#### Answer

(a) The engine's average power output is $3.64\times 10^5~Watts$
(b) The volume of gasoline that is consumed is 79 mL

#### Work Step by Step

(a) We can find the car's change in kinetic energy:
$KE = \frac{1}{2}mv^2 = \frac{1}{2}(1000.0~kg)(40.0~m/s)^2 = 8.00\times 10^5~J$
This energy is only 22% of the energy released by burning gasoline. We can find the total energy released:
$0.22~Energy = 8.00\times 10^5~J$
$Energy = \frac{8.00\times 10^5~J}{0.22}$
$Energy = 3.64\times 10^6~J$
This much energy is released in a time of 10.0 seconds. We can find the power output:
$Power = \frac{Energy}{Time} = \frac{3.64\times 10^6~J}{10.0~s} = 3.64\times 10^5~Watts$
The engine's average power output is $3.64\times 10^5~Watts$
(b) We can find the volume of gasoline that is consumed:
$V = \frac{3.64\times 10^6~J}{46\times 10^6~J/L} = 0.079~L = 79~mL$
The volume of gasoline that is consumed is 79 mL.