Recent content by Kittlinljd

Yay it did help, but you have to convert the radius to m first and then you get the right answer of 52.6 A. Thankyou, to the both of ya!

Wow, that would deffinetly help. Can't believe I forgot something so symple... okay I drew the picture wrong so that messed it up. Thankyou, now let me see if what I get is right now!

okay so I tried 1E-6, that didn't work so i looked again at what I did for the cross section, found out you can get 2 different values, if you use 4.2E-3 or just keep it in mm then convert you get 2 different numbers, the conversion to mm being the largest. So i tried all possible combinations...

I looked it up and it goes from 1.0E−6 to 1.5E−6ohm m, so I kinda had the right value. I guess i should try the first value as well, however I doubt if its right.

Sorry about the missing units! hee they are! A 60.0 cm long hollow nichrome tube of inner diameter 1.60mm , outer diameter 4.20mm is connected to a 4.00 V battery. The units for P is ohm m, and area should have been m^2. My bad! Oh and by the way the p value wasn't given in the problem...

Homework Statement A 60.0 hollow nichrome tube of inner diameter 1.60 , outer diameter 4.20 is connected to a 4.00 battery.What is the current in the tube? \rho = 1.5*10^-6 ohm for nichrome Homework Equations I=dletaV/R R=\rhoL/A -a should be the crossectional area The Attempt at a...

ahhhh, okay so now I see so I get: K* Q/(y^3) which is the answer masteringphysics took! Thanks So much!

so I change the constants to -s or s depending on which direction on the direction on the y axis? that makes sense. What I tried was V = K \sumQ/(y+s) -2Q/(y) + Q/(y-s) , and it said that the answer does not depend on the variable s. Would this have something to do with Q= 2qs^2, being the...

[b] 1. The arrangement of charges shown in the figure is called a linear electric quadrupole. The positive charges are located at +-s. Notice that the net charge is zero. Find an expression for the electric potential on the x-axis at distances y>>s. Homework Equations V= \sum1/4pi...