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masya89 [10]
4 years ago
13

A resistor, inductor, and capacitor are connected in series, each with effective (rms) voltage of 65 V, 140 V, and 80 V respecti

vely. What is the value of the effective (rms) voltage of the applied source in the circuit
Physics
1 answer:
Morgarella [4.7K]4 years ago
6 0

Answer:

The value of the effective (rms) voltage of the applied source in the circuit is 132 V

Explanation:

Given;

effective (rms) voltage of the resistor, V_R = 65 V

effective (rms) voltage of the inductor, V_L = 140 V

effective (rms) voltage of the capacitor, V_C = 80 V

Determine the value of the effective (rms) voltage of the applied source in the circuit;

V= \sqrt{V_R^2 + (V_L^2-V_C^2} )\\\\V= \sqrt{65^2 + (140^2-80^2} )\\\\V = \sqrt{4225+ 13200} \\\\V = \sqrt{17425} \\\\V = 132 \ V

Therefore, the value of the effective (rms) voltage of the applied source in the circuit is 132 V.

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Answer: 1.4 x 10^-8N

Explanation:

Given that,

Mass of Particle 1 (m1) = 12kg

Mass of Particle 2 (m2) = 25kg

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F = Gm1m2/r²

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F = (2.01 x 10^-8Nm²) / (1.44m²)

F = 1.396 x 10^-8N (Rounded to the nearest tenth as 1.4 x 10^-8N)

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Net force of 200 newtons is applied to a wagon for 3 seconds
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Momentum of the wagon increases by (200 x 3)

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When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.

During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.

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Answer:

V = 49.05 [m/s]

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y =y_{o}+(v_{o}*t)+(\frac{1}{2}*g*t^{2} )

where:

Yo = initial position = 0

y = final position [m]

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