be negative. This should fix the sign to be vo = 9.8 m/s. Note: When doing problems draw a picture and label it with a frame, i.e...

Is this how it's expressed: ±Cos{Θk} = e ± (Θk) ± Sin{Θk}, where k=1,2,3. Thanks.

I actually trired that method, but was unsure if it was correct since the terms were so messy. I'll continue with that method and post my solution. Thanks.

The problem states: Express Cos( Θ1 + Θ2 + Θ3) in terms of Sin(Θk) and Cos(Θk), k = 1, 2, 3, using the relation e+/-i*Θ = Cos(Θ) +/- i*Sin(Θ). [Hint: Use the product property of the exponential e.g., e(Θ1 + Θ2) =...

Interesting idea. I imagine the only thing you could do with the heat is heat water. Some how incorporate that heated water into a system with the water heater. Surprisingly that sounds like a practical idea, could save quite a bit of energy over a few months of use during the summer. But if...

I was bored earlier today, so I concluded the best way to elevate my boredom was to exercise my brain for next quarter. The best way to exercise your brain is to do a little physics, right? I started to think of a problem. I came up with a classic projectile problem: A cannon fires a...

I'll have to think about it further... Why does the B-field need to change in the z direction to impart a deflection on the Ag atoms? I haven't quite grasped that concept, maybe my B-field knowledge is a little rusty?

If Fz changes according to &muzdB/dz and the silver atoms are collimated into a beam how does the dipole magnetic moment of the atoms see and change in the B-field if the beam is cutting across the B-field perpendicularly? How does a changing B-field, in the verticle direction affect the Ag...

I get: t = 1.62 x 1012s Is this correct? Thanks Edit: Which is 51374 years. I just looked up how old C-14 dating is good too. It said 50,000 years. So it looks like I'm right. This problem was easy, I don't know why I struggled with it so much!

That's an interesting way to look at the problem. I'll see what I can do with that. Thanks

http://www.phy6.org/stargaze/Sfall.htm

Libby's observation that all the carbon in the world's living cycle is kept uniformly radioactive through the production of C-14 by cosmic radiation led to his development of the radioactive carbon dating method. Samples of carbon in the life cycle have been found to have a disintegration rate...

I get the K of electron after collision: 2/3mc^2 and for initial photon energy of: 4mc^2 If I didn't get those right, I'm quiting for the day.

Are you sure v isn't 8/9c, not .8c.

Yeah I found my error. Reworking it right now. I canceled &gamma, which you can't do. Since E of initial electron is only mc2

I like your way much better. I ended up with a &gamma2/[(&gamma +1)(&gamma-1)] = 4/9mc2, solving for v would have been fun. I bet I made some mistakes along the way, two pages of algebra and I write small. Plus that I tried to solve for K directly, then subsituted that and tried to go for v...

Is the algebra soposed to be really messy on this problem.[?] I just want to make sure I'm not going through all this for nothing.

Is this okay? Ephoton + moc2= 3moc2 + 3Ke Ephoton = energy of the photon: hc/&lambda moc2 = rest mass energy of electron Ke = kinetic energy of the electrons in motion (all three with equal linear momentum and kinetic energy) after the photon collides with the initially motionless electron.

Problem: A photon of energy E strikes an electron at rest and undergoes pair production, producing a positron and another electron. photon + e- --------> e+ + e- + e- The two electrons and the positron move off with identical linear momentum in the direction of the initial photon. All...

I worked out all of the kinks. I was wrong. I found: 1 = L/&pi A2[1/2&pi] The integral was more challenging than I thought. Thanks for your help. I would of never thought of the u substitution, I haven't done that sort of thing since Calculus II.

See my original post: I'm [zz)]

Yeah, that's what my fancy calculator told me, so what the heck did I do wrong?!

You're not understading: Let me give you all my work to alleaviate any confusion. Show that A = (2/L)1/2 &psi(x) = A Sin(&pi x/L) &psi2(x) = A2 Sin2(&pi x/L) [inte]0L &psi2dx = 1 A2[inte]0L Sin2(&pi x/L) dx = 1 Actually... I forgot to resubsitute... BTW: I only use a...

Here is the problem: A diatomic gas molecule consists of two atoms of mass m separated by a fixed distance d rotating about an axis as shown. Assuming that its angular momentum is quantized just as in the Bohr atom, determine a) the quantized angular speed, b) the quantized rotational energy...

Oh BTW: Thanks for your help Tom.

HAHHAHA! That's what happens when you do physics for 10 hours straight. Edit: BTW, I've never seen that type of unit used before so my brain must have dismissed it.:smile:

I just plugged it into my calculator. Since a=5.29x10^-11m the answer should be very large. This must not be the correct formula or something.

For some reason the answer I'm getting is no where close to the given answer: 2150(nm)-3 I get: 6.121e^9

So I want this: Ψ(r) = 1/(sqrt[[pi]a3/2) e-r/a Square that and evaulate, or can I just evaluate and squre?

Halliday gives: |Ψ|2(x) = A2Sin2(n[pi]/L * x), n = 1, 2, 3,... No radial coordinate.

Okay, what do I use for A, n, L and x? If I evaluate x=0 I get 0. But the answer is non-zero.

For the ground state of the hydrogen atom, evaluate the probabilty density psi^2(r) and the radial probability density of P(r) for the positions. a) r = 0 b) r = rb I confused how this probability function is used. What's the technique here? Thanks

Tom: You're correct, I got 1995MeV.:smile:

Calculate the smallest allowed energy of an electron were trapped inside an atomic nucleus of diameter 1.4x10^-14 meters. Compare this number with the several MeV of energy binding protons and neutrons inside the nucleus. On this basis, should we expect to find electrons within the nucleus...

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