Concept explainers
(a)
The brightness of the star Betelgeuse in terms of a fraction of the Sun’s brightness. It is given that Betelgeuse is transformed from a red supergiant to a Type II supernova at the distance of 425 ly from Earth.
(a)

Answer to Problem 62Q
Solution:
Explanation of Solution
Given data:
The distance of the star from Earth is
Formula used:
The expression for apparent magnitude of a supernova is,
Here,
The expression for ratio of brightness of two objects is,
Here,
Explanation:
Convert the distance from light years to parsec as follows:
Therefore, the provided distance of the star from Earth in parsecs is,
Write the formula for apparent magnitude of Type II supernova.
The absolute magnitude for Type II supernova is
The apparent magnitude of the Sun is
Write the expression for the brightness ratio of Betelgeuse and Sun.
Here, the subscript S refers to the corresponding quantities for the Sun and b is the brightness of Betelgeuse.
Substitute
Conclusion:
So, the supernova is
(b)
The comparison between the brightness of the supernova and that of Venus. It is given that it is transformed from a red supergiant to Type II supernova at the distance of 425 ly from Earth and the brightness of Venus is
(b)

Answer to Problem 62Q
Solution:
The ratio of the brightness of the supernova to that of Venus is 710.
Explanation of Solution
Given data:
The brightness of Venus is
The distance of the star from Earth is
Formula used:
The expression for apparent magnitude of a supernova is,
Here,
The expression for the ratio of brightness of two objects is,
Here,
Explanation:
Refer to part (a). The brightness of the star with respect to that of the Sun is
In order to compare the brightness of the star with that of Venus, determine the ratio of their respective brightness (relative to the Sun), that is,
Substitute
Conclusion:
So, the supernova is 710 times brighter than Venus.
Want to see more full solutions like this?
Chapter 20 Solutions
EBK LOOSE-LEAF VERSION OF UNIVERSE
- Sketch the harmonic.arrow_forwardFor number 11 please sketch the harmonic on graphing paper.arrow_forward# E 94 20 13. Time a) What is the frequency of the above wave? b) What is the period? c) Highlight the second cycle d) Sketch the sine wave of the second harmonic of this wave % 7 & 5 6 7 8 * ∞ Y U 9 0 0 P 150arrow_forward
- Show work using graphing paperarrow_forwardCan someone help me answer this physics 2 questions. Thank you.arrow_forwardFour capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μCarrow_forward
- In the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?arrow_forwardFour capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor µC 6.00 µF capacitor µC 3.00 µF capacitor µC capacitor C µCarrow_forwardTwo conductors having net charges of +14.0 µC and -14.0 µC have a potential difference of 14.0 V between them. (a) Determine the capacitance of the system. F (b) What is the potential difference between the two conductors if the charges on each are increased to +196.0 µC and -196.0 µC? Varrow_forward
- Please see the attached image and answer the set of questions with proof.arrow_forwardHow, Please type the whole transcript correctly using comma and periods as needed. I have uploaded the picture of a video on YouTube. Thanks,arrow_forwardA spectra is a graph that has amplitude on the Y-axis and frequency on the X-axis. A harmonic spectra simply draws a vertical line at each frequency that a harmonic would be produced. The height of the line indicates the amplitude at which that harmonic would be produced. If the Fo of a sound is 125 Hz, please sketch a spectra (amplitude on the Y axis, frequency on the X axis) of the harmonic series up to the 4th harmonic. Include actual values on Y and X axis.arrow_forward
- Stars and Galaxies (MindTap Course List)PhysicsISBN:9781337399944Author:Michael A. SeedsPublisher:Cengage LearningFoundations of Astronomy (MindTap Course List)PhysicsISBN:9781337399920Author:Michael A. Seeds, Dana BackmanPublisher:Cengage LearningStars and GalaxiesPhysicsISBN:9781305120785Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
- AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStaxCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College





