Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil
Answer:
6.169 μA
Explanation:
Formula for induced EMF is given by the equation;
EMF = M(di/dt). We are given;
di/dt = 2.5 A/s
M = 19μH = 19 × 10^(-6) H
Thus;
EMF = 19 × 10^(-6) × 2.5.
EMF = 47.5 × 10^(-6) V
Formula for current is;
i = EMF/R. R is resistance given as 7.7 ohms.
Thus; i = 47.5 × 10^(-6)/7.7
i = 6.169 μA
Answer:
Explanation:
Given
Potential Energy is given by

And Force is given by

Particle will be at equilibrium when Potential Energy is either minimum or maximum

i.e.

So angular Frequency of small oscillation is given by

for 
we get 


Answer:
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Given that
Velocity of missile (v) = 20 m/s ,
Angle of missile (Θ) = 53°
Determine , Vertical component = v sin Θ
= 20 sin 53°
= 15.97 m/s