solution_pdf - olayinka (to3377) Homework 3 Spurlock...

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Unformatted text preview: olayinka (to3377) Homework 3 Spurlock (44103) 1 This print-out should have 89 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 truck travels up a hill with a 6 . 2 incline. The truck has a constant speed of 25 m / s. What is the horizontal component of the trucks velocity? Correct answer: 24 . 8538 m / s. Explanation: Let : v = 25 m / s and = 6 . 2 . v y v x 2 5 m / s 6 . 2 v x = v sin = (25 m / s) cos6 . 2 = 24 . 8538 m / s . 002 (part 2 of 2) 10.0 points What is the vertical component of the trucks velocity? Correct answer: 2 . 69999 m / s. Explanation: v y = v sin = (25 m / s) sin6 . 2 = 2 . 69999 m / s . 003 10.0 points A hiker makes four straight-line walks A 11 km at 79 B 23 km at 282 C 33 km at 226 D 17 km at 145 in random directions and lengths starting at position (41 km , 41 km) , A B C D How far from the starting point is the hiker after these four legs of the hike? All angles are measured in a counter-clockwise direction from the positive x-axis. Correct answer: 39 . 4705 km. Explanation: Let : ( x , y ) = (41 km , 41 km) . a x = (11 km) cos79 = 2 . 09889 km , a y = (11 km) sin79 = 10 . 7979 km , b x = (23 km) cos282 = 4 . 78202 km , b y = (23 km) sin282 = 22 . 4974 km , c x = (33 km) cos226 = 22 . 9237 km , c y = (33 km) sin226 = 23 . 7383 km , d x = (17 km) cos145 = 13 . 9256 km , d y = (17 km) sin145 = 9 . 75078 km , olayinka (to3377) Homework 3 Spurlock (44103) 2 A B C D E Scale: 10 km = x = a x + b x + c x + d x = 2 . 09889 km + (4 . 78202 km) + ( 22 . 9237 km) + ( 13 . 9256 km) = 29 . 9684 km and y = a y + b y + c y + d y = 10 . 7979 km + ( 22 . 4974 km) + ( 23 . 7383 km) + (9 . 75078 km) = 25 . 687 km , so the resultant is E = radicalBig ( x ) 2 + ( y ) 2 = radicalBig ( 29 . 9684 km) 2 + ( 25 . 687 km) 2 = 39 . 4705 km . 004 (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 45 with the positive x axis. The second has a magnitude of 7 . 9 m and makes an angle of 152 with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 152 45 1 5 m 7 . 9 m Find the magnitude of the third displace- ment. Correct answer: 14 . 7688 m. Explanation: Let : bardbl vector A bardbl = 15 m , a = 45 , bardbl vector B bardbl = 7 . 9 m , and B = 152 . C A B A B C C vector A + vector B + vector C = 0 , so vector C = vector A vector B C x = A x B x = A cos A B cos b = (15 m) cos45 (7 . 9 m) cos152 olayinka (to3377) Homework 3 Spurlock (44103) 3 = 3 . 63132 m and C y = A y B x = A sin A B sin...
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solution_pdf - olayinka (to3377) Homework 3 Spurlock...

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