solution_pdf - alzaitoun (aa39383) Homework 3 Spurlock...

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Unformatted text preview: alzaitoun (aa39383) Homework 3 Spurlock (41003) 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 13 incline. The truck has a constant speed of 23 m / s. What is the horizontal component of the trucks velocity? Correct answer: 22 . 4105 m / s. Explanation: Let : v = 23 m / s and = 13 . v y v x 2 3 m / s 13 v x = v sin = (23 m / s) cos 13 = 22 . 4105 m / s . 002 (part 2 of 2) 10.0 points What is the vertical component of the trucks velocity? Correct answer: 5 . 17388 m / s. Explanation: v y = v sin = (23 m / s) sin13 = 5 . 17388 m / s . 003 10.0 points A hiker makes four straight-line walks A 23 km at 79 B 18 km at 172 C 33 km at 98 D 26 km at 22 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: 67 . 7742 km. Explanation: Let : ( x , y ) = (41 km , 41 km) . a x = (23 km) cos79 = 4 . 38859 km , a y = (23 km) sin79 = 22 . 5774 km , b x = (18 km) cos172 = 17 . 8248 km , b y = (18 km) sin172 = 2 . 50509 km , c x = (33 km) cos98 = 4 . 59274 km , c y = (33 km) sin98 = 32 . 6788 km , d x = (26 km) cos22 = 24 . 1068 km , d y = (26 km) sin22 = 9 . 73978 km , alzaitoun (aa39383) Homework 3 Spurlock (41003) 2 A B C e D E Scale: 10 km = x = a x + b x + c x + d x = 4 . 38859 km + ( 17 . 8248 km) + 4 . 59274 km + (24 . 1068 km) = 6 . 07781 km and y = a y + b y + c y + d y = 22 . 5774 km + (2 . 50509 km) + 32 . 6788 km + (9 . 73978 km) = 67 . 5011 km , so the resultant is E = radicalBig ( x ) 2 + ( y ) 2 = radicalBig (6 . 07781 km) 2 + (67 . 5011 km) 2 = 67 . 7742 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 29 with the positive x axis. The second has a magnitude of 8 . 2 m and makes an angle of 132 with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 132 29 1 5 m 8 . 2 m Find the magnitude of the third displace- ment. Correct answer: 15 . 3916 m. Explanation: Let : bardbl vector A bardbl = 15 m , a = 29 , bardbl vector B bardbl = 8 . 2 m , and B = 132 . 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) cos29 (8 . 2 m) cos132 = 7 . 63242 m and C y = A y B x = A sin A B sin b alzaitoun (aa39383) Homework 3 Spurlock (41003) 3 = (15 m) sin29 (8 . 2 m) sin132...
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This note was uploaded on 10/07/2010 for the course DD 1441 taught by Professor Spu during the Spring '10 term at University of Houston.

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solution_pdf - alzaitoun (aa39383) Homework 3 Spurlock...

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