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A merry-go-round revolves at a rate of 8.3 rev/min. A child who is seated at the outer edge of the merry-go-round travels at 6.9m/s. What is the diameter of the merry-go-round?

  • physics -

    v =ω•R = ω•D/2.
    ω = 2•π•f ,
    f = 8.3 rev/min = 8.3/60 =0.138 rev/s
    D = 2•v/ ω = 2•v/ 2•π•f ω =
    = 6.9/(3.14•0.138)=15.9 m

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