677_Physics ProblemsTechnical Physics

677_Physics ProblemsTechnical Physics - Chapter 23 P23.46...

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Chapter 23 17 P23.46 The acceleration is given by vv a xx fi 22 2 =+ di or va h f 2 02 =+ − a f . Solving a v h f =− 2 2 . Now Fa = m : −+ = mg q mv h f ± ± jE j 2 2 . Therefore q mv h mg f Ej + F H G I K J 2 2 ± . (a) Gravity alone would give the bead downward impact velocity 2 9 80 5 00 9 90 .. . ms m 2 ej af = . To change this to 21.0 m/s down, a downward electric field must exert a downward electric force. (b) q m E v h g f F H G I K J = × × F H G I K J L N M M O Q P P = 2 3 2 2 100 10 21 0 2500 980 343 . . . kg 1.00 10 N C Ns kg m m ±ms ±C 4 2 2 bg µ P23.47 (a) t x v x == × = 00500 450 10 111 10 111 5 7 . . . s n s (b) a qE m y ×× × 1602 10 960 10 167 10 921 10 19 3 27 11 . . e j 2 yyv t a t fiy i y −= + 1 2 2 : y f ×= −− 1 2 9 21 10 1 11 10 5 68 10 5 68 11 7 2 3 . . m m m (c) v x 5 m s vva t yf yi y =+= × × = × 9 21 10 111 10 1 02 10 11 7 5 . *P23.48 The particle feels a constant force: FE j j × = × q 1 10 2 000 2 10 63 C N C N ±± and moves with acceleration: a F j j ×⋅ ×
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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