Ch22_p47 - 47(a Using Eq 22-28 we find F = 8.00 105 C 3.00 103 N C i 8.00 105 C 600 N C j c hc = b0.240N g i b0.0480N g j F= 2 hc hb g Therefore

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G F × =− −− 8 00 10 8 00 10 600 0 240 0 0480 55 . # . # . # . # C 3.00 10 N C i C N C j Ni Nj . 3 c hc h c hb g b g b g Therefore, the force has magnitude equal to () ( ) 22 0.240N 0.0480N 0.245N. F =+= (b) The angle the force F G makes with the + x axis is 11 0.0480N tan tan 11.3 0.240N y x F F θ §· == = ° ¨¸ ©¹ measured counterclockwise from the + x axis. (c) With m = 0.0100 kg, the ( x , y ) coordinates at t = 3.00 s can be found by combining Newton’s second law with the kinematics equations of Chapters 2–4. The
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Fall '10 term at Kansas State University.

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