Solution Found!
A 65-kg trampoline artist jumps vertically upward from the
Chapter 17, Problem 37P(choose chapter or problem)
(II) A trampoline artist jumps vertically upward from the top of a platform with a speed of
.
(a) How fast is he going as he lands on the trampoline, below (Fig.
)?
(b) If the trampoline behaves like a spring with spring stiffness constant \(6.2 \times 10^{4} \mathrm{~N} / \mathrm{m}\) how far does he depress it?
Equation Transcription:
Text Transcription:
6.2 x 10^4N/m
Questions & Answers
QUESTION:
(II) A trampoline artist jumps vertically upward from the top of a platform with a speed of
.
(a) How fast is he going as he lands on the trampoline, below (Fig.
)?
(b) If the trampoline behaves like a spring with spring stiffness constant \(6.2 \times 10^{4} \mathrm{~N} / \mathrm{m}\) how far does he depress it?
Equation Transcription:
Text Transcription:
6.2 x 10^4N/m
ANSWER: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