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2 years ago

When a capacitor is fully charged, the current through the capacitor is.

Physics

1 answer:

Rainbow [258]2 years ago

7 0

Answer:

When a capacitor is fully charged, no current flows in the circuit. This is because the potential difference across the capacitor is equal to the voltage source. (i.e), the charging current drops to zero, such that capacitor voltage = source voltage.

Explanation:

Hope this helps

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1. Una bola de 0.510 kg de masa se mueve al este (dirección +x) con una rapidez de 4.80 m/s y choca frontalmente con una bola de

Grace [21]

Responder:

3,37 m / s, + ve x - dirección

Explicación:

Utilizando la ley de conservación de la cantidad de movimiento expresada por la fórmula;

m1u1 + m2u2 = (m1 + m2) v

m1 y m2 son las masas de los objetos

u1 y u2 son sus velocidades iniciales

v es su velocidad común

Dado

m1 = 0,519 kg

u1 = 4,80 m / s

m2 = 0,220 kg

u2 = 0 m / s (cuerpo en reposo)

Necesario

Velocidad común v

Sustituir en la fórmula los valores dados;

0,519 (4,8) + 0,22 (0) = (0,519 + 0,220) v

2,4912 + 0 = 0,739 v

2,4912 = 0,739v

Dividir ambos lados por 0,739

2,4912 / 0,739 = 0,739 / 0,739

<em>Por lo tanto, la rapidez de ambas bolas después de la colisión es de 3.37 m.s hacia la dirección x positiva, ya que m1> m2 y la velocidad común es positiva.</em>

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3 years ago

You will be colliding two carts in this lab. Select all the systems for which you predict the total momentum of the system will

baherus [9]

From the momentum conservation we know that the initial momentum is equal to the final momentum. The momentum in a singular way can be defined as the product between the mass and the velocity of an object. In the presented system, however, there are two objects, therefore the mass of both and the speed of both, before and after the collision must be taken into account. Mathematically we could describe this as

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

From here we can realize that it is necessary to use the system on both cars to be able to predict what will happen either with their masses, or their speeds.

The correct answer is C.

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4 years ago

How do the dark lines of an atom''s absorption spectrum relate to the bright lines of its emission spectrum?

tangare [24]

Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
energy level by absorbing a photon ! If there were some miraculous substance
that could do that, we'd have it made.

All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
connected to a tiny hand-crank generator.

Whenever we wanted some energy, like for cooking or heating the house, we'd
switch the light bulb on, point it towards the pile, and give the little generator a
little shove. It wouldn't take much to git 'er going.

The atoms in the pile would absorb some photons, raising their electrons to higher
energy levels. Then the electrons would fall back down to lower energy levels,
releasing more energy than they needed to climb up. We could take that energy,
use some of it to keep the light bulb shining on the pile, and use the extra to heat
the house or run the dishwasher.

The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).

Energy that wasn't either there in the atom to begin with or else pumped
into it from somewhere can't be created there.

You get what you pay for, or, as my grandfather used to say, "For nothing
you get nothing."

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3 years ago

The force applied when using a simple machine

jenyasd209 [6]

Answer:

effort

Explanation:

effort is the answer

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4 years ago

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