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HW7_Solutions

# HW7_Solutions - hiarker(srh959 homework 07 Turner(58220...

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hiarker (srh959) – homework 07 – Turner – (58220) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A particle undergoes three displacements. The frst has a magnitude oF 15 m and makes an angle oF 35 with the positive x axis. The second has a magnitude oF 8 m and makes an angle oF 149 with the positive x axis. (see the fgure below). AFter the third displacement the particle returns to its initial position. 149 35 15 m 8 m ±ind the magnitude oF the third displace- ment. Correct answer: 13 . 8341 m. Explanation: Given : b v A b = 15 m , θ a = 35 , b v B b = 8 m , and θ B = 149 . θ C θ A θ B A B C Since v A + v B + v C = 0 , we have v C = v A v B . The components oF the third displacement v C are C x = A x B x = A cos θ A B cos θ b = (15 m) cos 35 (8 m) cos149 = (12 . 2873 m) ( 6 . 85734 m) = 5 . 42994 m and C y = A y B x = A sin θ A B sin θ b = (15 m) sin 35 (8 m) sin149 = (8 . 60365 m) (4 . 1203 m) = 12 . 724 m . The magnitude oF v C is b v C b = r C 2 x + C 2 y = r ( 5 . 42994 m) 2 + ( 12 . 724 m) 2 = 13 . 8341 m . 002 (part 2 oF 2) 10.0 points ±ind the angle oF the third displacement (mea- sured From the positive x axis, with counter- clockwise positive within the limits oF 180 to +180 ).

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hiarker (srh959) – homework 07 – Turner – (58220) 2 Correct answer: 113 . 11 . Explanation: tan θ C = C y C x θ C = arctan b C y C x B = arctan b ( 12 . 724 m) ( 5 . 42994 m) B = 113 . 11 . 003 (part 1 of 2) 10.0 points Consider two vectors v A and v B and their re- sultant v A + v B . The magnitudes of the vectors v A and v B are, respectively, 15 . 8 and 6 . 7 and they act at 96 to each other.
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HW7_Solutions - hiarker(srh959 homework 07 Turner(58220...

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