Understanding Our Universe
3rd Edition
ISBN: 9780393614428
Author: PALEN, Stacy, Kay, Laura, Blumenthal, George (george Ray)
Publisher: W.w. Norton & Company,
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 14, Problem 1QAP
To determine
Whether the statement is true or false.
Expert Solution & Answer
Answer to Problem 1QAP
True
Explanation of Solution
The galaxies can be classified into so many ways like elliptical, spiral, irregular etc. Each and every galaxy has own spin and orientation. Galaxies having elliptical shape are known as elliptical galaxies and galaxies having spiral shape are spiral galaxies.
The galaxies which are not well organised in a particular manner, they are known as irregular galaxies. The galaxies have their own orientations so it is difficult to judge the type of galaxy by the orientation only.
Take an example of orientation of a coin. When a coin is thrown into the air it can be seen from different angles; the same is with the galaxies. When three dimensional pictures of galaxies from different angles are observed, the shape of galaxies looks different.
Conclusion:
Thus, the classification of galaxy does not only depend on the orientation of galaxies so sometimes it is difficult to classify them. Hence the statement is true.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A piece of copper originally 305mm long is pulled in tension
with a stress of 276MPa. If the deformation is elastic, what
will be the resultant elongation. E for copper is 110Gpa
Please solve and answer the problem correctly please. Be sure to give explanations on each step and write neatly please. Thank you!!
In the figures, the masses are hung from an elevator ceiling. Assume the velocity of the elevator is constant. Find the tensions in
the ropes (in N) for each case. Note that 0₁ = 35.0°, 0₂ = 55.0°, 03 = 60.0°, m₁ = 3.00 kg, and m2 = 7.00 kg. (Due to the
nature of this problem, do not use rounded intermediate values-including answers submitted in WebAssign-in your calculations.)
(a)
Τι
WY NY MY
T3
e₁
T₁
=
N
=
N
=
N
(b)
18
Τι
=
Τι
T3
=
|| || ||
=
T
T
Ts
m₂
N
N
N
02
T₂
T3
m₁
Chapter 14 Solutions
Understanding Our Universe
Ch. 14.1 - Prob. 14.1CYUCh. 14.2 - Prob. 14.2CYUCh. 14.3 - Prob. 14.3CYUCh. 14.4 - Prob. 14.4CYUCh. 14 - Prob. 1QAPCh. 14 - Prob. 2QAPCh. 14 - Prob. 3QAPCh. 14 - Prob. 4QAPCh. 14 - Prob. 5QAPCh. 14 - Prob. 6QAP
Ch. 14 - Prob. 7QAPCh. 14 - Prob. 8QAPCh. 14 - Prob. 9QAPCh. 14 - Prob. 10QAPCh. 14 - Prob. 11QAPCh. 14 - Prob. 12QAPCh. 14 - Prob. 13QAPCh. 14 - Prob. 14QAPCh. 14 - Prob. 15QAPCh. 14 - Prob. 16QAPCh. 14 - Prob. 17QAPCh. 14 - Prob. 18QAPCh. 14 - Prob. 19QAPCh. 14 - Prob. 20QAPCh. 14 - Prob. 21QAPCh. 14 - Prob. 22QAPCh. 14 - Prob. 23QAPCh. 14 - Prob. 24QAPCh. 14 - Prob. 25QAPCh. 14 - Prob. 27QAPCh. 14 - Prob. 28QAPCh. 14 - Prob. 29QAPCh. 14 - Prob. 30QAPCh. 14 - Prob. 31QAPCh. 14 - Prob. 32QAPCh. 14 - Prob. 33QAPCh. 14 - Prob. 34QAPCh. 14 - Prob. 35QAPCh. 14 - Prob. 36QAPCh. 14 - Prob. 37QAPCh. 14 - Prob. 38QAPCh. 14 - Prob. 39QAPCh. 14 - Prob. 40QAPCh. 14 - Prob. 41QAPCh. 14 - Prob. 43QAPCh. 14 - Prob. 44QAPCh. 14 - Prob. 45QAP
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
- You are working with a movie director and investigating a scene with a cowboy sliding off a tree limb and falling onto the saddle of a moving horse. The distance of the fall is several meters, and the calculation shows a high probability of injury to the cowboy from the stunt. Let's look at a simpler situation. Suppose the director asks you to have the cowboy step off a platform 2.55 m off the ground and land on his feet on the ground. The cowboy keeps his legs straight as he falls, but then bends at the knees as soon as he touches the ground. This allows the center of mass of his body to move through a distance of 0.660 m before his body comes to rest. (Center of mass will be formally defined in Linear Momentum and Collisions.) You assume this motion to be under constant acceleration of the center of mass of his body. To assess the degree of danger to the cowboy in this stunt, you wish to calculate the average force upward on his body from the ground, as a multiple of the cowboy's…arrow_forwardA box of mass m = 2.00 kg is released from rest at the top of an inclined plane as seen in the figure. The box starts out at height h =0.200 m above the top of the table, the table height is H = 2.00 m, and 0 = 41.0°. H m (a) What is the acceleration (in m/s²) of the box while it slides down the incline? m/s² (b) What is the speed (in m/s) of the box when it leaves the incline? m/s (c) At what horizontal distance (in m) from the end of the table will the box hit the ground? m (d) How long (in s) from when the box is released does it hit the ground? S (e) Does the box's mass affect any of your above answers? Yes Noarrow_forward(a) A sphere made of rubber has a density of 0.940 g/cm³ and a radius of 7.00 cm. It falls through air of density 1.20 kg/m³ and has a drag coefficient of 0.500. What is its terminal speed (in m/s)? m/s (b) From what height (in m) would the sphere have to be dropped to reach this speed if it fell without air resistance? marrow_forward
