Answer:
The current through the inductor at the end of 2.60s is 9.7 mA.
Explanation:
Given;
emf of the inductor, V = 41.0 mV
inductance of the inductor, L = 13 H
initial current in the inductor, I₀ = 1.5 mA
change in time, Δt = 2.6 s
The emf of the inductor is given by;
Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.
Answer:
(a) The spring constant is 59.23 N/m
(b) The total energy involved in the motion is 0.06 J
Explanation:
Given;
mass, m = 240 g = 0.24 kg
frequency, f = 2.5 Hz
amplitude of the oscillation, A = 4.5 cm = 0.045 m
The angular speed is calculated as;
ω = 2πf
ω = 2 x π x 2.5
ω = 15.71 rad/s
(a) The spring constant is calculated as;
(b) The total energy involved in the motion;
E = ¹/₂kA²
E = (0.5) x (59.23) x (0.045)²
E = 0.06 J