Answer:
Elastic potential energy, E = 200 J
Explanation:
It is given that,
Spring constant, K = 4 N/m
initial stretching in the spring, x = 5 m
Finally, it is stretched an additional 5 m i.e. x' = 5 m
Let E is the elastic energy in the spring after Varg stretches the spring. it is given by :
E = 200 J
So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.
Answer:
a.
b.
Explanation:
<u>Given:</u>
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.
At time t = 3 seconds,
<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>
<h2>(b):</h2>
The velocity of the particle at some is defined as the rate of change of the position of the particle.
For the time interval of 2 seconds,
The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,
It is the displacement of the particle in 2 seconds.