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Margarita [4]
2 years ago
6

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would pro

duce. Stars over this limit tend to be surrounded by massive nebulae of material blasted off of their surfaces.
Physics
1 answer:
Gnesinka [82]2 years ago
4 0

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of <u>Eddington luminosity</u>

<h3>What are stars?</h3>

Stars are a fixed luminous point in the sky which is a large and remote incandescent body

So therefore, the hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of Eddington luminosity

Learn more about stars:

brainly.com/question/13018254

#SPJ1

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denis-greek [22]

Answer:

Explanation:

conservation of momentum

initial momentum is zero

Lets say that m₁ moves along the x axis

and m₂ moves along the y axis

in the y direction

0 = m₁(0) + m₂(30) + m₃(vy)

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vy = -10 m/s

in the x direction

0 = m₁(30) + m₂(0) + m₃(vx)

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-3m₂(vx) = 2m₂(30)

vx = -20 m/s

v = √(-10² + -20²) = 10√5 m/s ≈ 22.36 m/s

θ = arctan(-10/-20) = 206.56505... ≈ 206.6°  CCW from the + x axis

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What is the minimum number of points necessary to determine a straight line?
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A 0.22 kg mass at the end of a spring oscillates 2.9 times per second with an amplitude of 0.13 m .
andre [41]

Answer:

2.3687599 m/s

0.91106 m/s

0.617213012 J

Explanation:

f = Frequency = 2.9\ Hz

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m = Mass of object = 0.22 kg

Angular velocity is given by

\omega=2\pi f\\\Rightarrow \omega=2\pi 2.9\\\Rightarrow \omega=18.22123\ rad/s

Velocity is given by

V=A\omega\\\Rightarrow V=0.13\times 18.22123\\\Rightarrow V=2.3687599\ m/s

Speed when it passes the equilibrium point is 2.3687599 m/s

Frequency is given by

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow k=m4\pi^2f^2 \\\Rightarrow k=0.22\times 4\pi^2\times 2.9^2\\\Rightarrow k=73.04296\ N/m

x = Displacement = 0.12 m

In this system the energies are conserved

\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{73.04296(0.13^2-0.12^2)}{0.22}}\\\Rightarrow v=0.91106\ m/s

The speed when it is 0.12 m from equilibrium is 0.91106 m/s

The energy in the system is given by

E=\dfrac{1}{2}kA^2\\\Rightarrow E=\dfrac{1}{2}\times 73.04296\times 0.13^2\\\Rightarrow E=0.617213012\ J

The total energy of the system is 0.617213012 J

4 0
4 years ago
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