Unintuituve naming convention: "para" denotes antiparallell?
I did a project a week ago that required studies of the helium atom. Depending on if the two electrons spin are parallell or antiparallell it is referred to as orthohelium (parallell) or parahelium (antiparallell).
Intuitively I...
In a way alxm is describing the flux pinning effect. I don't intuitively see it as an attractive force, more as just a local potential energy minimum for the system when the field lines line up. A small movement by either object would make the field lines want to bend close to the the boundaries...
The first and second bullet point follow from the bottom of page 58. Study the expression for \epsilon_f and see how it depends on m and N / V.
The third bullet point follow from the definition of k_B in a way. k_B relates energy to temperature, so the Fermi temperature associated with the...
Looking for info about "cluster orbitals"
Homework Statement
Why is the orbital 1d present in eg an Al2cluster but not in a Hatom?
Homework Equations

The Attempt at a Solution
This is a question for an assignment. The answer is probably not meant to be very elaborate, "Because...
When you use symmetry in an integral so that you don't have to integrate both parts you multiply the integral with 2 and only integrate one part since the other part will be equal to that one.
You should put the _middle_ of the rod at origo and let the rod lie along the xaxis. Draw the...
Maybe they are searching for the term used to describe satellites that are kept stationary above the Earth's surface. There is a special word for this. It is a bit of a strange question if they mean it like that though, but that would be my guess.
http://en.wikipedia.org/wiki/Adiabatic_process#Ideal_gas_.28reversible_case_only.29  The coefficient is denoted \gamma in the Wikipedia article, as is usually the case I believe.
You will have to motivate your answers more so we know why you came to them. If you just throw out an answer we can say "Wrong, try again" and nothing will be learnt.
2) is very simliar to 1), look at the motivations for the integral in 1) so you're really sure about why it looks exactly like...
Centripetal force doesn't have to be from gravity, it is just the case eg between the Earth and the sun. You have answered this in 1).
The image of a circulating ball in a string is a good one. Think of the ball  what force acts on the ball? Which direction has this force? Think of F = ma.
You are integrating over r, but at the same time you set r to a fixed value \sqrt{L^2 + a^2}. It has to be a variable if you're going to integrate over it if you want some meaning from it in this case.
And you don't have a term that describes the vertical component of the gravitational force...
I used the Minkowski diagrams incorrectly. The projection of l_2 onto l_1 wasn't perpendicular but with an angle corresponding to the timeaxis for l_2, not included in my image above. At first I thought this would lead to a contradiction when I tried to project l_2 onto x', but that was because...
No. The boundaries are wrong (the rod is of length 2L), you don't define which variable you are integrating over, what is your \theta and why do you use sine on it? The expression for r has to be more elaborate, and it should be squared.
M/2L is the mass per unit length, that is correct...
You will have to know the distance traveled by the bike to know the amount of work done by friction. Find that out from your picture. After that use that energy is conservated as suggested. Be wary of how you define your friction. Look at the dimensions of your terms, they should always...
I would abandon the circular coordinates due to the geometry of the problem. This problem requires a bit more difficult integration and even though the answer isn't especially complex it seems like more difficult work than is intended, maybe it's meant to be done another way, but what do I know...
Ok, I assumed you had the right answer since you said in your first post that the answer you got was wrong.
Read the Mathworld page again, you don't have a half disc in your problem, you have a semicircle. The centre of mass isn't the same.
I am not sure my integral is correct since I don't know the answer to the question, but I know the procedure I am describing is a working one. Was some time since I calculated these kind of problems though.
4r/3\pi isn't what you want, that is center of mass for a half disc.
Show the...
The radius is constant, it is the arc described by \phi I am integrating over. I add up every infinitesimal contribution to the force from every point of the arc, and I disregard all the contributions that cancel anyway because of the symmetry. I have pictured me a semicircle in the upper two...
Your problem text said that the angle between incoming and outgoing ray was 25°, but if you've deduced the relation 2dsinθ = nλ you should know that θ is defined as the angle against a perpendicular line against the surface, compare with the law of reflection.
1) I guess the integral would be something like
2 \int_0^{\pi/2} r d \phi \sin \phi G \frac{M}{L} \frac{m}{r^2} where r is the radius you had calculated, \phi is the integrating angle. \frac{M}{L} is the mass per length unit. At least it gives a simple expression, I have no means of checking...
1) Note the symmetry in the setup. The force must be directed towards the center of the wire, let's denote it A. All differentials of mass in the wire will exert gravitational force that can be split into one component of force directed towards A and one component that will cancel the component...
If i understand your question right, I would say the answer is yes. Do you mean that a beam would take the same path but in reverse direction if it came from the other "side"? The trigonometry makes the problems symmetric and thus the answer yes.
There are breaks in symmetry concerning the...
Well, it differs in time, but it is given linearly according to the problem description, starting at 0 going up to the maximum value after 8 s. From 2a) you can express energy given per time unit which should give you energy given after 4 s.
Yes, from Newton's second law the impulse equals the change in momentum as you write which gives the answer you provide.
Your thoughts on the indifference of the angle it is correct, you can think of it as that the impulse has "nowhere else to go" but the foot.
1a) F = ma gives you the acceleration, use it to find v(t).
1b) With 1a) you can get an expression for the work performed by the force on the object. Relate the work to the power and express p(t).
1c) Use 1b).
2a) You can express the cars kinetic energy at 100 km/h. This energy has been...
The θ used in your calculation is wrong. Check how it is defined in the 2dsinθ = nλ relation.
E = hc/λ is correct for a photon, but this is a particle with mass and the c has to be replaced with the neutrons velocity. Use the de Broglierelation λ = h/p and you should be able to get an...
Homework Statement
This problem is from "Relativity" by Rindler, second edition, problem 3.4:
Use a Minkowski diagram to establish the following result: Given two rods of equal length l_1 and l_2 (l_2 < l_1), moving along a common line with relative velocity v, there exists a unique inertial...