For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2
If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2
<span>So we want to know what happens to the momentum of the ball that rolls down hill and hits a box. So we need to use the law of conservation of momentum which states that the momentum must be conserved. It cant be transformed into inertia or mass. It can only be transferred to other object via some interactions like collisions. So it has to be a. transferred to the box and that is the correct answer. </span>
Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :

m is the mass of object


m = 0.25 kg


k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.