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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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If a car moves with a uniform speed of 2m/s in a circle of radius 0.4m, what is its angular speed?
REY [17]

Answer:

The car's angular speed is \frac{rad}{s}.

Explanation:

Angular velocity is usually measured with \frac{radians}{sec}, so I'm going to use that to answer your question.

The relationship between tangential velocity and angular velocity (ω)  is given by:

V = \omega R

Using the values from the question, we get:

2 \frac{m}{s} = \omega (0.4m)

\omega = 5 \frac{rad}{s}

Therefore, the car's angular speed is 5 \frac{rad}{s}.

Hope this helped!

3 0
3 years ago
A train traveling at 6.4 m/s accelerates at 0.10 m/s 2 over a distance of 100 m. How large is the train’s final velocity?
klasskru [66]

The final velocity of the train at the end of the given distance is 7.81 m/s.

The given parameters;

  • initial velocity of the train, u = 6.4 m/s
  • acceleration of the train, a = 0.1 m/s²
  • distance traveled, s = 100 m

The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;

v² = u² + 2as

v² = (6.4)² + (2 x 0.1 x 100)

v² = 60.96

v = √60.96

v = 7.81 m/s

Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.

Learn more here:brainly.com/question/21180604

4 0
3 years ago
.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum accelerat
AlexFokin [52]

Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg  

assume the greatest scale reading = 78.09 kg

<u>Determine the maximum acceleration for these elevators</u>

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N<u> </u>

<u> </u>At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

3 0
3 years ago
A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in
Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

\frac{GMm}{d^{2}}=\frac{mv^2}{d}

v=\sqrt{\frac{GM}{d}}

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}

v = 1.81 x 10^-4 m/s

7 0
4 years ago
The magnetic flux through a loop:
dexar [7]

Answer:

The magnetic flux through a loop is zero when the B field is perpendicular to the plane of the loop.

Explanation:

Magnetic flux are also known as the magnetic line of force surrounding a bar magnetic in a magnetic field.

It is expressed mathematically as

Φ = B A cos(θ) where

Φ is the magnetic flux

B is the magnetic field strength

A is the area

θ is the angle that the magnetic field make with the plane of the loop

If B is acting perpendicular to the plane of the loop, this means that θ = 90°

Magnetic flux Φ = BA cos90°

Since cos90° = 0

Φ = BA ×0

Φ = 0

This shows that the magnetic flux is zero when the magnetic field strength B is perpendicular to the plane of the loop.

4 0
3 years ago
Read 2 more answers
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