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katrin [286]
4 years ago
11

A star is in hydrostatic equilibrium when the outward push of pressure due to core burning is exactly in balance with the inward

pull of gravity. When the hydrogen in a star's core has been used up, burning ceases, and gravity and pressure are no longer in balance. This causes the star to undergo significant changes.
Which of the following evolutionary changes would bring a star back into hydrostatic equilibrium? Check all that apply:

a) A small increase in the star's internal pressure and temperature causes the star's outer layers to expand and cool.
b) A small decrease in the star's internal pressure and temperature causes the star's outer layers to contract and heat up.
c) A small increase in the star's internal pressure and temperature causes the star's outer layers to contract and heat up.
d) A small decrease in the star's internal pressure and temperature causes the star's outer layers to expand and cool.
Physics
1 answer:
viktelen [127]4 years ago
7 0

Answer:

a and b

Explanation:

Hydro static equilibrium holds a star steady and balanced. Whenever a star stops burning hydrogen in its center, there must be evolutionary improvements to maintain equilibrium for the star Of example, if a star's internal pressure and temperature fall, gravity will take over and force the star to contract and heat up, restoring stability. By contrast, if a star's internal pressure and temperature rises, the extra pressure causes the star to widen and cool, restoring balance.

so, according to above explanation options a and b both are true

a) A small increase in the star's internal pressure and temperature causes the star's outer layers to expand and cool.

b) A small decrease in the star's internal pressure and temperature causes the star's outer layers to contract and heat up.

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Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1
miv72 [106K]

Answer:

w = 4,786 rad / s ,  f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

       L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

      L₀ = 0 + I₂ w₂

Final after coupling

      L_{f} = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

      I = ½ M R²

How the moment is preserved

      L₀ = L_{f}

      I₂ w₂ = (I₁ + I₂) w

      w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

      d₁ = 60 cm = 0.60 m

      d₂ = 40 cm = 0.40 m

      f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

      w₂ = 2 π f₂

      w₂ = 2π 3.33

      w₂ = 20.94 rad / s

Let's replace

       w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

       w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

      w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

      w = 20.94 1.28 / 5.6

      w = 4,786 rad / s

Angular velocity and frequency are related.

      w = 2π f

      f = w / 2π

      f = 4.786 / 2π

      f = 0.76176 Hz

7 0
3 years ago
An airplane flies horizontally at constant speed in a straight­line direction. Its state of motion is unchanging. In other words
Alex_Xolod [135]

Answer:

sum of all forces on the air plane must be ZERO

So both forces must be of same magnitude

Explanation:

As we know that airplane is moving with uniform speed is horizontal plane is a straight line

so the motion of the air plane is uniform without any acceleration

So we will have

a = 0

acceleration must be zero

now by Newton's law

F_{net} = 0

F_1 + F_2 = 0

so sum of all forces on the air plane must be ZERO

7 0
3 years ago
Which explains how time of winds creates high waves?
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The smaller waves created by the constant winds gradually add up to form larger ones.
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3 years ago
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What is kinematics . Give two examples​
Wittaler [7]

Answer:

kinematics explains the terms like acceleration, velocity, and position of the objects while in motion.

Some important parameters in kinematics are displacement, velocity, and time.

Explanation:

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8 0
2 years ago
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Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the
lesya [120]

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

F_{net} = - W_x - f_s

The W_x is a x-component of force due to gravity (W) and, in this case, is given by: W_x = W.sin(14)

W is described as: W = m.g

Force due to friction (f_s) is given by: f_s = μs.N

N is the normal force and, in the system, is equivalent of W_y, so:

W_y = m.g.cos(14)

Therefore, the formula will be:

F_{net} = - W_x - f_s

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

3 0
3 years ago
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