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aleksandrvk [35]

4 years ago

5

th core of a star is the size of our sun with mass 2 times as great as the sun and is rotating at a frequency of 1 rev evru 100

days. If it were to undergo gravitational collapse to a neutron star of radius 10 kn what would its rtation frequency be

1 answer:

Rufina [12.5K]4 years ago

6 0

Answer:

3522.7 rad/s

Explanation:

From the law of conservation of angular momentum,

I₁ω₁ = I₂ω₂

I₁, ω₁ and I₂, ω₂ are the rotational inertia and angular speed of the sun and neutron star respectively.

ω₂ = I₁ω₁/I₂ I₁ = ²/₅M₁R₁² and I₂ = ²/₅M₂R₂² where M₁,M₂ and R₁,R₂ are the masses and radii of star and neutron star respectively.

ω₂ = I₁ω₁/I₂ = ²/₅M₁R₁²ω₁÷²/₅M₂R₂² = (M₁/M₂)(R₁/R₂)²ω₁

M₁ = M₂, R₁ = radius of sun = 6.96 10⁵ km and R₂ = 10 km, ω₁ = 1 rev/100days = 2π/(100 × 24 × 60 × 60 s) = 7.272 × 10⁻⁷ rad/s

ω₂ = (M₁/M₂)(R₁/R₂)²ω₁ = 1 × ( 6.96 10⁵ km/10 km)² × 7.272 × 10⁻⁷ rad/s

= 3522.7 rad/s

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xenn [34]

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Where,

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With our values we have that

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$\epsilon_1=0J$

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$\epsilon_3=1.6*10^{-21}J$

To make the calculations easier we can assume that the temperature and Boltzmann constant can be summarized as

$\beta = \frac{1}{kT}$

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$\beta = 2.9*10^{20}J$

Therefore the average energy would be,

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Replacing with our values we have

$\bar{\epsilon} = \frac{0e^{-0}+1.6*10^{-21}*e^{-\Beta(1.6*10^{-21})}+1.6*10^{-2-1}*e^{-(2.9*10^{20})(1.6*10^{-21})}}{1+2e^{-2.9*10^{20}*1.6*10^{-21}}}$

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3 0

3 years ago

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

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Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

$h=ut+\frac{1}{2}gt^2$

here u=0 and t=time taken

$h=\frac{1}{2}gt^2$

for $t=1 s$

$h_1=\frac{1}{2}g$

for $t=2 s$

$h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g$

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for $t=3 s$

$h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g$

distance traveled in 3 rd sec $=\frac{9}{2}g-2g=\frac{5}{2}g$

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5 0

3 years ago

Which law states that the pressure and absolute temperature of a fixed quantity of gas are directly proportional under constant

Brrunno [24]

<u>Gay Lussac’s law</u> state that the pressure and absolute temperature of a fixed quantity of a gas are directly proportional under constant volume conditions.

<h2>Further Explanation </h2><h3>Gay-Lussac’s law </h3>

It states that at constant volume, the pressure of an ideal gas I directly proportional to its absolute temperature.
Thus, an increase in pressure of an ideal gas at constant volume will result to an increase in the absolute temperature.

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This gas law states that the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
Therefore, when the volume of an ideal gas is increased at constant temperature then the pressure of the gas will also increase.

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It states that the volume of a fixed mass of a gas is directly proportional to absolute temperature at constant pressure.
Therefore, an increase in volume of an ideal gas causes a corresponding increase in its absolute temperature and vice versa while the pressure is held constant.

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It is also known as the Dalton’s law of partial pressure. It states that the total pressure of a mixture of gases is always equivalent to the total sum of the partial pressures of individual component gases.
Partial pressure refers to the pressure of an individual gas if it occupies the same volume as the mixture of gases.

Keywords: Gas law, Gay-Lussac’s law, pressure, volume, absolute temperature, ideal gas

<h3>Learn more about: </h3>

Gay-Lussac’s law: brainly.com/question/2644981
Charles’s law: brainly.com/question/5016068
Boyles’s law: brainly.com/question/5016068
Dalton’s law: brainly.com/question/6491675

Level: High school

Subject: Chemistry

Topic: Gas laws

Sub-topic: Gay-Lussac’s law

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More and more PE turns into KE as the ball falls, all the way down.

When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.

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4 years ago

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mihalych1998 [28]

Answer:

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3 years ago