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23 January, 02:40

A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at - 28.7° below horizontal. A friction force of 8.33 N pushes back against the motion.

What is the acceleration of the cart?

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  1. 23 January, 04:43
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    0.8214 m/s^2

    Explanation:

    Fnet = Fpushed - Ffriction

    Fpushed = 12.7N Ffriction = 8.33N

    Fnet = 12.7N - 8.33N = 4.37N

    Fnet = mass (acceleration)

    Fnet = 4.37N mass = 5.32 kg

    4.37N = 5.32 kg (acceleration)

    acceleration = 0.8214 m/s^2
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