An automobile engine is 15 percent efficient. How many joules of input work are required to produce 15,000 joules of output work to move the car?

15,000J/.15% = 100,000 J
If you want to check, Multiply 100,000 by .15 and you get 15,000 J

Answered by megeliz13 | 2024-06-10

Efficiency of machines is the ratio of work done by machine (work output) to the work put in the machine (work input)
Efficiency = Work output/Work input
Given efficiency = 15%
Work output = 15,000Joules
Work Input = ?
Substituting in the formula of efficiency
15% = 15,000/Work Input
15/100 = 15,000/Work Input
15× Work input = 15,000×100
15 × Work input = 1,500,000
Work input = 1,500,000/15
Work input = 100,000Joules

Answered by abidemiokin | 2024-06-12

An automobile engine with an efficiency of 15% requires 100,000 joules of input work to produce 15,000 joules of output work. This is calculated using the efficiency formula. We rearrange the formula to find the work input based on the output work and efficiency percentage.
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Answered by abidemiokin | 2024-10-06