The answer is Carbonic acid
For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:

where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus
Since the bus comes at rest, its final kinetic energy is zero:

, so the work done by the brakes to stop the bus is

And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.
Efficiency = Work Output / Work Input
92% = Work Output / 100
0.92 = Work Output / 100
Work Output = 0.92 * 100
Work Output = 92 joules.
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