a force of 5 N extends a spring of natural length 0.5m by 0.01, what will be the length of the spring when the applied force is 20 N

L_new =L+x^2  = L_new = 0.54_m.

Explanation:

Given data:

Force in the first case,  

F_1 = 5N

Force in the second case,  

F_2 = 20 N

Natural length of spring,  

L= 0.5

Extension in the first case,  

x_1 = 0.01m

Let the force constant of the spring be k.

Thus,

F_1=kx_1

5 = k × 0.01  

⇒ k = 500 N/m.

The extension in the spring in the second case can be given as,

F_2=kx_2

20 = 500x_2

⇒ x_2 = 0.04 m.

Thus, the effective length of the spring would be,

L_new =L+x^2

L_new = 0.5+0.04

L_new = 0.54_m.

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

laticia draws the diagram below to show earth’s magnetic field.

answer

the arrows to the right of earth should point up.