Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
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If one oscillation has 5.8 times the energy of the second one of equal frequency and mass, 2.4:1 is the ratio of their amplitudes.
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Answer:
Explanation:
Given,
Number of turns, N = 645 N
Area, A = 20.25 m²
Earth Magnetic field, B = 5 x 10⁻⁵ T
Maximum Emf = 1.25 V.
Angular velocity, ω = ?
Using Induced Emf formula
Angular velocity of the coil =
