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solniwko [45]
3 years ago
6

Please Help

Physics
1 answer:
Degger [83]3 years ago
8 0

Explanation:

The initial kinetic energy KE_0 is

KE_0 = \frac{1}{2}mv_0^2 = \frac{1}{2}(400\:\text{kg})(10\:\text{m/s})^2

\:\:\:\:\:\;\:= 2×10^4\:\text{J} = 20\:\text{kJ}

The final kinetic energy KE is

KE = \frac{1}{2}mv^2 = \frac{1}{2}(400\:\text{kg})(30\:\text{m/s})^2

\:\:\:\:\:\:\:= 1.8×10^5\:\text{J} = 180\:\text{kJ}

The work done W on the car is

W = \Delta{KE} = KE  - KE_0

\:\:\:\:\:\:\:= 180\:\text{kJ} - 20\:\text{kJ} = 1.6×10^5\:\text{J}

The power expended P is

P = \dfrac{W}{t} = \dfrac{1.6×10^5\:\text{J}}{15\:\text{s}} = 10667\:\text{Watts}

\:\:\:\:\:= 10.7\:\text{kW}

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