Assume the mass of a shuttlecraft is 2,200 kilograms [kg]. If the shuttlecraft requires 3 minutes [min] to rise from the surface of the Moon to an altitude of 19 kilometers [km], and the shuttle's engine is rated at 500 kilowatts [kW], what is the efficiency of the engines

The efficiency of the engine is 75.94 %

Explanation:

Given;

mass of the shuttlecraft, m = 2,200 Kg

time required by the shuttlecraft, t = 3 minutes = 180 s

height risen by the shuttle, h = 19 km = 19,000 m

Input power of the shuttle, Pi = 500 kW = 500,000 W

The potential energy due to height risen by the Shuttle on Moon surface is given as;

E = mgh

where;

g is the acceleration due to gravity on moon, = ¹/₆ x 9.81 m/s² = 1.635 m/s²

E = 2200 x 1.635 x 19,000

E = 68343000 J

Output power of the shuttle, is given as  Energy / time

Output power = (68343000) / (180)

Output power = 379,683.33 W

The efficiency of the shuttle is given as;

Efficiency = (Output power) / (Input power)

Efficiency = (379,683.33) / (500,000)

Efficiency = 0.7594

Efficiency(%) = 75.94 %

Therefore, the efficiency of the engine is 75.94 %

50 inches should be the resultant vector

uio

explanation: