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Yuri [45]
3 years ago
13

An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that appl

y.- As the capacitor is charging, the current is increasing.- The stored electric field energy can be greater than the stored magnetic field energy.- As the capacitor is discharging, the current is increasing.- The stored electric field energy can be less than the stored magnetic field energy.- The stored electric field energy can be equal to the stored magnetic field energy.
Physics
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

As the capacitor is discharging, the current is increasing

Explanation:

Lets take

C= Capacitance

L=Inductance

V=Voltage

I= Current

The total energy E given as

E=\dfrac{IL^2}{2}+\dfrac{CV^2}{2}

We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL²  will increases.

It means that when charge on the capacitor decreases then the current will increase.

As the capacitor is discharging, the current is increasing

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Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p
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Answer:

Energy density will be 14.73 J/m^3

Explanation:

We have given capacitance C=225\mu F=225\times 10^{-6}F

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm 0.2\times 10^{-3}m

We know that there is relation between electric field and potential

E=\frac{V}{d}, here E is electric field, V is potential and d is separation between the plates

So E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C

Energy density is given by E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3

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Some machines do not multiply the force that is applied to them
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Answer:

Machine Efficiency

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2 years ago
A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
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Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
2 years ago
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