Foundations of Astronomy, Enhanced
13th Edition
ISBN: 9781305980686
Author: Michael A. Seeds; Dana Backman
Publisher: Cengage Learning US
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 17, Problem 7DQ
To determine
Check whether a galaxy with red shift value 6 is nearer or far, young or old, universe at that time was young or at present age, and about the earth on observing the light started from that quasar.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
No chatgpt pls will upvote
Point charges of 6.50 μC and -2.50 μC are placed 0.300 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.)
(a) Where can a third charge be placed so that the net force on it is zero?
0.49
m to the right of the -2.50 μC charge
(b) What if both charges are positive?
0.49
xm to the right of the 2.50 μC charge
Find the electric field at the location of q, in the figure below, given that q₁ =9c9d = +4.60 nC, q = -1.00 nC, and the square is 20.0 cm on a side. (The +x axis is directed to the right.)
magnitude
direction
2500
x
What symmetries can you take advantage of? What charges are the same magnitude and the same distance away? N/C
226
×
How does charge sign affect the direction of the electric field? counterclockwise from the +x-axis
9a
9b
%
9
9d
Chapter 17 Solutions
Foundations of Astronomy, Enhanced
Ch. 17 - Prob. 1RQCh. 17 - Prob. 2RQCh. 17 - Prob. 3RQCh. 17 - Prob. 4RQCh. 17 - Prob. 5RQCh. 17 - Prob. 6RQCh. 17 - Prob. 7RQCh. 17 - Prob. 8RQCh. 17 - Prob. 9RQCh. 17 - Prob. 10RQ
Ch. 17 - Prob. 11RQCh. 17 - Prob. 12RQCh. 17 - Prob. 13RQCh. 17 - Prob. 14RQCh. 17 - Prob. 15RQCh. 17 - Prob. 16RQCh. 17 - Prob. 17RQCh. 17 - Prob. 18RQCh. 17 - Prob. 19RQCh. 17 - Prob. 20RQCh. 17 - Prob. 21RQCh. 17 - Prob. 1DQCh. 17 - Prob. 2DQCh. 17 - Prob. 3DQCh. 17 - Prob. 5DQCh. 17 - Prob. 6DQCh. 17 - Prob. 7DQCh. 17 - Prob. 1PCh. 17 - Prob. 2PCh. 17 - Prob. 3PCh. 17 - Prob. 4PCh. 17 - Prob. 5PCh. 17 - Prob. 6PCh. 17 - Prob. 7PCh. 17 - Prob. 8PCh. 17 - What is the change in the wavelength of the Balmer...Ch. 17 - Prob. 10PCh. 17 - Prob. 11PCh. 17 - Prob. 12PCh. 17 - Prob. 13PCh. 17 - Prob. 1LTLCh. 17 - Prob. 2LTLCh. 17 - Prob. 3LTLCh. 17 - Prob. 4LTLCh. 17 - Prob. 5LTL
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- would 0.215 be the answer for part b?arrow_forwardSuppose a toy boat moves in a pool at at a speed given by v=1.0 meter per second at t=0, and that the boat is subject to viscous damping. The damping on the boat causes the rate of speed loss to be given by the expression dv/dt=-2v. How fast will the boat be traveling after 1 second? 3 seconds? 10 seconds? Use separation of variables to solve this.arrow_forwardWhat functional form do you expect to describe the motion of a vibrating membrane without damping and why?arrow_forward
- If speed is tripled, how much larger will air drag become for an object? Show the math.arrow_forwardWhat does it tell us about factors on which air drag depends if it is proportional to speed squared?arrow_forwardWhat is the net charge on a sphere that has the following? x (a) 5.75 × 106 electrons and 8.49 × 106 protons 4.39e-13 What is the charge of an electron? What is the charge of a proton? C (b) 200 electrons and 109 protons 1.60e-10 What is the charge of an electron? What is the charge of a proton? Carrow_forward
- A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of 3.5 μC. A second spider with a charge of 4.2 μC rests in her own web exactly 2.1 m vertically below the first spider. (a) What is the magnitude of the electric field due to the charge on the second spider at the position of the first spider? 8.57e3 N/C (b) What is the tension in the silk fiber above the first spider? 0.125 How does the electric field relate to the force? How do you calculate the net force? Narrow_forwardPoint charges of 6.50 μC and -2.50 μC are placed 0.300 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.) (a) Where can a third charge be placed so that the net force on it is zero? 0.49 m to the right of the -2.50 μC charge (b) What if both charges are positive? 0.185 xm to the right of the 2.50 μC chargearrow_forwardc = ad Find the electric field at the location of q, in the figure below, given that q₁ = 9₁ = 9₁ = +4.60 nC, q=-1.00 nC, and the square is 20.0 cm on a side. (The +x axis is directed to the right.) magnitude direction N/C ° counterclockwise from the +x-axis 9a % 9 9barrow_forward
- A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of 3.5 μC. A second spider with a charge of 4.2 μC rests in her own web exactly 2.1 m vertically below the first spider. (a) What is the magnitude of the electric field due to the charge on the second spider at the position of the first spider? 8.57e3 N/C (b) What is the tension in the silk fiber above the first spider? 0.275 How does the electric field relate to the force? How do you calculate the net force? Narrow_forwardPlastic beads can often carry a small charge and therefore can generate electric fields. Three beads are oriented such that 92 is between 91 system of all three beads is zero. E field lines 91 92 93 X What charge does each bead carry? 91 = 92 = ?2.9 0 μC × What is the net charge of the system? What charges have to be equal? μC 93 2.9 με and 93. The sum of the charge on 91 and 92 is 91 +92 = -2.9 μC, and the net charge of thearrow_forwardAn electron has an initial speed of 5.26 x 100 m/s in a uniform 5.73 x 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? opposite direction to the electron's initial velocity same direction as the electron's initial velocity not enough information to decide × What is the direction of the force on the electron? How does it compare to the direction of the electric field, considering the sign of the electron's charge? (b) How far does the electron travel before coming to rest? 0.0781 × What kinematic equation is relevant here? How do you calculate the force due to the electric field? m (c) How long does it take the electron to come to rest? 5.27e8 What is the final velocity of the electron? s (d) What is the electron's speed when it returns to its starting point? 5.26e6 m/sarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Stars and GalaxiesPhysicsISBN:9781305120785Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStax
Foundations of Astronomy (MindTap Course List)PhysicsISBN:9781337399920Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
Stars and Galaxies (MindTap Course List)PhysicsISBN:9781337399944Author:Michael A. SeedsPublisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
Stars and Galaxies
Physics
ISBN:9781305120785
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Astronomy
Physics
ISBN:9781938168284
Author:Andrew Fraknoi; David Morrison; Sidney C. Wolff
Publisher:OpenStax
Foundations of Astronomy (MindTap Course List)
Physics
ISBN:9781337399920
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Stars and Galaxies (MindTap Course List)
Physics
ISBN:9781337399944
Author:Michael A. Seeds
Publisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
General Relativity: The Curvature of Spacetime; Author: Professor Dave Explains;https://www.youtube.com/watch?v=R7V3koyL7Mc;License: Standard YouTube License, CC-BY