{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern