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storchak [24]
2 years ago
8

Why are chemical processes unable to produce the same amount of energy flowing out of the sun as nuclear fusion?

Physics
1 answer:
julsineya [31]2 years ago
3 0

Answer:

Explanation:

One of the major differences between nuclear reactions and chemical reactions is that nuclear reactions involve larger amount of energy than chemical energy. <u>This is because the force between the protons and neutrons in the nucleus of an atom is much higher than the force of attraction between electrons and the positively charged nucleus, hence nuclear reactions involves/requires a larger amount of energy (because it's reactions involve the nucleus) than chemical reactions (because it's reactions involve the electrons)</u>.

Thus, during nuclear fusion, two light nuclei are bombarded against one another to produce a larger/heavier nuclei with the release of large amount of energy (because the forces between the protons and neutrons are much higher) unlike when two atoms/molecules are chemically combined together to form a new molecule with the rearrangement of electrons in the valence shells of the participating molecules.

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Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
2 years ago
The diagram shows movement of thermal energy. At bottom a fire has red curved lines labeled Y with arrowheads pointing upward to
Pepsi [2]

Answer:

X and Z

Explanation:

Conduction occurs through direct physical contact.  Heat transferred from the pot to the handle, and from the handle to the hand, are both examples of conduction.

6 0
2 years ago
Read 2 more answers
You are a particle physicist at the Large Hadron Collider who is tasked with designing an apparatus to separate annihilation pro
JulijaS [17]

Find solution in the attachments

4 0
2 years ago
A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
dlinn [17]

Answer:

a = 4.72 m/s²  

Explanation:

given,

mass of the box (m)= 6 Kg

angle of inclination (θ) = 39°

coefficient of kinetic friction (μ) = 0.19

magnitude of acceleration = ?

box is sliding downward so,

F - f = m a                        

f is the friction force

m g sinθ - μ N = ma                        

m g sinθ - μ m g cos θ = ma            

a = g sinθ - μ g cos θ                    

a = 9.8 x sin 39° - 0.19 x 9.8 x cos 39°

a = 4.72 m/s²                                      

the magnitude of acceleration of the box down the slope is a = 4.72 m/s²  

3 0
2 years ago
If a rock is thrown upward on the planet mars with a velocity of 14 m/s, its height (in meters) after t seconds is given by h =
crimeas [40]

<u>Answer:</u>

 Velocity of rock after 2 seconds = 6.56 m/s

<u>Explanation:</u>

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Here height of rock in meters, h = 14t-1.86t^2

Comparing both the equations

    We will get initial velocity = 14 m/s(already given) and \frac{1}{2} a = -1.86

     So,  Acceleration, a = -3.72 m/s^2

 Now we have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

 When time is 2 seconds we need to find final velocity.

     v = 14 - 3.72 * 2 = 6.56 m/s.

  So, Velocity of rock after 2 seconds = 6.56 m/s  

6 0
2 years ago
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