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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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Guessing you want the average speed. We can multiple each speed by the time we spent going that speed, and them all together and then divide by the total time we spent in traffic to get the average speed. We spent a total of 7.5 minutes in traffic, so average speed  = (12*1.5+0*3.5+15*2.5)/7.5 = 7.4 m/s
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A 5c charge experiences a force of 40n when put at a certain location in space. What is the electric field at that location?
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The 'strength' of the electric field is the force on 1C of charge at that point.

At this 'certain location', the field is 40/5 = 8 newtons per coulomb = <u>8 volts</u>
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3 years ago
If the voltage drop across the first resistor is 13.00V, then how many volts remain for the 2nd resistor?
Scorpion4ik [409]

Answer:

2+1

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2 years ago
Read 2 more answers
A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol
PIT_PIT [208]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula E = \dfrac{kq }{d}

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}

E = 1461.95 N/C

c) The electric field E is calculated as:

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}

E = 239.76 N/C

7 0
3 years ago
A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a
photoshop1234 [79]

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

f'=f(\frac{v-v_o}{v+v_s})    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

v=\lambda f\\\\

v: speed of sound

λ: wavelength = 0.700 m

f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz

Next, you replace the values of all parameters in the equation (1):

f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

I=\frac{P}{4\pi r^2}

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}

hence, the intensity is 8.12*10^⁻4 W/m^2

3 0
2 years ago
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