Stu

Hey, neither answer works, nor the step before the solution with the theta ^2 in side or outside the brackets.

Will let you know if the online test accepts either answer  j have a feeling it will and that I am really weak working with fractions like this. Also, there is a problem with mobile view/normal view on this site showing a large portion of the pagebin white and unable to zoom in to the section that is condesed to the lhs.

h

sintheta ^2

Can you simplfy the following please.
$$w=\frac{-1}{-1/2cos(theta)^2+(7/16)}$$

This problem has been resolved. 

Please explain this information for me--> 

Question 7 The figure shows four pairs of charged particles. For each pair, let  V = 0 at infinity and consider  V net at points on the x axis.

Your answer is incorrect.  Try again.     For which pairs is there a point at which Vnet = 0 between the particles? (1) (2) (3) (4) None       SHOW HINT

Your answer is incorrect.  Try again.     For which pairs is there a point at which Vnet = 0 to the right of the particles? (1) (2) (3) (4) None       SHOW HINT

Your answer is correct.     At such a point is  due to the particles equal to zero? Yes No

Your answer is correct.     For each pair, are there off-axis points (other than at infinity) where Vnet = 0? Mark the pairs where this is true. (1) (2) (3) (4) None of the pairs exhibit this quality.

Unsure about this. It is similar to last weeks question. I don't know what is being asked when I read it. 

A spherical conducting shell has a charge of -37 μC on its outer surface and a charged particle in its hollow. If the net charge on the shell is -31 μC, what is the charge (a) on the inner surface of the shell and (b) of the particle?

(a)

Number

Units

° (degrees) This answer has no units m kg s m/s m/s^2 N J W N/m A V μC N/C C/m C/m^2 C/m^3 N·m^2/C

(b)

Number

Units

° (degrees) This answer has no units m kg s m/s m/s^2 N J W N/m A V μC N/C C/m C/m^2 C/m^3 N·m^2/C

I entered it poorly. The equation should have been dR/dy +R equals 1 with the initial conditions as above. Is it the same now as your reply Alan?

I have tried again his morning and struggled with this one again.