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elena-s [515]
3 years ago
12

A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V . The rat

e at which electrical energy is dissipated in the resistor is 296 W Part A
What is the impedance Z of the circuit?
Express your answer to three significant figures and include the appropriate units.
Part B
What is the amplitude of the voltage across the inductor?
Express your answer to three significant figures and include the appropriate units
Part C
What is the power factor?
Express your answer using three significant figures.
Physics
1 answer:
max2010maxim [7]3 years ago
6 0

Answer:

Explanation:

 Rate of dissipation of energy at resistor

= I² R = 296

current I² = 296/R = 296 / 300

I = 0.9933 A.

Impedence Z = V / I = 500 / .9933

= 503 ohm.

B ) Let it be V

voltage across resistance = 300 x .9933 = 298 v

Now 500² = 298² + V²

V = 401 v

C)  Power factor = R / Z

=300 / 503

= 0.596

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Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs
Temka [501]
The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
a=g=-9.81 m/s^2
The following relationship holds for an uniformly accelerated motion:
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If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
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Now let's assume that v_i is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 2v_i. So the maximum height of the second ball is
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8 0
3 years ago
In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give
sveta [45]

Answer:

The current pass the R_2 is  I  = 0.25 A

Explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

    The voltage  is  V =  3V

     The first resistance is  R_1 = 7 \Omega

     The second resistance is  R_2 = 5 \Omega

Since the resistors are connected in series their equivalent resistance is  

       R_{eq} =  R_1 +R_2

Substituting values

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Since the resistance are connected in serie the current passing through the circuit  is the same current passing through R_2 which is mathematically evaluated as

        I  =  \frac{V}{R_{eq}}

Substituting values  

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      I  = 0.25 A

3 0
3 years ago
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