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17 August, 16:13

A 48.9 kg meteor is moving in outer space. If a 8.6 N force is applied opposite the direction of motion, what is the deceleration (in outer space we assume no friction)

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  1. 17 August, 18:33
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    The deceleration is 0.18 m/s²

    Explanation:

    Hi there!

    Using Newton's second law, we can calculate the deceleration:

    ∑F = m · a

    Where:

    ∑F = the sum of all forces in a given direction.

    m = mass of the object.

    a = acceleration.

    Solving for a:

    ∑F/m = a

    The only force acting on the meteor is the applied force of 8.6 N. So, the acceleration will be:

    8.6 N / 48.9 kg = a

    a = 0.18 m/s²

    The deceleration is 0.18 m/s² or, in other words, the acceleration is - 0.18 m/s²

    Have a nice day!
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