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# hw7(1) - lenoir(wml297 homework 07 Turner(58220 This...

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lenoir (wml297) – homework 07 – Turner – (58220) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A particle undergoes three displacements. The first has a magnitude of 15 m and makes an angle of 28 with the positive x axis. The second has a magnitude of 6 . 5 m and makes an angle of 143 with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 143 28 15 m 6 . 5 m Find the magnitude of the third displace- ment. Correct answer: 13 . 5956 m. Explanation: Given : bardbl vector A bardbl = 15 m , θ a = 28 , bardbl vector B bardbl = 6 . 5 m , and θ B = 143 . θ C θ A θ B A B C C Since vector A + vector B + vector C = 0 , we have vector C = vector A vector B . The components of the third displacement vector C are C x = A x B x = A cos θ A B cos θ b = (15 m) cos 28 (6 . 5 m) cos 143 = (13 . 2442 m) ( 5 . 19113 m) = 8 . 05308 m and C y = A y B x = A sin θ A B sin θ b = (15 m) sin 28 (6 . 5 m) sin 143 = (7 . 04207 m) (3 . 9118 m) = 10 . 9539 m . The magnitude of vector C is bardbl vector C bardbl = radicalBig C 2 x + C 2 y = radicalBig ( 8 . 05308 m) 2 + ( 10 . 9539 m) 2 = 13 . 5956 m . 002 (part 2 of 2) 10.0 points Find the angle of the third displacement (mea- sured from the positive x axis, with counter- clockwise positive within the limits of 180 to +180 ). Correct answer: 126 . 323 . Explanation: tan θ C = C y C x θ C = arctan bracketleftbigg C y C x bracketrightbigg = arctan bracketleftbigg ( 10 . 9539 m) ( 8 . 05308 m) bracketrightbigg = 126 . 323 . 003 (part 1 of 2) 10.0 points Consider two vectors vector A and vector B and their re- sultant vector A + vector B . The magnitudes of the vectors

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lenoir (wml297) – homework 07 – Turner – (58220) 2 vector A and vector B are, respectively, 14 . 6 and 8 . 3 and they act at 75 to each other.
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