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326_Physics ProblemsTechnical Physics

326_Physics ProblemsTechnical Physics - 328*P11.5 Angular...

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328 Angular Momentum *P11.5 τ = × = °− ° × = r F 0 450 0 785 90 14 0 343 . . sin . m N up east N m north a f a f FIG. P11.5 P11.6 The cross-product vector must be perpendicular to both of the factors, so its dot product with either factor must be zero: Does 2 3 4 4 3 0 ± ± ± ± ± ± i j k i j k + + = e j e j ? 8 9 4 5 0 = − No . The cross product could not work out that way. P11.7 A B A B × = = = AB AB sin cos tan θ θ θ 1 or θ = ° 45 0 . P11.8 (a) τ = × = = + = r F i j k i j k k ± ± ± ± ± ± . ± 1 3 0 3 2 0 0 0 0 0 2 9 7 00 a f a f a f a f N m (b) The particle’s position vector relative to the new axis is 1
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