Answer:
The compression of the spring is 24.6 cm
Explanation:
magnitude of the charge on the left, q₁ = 4.6 x 10⁻⁷ C
magnitude of the charge on the right, q₂ = 7.5 x 10⁻⁷ C
distance between the two charges, r = 3 cm = 0.03 m
spring constant, k = 14 N/m
The attractive force between the two charges is calculated using Coulomb's law;
The extension of the spring is calculated as follows;
F = kx
x = F/k
x = 3.45 / 14
x = 0.246 m
x = 24.6 cm
The compression of the spring is 24.6 cm
0.1 J of elastic potential energy will be stored by the spring when it gets compressed by 0.2 m if it has the spring constant of 5 N/m.
Elastic potential energy can be defined as the potential energy stored by the spring when it is stretched or compressed.
According to the question.
Compression in the spring, x = 0.2 m
Spring constant, k = 5 n/m
Elastic potential Energy, U =
U = 0.1 J
0.1 J of elastic potential energy will be stored by the spring.
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Answer:
5.5x 10^-11 C
Explanation:
Pls see attached file
