Recent content by Stefan Udrea

phyzguy, I didn't know that pressure can decrease the radius of a an atomic orbital.I thought pressure can only make atoms come closer to each others, not shrink them.

I don't remember where I've first read it. I've found the statement here, and it seems to be referenced by a serious book : http://www.answers.com/topic/potassium-argon-dating#cite_note-McD-H_10-0 EDIT: I also found it more exactly phrased here ...

Hello. I've heard that electron capture rate, which is used in the dating with potassium-argon, can vary at ultra-high pressures.How is it possible that this nuclear phenomenon is influenced by pressure?

I've read an article written in the 60's by the russian physicist Pavel Kapitza where he was saying that one of the field where physics was less successful was in finding laws to predict the properties of the metalic alloys,and thus the compositions of various alloys is determined by the...

You gave an intuitive answer.I didn't mean it as a "yes/no" question , I wrote "Show".

This wasn't my idea ; I found the problem and the solution in a book.

:smile : don't call me "sir" , I'm just a college student. jdavel, In the drawing, AB is the tunnel , the two cities are in A and B. T is a random position of the train in the tunnel, at the distance r from the center of the Earth ; r<R Because the volume of the Earth between r and R...

extreme_machinations, Because you were the closest to find the solution, you get to ask the next question.

ok, extreme_machinations, could you give it a try with the assumption that the tunnel doesn't go through the center of the Earth ?

jdavel, this is too simplistic.Where do you draw the line of zero potential energy ?

:approve: Ok,here's the problem : There are 2 towns on the surface of the Earth, the distance between them is thousands of kilometers. There is a *straight* tunnnel digged from one town to the other,and a railway in the tunnel. Show that the train ,which circulates throough the tunnel...

is this correct ?

R=d*sqrt(3) v= \frac {Bqd} {m} radius of trajectory of the first particle : R1=\frac {mv}{Bq}=\frac {m}{Bq} \frac{Bqd}{m}=d radius for the second particle : R2=\frac {2m(-v)} {B*(-2q)}=d The initial component of velocity on the X axis is the same in absolute value and of contrary...

nu_paradigm, Could you please explain how did you get to 2R/pi ?

ii) the normal reaction is -G_n Y = - 6 * 9.8 * \frac {\sqrt {3}} {2} \frac {\pi}{24}=-6.65 Newtons iii) the frictional force : I don't know.I know that it's miu * N , but I don't know how to calculate miu