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Answer:
The internal resistance of the cell is 0.051 ohm.
Explanation:
Given;
emf of the battery, E = 12 V
terminal voltage of the cell, V = 8.2 V
current in the circuit, I = 75 A
let the potential drop of the cell due to internal resistance (r) = Ir
The internal resistance of the cell is calculated from the equation below;
E = V + Ir
where;
r is the internal resistance of the cell
Therefore, the internal resistance of the cell is 0.051 ohm.
Explanation:
Given that,
Length of the spring, l = 50 cm
Mass, m = 330 g = 0.33 kg
(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :
(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.
(C) The frequency of the oscillation is given by :