SOLUTION: The force of the wind blowing on a verticle surface varies jointly as the area of the surface and the square of the velocity. If a wind blowing at 50 miles per hour exerts a force

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Question 180097: The force of the wind blowing on a verticle surface varies jointly as the area of the surface and the square of the velocity. If a wind blowing at 50 miles per hour exerts a force of 75 pounds on a surface of 500 ft squared, how much force will a wind of 75 mph place on a surface of 10 ft squared.

10 ft squared means 10 ft^ i think.
500 ft squared means 500 ft^.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The force of the wind blowing on a verticle surface varies jointly as the area of the surface and the square of the velocity.
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f = kAv^2
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If a wind blowing at 50 miles per hour exerts a force of 75 pounds on a surface of 500 ft squared,
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solve for "k":
50 = k*500*75^2
k = 50/(500*75^2)
k = 0.0000178
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how much force will a wind of 75 mph place on a surface of 10 ft squared.
f = (0.0000178)(10)*75^2
force = 1 lb
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Cheers,
Stan H.