How far must you stretch a spring with k = 1800 N/m to store 270 J of energy? Express your answer in meters.

0.55m

Explanation:

Given parameters:

Spring constant  = 1800N/m

Elastic potential energy  = 270J

Unknown:

extension = ?

Solution:

To solve this problem, we use the formula of the elastic potential energy;

     E. P  = k e²  

E.P is the elastic potential energy

k is the spring constant

e is the extension

            270  =   x 1800 x e²  

                e  = 0.55m

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s