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3 years ago

Name the three basic parts of a spiral galaxy and describe what is found in each.

Physics

2 answers:

ElenaW [278]3 years ago

7 0

Disk, the halo, and the nucleus or central bulge.

xeze [42]3 years ago

4 0

The three basic components of a spiral galaxy are the disk, the halo, and the nucleus or central bulge. The disk contains most of the gas and dust in the galaxy. The halo contains globular clusters of very old stars. The nucleus contains a high density of stars, but most of them are old stars.

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What is the density of a substance if 17 cm cubed has a mass of 35.7g

Arisa [49]

Density = (mass) / (volume)

Density = (35.7 g) / (17 cm³)

Density = 2.1 g/cm³

4 0

3 years ago

Why is it so important to follow classroom procedures?

Aleks [24]

Maybe because you will get in trouble if u don't follow them. I never follow classroom procedures and I'm in trouble all the time

7 0

3 years ago

At a distance of one centimeter from an electron, the electric field strength has a value

Burka [1]

We are asked in the problem to determine the <span>distance of an electron in which the electric field strength is equal to e/3. we use the gravitational formula of E = m1m2/d^2
hence,

e = m1m2/1 = k/1 =k
e/2 = k/d^2
e/2=e/d^2
d^2 = 2
d = sqrt 2 = </span><span>1.41 cm</span>

6 0

3 years ago

Sally and ner pet puagie go everywnere

FrozenT [24]

a)The total distance she walks will be 1100 m.

b)The magnitude and direction of the budgie's path will be 854.4 m at 69.44°.

c)Sally's average speed for the journey will be 2.82 m/sec

d)The average velocity of Sally's budgie willl be 2.19 m/sec North of East.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points.

Distance travelled in north direction,d₁ = 8.0 x 10² m

Time elapsed travelled in north direction,t₁ = 5.0 minutes

Distance travelled in east direction,d₂ = 0.30 km = 3.0 × 10² m

Time elapsed travelled in east direction,t₂= 1.5 minutes

The total distance she walks,d =? m

The magnitude and direction of the budgies path will be, D

Sally's average speed for the journey, $\rm S_{avg}$

The average velocity of Sally's budgie, $\rm V_{avg}$

The total distance she walks is found as;

d=d₁+d₂

d=800 m +300 m

d = 1100 m

The magnitude and direction of the budgies path will be;

$\rm D = \sqrt{d_1 ^2 + d_2^2 } \\\\ D = \sqrt{800^2 +300^2 } \\\\ D = 854.4 \ m$

The direction of the displacement is;

$\rm \theta = tan^{-1} \frac{800}{300} \\\\ \theta = tan^{-1}(2.6) \\\\ \theta = 69.44 ^ 0$

c)journey is found as;

Sally's average speed for the

$\rm S_{avg} = \frac{d}{t} \\\\ S_{avg} =\frac{100}{300+900} \\\\ S_{avg} =2.82 \ m/sec$ <em>
</em>

The average velocity of Sally's budgie is found as;

$\rm V_{avg} = \frac{854.4}{300+ 900 } \\\\ V_{avg} = 2.19 \ m/sec$ North of east

Hence, the total distance she walks, the magnitude and direction of the budgie's path, Sally's average speed, and the average velocity of Sally's budgie will be 1100 m,854.4 m at 69.44°,2.82 m/sec, and 2.19 m/sec North of East.

To learn more about displacement refer to the link;

brainly.com/question/10919017

#SPJ1

6 0

2 years ago

An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9000 m/s an

Sergio [31]

Answer:

The distance between earthquake center and the measuring station is 1350 kilometers.

Explanation:

Let the earthquake center be at a distance of 'S' meters from the recording station.

Now from the basic relation of distance, speed and time we know that

$Distance=Speed\times Time$

For a Primary wave (P wave) let us assume that it appraoches the measuring station after $t_{1}$ minutes

Thus making use of the above relation we have

$Distance=V_{p}\times Time\\\\\therefore D=9000\times t_{1}.......(i)$

Now since it is given that the secondary wave (S wave) reaches the measuring spot after 2 minutes or 120 seconds thus the time taken by secondary waves to reach recorder equals $t_{1}+120$ making use of the same relation we get

$Distance=V_{s}\times Time\\\\\therefore D=5000\times (t_{1}+120).......(ii)$

Solving equation 'i' and 'ii' we get

$D=5000\times (\frac{D}{9000}+120)\\\\\therefore \frac{D}{5000}=\frac{D}{9000}+120\\\\\frac{D}{5000}-\frac{D}{9000}=120\\\\\therefore D=\frac{120}{\frac{1}{5000}-\frac{1}{9000}}=1350000meters=1350kilomerers$

8 0

3 years ago