Answer:
elastic potential energy
You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.
Explanation:
Answer:
I cant see the provided answer next to the options there is nothing
Explanation:
<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116
So coefficient of static friction = 0.6116</span>
Answer:
Explanation:
Given that
F=2x³
Work is given as
The range of x is from x=0 to x=D
W=-∫f(x)dx
Then,
W=-∫2x³dx from x=0 to x=D
W=- 2x⁴/4 from x=0 to x=D
W=-2(D⁴/4-0/4)
W=-D⁴/2
W=1/2D⁴
The correct answer is F
