SOL23 - Thevalingam Donald – Homework 23 – Due midnight...

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Unformatted text preview: Thevalingam, Donald – Homework 23 – Due: Mar 21 2007, midnight – Inst: Eslami 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Three particles are located in a coordinate system as shown in the figure below. The acceleration of gravity is 9 . 8 m / s 2 . The mass 8 kg is located at the origin, the mass 3 kg is located on the x coordinate, and the mass 6 kg is located on the y coordinate. x y 6 kg 8 kg 3 kg 13 m 4 m Find the distance of the center of gravity from the origin. Correct answer: 2 . 69371 m. Explanation: The center of gravity of the x and y compo- nents from the origin respectively are x c = m 0 + m 1 x 1 m + m 1 + m 2 = (3 kg) (13 m) (8 kg) + (3 kg) + (6 kg) = 2 . 29412 m and y c = m 0 + m 2 y 2 m + m 1 + m 2 = (6 kg) (4 m) (8 kg) + (3 kg) + (6 kg) = 1 . 41176 m . Then the distance of the center of gravity from the origin is d = q x 2 c + y 2 c = q (2 . 29412 m) 2 + (1 . 41176 m) 2 = 2 . 69371 m ....
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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SOL23 - Thevalingam Donald – Homework 23 – Due midnight...

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