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inysia [295]
1 year ago
12

What main factor determines the stages a star will follow after the main sequence?.

Physics
1 answer:
Artyom0805 [142]1 year ago
5 0

The main factor that determines the stages a star will follow after the main sequence is the star's mass or size.

<h3>What is main sequence?</h3>

The main sequence of a star is a stage in the life cycle of that star. During the main sequence ( also called the zero age ), the star generates energy by nuclear fusion of Hydrogen atoms at the core of the star into Helium atoms. Eventually, the star runs out of hydrogen atoms, which concludes the main sequence. What happens afterward depends on the size of the star.

<h3>Low mass stars </h3>

For low-mass stars ( stars that are less than 0.1 times the mass of the sun), they slowly collapse into white dwarfs. These stars do not get hot enough to fuse helium atoms, instead hydrogen fusion continues until the whole star is filled with helium and slowly collapses into a white dwarf while it grows dimmer and colder.

<h3>Medium-sized stars</h3>

Medium-sized stars ( stars about 0.6 - 10 times the mass of our sun) become red giants. Stars similar in size to our sun are hot enough to fuse helium atoms, so towards the end of the main sequence it starts to fuse helium atoms, forming heavier elements like carbon and oxygen. The heavier elements move to the star's core due to gravity, while lighter elements like hydrogen form a shell around the core. This causes the sun to then grow in size, forming a red giant.  

<h3>Large stars</h3>

Massive stars ( greater than 10 times the mass of the sun) and super-massive stars ( more than 40 times the mass of the sun ) end up exploding into a supernova , while the dense core collapses into a neutron star or a black hole.

To know more about main sequence, check out;  

brainly.com/question/18141359

#SPJ4

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Hi There,

  
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3 0
3 years ago
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A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away
alekssr [168]

Answer:

31.2 m/s

Explanation:

f_{app} = Frequency of approach = 480 Hz

f_{aw} = Frequency of going away = 400 Hz

V = Speed of sound in air = 343 m/s

v = Speed of truck

Frequency of approach is given as

f_{app} = \frac{Vf}{V - v}                           eq-1

Frequency of moving awayy is given as

f_{aw} = \frac{Vf}{V + v}                          eq-2

Dividing eq-1 by eq-2

\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}

\frac{480}{400} = \frac{343 + v}{343 - v}

v = 31.2 m/s

7 0
3 years ago
A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
2 years ago
1. Determine the image distance in each of the following.
miskamm [114]

It is given that for the convex lens,

Case 1.

u=−40cm

f=+15cm

Using lens formula

v

1

−

u

1

=

f

1

v

1

−

40

1

=

15

1

v

1

=

15

1

−

40

1

v=+24.3cm

The image in formed in this case at a distance of 24.3cm in left of lens.

Case 2.

A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that

u=∞

f=15cm

Now, using mirror’s formula

v

1

+

u

1

=

f

1

v

1

+

∞

1

=

15

1

v=+15cm

The image is formed at a distance of 15cm in left of mirror

6 0
3 years ago
Why won’t anyone help me please anybody help me I really need help .
irga5000 [103]

Answer:

1➡️ this is the method of decomposition

2➡️ H2 and O2

3➡️ b

sorry if I am wrong

8 0
2 years ago
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