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23 May, 21:12

Bert is making a strawberry milkshake in his blender. A tiny, 5.0 g strawberry is rapidly spun around inside of the container with a speed of 14.0 m/s, held by a centripetal force of 10.0 N. What is the radius of the container at this location?

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  1. 23 May, 21:18
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    r = 0.098m

    the radius of the container at this location is 0.098m

    Explanation:

    Centripetal force F = (mv^2) / r

    Radius r = (mv^2) / F

    Where;

    r = radius

    F = centripetal force = 10N

    m = mass = 5g = 0.005kg

    v = velocity = 14m/s

    Substituting the values;

    r = (0.005 * 14^2) / 10

    r = 0.098m

    the radius of the container at this location is 0.098m
  2. 23 May, 21:31
    0
    Given that,

    Mass of strawberry

    m = 5g = 0.005kg

    Speed at which it is spun in a circular motion

    v = 14m/s

    Centripetal force that held it in motion

    Fc = 10N

    Radius of circular motion r?

    Centripetal force is given as

    Fc = m•ac

    Where m is mass

    And ac is centripetal acceleration.

    Then, centripetal acceleration is give as

    ac = v²/r.

    So,

    Fc = mv²/r

    Making r subject of formulas

    r = mv² / Fc

    Where,

    m is mass m = 0.005kg

    v is velocity v = 14m/s

    Fc is centripetal force Fc = 10N

    Then,

    r = mv² / Fc.

    r = 0.005 * 14² / 10

    r = 0.098m

    r = 9.8cm

    The radius of the path of circular motion is 9.8cm
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