#### Answer

The correct answer is:
D. $1.5\times 10^5~\times sin~10^{\circ}$

#### Work Step by Step

The extra power would be required to increase the gravitational potential energy of the car. We can find the increase in gravitational potential energy each second:
$\Delta U_g = mgh$
$\Delta U_g = mgd~sin~\theta$
$\Delta U_g = (1000~kg)(10~m/s^2)(15~m)~sin~10^{\circ}$
$\Delta U_g = 1.5\times 10^5~sin~10^{\circ}~J$
Since this amount of additional energy is required each second, the extra power that is required is $1.5\times 10^5~sin~10^{\circ}~W$
The correct answer is:
D. $1.5\times 10^5~\times sin~10^{\circ}$