Concept explainers
The reason due to which the helium fusion in a star’s core, that releases energy, actually lead to a decrease in its luminosity. Also, mention the name of these stars in the HR diagram.
Answer to Problem 1CC
Solution:
The energy released from core helium fusion enlarges the star's core, which decreases the
Explanation of Solution
Introduction:
Most stars have the same chemical composition but the difference in stars depend on the two primary properties like their surface temperature and luminosities. These two properties vary from one star to another. H-R diagrams or hertz sprung- Russell diagrams are a graphical tool used by astronomers for the classification of stars according to their spectral type, luminosity, temperature, color, evolutionary track, and temperature.
Explanation:
During helium fusion in core, the surrounding hydrogen fusion in the shell of the star provides most of the red giant’s luminosity and helium flash is not observed. The core expands when core helium fusion begins as a result of which the temperature of the core decreases. The cooling of the core indirectly cools the surrounding hydrogen fusing shell, due to which the shell liberates energy slowly. Hence, the luminosity decreases a bit after core helium fusion begins.
The contraction of star’s external layer is because of the slow energy release. While shrinking, there is an increase in the star’s surface temperature. So, the evolutionary track of the star moves to the left in the H-R diagram. In this stage, the changes in luminosity are very low due to which a horizontal shift can be seen in the evolutionary track, along the path which is referred as the horizontal branch.
Conclusion:
The fusion of helium enlarges the core which decreases the energy of both the hydrogen fusion in the shell and the helium fusion in core. This decrease in rate of energy decreases the luminosity of the stars. These stars are called horizontal branch stars.
Want to see more full solutions like this?
Chapter 20 Solutions
EBK LOOSE-LEAF VERSION OF UNIVERSE
- Required information Two speakers vibrate in phase with each other at 523 Hz. At certain points in the room, the sound waves from the two speakers interfere destructively. One such point is 1.45 m from speaker #1 and is between 2.00 m and 4.00 m from speaker #2. The speed of sound in air is 343 m/s. How far is this point from speaker #2? marrow_forwarda) Consider the following function, where A is a constant. y(x,t) = A(x — vt). Can this represent a wave that travels along? Explain. b) Which of the following are possible traveling waves, provide your reasoning and give the velocity of the wave if it can be a traveling wave. e-(a²x²+b²²-2abtx b.1) y(x,t) b.2) y(x,t) = = A sin(ax² - bt²). 2 b.3) y(x,t) = A sin 2π (+) b.4) y(x,t) = A cos² 2π(t-x). b.5) y(x,t) = A cos wt sin(kx - wt)arrow_forwardThe capacitor in (Figure 1) is initially uncharged. The switch is closed at t=0. Immediately after the switch is closed, what is the current through the resistor R1, R2, and R3? What is the final charge on the capacitor? Please explain all steps.arrow_forward
- Suppose you have a lens system that is to be used primarily for 620-nm light. What is the second thinnest coating of fluorite (calcium fluoride) that would be non-reflective for this wavelength? × nm 434arrow_forwardThe angle between the axes of two polarizing filters is 19.0°. By how much does the second filter reduce the intensity of the light coming through the first? I = 0.106 40 xarrow_forwardAn oil slick on water is 82.3 nm thick and illuminated by white light incident perpendicular to its surface. What color does the oil appear (what is the most constructively reflected wavelength, in nanometers), given its index of refraction is 1.43? (Assume the index of refraction of water is 1.33.) wavelength color 675 × nm red (1 660 nm)arrow_forward
- A 1.50 μF capacitor is charging through a 16.0 Ω resistor using a 15.0 V battery. What will be the current when the capacitor has acquired 1/4 of its maximum charge? Please explain all stepsarrow_forwardIn the circuit shown in the figure (Figure 1), the 6.0 Ω resistor is consuming energy at a rate of 24 J/s when the current through it flows as shown. What are the polarity and emf of the battery E, assuming it has negligible internal resistance? Please explain all steps. I know you need to use the loop rule, but I keep getting the answer wrong.arrow_forwardIf you connect a 1.8 F and a 2.6 F capacitor in series, what will be the equivalent capacitance?arrow_forward
- Suppose that a particular heart defibrillator uses a 1.5 x 10-5 Farad capacitor. If it is charged up to a voltage of 7300 volts, how much energy is stored in the capacitor? Give your answer as the number of Joules.arrow_forwardThe voltage difference across an 8.3 nanometer thick cell membrane is 6.5 x 10-5volts. What is the magnitude of the electric field inside this cell membrane? (Assume the field is uniform, and give your answer as the number of Volts per meter... which is the same as the number of Newtons per Coulomb.)arrow_forwardThree identical capacitors are connected in parallel. When this parallel assembly of capacitors is connected to a 12 volt battery, a total of 3.1 x 10-5 coulombs flows through the battery. What is the capacitance of one individual capacitor? (Give your answer as the number of Farads.)arrow_forward
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
AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStax
Stars and GalaxiesPhysicsISBN:9781305120785Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning