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diamong [38]
2 years ago
12

An electron in an atom's orbital shell, labeled X in the model below, released enough energy to move to a different orbital shel

l. Which of the two orbital shells would the electron have the possibility of moving to
Physics
1 answer:
Delicious77 [7]2 years ago
7 0

Answer:

Lower energy shell which will be nearer to the nucleus.

Explanation:

When electron move from one energy level to another, an electron must gain or lose just the right amount of energy.

When atoms releases energy, electrons move into lower energy levels.  The electrons in the shells aways from the nucleus have more energy as compared to the electrons in the nearer shells.

Electrons with the lowest energy are found closest to the nucleus, where the attractive force of the positively charged nucleus is the greatest. Electrons that have higher energy are found further away

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What is the acceleration of a 24 kg mass pushed by a 6N force?
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Use the Venn Diagram to compare and contrast solar and lunar eclipses.
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Use the law of conservation of energy (assume no friction nor air resistance) to determine the kinetic and potential energy at t
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Answer:

Part A

1) At the starting point, we have;

PE = 40,000 J

2) PE = 0 J, KE = 40,000 J

3) KE = 20,000 J

4) PE = 15,000 J

5) KE = 32,500 J

6) KE = 40,000 J, PE = 0 J

7) KE = 35,000 J

8) KE = 40,000 J, PE = 0 J

Part B

The total Mechanical Energy = ME = 40,000 J

At the final point, we have;

ME = KE + PE = 40,000 J + 0 J = 40,000 J

Explanation:

Part A

By the law of conservation of energy, we have;

ME = PE + KE

Where;

ME = The total Mechanical Energy of the system

PE = The Potential Energy of the system

KE = The Kinetic Energy of the system

Where there is no friction, we have;

At the final stage, KE = 40,000 J. PE = 0 J

Therefore, ME = PE + KE = 40,000 J + 0 J = 40,000 J

1) At the starting point, we have;

KE = 0 J, therefore, PE = ME - KE = 40,000 J - 0 J = 40,000 J

2) At the bottom of the roller coaster, at the same level the PE is taken as PE = 0 J at the final stage, we have;

PE = 0 J, therefore, KE = ME - PE = 40,000 J - 0 J = 40,000 J

3) Where PE = 20,000 J, KE = ME - PE = 40,000 J - 20,000 J = 20,000 J

4) Where KE = 25,000 J, PE = ME - KE = 40,000 J - 25,000 J = 15,000 J

5) Where PE = 7,500 J, KE = ME - PE = 40,000 J - 7,500 J = 32,500 J

6) At the bottom KE = 40,000 J, PE = 0 J

7) Where PE = 5,000 J, KE = ME - PE = 40,000 J - 5,000 J = 35,000 J

8) KE = 40,000 J, PE = 0 J

Part B

The given that there is no friction nor air resistance, the total Mechanical Energy, ME, is constant and equal to the sum of the Potential Energy, PE and the Kinetic Energy, KE, as follows;

ME = KE + PE

At the final point, we have;

ME = 40,000 J + 0 J = 40,000 J

The total Mechanical Energy = ME = 40,000 J

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Why do the earth have many states
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