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Ostrovityanka [42]
3 years ago
5

A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr

ent leads the emf of the generator by 0.850 rad. A) What is the impedance of the circuit? B) What is the resistance of the circuit? C) Is the circuit predominantly inductive, or capacitive?
Physics
1 answer:
weqwewe [10]3 years ago
3 0

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

\phi = 0.85 rad = 48.72°

(a). Impedance of the circuit is given by

Z = \frac{V}{I}

Z = \frac{150}{2.75}

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }

Put the values of Z & \phi in above formula we get

R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

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Given:

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A plastic rod 1.3 m long is rubbed all over with wool, and acquires a charge of -3e-08 coulombs. we choose the center of the rod
Anestetic [448]
(a) The plastic rod has a length of L=1.3m. If we divide by 8, we get the length of each piece:
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c_6 = 0.0812m+0.1625m=0.2437 m

(c) The total charge is Q=-3 \cdot 10^{-8}C. To get the charge on each piece, we should divide this value by 8, the number of pieces:
Q/8=-3\cdot 10^{-8}C/8=-3.75\cdot 10^{-9}C

(d) We have to calculate the electric field at x=0.7 generated by piece 6. The charge on piece 6 is the value calculated at point (c):
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If we approximate piece 6 as a single  charge, the electric field is given by
E=k_e  \frac{q}{d^2}
where k_e=8.99\cdot 10^9Nm^2C^{-2} and d is the distance between the charge (center of piece 6, located at 0.2437m) and point a (located at x=0.7m). Therefore we have
E= 8.99\cdot 10^9 Nm^2C^{-2} \frac{-3.75\cdot 10^9 C}{(0.2437m-0.7m)^2} =-161.9 V/m
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(e) missing details on this question.
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3 years ago
Pls hurry worth 28 points <3
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Answer:

Explanation:

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What color of light is scattered the most by the molecules in earth’s atmosphere?.
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Answer:

Blue light

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Read 2 more answers
Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards​ Vernonville, and another car leaves Verno
iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is v_{A} =\frac{x_{A} }{t_{A} } (Equation A) and Average speed of the car B is v_{B} =\frac{x_{B} }{t_{B} } (Equation B), where x_{A} and x_{B} are the distances and t_{A} and t_{B} are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

96 minutes . \frac{1 hour}{60 minutes} = 1.6 h

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h} (Equation C)

v_{B} = \frac{208-x}{1.6h} (Equation D)

From Equation A and C, we have:

\frac{208-x}{1.6}+10 = \frac{x}{1.6}

208-x+16 = x

208 + 16 = 2x

x = \frac{224}{2}

x = 112 miles

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v_{A}  = \frac{112miles}{1.6h}

v_{A} = 70 miles per hour

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v_{B}  = \frac{208miles-112miles}{1.6h}

v_{B}  = \frac{96miles}{1.6h}

v_{B}  = 60 miles per hour

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3 years ago
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