soln11.1(2)

# soln11.1(2) - Solution(Individual Exercise 1 1 Fd = Cd Av 2...

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Solution (Individual Exercise 1) ( 29 2 2 3 3 2 2 2 s m m N 61 . 0 m kg 61 . 0 m kg 23 . 1 2 1 2 1 2 1 = = = = = = = Av C F k Av kC F Av C F d d d d d d ρ

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Solution (Individual Exercise 2) ( 29 ( 29 2 2 f 2 2 2 f 2 2 2 h mi ft lb 00256 . 0 s 3600 h mi m 1609 ft 3.281 m N 4.448 lb s m m N 61 . 0 2 1 = × × × × = = = Av C F k d d ρ
Pairs Exercise 1 The drag force due to wind (air) acting on an object  can be found by: F D  = 0.00256 C D A V 2 where:  F D  = drag force (lb f ) C D  = drag coefficient (no units) A = projected area (ft 2 ) V  = velocity of object (mph) convert the proportionality factor in the drag force  equation on the previous slide if the units of velocity are ft/s, and  the units of area are in 2

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Pairs Exercise 1 Step 1 - Solve for the constant
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soln11.1(2) - Solution(Individual Exercise 1 1 Fd = Cd Av 2...

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