In 1889, at Jubbulpore, India, a tug-of-war was finally won after 2 h 41 min, with the winning team displacing the center of the rope 3.7 m.

Question

In 1889, at Jubbulpore, India, a tug-of-war was finally won after 2 h 41 min, with the winning team displacing the center of the rope 3.7 m. What was the magnitude of the average velocity of that center point during the contest

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Khải Quang 3 years 2021-08-06T17:09:32+00:00 1 Answers 23 views 0

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  1. Farah
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    2021-08-06T17:11:16+00:00
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    Answer:

    The magnitude of the average velocity of that center point during the contest is 0.0229 \frac{m}{min}

    Explanation:

    Given :

    Time interval\Delta t = 161 min

    Displacement of the rope \Delta x = 3.7 m

    For calculating average velocity of the rope,

      v = \frac{\Delta x}{\Delta t}

      v = \frac{3.7}{161}

      v = 0.0229 \frac{m}{min}

    Therefore, the magnitude of the average velocity of that center point during the contest is 0.0229 \frac{m}{min}

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