P23_042

# P23_042 - 42. (a) Using Eq. 23-28, we ﬁnd F = = (8.00 ×...

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Unformatted text preview: 42. (a) Using Eq. 23-28, we ﬁnd F = = (8.00 × 10−5 C)(3.00 × 103 N/C)ˆ + (8.00 × 10−5 C)(−600 N/C)ˆ i j (0.240 N)ˆ − (0.0480 N)ˆ . i j Therefore, the force has magnitude equal to F= (0.240 N)2 + (0.0480 N)2 = 0.245 N , and makes an angle θ (which, if negative, means clockwise) measured from the +x axis, where θ = tan−1 Fy Fx = tan−1 −0.0480 N 0.240 N = −11.3◦ . (b) With m = 0.0100 kg, the coordinates (x, y ) at t = 3.00 s are found by combining Newton’s second law with the kinematics equations of Chapters 2-4: x= y = 1 Fx t 2 (0.240)(3.00)2 ax t 2 = = = 108 m , 2 2m 2(0.0100) Fy t 2 (−0.0480)(3.00)2 12 ay t = = = −21.6 m . 2 2m 2(0.0100) ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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