Answer :
Part (A)
The final speed of the car can be expressed as,
[tex]v=u+at[/tex]Plug in the known values,
[tex]\begin{gathered} 40.0\text{ m/s=0 m/s+a(10.0 s)} \\ a=\frac{40.0\text{ m/s}}{10.0\text{ s}} \\ =4.00m/s^2 \end{gathered}[/tex]The force acting on the car can be given as,
[tex]F=ma[/tex]Substitute the known values,
[tex]\begin{gathered} F=(1034kg)(4.00m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =4136\text{ N} \end{gathered}[/tex]The average speed of the car can be given as,
[tex]v_a=\frac{v+u}{2}[/tex]Substitute the known values,
[tex]\begin{gathered} v_a=\frac{40.0\text{ m/s+0 m/s}}{2} \\ =\frac{40.0\text{ m/s}}{2} \\ =20.0\text{ m/s} \end{gathered}[/tex]The average mechanical output of the engine is,
[tex]P_a=Fv_a[/tex]Substitute the known values,
[tex]\begin{gathered} P_a=(4136\text{ N)(20.0 m/s)(}\frac{1\text{ W}}{1\text{ Nm/s}}) \\ =82720\text{ W} \end{gathered}[/tex]Thus, the average mechanical power output of engine is 82720 W.
Part (B)
The power generated by gasoline can be given as,
[tex]P=\frac{P_a}{e}[/tex]Substitute the known values,
[tex]\begin{gathered} P=\frac{82720\text{ W}}{(\frac{22.0}{100})} \\ =376000\text{ W} \end{gathered}[/tex]The energy generated by gasoline can be given as,
[tex]E=Pt[/tex]Substituting known values,
[tex]\begin{gathered} E=(376000\text{ W)(10.0 s)(}\frac{1\text{ J}}{1\text{ Ws}}) \\ =3760000\text{ J} \end{gathered}[/tex]The volume consumed by gasoline can be calculated as,
[tex]\begin{gathered} V=\frac{3760000\text{ J}}{(46.0\text{ MJ/L)}(\frac{10^6\text{ J}}{1\text{ MJ}})} \\ \approx0.08\text{ L} \end{gathered}[/tex]Thus, the volume of gasoline consumed is 0.08 L.