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Tpy6a [65]
3 years ago
14

Which statements describe a fully charged capacitor? Check all that apply.

Physics
2 answers:
Oksanka [162]3 years ago
4 0
Well, first of all, there's no such thing as "fully charged" for a capacitor.

A capacitor has a "maximum working voltage", because of mechanical
or chemical reasons, just like a car has a maximum safe speed.  But
anywhere below that, cars and capacitors do their jobs just fine, without
any risk of failing.

So we have a capacitor that has some charge on it, and therefore some
voltage across it.  From the list of choices above . . .

<span>-- Both plates have the same amount of charge.
Yes.  And both plates have opposite TYPES of charge.
One plate is loaded with electrons and is negatively charged.
The other plate is missing electrons and is positively charged.

-- There is a potential difference between the plates.
Yes.  That's the "voltage" mentioned earlier.
It's a measure of how badly the extra electrons want to jump
from the negative plate to the positive plate.

-- Electric potential energy is stored.
Yes.  It's the energy that had to be put into the capacitor
to move electrons away from one plate and cram them
onto the other plate.


</span>
Elza [17]3 years ago
3 0

composed of conducting plates

includes a dielectric between the plates

regulates the flow of current

stores both charge and energy

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Which one of the following temperatures is equal to 5°C?
natali 33 [55]

Answer : The correct option is, (D) 278 K

Explanation :

We are given temperature 5^oC.

Now the conversion factor used for the temperature is,

K=^oC+273

where, K is kelvin and ^oC is Celsius.

Now put the value of temperature, we get

K=5^oC+273=278K

Therefore, the temperature 278 K is equal to the 5^oC


7 0
3 years ago
A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
4 years ago
What is the build up of electrons on a surface
Minchanka [31]

Answer:

Static energy

Explanation:

Think of it as a balloon rubbing against your hair, the two attractions of friction causes Static energy.

5 0
3 years ago
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