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elena55 [62]
3 years ago
14

If you adjusted the frequency to be lower than the resonant frequency, would the voltage have been leading ahead of or lagging b

ehind the current? Explain

Physics
1 answer:
tatyana61 [14]3 years ago
4 0

Answer:

current will lead the voltage

Explanation:

From the bellow figure we can see the plot between the frequency and impedance

From the figure it is clear that when the frequency is less than the resonance frequency then the impedance decreases as the frequency increases it is possible only when impedance is capacitive in nature.

As the impedance is capacitive in nature so current will lead than the voltage

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The Earth is a sphere is a statement to describe a scientific law

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Select the correct answer.
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Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

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We’ll see the bell move, but we won’t hear it ring.

B.

We won’t see the bell move, but we’ll hear it ring.

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We’ll see the bell move and hear it ring.

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We won’t see the bell move or hear it ring.

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We’ll see the sound waves exit the vacuum pump.

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so, the answer to the question is

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2 years ago
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a basketball is tossed upwards with a speed of 5.0 m/s What is the maximum height reached by the basketball from its release poi
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Answer:

1.275 m

Explanation:

Let the maximum height reached be h.

Here initial velocity, u = 5 m/s

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0 = 25 - 2 × 9.8 × h

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3 years ago
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A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

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2 years ago
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