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Unformatted text preview: johnson (mjj622) – 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 16 km at 301 ◦ B 19 km at 359 ◦ C 31 km at 81 ◦ 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 = 32 . 2631 km , θ d = 273 . 996 ◦ 2. bardbl vector D bardbl = 26 . 5131 km , θ d = 172 . 211 ◦ 3. bardbl vector D bardbl = 22 . 5473 km , θ d = 301 . 08 ◦ 4. bardbl vector D bardbl = 60 . 135 km , θ d = 104 . 162 ◦ 5. bardbl vector D bardbl = 12 . 3751 km , θ d = 290 . 007 ◦ 6. bardbl vector D bardbl = 36 . 114 km , θ d = 207 . 315 ◦ cor rect 7. bardbl vector D bardbl = 32 . 5179 km , θ d = 206 . 246 ◦ 8. bardbl vector D bardbl = 4 . 4293 km , θ d = 167 . 501 ◦ 9. bardbl vector D bardbl = 20 . 6618 km , θ d = 22 . 4451 ◦ 10. bardbl vector D bardbl = 45 . 172 km , θ d = 131 . 966 ◦ Explanation: Δ a x = (16 km) cos 301 ◦ = 8 . 24064 km , Δ a y = (16 km) sin 301 ◦ = − 13 . 7147 km , Δ b x = (19 km) cos 359 ◦ = 18 . 9971 km , Δ b y = (19 km) sin 359 ◦ = − . 331545 km , Δ c x = (31 km) cos 81 ◦ = 4 . 84945 km , Δ c y = (31 km) sin 81 ◦ = 30 . 6183 km , A B θ d C D Scale: 10 km = Δ x = Δ a x + Δ b x + Δ c x = 8 . 24064 km + (18 . 9971 km) johnson (mjj622) – HW 2 – Coker – (56625) 2 + (4 . 84945 km) Δ y = Δ a y + Δ b y + Δ c y = − 13 . 7147 km + ( − . 331545 km) + (30 . 6183 km) The resultant is D = radicalBig (Δ x ) 2 + (Δ y ) 2 = radicalBig (73 . 0872 km) 2 + (57 . 5721 km) 2 θ D = arctan vextendsingle vextendsingle vextendsingle vextendsingle 57 . 5721 km − (41 km) 73 . 0872 km − (41 km) vextendsingle vextendsingle vextendsingle vextendsingle = 207 . 315 ◦ ....
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This note was uploaded on 02/10/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner

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