Asked by Cassidy
How long does it take a 19 kW steam engine to do 6.8x10^7 of work?
I have no idea how to solve this... Would someone please help?
I have no idea how to solve this... Would someone please help?
Answers
Answered by
Matt
well 1 W = J/s So, 19 kW is 19000 J/s. But you want to find time, so you want to have seconds in the denominator, you you would multiply the inverse of the watts times the amount of work.
so (1 sec/19000 J) * (6.8 X 10^7 J)
Joules should cancel out with joules and you should be left with seconds.
so (1 sec/19000 J) * (6.8 X 10^7 J)
Joules should cancel out with joules and you should be left with seconds.
Answered by
Damon
power = 19*10^3 Watts = 19*10^3 Joules/s
work = power * time
6.9*10^7 J = 19 * 10^3 J/s * t
so
t = 6.9*10^7 / 19*10^3
work = power * time
6.9*10^7 J = 19 * 10^3 J/s * t
so
t = 6.9*10^7 / 19*10^3
Answered by
drwls
You omitted the dimensions of the work in your quetion. We assume it is in Joules
I'm pretty sure Damon meant 6.8 instead of 6.9 in his number for the work done. With that change, both previous answers agree.
t = 3579 s
With only 2 significant figures in the input numbers, the answer should be stated as 3600 s or one hour.
I'm pretty sure Damon meant 6.8 instead of 6.9 in his number for the work done. With that change, both previous answers agree.
t = 3579 s
With only 2 significant figures in the input numbers, the answer should be stated as 3600 s or one hour.
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