We could take the easy way out and just say
(110 kW) x (3 hours) = 330 kilowatt hours .
But that's cheap, and hardly worth even 5 points.
If we want to talk energy, let's use the actual scientific unit of energy.
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" 110 kw " means 110,000 watts = 110,000 joules/second .
(3 hours) x (3600 sec/hour) = 10,800 seconds.
(110,000 joules/second) x (10,800 seconds) = 1.188 x 10⁹ Joules
That's
==> 1,188,000,000 joules
==> 1,188,000 kilojoules
==> 1,188 megajoules
==> 1.188 gigajoules
Atsa nawfulotta energy !
It goes back to that "110 kw appliance" that we started with.
That's no common ordinary household appliance. 110 kw is something like
147 horsepower. In order to bring 110 kw into your house, you'd need to
take 458 Amperes through the 240-volt line from the pole. Most houses
are limited to 100 or 200 Amperes, tops. And the TRANSFORMER on
the pole, that supplies the whole neighborhood, is probably a 50 kw unit.
option c...have the waves move thru the vacuum or space
Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is 
Explanation:
From the question we are told that
The mass of the steel ball is 
The length of arm is 
The mass of the arm is 
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

=>
=>
Generally the magnitude of torque about the athlete shoulder is mathematically represented as

=> 
=> 
Let say for every 5 s of time interval the speed will remain constant
so it is given as
v(mi/h) 16 21 23 26 33 30 28
now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s
so here we will have
v(ft/s) 23.5 30.8 33.73 38.13 48.4 44 41.1
now for each interval of 5 s we will have to find the distance cover for above interval of time



so here it will cover 1298.1 ft distance in 30 s interval of time