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A novice skier, starting from rest, slides down a

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

Solution for problem 34P Chapter 6

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 34P

A novice skier, starting from rest, slides down a frictionless 35.0° incline whose vertical height is 185 m. How fast is she going when she reaches the bottom?

Step-by-Step Solution:

Solution Step 1 of 3 The acceleration due to gravity at the surface is g and it is given by, GM g = 2 R Where G is the universal gravitational constant, M is the mass and R is the radius. Similarly for the planet The acceleration due to gravity at the surface of the newly discovered planet using above expression, GM planet g planet R 2 planet Step 2 of 3 (a) Mars The acceleration due to gravity at the surface of the mars is g and it is given by, GM m g m R m Where G is the universal gravitational constant, M m is the mass and R ms the radius of the mars. Given data, 23 Mass of mars, M = 6.4m × 10 kg 6 Radius of the marst, R =3.3m × 10 m G=6.67 × 10 11 Nm /kg 2 To find the acceleration due to gravity on the surface of mars,g = m Using this data in above equation, (6.67 ×101Nm /kg )(6.42×10 kg) g m 2 (3.37×10 m) g m3.77 m/s 2 Therefore, the freely falling acceleration on the surface of mars is 3.77 m/s . 2

Step 3 of 3

Chapter 6, Problem 34P is Solved
Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

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