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leonid [27]
2 years ago
14

Lc circuit consists of a 20.0 mh inductor and a 0.150 µf capacitor. if the maximum instantaneous current is 0.400 a, what is the

greatest potential difference across the capacitor?
Physics
1 answer:
slavikrds [6]2 years ago
5 0
The following expression is applicable:

Max. inductor energy = Max. capacitor energy

Where;
Max. inductor energy = LI^2/2, with L = 20.0 mH, I = 0.400 A
Max. capacitor energy = CV_max^2/2, C = 0.150 micro Faraday, V_max = Max. potential difference

Substituting;
LI^2/2 = CV^2/2
LI^2 = CV^2
V^2 = (LI^2)/C
V_max = Sqrt [(LI^2)/C] = Sqrt [(20*10^-3*0.4^2)/(0.15*10^-6)] = 146.06 V
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To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

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Where,

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PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

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KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

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