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2 years ago

In spiral galaxies, the relationship between nuclear bulge and tightness of spiral arms seems to be that

1 answer:

7 0

In spiral galaxies, the relationship between nuclear bulge and tightness of spiral arms seems to be that the larger the nuclear bulge of the galaxy, tighter the spiral arms are wound together.

What are spiral galaxies?

Spiral galaxies form a class of galaxy originally described by Edwin Hubble in his 1936 work The Realm of the Nebulae and, as such, form part of the Hubble sequence.

Most spiral galaxies consist of a flat, rotating disk containing stars, gas and dust, and a central concentration of stars known as the bulge. These are often surrounded by a much fainter halo of stars, many of which reside in globular clusters.

Spiral galaxies are named by their spiral structures that extend from the center into the galactic disc.

The spiral arms are sites of ongoing star formation and are brighter than the surrounding disc because of the young, hot OB stars that inhabit them.

Roughly two-thirds of all spirals are observed to have an additional component in the form of a bar-like structure, extending from the central bulge, at the ends of which the spiral arms begin.

The proportion of barred spirals relative to bar less spirals has likely changed over the history of the universe, with only about 10% containing bars about 8 billion years ago, to roughly a quarter 2.5 billion years ago, until present, where over two-thirds of the galaxies in the visible universe (Hubble volume) have bars.

To learn more about spiral galaxies: brainly.com/question/14243370

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mash [69]

B. Frequency divided by wavelength

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4 years ago

An 18-cm-long bicycle crank arm, with a pedal at one end, is attached to a 20-cm-diameter sprocket, the toothed disk around whic

Tpy6a [65]

Answer:

Part a)

$a = 0.056 m/s^2$

Part b)

$L = 7.85 m$

Explanation:

Part a)

Angular speed of the pedal is changing from 60 rpm to 90 rpm in 10 s

so the angular acceleration is given as

$\alpha = \frac{\omega_2 - \omega_1}{\Delta t}$

so we will have

$\alpha = \frac{2\pi(\frac{90}{60}) - 2\pi(\frac{60}{60})}{10}$

$\alpha = 0.314 rad/s^2$

now the tangential acceleration of the pedal is given as

$a = r \alpha$

$a = 0.18 \times 0.314$

$a = 0.056 m/s^2$

Part b)

Total angular displacement made by the sprocket in the interval of 10 s is given as

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{2\pi (\frac{90}{60}) + 2\pi (\frac{60}{60})}{2}(10)$

$\theta = 78.5 radian$

now length of the chain passing over it is given as

$L = R\theta$

$L = 0.10 \times 78.5$

$L = 7.85 m$

6 0

4 years ago

Which of the following is NOT one of the four Grand Slam Tournaments?

postnew [5]

Answer:

Davis Cup

Explanation:

Hope this Helps!

6 0

4 years ago

A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before co

alina1380 [7]

Answer:

The force exerted by the wood on the bullet is 399.01 N

Explanation:

Given;

mass of bullet, m = 0.0021 kg

initial velocity of the bullet, u = 497 m/s

final velocity of the bullet, v = 0

distance traveled by the bullet, S = 0.65 m

Determine the acceleration of the bullet which is the deceleration.

Apply kinematic equation;

V² = U² + 2aS

0 = 497² - (2 x 0.65)a

0 = 247009 - 1.3a

1.3a = 247009

a = 247009 / 1.3

a = 190006.92 m/s²

Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;

F = ma

F = 0.0021 x 190006.92

F = 399.01 N

Therefore, the force exerted by the wood on the bullet is 399.01 N

6 0

3 years ago

Compare the magnitude of the electromagnetic and gravitational force between two electrons separated by a distance of 2.00 m. As

Lesechka [4]

First you need to know about two laws, which are:
1) Coulomb's law
2) Newton's law of gravitation

1.
According to Coulomb's law, Electric force between TWO charges is:
$F_{e} = \frac{k*q_{1}*q_{2}}{r^{2}}$ -- (A)

Where, k = 1/(4*π*epsilon_not) = $9 * 10^{9}$ $\frac{Nm^{2}}{C^{2}}$

Both $q_{1}$ and $q_{2}$ = -1.61 x $10^{-19}$ C
r = Distance between the two charges = 2.00m

Plug-in the above values in (A), you would get:

$F_{e}$ = (9 * $10^{9}$ ) (1.61 * $10^{-19}$ * 1.61 * $10^{-19}$ ) / (2*2)

$F_{e}$ = $5.83* 10^{-29}$ N

2.
According to Newton's law of gravitation:
$F_{g} = \frac{Gm_{1}m_{2}}{r^{2}}$ -- (B)

Where G = Gravitational constant = 6.674 * $10^{-11} m^{2} kg^{-1}s^{-2}$

m1 = m2 = Mass of the electron = $9.11 * 10^{-31}$ kg
r = 2.0 m

Plug-in the above values in (B), you would get:

$F_{g}$ = (6.674 * $10^{-11}$ ) ( $9.11 * 10^{-31}$ * $9.11 * 10^{-31}$ }[/tex]) / (2*2)

$F_{e}$ = $5.83* 10^{-29}$ N

$F_{e}$ = $1.38 * 10^{-71} N$

Now do Fe over Fg, you would get:
$\frac{F_{e}}{F{g}} = 4.23 * 10^{42}$

Ans: So the blanks are:
1) 5.82
2) 1.38
3) 4.23

-i

6 1

3 years ago

Read 2 more answers