EBK LOOSE-LEAF VERSION OF UNIVERSE
11th Edition
ISBN: 9781319227975
Author: KAUFMANN
Publisher: VST
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Chapter 13, Problem 1CC
To determine
The interpretation that can be drawn when the speed of a Moon or Satellite is given (Kepler’s third law), and the time used by the satellite to disappear during its orbit behind its planet is measured.
Expert Solution & Answer
Answer to Problem 1CC
Solution:
The diameter of a satellite can be estimated when it is screened by a planet and the visibility of satellite reflected light is less. So, the time used by the satellite to fade away reveals its transit time and speed, through which the diameter of the satellite can be estimated.
Explanation of Solution
Introduction:
Satellite is an artificial device that is used to visualize distant objects like stars, galaxies, and other astronomical objects from the Earth’s atmosphere. It provides a signal from one side of the Earth to the other side of the Earth, and acts like a giant mirror in space. It gives clear and sharp images of the Earth, planets, and other astronomical objects.
Explanation:
When the satellite passes between humans and planets, they emerge from the planet’s shadow. There is a brief time interval in which the satellite increases, as more of its surface exposed under sunlight. During this period, the diameter of the satellite and its transit time can be calculated by this interval of its appearance and disappearance.
Conclusion:
Therefore, it is concluded that the transit time and diameter of a satellite are calculated by its appearance and disappearance.
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Please solve and answer the problem correctly please.Thank you!!
Problem Eight. A snowmobile is originally at the point with position vector 31.1 m at 95.5°
counterclockwise from the x-axis, moving with velocity 4.89 m/s at 40.0°. It moves with constant
acceleration 1.73 m/s² at 200°. After 5.00 s have elapsed, find the following.
9.) The velocity vector in m/s.
(A)=-4.38+0.185ĵ
(D) = 0.185 +4.38ĵ
(B)=0.1851-4.38ĵ
(E) = 4.38 +0.185ĵ
(C) v=-0.1851-4.38ĵ
(A)=-39.3-4.30ĵ
10.) The final position vector in meters.
(B)=39.3-4.30ĵ
(C) = -4.61 +39.3ĵ
(D) = 39.31 +4.30ĵ
(E) = 4.30 +39.3ĵ
Problem Seven. A football
receiver
running
straight
downfield at 5.60 m/s is 11.5 m
in front of the quarterback when
a pass is thrown downfield at an
angle of 35.0° above the
horizon.
8.) If the receiver never changes speed and the ball is caught at the same height from which it was
thrown, find the distance between the quarterback and the receiver when the catch is made.
(A) 21.3
(B) 17.8
(C) 18.8
(D) 19.9
(E) 67.5
Chapter 13 Solutions
EBK LOOSE-LEAF VERSION OF UNIVERSE
Ch. 13 - Prob. 1CCCh. 13 - Prob. 2CCCh. 13 - Prob. 3CCCh. 13 - Prob. 4CCCh. 13 - Prob. 5CCCh. 13 - Prob. 6CCCh. 13 - Prob. 7CCCh. 13 - Prob. 8CCCh. 13 - Prob. 9CCCh. 13 - Prob. 10CC
Ch. 13 - Prob. 11CCCh. 13 - Prob. 12CCCh. 13 - Prob. 13CCCh. 13 - Prob. 14CCCh. 13 - Prob. 1QCh. 13 - Prob. 2QCh. 13 - Prob. 3QCh. 13 - Prob. 4QCh. 13 - Prob. 5QCh. 13 - Prob. 6QCh. 13 - Prob. 7QCh. 13 - Prob. 8QCh. 13 - Prob. 9QCh. 13 - Prob. 10QCh. 13 - Prob. 11QCh. 13 - Prob. 12QCh. 13 - Prob. 13QCh. 13 - Prob. 14QCh. 13 - Prob. 15QCh. 13 - Prob. 16QCh. 13 - Prob. 17QCh. 13 - Prob. 18QCh. 13 - Prob. 19QCh. 13 - Prob. 20QCh. 13 - Prob. 21QCh. 13 - Prob. 22QCh. 13 - Prob. 23QCh. 13 - Prob. 24QCh. 13 - Prob. 25QCh. 13 - Prob. 26QCh. 13 - Prob. 27QCh. 13 - Prob. 28QCh. 13 - Prob. 29QCh. 13 - Prob. 30QCh. 13 - Prob. 32QCh. 13 - Prob. 33QCh. 13 - Prob. 34QCh. 13 - Prob. 35QCh. 13 - Prob. 36QCh. 13 - Prob. 37QCh. 13 - Prob. 38QCh. 13 - Prob. 39QCh. 13 - Prob. 40QCh. 13 - Prob. 41QCh. 13 - Prob. 42QCh. 13 - Prob. 43QCh. 13 - Prob. 44QCh. 13 - Prob. 45QCh. 13 - Prob. 46QCh. 13 - Prob. 47QCh. 13 - Prob. 48QCh. 13 - Prob. 49QCh. 13 - Prob. 50QCh. 13 - Prob. 51QCh. 13 - Prob. 52QCh. 13 - Prob. 53QCh. 13 - Prob. 54QCh. 13 - Prob. 55QCh. 13 - Prob. 56QCh. 13 - Prob. 58QCh. 13 - Prob. 59QCh. 13 - Prob. 60QCh. 13 - Prob. 61QCh. 13 - Prob. 62QCh. 13 - Prob. 63QCh. 13 - Prob. 64QCh. 13 - Prob. 65QCh. 13 - Prob. 66Q
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