a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the fr
1 answer:
Answer:
the inductive reactance of the coil is 1335.35 Ω
Explanation:
Given;
inductance of the coil, L = 250 mH = 0.25 H
effective current through the coil, I = 5 mA
frequency of the coil, f = 850 Hz
The inductive reactance of the coil is calculated as;
Therefore, the inductive reactance of the coil is 1335.35 Ω
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Answer:
141.78 ft
Explanation:
When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.
Calculating the acceleration using one of Newton's equations of motion:
Note: The negative sign denotes deceleration.
When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2
Hence, we can find the minimum stopping distance using:
The minimum stopping distance is 141.78 ft.
Answer:
The magnitude of the magnetic field is .
Explanation:
Given that,
Charge,
Speed of the charged particle,
The angle between the velocity of the charge and the field is 56°.
The magnitude of force,
We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :
B is the magnetic field.
So, the magnitude of the magnetic field is . Hence, this is the required solution.