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Vladimir79 [104]
3 years ago
10

Determine the inductive reactance for a 50 mH inductor that is across a 15 volt, 400 Hz source.

Physics
1 answer:
tresset_1 [31]3 years ago
8 0

Answer:

Inductive reactance is 125.7 Ω

Explanation:

It is given that,

Inductance, L=50\ mH=0.05\ H

Voltage source, V = 15 volt

Frequency, f = 400 Hz

The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :

X_L=2\pi fL

X_L=2\pi \times 400\ Hz\times 0.05\ H

X_L=125.66\ \Omega

X_L=125.7\ \Omega

So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.

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During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
Jlenok [28]

Well, I guess you can come close, but you can't tell exactly.

It must be presumed that the seagull was flying through the air
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That impact velocity of 98.5 m/s is actually the resultant of
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was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.

  98.5 m/s  =  √ [ (horizontal component)² + (vertical component)² ].

The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
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If the impact velocity were comprised solely of its vertical
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                  Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
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8 0
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