An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J. (a) If the

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An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J. (a) If the electric motor draws 8.80 kW as the car moves at a steady speed of 20.0 m/s, what is the current (in A) delivered to the motor? A (b) How far (in km) can the car travel before it is “out of juice”? km (c) What If? The headlights of the car each have a 55.0 W halogen bulb. If the car is driven with both headlights on, how much less will its range be (in m)? m

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Diễm Thu 4 years 2021-08-18T19:56:12+00:00 1 Answers 4 views 0

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  1. Nem
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    2021-08-18T19:57:25+00:00
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    Answer:

    a) I=733.33\ A

    b) d=52272.7273\ m

    c) d'=51948.0519\ m

    Explanation:

    Given:

    • voltage of the battery, V=12\ V
    • energy storage capacity of the battery, E=2.3\times 10^7\ J
    • speed of the car, v=20\ m.s^{-1}

    a)

    power drawn by the car, P=8.8\ kW

    Now the Current delivered to the motor:

    we the relation between the power and electrical current,

    P=V.I

    8800=12\times I

    I=733.33\ A

    b)

    Distance travelled before battery is out of juice:

    we first find the time before the battery runs out,

    t=\frac{E}{P}

    t=\frac{2.3\times 10^7}{8800}

    t=2613.636\ s

    Now the distance:

    d=v.t

    d=20\times 2613.636

    d=52272.7273\ m

    c)

    When the head light of 55 W power is kept on while moving then the power   consumption of the car is:

    P'=P+55

    P'=8800+55

    P'=8855\ W

    Now the time of operation of the car:

    t'=\frac{E}{P'}

    t'=\frac{2.3\times 10^7}{8855}

    t'=2597.4026\ s

    Now the distance travelled:

    d'=v.t'

    d'=20\times 2597.4025

    d'=51948.0519\ m

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