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Feliz [49]
3 years ago
15

What do an electron and a neutron have in common?

Physics
2 answers:
Harman [31]3 years ago
4 0

The correct answer to this question is B) Each particle exists inside an atom.

EXPLANATION:

Before going to answer this question , first we have to understand the nature of electron and neutron.

Electrons and neutrons are the subatomic particles which are present inside the atom.

Electron is that subatomic particle which is present in the extra nuclear part of the atom, and it moves around the nucleus in different permitted, stationary orbits. The charge of electron is negative.

The neutron is that subatomic particle which is present in the nuclear part of the atom along with protons. The charge of proton is positive and the neutron is neutral in nature.

Hence, the correct option of this question is that both electrons and neutrons are present inside the atom.

KIM [24]3 years ago
4 0

Answer: It's B

Explanation:

I just took the test

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Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o
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Answer:

a. q_C experiences the greatest net force and q_B experiences the smallest net force

b. Ratio of the greatest to the smallest net force= 9

Explanation:

<u>Electrostatic Forces </u>

Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude

\displaystyle F=K\frac{q_1q_2}{d^2}

If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.

Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.

a.

The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is

F_{TA}=-F_B-F_C+F_D

Computing the individual forces we have

\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}

\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}

\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}

The total force on qA is:

\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})

\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}

Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus

\displaystyle F_{TB}=F_A-F_C+F_D

\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}

\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}

Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus

\displaystyle F_{TC}=F_A+F_B+F_D

\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}

\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}

Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus

\displaystyle F_{TD}=-F_A-F_B-F_C

\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}

\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}

\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}

Comparing the relative values of all the forces

\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|

This means that qc experiences the greatest net force and qB experiences the smallest net force

b.

The ratio of the greatest to the smallest forces is

\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9

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