Recent content by DaVinci

And after those few hours it comes down to a stupid arithmitic mistake... --slap forhead-- Got it soved.

Few hours later and this one is still driving me nuts. I tried to take the F vector projections on each plane and came up with 3 equations there and used the cosine^2alpha + cos^2beta + cos^2gamma = 1 equation and solve but the numbers don't come out right.

Oh, I thought is was convention that alpha is the angle between the x-axis and the vector, beta is the angle between the y-axis and the vector, and gamma is the angle between the z-axis and the vector. So much for standardization amongst science. :smile: Will remember that for the future.

Im stuck on a 3d vector problem. I know Fx and Fz and angle Beta. I've looked at all the equations in the book (both my statics and physics book) and just cannot seem to put the given data together in a manner to give me another value. If I can just get either Fy or one of the other angles, I...

Yes, that did the trick. Thanks!

Quick question that is throwing me off... If you have a dependant current source in a parallel circuit, how do you represent that in your KCL equation? The one in this problem is Vo/2000. I took that as I=V/R so R=2000. Since it is in a parallel circuit on its own 'branch' I used Vs/2000...

Easy enough... "Duh! Einstein was a physicist!" Then they will just think you are on the level of genious and feel stupid for asking the question.

Tiplers / LLewellen's is not that good at all. I think the only reason my University uses it is because Dr. Llewellen is the head of the Physics Department. :biggrin:

Strongly agree on this book as well.

1-v^2/c^2=1/9-1 Is an incorrect statement. You need to have this: 1-v^2/c^2=1/9 Then subtract 1 from both sides simultaniously. You got the right answer though ... good job.

1-v^4/c^4=1/9 That is incorrect. \sqrt{(1-v^2/c^2)}^2 Take a look at that step...

Simple algebra... what do you do to remove a square root sign?

Thats the hard way. Start with the original equation. Multiply both sides of the equation by the denominator on the left hand side. This will move the entire square root to the left hand side with the 3Mo. Then, divide both sides of the equation by 3Mo. So on the right hand side you have Mo /...

What do you get after you do the steps I had listed in my previous post? Let's make sure you got that far and then we will look at it from there. Post the answer you get after you square both sides.

m becomes 3m_0. The m_0 on the right hand side stays the same. Then multiply both sides by the denominator on the left hand side and them divide both sides by 3m_0. Here the two rest masses cancel leaving you with 1/3 on the right hand side. Then square both sides and continue to use...