An isolated conducting sphere has a 12 cm radius. One wire carries a current of 1.0000049 A into it while another wire carries a current of 1.0000000 A out of it. How long in seconds would it take for the sphere to increase in potential by 1500 V

it will take for the sphere to increase in potential by 1500 V, 503.71 s.

Explanation:

The charge on the sphere after t seconds is:

q = (1.0000049 - 1.0000000) t = 0.0000049 t

The voltage on the surface is

V = k * = k 0.0000049 t / R

solve for t

t = (R*V) / (0.0000049 k) = (0.12 * 1500) / (0.0000049 * ) = 503.71 s

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .