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Y_Kistochka [10]
2 years ago
10

The energy (such as light and heat) released by the Sun is produced by nuclear reactions in the core of the Sun, which convert h

ydrogen into helium. According to Einstein's equation E = mc2, the mass of one helium nucleus should be ___________.
Physics
1 answer:
Yanka [14]2 years ago
3 0

Answer:

<em>the mass of one helium nucleus should be</em> <em>less than the mass of four hydrogen nuclei.</em>

Explanation:

Deep inside the core of the Sun, enough protons can collide into each other with enough speed that they stick together to form a helium nucleus and generate a tremendous amount of energy at the same time. This process is called nuclear fusion.

The mass-to-energy conversion is described by Einstein's famous equation:

E = mc2, or, in words, energy equals mass times the square of the velocity of light. Because the velocity of light is a very large number, this equation says that lots of energy can be gained from using up a modest amount of mass.

Photons In the proton-proton chain reaction, hydrogen nuclei are converted to helium nuclei through a number of intermediates. The reactions produce high-energy photons (gamma rays) that move through the "radiative layer" surrounding the core. This layer takes up 60 percent of the radius of the Sun. It takes a million years for energy to get through this layer into the "convective layer", because the photons are constantly intercepted, absorbed and re-emitted. In the core, the helium nuclei make up 62% of the mass (the rest is still hydrogen). The radiative and convective layers have about 72% hydrogen, 26% helium, and 2% heavier elements (by mass). The energy produced by fusion is then transported to the solar surface and emitted as light or ejected as high-energy particles.

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How far does a car travel in 90 seconds if it’s traveling 55 m/s? Show equation
podryga [215]

You just said the car is traveling at the speed of 55 m/s.  If I understand this  correctly, that means the car will cover:

55 meters in the first second,

55 meters in the 2nd second,

55 meters in the 3rd second,

55 meters in the 4th second,

55 meters in the 5th second,

.

.

.

55 meters in the 87th second,

55 meters in the 88th second,

55 meters in the 89th second, and

55 meters in the 90th second.

That's 55 meters 90 times.  If you just move these words around a little bit, it says "90 times 55 meters" . . . a pretty simple arithmetic problem.

The equation is . . . <em>Distance = (55 m/s) times (time, in seconds)</em> .

I get <em>4,953 meters</em>.  You should check me on this.

8 0
3 years ago
The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total
pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass m travelling at a velocity \vec{v}. The momentum \vec{p} of this object would be:

\vec{p} = m \cdot \vec{v}.

For the law of conservation of momentum, consider two objects: object \rm a and object \rm b. Assume that these two objects collided with each other.

  • Let m_{\rm a} and m_{\rm b} denote the mass of the two objects.
  • Let \vec{v}_{\rm a}(\text{initial}) and \vec{v}_{\rm b}(\text{initial}) denote the velocity of the two object right before the interaction.
  • Let \vec{v}_{\rm a}(\text{final}) and \vec{v}_{\rm b}(\text{final}) denote the velocity of the two objects right after the interaction.
  • The momentum of the two objects right before the collision would be m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) and m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}), respectively.
  • The momentum of the two objects right after the collision would be m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) and m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}), respectively.

The sum of the momentum of the two objects would be:

  • m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) right before the collision, and
  • m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}) right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final}).

4 0
3 years ago
The diagram shows an example of convection.
const2013 [10]

Answer: Option (C) is the correct answer.

Explanation:

When we heat a fluid then the movement within the fluid makes hot (less dense) material to rise and cooler (more denser) material to sink at the bottom. This process is known as convection.

Thus, in the diagram hot (less dense) water will rise and cooler (more dense) water sinks at the bottom.

Therefore, we can conclude that according to the arrow the label belongs to cooler water sinks.

5 0
3 years ago
Read 2 more answers
PLEASE HELP!
Harman [31]
Hey user!

your answer is here..

correct option is A. steel

we know that sounds travel faster in solid as compared to gas and liquids. in gas the molecules are very loosely packed and there is lot of space between so it takes more time to pass sound from each other. and in liquid, the molecules are closer as compared to gas hence it will be little faster and in solid, the molecules are very tightly packed so it will be the fastest. and among these options, steel is the only solid so the speed of sound in steel will be the fastest.

and note that the closer the molecules are to each other ( tightly packed ) makes the bond also tighter and less time to pass sound.

cheers!!
8 0
3 years ago
Define potential energy. Give its mathematical formula and SI unit.
Stels [109]

Answer:

The type of mechanical energy that is possessed due to the virtue of motion or state of a body is known as potential energy.

Its formula: PE= mgh

Its SI unit is joule.

Hope it helps you..

3 0
3 years ago
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