Awind turbine is initially spinning at a constant angular speed. as the wind's strength gradually increases, the turbine experiences a constant angular acceleration 0.179 rad/s2. after making 2870 revolutions, its angular speed is 140 rad/s.
(a) what is the initial angular velocity of the turbine?
(b) how much time elapses while the turbine is speeding up?

(a) 114.6 rad/s

(b) 142 second

Explanation:

angular acceleration, α = 0.179 rad/s^2

angular displacement,θ = 2 x π x 2870 rad = 18023.6 rad

ω = 140 rad/s

Let the initial angular velocity is ωo.

(a) use third equation of motion

ω² = ωo² + 2αθ

140 x 140 = ωo² + 2 x 0.179 x 18023.6

19600 = ωo² + 6452.45

ωo = 114.6 rad/s

(b) Let t be the time elapsed.

use first equation of motion

ω = ωo + αt

140 = 114.6 + 0.179 x t

25.4 = 0.179 t

t = 141.9 second

t = 142 second

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

grav potential energy