A freight train leaving a yard must exert a force of 2530000 N in order to increase its speed from rest to 17 m/s. During this process, the

Question

A freight train leaving a yard must exert a force of 2530000 N in order to increase its speed from rest to 17 m/s. During this process, the train must do 1110000000 J of work. How far does the train travel? *

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Thu Thảo 4 years 2021-07-30T23:28:23+00:00 2 Answers 25 views 0

Answers ( )

  1. RuslanHeatt
    0
    2021-07-30T23:30:00+00:00
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    Answer: The distance traveled in other to increase its speed from rest to 17m/s is 438.74m

    Explanation:

    GIVEN the following :

    Workdone = 1110000000 J

    Force = 2530000 N

    Workdone = force × distance

    1110000000 = 2530000 × distance

    Distance = 1110000000 ÷ 2530000

    Distance = 438.735m

    Distance = 438.74m

    The distance traveled by the freight train in other to increase it’s speed from rest to 17m/s is 438.74m

  2. thienan
    0
    2021-07-30T23:30:04+00:00
    Reply

    Answer:

    \Delta s = 438.735\,m

    Explanation:

    Given that force exerted is constant and parallel to the railroad during acceleration, the equation of work is equal to:

    W = F\cdot \Delta s

    The travelled distance of the freight train is:

    \Delta s = \frac{W}{F}

    \Delta s = \frac{1110000000\,J}{2530000\,N}

    \Delta s = 438.735\,m

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