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OH-09-preview

# OH-09-preview - Version PREVIEW OH-09 Bradley(191 This...

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Version PREVIEW – OH-09 – Bradley – (191) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No. 09. It is due before 06:00 AM on Tuesday, 28 October, UAE time (09:00 PM on Monday, 27 October, in Texas), unless otherwise notified. Walking in a Boat 001 10.0 points A(n) 47 kg boat that is 7 . 2 m in length is initially 7 . 4 m from the pier. A 28 kg child stands at the end of the boat closest to the pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle. Assume: There is no friction between boat and water. 7 . 4 m 7 . 2 m How far is the child from the pier when she reaches the far end of the boat? Correct answer: 11 . 912 m. Explanation: Let D = distance of the boat from the pier , = 7 . 4 m , L = length of the boat , = 7 . 2 m , M = mass of the boat , = 47 kg , and m = mass of the child , and = 28 kg , X = change in position of the boat . = final distance of child from pier . d = final distance of boat from pier . D L X d X We will use the pier as the origin of the x -coordinate. Initially, the center of mass of the boat-child system is x cm = parenleftbigg D + L 2 parenrightbigg M + D m M + m Finally, the center of mass of the boat-child system is x cm = parenleftbigg D + L 2 - X parenrightbigg M + ( D + L - X ) m M + m , where X is the change in position of the center of mass of the boat. Since the center of mass of the system does not move, we can equate the above two expressions for x cm parenleftbigg D + L 2 parenrightbigg M + D m M + m = parenleftbigg D + L 2 - X parenrightbigg M + ( D + L - X ) m M + m and, solving for X , we have parenleftbigg D + L 2 parenrightbigg M + D m = parenleftbigg D + L 2 - X parenrightbigg M + ( D + L - X ) m 0 = -X M + ( L - X ) m X ( M + m ) = L m

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Version PREVIEW – OH-09 – Bradley – (191) 2 X = m m + M L = (28 kg) (28 kg) + (47 kg) × (7 . 2 m) = 2 . 688 m . The final distance of the child from the pier is = D + L - X = (7 . 4 m) + (7 . 2 m) - (2 . 688 m) = 11 . 912 m . The final distance d of the near end of the boat to the pier is d = D - X = (7 . 4 m) - (2 . 688 m) = 4 . 712 m . Water Fills a Bucket 002 10.0 points Water ( ρ = 1000 kg / m 3 ) falls without splash- ing at a rate of 0 . 191 L / s from a height of 76 . 8 m into a 1 . 06 kg bucket on a scale. The acceleration of gravity is 9 . 8 m / s 2 . If the bucket is originally empty, what does the scale read after 3 . 08 s ? Correct answer: 23 . 5636 N. Explanation: From conservation of energy, the velocity is m g h = 1 2 m v 2 v = radicalbig 2 g h = radicalBig 2 (9 . 8 m / s 2 ) (76 . 8 m) = 38 . 7979 m / s . u is the volume rate of the flow, so the mass rate of the flow is u m = ρ u = (1000 kg / m 3 )(0 . 191 L / s) parenleftbigg 1 m 100 cm parenrightbigg 3 × 1000 cm 3 1 L = 0 . 191 kg / s . The force exerted by the water hitting the bucket is F w = m v t = u m t v t = u m v = (0 . 191 kg / s) (38 . 7979 m / s) = 7 . 41041 N .
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