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52_Physics ProblemsTechnical Physics

52_Physics ProblemsTechnical Physics - 53 Chapter 2 P2.75...

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Chapter 2 53 P2.75 The distance x and y are always related by x y L 2 2 2 + = . Differentiating this equation with respect to time, we have 2 2 0 x dx dt y dy dt + = Now dy dt is v B , the unknown velocity of B ; and dx dt v = . From the equation resulting from differentiation, we have dy dt x y dx dt x y v = F H G I K J = ( ) . B O y A α x L v x y FIG. P2.75 But y x = tan α so v v B = F H G I K J 1 tan α . When α = ° 60 0 . , v v v v B = ° = = tan . . 60 0 3 3 0 577 . ANSWERS TO EVEN PROBLEMS P2.2 (a) 2 10 7 × m s; 1 10 6 × m s; P2.24 (a) 1.88 km; (b) 1.46 km; (c) see the solution; (b) 5 10 8 × yr (d) (i) x t 1 2 1 67 = . m s 2 e j ; P2.4 (a) 50 0 . m s ; (b) 41 0 . m s (ii) x t 2 50 375 = m s m b g ; (iii) x t t 3 2 250 2 5 4 375 = m s m s m 2 b g e j . ; P2.6 (a) 27 0 . m; (e) 37 5 . m s (b) 27 0 18 0 3 00 2 . . . m m s m s 2 + + b g e j a f t t ; (c) 18 0 . m s P2.26 958 m P2.8 (a), (b), (c) see the solution; 4 6 . m s 2 ; (d) 0 P2.28 (a) x t t f = 30 0 2 . e j m; v t f = 30 0 2 . a f m s ; (b) 225 m P2.10 5.00 m P2.12 (a) 20 0 . m s; 5 00 . m s ; (b) 262 m P2.30 x x v t a t f i xf x = 1 2 2 ; 3 10 . m s P2.14 (a) see the solution;
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