A 65-kg trampoline artist jumps vertically upward from the

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

Problem 37P 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 37P

A 65-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0 m/s. (a) How fast is he going as he lands on the trampoline, 3.0 m below (Fig. 6–38)? (b) If the trampoline behaves like a spring with spring stiffness constant 6.2 × 104N/m, how far does he depress it?

FIGURE 6–38

Step-by-Step Solution:

Solution Step 1 of 5 The quantities like orbital period, speed and acceleration of the space shuttle revolving around the earth, does not depend on the mass of the space shuttle. It depends on the mass of the central body around which the shuttle is revolving and the radius of the orbit. So we can calculate the above physical quantities using the given information; height of the orbit from earth surface. Orbital time period The time period in the given orbit is the time taken by the space shuttle to complete one full revolution around earth. The relation between the orbital time period T and the distance of space shuttle r from the center of the earth, 2 23 T = GM r Where T is the orbital time period, r is the radius of the orbit, G is gravitational constant and M is the mass of the earth. From the above equation, the orbital time period T is, 3/2 T = 2r GM Step 2 of 5 Space shuttle is revolving at 250 miles from the earth surface, so the total distance from the earth center will be the sum of the radius of earth and the distance of space shuttle from the surface. That is, r = R +h e Where r is the distance of shuttle from the earth center, R isethe radius of the earth and h is the height at which the shuttle is revolving around the earth. Using this in above equation, 2(Re+h)/2 T = GM Given data Height, h = 250 miles To convert height from miles to meter Using 1 mile = 1609 m h = 250 (1609)=0.40225×10 m 6 6 Radius of the earth, R =6e37×10 m 24 Mass of the earth, M=5.98×10 kg 11 2 2 G=6.67 ×10 Nm /kg To find, Orbital time period, T= Step 3 of 5 Substituting above given data in time period equation, 3/2 2(3.14)[(6.37×10 m)+(0.40225×10 m)] T = (6.67 ×101Nm /kg )(5.98×10 kg) 3 T = 5.5445×10 s Using 1 min = 60 sec T = 5.5445×10 s( 3 1 mi) 60 s T = 92.41 min Therefore, the orbital time period of the space shuttle around the earth is 92.41 min.

Step 4 of 5

Chapter 6, Problem 37P is Solved
Step 5 of 5

Textbook: Physics: Principles with Applications
Edition: 6th
Author: Douglas C. Giancoli
ISBN: 9780130606204

The answer to “A 65-kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0 m/s. (a) How fast is he going as he lands on the trampoline, 3.0 m below (Fig. 6–38)? (b) If the trampoline behaves like a spring with spring stiffness constant 6.2 × 104N/m, how far does he depress it?FIGURE 6–38” is broken down into a number of easy to follow steps, and 58 words. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. Since the solution to 37P from 6 chapter was answered, more than 272 students have viewed the full step-by-step answer. This full solution covers the following key subjects: shuttle, orbital, mile, minutes, orbit. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6th. The full step-by-step solution to problem: 37P from chapter: 6 was answered by , our top Physics solution expert on 03/03/17, 03:53PM.

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