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jonny [76]
2 years ago
13

Since nuclear fusion in the sun creates energy from matter, why doesn't it violate the law of conservation of energy?

Physics
1 answer:
alexandr402 [8]2 years ago
3 0

Since nuclear fusion in the sun creates energy from matter, Einstein's formula E=mc² states that matter and energy are equal.

To find the answer, we have to know more about the nuclear fusion.

<h3>What is nuclear fusion?</h3>
  • We are aware that the sun can achieve nuclear fusion by the fusion of hydrogen atoms.
  • These atoms need to get closer to one another in order to fuse.
  • Since both protons inside each nucleus are positively charged, they attempt to repel one another as they get closer to one another.
  • If this issue cannot be solved, nuclear fusion in the sun cannot occur.

Thus, we can conclude that, since nuclear fusion in the sun creates energy from matter, Einstein's formula E=mc² states that matter and energy are equal.

Learn more about nuclear fusion here:

brainly.com/question/25663405

#SPJ4

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Answer:

[13,101 RPM]

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Explanation:

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In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

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Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

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Vesnalui [34]

Answer:

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