When you do 100 J of work on the handle of a bicycle pump, the pump does 40 J of work pushing the air into the tire. What is the efficiency of the pump?
Efficiency is "Efficiency = (useful work output / work input)"
So the problem would look like:
EC = (40/100)
Divide 40 by 100, and you get 0.4 - which is 40%.
(A) 12% of 6500 J
That would be 780 J (per second).
(B)780 J/s (Watts)
A river does 6,500 J of work on a water wheel every
second. The wheel’s efficiency is 12%.
a. How much work in joules can the axle of the
wheel do?
b. What is the power output of the wheel in 1 s?
To find the efficiency of the pump, we can use the formula:
Efficiency = Useful Output / Input
In this case, the useful output is the work done by the pump pushing air into the tire, which is 40 J. The input is the work done on the handle, which is 100 J.
Substituting the values into the formula:
Efficiency = 40 J / 100 J
To simplify this fraction, we divide both the numerator and denominator by their greatest common divisor, which is 20:
Efficiency = 2 J / 5 J
The units of "J" cancel out, and we are left with a simplified fraction:
Efficiency = 2/5
So, the efficiency of the pump is 2/5 or 40%.