Recent content by xnick

@Ulagatin: You can't use just any kind of rubber chicken to answers questions in physics. It must be a http://www.miwiki.net/Rubber_chicken_with_a_pulley_in_the_middle". It's getting hot here. Why hasn't no one yet come up with a "microwave freezer"?

The real Evo imprinted his mind on the net a long time ago. So he/it still lurks the Internet in the future. Won't it bother the users the interplanetary lag in the comunications? Think about an earthling chatting with someone in Saturn.

I looks good to me!

Your equations are fine, but you have to prove an inequality. Have you noticed x^n > x^{n+1} ? (this doesn't hold if you let x have any value :wink:) If it works using that, it may be easier than using binomial coeff. There is a link above pointing to an explanation of the subject... in...

I should work using binomial coefficients. But I first would examine the relationship between x^n and x^{n+1} The fact that x belongs to (0,1) should matter somewhere... ;-) (it's easy to check with some examples that it doesn't work if that condition is not verified)

I stress that we can't predict the outcome of ONE measure. Examples: 1) A lottery game has 1000 numbers. I buy 500 numbers so i have a probability of winning p=0.5 . Some guy buys only one number, and so he has probability p=0.001, and I laugh at him saying "you're so screwed"... Yet he wins...

Some observations: 1) A high magnetic intensity is not the same as a magnetic bubble as defined before. 2) I'm not sure about the question of the collimation, but I think that you have to take into account the Electric and Magnetic field altogether (i.e. a light beam) in the case mentioned. 3)...

I think that there's only a mathematical significance to the question, which relates to what probability really is. If for a given measurement our knowledge is probabilistic, we can't predict what will occur, but we can predict the proportions of different results if we practice the measurement...

I think the description of the problem is correct. Most of the anwers were taking de Lorentz force for granted... \vec F = q (\vec E + \vec v \times \vec B ) . Your problem is that you are considering momentum to be just m \vec v . Remember that the EM field carries momentum as well... ;)

No can do using only magnetic dipoles. You have to have a magnetic monopole (yet to have discovered). Hint: apply Gauss's law to the bubble.

Same here :rolleyes: I smell a problem in setting up the original configuration... (i.e. related to the unbounded density of the progressive balls and... oh yeah! we need infinity balls! If they start packed together, how could we unpack them? If they start separated, how do we move them to...

That formula says that the force that counts for the pressure is only the normal (perpendicular) part of the vector, and that only its modulus if of interest. Side commentary, IMO, the given definition of pressure at a point in most textbooks lacks an explanation of why the orientation of the...

I guess the sign is conventional (although i don't think anyone's going to change it now) and also you can add an arbitrary constant to the potential energy and get the same physical results, since the quantities of interest are energy differences. But the force MUST be conservative.

In addition, the potential energy can only be defined for the (minus) work of a conservative force. The key is that the work depends ONLY on the starting and ending points, and not on the trajectory. Otherwise (for a non-conservative force) the path has to be specified, so it makes no sense to...

My native language is Spanish. I speak/read English, and read French and Portuguese.