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A mass of 500 g is attached to a spring with elastic constant coefficient of elasticity = 50.0 N / m. The spring is lengthened by 10 cm and then left free. Calculate the return force exerted by the spring.

Force, F = -5 N                  

Explanation:

We have,

Mass, m = 500 g = 0.5 kg

Spring constant of the spring, K = 50 N?M

The spring is lengthened by 10 cm and then left free.

It is required to find the return force exerted by the spring. It is based on Hooke's law. The force on the spring is given by :

So, the return force exerted by the spring is (-5 N)

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

a sonic boom is a loud boom that breaks the sound barrier

explanation: