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irakobra [83]
3 years ago
13

The state in which a star's gravitational collapse is balanced between gravity pushing inward and gas pressure pushing outward i

s called
Physics
2 answers:
Andrew [12]3 years ago
5 0
The relation between temperature and pressure is called the "equation of state of the gas". or "Hydrostatic equilibrium in ordinary star". Take for example a balloon, it will have a larger spherical shape, if the pressure inside exerted by the gas on a wall of a balloon balance the inward force exerted by the outside atmospheric pressure.  In a dying star which is being compressed by gravity, the gas is being squeezed so the molecules is moving rapidly, resulting to a very high temperature, and this provide a balance that counteract or balances the compressive force of gravity. The very high temperature inside the star is needed to balance the force of gravity, and it is provide by "nuclear fusion energy" or else the star would collapse under the force of gravity. Depending on the size or mass of the star, it will either become, a "neutron star" or a "black hole".                 
joja [24]3 years ago
4 0

Answer:  Hydro-static Equilibrium

Explanation:

A fluid in its stable rest state is said to have hydro-static equilibrium. A star is composed of gases mainly hydrogen and helium. It is huge in size and mass. It does not collapses under its own weight because of the pressure exerted outwards by the ongoing nuclear fusion reaction in its core which balances the gravitational force. Thus, a star maintains a state of hydro-static equilibrium.

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Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0
2 years ago
When an unbalanced force acts on an object,
Gwar [14]
When balanced forces follow up on an object, the object won't move. If you push against a wall, the wall pushes back with an equal but opposite force. Neither you nor the wall will move. Forces that cause a change in the motion of an object are unbalanced forces.
6 0
2 years ago
Read 2 more answers
Earth, the Sun, and billions of stars are contained within
Nitella [24]
I think it is the Milky Way.
3 0
3 years ago
Read 2 more answers
The angle between incident ray and reflected ray is 80 degrees. What is the value of angle of incidence?
saul85 [17]

Answer:

The angle of incident ray is 40°.

Explanation:

Given that the angle of incident and reflected ray are the same. In this question, we had given that both angles added up will gives you 80° so you have to divide it by 2 :

incident + reflected = 80°

Let incident = reflected = θ

θ + θ = 80°

2θ = 80°

θ = 80° ÷ 2

= 40°

8 0
3 years ago
A man applies a force of 100 N to a rock for 60 seconds, but the rock does not move what is the amount of work done by the man o
Art [367]
600. I forgot the measurement. but 600 is correct
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