64_Physics ProblemsTechnical Physics

64_Physics ProblemsTechnical Physics - Chapter 3 P3.33 e j...

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Chapter 3 65 P3.33 d 1 350 =− . ± j ej m d 2 820 450 580 = + .c o s. ± .s i n. ± . ± . ± ij i j m d 3 15 0 . ± i m Ri j i j =++= − + + + ddd 123 15 0 5 80 5 80 3 50 9 20 2 30 .. ± ± . ± . ± a f a f m (or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is R =+ = + = RR xy 22 920 230 948 . af a f m . The direction is θ = F H G I K J arctan . . 166 . P3.34 Refer to the sketch RABC i j i R + = 10 0 15 0 50 0 40 0 15 0 40 0 15 0 42 7 12 . ± . ± . ± . ± . ± . a f yards A = 10 0 . B = 15 0 . C = 50 0 . R FIG. P3.34 P3.35 (a) FF F Fijij Fi jij i j F ° −+= + = = F H G I K J 1 120 60 0 120 60 0 80 0 75 0 80 0 75 0 60 0 104 20 7 77 3 39 3 181 39 3 181 185 181 39 3 77 8 cos
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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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