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inna [77]
3 years ago
13

Appropriate word inside the parentheses.

Physics
1 answer:
Taya2010 [7]3 years ago
6 0

Answer:

<em>At constant mass, the  acceleration of an object varies (</em><em>directly</em><em>) with the net external force  applied. That is to say, that an object's acceleration increases as the force applied is (</em><em>increased</em><em>),  but its acceleration decreases if the force applied is (</em><em>decreased</em><em>).</em>

Explanation:

<u>Mechanical Force </u>

According to the second Newton's law, the acceleration of an object varies directly proportional to the external net force applied and inversely proportional to the mass of the object.

If the mass is constant, then the acceleration will vary in the same way as the force does.

Completing the sentences:

At constant mass, the  acceleration of an object varies (directly) with the net external force  applied. That is to say, that an object's acceleration increases as the force applied is (increased),  but its acceleration decreases if the force applied is (decreased).

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A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located
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The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁  is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241  \times 10^{-19 } \ C

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

#SPJ2

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