QTest #3-solutions - Version 127/ABDDD QTest#3...

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Version 127/ABDDD – QTest #3 – Antoniewicz – (56445) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charged particle is acted upon by a force of magnitude 120N over a time t. Given that the initial momentum is (15 , 10 , 0) kg.m/s and that the final momen- tum is (20 , 20 , - 10) kg.m/s , find t and the force F (in Newtons). 1. 0.125s, (40 , 80 , 80) N 2. 0.25s, (20 , 40 , - 40) N 3. 0.125s, (80 , 40 , 40) N 4. 0.25s, (20 , - 40 , 40) N 5. 0.125s, ( - 80 , 40 , 80) N 6. 0.125s, ( - 40 , - 80 , 80) N 7. 0.125s, (40 , 80 , - 80) N correct Explanation: Given that p i = (15 , 10 , 0) kg.m/s p f = (20 , 20 , - 10) kg.m/s | F | = 120 N We use the formula for the magnitude of the impulse = | p f - p i | = | F | t t = ( | p f - p i | ) | F | = 0 . 125 s F = ( p f - p i ) t = (40 , 80 , - 80) N 002 10.0 points We are given a velocity vector v = ( - 22 . 3 , 0 . 42 , - 19 . 5) m/s Calculate the unit vector for the velocity. 1. ( - 0 . 891 , 0 . 045 , - 0 . 452) m/s 2. ( - 0 . 761 , 0 . 032 , - 0 . 648) m/s 3. ( - 0 . 753 , 0 . 014 , - 0 . 658) m/s correct 4. ( - 0 . 852 , 0 . 063 , +0 . 537) m/s 5. ( - 0 . 891 , 0 . 045 , +0 . 452) m/s 6. ( - 0 . 753 , 0 . 014 , +0 . 658) m/s 7. (0 . 761 , - 0 . 032 , - 0 . 648) m/s Explanation: Let : x = - 22 . 3 , y = 0 . 42 , and z = - 19 . 5 .
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