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RECITATION-CH 22-2

# RECITATION-CH 22-2 - RECITATION 2 11(a With a understood to...

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RECITATION 2 11. (a) With a understood to mean the magnitude of acceleration, Newton’s second and third laws lead to ( 29 ( 29 7 2 7 2 6.3 10 kg 7.0 m s 4.9 10 kg. 9.0 m s B B A A B m a m a m - - × = = = × (b) The magnitude of the force on the first particle (A) is ( 29 ( 29 ( 29 2 2 2 2 2 B A 7 9 2 2 2 11 so ( ) 6.3 10 kg 7.0 m s 8.99 10 N m C (0.0032 m) 7.1 10 C. A B B A A A q q q F m a k k r r q F q k - - = = = = = × = × = × Therefore, | q | = 7.1 x 10 –11 C 30. We take rightwards to be the positive x -direction and label the particles: A (for the one with the +2 e charge), B ( -e ), C (+ e ) and D (+4 e ). (a) We consider the net force on A . B A F r points rightward (and therefore has a positive x -component) since A is “attracted” to B . C A F r and D A F r both point leftward (with a negative x -component) since A is “repelled” by C and D . Thus, using d = 0.0200 m, the x -component of the net force, ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 net 25 2 2 2 2 2 2 4 3.52 10 N. 2 3 A x B A x C A x D A x e e e e e e F F F F k k k d d d - = + + = - - = + × (b) We now consider the net force on B . We note that A B F r

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