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2 years ago

As capacitor was discharging, what did you observe about q on its plates and the motion of charges in the external circuit?

1 answer:

8 0

As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.

The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.

The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.

A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.

Learn more about capacitor here:-brainly.com/question/14883923

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All galaxies follow the law of gravity. True or False

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The answer is true. All the galaxies in the universe follow the law of gravity.

<span>Based from the book, It's about Time: the Illusion of Einstein’s Time Dilation Explained, </span>

Einstein had explained that all the heavenly bodies in the universe follow the same scientific laws that are similar to our solar system. The stars and planets are held by the principles of inertia and gravity

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3 years ago

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms a

lawyer [7]

Answer:

This is because The energies of atoms are quantized.

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed

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3 years ago

One major disadvantage of breeder reactors is that they can do which of the following? Select one:a.explodeb.leak radioactivityc

zimovet [89]

definition of breeder reactors.

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1 year ago

An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration

umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

$F=qvB$ (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

$F=qvB=ma$ (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

$B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T$

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

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3 years ago

If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?

andreyandreev [35.5K]

Answer:

v = -6m/s

Explanation:

$x=4-12t+3t^2$