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A 78-kg person has an apparent mass of 54 kg (because of

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 26P Chapter 10

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 26P

Problem 26P

A 78-kg person has an apparent mass of 54 kg (because of buoyancy) when standing in water that comes up to the hips. Estimate the mass of each leg. Assume the body has SG = 1.00.

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ANSWER:The height of the window or from where the ball was thrown is 20 meters above the ground.The initial speed given to the ball is 10 m/s.So the initial kinetic energy is, 2 K.E initial 1/2 m v = 1/2 × m × 10 = 50 m.Initial potential energy of the block was P.E = mgh = m × 9.8 × 20 m = 196 m. initialFinal kinetic energy K.E final= 0, as the velocity will be zero there.. The final potential energy is P.E final= mg(20 + h), where h is the height above the window. a) The ball’s maximum height above the ground.The conservation of mechanical energy tells that, P.E initial K.E initial P.E final+ K.E final 196 m + 50 m = mg(20 + h) + 0 246 m = mg(20 + h) + 0 246 = g(20 + h) (20 + h) = 246/g = 246/9.8 = 25.10 = 25meters (approximately). h = 25 20 = 5meters.So, the ball’s maximum height above the ground will be (20 + h) = 25meters. \nb) The ball’s speed as it passes the window on its way down.While coming down again the same laws and rules we will use.The initial speed given to the ball is 10 m/s.So the initial kinetic energy is, K.E initial 0.Initial potential energy of the block was P.E = mgh = m × 9.8 × 25 = 254.8 m. initial Final kinetic energy while crossing the window, K.E final= 1/2 m v , as the velocity will be v there.The final potential energy is P.E final= mgh = m × 9.8 × 20 = 196 m , where h is the heightof the window.The conservation of mechanical energy tells that, P.E + K.E = P.E + K.E initial initial final final 2 254.8 m + 0 = 196 m + 1/2 m v 1/2 mv = 254.8 m 196 m = 58.8 m v = 2 × 58.8 = 117.6 v = 117.6 = 10.84 = 11 m/s(approximately)So, while crossing down the window the speed of the ball will be 11 m/s.c) The speed of impact on the ground\nWhile reaching the ground, the speed will be,Again the same rules we are going to follow.The conservation of mechanical energy tells that, P.E + K.E = P.E + K.E initial initial final final 2 254.8 m + 0 = 0 + 1/2 m v 2 1/2 mv = 254.8 m v = 2 × 254.8 = 509.6 v = 509.6 = 22.57 m/s.So, just before hitting the ground the speed of the ball will be 22.57 m/s.

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Chapter 10, Problem 26P is Solved
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Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

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