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Unformatted text preview: morris (gmm643) – HW 2 – Coker – (56625) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. In class we did not cover the part of Ch. 3 dealing with vector products, and vector products will not be covered on this home work or on Quiz 1. Vector products will be in troduced in later chapters as they are needed for the concepts of work and torque. 001 10.0 points A hiker makes three straightline walks A 18 km at 249 ◦ B 24 km at 32 ◦ C 27 km at 126 ◦ in random directions and lengths starting at position (41 km , 41 km) , listed below and shown below in the plot. A B C Scale: 10 km = Figure: Drawn to ok. Which vector will return the hiker to the starting point? All angles are measured in a counterclockwise direction from the positive xaxis. 1. bardbl vector D bardbl = 29 . 8432 km , θ d = 129 . 242 ◦ 2. bardbl vector D bardbl = 48 . 5917 km , θ d = 262 . 812 ◦ 3. bardbl vector D bardbl = 13 . 4911 km , θ d = 287 . 152 ◦ 4. bardbl vector D bardbl = 17 . 8657 km , θ d = 276 . 323 ◦ cor rect 5. bardbl vector D bardbl = 37 . 383 km , θ d = 353 . 36 ◦ 6. bardbl vector D bardbl = 43 . 2297 km , θ d = 73 . 2102 ◦ 7. bardbl vector D bardbl = 66 . 4533 km , θ d = 90 . 7368 ◦ 8. bardbl vector D bardbl = 64 . 1863 km , θ d = 192 . 747 ◦ 9. bardbl vector D bardbl = 38 . 1275 km , θ d = 77 . 7664 ◦ 10. bardbl vector D bardbl = 60 . 3069 km , θ d = 64 . 3143 ◦ Explanation: Δ a x = (18 km) cos249 ◦ = − 6 . 45059 km , Δ a y = (18 km) sin249 ◦ = − 16 . 8045 km , Δ b x = (24 km) cos32 ◦ = 20 . 3532 km , Δ b y = (24 km) sin32 ◦ = 12 . 7181 km , Δ c x = (27 km) cos126 ◦ = − 15 . 8702 km , Δ c y = (27 km) sin126 ◦ = 21 . 8434 km , A B θ d C D Scale: 10 km = Δ x = Δ a x + Δ b x + Δ c x morris (gmm643) – HW 2 – Coker – (56625) 2 = − 6 . 45059 km + (20 . 3532 km) + ( − 15 . 8702 km) Δ y = Δ a y + Δ b y + Δ c y = − 16 . 8045 km + (12 . 7181 km) + (21 . 8434 km) The resultant is D = radicalBig (Δ x ) 2 + (Δ y ) 2 = radicalBig (39 . 0323 km) 2 + (58 . 7571 km) 2 θ D = arctan vextendsingle vextendsingle vextendsingle vextendsingle 58 . 7571 km − (41 km) 39 . 0323 km − (41 km) vextendsingle vextendsingle vextendsingle vextendsingle = 276 . 323 ◦ ....
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This note was uploaded on 11/17/2010 for the course PHY 56630 taught by Professor Coker during the Spring '10 term at University of Texas.
 Spring '10
 COKER

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