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Roman55 [17]
3 years ago
6

What kind of stars make up the galactic nucleus?

Physics
2 answers:
Solnce55 [7]3 years ago
5 0
The galactic nucleus or the active galactic nucleus (AGN) is the region of the galaxy that is considered compact. It's luminosity is higher than normal which is not generated by stars. Therefore, the kind of stars that composes the galactic nucleus are the old, metal-rich stars.
Hatshy [7]3 years ago
4 0
I think the correct answer would be old and metal poor stars are found in the galactic nucleus. This nucleus us a region in the center of a galaxy which contains a higher luminosity than other parts. It produces very high amounts of energy. Hope this helps.
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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
A car goes from point A to point B then back again. Which statement about its speed and velocity is true? A. They are the same.
pav-90 [236]

because the car returned to A the velocity is 0

D. The velocity is zero but the speed

4 0
4 years ago
One system collects waste products and turns them into urine.
Illusion [34]

Answer:

E but it is really B

Explanation:

4 0
3 years ago
Read 2 more answers
A force of 70 N is inclined at an angle
Bad White [126]

Answer:

The horizontal component of the 70 N force = 35·√3 N

The vertical component of the 70 N force = 35 N

Explanation:

The magnitude of the given force, F = 70 N

The angle of inclination of the given force to the horizontal, θ = 30°

By the resolution of forces, we can resolve the given force, F, into its horizontal component, Fₓ, and vertical components, F_y, as follows;

The horizontal component of the given force = Fₓ = F × cos(θ)

Substituting the values, we have;

The horizontal component of the 70 N force, Fₓ = 70 × cos(30°) = 35·√3 N

The vertical component of the given force = F_y = F × sin(θ)

Substituting the values, we have;

The vertical component of the 70 N force, F_y = 70 × sin(30°) = 35 N.

5 0
3 years ago
Calculate the momentum of the following objects:
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I think/ believe it’s A
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3 years ago
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