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nikitadnepr [17]
3 years ago
10

What can you say about the impedance of a series RLC circuit at the resonant frequency? The impedance of a series RLC circuit is

maximum in magnitude at the resonant frequency and is equal to the resistance.
Physics
2 answers:
Svet_ta [14]3 years ago
7 0

Answer:

At resonance impedance of series RLC circuit is minimum and equal to resistance of the circuit

Explanation:

Let the resistance of the series RLC circuit is R

Capacitive reactance of the circuit is X_C and inductive reactance of the circuit is X_l

We know that impedance of the series RLC circuit is given by Z=\sqrt{R^2+(X_L-X_C)^2}, here R is the resistance X_l is inductive reactance and X_C is capacitive reactance

At resonance capacitive reactance will be equal to inductive reactance

So X_C=X_l

So impedance will become Z=R

So from above calculation we can say that impedance of series RLC circuit is minimum and equal to resistance of the circuit

Katyanochek1 [597]3 years ago
6 0

Answer:

Explanation:

Let R is the resistance of the circuit, L is the inductance of the circuit and c is the capacitance.

Inductive reactance= XL

capacitive reactance = Xc

The impedance of the RLC circuit is

Z=\sqrt{R^{2}+\left ( X_{L}^{2}-X_{C}^2 \right )}

As the circuit is in resonance, So, XL = Xc

In this condition, Z = R

So, the impedance is equal to the resistance of the circuit.  

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A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga
tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

               d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

 A₁ v₁ = A₂ v₂

 π r₁² v₁ = π r₂² v₂

 v_1= \dfrac{r_2^2}{r_1^2} v_2

 v_1= \dfrac{2.25^2}{3.75^2} v_2

 v_1= 0.36 v_2

Applying Bernoulli's equation

 \Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)

 P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)

 32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)

 v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}

 v_2=\sqrt{16.084}

       v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)²  x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0
3 years ago
Calculated the measurement uncertainty for Kinetic Energy when :mass = 1.3[kg] +/- 0.4[kg]velocity= 5.2 [m/s] +/- 0.2 [m/s]KE= 1
andriy [413]

Answer:

\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.

Explanation:

<u>Given:</u>

  • Mass, \rm m\pm\Delta m = 1.3\pm 0.4\ kg.
  • Velocity, \rm v\pm \Delta v = 5.2\pm 0.2\ m/s.

where,

\rm \Delta m,\ \Delta v are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.

The uncertainty in kinetic energy is given as:

\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.

7 0
3 years ago
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lora16 [44]
The change in gravitational potential energy due to change in position must be the change in it's kinetic energy as the system is isolated! so find out the potential energies of the two different points!

<span>PE=−[G<span>M1</span><span>M2</span>]÷R

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3 years ago
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Zigmanuir [339]

Answer:

Explanation:

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Here T₀ = 1 second

T = 1/√ 1-0.9² = 1/.4358 = 2.3 second

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6 0
3 years ago
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Semmy [17]

Answer:

numbers

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