Recent content by jra_1574

It was actually in my exam a few years ago.. i solved the problem in class tho... i don't remember the answer, but the equation given here was very useful... :)

Look for "Physics for Scientists and engineerings" 6th edition.. there are two volumes to this those books are really good.. the cover is blue

OK thanks so much for your help!

How would i type that in maple? f:=sin(x)^3*cos(y)? or f:=sin^3(x)*cos(y)? anyways i pulled the cos(y) out of the integral, then change sin^3(x) to sin(x)*(1-cos^2(x))... i think that led me to sin(x)-cos^2(x)*sin(x) so integrate sin(x)= cos(x) minus integration of cos^2(x)*sin(x)...

f:=sin^3(x)*cos(y); Integrate f with respect to x... from y to Pi/2

Homework Statement f:=sin^3(x)cos(y) Integrate f, from x=y...pi/2) Homework Equations I can do it if the two letters were y... But i have no idea how to solve when they are combined..

ok i did the calculation Eo = 8.85 x10^-12 A= 8.1x10^-3 d = 2.9x10^-4 so the final answer is C = 2.4719x10^-10 Now what should i do?

The full question is like this: The square plates of a 9000pF capacitor measure 90mm by 90mm and are separated by a dielectric which is 0.29mm thick. The voltage rating of the capacitor is 300V. THe dielectric strength, in KV/m is closest to: A. 1300 B. 930 C. 1200 D. 1100 E. 1000 To...

Hello, everyone! I have a question, and i don't know how to solve it... They ask me for the dielectric strength! How do i find it?

Thanks for your Help! I really appreciate it

can anyone give me a hint?

My accent is well noticed! but i try my best when speaking and writing..

Hello Everyone! i need some guide and hints about this... Here is the link for the picture... From the figure, the electric Field lines arising from two charges Q1 and Q2 are shown. From this drawing we can see that A) The electric field could be zero at P2 B) Both Q1 and Q2 have the...