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Unformatted text preview: This printout should have 12 questions. Multiplechoice 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 38 with the positive x axis. The second has a magnitude of 7 m and makes an angle of 144 with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 144 38 1 5 m 7 m Find the magnitude of the third displace ment. Correct answer: 14 . 7009 m. Explanation: Given : bardbl vector A bardbl = 15 m , a = 38 , bardbl vector B bardbl = 7 m , and B = 144 . C A 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 A B cos b = (15 m) cos 38 (7 m) cos144 = (11 . 8202 m) ( 5 . 66312 m) = 6 . 15704 m and C y = A y B x = A sin A B sin b = (15 m) sin 38 (7 m) sin144 = (9 . 23492 m) (4 . 1145 m) = 13 . 3494 m . The magnitude of vector C is bardbl vector C bardbl = radicalBig C 2 x + C 2 y = radicalBig ( 6 . 15704 m) 2 + ( 13 . 3494 m) 2 = 14 . 7009 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 homework 07 Turner (58185) 1 hinojosa (jlh3938) homework 07 Turner (58185) 2 to +180 ). Correct answer: 114 . 76 . Explanation: tan C = C y C x C = arctan bracketleftbigg C y C x bracketrightbigg = arctan bracketleftbigg ( 13 . 3494 m) ( 6 . 15704 m) bracketrightbigg = 114 . 76 . 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 vector A and vector B are, respectively, 18 . 9 and 8 . 3 and they act at 110 to each other....
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 Fall '08
 Turner
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