If the maximum emf of the ac generator is 20 V and the maximum potential difference across the resistor is 16 V Then the maximum potential difference across the inductor is 4 V.
Calculation:
Step-1:
It is given that the RL circuit is connected to a 20 V ac generator. The maximum potential difference across the resistor is 16 V. It is required to find the maximum potential drop across the inductor.
Step-2:
The maximum emf of the generator is equal to the sum of the maximum potential difference across the resistor and the maximum potential difference across the inductor.
Therefore,
The maximum potential difference across the inductor + Maximum maximum potential difference across the resistor = Maximum emf of the generator
Thus,
Maximum maximum potential difference across the inductor + 16 V = 20 V
Therefore,
Maximum maximum potential difference across the inductor = 20 V - 16 V = 4 V
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Answer:
The maximum velocity is 1.58 m/s.
Explanation:
A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.
Spring constant, K = 100 N/m
mass, m = 0.1 kg
Amplitude, A = 5 cm = 0.05 m
Let the angular frequency is w.

The maximum velocity is

The area of the circle with radius r is
A = πr²
The rate of change of area with respect to time is

The rate of change of the radius is given as

Therefore

When r = 10 ft, obtain

Answer: - 40π ft²/s (or - 127.5 ft²/s)
Answer:
6
Explanation:
Number of lines emanate from + 5 micro coulomb is 15 .
They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.
the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.
So the lines terminating at - 3 micro coulomb
= 
So the lines terminating at - 2 micro coulomb
= 
So, the number of filed lines terminates at - 2 micro Coulomb are 6.