HW3-solutions - Agha (sa9896) – HW3 – bohm – (59970)...

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Unformatted text preview: Agha (sa9896) – HW3 – bohm – (59970) 1 This print-out should have 11 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 pedestrian moves 4 km east and then 17 km north. Find the magnitude of the resultant dis- placement vector. Correct answer: 17 . 4642 km. Explanation: Using the Pythagorean theorem, we find the magnitude of the displacement vector: R = radicalbig x 2 + y 2 The direction is given by: θ = arctan parenleftBig y x parenrightBig 002 (part 2 of 2) 10.0 points What is the direction of the displacement vec- tor (using the counter-clockwise angular di- rection to be positive, within the limits of − 180 ◦ to +180 ◦ )? Correct answer: 76 . 7595 ◦ . Explanation: 003 10.0 points All angles are measured in a counter- clockwise direction from the positive x-axis. A hiker makes four straight-line walks ( A , B , C , and D ) in random directions and lengths starting at position (41 km , 41 km) , listed below and shown below in the plot. A 24 km at 62 ◦ B 33 km at 180 ◦ C 23 km at 90 ◦ D 16 km at 30 ◦ A B C D Figure: Drawn to scale. How far from the starting point is the hiker after these four legs of the hike? Correct answer: 52 . 7817 km. Explanation: A B C θ e D E Scale: 10 km = Note: bardbl vector E bardbl = 52 . 7817 km and θ e = 278 . 582 ◦ . Δ a x = (24 km) cos62 ◦ = 11 . 2673 km , Δ a y = (24 km) sin62 ◦ = 21 . 1907 km , Δ b x = (33 km) cos180 ◦ = − 33 km , Δ b y = (33 km) sin180 ◦ = − 4 . 44315 × 10 − 5 km , Agha (sa9896) – HW3 – bohm – (59970) 2 Δ c...
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This note was uploaded on 11/03/2008 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.

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HW3-solutions - Agha (sa9896) – HW3 – bohm – (59970)...

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