Given: Mass m = 44 Kg; Velocity v = 10 m/s
Required: Kinetic energy K.E = ?
Formula: K.E = 1/2 mv²
K.E 1/2 (44 Kg)(10 m/s)²
K.E = 2,200 Kg.m²/s²
K.E = 2,200 J Answer is A
<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>
<span>hydrocarbon (but im not 100% sure)</span>
Answer:
Density =mass/volume 20/10=2
Awnser:
Elastic Potential Energy. Elastic potential energy is Potential energy stored as a result of deformation of an elastic object,
Explanation:
