hw7(1) - lenoir(wml297 homework 07 Turner(58220 This print-out should have 12 questions Multiple-choice questions may continue on the next column or

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