Meteorite On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, creating a dent about 22 cm deep. If the initial speed of

Question

Meteorite On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, creating a dent about 22 cm deep. If the initial speed of the meteorite was 550 m/s, what was the average force exerted on the meteorite by the car?

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Phúc Điền 2 months 2021-07-28T08:58:43+00:00 2 Answers 1 views 0

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    0
    2021-07-28T08:59:59+00:00

    Answer:

    The average force exerted on the meteorite by the car is 8.435×10⁶N

    Explanation:

    Given:

    m = 27 lb = 12.27 kg

    D = 22 cm = 0.22 m

    v1 = 550 m/s

    The acceleration of the meteorite is:

    v_{2} ^{2} =v_{1} ^{2}+2aD\\0=550^{2} +2a(0.22)\\a=-\frac{550^{2} }{2*0.22} =-687500m/s^{2}

    Negative sign indicates opposite to the initial velocity. The average force is equal to:

    Magnitude of a = 687500 m/s²

    F = m*a = 12.27 * 687500 = 8.435×10⁶N

    0
    2021-07-28T09:00:11+00:00

    Answer:

    8427375 N

    Explanation:

    From newton’s equation of motion,

    F = ma…………….. Equation 1

    Where F = average force exerted on the meteorite by the car, m = mass of the meteorite, a = acceleration.

    We can find the value of acceleration from newton’s equation of motion

    using,

    v² = u²+2as……………… Equation 2

    Where v = final velocity, u = initial velocity, a = acceleration, s = distance.

    Make a the subject of the equation

    a = (v²-u²)/2s…………… Equation 3

    Note: When the meteorite struck the car, it was brought to rest

    Given: u = 550 m/s, v = 0 m/s ( brought to rest), s = 22 cm = 0.22 m

    Substitute into equation 3

    a = (0²-550²)/(2×0.22)

    a = -302500/0.44

    a = -687500 m/s²

    Also given: m = 27 pound = (27×0.454) = 12.258 kg

    Substitute into equation 1

    F = 12.258(-687500)

    F = -8427375 N

    Note: The negative sign tells that the force act in opposite direction to the motion of the meteorite

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