Recent content by kidi3

Ah.. they all point in the same direction, and since change is equal to the one in situation A, they must be equal in magnitude..

It's both in and out.. but wouldn't the dec. part go to 0, and lead the B to become ½A.. I can see B would be as high as A in a short moment,should it not be ½A thereby lead B<A

Hey Pf.. I am trying to understand Lenz's law, but somehow it doesn't make sense. In my book there is some checkpoints execise to test wheather you've understood what you read about, one those checkpoints looks like this. http://snag.gy/NCNLh.jpg I do understand why the current in...

Could you tell me what i am doing wrong with this one.. Think i got it..

Ahh.. It makes sense now.

Homework Statement Find the Laplace transform of the following function t[SUP]2[SUP] - 2t The Attempt at a Solution \[\begin{gathered} f\left( t \right) = {t^2} - 2t \hfill \\ l\left( {f\left( t \right)} \right) = F\left( s \right) = \int_0^\infty {{e^{ - st}} \cdot \left( {{t^2} - 2t}...

I am getting confused.. wasn't F_x i shoud have used.. And the error, I really don't understand how there can be a error in the equation i wrote..

...' I still get an incorrect answer.. http://snag.gy/veJcm.jpg

#30 Well.. would it make any difference? #29 yeah i may have understood your previous post.. Now is the result 1.89577 http://snag.gy/E71gm.jpg

These are my steps... http://snag.gy/3oGJ5.jpg

But isn't what i have already done before..

How do you get this ? 2q3cos3θ/d2 = q2/R2

D has to be 2,266 cm But i get 1471,4 m...

Hmm.. but i still get an incorrect answer.. :( http://snag.gy/LDwAe.jpg When i Tries to calculate D i get complex number..

Ahh.. I see what you mean now.. My drawings weren't correct.. But by using youre drawing I see that Cos(theta) = F_x/F <=> F_x = Fcos(theta) So the expression for F_X = (q1q2/(4∏εo(d^2))) cos(theta)