How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp equals 746 W) to reach a speed of 15.0 m/s, neglecting frict

Question

How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp equals 746 W) to reach a speed of 15.0 m/s, neglecting friction if the car also climbs a 3.00 m high hill in the pro?

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Euphemia 2 weeks 2021-09-05T08:21:07+00:00 1 Answers 0 views 0

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    2021-09-05T08:22:17+00:00

    Answer:

    The car will take 4 seconds.

    Explanation:

    First of all, we convert the power from hp to W:

    40.0hp=29840W

    The work done by the car is equal to the change in its mechanical energy. This means that:

    P=\frac{W}{\Delta t}=\frac{\Delta E}{\Delta t}\implies \Delta t=\frac{\Delta E}{P}

    Now, the change in mechanical energy in terms of the kinetic and potential energy is:

    \Delta E= \frac{1}{2} mv_f^{2} +mgh_f-\frac{1}{2} mv_o^{2} +mgh_o

    Since the initial height and the initial velocity are zero, we have:

    \Delta E= \frac{1}{2} mv_f^{2} +mgh_f

    Finally,

    \Delta t=\frac{\frac{1}{2} mv_f^{2} +mgh_f}{P}

    Plugging in the given values, we obtain:

    \Delta t=\frac{\frac{1}{2} (850kg)(15.0m/s)^{2} +(850kg)(9.8m/s^{2})(3.0m)}{29840W}=4.0s

    In words, the car will take 4 seconds.

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