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# solution_pdf - patel (np6453) – Homework 3 – Spurlock...

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Unformatted text preview: patel (np6453) – 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 9 . 7 ◦ incline. The truck has a constant speed of 19 m / s. What is the horizontal component of the truck’s velocity? Correct answer: 18 . 7284 m / s. Explanation: Let : v = 19 m / s and θ = 9 . 7 ◦ . v y v x 1 9 m / s 9 . 7 ◦ v x = v sin θ = (19 m / s) cos9 . 7 ◦ = 18 . 7284 m / s . 002 (part 2 of 2) 10.0 points What is the vertical component of the truck’s velocity? Correct answer: 3 . 2013 m / s. Explanation: v y = v sin θ = (19 m / s) sin9 . 7 ◦ = 3 . 2013 m / s . 003 10.0 points A hiker makes four straight-line walks A 24 km at 35 ◦ B 14 km at 284 ◦ C 21 km at 31 ◦ D 15 km at 319 ◦ 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: 52 . 3805 km. Explanation: Let : ( x , y ) = (41 km , 41 km) . Δ a x = (24 km) cos 35 ◦ = 19 . 6596 km , Δ a y = (24 km) sin35 ◦ = 13 . 7658 km , Δ b x = (14 km) cos 284 ◦ = 3 . 38694 km , Δ b y = (14 km) sin284 ◦ = − 13 . 5841 km , Δ c x = (21 km) cos 31 ◦ = 18 . 0005 km , Δ c y = (21 km) sin31 ◦ = 10 . 8158 km , Δ d x = (15 km) cos 319 ◦ = 11 . 3207 km , Δ d y = (15 km) sin319 ◦ = − 9 . 84086 km , patel (np6453) – Homework 3 – Spurlock – (41003) 2 A B C θ e D E Scale: 10 km = Δ x = Δ a x + Δ b x + Δ c x + Δ d x = 19 . 6596 km + (3 . 38694 km) + 18 . 0005 km + (11 . 3207 km) = 52 . 3678 km and Δ y = Δ a y + Δ b y + Δ c y + Δ d y = 13 . 7658 km + ( − 13 . 5841 km) + 10 . 8158 km + ( − 9 . 84086 km) = 1 . 15665 km , so the resultant is E = radicalBig (Δ x ) 2 + (Δ y ) 2 = radicalBig (52 . 3678 km) 2 + (1 . 15665 km) 2 = 52 . 3805 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 35 ◦ with the positive x axis. The second has a magnitude of 7 . 9 m and makes an angle of 140 ◦ with the positive x axis. (see the figure below). After the third displacement the particle returns to its initial position. 140 ◦ 35 ◦ 1 5 m 7 . 9 m Find the magnitude of the third displace- ment. Correct answer: 15 . 0356 m. Explanation: Let : bardbl vector A bardbl = 15 m , θ a = 35 ◦ , bardbl vector B bardbl = 7 . 9 m , and θ B = 140 ◦ . θ C θ A θ 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 θ A − B cos θ b = − (15 m) cos35 ◦ − (7 . 9 m) cos140 ◦ = − 6 . 23553 m and C y = − A y − B x patel (np6453) – Homework 3 – Spurlock – (41003) 3 = − A sin θ A − B sin θ b = − (15 m) sin35 ◦ − (7 . 9 m) sin140...
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## This note was uploaded on 10/25/2010 for the course PHYSICS 1441 taught by Professor Spurlock during the Fall '10 term at UT Arlington.

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solution_pdf - patel (np6453) – Homework 3 – Spurlock...

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