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Math Help - Revolution problem, need help!

  1. #1
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    Arrow Revolution problem, need help!

    A simple model of the core of a tornado is a right circular cylinder that rotates about its axis. If a tornado has a core diameter of 280 feet and maximum wind speed of 150 mi/hr (or 220 ft/sec) at the perimeter of the core, approximate that number of revolutions the core makes each minute.

    Can anyone help me???
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by TinkerBell2175 View Post
    A simple model of the core of a tornado is a right circular cylinder that rotates about its axis. If a tornado has a core diameter of 280 feet and maximum wind speed of 150 mi/hr (or 220 ft/sec) at the perimeter of the core, approximate that number of revolutions the core makes each minute.

    Can anyone help me???
    First calculate the perimeter of the circle.

    The radius is 140 if the diameter is 280.


    2 \times \pi \times 140 = P

    Thus P = 879,645943 feet

    The tornado's perimeter spins at 220 feet per second, so times that by 60 to get 13200 feet per minute.

    13200 \div 879,645943 = 15,0060 times per minute.
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