326_Physics ProblemsTechnical Physics

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

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328 Angular Momentum *P11.5 τ =× = ° − ° × =⋅ rF 0 450 0 785 90 14 0343 .. s i n . m N u p e a s t Nm no r th 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 ±±± ±± ± ijk ij k −+ ⋅ +−= ej e j ? 894 50 −−= −≠ No . The cross product could not work out that way. P11.7 AB AB ×=⋅⇒ = = AB AB sin cos tan θθ θ 1 or 45 0 . P11.8 (a) =×= = −− −+ −=− k ±±± . ± 130 320 00 29 7 0 0 a f a f a f a f (b) The particle’s position vector relative to the new axis is 1
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