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.
They create energy ……………….
The horizontal velocity<span> of a projectile is </span>constant<span> (a never </span>changing<span> in value), There is a </span>vertical<span>acceleration caused by gravity; its value is 9.8 m/s/s, down, The </span>vertical velocity<span> of a projectile </span>changes<span> by 9.8 m/s each second, The </span>horizontal<span> motion of a projectile is independent of its </span>vertical<span> motion.</span>