Question a robotic rover on mars finds a spherical rock with a diameter of 1313 centimeters. the rover picks up the rock and lifts it 1515 centimeters straight up. the rock has a specific gravity of 5.255.25. the gravitational acceleration on mars is 3.73.7 meters per second squared. if the robot's lifting arm has an efficiency of 3535% and required 1212 seconds to raise the rock 1515 centimeters, how much power (in watts) did the arm use?

Due to apparent doubling of numbers in the question, assume
diameter=13 cm/2=0.065m
h=15 cm=0.15m
rho=5.25*1000=5250 kg/m^3
eta=35%=0.35 [efficiency]
t=12 s
g=3.7 m/s^2

mass, m=rho*V=rho*(4pi/3)(r^3)
=5250 kg/m^3 * (4pi/3)(0.065)^3 m^3
=1.922 kg

Work done in lifting h=0.15m
W=mgh
=1.922pi kg * 3.7 m/s^2 * 0.15 m
=1.067pi kg (m/s)^2
=1.067pi J

Average power required, with efficiency eta=0.35
P=(W/t)/eta
=(1.067pi J )/ (12 s) /0.35
=0.254pi J/s
=0.798W (approx.)

Idrk but i think u should figure out the n dia n then the units it will make it way more easier