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Neko [114]
3 years ago
11

An RLC circuit is composed of a 3.5 kΩ resistor, 1.1 nF capacitor and 7.6 mH inductor. If an AC 6.8 V rms is applied, what frequ

ency (in Hz) is necessary for the maximum power to be supplied to the circuit?
Physics
1 answer:
topjm [15]3 years ago
7 0

Answer:

5.5\times 10^{4} Hz

Explanation:

we know that the power is maximum in RLC circuit when inductive reactance and capacitive reactance are equal.

C = Capacitance of the capacitor = 1.1 x 10⁻⁹ F

L = Inductance of the inductor = 7.6 x 10⁻³ H

f = frequency necessary for maximum power

X_{C} =  capacitive reactance = \frac{1}{2\pi fC}

X_{L} =  inductive reactance = 2\pi fL

For maximum power :

X_{L} = X_{C}

2\pi fL = \frac{1}{2\pi fC}

f = \frac{1}{2\pi \sqrt{LC}}

f = \frac{1}{2(3.14) \sqrt{(7.6\times 10^{-3})(1.1\times 10^{-9})}}

f =5.5\times 10^{4} Hz

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valentinak56 [21]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>
6 0
3 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
Which<br> factors will increase the speed of a sound wave in the air?
Dafna11 [192]
A higher temperature, stiffer materials, and less dense materials increase the speed of sound.
7 0
4 years ago
A wheel is rotating about a fixed axis with constant angular acceleration 3 rad/s². At different moments, its angular speed is -
meriva

b and e are the largest and equal in magnitude. w*2R = (2)(2)R = 4R

A and d are next. aR = (3rad/s2)R = 3R

c is zero. wR = v = 0; Angular acceleration is zero.

<h3>What is angular acceleration?</h3>
  • The temporal rate at which angular velocity changes is known as angular acceleration. The standard unit of measurement is radians per second per second. Therefore, = d d t. Rotational acceleration is another name for angular acceleration.
  • Angular velocity divided by acceleration time can be used to define angular acceleration. (t). As an alternative, use pi times the drive speed (n) divided by the acceleration time (t) times 30. Radians per second squared (Rad/sec2) is the standard SI unit for rotational acceleration resulting from this equation.
  • To calculate angular velocity, we can use one of three formulas. The definition itself provides the first. Theta = position angle, t = time, and w = angular velocity, where w = angular velocity, theta = position angle, and t = time. Angular velocity is the rate of change of an object's position angle with respect to time.
  • The symbol for angular acceleration is, and it is measured in rad/s2, or radians per second square.

If two items are equal, show them as equal in your ranking. If a quantity is equal to zero, show that fact in your ranking:

b and e are the largest and equal in magnitude. w*2R = (2)(2)R = 4R

A and d are next. aR = (3rad/s2)R = 3R

c is zero. wR = v = 0; Angular acceleration is zero.

To learn more about angular acceleration, refer to:

brainly.com/question/20912191

#SPJ4

8 0
2 years ago
An object exhibits SHM with an angular frequency w = 4.0 s-1 and is released from its maximum displacement of A = 0.50 m at t =
vivado [14]

Explanation:

It is given that,

Angular frequency, \omega=4\ s^{-1}

Maximum displacement, A = 0.5 m at t = 0 s

We need to find the time at which it reaches its maximum speed. Firstly, we will find the maximum velocity of the object that is exhibiting SHM.

v_{max}=A\times \omega

v_{max}=0.5\times 4

v_{max}=2\ m/s............(1)

Acceleration of the object, a=\omega^2A

a=4^2\times 0.5

a=8\ m/s^2...............(2)

Using first equation of motion we can calculate the time taken to reach maximum speed.

v=u+at

t=\dfrac{v-u}{a}

t=\dfrac{2-0}{8}

t = 0.25 s

So, the object will take 0.25 seconds to reach its maximum speed. Hence, this is the required solution.

4 0
3 years ago
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