Power = energy used/time to use it
Power of this CD player = 60,000/(30 x 60) = 331/3 watts
That solution doesn't depend on the player's efficiency ... you specified
only the energy it uses.
But since you did mention the concept of efficiency, I'm compelled to add a
thought here: If you're describing the kind of CD player that you can carry
in your pocket and listen to it on the bus, then 33 watts is an astronomical
power consumption for it. You'd need to carry a back-pack full of batteries,
and recharge them every few hours.
If these are real figures, and the thing really does consume 33 watts in normal
use, then I hope it plugs into the wall, and drives a nice pair of speakers.
.33
Assuming that juels is your attempt to spell joules, the answer is 1 newton-metre.
The box acquires (75 x 2.5) = 187.5 joules of gravitational potential energy.The efficiency of your effort is (187.5 / 425) = 44.1% (rounded)
3 Kilowatts is a measure of electrical power and means the wind turbine produces 3,000 joules per second. So the question is fairly meaningless, assuming the wind is at the right speed it will be producing 3kW every second, every hour and every day!
Power input = 700 WEfficiency = 75% so power output = 75% of 700 = 525 W Time = 30 seconds => Work done = 525W * 30s = 15750 W.s = 15750 Joules = 15.75 kiloJoules.
The machine efficiency is 35 percent (35/100).
25 percent
25 percent
25 percent
25%
Efficiency is measured in joules.
The efficiency is 80%. To find the efficiency, 400/500 = 80%.
The efficiency is 80%. To find the efficiency, 400/500 = 80%.
If the input energy is 210 joules and the efficiency of the system is 30%,then the output energy is30% of 210 = (0.3 x 210) = 63 joules.
The efficiency is 80%. To find the efficiency, 400/500 = 80%.
.33
Just divide the useful work done (the 90J) by the input work. If you want the result in percent, multiply by 100.