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expeople1 [14]
3 years ago
5

Which of the following is an effect of the electric force?

Physics
2 answers:
Zolol [24]3 years ago
7 0
<h2>Answer</h2>

Option A - A positive and negative charge attract each other.

<u>Explanation</u>

The electric force is the result of the interaction between electric charges. An electric force is a force that exists between all charged particles. There are two types of charge, positive and negative. Like charges repel each other while unlike charges attract each other. This means that if two objects have the same charges that is positive or negative then they will repel each other. If a negatively charged particle comes in contact with a positively charged particle they will attract.

stich3 [128]3 years ago
6 0

As we know that two charges exert force on each other when they are placed near to each other

The force between two charges is given as

F = \frac{kq_1q_2}{r^2}

here we know that

q_1, q_2 = two different point charges

r = distance between two point charges

also we know that two similar charges always repel each other while two opposite charges always attract each other

so here correct answer would be

<em>A. A positive and negative charge attract each other.</em>

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two long parrallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction.Find the magnitude of
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Answer: The magnitude of force per length that each wire exert on the other wire is 2.67×10^-5 N/m.

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3 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
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Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

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Z = atomic number

Here: Z = 1 (hydrogen atom)

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E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

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Energy of the seventh orbit in H atom .

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1) n = 2 → n = 3

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E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0
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