210_Physics ProblemsTechnical Physics

210_Physics ProblemsTechnical Physics - 212 P7.65 Energy...

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212 Energy and Energy Transfer P7.65 If positive F represents an outward force, (same as direction as r ), then Wd F r F r d r W Fr F r W Frr F r r F rr F Wr r r r W i f r r r r fi f i f i f i i f i f =⋅= = = −− + =− −−× × zz 2 2 12 6 66 6 6 1 03 10 1 89 10 1 03 10 1 88 10 2 0 13 13 0 77 0 13 12 0 76 0 13 12 12 0 76 6 0 7 0 13 12 12 77 6 6 134 12 12 77 6 σσ ej e j .. . . . 44 10 10 1 89 10 3 54 10 5 96 10 10 2 49 10 112 10 1 37 10 6 60 134 12 8 120 21 21 21 ×− × × × × + × = − × W J J J P7.66 P ∆∆ tW K mv == = a f 2 2 The density is ρ == mm Ax vol . Substituting this into the first equation and solving for , since x t v = , for a constant speed, we get = Av 3 2 . FIG. P7.66 Also, since = Fv , F Av = 2 2 . Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for the drag coefficient. Air actually slips around the moving object, instead of accumulating in front of it. For this reason, the drag coefficient is not necessarily unity. It is typically less than one for a streamlined object and can be greater than one if the airflow around the object is complicated.
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