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Stars and Galaxies (MindTap Course List)
10th Edition
ISBN: 9781337399944
Author: Michael A. Seeds
Publisher: Cengage Learning
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Concept explainers
Question
Chapter 1, Problem 1RQ
To determine
The changes in height and area.
Expert Solution & Answer
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Answer to Problem 1RQ
The area can be changed by zooming in to the field of view, width changes by a factor of 100 and length also changes.
Explanation of Solution
Area can be changed by zooming in and out of the field of view. Corresponding to every step size, there is an increase by a factor of 100.
The width of the picture increases by a factor of 100. Hence, the final width will be increased to 1600 m from 16 m. The length also gets altered with respect to the width.
Conclusion:
The area can be changed by zooming in to the field of view, width changes by a factor of 100 and length also changes.
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Please help me on the following question (Please refer to image below)An Olympic lifter (m = 103kg) is holding a lift with a mass of 350 kg. The barexerts a purely vertical force that is equally distributed between both hands. Each arm has amass of 9 kg, are 0.8m long and form a 40° angle with the horizontal. The CoM for each armis 0.5 m from hand. Assuming the lifter is facing us in the diagram below, his right deltoidinserts 14cm from the shoulder at an angle of 13° counter-clockwise from the humerus.A) You are interested in calculating the force in the right deltoid. Draw a free body diagramof the right arm including the external forces, joint reaction forces, a coordinate system andstate your assumptions.B) Find the force exerted by the right deltoidC) Find the shoulder joint contact force. Report your answer using the magnitude and directionof the shoulder force vector.
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Question 6:
Chlorine is widely used to purify municipal water supplies and to treat swimming pool
waters. Suppose that the volume of a particular sample of Cl₂ gas is 8.70 L at 895 torr
and 24°C.
(a) How many grams of Cl₂ are in the sample?
⚫ Atomic mass of CI = 35.453 g/mol
• Molar mass of Cl₂ = 2 x 35.453 = 70.906 g/mol
Solution:
Use the Ideal Gas Law:
Step 1: Convert Given Values
• Pressure: P = 895 torr → atm
PV=
= nRT
1
P = 895 ×
= 1.1789 atm
760
•
Temperature: Convert to Kelvin:
T24273.15 = 297.15 K
• Gas constant: R = 0.0821 L atm/mol. K
Volume: V = 8.70 L
Step 2: Solve for n
.
PV
n =
RT
n =
(1.1789)(8.70)
(0.0821)(297.15)
10.25
n =
= 0.420 mol
24.405
Step 3: Calculate Mass of Cl₂
Final Answer: 29.78 g of Cl₂.
mass nx M
mass=
(0.420)(70.906)
mass=
29.78 g
Chapter 1 Solutions
Stars and Galaxies (MindTap Course List)
Ch. 1 - Prob. 1RQCh. 1 - Prob. 2RQCh. 1 - Prob. 3RQCh. 1 - What is the difference between the Moon and a...Ch. 1 - Prob. 5RQCh. 1 - Why are light-years more convenient than miles,...Ch. 1 - Prob. 7RQCh. 1 - Prob. 8RQCh. 1 - Prob. 9RQCh. 1 - Prob. 10RQ
Ch. 1 - What are the largest known structures in the...Ch. 1 - Prob. 12RQCh. 1 - Prob. 13RQCh. 1 - Prob. 14RQCh. 1 - Prob. 15RQCh. 1 - Prob. 16RQCh. 1 - Prob. 1PCh. 1 - The equatorial diameter of the Moon is 3476...Ch. 1 - Prob. 3PCh. 1 - A typical galaxy is shown on the first page of the...Ch. 1 - Prob. 5PCh. 1 - Prob. 6PCh. 1 - Prob. 7PCh. 1 - Prob. 8PCh. 1 - If the speed of light is 3.0 105 km/s, how many...Ch. 1 - Prob. 10PCh. 1 - How long does it take light to cross the diameter...Ch. 1 - Prob. 12PCh. 1 - Prob. 13PCh. 1 - Prob. 1SPCh. 1 - Prob. 2SPCh. 1 - Prob. 3SPCh. 1 - Prob. 4SPCh. 1 - Prob. 1LLCh. 1 - Prob. 2LLCh. 1 - Prob. 3LLCh. 1 - Prob. 4LLCh. 1 - Prob. 5LLCh. 1 - Prob. 6LL
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