HW_2_Solutions_S06 - Physics 112 HW#2 Solutions 1-47[Vector Components(a | A |= | B |=(b(c(4.00)2(3.00)2 = 5.00(5.00)2(2.00)2 = 5.39(d Spring 2006

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Physics 112 - HW #2 Solutions Spring 2006 1-47 [Vector Components] (d) (a) | A |= (4.00) 2 + (3.00) 2 = 5.00 | B |= (5.00) 2 + (2.00) 2 = 5.39 (b) A - B = -1.00 i + 5.00 j (c) | A - B |= (1.00) 2 + (5.00) 2 = 5.10 1-51 [Dot Product & Angles] (a) A B =(4.00 × 5.00+3.00 × (-2.00))=14.0 (b) A B =| A || B |cos( θ ) cos θ = A B | A || B | = 14.0 (5.00) (5.39) = 0.519 θ = 59° - 1 -
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1.66 [Vector Addition] () ( ) 344 . 0 km 311.5 km 2 . 107 tan km 330 km 2 . 107 km 5 . 311 km 2 . 107 48 sin km) 230 ( 68 cos km) 170 ( km 5 . 311 48 cos km) 230 ( 68 sin km) 170 ( 2 2 2 2 = = = = + = + = = = + = = + = + = x y R y x y y y x x x R R θ R R R B A R B A R D D D D east of south 19 D = R θ 2.19 [Motion Diagrams] - 2 -
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b) 2.35: a) b) From the graph (Fig. (2.35)), the curves for A and B intersect at t = 1 s and t = 3 s. c) d) From Fig. (2.35), the graphs have the same slope at t = 2 s . e) Car A passes car B when they have the same position and the slope of curve A is greater than that of curve B in Fig. (2.30); this is at t = 3 s. f) Car B passes car A when they have the same position and the slope of curve B is greater than that of curve A ; this is at t = 1 s. - 3 -
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N = +y S E = +x W 3-81 Let "p" = plane, "a" = air, and "g" = ground: (a) In x-y (E-N) components: v p/a = (-220 km/h , 0) v p/g = -120 km 0.50 h , -20 km 0.50 h = (-240 km/h , -40 km/h) v p/a v p/g v a/g v p/a + v a/g = v p/g v a/g = v p/g - v p/a = (-240 km/h - (-220 km/h) , -40 km/h) = (-20 km/h , -40 km/h) wind speed = | v a/g | = (-20 km/s) 2 + (-40 km/h) 2 = 45 km/h θ v a/g tan θ = 20 km/h 40 km/h = 0.50 θ v a/g v p/a v p/g θ = 27° W of S (b) sin θ = v a/g v p/a = 40 km/h 220 km/h = 0.182 θ = 10.5° N of W - 4 -
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- 5 - 3-93 Relative Velocity/Reference Frames (a) The easy way to approach this problem is in the reference frame of the river (water). You can imagine the river to act like a conveyor belt carrying the bottle along with it at a steady speed. In this reference frame the bottle is stationary, so the canoe travels upstream past the bottle the same distance relative to the river that it travels downstream relative to the river to reach the bottle again. Since the canoe always travels at the same constant speed relative to the river , it must take the same amount of time to travel upstream beyond the bottle as it does to get downstream back to the bottle, or 60 minutes each way.
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This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

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HW_2_Solutions_S06 - Physics 112 HW#2 Solutions 1-47[Vector Components(a | A |= | B |=(b(c(4.00)2(3.00)2 = 5.00(5.00)2(2.00)2 = 5.39(d Spring 2006

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