Answer:
.7934
Explanation:
Acceleration = change in velocity / change in time
A = 10.98
/ 13.84
A = .7934
Answer:
The same amount of energy is required to either stretch or compress the spring.
Explanation:
The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

where
k is the spring constant
is the stretch/compression of the spring
In the first case, the spring is stretched from x=0 to x=d, so

and the amount of energy required is

In the second case, the spring is compressed from x=0 to x=-d, so

and the amount of energy required is

so we see that the amount of energy required is the same.
Answer:
31.32 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
Let us assume the height of the Disque hall is 50 m

In order to make the jump Superman's initial velocity must be greater than or equal to 31.32 m/s