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What happens to the effort force as the radius increases?

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zhannawk [14.2K]3 years ago

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time of motion, t = 3.47 s

The distance traveled by Charlotte in feet is calculated as;

$Distance = Speed \ \times \ time \\\\D = ut\\\\D = (\frac{66.5 \ mi}{h} \times \frac{5280 \ ft}{1 \ mi} \times \frac{1 \ h}{3600 \ s} )(3.47 \ s)\\\\D = 338.44 \ ft$

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