122_Physics ProblemsTechnical Physics

# 122_Physics ProblemsTechnical Physics - 124 P5.14 The Laws...

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124 The Laws of Motion P5.14 F a = m reads −++−− = 2 00 2 00 5 00 3 00 45 0 3 75 . ± . ± . ± . ± . ± . ± ijiji a ej e j N m s 2 m where ± a represents the direction of a −− = 42 0 1 00 3 75 . ± . ± . ± ij a e j m s 2 m F =( )+ ( ) 42 0 1 00 22 .. N at tan . . F H G I K J 1 100 42 0 below the – x -axis Fa = 42 0 3 75 ± N at 181 m s 2 m ch . For the vectors to be equal, their magnitudes and their directions must be equal. (a) ° ± a is at 181 counterclockwise from the x -axis (b) m == 42 0 11 2 . . N 3.75 m s kg 2 (d) vv a fi t =+=+ ° 0 3 75 10 0 ms a t 181 s 2 so v f 37 5 . ms a t 1 8 1 vi j f 37 5 181 37 5 181 .c o s ± .s i n ± m s so j f =− 37 5 0 893 . ± . ± ms (c) v f =+ = 37 5 0 893
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