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Unformatted text preview: Patel, Kinal – Homework 4 – Due: Sep 14 2007, 7:00 pm – Inst: D Weathers 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A particle undergoes three consecutive dis placements: ~ d 1 = d 1 x ~ i + d 1 y ~ j + d 1 z ~ k ~ d 2 = d 2 x ~ i + d 2 y ~ j + d 2 z ~ k ~ d 3 = d 3 x ~ i + d 3 y ~ j , where d 1 x = 1 . 39 cm , d 1 y = 2 . 19 cm , d 1 z = 1 . 94 cm , d 2 x = 3 cm , d 2 y = 1 . 58 cm , d 2 z = 4 . 51 cm , d 3 x = 2 . 37 cm , d 3 y = 2 . 29 cm . Find the x component of the resultant dis placement. Correct answer: 2 . 02 cm. Explanation: The resultant vector ~ R is given by ~ R = ~ d 1 + ~ d 2 + ~ d 3 = ( d 1 x + d 2 x + d 3 x ) ~ i + ( d 1 y + d 2 y + d 3 y ) ~ j + ( d 1 z + d 2 z + d 3 z ) ~ k = [(1 . 39 cm) + (3 cm) + ( 2 . 37 cm)] ~ i + [(2 . 19 cm) + ( 1 . 58 cm) + (2 . 29 cm)] ~ j + [( 1 . 94 cm) + ( 4 . 51 cm) + (0 cm)] ~ k = [2 . 02 cm] ~ i + [2 . 9 cm] ~ j + [ 6 . 45 cm] ~ k so the answer to this part of the problem is R x = 2 . 02 cm . 002 (part 2 of 2) 10 points Find the magnitude of the resultant displace ment. Correct answer: 7 . 35479 cm. Explanation: The magnitude of ~ R is R = q R 2 x + R 2 y + R 2 z = q (2 . 02 cm) 2 + (2 . 9 cm) 2 + ( 6 . 45 cm) 2 = 7 . 35479 cm . keywords: 003 (part 1 of 2) 10 points Consider four vectors ~ F 1 , ~ F 2 , ~ F 3 , and ~ F 4 , where their magnitudes are F 1 = 57 N, F 2 = 39 N, F 3 = 30 N, and F 4 = 40 N. Let θ 1 = 150 ◦ , θ 2 = 150 ◦ , θ 3 = 37 ◦ , and θ 4 = 57 ◦ , measured from the positive x axis with the counterclockwise angular direction as positive. What is the magnitude of the resultant vec tor ~ F , where ~ F = ~ F 1 + ~ F 2 + ~ F 3 + ~ F 4 ?...
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This note was uploaded on 10/22/2010 for the course PHYS PHYS 1710 taught by Professor Weathers during the Fall '07 term at North Texas.
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