g 12. Spring A has a spring constant of 100 N/m and spring B has a spring constant of 200 N/m, and both have the same displacement of 0.20 m. The spring potential energy of spring A is that of spring B. twice half the same as None of the above

Explanation:

Spring potential Energy is expressed as E = 1/2 ke²

k = spring constant

e = displacement of the string

If  Spring A has a spring constant of 100 N/m and displacement of 0.2m, the spring potential energy will be;

= 1/2 * 100 * 0.2²

= 50 * 0.04

= 2Joules

If spring B has a spring constant of 200 N/m and displacement of 20m, the spring potential energy will be;

= 1/2 * 200 * 0.2²

= 100 * 0.04

= 4Joules

From the values gotten, it can be seen that the spring potential energy of spring B is twice that of A

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 .

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach