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hw2 - cerutti(cpc566 HW02 TSOI(58160 This print-out should...

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cerutti (cpc566) – HW02 – TSOI – (58160) 1 This print-out should have 25 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 car travels 12 . 2 km due north and then 35 . 1 km in a direction φ = 45 . 2 west of north. β θ φ A B R x y N E S W Find the magnitude of the car’s resultant displacement. Correct answer: 44 . 5458 km. Explanation: Let : A = 12 . 2 km , B = 35 . 1 km , and φ = 45 . 2 . θ = 180 - φ = 180 - 45 . 2 = 134 . 8 , applying the Law of Cosines, R 2 = A 2 + B 2 - 2 A B cos θ . Since - 2 AB cos θ = - 2 (12 . 2 km)(35 . 1 km) cos 134 . 8 = 603 . 477 km 2 , then R = radicalBig (12 . 2 km) 2 + (35 . 1 km) 2 + 603 . 477 km 2 = 44 . 5458 km . 002 (part 2 of 2) 10.0 points Calculate the direction of the car’s resul- tant displacement, measured counterclock- wise from the northerly direction. Correct answer: 33 . 9941 . Explanation: Applying the Law of Sines, sin β B = sin θ R sin β = B R sin θ = 35 . 1 km 44 . 5458 km sin 134 . 8 = 0 . 559108 β = arcsin 0 . 559108 = 33 . 9941 . 003 (part 1 of 2) 10.0 points A river flows at a speed v r = 5 . 36 km / hr with respect to the shoreline. A boat needs to go perpendicular to the shoreline to reach a pier on the river’s other side. To do so, the boat heads upstream at an angle θ = 39 from the direction to the boat’s pier. Find the ratio of v b to v r , where v r is defined above and v b is the boat’s speed with respect to the water. 1. v b v r = sin 2 θ 2. v b v r = tan θ 3. v b v r = cos 2 θ 4. v b v r = 1 tan θ 5. v b v r = 1 sin θ correct 6. v b v r = 1 cos θ 7. v b v r = 1 cos 2 θ 8. v b v r = sin θ 9. v b v r = cos θ

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cerutti (cpc566) – HW02 – TSOI – (58160) 2 10. v b v r = 1 sin 2 θ Explanation: Let : v bs =? , boat relative to shore v ws = v r , water relative to shore v bs = v b , boat relative to water . This is a problem about relative velocity. See the figure below. L v bs v bw v ws θ The velocity of the boat relative to the shore is given by -→ v bs = -→ v bw + -→ v ws . It is easy to see from the figure above that v ws , v bw and v bs form a right triangle if the boat moves northward relative to the earth. Therefore, v ws = v bw sin θ = v bw v ws = 1 sin θ . 004 (part 2 of 2) 10.0 points If the time taken for the boat to cross the river is 13 . 9 min, determine the width of the river. Correct answer: 1 . 53342 km. Explanation: By similar reasoning, we know relative to the shore, the velocity of the boat is v bs = v ws tan θ = 6 . 61905 km / hr . Therefore the time taken for the boat to cross the river is given by t = L v bs = L v ws tan θ = L tan θ v ws . = L = v ws t tan θ = (5 . 36 km / hr) (13 . 9 min) (1hr / 60 min) tan 39 = (5 . 36 km / hr) (0 . 231667 hr) tan 39 = 1 . 53342 km . 005 (part 1 of 4) 10.0 points Denote the initial speed of a cannon ball fired from a battleship as v 0 . When the initial projectile angle is 45 with respect to the horizontal, it gives a maximum range R . y x θ 45 R R/ 2 The time of flight t max of the cannonball for this maximum range R is given by 1. t max = 2 v 0 g 2. t max = 3 v 0 g 3. t max = 1 2 v 0 g 4. t max = 1 2 v 0 g 5. t max = 1 3 v 0 g 6. t max = 2 v 0 g correct 7. t max = 2 3 v 0 g
cerutti (cpc566) – HW02 – TSOI – (58160) 3 8. t max = 4 v 0 g 9. t max = v 0 g 10. t max = 1 4 v 0 g Explanation:

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hw2 - cerutti(cpc566 HW02 TSOI(58160 This print-out should...

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