- The systems shown below are in equilibrium. If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the incline are frictionless. (Let m = 2.19 kg and € = 29.0°.) scale in (a) N N scale in (b) scale in (c) N scale in (d) N a C m m m m m b d m Ꮎarrow_forwardAn elevator car has two equal masses attached to the ceiling as shown. (Assume m = 3.10 kg.) m m T₁ T2 (a) The elevator ascends with an acceleration of magnitude 2.00 m/s². What are the tensions in the two strings? (Enter your answers in N.) = N T₁ Τι = N (b) The maximum tension the strings can withstand is 78.8 N. What is the maximum acceleration of the elevator so that a string does not break? (Enter the magnitude in m/s².) m/s²arrow_forward(a) At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.85 x 100 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.20 x 10-5 T? 4.27e3 m/s (b) What would the radius (in m) of the path be if the proton had the same speed as the electron? 7.85e6 x m (c) What would the radius (in m) be if the proton had the same kinetic energy as the electron? 195.38 x m (d) What would the radius (in m) be if the proton had the same momentum as the electron? 3.7205 marrow_forward
- ! Required information The block shown is made of a magnesium alloy, for which E = 45 GPa and v = 0.35. Know that σx = -185 MPa. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 25 mm B D 40 mm 100 mm Determine the magnitude of Oy for which the change in the height of the block will be zero. The magnitude of Oy is MPa.arrow_forwardThe rigid bar ABC is supported by two links, AD and BE, of uniform 37.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σy= 250 MPa. The magnitude of the force Q applied at B is gradually increased from zero to 265 kN and a = 0.640 m. 1.7 m 1 m D A B 2.64 m E Determine the value of the normal stress in each link. The value of the normal stress in link AD is The value of the normal stress in link BE is 250 MPa. MPa.arrow_forwardTwo tempered-steel bars, each 16 in. thick, are bonded to a ½ -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E= 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is gradually increased from zero until the deformation of the bar reaches a maximum value dm = 0.04 in. and then decreased back to zero. Take L = 15 in. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 2.0 in. in. 3 in. 3 16 in. Determine the maximum stress in the tempered-steel bars. The maximum stress in the tempered-steel bars is ksi.arrow_forward
- Ammonia enters the compressor of an industrial refrigeration plant at 2 bar, -10°C with a mass flow rate of 15 kg/min and is compressed to 12 bar, 140°C. Heat transfer from the compressor to its surroundings occurs at a rate of 6 kW. For steady-state operation, calculate, (a) the power input to the compressor, in kW, Answer (b) the entropy production rate, in kW/K, for a control volume encompassing the compressor and its immediate surroundings such that heat transfer occurs at 300 K.arrow_forwardNo chatgpt pls will upvotearrow_forwardShown to the right is a block of mass m=5.71kgm=5.71kg on a ramp that makes an angle θ=24.1∘θ=24.1∘ with the horizontal. This block is being pushed by a horizontal force, F=229NF=229N. The coefficient of kinetic friction between the two surfaces is μ=0.51μ=0.51. Enter an expression for the acceleration of the block up the ramp using variables from the problem statement together with gg for the acceleration due to gravity. a=arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Stars and Galaxies (MindTap Course List)PhysicsISBN:9781337399944Author:Michael A. SeedsPublisher:Cengage Learning
Stars and GalaxiesPhysicsISBN:9781305120785Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
Foundations of Astronomy (MindTap Course List)PhysicsISBN:9781337399920Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStax
An Introduction to Physical SciencePhysicsISBN:9781305079137Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar TorresPublisher:Cengage Learning
Stars and Galaxies (MindTap Course List)
Physics
ISBN:9781337399944
Author:Michael A. Seeds
Publisher:Cengage Learning
Stars and Galaxies
Physics
ISBN:9781305120785
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Foundations of Astronomy (MindTap Course List)
Physics
ISBN:9781337399920
Author:Michael A. Seeds, Dana Backman
Publisher:Cengage Learning
Astronomy
Physics
ISBN:9781938168284
Author:Andrew Fraknoi; David Morrison; Sidney C. Wolff
Publisher:OpenStax
An Introduction to Physical Science
Physics
ISBN:9781305079137
Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
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