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Ksenya-84 [330]
2 years ago
6

Which statement correctly lists the four forces from the weakest to the strongest?

Physics
2 answers:
harkovskaia [24]2 years ago
4 0
The strongest of the natural forces is Strong Nuclear, followed by Electromagnetic, then Weak Nuclear and lastly Gravitational.
So from weakest to strongest: gravitational --> weak nuclear --> electromagnetic --> strong nuclear
Answer C)
mash [69]2 years ago
3 0

The correct answer of this question is : C i.e Gravitational < Weak nuclear force < Electromagnetic force < Strong nuclear force.

EXPLANATION:

There are four fundamental forces in nature which are known as gravitational force, weak nuclear force, electromagnetic force and strong nuclear force.

Out of these four fundamental forces, the gravitational force is the weakest forces in nature,but has infinite range.

Weak nuclear force is produced due to W/Z bosons. Hence, it is stronger than gravitational force but weaker as compared to electromagnetic and strong nuclear force. It is a short range force.

Electromagnetic force is the third one which is weaker as compared to strong nuclear force but has infinite range.

Strong nuclear force is a short range, spin dependent strong attractive force that arises between nucleons due to exchange of mesons.

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  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

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