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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;
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;
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Answer:
n=6.56×10¹⁵Hz
Explanation:
Given Data
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
To find
Revolutions per second
Solution
Let F be the force of attraction
let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by
F=mω²r......................where ω=2π n
F=m4π²n²r...............eq(i)
as the values given where
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
we have to find n from eq(i)
n²=F/(m4π²r)