1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
svetlana [45]
3 years ago
12

M84, M87, and NGC 4258 all have accretion disks around their central black holes for which the rotational velocities have been m

easured in HST spectra. In M84, the disk extends 8 pc from the center and exhibits Doppler velocities as large as ±400 km/s with respect to the galaxy's overall radial velocity. In M87, the corresponding figures are 20 pc and 500 km/s. In NGC 4258, the figures are 0.5 ly and 1000 km/s. Calculate the black hole masses to two significant figures. Comment also on the assumptions under which you did your calculation (e.g., orbital plane viewed edge-on; note the appearance of the galaxies and their disks in the notes) and the effect this may have on the accuracy of the answers.
Physics
1 answer:
givi [52]3 years ago
3 0

Answer:

<u>For M84:</u>

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

M = 2307.46 * 10³⁶ kg

Explanation:

1 parsec, pc  = 3.08 * 10¹⁶ m

The equation of the orbit speed can be used to calculate the doppler velocity:

v = \sqrt{\frac{GM}{r} }

making m the subject of the formula in the equation above to calculate the mass of the black hole:

M = \frac{v^{2} r}{G}.............(1)

<u>For M84:</u>

r = 8 pc = 8 * 3.08 * 10¹⁶

r = 24.64 * 10¹⁶ m

v = 400 km/s = 4 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }

M = 2307.46 * 10³⁶ kg

The mass of the black hole in the galaxies is measured using the doppler shift.

The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.

You might be interested in
If an area has all the wolves that it can support the wolf population has reached is what
Liono4ka [1.6K]
The wolf population in that area has reached its carrying capacity.
7 0
3 years ago
Read 2 more answers
3. What are the challenges of looking for Dyson spheres?
S_A_V [24]

1. it is difficult to search for it . Because infrared rays will never penetrate through earth atmosphere.

2. we are unaware of how it looks like and we only know it is red and will glow . A damaged star also looks like this.

3. Dust also makes is hard to detect Dyson spheres . So we will get confused between Dyson sphere and a star surrounded by dust.

5 0
3 years ago
Read 2 more answers
31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r
posledela

Answer:

See answers below

Explanation:

a.

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

b.

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

7 0
2 years ago
A. A light wave moves through glass (n=1.5) at an angle of 25°. What angle will it have when it moves from the glass into air (n
torisob [31]

PART A)

By Snell's law we know that

n_1sin i = n_2 sin r

here we know that

n_1 = 1.5

i = 25 degree

n_2 = 1

now from above equation we have

1.5 sin25 = 1 sin r

r = 39.3 degree

so it will refract by angle 39.3 degree

PART B)

Here as we can see that image formed on the other side of lens

So it is a real and inverted image

Also we can see  that size of image is lesser than the size of object here

Here we can use concave mirror to form same type of real and inverted image

PART C)

As per the mirror formula we know that

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20}

d_i = 30 cm

so image will form at 30 cm from mirror

it is virtual image and smaller in size

3 0
4 years ago
The diagram below shows different weights on the see-saw. Will the see-saw move?
oee [108]

Answer:

yes it will

Explanation:

8 0
3 years ago
Other questions:
  • What are the main agents of metamorfic rock?
    12·1 answer
  • a particle with charge Q is on the y axis a distance a from the origin and a particle with charge qi is on the x axis at a dista
    5·1 answer
  • You want to close an open door by throwing either a 400-g lump of clay or a 400-g rubber ball toward it. you can throw either ob
    9·1 answer
  • Substance of an atom of 2 or more elements
    14·1 answer
  • What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me
    5·1 answer
  • A remote control car is traveling at a velocity of 0.50 m/s when it hits a wall and comes to a stop in 0.050 seconds. What is th
    13·1 answer
  • Can anyone answer this question for me please..​
    13·1 answer
  • A car traveling at 26 m/s skids to a stop in 3 seconds. Determine the skidding distance of the car.
    5·1 answer
  • Help ASAP I’ll mark you as brainlister
    13·1 answer
  • The value of acceleration due to gravity (g) on Pluto is about 0.61 meters/second2. How much will an object that weighs 250 newt
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!