Suppose that 2 J of work is needed to stretch a spring from its natural length of 24 cm to a length of 42 cm. (a) How much work is needed to stretch the spring from 32 cm to 34 cm

Workdone = 0.025 Joules

Explanation:

Given the following data;

Workdone = 2J

Extension = 42 - 24 = 18 cm to meters = 18/100 = 0.18m

The workdone to stretch a string is given by the formula;

Workdone = ½ke²

Where;

k is the constant of elasticity.

e is the extension of the string.

We would solve for string constant, k;

2 = ½*k*0.18²

2 = ½*k*0.0324

Cross-multiplying, we have;

4 = 0.0324k

k = 4/0.0324

k = 123.46 N/m

a. To find the workdone when e = 32, 34.

Extension = 34 - 32 = 2 to meters = 2/100 = 0.02m

Workdone = ½*123.46*0.02²

Workdone = 61.73 * 0.0004

Workdone = 0.025 Joules

Therefore, the amount of work (in J) needed to stretch the spring from 32 cm to 34 cm is 0.67.

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

it is very well related to a lot of things but with my limited knowledge i can confidently say that the answer is space.