A700 kg elevator starts from rest. it moves upward for 4.50 s with constant acceleration until it reaches its cruising speed, 1.75 m/s. (a) what is the average power of the elevator motor during this period? w (b) how does this power compare with the motor power when the elevator moves at its cruising speed? pcruising = w

(a). The average power of the elevator motor during this period is 6235.25 W.

(b). The power input from motor is

Explanation:

Given that,

Mass of elevator =700 kg

Initial velocity = 1.75 m/s

Time = 4.50 s

We need to calculate the acceleration of the elevator

Using formula of acceleration

Put the value into the formula

We need to calculate net force on elevator

Using formula of net force

Put the value into the formula

We need to calculate the average velocity

Using formula of average velocity

Put the value into the formula

(a). We need to calculate the average power of the elevator motor during this period

Using formula of power

Put the value into the formula

(b). We need to calculate power input from motor

Using formula of power

Put the value into the formula

Hence, (a). The average power of the elevator motor during this period is 6235.25 W.

(b). The power input from motor is

convex lenses are thicker at the middle. rays of light that pass through the lens are brought closer together (they converge). a convex lens is a converging lens. when parallel rays of light pass through a convex lens the refracted rays converge at one point called the principal focus.

hey , the answer is c