Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

2N/cm

Step-by-step explanation:

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

where, is the force which is stretching or compressing the spring,

is the spring constant; and

is the distance the spring is stretched.

Substituting the given values to find the elastic constant   to get:

Therefore, the elastic constant is 2 Newton/cm.

(75°f − 32) × 5/9 = 23.889°c

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete