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3 November, 04:20

Superman leaps in front of Lois Lane to save her from a volley of bullets. In a 1-minute interval, an automatic weapon fires 147 bullets, each of mass 8.0 g, at 350 m/s. The bullets strike his mighty chest, which has an area of 0.74 m2. Find the average force exerted on Superman's chest if the bullets bounce back after an elastic, head-on collision.

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  1. 3 November, 04:43
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    F/Area = 9.27 N / m^2

    Explanation:

    Given:

    - mass of 1 bullet m = 0.008 kg

    - number of bullets fired n = 147

    - time interval dt = 60 s

    - Initial velocity v_i = 350 m/s

    - Area of chest A = 0.74 m^2

    Find:

    Average force imparted on the chest F

    Solution:

    The number of bullets fired per second:

    - 147 / 60 s = 2.45 bullets / s

    Momentum of single bullet:

    - mass of one bullet * speed of bullet = 0.008 * 350 = 2.8 kg m/s.

    Total momentum per second:

    - number of bullets fired per second * Momentum of single bullet

    = 2.8 * 2.45 = 6.86 kg m/s

    Momentum = Force * time

    - f = 6.86*1 = 6.86 N

    Superman experiences a constant average force of 6.86 N for 1 minute

    Assuming the bullets are spread out evenly over the chest:

    - (Force / Area) = 6.86 / 0.74 = 9.27 N / m^2

    So the average force experienced by superman per unit area is 9.27 N/m^2
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