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Masteriza [31]
2 years ago
12

Find the equivalent impedance Zeq seen by the source when Vs = 2 cos (5t) v, C = 0.2 F, R = 1 Ω and L = 0.1 H. (Give angles in d

egrees and round final answers to two decimal places.) Calculate the voltage across the resistor. (Give angles in degrees and round final answers to two decimal places.) complex impedance The equivalent impedance seen by the source is + j Ω = at an angle of Ω. The voltage across the resistor is at an angle of v.

Engineering
1 answer:
Yanka [14]2 years ago
6 0

Answer:

0.89 cos (st +116.57°v)

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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How to stop toilets  

Explanation:

I think

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Table 1(a) shows the marks obtained by 40 students in an examination
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I dont know the answer to this
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Which of the following is described as a way engineers can test and investigate how things should be under certain circumstances
goblinko [34]

Answer:

The option that is best described as a way engineers can test and investigate how things should be under certain circumstances is;

  • Modeling

Explanation:

Modeling is a tool an engineer can use for the physical representation of a system that will facilitate the definition, testing and analysis, communication, data generation, data verification and data validation of given concepts

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A friend would like you to build an "electronic eye" for use as a fake security device. The device consists of three lights line
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Answer and explanation:

The graphical representation of the electronic eye

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8 0
3 years ago
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A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3
labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass co_2 = 1 kg

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of co_2 = 44

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

V_1 = \frac{m RT_1}{P_1}

       =\frac{1*0.189*298}{104}

V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

     =\frac{1*0.189*584}{1068}

V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

HEAT TRANSFER

Q = \Delta U  +W

   = 187.902 +(-256.46)

Q = -68.859 kJ

7 0
3 years ago
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