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PilotLPTM [1.2K]

2 years ago

14

A series circuit contains an 80-μF capacitor, a 0.020-H inductor, and a switch. The resistance of the circuit is negligible. Ini

tially, the switch is open, the capacitor voltage is 50 V, and the current in the inductor is zero. At time t = 0 s, the switch is closed. A) The time t at which the magnetic energy of the inductor first reaches its maximum value is?
B) The maximum current in the circuit is?
C) When the current is 1.5 A, the charge on the capacitor, in µC, is?
D) When the capacitor voltage is 30 V, the rate of change of the current is?
E) At a given instant, the magnetic energy of the inductor is twice the electrostatic energy of the capacitor. At that instant, the current is?

1 answer:

Hunter-Best [27]2 years ago

7 0

Answer:

Explanation: see attachment below

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2 years ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

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The heat released by the water when it cools down by a temperature difference $\Delta T$ is
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2 years ago

Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp

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Answer:

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To answer the final statements, let's pose the solution of the exercise

Energy is conserved

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For ball B

h_B = (2v)² / 2g

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Let's review the claims

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1 year ago

A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t

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Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

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2 years ago

Read 2 more answers

Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner fac

dangina [55]

Explanation:

Relation between electric field and charge density is as follows.

E = $\frac{\sigma}{2 \epsilon}$

where, $\sigma$ = charge density

$\epsilon$ = permittivity of free space = $8.85 \times 10^{-12}$

So, $E_{\text{outside}}$ = 0

$E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}$

or, $E_{inside} = \frac{\sigma}{\epsilon}$

Now, formula to calculate the potential difference of two conductors is as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

It is given that,

d = 6.0 mm = $6 \times 10^{-3} m$

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Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

= $\frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}$

= 0.0271 volts

thus, we can conclude that value of the magnitude of the electric potential differences between the two conductors is 0.0271 volts.

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2 years ago