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11 February, 12:36

An electrical motor spins at a constant 2662.0 rpm. If the armature radius is 6.725 cm, what is the acceleration of the edge of the rotor? O 524,200 m/s O 29.30 m/s O292.7 m/s O 5226 m/s2

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  1. 11 February, 12:55
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    18.73 m/s^2

    Explanation:

    f = 2662 rpm = 2662 / 60 rps

    r = 6.725 cm = 0.06725 m

    Acceleration, a = r w

    a = r x 2 x pi x F

    a = 0.06725 * 2 * 3.14 * 2662 / 60

    a = 18.73 m/s^2
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